A Minimum Exact Disjunctive Basis for Two-Robot Solvability Profiles

A Minimum Exact Disjunctive Basis for Two-Robot Solvability Profiles Tsukasa Ninomiya1 , Shota Naito1 , Naoki Kitamura2 , Yuichi Sudo1 , and Koichi Wada1 1

arXiv:2610.17677v1 [cs.DC] 13 Oct 2026

2

Hosei University, Tokyo, Japan The University of Osaka, Osaka, Japan

Abstract We completely characterize the solvability profiles of tasks for two autonomous mobile robots in the plane under variable disorientation, chirality, and rigid movements. The four standard capability models and the schedulers considered here have a known quotient order Q of thirteen computational classes. A task’s solvability profile is the set of classes in which it is solvable, and is necessarily an upset of Q. We prove that every upset is realizable and construct a minimum exact disjunctive basis of thirteen concrete tasks. The basis contains one task for each principal upset. For every antichain of Q, the disjunction of the corresponding tasks has profile exactly equal to the union of their profiles. This exactness is not implied by the individual profiles: an algorithm may satisfy different disjuncts in different executions. We establish it by constructing, for every algorithm in each maximal class outside the target union, a single admissible fair execution violating all selected predicates simultaneously. Consequently, the realizable profiles form the entire finite distributive lattice Up(Q): there are exactly 54 profiles, counted by the antichain polynomial 1 + 13z + 24z 2 + 16z 3 , and each has a canonical disjunctive representation using at most three principal tasks. We also identify six tasks whose solvability signatures recover every inclusion and incomparability in Q, and prove that six is minimum even among arbitrary separating tasks. Their profiles are precisely the principal upsets generated by the join-irreducible elements of Q. In contrast, any task family generating every profile by exact disjunctions must contain all thirteen principal profiles. Thus our exact disjunctive basis has minimum size and is unique at the profile level.

1

Introduction

1.1

Computational landscapes of mobile robots

The Look–Compute–Move (LCM ) model is a standard abstraction for studying distributed computation by autonomous mobile robots. Robots are anonymous, identical points in the Euclidean plane, each repeatedly observing the configuration, computing a destination, and moving toward it. The classical model, OBLOT , provides neither persistent memory nor explicit communication [SY99]. The luminous model, LUMI, equips robots with persistent visible lights, providing both resources [DFP+ 16]. Two intermediate models isolate their roles: FSTA robots observe only their own lights and have finite private memory, whereas F COM robots observe only other robots’ lights and have finite communication without private memory [FSW20]. The activation scheduler supplies a second computational resource. Under full synchrony, all robots execute a cycle in every round; under semi-synchrony, an arbitrary nonempty subset is activated. Restricted synchrony limits consecutive activations, while asynchrony permits arbitrary finite delays between phases. Partially atomic variants distinguish delayed computation from observation during movement. Cross-model studies have characterized how these timing assumptions interact with memory

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and communication, refining the synchronous landscape and extending it to asynchronous and energyconstrained settings [FSW20, BFK+ 21, FSSW24, BFK+ 25]. Related computational-landscape studies: The computational-landscape viewpoint has been developed along several orthogonal dimensions, including synchrony, memory, communication and energy constraints [DFP+ 16, FSW20, BFK+ 21, BFK+ 25, FSSW24], and more recently beyond the Fsynch–Ssynch–Asynch axis: sequential schedulers activating one robot per round [FFS26b, FFS26a], robots on graphs [DDFN18], and limited-visibility robots, where visibility itself is a computational resource [DGS+ 24]. Why two robots are structurally exceptional: Two robots form the smallest nontrivial population, but their computational landscape is not obtained merely by restricting separators designed for larger populations. Their initial geometry is a single segment, with no third robot to provide an asymmetric neighborhood or a richer geometric reference. Some separations therefore require new witnesses, while some distinctions between internal capabilities disappear. Kitamura, Sudo, and Wada [KSW25] established the complete pairwise computational landscape for two robots under variable disorientation, chirality, and rigid movements. In particular, they proved the full-synchrony collapse F STAF ≡ FCOMF ≡ LUMI F , including its color- and time-preserving simulation, and introduced Dyadic-Multiple Stopped-Distance (DMSD) to separate atomic from unrestricted asynchrony. We use these known results as ingredients. From a Venn diagram to a profile lattice: Naito, Ninomiya, and Wada [NNW26] developed this global perspective through a Venn-diagram representation of solvable task sets. A problem’s position records all models that can solve it, exposing interactions that pairwise comparisons do not capture. Their study populated additional regions of the general landscape but left other regions unresolved. We complete this regional classification for the two-robot specialization, after identifying the settings that become computationally equivalent for two robots. Let Q be the resulting thirteen-class order, with X ≤ Y when every task solvable in X is solvable in Y . The solvability profile of a task P is Prof(P ) = {X ∈ Q : P ∈ Sol(X)}. Every profile is an upset of Q. Thus the finite distributive lattice Up(Q) describes all membership patterns consistent with the model order. A Venn region with pattern U consists of tasks whose profile is exactly U ; the region is populated only if such a task exists. Counting abstract upsets therefore does not establish that all regions are populated. The central question is whether every upset can be realized by a concrete two-robot task. A task is principal at X if its profile is the principal upset ↑X = {Y ∈ Q : X ≤ Y }. Every upset is the union of the principal upsets generated by its minimal elements, suggesting a task basis indexed by model classes. However, disjoining task specifications need not merely take the union of their profiles. An algorithm can potentially satisfy different disjuncts in different executions without solving any fixed disjunct. Consequently, constructing one principal task per class is not enough: exact realization by disjunction requires a separate common-failure argument.

1.2

Contributions

Under variable disorientation, chirality, and rigid movements, we construct a minimum exact disjunctive basis for the complete solvability-profile landscape of two robots. All basis tasks share the initial-instance space of two distinct positions. Our results combine an optimal principal representation of the known model order with a complete realization theorem for its profile lattice. An optimal principal representation of the model order: We identify six tasks B6 = {EPG∗ , SRO, MCv, CFSM, VSS, DMSD} whose solvability signatures recover every inclusion and incomparability among the thirteen classes. SRO is inherited from earlier landscape studies [FSW20], MCv from the asynchronous landscape [KSW25], and DMSD from the complete two-robot characterization [KSW25]. The other three are new or revised here: EPG∗ (Perpetual Expansion Preserving Gravity) replaces Center of Gravity Expansion [FSW20], whose coordinatewise floor refers to the adversary’s local coordinate systems and therefore does not define a predicate on trajectories under variable disorientation; CFSM is a collision-free refinement of Single Move [KSW25]; and VSS (Visit Swap and Stop) is introduced for the present construction. We prove that six tasks are necessary, even 2

when separating tasks need not be principal. Moreover, Q is a lattice with six nonleast join-irreducibles, and these tasks realize exactly the principal profiles generated by them. Those profiles form the unique minimum principal separating family. A principal task for every class: We extend B6 to a family B = {BX : X ∈ Q} of thirteen tasks satisfying Prof(BX ) = ↑X for every X ∈ Q. Three tasks come from earlier landscape studies (SRO, MCv and Cv), three from SSS 2025 (HET, ZCC and VEC) [NNW26], and DMSD from the complete two-robot characterization [KSW25]. The remaining six are new or revised here: EPG and EPG∗ , the trajectory-only revisions of Center of Gravity Expansion and its perpetual variant; CFSM, a collisionfree refinement of Single Move; OSPO, a strengthened version of Oscillating Points [DFP+ 16] requiring exactly one nonzero Move per robot per phase; S-RDV (Swap Rendezvous); and VSS. ExactW disjunctions
S via common-failure executions: For every antichain A of Q, we prove B Prof = X X∈A X∈A ↑X. The substantive step is the reverse inclusion. For each maximal class outside the target union and each algorithm in that class, we construct a single admissible fair execution violating all selected task predicates. Separate counterexamples to the individual tasks would not suffice. We organize these common-failure executions into families sharing expansion, rotation, or exchange obstructions, combining compatible finite violations with appropriate fair continuations. The resulting exactness is a property of the chosen tasks and their composition, not a consequence of the model order alone. Complete W realization and minimum basis size: Every upset U is realized by the canonical disjunction BU = X∈Min(U ) BX , so Real(Q) = Up(Q). The antichain polynomial is 1 + 13z + 24z 2 + 16z 3 . Hence there are exactly 53 nonempty profiles: 13 principal profiles, 24 with two minimal classes, and 16 with three. Including the empty profile gives 54. Every nonempty profile has an exact canonical representation using at most three principal tasks, and three are sometimes necessary. Any task family supplying exact disjunctive representations of every profile must contain all thirteen principal profiles. Our basis therefore has minimum size and is unique at the profile level among minimum exact generating families.

2

Models, Tasks and Solvability Profiles

2.1

Robots and Internal Capabilities

We consider exactly two anonymous, identical, and homogeneous robots R = {a, b} moving in the Euclidean plane R2 . The names a and b are used only for analysis; for r ∈ R, r̄ denotes the other robot. The position of r at time t ≥ 0 is denoted by r(t). Each robot has a local coordinate system centred at its current position. Two attributes of such a system may differ between the robots, namely the orientation of its axes and its unit distance, and both may also change from one activation to the next; only the handedness is fixed, since the robots share chirality. Thus a robot’s local view is obtained from the global frame by a translation, a rotation, and a positive scaling, with the rotation angle and the scaling factor chosen independently for each robot and each activation (variable disorientation, VD). In particular, no robot can rely on a common direction, on a common unit of length, or on the persistence of either across its own activations. When activated, a robot executes a Look–Compute–Move (LCM ) cycle: it takes an instantaneous snapshot in its local coordinates, computes a destination by a deterministic common algorithm, and moves continuously along the corresponding segment to that destination. Movements are rigid. We assume no general multiplicitydetection capability. We consider four standard capability models. In OBLOT , robots are oblivious and have no explicit communication. In LUMI, each robot has a persistent finite-valued visible light, providing both memory and communication. The intermediate models FSTA and FCOM retain only one of these capabilities: an F STA robot observes only its own light, which acts as finite private memory, whereas an FCOM robot observes only the peer’s light, providing finite communication but no readable private state. Multiple finite lights, when convenient, are identified with one light whose color set is their Cartesian product.

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For the two-robot setting, let pr be the position of r̄ in the current local coordinates of r, and let cr , cr̄ be their current colors. The information available at a Look is ObsOBLOT (r) = pr ,

ObsF STA (r) = (pr , cr ),

ObsF COM (r) = (pr , cr̄ ),

ObsLU MI (r) = (pr , cr , cr̄ ).

These definitions also apply when pr = 0: in the fixed two-robot setting the robots can therefore detect coincidence, and an externally visible peer light remains observable there. This does not amount to general multiplicity detection. In FCOM, the robot never reads its own light; at Compute it may either keep its current color or replace it by a color determined solely by its observation.1

2.2

Schedulers and Executions

All schedulers are fair: each robot is activated infinitely often. In synchronous executions, time is divided into rounds and each round activates a nonempty subset of the two robots. Under Fsynch both robots are activated in every round, whereas Ssynch allows any nonempty activation set. Under Rsynch, an execution is either fully synchronous throughout, or has a finite fully synchronous prefix followed by alternating singleton activations. Under Asynch, the Look, Compute, and Move phases of different robots may overlap arbitrarily, subject to their local phase order, finite operation durations, and fairness. Movements are continuous and rigid. We also use partially atomic subclasses of Asynch, distinguished by which phase of one robot is protected from a Look of the other: LC-atomic protects the Look–Compute interval, M -atomic protects the Move interval, and CM -atomic protects the Compute–Move interval. The strongest variant, LCM atomic Asynch, protects the whole cycle and is equivalent to Ssynch. Their formal timing definitions are given in Appendix A.1. For the models considered here, S ≥ ALC ≥ A, S ≥ ACM ≥ AM ≥ A. Moreover, AM and ACM are equivalent for OBLOT and FSTA, and also for FCOM [FSSW24]; hence ACM is omitted from the quotient landscape. For compactness, we write F, R, S, A, ALC , AM for Fsynch, Rsynch, Ssynch, Asynch, LCatomic Asynch, and M -atomic Asynch, respectively. Throughout the paper, unless explicitly stated otherwise, we assume variable disorientation, chirality, and rigid movements.

2.3

Tasks, Solvability, and Profiles

Let M = {LUMI, FCOM, FSTA, OBLOT }, S = {F, R, S, A, ALC , AM }. Tasks and executions. A task P = (IP , ΦP ) consists of a set of admissible initial geometric configurations and a temporal predicate on the robots’ geometric motion execution, independent of internal colors, states, and local coordinate systems. All basis tasks use the common initial-instance space I⋆ = {(pa , pb ) ∈ (R2 )2 : pa ̸= pb }, and any auxiliary geometric objects are determined from the initial positions. We use Move r (t) to mean that r is inside a nonzero Move at time t, that is, that the Move phase containing t has a destination distinct from the position from which it starts. Throughout such a phase the robot is in motion at every instant: it travels along the segment without ever pausing, and, movements being rigid, it is not halted before the destination either. A robot is therefore at rest at time t precisely when t lies in no nonzero Move; we write Stopr (t) for this, so that Stopr (t) ⇐⇒ ¬Move r (t), and a Move whose destination coincides with the current position leaves r at rest throughout. We write moves(r) for the number of its nonzero Moves and Stopped(r) for eventual permanent stationarity. Their 1 All previously established inclusions, equivalences, and separations used in this paper remain valid for exactly two

robots under this observation convention and the initial-instance and movement assumptions adopted here [KSW25].

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formal definitions, together with the phase-time conventions used in the appendix proofs, are given in Appendix A.2. For a deterministic algorithm A in model M and scheduler K, let ExecM,K (A, I) be the set of all admissible fair, time-divergent executions from I, including all permitted choices of schedules, local coordinate systems, phase timings, and movement speeds. Both robots start in the designated initial state or color, with no pending cycle. An algorithm A ∈ Alg(M ) solves P = (IP , ΦP ) under (M, K) if ∀I ∈ IP ∀ρ ∈ ExecM,K (A, I) : πgeom (ρ) |= ΦP , where πgeom (ρ) denotes the geometric motion execution obtained from ρ by forgetting internal states, colors, and local coordinate systems. The task P is solvable under (M, K) if such an algorithm exists: P ∈ Sol(M, K) ⇐⇒ ∃A ∈ Alg(M ) : A solves P under (M, K). We abbreviate πgeom (ρ) |= ΦP by ρ |= ΦP , and write Φi for ΦPi when the indexing family of tasks is clear from the context. For a setting µ = (M, K), also written M K , we write Sol(µ) = Sol(M, K).

2.4

Computational Order, Solvability Profiles and Disjunction of Tasks

For settings µ, ν, define µ ≥ ν ⇐⇒ Sol(µ) ⊇ Sol(ν), µ ≡ ν ⇐⇒ Sol(µ) = Sol(ν). Strict comparison and incomparability have their usual meanings. Let M = M × S be the set of settings, and let Q = M/≡ be the quotient poset of computational equivalence classes, ordered by X ≤ Y ⇐⇒ Sol(X) ⊆ Sol(Y ). For S ⊆ Q, let Min(S) and Max(S) denote its minimal and maximal elements, and let Ant(Q) be the family of antichains. For a task P , its solvability profile is Prof(P ) = {X ∈ Q : P ∈ Sol(X)}. Every solvability profile is an upset of Q. For X ∈ Q, write ↑X = {Y ∈ Q : X ≤ Y }. A task is principal at X if Prof(P ) = ↑X. A family B = {BX : X ∈ Q} is a principal task W basis if
Prof(B S X ) = ↑X for every X. It is an exact disjunctive basis if, for every A ∈ Ant(Q), Prof B = X∈A X X∈A ↑X. Proposition 2.1. For every task P , Prof(P ) is an upset of Q. The family of abstract profiles is therefore the finite distributive lattice Up(Q) under union and intersection. We write Real(Q) = {Prof(P ) : P is a two-robot task} ⊆ Up(Q). S Proposition 2.2. For every U ∈ Up(Q), U = X∈Min(U ) ↑X, and Min(U ) is the unique antichain giving an irredundant representation of U by principal upsets. Consequently, antichains and upsets are in bijection. Lemma 2.3 (Boundary certificate). Let P be a task and U ∈ Up(Q). Then Prof(P ) = U if and only if P ∈ Sol(X)

for every X ∈ Min(U ),

P ∈ / Sol(Y ) for every Y ∈ Max(Q \ U ). P The antichain polynomial is AQ (z) = A∈Ant(Q) z |A| , so AQ (1) = | Up(Q)|.
W W For tasks Pi = (I⋆ , Φi ) over the common initial space, define i∈I Pi = I⋆ , i∈I Φi .
S W Proposition 2.4. For every finite family, i∈I Prof(Pi ) ⊆ Prof i∈I Pi . The disjunction is exact when equality holds. The reverse inclusion is not automatic; Section 6 establishes it for the antichain-indexed disjunctions of our basis. Proofs of the elementary results above are given in Appendix A.3.

3

The Two-Robot Quotient and Minimum Separation

3.1

Basic Inclusions and Known Equivalences

Proposition 3.1. For every scheduler K and model M , LUMI K ≥ FSTAK ≥ OBLOT K ,

LUMI K ≥ FCOMK ≥ OBLOT K , 5

and M F ≥ M R ≥ M S , M S ≥ M ALC ≥ M A , M S ≥ M ACM ≥ M AM ≥ M A . Proposition 3.2. FSTAA ≡ FSTAALC , OBLOT A ≡ OBLOT ALC , FSTAAM ≡ FSTAACM , OBLOT AM ≡ OBLOT ACM . We also use the known equivalences LUMI R ≡ LUMI A ≡ FCOMR [DFP+ 16, FSW20, KSW25], and FCOMA ≡ FCOMAM , FCOMALC ≡ FCOMS [KSW25]. The elementary proofs of the two propositions are deferred to Appendix B.1. The following two-robot full-synchrony collapse follows from the stronger color- and round-preserving simulation of [KSW25], recalled in Appendix B.2. Theorem 3.3 (Computational collapse under full synchrony). For exactly two robots under the assumptions above, LU MI F ≡ FCOMF ≡ FSTAF . The proof is deferred to Appendix B.2.

3.2

The Thirteen Classes and Their Signatures

Combining the basic capability and scheduler inclusions with the known equivalences and the two-robot full-synchrony collapse above yields the thirteen computational classes considered below. To distinguish these classes, we use the six principal tasks B6 = {EPG∗ , SRO, MCv, CFSM, VSS, DMSD}. Three tasks deserve explicit specification here. EPG∗ and VSS are introduced in this paper, while CFSM is a collision-free refinement of Single Move. The remaining separators SRO [FSW20], MCv [KSW25], and DMSD [KSW25] are taken from earlier landscape results; their specifications and references are recalled in Appendix C. Perpetual Expansion Preserving Gravity (EPG∗ ): Let c0 = a(0)+b(0) , and let Ci = (ai , bi ) be the 2 sequence of jointly stationary configurations with consecutive repetitions suppressed. The task requires i+1 = c0 , bi+1 − ai+1 = 2(bi − ai ). this sequence to be infinite and, for every i ≥ 0, ai+1 +b 2 Collision-Free Single Move (CFSM): CFSM strengthens Single Move [FSSW24]: each robot must perform exactly one nonzero Move, and the robots must never collide. Thus moves(a) = moves(b) = 1, a(t) ̸= b(t) for all t ≥ 0. Visit Swap and Stop (VSS): Let a0 = a(0) and b0 = b(0), and set A0 = 2a0 − b0 , B0 = 2b0 − a0 . VSS requires the prescribed exchange phase to reach one of (B0 , b0 ), (a0 , A0 ), (B0 , A0 ), after which the robots must eventually remain permanently stationary at one of the four designated configurations determined by A0 , a0 , b0 , B0 . Every Move ends in {A0 , a0 , b0 , B0 }; the complete move-count and terminal conditions are given in Appendix D.7. Table 1 records, for each class, which of these six tasks are solvable. Thus each row is a six-bit solvability signature of that class. Since the thirteen signatures are distinct, B6 separates all thirteen computational classes. Lemma 3.4 (Bounds for the six separators). Under the assumptions of this paper, the following hold. (1) EPG∗ ∈ Sol(OBLOT F ) and EPG∗ ∈ / Sol(LUMI R ). (2) SRO ∈ Sol(OBLOT R ) and SRO ∈ / Sol(F COMS ) ∪ Sol(F STAS ). (3) MCv ∈ Sol(OBLOT S ) and MCv ∈ / Sol(FCOMA ) ∪ Sol(F STAAM ). (4) CFSM ∈ Sol(F STAA ) and CFSM ∈ / Sol(OBLOT F ) ∪ Sol(F COMS ). (5) VSS ∈ Sol(F COMA ) and VSS ∈ / Sol(OBLOT F ) ∪ Sol(F STAR ). (6) DMSD ∈ Sol(OBLOT AM ) and DMSD ∈ / Sol(FSTAA ). Proof. Items (1)–(6) are proved in Appendices D.2, C.1, C.2, D.4, D.7 and C.6, respectively.

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Theorem 3.5 (The thirteen-class order from six signatures). Write σ(µ) = {P ∈ B6 : P ∈ Sol(µ)}. The settings under consideration form exactly thirteen computational equivalence classes, represented by the rows of Table 1, and for any two representatives µ, ν, µ≥ν

⇐⇒

σ(µ) ⊇ σ(ν).

The certificate is given in Appendix H.1. Theorem 3.6 (Six principal separators). The tasks in B6 are principal, with Prof(EPG∗ ) = ↑OBLOT F ,

Prof(SRO) = ↑OBLOT R ,

Prof(MCv) = ↑OBLOT S ,

Prof(CFSM) = ↑FSTAA ,

Prof(VSS) = ↑FCOMA ,

Prof(DMSD) = ↑OBLOT AM .

Proof. For each task, the positive and maximal negative bounds stated above, together with upward closure of solvability, determine exactly the corresponding principal upset.

3.3

Minimum Separation

Theorem 3.7. Any family of tasks separating all thirteen classes of Q contains at least six tasks. Hence B6 is minimum. Proof. The quotient contains the chain OBLOT A < OBLOT AM < OBLOT S < OBLOT R < FSTAR < LUMI R < LUMI F . Along a chain, the solvability set of each task is an upper interval and hence can create at most one cut. Distinguishing the seven elements therefore requires at least six tasks. Table 1 shows that B6 achieves this bound.

3.4

The Minimum Principal Separator Family

Among principal profiles, the minimum separating family is unique: its generators are precisely the join-irreducible elements of Q. Proposition 3.8. The unique minimum family of principal upsets separating all elements of Q is {↑OBLOT F , ↑OBLOT R , ↑OBLOT S , ↑F COMA , ↑F STAA , ↑OBLOT AM }. Equivalently, the profiles of the six tasks in B6 form the unique minimum principal separating family. The lattice-theoretic characterization of Q, including its join-irreducible elements and width, is given in Appendix H.

4

A Principal Task Basis for the Thirteen Classes

Section 3 showed that six principal tasks suffice, and are necessary, to separate the thirteen computational classes. We now extend this family to one task BX for every X ∈ Q and prove that each BX is principal at X: (Prof(BX ) = ↑X). Thus the thirteen tasks realize all principal upsets of the quotient. The required solvability and impossibility bounds are summarized in Table 2; full task specifications and proofs are deferred to the appendices.

4.1

Principal Tasks and Boundary Certificates

For X ∈ Q, let NX = Max(Q\↑X). By Lemma 2.3, a task BX is principal at X if and only if BX ∈ Sol(X) and BX ∈ / Sol(Y ) for every Y ∈ NX . For principal profiles these negative bounds may be witnessed separately; simultaneous common-failure executions are needed only for disjunctions and are treated in Section 6. 7

Table 1: Six-task solvability signatures of the thirteen computational classes. Each row gives the sixbit signature with respect to B6 . An entry 1 indicates solvability and 0 indicates impossibility. Each representative denotes its entire computational equivalence class. Representative

EPG∗

SRO

MCv

CFSM

VSS

DMSD

1 1 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 0 0 0 0 0 0 0 0

1 1 1 1 1 1 0 1 0 0 1 0 0

1 0 1 1 0 0 0 1 1 1 0 0 0

1 0 1 0 0 1 1 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 0 1 1 0

LUMIF OBLOTF LUMIR FSTAR OBLOTR FCOMS FCOMA FSTAS FSTAAM FSTAA OBLOTS OBLOTAM OBLOTA

Table 2: Thirteen principal tasks and their boundary proof obligations. The third column lists the maximal classes outside ↑ X. Minimal profile class X

Principal task BX

Max(Q\ ↑ X)

Proof reference

LU MI F OBLOT F LU MI R F STAR OBLOT R F COMS F STAS OBLOT S F COMA F STAAM OBLOT AM F STAA OBLOT A

EPG EPG∗ HET OSPO SRO S-RDV ZCC MCv VSS VEC DMSD CFSM Cv

{OBLOT F , LU MI R } {LU MI R } {OBLOT F , F STAR , F COMS } {OBLOT F , F COMS , F STAS } {F COMS , F STAS } {OBLOT F , F STAR , F COMA } {OBLOT F , F COMS , F STAAM } {F COMA , F STAAM } {OBLOT F , F STAR } {OBLOT F , F COMS , F STAA } {F STAA } {OBLOT F , F COMS } ∅

Cor. D.7 Thm. 3.6 [NNW26] Lem. 4.1 Thm. 3.6 Lem. 4.2 [NNW26],App. C.7 Thm. 3.6 Thm. 3.6 [NNW26] Thm. 3.6 Thm. 3.6 Trivial, App. C.7

4.2

The Principal Task Basis

Write X6 = {OBLOT F , OBLOT R , OBLOT S , FSTAA , FCOMA , OBLOT AM } for the generators of B6 and X7 = Q \ X6 . Table 2 gives the complete assignment and the corresponding boundary certificates. Boundary results for the seven additional tasks. All thirteen tasks use I⋆ . Their specifications and boundary proofs are collected in Appendices C and D, with references in Table 2. For Cv, the entry “Trivial” refers only to the empty negative boundary of the least class OBLOT A ; the Go-to-midpoint algorithm is proved correct in Appendix C.7. Expansion Preserving Gravity (EPG): EPG is the one-shot version of EPG∗ : preserving the initial center of gravity at joint stops, the robots double the initial segment once and then stop forever. Its formal specification and bounds are given in Appendix D.1.

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Oscillating Points with One Move per Robot per Phase (OSPO): OSPO is a strengthened version of Oscillating Points [DFP+ 16]. Starting from distinct positions, the robots must alternate approach and retreat forever without ever colliding. The first approach phase starts at time 0, and every phase is delimited by joint stops. Within each phase, the distance is monotone and changes strictly between its endpoints, and each robot completes exactly one nonzero Move. The phase boundaries must tend to infinity. The full specification and boundary proofs are given in Appendix D.3. Lemma 4.1 (Bounds for OSPO). Under the assumptions of this paper, OSPO ∈ Sol(F STAR ), whereas OSPO ∈ / Sol(OBLOT F ) ∪ Sol(F STAS ) ∪ Sol(F COMS ). Proof. The solvability algorithm in FSTAR and the three impossibility arguments are given in Appendix D.3. Swap–Rendezvous (S-RDV): The two robots must achieve the common exchange first phase of Appendix D.5, namely form one of the three configurations in which their order on the initial line has been reversed, and must then gather at a single point and stay there, never leaving the segment [A, B] determined by the initial positions. Its exchange phase is shared with VSS; their subsequent requirements differ. The full specification is Definition D.24 in Appendix D.6. Lemma 4.2 (Bounds for S-RDV). Under the assumptions of this paper, S-RDV ∈ Sol(FCOMS ), whereas S-RDV ∈ / Sol(OBLOT F ) ∪ Sol(F STAR ) ∪ Sol(F COMA ). Proof. The solvability algorithm in FCOMS and the three impossibility arguments are given in Appendix D.6.

4.3

Principal-Basis Theorem

Theorem 4.3 (A principal task basis for the thirteen classes). Let B = {BX : X ∈ Q} be the family specified in Table 2. Under the model assumptions and observation convention of this paper, every task BX has initial-instance space I⋆ and satisfies Prof(BX ) = ↑X. Consequently, B is a principal task basis consisting of thirteen tasks and extending the six-task separating family B6 . Proof. If X ∈ X6 , the identity Prof(BX ) = ↑X is exactly one of the six statements of Theorem 3.6. If X ∈ X7 , the task-specific results referenced in Table 2, including Corollary D.7, establish solvability at X and impossibility at every class in NX . Applying Lemma 2.3 gives Prof(BX ) = ↑X. This also covers X = OBLOT A , where NX = ∅. Since Q = X6 ∪˙ X7 and all tasks use I⋆ , the conclusion follows. The six-task family determines the computational order, whereas the thirteen-task basis realizes every principal upset. These are distinct roles: the minimality of the former does not make the seven additional principal profiles redundant. Nor do the individual identities above establish exactness of task disjunctions. Section 5 presents representative common-failure constructions, and W Section 6
establishes the required property for every antichain-indexed family from B, yielding Prof X∈A BX = S ↑X for every A ∈ Ant(Q). X∈A

5

Representative Common-Failure Construction

Section 4 established each principal profile individually. Refuting a disjunction requires one admissible fair execution on which all selected predicates fail. We fix an arbitrary deterministic algorithm A and an initial instance ι ∈ I⋆ ; all predicates refer to this same instance, and execution blocks preserve the reached states and pending operations without reset. For the FSTA constructions below, normalization means presenting the peer at e = (1, 0) at every Look with distinct positions, using the rotations and rescalings permitted by VD. The destination in these normalized coordinates and the next private state then depend only on the current private state. Finite safety violations persist under extension; liveness violations require a suitable fair infinite continuation. 9

5.1

Disjunctions with Perpetual Expansion

We group the pairs EPG∗ ∨ P with P ∈ Cexp := {HET, OSPO, S-RDV, ZCC, VSS, VEC, CFSM}. Let C0 , C1 , . . . be the joint-stop configurations in chronological order with consecutive repetitions removed, and put c0 = 21 (a(0) + b(0)): EPG∗ asks this sequence to be infinite and every step of it to keep the center of gravity at c0 while doubling the segment. Keeping the center of gravity is exactly what rules out a solo expansion, and this is the component shared by all seven pairs. Lemma 5.1. Let ρ be an admissible execution with joint-stop sequence (Ci )i≥0 . 1. If one of the two robots completes no nonzero Move between the joint stop realizing Ci and the one realizing Ci+1 , then 12 (ai+1 + bi+1 ) ̸= 12 (ai + bi ). 2. Let K ∈ {R, S, ALC , AM , A}. For every algorithm there is a fair execution in which the LCM cycles of the two robots are pairwise disjoint in time. In such an execution either no nonzero Move ever occurs, in which case (Ci ) has the single term C0 and the perpetuity condition fails, or 1 ∗ 2 (a1 + b1 ) ̸= c0 and the expansion condition fails at i = 0. In both cases ΦEPG fails, and so does ΦEPG . For (1), only one endpoint changes between two distinct joint-stop configurations, so their midpoints differ. For (2), alternate complete cycles of a and b, assigning each cycle unit duration. The execution is fair and time-divergent, belongs to R and S, and respects every atomicity constraint because cycles do not overlap. Either no robot moves and progress fails, or the first nonzero Move changes the midpoint by (1). Formal timing details are in Appendix E.6. Combining the obstruction with another failure. A solo first nonzero Move suffices to violate EPG∗ ; later cycles need not remain serialized. To refute EPG∗ ∨ P , we must also violate P on that same execution, from the state actually reached. An admissible fair tail preserves finite safety violations, whereas a liveness failure requires a specifically constructed fair continuation. The next proposition supplies a common execution for both exchange tasks. Proposition 5.2 (Perpetual expansion and exchange tasks). For every deterministic algorithm in FSTAR , there is an admissible fair execution that violates EPG∗ , S-RDV, and VSS simultaneously. In particular, for each E ∈ {S-RDV, VSS}, EPG∗ ∨ E ∈ / Sol(F STAR ). Proof. Use coordinates on the initial line with a0 = d, b0 = 2d, d > 0, so the outer points are 0, 3d and c0 = 3d/2. Activate a, b alternately, completing each cycle before the next one starts; this is an admissible fair R execution with an empty fully synchronous prefix. Under normalized observations, the two robots traverse the same private state sequence until the first nonzero Move. If no such Move occurs, all three tasks fail their progress requirements. Otherwise a performs the first nonzero Move alone, changing the center of gravity at the next joint stop and violating EPG∗ . The shared exchange phase requires one of (3d, 2d), (d, 0), (3d, 0) with cumulative nonzero-move counts (1, 0), (0, 1), (1, 1), respectively. Thus this first solo Move of a must land at 3d: any other landing would require another nonzero Move of a to reach an admissible exchange configuration, exceeding its allowed count. If a lands at 3d, then b, at its next activation, has the same private state as a had before that Move and receives the same normalized input. Hence it performs the same normalized displacement, which in its current physical frame is twice its current distance to the peer, and therefore lands at 2d + 2d = 4d. This lies outside the segment [0, 3d] required by S-RDV and outside the allowed endpoint set {0, d, 2d, 3d} of VSS. In either case both exchange tasks fail irreversibly. Continue with fair alternating activations; all three violations persist.

10

The remaining companions in Cexp use the same solo expansion obstruction, with task- and classspecific prefixes or fair tails. These applicability and continuation arguments are recorded in Appendix F.1. The complete boundary coverage and the resulting profile equalities are assembled in Section 6; the obstruction alone does not supply those companion-task arguments.

5.2

Shrinking Rotation and Exchange Tasks

Let E = {S-RDV, VSS}. The two tasks share the exchange phase, so a construction that defeats the shared phase defeats both at once. Proposition 5.3 (Shrinking rotation and exchange tasks). For every deterministic algorithm in F STAS there is an admissible fair execution that violates SRO and both tasks in E simultaneously. Consequently SRO ∨ E ∈ / Sol(FSTAS ) for every E ∈ E. ◦ Proof sketch. We first recall the geometric obstruction. √ SRO permits a clockwise 90 rotation of the segment, or a clockwise 45◦ rotation with scale 1/ 2. With the peer fixed at (0, 0) and the moving robot initially at (−1, 0), the four possible two-step choices place the moving robot at one of (1, 0), (1/2, 1/2), (0, 1/2). All lie outside the square with the initial segment as diagonal, contrary to SRO’s containment condition. Thus two consecutive solo nonzero moves defeat SRO (Lemma E.19). Use the same initial positions and normalized observations as in Proposition 5.2. If the initial state orbit contains no nonzero output, a fair alternating execution never moves and violates all three tasks. Otherwise activate a alone until its first nonzero Move; S permits any finite number of solo activations. If a lands at 3d, the segment reverses without changing length, violating SRO. Activating b through its preceding zero-output states then reproduces the move to 4d, violating both exchange tasks. For any other landing, both exchange tasks have already failed. Unless SRO also fails immediately, the robots remain distinct. If a has another nonzero output, execute it with b fixed and apply the twomove obstruction. Otherwise let b perform its corresponding first nonzero Move. Either this violates SRO, or both robots now follow the same zero-output state suffix. Fair alternating activations then leave the configuration stationary forever, violating SRO’s perpetuity. In the finite-violation cases, append a fair alternating suffix. Appendix F.2 gives the complete argument.

This covers the entire maximal negative boundary of SRO ∨ VSS and the FSTAS boundary of SRO ∨ S-RDV; the latter pair also has the boundary class FCOMA , treated in Appendix F.2.

6

Exact Disjunctions via Common-Failure Executions

Section 4 established the individual principal profiles, and Section 5 exhibited representative executions violating several task predicates simultaneously. We now combine those constructions with the remaining boundary cases in the appendix to prove exactness for every antichain-indexed disjunction. The argument separates the execution constructions from the order-theoretic step that identifies the resulting profiles.

6.1

Reduction to Maximal Unsolvable Classes

W S For every A ∈ Ant(Q), write BA = X∈A BX , UA = X∈A ↑X, NA = Max(Q \ UA ). Principality immediately gives UA ⊆ Prof(BA ): a class in UA can solve one of the disjuncts. The converse does not follow from the individual impossibility results. It requires one execution on which all the selected predicates fail. Lemma 6.1 (Exactness from boundary common failure). Let A ∈ Ant(Q). Suppose that, for every Y ∈ NA and every deterministic algorithm A in Y , there are an initial instance ι ∈ I⋆ and an admissible fair execution ρ of A from ι such that ρ ̸|= ΦBX for every X ∈ A, where ΦBX denotes the temporal predicate of the principal task BX of Table 2. We then say that {BX : X ∈ A} has the common-failure property in Y . Under this hypothesis, Prof(BA ) = UA . 11

Table 3: Coverage of nonprincipal profiles by construction family. The last column counts the boundary P obligations (A, Y ), namely A |NA | within the corresponding row. Full boundaries and construction references are listed in Tables 4 and 5. |A| Disjunction family 2 2 2 2 2

3 3 3

No. of A No. of (A, Y )

EPG∗ ∨ P , P ∈ Cexp SRO ∨ P , P ∈ E ∪ P3 E ∨ P , E ∈ E, P ∈ P4 MCv ∨ P , P ∈ {VSS, VEC, CFSM} DMSD ∨ CFSM

7 5 8 3 1

16 8 18 4 1

Two-element total

24

47

EPG∗ ∨ E ∨ P , E ∈ E, P ∈ P4 SRO ∨ E ∨ P , E ∈ E, P ∈ P3 MCv ∨ VSS ∨ P , P ∈ {VEC, CFSM}

8 6 2

18 13 3

Three-element total

16

34

Proof. If Z ∈ UA , then Z solves some BX with X ∈ A, and the same algorithm solves BA . Conversely, the assumed executions show that BA ∈ / Sol(Y ) for every Y ∈ NA . Every Z ∈ / UA lies below some such Y , by finiteness of Q. If BA were solvable in Z, it would also be solvable in Y , a contradiction. Hence Prof(BA ) ⊆ UA . The witness instance and execution may depend on A, Y , and A, but not on the choice of a disjunct within A. Different boundary classes may use different executions. Thus, the remaining proof obligations are indexed by pairs (A, Y ) with Y ∈ NA , rather than by all classes outside UA .

6.2

Coverage of Two- and Three-Element Antichains

The thirteen-class order has twenty-four two-element antichains and sixteen three-element antichains, with no larger antichains. Table 3 groups them by the construction families introduced in Section 5. For compactness, set E = {S-RDV, VSS}, P3 = {ZCC, VEC, CFSM},

P4 = {OSPO} ∪ P3 .

We retain Cexp from Section 5.1. Each displayed task family determines an antichain through the principal generators in Table 2. The groups are disjoint and exhaustive. Coverage. For the expansion pairs, Lemma 5.1 supplies the solo-expansion obstruction, combined with a compatible prefix or fair tail for the companion task; Proposition 5.2 discharges the two FSTAR cases directly. For shrinking rotation, Proposition 5.3 discharges the two F STAS cases. The remaining pairs use the exchange obstruction of Appendix D.5, delayed movements against MCv, or an observation during movement against DMSD, as recorded row by row in Table 4. For three-element antichains, a two-task construction is reused wherever one exists and the third predicate is checked on that same execution, with every earlier violation preserved; the cases whose maximal boundary belongs to no constituent pair, namely the OBLOT S and OBLOT AM cases, get their own constructions. Table 5 indexes all thirty-four. Proposition 6.2 (Common failure at all nonprincipal boundaries). For every A ∈ Ant(Q) with 2 ≤ |A| ≤ 3 and every Y ∈ NA , the family {BX : X ∈ A} has the common-failure property in Y . Proof. The five two-task rows of Table 3 exhaust the twenty-four two-element antichains and the three remaining rows exhaust the sixteen three-element ones. Table 4 and Table 5 name, for each listed Y ∈ NA , the construction that violates every predicate of A on one admissible fair execution.

12

6.3

Exact Disjunctive Basis

Theorem 6.3 (An exact disjunctive basis). The
task W principal S basis B = {BX : X ∈ Q} is an exact disjunctive basis. For every A ∈ Ant(Q), Prof X∈A BX = X∈A ↑X. Proof. For A = ∅, both sides are empty by the convention for the empty disjunction. For |A| = 1, the claim is Theorem 4.3. For 2 ≤ |A| ≤ 3, apply Proposition 6.2 and Lemma 6.1. By Lemma H.3 these cases exhaust all antichains.

7

Complete Realization and a Minimum Exact Disjunctive Basis

7.1

Complete Realization and Minimality

Theorem 7.1 (Realization and canonical disjunctive normal form). For every nonempty upset U ∈ W Up(Q), define BU := X∈Min(U ) BX . Then Prof(BU ) = U , and Min(U ) is the unique irredundant set of principal generators of U ; the expression above is therefore the canonical disjunctive normal form of U at the profile level. With B∅ = (I⋆ , false) the same identity holds for U = ∅. S Proof. Min(U ) is an antichain, so Theorem 6.3 and Proposition 2.2 give Prof(BU ) = X∈Min(U ) ↑X = U . S If U = X∈S ↑X irredundantly, then every Y ∈ Min(U ) lies in some ↑X with X ∈ S ⊆ U , and minimality forces X = Y ; any further generator lies above a minimal element and is redundant. Hence S = Min(U ). Theorem 7.1 realizes every upset by a task over I⋆ ; together with upward closure, this gives Real(Q) = Up(Q).

(1)

Every task profile thus has the canonical representative BProf(P ) . This is equality of solvability profiles, not of task specifications. The matching lower bound is a property of the profile lattice alone and does not assume that a competing family consists of principal tasks. Theorem 7.2 (Minimality of the principal-upset basis). Let G ⊆ Up(Q) be a family whose unions generate every element of Up(Q). Then ↑X ∈ G for every X ∈ Q. Consequently {↑X : X ∈ Q} is the unique least union-generating family of profiles, of cardinality thirteen. S Proof. Write ↑X = G∈GX G with GX ⊆ G. Some G ∈ GX contains X, and being an upset it contains ↑X; the displayed equality gives the reverse inclusion, so G = ↑X. Conversely the principal upsets generate every upset by canonical decomposition, and they are distinct. Corollary 7.3 (Minimum size and uniqueness at the profile level). Any task family D supplying an exact representation2 of every upset of Q contains at least thirteen tasks. The family B attains this bound, and if |D| = 13 its task profiles are exactly {↑X : X ∈ Q}, each occurring once. Proof. The profiles {Prof(P ) : P ∈ D} generate every upset by union, so Theorem 7.2 forces all thirteen principal upsets to occur; B attains the bound by Theorem 7.1. Uniqueness here concerns the profiles, not the predicates realizing them: a task may be replaced by another with the same profile, though this need not preserve exactness of disjunctions. Nor does the lower bound claim minimality among families whose disjunctions may have profiles strictly larger than the union of their components. 2 A finite disjunction from D is an exact representation of U when the selected subfamily D U ⊆ D satisfies W S Prof( P ∈D P ) = Prof(P ) = U ; the empty disjunction is allowed, and only the representations used to genP ∈DU U erate the profiles need be exact.

13

7.2

Structural Consequences

By the antichain–upset correspondence and Theorem 7.1, the antichain polynomial 1 + 13z + 24z 2 + 16z 3 counts the realizable profiles: thirteen principal, twenty-four with two minimal classes, sixteen with three, and the empty one, so |Real(Q)| = | Up(Q)| = 54. Every nonempty profile therefore has an exact canonical expression using at most three tasks from B, and three are sometimes S necessary: for the threeelement antichain A0 = {OBLOT F , FSTAR , FCOMS } the upset U0 = X∈A0 ↑X has exactly these three minimal elements and canonical representative BU0 = EPG∗ ∨OSPO∨S-RDV, and no union of two principal profiles equals U0 by uniqueness of the irredundant decomposition. This bounds the number of principal terms, not the number of arbitrary tasks needed to realize a given profile. Figure 1 displays the 53 nonempty profiles, labelled by their canonical principal-task representations. LU MI F LU MI R F COMS F STAA F STAAM S F STA F STAR

· CF SM · V EC SC SX SZ OS · ZCC · S-RDV ESC · OSP O ESX · HET EZ EX EC ESZ RSX · EP G EOS RSC EO MX A RSZ F COM · M Cv MC OBLOT F RC RS OBLOT R RX A OBLOT RZ OBLOT S EH ES DC · Cv · DM SD · SRO OBLOT AM RV · EP G∗ MV MVC MVX EV RZV EO EOV RVC RVX EZ EZV EX EC · HET EVX EVC · OSP O · V SS · ZCC · V EC VX ZV OV · CF SM VC

VD, Chirality, Rigidity

Figure 1: Complete solvability-profile diagram for the thirteen classes. Colors distinguish principal, twominimal, and three-minimal profiles; labels give their canonical principal-task representations (e.g., EO and EOS). The task codes are E = EPG∗ , R = SRO, M = MCv, C = CFSM, V = VSS, D = DMSD, G = EPG, H = HET, O = OSPO, S = S-RDV, Z = ZCC, X = VEC, and Q = Cv.

8

Conclusion

We completely characterize the two-robot solvability-profile landscape: all 54 upsets of Q are realizable. Minimum separation requires six principal tasks, whereas minimum exact disjunctive generation requires thirteen.

Acknowledgment This work was supported by JSPS KAKENHI Grant Numbers JP20KK0232, JP25K15057, JP25K03078, JP25K03079, JP26K02865, and by JST FOREST Program JPMJFR226U.

14

References [BFK+ 21] K. Buchin, P. Flocchini, I. Kostitsyna, T. Peters, N. Santoro, and K. Wada. Autonomous mobile robots: Refining the computational landscape. In 23rd IPDPS Workshop on Advances in Parallel and Distributed Computational Models (APDCM), pages 576–585, 2021. [BFK+ 25] K. Buchin, P. Flocchini, I. Kostitsyna, T. Peters, N. Santoro, and K. Wada. On the computational power of energy-constrained mobile robots. Information and Computation, 303:105280, 2025. [CP05]

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[DDFN18] M. D’Emidio, G. Di Stefano, D. Frigioni, and A. Navarra. Characterizing the computational power of mobile robots on graphs and implications for the Euclidean plane. Information and Computation, 263:57–74, 2018. [DFP+ 16] S. Das, P. Flocchini, G. Prencipe, N. Santoro, and M. Yamashita. Autonomous mobile robots with lights. Theoretical Computer Science, 609:171–184, 2016. [DGS+ 24] A. Das, S. Ghosh, A. Sharma, P. Goswami, and B. Sau. The computational landscape of autonomous mobile robots: The visibility perspective. In Distributed Computing and Intelligent Technology (ICDCIT), volume 14501 of Lecture Notes in Computer Science, pages 85–100, 2024. [FFS26a]

C. Feletti, P. Flocchini, and N. Santoro. Computational power of energy-constrained autonomous robots under sequential schedulers. In 51st International Symposium on Mathematical Foundations of Computer Science (MFCS), volume 386 of Leibniz International Proceedings in Informatics (LIPIcs), pages 22:1–22:17, 2026.

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C. Feletti, P. Flocchini, and N. Santoro. On the computational power of mobile robots under sequential schedulers. Theoretical Computer Science, 1083:116172, 2026.

[FSSW24] P. Flocchini, N. Santoro, Y. Sudo, and K. Wada. On asynchrony, memory, and communication: Separations and landscapes. In 27th International Conference on Principles of Distributed Systems (OPODIS), volume 286 of Leibniz International Proceedings in Informatics (LIPIcs), pages 28:1–28:23, 2024. [FSW20]

P. Flocchini, N. Santoro, and K. Wada. On memory, communication, and synchronous schedulers when moving and computing. In 23rd International Conference on Principles of Distributed Systems (OPODIS), volume 153 of Leibniz International Proceedings in Informatics (LIPIcs), pages 25:1–25:17, 2020.

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[MSW26] T. Masuzawa, Y. Sudo, and K. Wada. Rendezvous under variable disorientation: The algorithmic power of fixed unit distance. CoRR, abs/2610.09713, 2026. [NNW26] S. Naito, T. Ninomiya, and K. Wada. Beyond pairwise comparisons: Unveiling structural landscape of mobile robot models. In Stabilization, Safety, and Security of Distributed Systems (SSS 2025), Lecture Notes in Computer Science, pages 408–424, 2026. [SY99]

I. Suzuki and M. Yamashita. Distributed anonymous mobile robots: Formation of geometric patterns. SIAM Journal on Computing, 28(4):1347–1363, 1999.

15

Appendix A

Model Conventions and Properties of the Lattice

A.1

Formal Definition of Schedulers

We define the schedulers considered in this paper for the two-robot system R = {a, b}. All schedulers are assumed to be fair: each robot performs infinitely many LCM cycles. For a robot r ∈ R and i ≥ 1, let tL (r, i), tC (r, i), tB (r, i), and tE (r, i) denote the times at which r performs its i-th Look, Compute, begins its i-th Move, and ends its i-th Move, respectively. We assume tL (r, i) < tC (r, i) < tB (r, i) < tE (r, i) < tL (r, i + 1). The Look and Compute operations are instantaneous, while every Move operation has finite duration. Movements are continuous and rigid: if r computes destination pE from its current position pB , then it moves along the segment [pB , pE ] during [tB (r, i), tE (r, i)] and reaches pE at time tE (r, i).  A synchronous execution is described by an infinite activation sequence E1 , E2 , . . ., where each Ej ∈ {a}, {b}, {a, b} is the set of robots activated in round j. Robots in Ej perform one complete LCM cycle synchronously, while robots not in Ej remain inactive. The fully synchronous scheduler Fsynch is the case where Ej = {a, b} for all j. The semi-synchronous scheduler Ssynch allows any nonempty activation set in each round, subject only to fairness. The restricted synchronous scheduler Rsynch consists of Fsynch or an optional finite prefix of fully synchronous rounds followed by alternating singleton activations: {a}, {b}, {a}, {b}, . . . or {b}, {a}, {b}, {a}, . . . . In the asynchronous scheduler Asynch, there are no rounds. The adversary chooses the times tL , tC , tB , tE and the duration of each Move operation arbitrarily, subject only to the ordering constraint above, finite duration of operations, and fairness. We also consider atomic subclasses of Asynch. Since there are only two robots, these subclasses can be defined by forbidding one robot from performing a Look operation during certain intervals of the other robot’s cycle. For all r ∈ R and all i, j ≥ 1, we define: / (tL (r, i), tC (r, i)]. • LC-atomic Asynch: tL (r, j) ∈ • M -atomic Asynch: tL (r, j) ∈ / [tB (r, i), tE (r, i)]. • CM -atomic Asynch: tL (r, j) ∈ / [tC (r, i), tE (r, i)]. • LCM -atomic Asynch: tL (r, j) ∈ / (tL (r, i), tE (r, i)]. The LCM -atomic Asynch scheduler is equivalent to Ssynch. In the OBLOT and FSTA models, M -atomic Asynch and CM -atomic Asynch are equivalent because robots cannot observe the other robot’s color. The same equivalence is known for F COM robots [FSSW24]. Hence we focus on M atomic Asynch and omit CM -atomic Asynch from the landscape.

A.2

Formal Definition of Temporal Predicates

A colored configuration at time t is C(t) = (a(t), colora (t)), (b(t), colorb (t)) , where the ordering is used only for analysis. Its geometric and color components are (a(t), b(t)) and (colora (t), colorb (t)), respectively. In OBLOT , the color set is a singleton. For a fixed execution, we use the following predicates, suppressing its name from the notation. For each robot r ∈ {a, b}, its moving-time set and moving predicate are [
MoveSet(r) := tB (r, i), tE (r, i) , Move r (t) :⇐⇒ t ∈ MoveSet(r), i≥1: r(tB (r,i))̸=r(tE (r,i))

16

the union being taken over the Move operations of r whose starting and ending positions differ. A robot is stationary exactly when it is not inside such a Move: Stopr (t)

:⇐⇒

t∈ / MoveSet(r).

Since MoveSet(r) is a union of open intervals, the endpoints of nonzero Move intervals are included among the stopping times. Writing the open set MoveSet(r) as the disjoint union of its maximal open intervals, we define the number of nonzero moves by MoveSet(r) =

Nr G

Ir(k) , moves(r) := Nr ∈ Z≥0 ∪ {∞},

k=1

and, for t ≥ 0, the number of nonzero Moves of r completed by time t by mr (t) := { k ≤ Nr : sup Ir(k) ≤ t } . Thus mr is non-decreasing with mr (0) = 0, and moves(r) = limt→∞ mr (t). The strict separation between successive Move operations ensures that each such interval corresponds to one nonzero Move operation. Permanent stopping is expressed by Stopped(r)

:⇐⇒

∃T ≥ 0 ∀t ≥ T : Stopr (t).

Finally, the joint stops of the execution are the times at which neither robot is moving:
Σab := { t ≥ 0 : Stopa (t) ∧ Stopb (t) } = [0, ∞) \ MoveSet(a) ∪ MoveSet(b) . This set is closed in [0, ∞) and contains 0, since no cycle is pending at time 0. The geometric configuration (a(t), b(t)) is constant on each maximal subinterval of Σab . Listing these configurations in chronological order and suppressing consecutive repetitions gives the joint-stop sequence C0 , C1 , . . . of the execution, with Ci = (ai , bi ) and di = bi − ai .

A.3

Proofs of Results in Section 2.4

Proof of Proposition 2.1. Let X ∈ Prof(P ) and X ⪯ Y . Then P ∈ Sol(X) ⊆ Sol(Y ), and hence Y ∈ Prof(P ). Proof S of Proposition 2.2. Since Q is finite, every element of U lies above S a minimal element of U . Thus U = X∈Min(U ) ↑X. Conversely, let A be an antichain with U = X∈A ↑X. Every element of A is minimal in U , since A is an antichain, and every minimal element of U lies above, hence equals, some element of A. Thus Min(U ) = A. The representation is irredundant because each X ∈ Min(U ) belongs to no other principal upset in the union. Proof of Lemma 2.3. Necessity is immediate. For sufficiency, every Z ∈ U lies above some X ∈ Min(U ), since Q is finite. Thus P ∈ Sol(Z) by Proposition 2.1. Every Z ∈ / U lies below some Y ∈ Max(Q \ U ). If P were solvable in Z, it would also be solvable in Y , since Z ≤ Y , contradicting the assumed impossibility at Y . Hence Prof(P ) = U . Proof of Proposition W 2.4. If a model class solves Pj , the same algorithm always satisfies Φj and therefore always satisfies i∈I Φi .

17

B

Model Comparison

B.1

Basic inclusions

Proof of Proposition 3.1. The cross-model inclusions follow by ignoring capabilities: a LU MI robot can execute an FSTA or FCOM algorithm, and either intermediate model can execute an OBLOT algorithm. The scheduler inclusions follow by restricting admissible executions. If every schedule and timing choice permitted by K1 is also permitted by K2 , then any algorithm correct under K2 remains correct under K1 . Thus, Sol(M, K2 ) ⊆ Sol(M, K1 ). Applying this observation to the scheduler restrictions in the model definitions proves the stated inclusions. Proof of Proposition 3.2. These equivalences follow directly from the observation capabilities of FSTA and OBLOT and from the definitions of the corresponding atomic scheduler classes.

B.2

Correctness of the Color- and Round-Preserving Simulation

The following color- and round-preserving full-synchrony simulation for two robots was established in [KSW25]. Theorem B.1 ([KSW25]). Let A be a deterministic two-robot LU MI F algorithm with a palette C of k colors and a designated initial color c0 . There are algorithms AS in F STAF and AC in F COMF , both using the palette C, such that, for every initial geometric configuration I and every execution ρ of either simulator from I, there is an admissible execution ρb of A from I with πgeom (ρ) = πgeom (b ρ). The matching execution has the same round boundaries and movement timings. Each simulator round implements exactly one round of A, without auxiliary colors or initialization rounds. Proof. Write

A(p, cr , cr̄ ) = (q, c′ ).

Fix e = (0, 1). For p ̸= 0, let Np be the orientation-preserving similarity mapping p to e, and let N0 = id. Construction. Both simulators start with c0 . In each round, a robot with peer position p reads the available color c: its own color in FSTAF , or its peer’s color in F COMF . It computes (q, c′ ) = A(Np (p), c, c), sets its state or external light to c′ , and moves to Np−1 (q) in its actual local coordinates. This uses only permitted observations and the palette C; Np is computed from the current snapshot and is not stored between rounds. Correctness. Fix an arbitrary execution ρ of either simulator. We construct a matching execution ρb of the LUMI F algorithm A by induction over rounds. At the beginning of round j, we maintain the following invariant: the corresponding robots in ρ and ρb occupy the same positions, and the two robots have a common color cj . The invariant holds initially by the designated common initial color. Let ψr,j be the map from global coordinates to robot r’s actual local coordinates at its Look in round j, and let pr,j be the observed peer position in these coordinates. In the corresponding execution of A, choose the virtual local coordinate system ψbr,j = Npr,j ◦ ψr,j . Because Npr,j is an orientation-preserving similarity, ψbr,j is admissible under variable disorientation and preserves chirality.

18

By the induction hypothesis, the geometric configuration and the two colors at the beginning of the round coincide in the two executions. Hence, under the virtual coordinate system, each source robot receives exactly the normalized input used by the simulator. In particular, when the robots are at distinct positions, both receive (e, cj , cj ), whereas at coincidence both receive

(0, cj , cj ).

Therefore, determinism of A implies that both corresponding computations produce the same next color cj+1 . Let qj be the destination computed by A in the normalized coordinates. The global destination of robot r in ρb is
−1 −1 ψbr,j (qj ) = ψr,j Np−1 (qj ) , r,j which is exactly the destination used by the simulator. We now choose the same round boundaries, phase timings, and progress along each rigid movement segment in ρb as in ρ. Thus the two executions have identical geometric trajectories throughout round j, and the common-color invariant is preserved for round j + 1. Induction over all rounds therefore yields an admissible infinite execution ρb of A satisfying πgeom (ρ) = πgeom (b ρ). Moreover, each simulator round implements exactly one round of A, and no auxiliary colors or initialization rounds are introduced. Hence the simulator uses the same palette and preserves the round and movement timings claimed in the theorem. Scope of the construction. The correspondence constructed above maps every simulator execution to an admissible execution of the original LU MI F algorithm with the same geometric trajectory. The virtual local coordinate systems of the source execution need not coincide with those of the simulator. The construction also relies on the designated common initial color; it does not address arbitrary initial color assignments. Proof of Theorem 3.3. The quantitative simulation theorem (Theorem B.1) shows that every LU MI F algorithm can be simulated in both FSTAF and F COMF , preserving the geometric execution. Hence Sol(LUMI F ) ⊆ Sol(FSTAF ) ∩ Sol(F COMF ). The reverse inclusions follow immediately because FSTA and F COM are submodels of LUMI.

C

Definitions and Boundary Proofs of the Geometric Basis Tasks

This appendix collects the formal specifications of the geometric basis tasks and the boundary proofs cited in Section 4.2.

C.1

Shrinking Rotation (SRO)

This subsection records the formal specification of SRO and proves Lemma 3.4 (2). Definition C.1 (Shrinking Rotation). Two robots a, b start from distinct positions. Let Σab be the set of times at which neither robot is inside a nonzero Move operation, list the configurations at the times of Σab in chronological order, and remove consecutive repetitions; call the resulting sequence C0 , C1 , C2 , . . . with Ci = (ai , bi ). Put di := bi − ai , let Qi be the square with diagonal ai bi , interior included, and set 19

Q−1 := R2 . Let Rϕ denote the counterclockwise rotation by ϕ. The first condition (perpetuity) requires the sequence (Ci ) to be infinite. The second condition (rotation) requires that for every √ i the vector di+1 be the clockwise 90◦ rotation of di (C1), or its clockwise 45◦ rotation scaled by 1/ 2 (C2), and that ai+1 , bi+1 ∈ Qi−1 (C3).  ΦSRO ≡ { (Ci )i≥0 is infinite }
and { ∀i ≥ 0 : di+1 = R−90◦ di or di+1 = √12 R−45◦ di
 ∧ ai+1 , bi+1 ∈ Qi−1 } Here Σab is the set of joint stops of Definition C.24. The reference [FSW20] does not make the sequence (Ci ) explicit. We define (Ci ) as the configurations at the times at which neither robot is inside a nonzero Move, so that the intermediate configurations in which one robot has landed while the other is still in flight are excluded. With this convention each term of (Ci ) has both robots at their initial positions or at the landing points of their last completed nonzero Move operations, and the sequence does not depend on how the adversary distributes the movement speed inside a Move operation. In the proofs below we write c1 (γ) and c2 (γ) for the solo and simultaneous contraction ratios of Lemma E.2; these are not to be confused with the configurations C1 , C2 of the sequence (Ci ) or with the conditions C1, C2. Impossibility Lemma C.2 ([FSW20, Lemma 23]). SRO ∈ / Sol(FCOMS ). Proof. Run a fully synchronous prefix, which is admissible under Ssynch. By Lemma E.14 the two robots always carry the same color. If λγ = 0 for every color γ occurring in that common sequence, this fair execution contains no nonzero Move, the configuration never changes, the sequence (Ci ) reduces to C0 alone, and the first condition fails. Otherwise let γ be the first color with λγ > 0 and truncate the prefix at the round in which both robots carry γ; the configuration is still the initial one. Now activate a alone twice in a row (Lemma E.4; under Ssynch the robot that is not activated has no pending operation), and write Ci for the configuration at the beginning of this interval. The visible state of an F COM robot is the color of its peer, and b is not activated, so that color is constant; by Lemma E.16(b) the displacement rule of a is the single pair (λγ , θγ ), and λγ > 0 makes every activation a nonzero Move. If c1 (γ) = 0 and a reaches the position of b, then d = 0 in that configuration and neither C1 nor C2 can hold, so the second condition fails. Otherwise d > 0 is preserved. For the first move to satisfy the second condition it must satisfy the rotation condition, and with the peer at rest only two displacements do so, one of type C1 and one of type C2 (Lemma E.19). The rule being single, the second move is of the same type, and among the four computations of that lemma the two relevant ones both give ai+2 ∈ / Qi , so C3 fails. Lemma C.3 ([FSW20, Lemma 23]). SRO ∈ / Sol(FSTAS ). Proof. Under Ssynch the robot that is not activated has no pending operation, so the hypotheses of Lemma E.4 hold at every round boundary. The visible state of an FSTA robot is its own color, so in the normalized frame of Lemma E.1(1) the color sequence of each robot is c0 , g(c0 ), g 2 (c0 ), . . . independently of the schedule, and it is eventually periodic with preperiod p and period q (Lemma E.11). The displacement of each activation is likewise determined by the color through (λγ , θγ ). The periodic part contains a color with λγ > 0. Activating a alone p + 2q times (Lemma E.4) makes a perform at least two nonzero Move operations while b stays at rest; write Ci for the configuration at the beginning of this interval. Zero Move operations do not advance the sequence (Ci ), so these correspond to two consecutive terms of (Ci ). If at some point c1 (γ) = 0 and a reaches the position of b, then d = 0 in that configuration and neither C1 nor C2 can hold, so the second condition fails. Otherwise d > 0 is preserved and the same robot performs two consecutive nonzero Move operations with its peer at rest, so by Lemma E.19 the rotation condition and C3 cannot both be satisfied and the second condition fails. 20

Suppose now that the periodic part contains no color with λγ > 0. After the preperiod the color sequence stays in the periodic part, where λγ = 0, so each robot performs at most p nonzero Move operations whatever the schedule. Hence the configuration changes at only finitely many times, the sequence (Ci ) is finite, and the first condition fails. Solvability Lemma C.4 ([BFK+ 25, Lemma 2]). SRO ∈ Sol(OBLOT R ). Proof. Each robot moves to the point obtained by rotating its own position clockwise by 90◦ about the midpoint of the two robots. If both robots are activated simultaneously,√the segment is rotated by 90◦ and C1 holds; if only one robot is activated, the segment is scaled by 1/ 2 and rotated by 45◦ , and C2 holds. An Rsynch execution is either fully synchronous throughout, which is the case treated in [FSW20, Lemma 22], or a finite fully synchronous prefix followed by alternating singleton activations (Lemma E.8). Fully synchronous rounds preserve the square. If such a round is followed by a singleton round, both new positions lie in the same square. For two consecutive alternating singleton rounds, consider the square immediately before the first round: the first mover reaches one of its vertices, and the second reaches its centre. Thus ai+1 , bi+1 ∈ Qi−1 for every i ≥ 1; the case i = 0 is vacuous since Q−1 = R2 . Hence C3 holds as well. The destination is determined by the midpoint and the rotation alone and requires no unit of length, so it is computable in each robot’s own local frame. Lemma C.2, Lemma C.3 and Lemma C.4 together prove Lemma 3.4 (2).

C.2

Monotone Convergence (MCv)

This subsection records the formal specification of MCv and proves Lemma 3.4 (3). Definition C.5 (Monotone Convergence). Two robots a, b start from distinct positions, and we write d(t) := ∥a(t) − b(t)∥. The first condition (convergence) requires the two robots to converge to a common point l. The second condition (monotonicity) requires their distance never to increase.   ΦMCv ≡ { ∃l ∈ R2 : lim a(t) = lim b(t) = l } and { ∀t, t′ ≥ 0 : t ≤ t′ ⇒ d(t′ ) ≤ d(t) } t→∞

t→∞

Monotonicity is required at every instant of the continuous execution, not only at the joint stops. This is Monotone Line Convergence of [FSSW24] with the two conditions confining the robots to the segment joining their initial positions and forbidding overtaking removed. Impossibility Lemma C.6 ([KSW25, Theorem 8], [FSSW24, Lemma 13]). MCv ∈ / Sol(FCOMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch. Run fully synchronous cycles with normalized frames. By Lemma E.14, the two robots carry the same color at every round boundary. If every output along this execution is zero, the resulting fair, time-divergent execution keeps the robots at their distinct initial positions, violating convergence. Otherwise, stop the zero-Move prefix just before the first round with a nonzero output. Let γ be the common color at this point, so that λγ > 0, and put c := c1 (γ). Let b perform its Look, and delay its Comp. This finite delay is permitted under Asynch by Lemma E.5. The saved snapshot determines the nonzero displacement v = λγ Rθγ (wold ),

21

where wold is the vector from b to a at that Look. Although b has not yet performed its Comp, determinism ensures that it will compute this displacement from the saved snapshot when its Comp is eventually performed. Throughout the delay, both the position and the visible color γ of b remain unchanged. Since an FCOM robot does not read its own color, we may therefore activate a for any prescribed finite number of complete cycles, using normalized frames and the same nonzero displacement rule (λγ , θγ ) in each cycle. If c > 1, a single such Move increases the distance between its starting and ending times, violating monotonicity. If c = 1, the starting and ending positions of this nonzero Move are distinct points at the same distance from the stationary robot b. The midpoint of the movement segment is strictly closer to b than either endpoint. Thus the distance increases between the time a passes that midpoint and the completion of its Move, again violating monotonicity. It remains to consider 0 ≤ c < 1. Choose a finite k ≥ 1 such that, after k complete cycles of a, ∥wk ∥ = ck ∥wold ∥ <

∥v∥ , 2

where wk is the vector from b to a at that time. For c = 0, take k = 1; otherwise every cycle starts with distinct robot positions, so the normalized rule can be applied repeatedly. Keep a stationary and let b complete its Comp using the saved snapshot, update its color, and perform the resulting Move with displacement v. The distance immediately before that Move is ∥wk ∥, whereas at its completion it satisfies ∥wk − v∥ ≥ ∥v∥ − ∥wk ∥ > ∥wk ∥. Hence monotonicity fails. In every finite-violation case, complete any pending cycle and append an arbitrary admissible fair, time-divergent continuation. The violation persists, so no such algorithm solves MCv. Lemma C.7 ([KSW25, Theorem 8], [FSSW24, Lemma 20]). MCv ∈ / Sol(FSTAAM ). Proof. We prove the following stronger statement: from any configuration with distinct robot positions, a common internal color, and no pending operation, there is an admissible fair, time-divergent continuation violating ΦMCv . Use normalized frames whenever the robots are distinct. At a configuration with common color s, write zs = λs eiθs , cs = |1 − zs |. We advance the common color by activating a and then b, one complete cycle at a time, except in the special case specified below. If zs = 0, both cycles are zero moves and both robots reach color g(s). If cs = 0, then zs = 1. Instead, let both robots Look and Compute before either moves, and let them execute their moves simultaneously with equal fractional progress. They exchange positions, so their distance becomes zero and then positive again. Monotonicity fails. If zs ̸= 0 and cs ≥ 1, a solo move of a already violates monotonicity. For cs > 1 its landing distance is larger than its starting distance. For cs = 1, the endpoints are distinct points at the same distance from the stationary peer; the interior of the movement segment contains points strictly closer to that peer, so the distance increases before landing. It remains to consider 0 < cs < 1. Execute the two singleton cycles. If either move increases the distance at any time, monotonicity has already failed. Otherwise both robots finish at distinct positions with common color g(s). Continue in this way through the transient and one period of the common normalized color orbit. Unless a violation has already occurred, every nonzero rule in the period has 0 < cs < 1. If every rule in the period is zero, an infinite fair continuation leaves the robots at distinct fixed positions, so convergence fails.

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Otherwise, the product Γ of the solo distance ratios over one period satisfies 0 < Γ < 1. Advance both robots to a common periodic color with a nonzero rule. Let b Look and Compute a nonzero displacement v, and delay its Move until after a prescribed finite number of complete cycles of a. This is admissible under AM by Corollary E.6. During the delay, b remains fixed, while the private color of a follows its periodic orbit. Consequently, after sufficiently many periods the vector w from b to a satisfies ∥w∥ < ∥v∥/2. Keep a stationary and let b complete its pending Move. The distance increases, since ∥w − v∥ ≥ ∥v∥ − ∥w∥ > ∥w∥. After every finite violation, complete any pending cycle and append an arbitrary fair, time-divergent continuation. The construction uses neither the designated initial color nor the original positions, provided that its starting configuration satisfies the three stated conditions. Solvability Lemma C.8 ([KSW25, Theorem 8], [FSSW24, Lemma 4]). MCv ∈ Sol(OBLOT S ). Proof. Each robot moves to the midpoint of the two observed positions. Under Ssynch every round is atomic, so it suffices to follow the configuration at the round boundaries. If only one robot is activated in a round, it lands at the midpoint and the distance is halved; if both are activated, they land on the same midpoint and the distance becomes 0. In either case the distance never increases, so the monotonicity condition holds, and d is at least halved in every round, so d(t) → 0. Moreover each landing point lies on the segment joining the two current positions, so the segments [a(t), b(t)] form a nested family of compact sets whose diameters tend to 0. Their intersection is a single point l, and both robots converge to it, which is the convergence condition. Lemma C.6, Lemma C.7 and Lemma C.8 together prove Lemma 3.4 (3).

C.3

Hexagonal Edge Traversal (HET)

HET [NNW26] places the two robots at opposite vertices of a regular hexagon and requires each of them to move to an adjacent vertex along an edge and back, alternating left and right forever. Definition C.9 (Hexagonal Edge Traversal). Two robots a, b start from distinct positions. Let v0 , . . . , v5 be the vertices of the regular hexagon having a(0) and b(0) as opposite vertices, labelled counterclockwise from v0 = a(0) and v3 = b(0), the indices being read modulo 6. The cycles of the two robots are wa := (v0 , v−1 , v0 , v1 ),

wb := (v3 , v2 , v3 , v4 ).

The first condition (perpetuity) requires each robot to perform infinitely many nonzero Move operations. The second condition (edge traversal) requires each robot r, at every time at which it is not moving, to stand at the vertex of its own cycle indexed by the number mr (t) of nonzero Move operations it has performed so far.  ΦHET ≡ { ∀r ∈ {a, b} : moves(r) = ∞ }
 and { ∀r ∈ {a, b}, ∀t ≥ 0 : ¬ Mover (t) ⇒ r(t) = wr mr (t) mod 4 } The reference [NNW26] states the same requirement by exhibiting the time intervals spent at each vertex and the monotone progress along each edge. Since movements here are rigid and follow the segment joining the two endpoints, we have rewritten it by prescribing the sequence of landing points of the nonzero Move operations. The hexagon is uniquely determined by a(0) and b(0), and the two cycles are carried onto each other by the rotation by π about the midpoint.

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Impossibility Lemma C.10 ([NNW26, Lemma 8]). HET ∈ / Sol(OBLOT F ). Proof. Let A be any OBLOT algorithm and consider the fully synchronous execution of Lemma E.13, in which the two robots apply one and the same pair (λ, θ) in every round. If λ = 0 then moves(a) = 0 and the perpetuity condition fails. Assume λ > 0. If either robot ever leaves its prescribed cycle, the edge-traversal condition fails. Otherwise, normalizing the circumradius of the hexagon to 1, the configuration at the beginning of round 1 is (v0 , v3 ) and at the beginning of round 2 it is (v−1 , v2 ). In round 1 the robot a sees its peer at v3 − v0 , of length 2, and must perform the displacement v−1 − v0 , of length 1, so the required pair is (λ, θ) = (1/2, +60◦ ). In round 2 it sees its peer at v2 − v−1 , again of length 2, and must perform the displacement v0 − v−1 , of length 1, so the required pair is (1/2, −60◦ ); the same two pairs are required of b by the symmetry of the two cycles. The two rounds therefore demand opposite rotations relative to the peer, which a single pair (λ, θ) cannot deliver, so the edge-traversal condition fails in one of the two rounds. Lemma C.11 ([NNW26, Lemma 7]). HET ∈ / Sol(F STAR ). Proof. The adversary chooses the alternating singleton activation {a}, {b}, {a}, . . . with an empty fully synchronous prefix (Lemma E.8). First stage. The configuration is unchanged until the first nonzero Move. The hexagon and the two cycles of Definition C.9 are carried onto each other by the rotation by π about the midpoint combined with exchanging a and b, and this rotation preserves chirality. The local input of an FSTA robot is the pair consisting of the relative position of the peer and its own state, and does not contain the state of the peer, so in the normalized frame of Lemma E.1(1) both local inputs have the form (e, own state) and are isomorphic. The two robots start from the same initial state s0 , so a in round 2k − 1 and b in round 2k both pass from sk−1 to sk . If no state occurring in s0 , s1 , . . . produces a nonzero Move, then moves(a) = 0 and the perpetuity condition fails. Let j be the least index that produces a nonzero Move; then a performs the first nonzero Move in round 2j + 1, from state sj . Second stage. If the landing point of a is not v−1 , the edge-traversal condition a(t) = wa (1) = v−1 fails at the time ma = 1. So assume a lands at v−1 . Writing R for the circumradius of the hexagon, the edge length is R and the diagonal length is 2R, so the constant of Lemma E.1(1) is λsj =

∥v−1 − v0 ∥ 1 = . ∥v3 − v0 ∥ 2

Third stage. The robot b, activated in round 2j + 2, has not yet moved and is in state sj . The robots have no common unit of length, so in the normalized frame the local input of b is the same√(e, sj ) as that of a in the second stage: the change of the observed distance from 2R to ∥v−1 − v3 ∥ = 3 R does not appear in the local input. Hence b applies the same λsj = 12 to its own peer vector v−1 − v3 , and √ the length of its displacement is 23 R. The landing point required of b by the edge-traversal condition is √ wb (1) = v2 , with ∥v2 − v3 ∥ = R. Since 23 R ̸= R, the robot b cannot land at v2 and the edge-traversal condition fails. Lemma C.12 ([NNW26, Lemma 9]). HET ∈ / Sol(F COMS ). Proof. Run a fully synchronous prefix, which is admissible under Ssynch. By Lemma E.14 the two robots always carry the same color. If λγ = 0 for every color γ occurring in that common sequence, this fair execution contains no nonzero Move, so moves(a) = 0 and the perpetuity condition fails. Otherwise let γ be the first color with λγ > 0 and truncate the prefix at the round in which both robots carry γ; no nonzero Move has occurred, so the configuration is still the initial one. Now activate a alone repeatedly (Lemma E.4; under Ssynch the robot that is not activated has no pending operation). The visible state of an F COM robot is the color of its peer, and b is not activated, so that color stays γ; by Lemma E.16(b) the displacement rule of a is the single pair (λγ , θγ ), and λγ > 0 24

makes every activation a nonzero Move. The cycle wa = (v0 , v−1 , v0 , v1 ) requires the two successive steps v0 → v−1√and v−1 → v0 . With b standing still at v3 , the first of them approaches the peer, since ∥v−1 − v3 ∥ = 3 R < 2R = ∥v0 − v3 ∥, and the second retreats from it. By Lemma E.16 a single pair (λγ , θγ ) cannot realize both, so the edge-traversal condition fails. Solvability Lemma C.13 ([NNW26, Lemma 10]). HET ∈ Sol(LU MI R ). Proof. Each robot records its position in its own cycle with a light of four colors. Since a LUMI robot sees both lights, it learns from the color of its peer which vertex the peer currently occupies, so even under the alternating singleton activations of an Rsynch execution it can reconstruct the hexagon from the observed relative position and move to the next vertex of its cycle. The destination is determined by the relative position of the peer and by chirality alone and requires no unit of length, so it is computable in each robot’s own local frame. Lemma C.10, Lemma C.11, Lemma C.12 and Lemma C.13 supply the boundary certificate of Table 2 for HET, which by Lemma 2.3 gives Prof(HET) = ↑LUMI R as asserted in Theorem 4.3.

C.4

Zig-Zag Collision-Free Convergence (ZCC)

ZCC [NNW26] requires the two robots to converge to a single point without overtaking each other, while each of them perpetually repeats a zig-zag motion that alternately increases and decreases its distance from its own initial position. Definition C.14 (Zig-Zag Collision-Free Convergence). Two robots a, b start from distinct positions. The first condition (convergence) requires the two robots to converge to a point l. The second condition (collision freedom) requires each robot never to go, as seen from its own initial position, farther than its peer. The third condition (zig-zag) requires, with fr (t) := ∥r(t) − r(0)∥, that for each robot r there be times 0 = t−1 < t0 < t1 < · · · at which fr takes maxima fr (t2i ) and minima fr (t2i+1 ) alternately, the minima increasing.  ΦZCC ≡ { ∃l ∈ R2 : lim a(t) = lim b(t) = l } t→∞

t→∞

and { ∀r ∈ {a, b}, ∀t ≥ 0 : fr (t) ≤ ∥r̄(t) − r(0)∥ } and { ∀r ∈ {a, b}, ∃ 0 = t−1 < t0 < t1 < · · · , ∀i ≥ 0 : fr (t2i−1 ) < fr (t2i+1 ) < fr (t2i ) }



There are two differences from [NNW26]. The quantification in the first condition is ∀ϵ ≥ 0 there, but including ϵ = 0 would require the two robots to coincide at l and stay there from some time on, which is incompatible with the third condition; we therefore use ∀ϵ > 0. Also, that paper imposes the third condition on one robot only, whereas we impose it on both. Along a single Move the position of r traverses a segment, so fr is a convex function of the movement progress and hence non-increasing and then non-decreasing in time; outside the Move intervals fr is constant. Each nonzero Move therefore contributes at most one turning point of fr , and since executions are time-divergent, only finitely many Moves occur in any bounded time interval. Consequently the times ti of the zig-zag condition satisfy ti → ∞. The three conditions differ in nature. Collision freedom is a safety condition: it is violated as soon as fr (t) > ∥r̄(t)−r(0)∥ at a single time t, and no extension can repair this. The zig-zag condition is a liveness condition: beyond any T the sequence (ti ) still provides a strictly decreasing step fr (t2i ) > fr (t2i+1 ) and a strictly increasing one fr (t2i+1 ) < fr (t2i+2 ). Hence, if for some T the restriction of fr to [T, ∞) is monotone for one of the two robots, then ΦZCC fails. In particular, the monotonicity of fr along a finite solo activation interval does not by itself establish failure.

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Impossibility Lemma C.15. ZCC ∈ / Sol(OBLOT F ). Proof. An OBLOT F robot applies the same rule in every round, so it cannot switch between the two directions that the zig-zag condition demands. Consider the fully synchronous execution of Lemma E.13, let m be the midpoint and v := a(0) − m. The difference vector is mapped in every round by the same similarity I − 2λRθ , of ratio c2 and rotation angle φ, and the midpoint is invariant (Lemma E.2(ii)), so a(k) − m = c2k Rφk v and b(k) − m = −(a(k) − m). If λ = 0 the robots never move and fr ≡ 0. If c2 = 0 the two robots coincide at m at the end of round 1, and by Lemma E.12 the coincidence is maintained. While they coincide the observed peer position is 0. If the local output is then zero, the robots never move again and fr is constant from round 1 on. Otherwise the adversary chooses their common global displacement (Lemma E.1(2)); choosing in every round one and the same direction w with ⟨w, v⟩ < 0 makes fa (t) = ∥a(t) − a(0)∥ strictly increasing from round 1 on. In both cases the restriction of fa to a tail is monotone, so the zig-zag condition fails by the criterion stated after Definition C.14. Assume λ ̸= 0 and c2 ̸= 0. If φ ̸≡ 0 (mod 2π), then by Lemma E.2(iii) applied to the nonzero angle φ there is k ≥ 1 with cos(kφ) ≤ − 21 < 0, whence ⟨v, a(k) − m⟩ = c2k cos(kφ)∥v∥2 < 0. Since b(k) − m = −(a(k) − m) and a(0) = m + v, ∥a(k) − a(0)∥2 − ∥b(k) − a(0)∥2 = −4 ⟨v, a(k) − m⟩ > 0, so at the end of round k we have fa > ∥b(k) − a(0)∥ and the collision-freedom condition fails. This is a safety violation at a finite time. If φ ≡ 0, then Rφ = I and a(k) = m+c2k v, so all the positions of a lie on the ray from m through a(0) and advance monotonically along it, towards m when c2 < 1 and away from a(0) when c2 > 1; the value c2 = 1 would force λ = 0. Since movements are rigid along segments, fa (t) = | c2· − 1 | ∥v∥ interpolates monotonically within each round, so fa is monotone on the whole execution and the zig-zag condition fails by the same criterion. This is a liveness violation, witnessed by the infinite fully synchronous execution itself. Lemma C.16 ([NNW26, Lemma 26]). ZCC ∈ / Sol(F COMS ). Proof. We prove a continuation property that will also be used in the common-failure arguments. Fix the original initial positions a0 , b0 . Suppose that a finite prefix ends with no pending operation and with the robots lying in [a0 , b0 ] in their original order. Their current colors need not be equal. There is a fair, time-divergent continuation violating ΦZCC , with a0 , b0 kept fixed. First consider a configuration in which the robots coincide. Activate them alternately, one complete cycle at a time. If every output is zero, both positions remain constant forever, so the zig-zag condition fails by the criterion stated after Definition C.14. Otherwise, at the first nonzero output, choose the local frame so that the moving robot increases its distance from its own original initial position; this is possible by Lemma E.1(2). Its peer remains at the common position, so the collision-freedom inequality is violated. Append an arbitrary fair tail. We call this construction the coincidence continuation. Now suppose that the robots are distinct. Use alternating singleton activations as the default schedule, with normalized frames at every Look. For the currently activated robot r, put σ = 1 − λeiθ . If σ ∈ (0, 1], let the cycle complete and continue with the default schedule. Such a cycle either leaves r stationary or moves it towards its peer without reaching it. Thus both robots remain in the initial segment in their original order. At the first output with σ ∈ / (0, 1], freeze the peer and repeatedly activate r. Its peer’s color is fixed, so the same normalized rule is used until a coincidence occurs. If σ = 0, the first cycle reaches the peer and the coincidence continuation above applies. 26

Assume σ ̸= 0, and write σ = ceiψ . At the start of this repetition, let r̄ denote the peer and L = ∥r̄ − r(0)∥,

h = ∥r̄ − r∥.

The maintained order gives L ≥ h > 0. After j repetitions, ∥rj − r(0)∥2 = L2 + c2j h2 − 2Lcj h cos(jψ). If ψ ̸≡ 0 (mod 2π), some finite j satisfies cos(jψ) ≤ 0 (Lemma E.2(iii)), so this quantity exceeds L2 . If ψ ≡ 0, then necessarily c > 1, and for sufficiently large finite j we again have ∥rj − r(0)∥ > L. In either case the stationary peer remains at distance L from r(0), so the collision-freedom condition fails. Append an arbitrary fair tail. Finally, if no output outside (0, 1] ever occurs, the default alternating execution is fair and each fr (t) = ∥r(t) − r(0)∥ is non-decreasing throughout. The zig-zag condition therefore fails. All infinite executions can be scheduled to be time-divergent, and every exceptional repetition above is finite. Applying this continuation property to the initial configuration proves the claim. Lemma C.17 ([NNW26, Lemmas 24–25]). ZCC ∈ / Sol(F STAAM ). Proof. We prove a continuation statement with the original initial positions a0 , b0 kept fixed. Suppose that no operation is pending and that either the robots coincide, or their current normalized color orbits have the same eventual cycle. There is an admissible fair, time-divergent continuation violating ΦZCC . If a safety violation has already occurred, append any fair tail. If the robots coincide, use the coincidence continuation from the proof of Lemma C.16. Hence assume that they are distinct and that safety has not yet failed. If a state with solo ratio c1 = 0 is reachable along the normalized orbit of either robot, activate that robot alone until it first lands on its stationary peer. All preceding Looks see distinct positions. At the resulting coincidence neither robot has a pending operation, so the same coincidence continuation applies. We may therefore assume that all future solo ratios are positive. Let Γ > 0 be their product over the common eventual color cycle. Consider the fair execution E of alternating singleton activations under normalized frames. If Γ ≥ 1, then, after both color sequences have entered their periodic parts, every block of one full period of each robot multiplies the distance by Γ2 ≥ 1. The distance thus has a positive subsequence bounded away from zero, and convergence fails. Assume 0 < Γ < 1. If E already violates ΦZCC , use E. Otherwise E satisfies both safety and convergence. Put b0 − a0 , xr (t) = ⟨r(t), e⟩. e= ∥b0 − a0 ∥ Adding the squared safety inequalities gives 2⟨a(t) − b(t), b0 − a0 ⟩ ≤ 0, hence xa (t) ≤ xb (t) at every time. The execution E must contain a Move of a with positive displacement along e, or a Move of b with negative displacement along e. Indeed, otherwise xa would be non-increasing and xb non-decreasing. Convergence would then require both projections to be equal and constant. Equality of the projections, together with safety, forces both squared safety inequalities to be equalities. During a singleton Move, the moving robot would consequently have to remain on the intersection of two fixed circles centered at a0 and b0 . This finite set admits no nonconstant continuous movement. Thus the distinct robots could not converge, a contradiction. Fix the first such Move in E. Suppose first that the selected Move is a Move of b with displacement v satisfying ⟨v, e⟩ < 0. Follow E through its Look and Comp, but delay the start of its Move. While b stays fixed, activate a through

27

any remaining transient part of its color orbit and then for sufficiently many complete periods. Since Γ < 1, the vector w from b to a can be made to satisfy ∥w∥ < −⟨v, e⟩/2. If safety fails during this finite interval, we are done. Otherwise keep a stationary and release the pending Move of b. At its completion, xa − xb = ⟨w − v, e⟩ > 0, contradicting the necessary projection order. If the selected Move is a Move of a with ⟨v, e⟩ > 0, delay that Move and instead activate b through any remaining transient part of its color orbit and then for sufficiently many complete periods, bringing it arbitrarily close to the fixed position of a. Choose the resulting vector w from a to b with ∥w∥ < ⟨v, e⟩/2. Releasing the Move of a gives xa − xb = ⟨v − w, e⟩ > 0, again violating safety. All delays are finite and occur before the delayed Move starts, as permitted by AM and Corollary E.6. No moving robot is stopped inside its Move. After a finite violation, complete any pending operation and append an arbitrary fair, time-divergent tail. The initial configuration satisfies the continuation hypotheses, which proves the stated impossibility. Solvability Algorithm 1 ZCC for F STA under Ssynch (robot r) Model and scheduler: FSTA, Ssynch. State: r.status ∈ {A, B}, initially A. Look: observe the peer r and obtain the distance D = ∥r − r∥. Comp: 1: if r.status = A then 2: r.des ← the point at distance D/4 from r in the direction of r 3: r.status ← B 4: else 5: r.des ← the point at distance D/8 from r in the direction away from r 6: r.status ← A 7: end if Move: move to r.des.

▷ approach ▷ retreat

Lemma C.18. ZCC ∈ Sol(F STAS ), as witnessed by Algorithm 1. Proof. The algorithm of [NNW26, Lemma 27], which approaches by D/2 and retreats by D/4, makes the two robots coincide at the midpoint when they are activated simultaneously, after which every Move is zero and the zig-zag condition fails. We therefore approach by D/4 and retreat by D/8. Every movement takes place on the line through a(0) and b(0), so we put a coordinate x on that line with xa (0) = 0 and xb (0) = 1, and write D := xb − xa . (i) The order is preserved. Under Ssynch each round is atomic, so the destinations of a round are computed from the configuration at its beginning. The factor by which D changes in one round is 1/2 when both approach, 3/4 when only one approaches, 7/8 when one approaches and the other retreats, 9/8 when only one retreats, and 5/4 when both retreat; all of these are positive. Moreover the two movement segments of a round are disjoint, since a never goes beyond xa + D/4 while b never goes below xb − D/4 = xa + 3D/4. Hence xa (t) < xb (t) at every time, movements included. (ii) No robot leaves the outer side of its own initial position. Let D1 be the distance at the Look of some approach of a and D2 the distance at the Look of its next retreat. After the approach round the 28

distance is at most 7D1 /8, and while a is not activated the robot b alternates approach and retreat, so the distance increases by a factor of at most 9/8. Hence D2 ≤ 63 64 D1 , and the retreat amount D2 /8 is smaller than the preceding approach amount D1 /4. Therefore the position of a after each retreat is larger than its position before the preceding approach, and inductively xa (t) ≥ 0. Symmetrically xb (t) ≤ 1. (iii) The collision-freedom condition. By (i) and (ii) we have 0 ≤ xa (t) < xb (t) ≤ 1, so fa (t) = xa (t) < xb (t) = ∥b(t) − a(0)∥, and symmetrically for b. (iv) The convergence condition. Give weight 1 to the state A and weight 6/5 to the state B, and put 9 when only one approaches, H := D · wa · wb . The factor by which H changes in one round is 34 · 65 = 10 9 5 15 1 36 18 5 25 125 · = when only one retreats, · = when both approach, · 8 6 16 2 25 25 4 36 = 144 when both retreat, 15 7 and 8 when one approaches and the other retreats; all of these are at most 16 . At least one robot is activated in every round, and wa , wb ∈ [1, 6/5], so D → 0 geometrically. Each displacement is at most D/4, so the total length travelled by each robot is finite and both converge to one and the same point l. (v) The zig-zag condition. By (i) we have D > 0, so every activation produces a nonzero Move, and by fairness each robot performs infinitely many approaches and retreats alternately. By (ii) we have xa ≥ 0, so an approach of a increases fa and a retreat decreases it. Let tk−1 be the landing time of the k-th Move of a, with t−1 := 0. Then [t2i−1 , t2i ] is an approach and [t2i , t2i+1 ] a retreat, and each Move progresses monotonically towards its destination, so fa takes the maxima fa (t2i ) and the minima fa (t2i+1 ) alternately. By (ii) the position after each retreat is larger than the position after the previous retreat, so fa (t2i+1 ) > fa (t2i−1 ). The same holds for b. Finally the displacement is a fixed ratio of the observed distance, so it requires no unit of length, and the state is private, so the algorithm is admissible in FSTA. Lemma C.15, Lemma C.16, Lemma C.17 and Lemma C.18 supply the boundary certificate of Table 2 for ZCC, which by Lemma 2.3 gives Prof(ZCC) = ↑F STAS as asserted in Theorem 4.3.

C.5

Vertical Configuration (VEC)

VEC [NNW26] requires the two robots to transform the segment joining them by the same rotations as in SRO and, once a configuration orthogonal to the initial segment has been reached, to stop there forever. Definition C.19 (Vertical Configuration). Two robots a, b start from distinct positions. The sequence of configurations Ci = (ai , bi ), the vectors di := bi − ai , the squares Qi with Q−1 := R2 , and the rotations Rϕ are defined as in Definition C.1. A vector u is orthogonal to d0 , written u ⊥ d0 , if u ̸= 0 and ⟨u, d0 ⟩ = 0. The first condition (attainment) requires di ⊥ d0 for some i. The second condition (transition) requires that for every i the vector di+1 be equal to di , or its clockwise 90◦ rotation, or its clockwise 45◦ rotation scaled by 2∓1/2 , and that ai+1 , bi+1 ∈ Qi−1 . The third condition (stopping) requires that if at a time at which both robots are at rest the segment is orthogonal to d0 , then neither robot ever moves again.  ΦVEC ≡ { ∃i ≥ 0 : di ⊥ d0 } n and ∀i ≥ 0 : di+1 ∈ {di , R−90◦ di , 2−1/2 R−45◦ di , 21/2 R−45◦ di }
o ∧ ai+1 , bi+1 ∈ Qi−1  and { ∀t ∈ Σab : b(t) − a(t) ⊥ d0 ⇒ ∀t′ ≥ t : a(t′ ) = a(t) ∧ b(t′ ) = b(t) } Impossibility Lemma C.20. VEC ∈ / Sol(OBLOT F ). Proof. Once an OBLOT F robot moves it keeps moving in every round and cannot stop when the orthogonal configuration is reached. Consider the fully synchronous execution of Lemma E.13. If λ = 0 the robots do not move, the sequence reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition fails. If 29

c2 = 0 the two robots coincide in round 1 and d1 = 0, while all the transitions permitted by the second condition are nonzero, so that condition fails. If c2 > 0 both robots perform a nonzero Move in every round, so even if dk ⊥ d0 holds at the end of some round they move again in the next one and the stopping condition fails; and if no such round exists the attainment condition fails. Lemma C.21 ([NNW26, Lemma 21]). VEC ∈ / Sol(F COMS ). Proof. Run a fully synchronous prefix, which is admissible under Ssynch. By Lemma E.14 the two robots always carry the same color. If λγ = 0 for every color γ occurring in that sequence, this fair execution contains no nonzero Move, the sequence reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition fails. Otherwise let γ be the first color with λγ > 0 and truncate the prefix at the round in which both robots carry γ; the configuration is still the initial one. Now activate a alone repeatedly (Lemma E.4; under Ssynch the robot that is not activated has no pending operation). The color of b stays γ, so by Lemma E.16(b) the robot a applies the same pair (λγ , θγ ) at every activation. With the peer at rest, a single nonzero Move can satisfy the transition condition only by making di+1 equal to R−90◦ di or to 2∓1/2 R−45◦ di , and each of these applies the same rotation to d at every activation. Hence a reaches a configuration with di ⊥ d0 in at most two moves, and since λγ > 0 it moves again at the very next activation, so the stopping condition fails. Lemma C.22 ([NNW26, Lemma 22]). VEC ∈ / Sol(F STAA ). Proof. By Proposition 3.2 we have F STAA ≡ F STAALC , so it suffices to exhibit an admissible ALC execution. Run a fully synchronous prefix, which is admissible under ALC . By Lemma E.14 the two robots always carry the same state. If λs = 0 for every state s occurring in that common sequence, this fair execution contains no nonzero Move, the sequence reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition fails. Otherwise let s be the first state with λs > 0 and truncate the prefix at the round in which both robots carry s; no nonzero Move has occurred, so the configuration is still the initial one. Activate a alone. It performs a nonzero Move from pB := a0 to some pE ̸= pB . The robot b has finished its cycle and has no pending operation, so the hypotheses of Lemma E.7 are met and, for each µ ∈ (0, 1), there is an execution in which b observes a at pµ = (1 − µ)pB + µpE . Let the Move of b begin before the Move of a ends; then the two Move intervals overlap, so no time of Σab occurs between the two landings and the configuration after both landings is C1 . In state s the robot b lands, by Lemma E.1(1), at b0 + λs Rθs (pµ − b0 ), so d1 (µ) = b0 + λs Rθs (pµ − b0 ) − pE is an affine function of µ whose slope λs Rθs (pE − pB ) is nonzero because λs > 0. Hence a suitable coordinate h(µ) of d1 is continuous and non-constant on (0, 1). For i = 0 we have Q−1 = R2 , so the transition condition only requires d1 ∈ {d0 , R−90◦ d0 , 2−1/2 R−45◦ d0 , 21/2 R−45◦ d0 }, and the corresponding values of h are only four. By Corollary E.22 there is a µ avoiding those four values, and for it the transition condition fails. Solvability Lemma C.23 ([NNW26, Lemma 23]). VEC ∈ Sol(F STAAM ), as witnessed by Algorithm 2. Proof. Since AM forbids a Look of one robot during a Move of the other, every configuration observed at a Look is one of the terms of the sequence (Ci ). Suppose first that the Look of b precedes the landing of a. Then both robots take as destinations the two remaining vertices of the square Q0 . If there is a time at which both robots are at rest between the two landings, the intermediate configuration has d1 = 2−1/2 R−45◦ d0 , and the next one has d2 = R−90◦ d0 = 21/2 R−45◦ d1 ; if there is no such time, the two landings produce a single term with d1 = R−90◦ d0 . Suppose instead that the Look of b follows the landing of a. Then d1 = 2−1/2 R−45◦ d0 , and b moves to a vertex of Q1 , giving d2 = 2−1/2 R−45◦ d1 . In every case, each transition has one of the permitted forms. In the first case, the final positions are the two remaining vertices of Q0 ; in the second case, they are a vertex and the centre of Q0 . Thus, when 30

Algorithm 2 VEC for FSTA under AM (robot r) Model and scheduler: FSTA, AM . State: r.status ∈ {1, 2}, initially 1. Look: observe the peer r. Comp: 1: if r.status = 1 then 2: r.des ← the image of r under the clockwise 90◦ rotation about the midpoint of r and r 3: r.status ← 2 4: else 5: r.des ← the current position of r ▷ no further movement 6: end if Move: move to r.des. an intermediate configuration exists, both positions of C2 belong to Q0 , as required. The containment condition for the first transition is vacuous since Q−1 = R2 . The final segment is orthogonal to d0 , which is the attainment condition, and both robots are then in state 2 and never move again; no earlier segment is orthogonal to d0 , since the only intermediate one is at 45◦ . Hence the stopping condition holds as well. Lemma C.20, Lemma C.21, Lemma C.22 and Lemma C.23 supply the boundary certificate of Table 2 for VEC, which by Lemma 2.3 gives Prof(VEC) = ↑FSTAAM as asserted in Theorem 4.3.

C.6

Dyadic-Multiple Stopped-Distance (DMSD)

Definition C.24 (Dyadic-Multiple Stopped-Distance). Let D0 = ∥a(0) − b(0)∥ > 0, and let Z[1/2] = {m/2k | m ∈ Z, k ∈ Z≥0 } be the set of dyadic rationals. Using the moving predicate of Section 2.3, define the set of joint stops Σab = {t ≥ 0 | ¬ Movea (t) ∧ ¬ Moveb (t)}, the set of times at which neither robot is inside a nonzero Move operation. Every execution must satisfy ∀t ∈ Σab : ∃t ∈ Σab :

∥a(t) − b(t)∥ ∈ Z[1/2], D0 ∥a(t) − b(t)∥ < D0 .

(D1) (D2)

Impossibility Lemma C.25. DMSD ∈ / Sol(F STAA ). Proof. The argument extends [KSW25, Lemma 6] to a general displacement (λ, θ) including a rotation, and to the internal states of FSTA. By Proposition 3.2 we have F STAA ≡ FSTAALC , so it suffices to exhibit, for an arbitrary FSTA algorithm A, an admissible ALC execution that violates DMSD. In the normalized frame of Lemma E.1(1), a robot in state s has displacement λs Rθs (w), where w is the vector to its peer, and its next state is determined by s alone. The local coordinate system may be chosen anew at each activation, and we supply this frame independently at every activation below. Normalize a(0) = 0, b(0) = e1 and D0 = 1. First stage. Run a fully synchronous prefix, which is admissible under ALC . By Lemma E.14 the two robots always carry the same state; let s0 , s1 , . . . be that common state sequence. If λs = 0 for every state s occurring in the sequence, this fair execution contains no nonzero Move, so a and b stay at their initial positions and ∥a(t) − b(t)∥ = D0 for every t ∈ Σab , which violates (D2). Otherwise let s be the first state with λs > 0 and truncate the prefix at the round in which both robots carry s; write λ := λs > 0 and R := Rθs . No nonzero Move has occurred, so the robots are at their initial positions and the ratio at every earlier joint stop is 1 ∈ Z[1/2]. Second stage. Activate b alone. It performs a nonzero Move to pE = e1 − λRe1 ̸= b(0), 31

and for µ ∈ (0, 1) we put pµ := (1 − µ)b(0) + µpE = e1 − µλRe1 . The robot a has finished its cycle and has no pending operation, so the hypotheses of Lemma E.7 are met and there is an execution in which a performs a Look at a time when b is at pµ . In state s the robot a then lands at λRpµ = λRe1 − µλ2 R2 e1 . Let both Move operations complete, delay the next Look of each robot by a finite amount, and append a fair tail. Throughout that delay neither robot is inside a nonzero Move, so the whole interval is contained in Σab , and mb ≥ 1 there. Third stage. On that interval the distance equals h(µ) = ∥c + µw∥,

c := e1 − 2λRe1 ,

w := λ2 R2 e1 ̸= 0.

Indeed ∥c + µw∥2 is a quadratic polynomial in µ with leading coefficient ∥w∥2 > 0, so h is continuous and non-constant on (0, 1). By Corollary E.22 there is a µ with h(µ)/D0 ∈ / Z[1/2], and the corresponding execution violates (D1). Since (D1) is of the form ∀t ∈ Σab , a single violating time suffices, and the joint stops produced in the first two stages need not be inspected. Corollary C.26 ([KSW25, Lemma 6]). DMSD ∈ / Sol(OBLOT A ). Proof. In Q the class OBLOT A lies below FSTAA , so this follows from Lemma C.25 and Proposition 2.1.

Solvability Lemma C.27 ([KSW25, Lemma 5]). DMSD ∈ Sol(OBLOT AM ), as witnessed by the Go-to-midpoint algorithm. Proof. Each robot moves to the midpoint of its own position and the observed position of its peer. By Proposition 3.2 we have OBLOT ACM ≡ OBLOT AM , so it is enough to argue under AM . Observed positions. Under AM a Look of one robot never falls inside the closed interval [tB , tE ] of a Move operation of the other, so every observed position is an initial position or the landing point of a completed nonzero Move. Moreover a robot is itself at rest at its own Look, since tE (r, k −1) < tL (r, k) < tB (r, k). Consequently every Look time belongs to Σab . Condition (D1). We claim that at every time the position of each robot is of the form (1 − α)a(0) + αb(0) with α ∈ Z[1/2], whenever that robot is not inside a Move operation. This holds initially with α = 0 and α = 1. A robot lands at the midpoint of its own position and an observed position, and by the previous paragraph both are of the stated form; the midpoint of (1−α)a(0)+αb(0) and (1−β)a(0)+βb(0) is the point with parameter (α + β)/2 ∈ Z[1/2]. Since the destination is computed from the snapshot taken at the Look, a delayed Move does not change the landing point (Corollary E.6), so the claim is preserved even when a snapshot is stale. Hence for t ∈ Σab we get ∥a(t) − b(t)∥ = |β − α| D0 with β − α ∈ Z[1/2], which is (D1). Condition (D2). Every landing point is a midpoint of two points of the segment [a(0), b(0)], so by induction both robots always lie in that segment. By fairness some robot, say a, is activated first; at its Look the two robots are still at their initial positions, so the observed distance is D0 > 0 and a performs a nonzero Move, landing at the midpoint m of a(0) and b(0). Let t := tL (a, 2) be the next Look of a, which exists by fairness and satisfies tE (a, 1) < t < tB (a, 2), so a(t) = m; by the first paragraph t ∈ Σab . Since b(t) lies in [a(0), b(0)] and m is the midpoint of that segment, ∥a(t) − b(t)∥ ≤ D20 < D0 , which is (D2). Lemma C.25 and Lemma C.27 together prove Lemma 3.4 (6).

32

C.7

Convergence (Cv)

Cv requires nothing but that the two robots converge to a single point. It occupies the row of the least class of Q, so its profile is the whole of Q. Definition C.28 (Convergence). Two robots a, b start from distinct positions and converge to a point l. ΦCv ≡ { ∃l ∈ R2 : lim a(t) = lim b(t) = l } t→∞

t→∞

Since OBLOT is the least element of Q, we have NOBLOT A = Max(Q \ ↑OBLOT A ) = ∅, so the negative half of the condition in Lemma 2.3, applied with U = ↑OBLOT A = Q, is vacuous for Cv. This is what the entry “Trivial” of Table 2 records. It concerns only the impossibility side: the positive half still has to be established, and we do so here. A

Solvability Algorithm 3 Cv for OBLOT under Asynch (robot r) Model and scheduler: OBLOT , Asynch. Look: observe the peer r at the local coordinate v. Comp: 1: r.des ← v/2 Move: move to r.des.

▷ the midpoint of the two observed positions

Lemma C.29 ([CP05]). Cv ∈ Sol(OBLOT A ), as witnessed by Algorithm 3. Proof. Algorithm 3 is the two-robot case of the gravitational algorithm of [CP05]: each robot moves to the center of gravity of the configuration it has observed, which for two robots is the midpoint of the two observed positions. The destination is determined by the midpoint alone, so it requires neither a unit of length nor chirality, and it is computed from the current snapshot alone, so the algorithm is admissible in OBLOT . All the motion takes place on the line through a(0) and b(0), since the midpoint of two points of that line lies on it. Let H(t) be the convex hull, an interval of that line, of the current positions of the two robots together with the destinations determined by all Looks whose cycles have not yet completed at time t, including cycles awaiting Compute We claim that H is non-increasing. Indeed, the destination determined by a Look is the midpoint of two positions occupied at that time, both of which belong to H at that time, so the destination belongs to H as well; and every movement is a rigid motion along a segment whose two endpoints belong to H, so it never leaves H. In particular H(t) ⊆ H(0) = [a(0), b(0)] for every t, which is the containment hypothesis of [CP05]. Executions are fair, so each robot performs infinitely many LCM cycles, and the adversary’s freedom in Asynch is exactly the one allowed there: the destination carried into a Move may have been computed from a stale snapshot, and the two robots may be in flight simultaneously. Under these hypotheses the gravitational algorithm converges [CP05], so there is a point l with a(t), b(t) → l, which is ΦCv . Lemma C.29 is the positive half of the boundary certificate for Cv; the negative half is empty, as noted above. By Lemma 2.3 we obtain Prof(Cv) = ↑OBLOT A = Q, as asserted in Theorem 4.3.

D

The New and Revised Tasks

D.1

Expansion Preserving Gravity (EPG)

This subsection and Appendix D.2 read their requirements off the same sequence of configurations, the one already used for SRO in Definition C.1 and for VEC in Definition C.19. Let Σab := { t ≥ 0 | ∀r ∈ {a, b} : ¬Move r (t) } 33

Algorithm 4 EPG for FSTA under Fsynch (robot r) Model and scheduler: FSTA with C = {0, 1} and initial color 0, Fsynch. Look: observe the peer r̄ at the local coordinate v, and read s := r.color. Comp: 1: if s = 0 then 2: r.des ← −v/2 ▷ away from the observed midpoint, by the observed distance 3: r.color ← 1 4: else 5: r.des ← 0 6: end if Move: move to r.des. be the set of joint stops, the times at which neither robot is inside a nonzero Move operation. List the configurations taken at the times of Σab in chronological order, remove consecutive
repetitions, and write the resulting sequence as C0 , C1 , C2 , . . . with Ci = (ai , bi ); thus C0 = a(0), b(0) . Put
di := bi − ai . c0 := 21 a(0) + b(0) , Since MoveSet(r) collects only the open intervals of the nonzero Move operations of r, at every t ∈ Σab each robot stands either at its initial position or at the landing point of its last completed nonzero Move; in particular 0 ∈ Σab and the sequence is well defined. Remark D.1 (Preserving the center of gravity forces simultaneity). Suppose 21 (ai+1 + bi+1 ) = c0 = 1 2 (ai + bi ). Then ai+1 + bi+1 = ai + bi , so bi+1 = bi would force ai+1 = ai and hence Ci+1 = Ci . Consecutive terms of the sequence being distinct, every step of the sequence therefore moves both robots. Equivalently, a nonzero Move performed by one robot while the other is stationary across the surrounding joint stops displaces the center of gravity by half that displacement. Definition D.2 (Expansion Preserving Gravity). Two robots a, b start from distinct positions. The first condition (one expansion) requires the joint-stop sequence to consist of exactly the two terms C0 and C1 . The second condition (expansion preserving gravity) requires the step from C0 to C1 to keep the center of gravity at c0 and to double the segment. The third condition (stopping) requires both robots to remain stationary from some time on.  ΦEPG ≡ { (Ci )i≥0 has exactly the two terms C0 , C1 }  and 12 (a1 + b1 ) = c0 and d1 = 2d0  and { ∀r ∈ {a, b} : Stopped(r) } The two equalities determine C1 outright: a1 = c0 − d0 and b1 = c0 + d0 , that is, each robot ends at twice its initial distance from c0 each robot ends at twice its initial distance from c0 . The predicate mentions only the trajectory and γ(0): no floor function, no reference to the local coordinate systems, and no reference to activation counts, since the terms of (Ci ) are determined by the configurations at the joint stops and not by how many cycles produced them. Lemma D.3 (Bounds for EPG). Under the assumptions of this paper, EPG ∈ Sol(LU MI F ),

EPG ∈ / Sol(OBLOT F ) ∪ Sol(LU MI R ).

Proof. The positive statement is Lemma D.4, and the two negative ones are Lemmas D.5 and D.6 below. Solvability Lemma D.4. EPG ∈ Sol(LU MI F ), as witnessed by Algorithm 4 in FSTAF . 34

Proof. Both robots start with color 0, so in round 1 both execute the first branch. A robot at p observing its peer at the local coordinate v lands at p + L−1 (−v/2) = p − 21 (pr̄ − pr ), that is, it moves away from the peer by half the observed distance; the displacement is determined by the snapshot alone and needs neither a unit of length nor chirality, so Algorithm 4 is admissible in F STA. The two displacements are 1 1 2 (a0 − b0 ) and 2 (b0 − a0 ), opposite and of equal norm, so after round 1 1 1 2 (a1 + b1 ) = 2 (a0 + b0 ) = c0 ,

d1 = 2d0 ,

which is the second condition. Both displacements are nonzero because d0 ̸= 0, and under Fsynch the two Move operations occupy the same round, so no time of Σab lies strictly between the two boundaries of round 1 and C1 = (a1 , b1 ) is the term of the sequence following C0 . Both robots end round 1 with color 1 and thereafter execute the second branch, so every later Move is zero. The configuration never changes again, the joint-stop sequence consists of exactly C0 and C1 , and both robots are permanently stationary. Hence ΦEPG holds, and F STAF is the class LUMI F by Theorem 3.3. Impossibility Lemma D.5. EPG ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch. An OBLOT robot has no color and no memory, so its output at an activation is a function of the observed local position v of the peer alone; write it f (v). The adversary may give a robot a frame in which the peer appears at any prescribed nonzero local position (Lemma E.1(1)). Fix frames for round 1 and let va , vb be the resulting local inputs. If C1 , the configuration at the end of round 1, violates 21 (a1 + b1 ) = c0 or d1 = 2d0 , the second condition fails, provided C1 ̸= C0 . If C1 = C0 , both f (va ) and f (vb ) are zero; repeating the same frames in every round leaves the configuration unchanged forever, the joint-stop sequence has the single term C0 , and the first condition fails. So assume C1 ̸= C0 and the second condition holds at i = 0. Then d1 = 2d0 ̸= 0, so in round 2 the adversary may choose a frame for a with L′a (b1 − a1 ) = va and one for b with L′b (a1 − b1 ) = vb . Both robots receive the local input they received in round 1, and by determinism they output f (va ) and f (vb ) again. At least one of these is nonzero, since C1 ̸= C0 , so the configuration at the end of round 2 differs from C1 and contributes a third term to the joint-stop sequence. The first condition fails. Lemma D.6. EPG ∈ / Sol(LU MI R ). Proof. Apply Lemma 5.1(2) with K = R. Corollary D.7. Prof(EPG) = ↑LUMI F . Proof. Table 2 gives Max(Q \ ↑LU MI F ) = {OBLOT F , LUMI R }. Apply Lemma 2.3 with Lemma D.4 on the positive side and Lemmas D.5 and D.6 on the negative side.

D.2

Perpetual Expansion Preserving Gravity (EPG∗ )

Definition D.8 (Perpetual Expansion Preserving Gravity). Two robots a, b start from distinct positions, and Σab , the sequence (Ci )i≥0 , c0 and di are as in Appendix D.1. The first condition (perpetuity) requires the sequence to be infinite. The second condition (expansion preserving gravity) requires every step of the sequence to keep the center of gravity at c0 and to double the segment.  ΦEPG∗ ≡ { (Ci )i≥0 is infinite } n o and ∀i ≥ 0 : 12 (ai+1 + bi+1 ) = c0 and di+1 = 2di

35

Algorithm 5 EPG∗ for OBLOT under Fsynch (robot r) Model and scheduler: OBLOT , Fsynch. Look: observe the peer r̄ at the local coordinate v. Comp: 1: r.des ← −v/2 ▷ away from the observed midpoint, by the observed distance Move: move to r.des.
The two conditions give Ci = c0 − 2i−1 d0 , c0 + 2i−1 d0 for i ≥ 1, so the configuration at the i-th joint stop of the sequence is determined by γ(0) alone. EPG is the one-shot version: Definition D.2 keeps the second condition at i = 0 and replaces perpetuity by a single expansion followed by permanent stopping. Remark D.9 (Relation to CGE and CGE∗ ). Center of Gravity Expansion and its perpetual variant [BFK+ 25, FSW20] prescribe each landing point by a coordinatewise floor, which makes the expansion irregular and moves the center of gravity. That formulation cannot be carried over to the present model. Evaluated on a lattice common to the two robots, as in the global formula of [BFK+ 25], the destination is not computable under variable disorientation, so the profile would be empty; evaluated on each robot’s own lattice, the predicate refers to the local frames chosen by the adversary and is therefore no longer a predicate on the trajectory, contrary to Section 2.3. We keep the exact doubling, drop the floor, and recover the same boundary by requiring the center of gravity to be preserved at every joint stop. By Remark D.1 that requirement rules out a solo expansion, which is the role the floor played in [FSW20]. Solvability Lemma D.10. EPG∗ ∈ Sol(OBLOT F ), as witnessed by Algorithm 5. Proof. The destination is computed from the current snapshot alone and requires neither a unit of length nor chirality, so the algorithm is admissible in OBLOT . We show by induction on k ≥ 0 that at the start of round k + 1 the two robots are stationary with no pending operation in the configuration
Γk := c0 + 2k (a0 − c0 ), c0 + 2k (b0 − c0 ) . This holds for k = 0. Assume it for k and let δ := 2k d0 ̸= 0 be the distance. Under Fsynch both robots are activated in round k + 1 and compute from the same configuration; each moves away from the peer by δ/2, so the two displacements are opposite and of equal norm. The midpoint is therefore unchanged and the difference vector is doubled, which is Γk+1 , and both Move operations are nonzero since δ ̸= 0. The two Move operations of a round occupy the same interval, so no time of Σab lies strictly between consecutive round boundaries at which a Move occurs, and the joint-stop sequence is exactly (Γk )k≥0 . It is infinite, consecutive terms are distinct because δ ̸= 0, and every step satisfies 21 (ak+1 + bk+1 ) = c0 and dk+1 = 2dk . Hence ΦEPG∗ holds. Impossibility Lemma D.11. EPG∗ ∈ / Sol(LUMI R ). Proof. Apply Lemma 5.1(2) with K = R. Lemma D.11 is the negative half of Lemma 3.4 (1), whose positive half is Lemma D.10. Since Max(Q \ ↑OBLOT F ) = {LUMI R }, Lemma 2.3 gives Prof(EPG∗ ) = ↑OBLOT F , as already recorded in Theorem 3.6.

D.3

Oscillating Points with One Move per Robot per Phase (OSPO)

We introduce OSPO (Oscillating Points with One Move per Robot per Phase), a strengthened version of OSP [DFP+ 16]. The first approach phase starts at time 0, all phase boundaries are joint stops, and each robot completes exactly one nonzero Move in each phase. 36

Definition D.12 (Oscillating Points with One Move per Robot per Phase). Two robots a, b start from distinct positions, and d(t) is their distance as in Definition C.5. For r ∈ {a, b}, let mr (t) denote the number of nonzero Move operations completed by r by time t. The task OSPO requires the robots to remain at distinct positions throughout the execution and to alternate approach and retreat forever, starting with approach at time 0. Each phase is delimited by joint stops, and each robot completes exactly one nonzero Move during that phase. More precisely, there is a strictly increasing sequence (tj )j≥0 of joint stops, with t0 = 0 and tj → ∞, such that on each interval [t2i , t2i+1 ] the distance is monotonically non-increasing and decreases overall, and on each interval [t2i+1 , t2i+2 ] it is monotonically non-decreasing and increases overall. The move-count condition is mr (tj+1 ) − mr (tj ) = 1 for every j ≥ 0 and r ∈ {a, b}. ΦOSPO ≡ { ∀t ≥ 0 : d(t) > 0 } ∧ ∃ (tj )j≥0 ⊆ Σab :

h t0 = 0,

tj → ∞,

∀j ≥ 0 : tj < tj+1 , ∀j ≥ 0 ∀r ∈ {a, b} : mr (tj+1 ) − mr (tj ) = 1,  ∀i ≥ 0 : d(t2i+1 ) < d(t2i ) ∧ ∀h, h′ ∈ [t2i , t2i+1 ] : h < h′ ⇒ d(h′ ) ≤ d(h)

∧ d(t2i+2 ) > d(t2i+1 ) ∧ ∀h, h′ ∈ [t2i+1 , t2i+2 ] : h < h′ ⇒ d(h′ ) ≥ d(h)

i .

The specification ΦOSPO combines safety constraints with a requirement of infinite progress. Collision freedom is violated as soon as d(t) = 0 for some t. The move-count requirement also admits finite witnesses of failure: since t0 = 0 and each phase contains exactly one completed nonzero Move by each robot, every satisfying execution must have ma (tj ) = mb (tj ) = j and hence |ma (t) − mb (t)| ≤ 1 for every t ≥ 0. Thus, if one robot completes two nonzero Moves before the other completes its first, no extension can satisfy ΦOSPO . After either a collision or such a move-count violation, any admissible fair continuation witnesses failure. By contrast, monotonicity of d along a finite solo activation interval does not by itself establish failure: such an interval may lie within one phase, subject to the move-count requirement. Requiring phase boundaries to be joint stops also prevents a decrease and subsequent increase within a single simultaneous Move from being counted as two phases, since there is no intervening joint stop. For infinite executions, the divergence tj → ∞ provides another sufficient criterion for failure. If, for some T , the restriction of d to Σab ∩ [T, ∞) is monotone (non-increasing or non-decreasing), then ΦOSPO fails. Indeed, any witnessing sequence would eventually have both a strictly decreasing phase and a strictly increasing phase with endpoints in Σab ∩ [T, ∞), contradicting that monotonicity. This includes the case where d is eventually constant at joint stops. Impossibility Lemma D.13. OSPO ∈ / Sol(OBLOT F ). Proof. Fix any deterministic OBLOT algorithm and take the fully synchronous execution of Lemma E.13, with normalized local frames at every Look while the robots are distinct and with time-divergent round timings. If the two robots ever occupy the same point, the collision-freedom condition of ΦOSPO fails at a finite time. This covers c2 = 0, where they coincide at the end of round 1, and also the case in which they meet while travelling inside a round. Otherwise d(t) > 0 throughout. By Lemma E.13 every joint stop of the execution carries one of the round-boundary distances d(k) = c2k d(0), so the restriction of d to Σab is monotone. The turning times

37

required by ΦOSPO are joint stops, so by the criterion stated after Definition D.12 their distances cannot alternate between strict decrease and strict increase. This admissible fair execution violates ΦOSPO . Lemma D.14. OSPO ∈ / Sol(F COMS ). Proof. Fix any deterministic FCOM algorithm. Use normalized local frames and activate both robots synchronously as long as their computed displacements are zero. Their colors remain equal during this prefix. If no nonzero displacement ever occurs, continuing synchronously gives an admissible fair execution with no movement, which violates OSPO. Otherwise, let γ be the common color immediately before the first round that would produce nonzero displacements. Instead of activating both robots in that round, activate a alone for two complete consecutive cycles, keeping b inactive. If a collision occurs, OSPO already fails. Otherwise, b remains at its initial position with color γ. At both Looks of a, normalization presents the same relative position and the same peer color. Since an FCOM robot cannot read its own color, it computes the same nonzero normalized displacement in both cycles. At the completion time T of the second Move, ma (T ) = 2,

mb (T ) = 0.

In any execution satisfying OSPO, the first phase starts at t0 = 0 and ends at a joint stop t1 with ma (t1 ) = mb (t1 ) = 1. Since b has not moved by time T , such a t1 would have to satisfy t1 > T ; but then ma (t1 ) ≥ 2, a contradiction. Thus the finite prefix already makes OSPO impossible. Append any admissible fair, time-divergent Ssynch continuation. Lemma D.15. OSPO ∈ / Sol(F STAS ). Proof. Fix any deterministic F STA algorithm and use normalized local frames whenever the robots are distinct. Under these observations, a robot’s next private state and normalized displacement depend only on its current private state. Let s0 , s1 , s2 , . . . be the resulting state sequence from the designated initial state, and let λj ≥ 0 be the displacement magnitude associated with state sj . Suppose first that this sequence contains at least two nonzero displacements. Activate a alone until it completes its second nonzero Move, while keeping b inactive. This takes finitely many cycles. If a collision occurs, the collision-freedom condition fails. Otherwise, at the completion time T , ma (T ) = 2,

mb (T ) = 0.

No endpoint t1 of the first phase can then satisfy ma (t1 ) = mb (t1 ) = 1: it cannot occur by time T because b has not moved, and it cannot occur later because a has already moved twice. The violation persists under any admissible fair, time-divergent continuation. Suppose instead that the state sequence contains at most one nonzero displacement. Choose an infinite fully synchronous execution with normalized frames. If the robots collide, OSPO fails. Otherwise, each robot follows the state sequence above and completes at most one nonzero Move. This is incompatible with infinitely many phases, each requiring one nonzero Move from each robot. Hence this fair, time-divergent execution also violates OSPO. Solvability Lemma D.16. OSPO ∈ Sol(F STAR ). Proof. The algorithm. Each robot keeps two private states, approach and retreat, initially approach. Upon an activation, let p be its own position and q the observed position of its peer. In state approach its destination is p + 14 (q − p), in state retreat it is p − 14 (q − p); in both cases it then switches to the other state. The destination is a fixed fraction of the observed displacement, so it depends neither on the 38

orientation nor on the unit distance of the local frame and is admissible under variable disorientation. The two robots are always at distinct positions, as shown below, so q ̸= p and every Move is nonzero; hence every activation increases mr by exactly one. One round. Let D > 0 be the distance at the start of a round, and for each robot r let ur ∈ [0, 1] be the fraction of its movement segment covered so far, set to 0 before its Move begins and to 1 after its completion; each ur is continuous and non-decreasing, and ua + ub runs from 0 to the number of robots activated in the round. If both robots are activated and share their state, the distance during the round is   ua + ub  ua + ub  (approach), D 1+ (retreat), D 1− 4 4 since both displacements are collinear with the segment. Here ua + ub ∈ [0, 2], so the approach factor lies in [ 21 , 1]: the distance is monotone in the required direction, stays positive, and ends at D/2 or 3D/2. If a single robot is activated, the same expressions hold with ua + ub ∈ [0, 1], the approach factor lies in [ 34 , 1], and the final distance is 3D/4 or 5D/4. Phases. If the execution is fully synchronous forever, both robots always share their state and successive rounds alternate between approach and retreat; take each round as one phase. Otherwise let ℓ ≥ 0 be the length of the fully synchronous prefix, after which the activations are alternating singletons (Lemma E.8). Both robots share their state throughout the prefix, and round k ≤ ℓ is an approach round exactly when k is odd; at the end of the prefix their common state is approach exactly when ℓ is even. Partition the singleton tail into consecutive pairs of rounds. In a pair the two robots are activated once each; the second one has not yet switched, so both use the state held in common at the start of the pair, and at the end of the pair both hold the switched state. Take each such pair as one phase. Consequently the first tail phase is an approach phase exactly when ℓ is even, which for ℓ ≥ 1 is the type opposite to round ℓ; the tail phases alternate, so the phase types alternate across the whole execution, beginning with approach. Within a phase. A phase of the first kind is a single round, treated above. A phase of the second kind is two singleton rounds of the same type. Each of them is monotone in the same direction and they agree at the round boundary, so the distance is monotone on the whole phase; starting from D it ends 9 D in an approach phase and at 25 at 16 16 D in a retreat phase, so it also changes strictly. In every case the distance stays positive, so d(t) > 0 for all t ≥ 0. The times tj . Take t0 = 0 and let tj+1 be the end of the (j + 1)-st phase. Each phase ends at a round boundary, where neither robot is inside a Move, so tj ∈ Σab for every j; the same holds at t0 = 0, where no cycle is pending. Each phase contains exactly one nonzero Move of each robot, so mr (tj+1 ) − mr (tj ) = 1 for every j ≥ 0 and r ∈ {a, b}. The phases exhaust the rounds of the execution and each contains at least one round, so tj → ∞ by time divergence. Together with the monotonicity, strictness and positivity established above, and with the first phase being an approach phase, all conditions of Definition D.12 hold. Lemma D.13, Lemma D.14, Lemma D.15 and Lemma D.16 supply the boundary certificate of Table 2 for OSPO, which by Lemma 2.3 gives Prof(OSPO) = ↑FSTAR as asserted in Theorem 4.3.

D.4

Collision-Free Single Move (CFSM)

Definition D.17 (Collision-Free Single Move). Two robots a and b, initially at distinct positions, must each perform exactly one nonzero Move and remain stationary before that Move begins and after it ends. They must never collide, including during their movements. With moves(r) denoting the number of nonzero Moves of robot r, the specification is


ΦCFSM ≡ moves(a) = 1 ∧ moves(b) = 1 ∧ ∀t ≥ 0 : a(t) ̸= b(t) . Impossibility Lemma D.18. CFSM ∈ / Sol(OBLOT F ). 39

Proof. Let A be any OBLOT algorithm and consider the fully synchronous execution of Lemma E.13. Since an OBLOT robot has no visible state and the normalized local input of each robot is the same in every round, the two robots apply one and the same pair (λ, θ) in every round, and the configuration is updated by the simultaneous contraction ratio c2 = c2 (λ, θ) of Lemma E.2(ii). If λ = 0, no robot ever performs a nonzero Move, so moves(a) = moves(b) = 0 and the single-move condition fails. If c2 = 0, that is (λ, θ) = (1/2, 0), both robots land on the common midpoint in round 1, so a(t) = b(t) at that time and the collision-free condition fails. If c2 > 0, the distance stays positive, hence λ > 0 makes every round a nonzero Move for both robots and moves(a) = moves(b) = ∞, so the single-move condition fails again. The execution is fair in all three cases. Lemma D.19. CFSM ∈ / Sol(F COMS ). Proof. Let A be any deterministic FCOM algorithm. First stage. Run a fully synchronous prefix, which is admissible under Ssynch. By Lemma E.14 the two robots always carry the same color, and we write γ0 , γ1 , . . . for that common color sequence. If λγ = 0 for every color γ occurring in the sequence, then this fair execution contains no nonzero Move and moves(a) = 0, so the single-move condition fails. Otherwise let γ be the first color of the sequence with λγ > 0 and truncate the prefix at the round in which both robots carry γ. Nothing has moved so far, so the robots are still at their initial positions and the distance is d > 0. Second stage. From that configuration, activate a alone repeatedly (Lemma E.4; under Ssynch the robot that is not activated has no pending operation, and the activations of b are placed far enough in the future to keep the execution fair). The robot b is never activated, so its color remains γ, and the visible state of a, which is the color of its peer, is constant along the interval. By Lemma E.16(b) the robot a therefore applies the same pair (λγ , θγ ) at every activation. Since λγ > 0 and the distance is positive, the first activation is a nonzero Move. If it lands on b(0), the two robots occupy the same point and the collision-free condition fails. Otherwise the distance remains positive, so the second activation is a nonzero Move as well and moves(a) ≥ 2, which violates the single-move condition. Solvability Algorithm 6 uses one private bit to make each robot move exactly once. The move is directed away from the observed peer, so the two trajectories remain on opposite sides of the initial segment, even under Asynch. Algorithm 6 CFSM for FSTA under Asynch (robot r) Model and scheduler: FSTA, Asynch. State: r.status ∈ {0, 1}, initially 0. Look: observe the peer r and obtain the distance D = ∥r − r∥. Comp: 1: if r.status = 0 then 2: r.des ← the point at distance D/2 from r in the direction away from r 3: r.status ← 1 4: else 5: r.des ← the current position of r 6: end if Move: move to r.des.

▷ no further movement

Lemma D.20. CFSM ∈ Sol(F STAA ), as witnessed by Algorithm 6. Proof. Put a coordinate on the line through a(0) and b(0) so that b(0) < a(0). We claim the invariants a(t) ≥ a(0)

and b(t) ≤ b(0)

40

(∀t ≥ 0),

which give b(t) ≤ b(0) < a(0) ≤ a(t) and hence the collision-free condition at every instant, during movements included. Each robot moves at most once, namely at the activation at which its status is still 0; at that moment it stands at its initial position. Let p be the position of b recorded by the Look of a. If b has not yet moved then p = b(0); if b is moving or has already landed then p ≤ b(0) by the invariant for b. In both cases p < a(0), so D = a(0) − p > 0 and a moves to
a(0) + 21 a(0) − p > a(0), and it stays there because all its later Move operations are zero. The argument for b is symmetric. The two inductions use only the fact, valid from time 0 on, that neither trajectory has ever crossed the initial position of the other, so they are not circular. For the single-move condition, fairness guarantees that each robot is activated at least once. At its first activation the observed distance is positive, so its displacement D/2 is nonzero; every later activation produces a zero Move. Hence moves(r) = 1 for both robots. Lemma D.18, Lemma D.19 and Lemma D.20 together prove Lemma 3.4 (4).

D.5

The Shared Exchange Phase of S-RDV and VSS

S-RDV (Definition D.24) and VSS (Definition D.29) differ in their terminal requirements, one demanding rendezvous and the other permanent stopping, but they impose the same initial instance and the same exchange phase. This subsection isolates that common part and proves the four statements that Appendices D.6, D.7, F and G use. The common setting Let a0 = a(0) and b0 = b(0) be the distinct initial positions, let d = ∥b0 − a0 ∥ > 0, and let A = 2a0 − b0 ,

B = 2b0 − a0 ,

so that A, a0 , b0 , B occur in this order at equal spacing d on the initial line. Throughout this appendix we place a coordinate on that line with A = 0,

a0 = d,

b0 = 2d,

B = 3d,

and write P := {0, d, 2d, 3d} for the set of designated points. The exchange configurations are C1 = (B, b0 ),

C2 = (a0 , A),

C3 = (B, A).

In the notation of Definition D.29 these are A = A0 , B = B0 and Ci = γi for i = 1, 2, 3, while γ0 = (a0 , b0 ) is the initial configuration. We write mr (T ) for the number of nonzero Move operations performed by r up to time T . Both tasks require the following, which we refer to collectively as condition (P1). (P1-a) In the initial configuration the points A, a0 , b0 , B lie on a line in this order at equal spacing d. (P1-b) Every nonzero Move ends at a point of P : before T1 for S-RDV, and throughout the execution for VSS. (P1-c) At some time
T1 one of C1 , C2 , C3 is formed, with the corresponding cumulative counts ma (T1 ), mb (T1 ) equal to (1, 0), (0, 1) or (1, 1).

41

γ0

A

a0

b0

a

b

C1

b

C2 C3

b

B

a

a a

b

Figure 2: The initial configuration γ0 and the three exchange configurations C1 –C3 admitted in the first phase. The first nonzero Move Lemma D.21 (A nonzero Move must occur). In every execution, starting from the initial configuration, of a deterministic algorithm that solves a task containing (P1), at least one robot performs a nonzero Move. Proof. If neither robot ever performs a nonzero Move, then ma (T ) = mb (T ) = 0 for every T . Each of the three clauses of (P1-c) requires ma (T1 ) + mb (T1 ) ≥ 1, so no valid T1 exists, a contradiction. Lemma D.22 (Uniqueness of the first Move). In every execution satisfying S-RDV or VSS, each nonzero Move with earliest completion time lands at B if performed by a, and at A if performed by b. If the Look associated with that Move observes the initial geometric configuration, its normalized displacement is (λ, θ) = (2, 0). Proof. Such a Move exists by Lemma D.21. Suppose that a completes one at the earliest completion time t∗ , and let p be its landing point. Then ma (t∗ ) = 1; the peer may complete its first nonzero Move simultaneously. Any valid exchange time T1 satisfies T1 ≥ t∗ , since both completed-move counts are zero before t∗ . As ma (T1 ) ≥ 1, the witness cannot be C2 . The remaining possibilities, C1 and C3 , both require a(T1 ) = B,

ma (T1 ) = 1.

Suppose that p ̸= B. Then T1 > t∗ , and reaching B requires another nonzero Move of a. In S-RDV, every movement remains in [A, B]. In VSS, every movement is a straight segment with both endpoints in P ⊆ [A, B]. In either case, the exterior endpoint B cannot be reached in the interior of a nonzero movement. Thus reaching B requires completing another nonzero Move, giving ma (T1 ) ≥ 2, a contradiction. Therefore p = B. The argument for a first completion by b is symmetric and gives the landing point A. Now assume that the Look associated with this Move observes the initial geometric configuration. For a Move of a, the observed positions are a0 , b0 , and B − a0 = 2(b0 − a0 ). Hence the displacement is twice the observed distance in the direction of the peer, so its normalized form is (λ, θ) = (2, 0). For a Move of b, the same conclusion follows from A − b0 = 2(a0 − b0 ). The proof uses the fact that every Move is contained in [A, B]. For S-RDV, this is an explicit requirement. For VSS, it follows from straight-line motion and the requirement that every Move ends in P ⊆ [A, B], since the initial positions also belong to P . Replaying the first Move The next lemma is the obstruction used whenever an exchange task is refuted in a model without peerstate information. It says that in FSTA and OBLOT the required first Move is necessarily imitated by the peer, which carries the pair out of the admissible region. 42

Lemma D.23 (The peer replays the first Move). Let M ∈ {FSTA, OBLOT } and consider an execution that starts from the initial configuration, consists of alternating singleton activations (Lemma E.8), and is given normalized frames (Lemma E.1(1)) up to and including the round following the first nonzero Move. If that first nonzero Move complies with Lemma D.22, then the robot activated in the next round receives an isomorphic local input, applies the same displacement, and lands at 4d if it is b, and at −d if it is a. In particular the landing point lies outside P and outside the segment [A, B]. Proof. As long as no nonzero Move has occurred, the configuration is unchanged, so under alternating singleton activation the two robots traverse the same sequence of states. In F STA and OBLOT the internal state of the peer is not part of the local input, which consists of the relative position of the peer together with the robot’s own color in FSTA and of the relative position alone in OBLOT . Suppose the first nonzero Move is performed by a. By Lemma D.22 it lands at B = 3d, while b is still at b0 = 2d, so the distance is again d. In the next round b is activated, and since the adversary may supply the normalized frame, the local input of b is isomorphic to the one that a received. By determinism b applies the same displacement (λ, θ) = (2, 0) and lands at 2d + 2d = 4d. We have 4d ∈ /P and 4d ∈ / [A, B] = [0, 3d]. The case in which b moves first is symmetric: b lands at A = 0, a is still at d, and a lands at d − 2d = −d ∈ / P ∪ [A, B]. Lemma D.23 refutes the endpoint requirement (P1-b) of VSS and the interval requirement of S-RDV in a single step, and it is a safety violation, so it may be placed before any liveness argument in a concatenated execution (Lemma E.18(1)).

D.6

Swap–Rendezvous (S-RDV)

This subsection gives the formal specification of S-RDV, informally described in Section 4.2, and proves Lemma 4.2. Recall from Appendix D.5 the auxiliary points A = 2a0 − b0 and B = 2b0 − a0 , the exchange configurations C1 = (B, b0 ), C2 = (a0 , A), C3 = (B, A), the coordinate A = 0, a0 = d, b0 = 2d, B = 3d on the initial line, and the set P = {0, d, 2d, 3d} of designated points. Definition D.24 (Swap–Rendezvous). An execution solves S-RDV if the following conditions hold. 1. Throughout the execution, both robots remain in the segment [A, B]. 2. There exists a time T1 > 0 at which one of C1 , C2 , C3 is formed, with the corresponding cumulative nonzero-move counts (1, 0), (0, 1), or (1, 1). Before T1 , every robot that is not moving is located at one of A, a0 , b0 , B. 3. There exists T2 > T1 such that a(t) = b(t) for every t ≥ T2 . Conditions 1 and 2 are condition (P1) of Appendix D.5, with (P1-b) imposed before T1 , and condition 3 is the terminal rendezvous requirement that distinguishes S-RDV from VSS. Consequently Lemmas D.21, D.22, and Lemma D.23 all apply to S-RDV. Impossibility Lemma D.25. S-RDV ∈ / Sol(F COMA ). Proof. Stable Rendezvous is impossible in FCOMA under variable disorientation, regardless of the number of colors and without restricting the algorithm to the L-class [MSW26]. Suppose, for contradiction, that an algorithm A solves S-RDV in FCOMA . Every initial configuration of Rendezvous with distinct robot positions is an instance of S-RDV, since the auxiliary points A and B are determined solely from the initial positions and are not landmarks available to the robots. Moreover, every execution satisfying S-RDV eventually satisfies a(t) = b(t)

for all sufficiently large t,

which is exactly the stable Rendezvous requirement. Hence A would solve Rendezvous in F COMA , a contradiction. 43

Lemma D.26. S-RDV ∈ / Sol(FSTAR ). Proof. Suppose an FSTAR algorithm A solves S-RDV. The local input of an FSTA robot is the pair consisting of the relative position of the peer and its own state; the state of the peer is not included. The adversary chooses the alternating singleton activation {a}, {b}, {a}, . . . with an empty fully synchronous prefix (Lemma E.8). First stage. The configuration is unchanged until the first nonzero Move. The initial configuration is symmetric under the rotation by π about the midpoint combined with exchanging a and b, and this rotation preserves chirality, so in the normalized frame (Lemma E.1(1)) both local inputs have the form (e, own state) and are isomorphic. The two robots start from the same initial state s0 , so a in round 2k − 1 and b in round 2k both pass from sk−1 to sk . If no state occurring in s0 , s1 , . . . produces a nonzero Move, the configuration never changes, contradicting Lemma D.21. Let j be the least index that produces a nonzero Move; then a performs the first nonzero Move of the whole execution in round 2j + 1, from state sj . Second and third stages. Since A is assumed to solve S-RDV, Lemma D.22 forces that Move to land at B = 3d. The execution is an alternating singleton activation from the initial configuration under normalized frames whose first nonzero Move complies with Lemma D.22, which is exactly the hypothesis of Lemma D.23. Hence the robot b, activated in the next round, receives an isomorphic local input, performs the same displacement and lands at 4d. Since 4d ∈ / [A, B], condition 1 of Definition D.24 is violated. Lemma D.27. S-RDV ∈ / Sol(OBLOT F ). Proof. Suppose an OBLOT F algorithm A solves S-RDV. The local input of an OBLOT robot is its current snapshot alone. In round 1 the local input of both robots is “the peer at distance d”. A zero Move would leave all later inputs unchanged and the configuration fixed forever, so by Lemma D.21 the Move is nonzero, and by Lemma D.22 the robot a lands at B and b lands at A, forming C3 at distance 3d. The local input in round 2, “the peer at distance 3d”, coincides with the round-1 local input of the initial instance of initial distance 3d, namely A′ = 0, a′ (0) = 3d, b′ (0) = 6d, B ′ = 9d, and an OBLOT robot cannot distinguish the two. In that instance the same argument as above shows that the first Move is nonzero, so by Lemma D.22 the output is twice the observed distance in the direction of the peer. Applying the same local rule in round 2 of the actual execution, a lands at 3d − 6d = −3d and b lands at 0 + 6d = 6d. Both lie outside [A, B] = [0, 3d], which violates condition 1 of Definition D.24. Solvability Lemma D.28. S-RDV ∈ Sol(FCOMS ), as witnessed by Algorithm 7. Proof. Each robot carries a light with color set {0, 1, 2, 3}, initially 0. Apart from the coincidence test, every rule depends on the color of the peer alone. (i) Both robots always lie in [A, B]. Under Ssynch, all active robots compute from the configuration at the beginning of the round. Rules R and 2 do not move, while rules 1 and 3 choose the midpoint and the peer’s position, respectively. These destinations belong to [A, B] whenever both current positions do. For rule 0, the peer still has color 0 and remains at its initial position. While that color remains 0, successive applications of rule 0 alternate the observing robot between its initial position and the opposite outer point. The possible configurations and destinations are (a, b) = (d, 2d) :

a → 3d = B,

(a, b) = (3d, 2d) :

a → d = a0 ,

(a, b) = (d, 2d) :

b → 0 = A,

(a, b) = (d, 0) :

b → 2d = b0 .

Thus every destination lies in [A, B]. Since each Move follows a straight segment, both robots remain in [A, B] throughout the execution. 44

Algorithm 7 S-RDV for F COM under Ssynch (robot r) Model and scheduler: FCOM, Ssynch. Light: r.color ∈ {0, 1, 2, 3}, initially 0. Look: observe the peer r and obtain the distance D = ∥r − r∥ and the color r.color. Comp: 1: if D = 0 then 2: r.des ← the current position of r ▷ rule R: keep the rendezvous 3: else if r.color = 0 then ▷ rule 0: jump over 4: r.des ← the point at distance 2D from r in the direction of r 5: r.color ← 1 6: else if r.color = 1 then 7: r.des ← the midpoint of r and r ▷ rule 1: to the midpoint 8: r.color ← 2 9: else if r.color = 2 then 10: r.des ← the current position of r ▷ rule 2: stay 11: r.color ← 3 12: else ▷ rule 3: to the peer 13: r.des ← the current position of r 14: r.color ← 1 15: end if Move: move to r.des. (ii) The first phase. Let T1 be the end of the first round. If both robots are activated, C3 is formed with cumulative nonzero-move counts (1, 1). If only a is activated, C1 is formed with counts (1, 0); if only b is activated, C2 is formed with counts (0, 1). Moreover, up to T1 , every stationary robot is at its initial position or at its first destination, all of which belong to P . Hence T1 is a valid witness for the exchange phase. (iii) The second phase. By fairness, both robots are activated within finite time. At the end of the first round by which this has happened, the color pair is (1, 1), (1, 2), or (2, 1). From (1, 1), simultaneous activation brings both robots to their common midpoint; a singleton activation produces (1, 2) or (2, 1). Consider (1, 2), the other case being symmetric. Until a is activated, only b moves, repeatedly halving the distance while preserving the color pair. By fairness, a is eventually activated: it sees color 2, stays in place, and changes its color to 3. Whether or not b is activated in that round, the resulting color pair is (3, 2). Until the next activation of b, robot a remains stationary. When b is activated, it sees color 3 and moves to a; if a is activated simultaneously, it sees color 2 and stays. Thus the robots occupy the same point at the end of this round. Fairness ensures that this occurs at some finite time T2 > T1 . (iv) Stability of the rendezvous. At the round boundary T2 , both robots have completed their Moves and occupy the same point. Every subsequent Look therefore observes coincidence, so rule R applies and neither robot moves again. Hence a(t) = b(t) for every t ≥ T2 . Lemma D.25, Lemma D.26, Lemma D.27 and Lemma D.28 together prove Lemma 4.2.

D.7

Visit Swap and Stop (VSS)

Definition D.29 (VSS). Let a0 = a(0) and b0 = b(0) be distinct, and define A0 = 2a0 − b0 , B0 = 2b0 − a0 . The four points A0 , a0 , b0 , B0 occur in this order at equal spacing on the initial line. Write γ(t) = (a(t), b(t)) and set γ0 = (a0 , b0 ), γ1 = (B0 , b0 ), γ2 = (a0 , A0 ),

γ3 = (B0 , A0 ).

Every completed Move must end at one of the four designated points. The execution must first reach one of γ1 , γ2 , γ3 , with cumulative nonzero move counts (1, 0), (0, 1), or (1, 1), respectively. After this exchange configuration has been formed, the robots must eventually remain permanently stationary in a 45

configuration belonging to {γ0 , γ1 , γ2 , γ3 }. The restriction to designated points concerns Move endpoints; straight-line travel between them, including passage through the peer’s position, is allowed. Impossibility Lemma D.30. VSS ∈ / Sol(F STAR ). Proof. Use the alternating singleton execution of the proof of Lemma D.26. Its first stage holds word for word, since it uses only (P1), so the first nonzero Move of the whole execution is performed by a in round 2j + 1 under a normalized frame. Since the algorithm is assumed to solve VSS, Lemma D.22 forces that Move to land at B = 3d, and Lemma D.23 then applies: the robot b, activated in the next round, receives an isomorphic local input, performs the same displacement and lands at 4d ∈ / P . This contradicts the requirement of Definition D.29 that every completed Move end at one of the four designated points. Lemma D.31. VSS ∈ / Sol(OBLOT F ). Proof. As in the proof of Lemma D.27, in round 1 the robot a lands at B and b lands at A, forming C3 , and in round 2 the robots cannot distinguish their input from the round-1 input of the initial instance of initial distance 3d, so a lands at −3d and b lands at 6d. Neither point belongs to P , so the endpoint requirement of Definition D.29 is violated. Solvability Algorithm 8 VSS for FCOM under Asynch (robot r) Model and scheduler: FCOM, Asynch. Light: r.color ∈ {0, 1}, initially 0. Look: observe the peer r and obtain the distance D = ∥r − r∥ and the color r.color. Comp: 1: if r.color = 0 then 2: r.des ← the point at distance 2D from r in the direction of r ▷ rule 0: jump over 3: r.color ← 1 4: else 5: r.des ← the current position of r ▷ rule 1: stop 6: r.color ← 1 7: end if Move: move to r.des. Lemma D.32. VSS ∈ Sol(F COMA ), as witnessed by Algorithm 8. Proof. Here D is the distance recorded at the robot’s own Look. (i) A robot of color 0 is at its initial position and is not moving. The color of a robot is 0 until its first Comp, and since tL < tC < tB < tE , its first Move starts after that Comp. Hence an observation of a peer of color 0 is an observation of a robot that has not begun any Move. In particular the observations during a peer’s movement that Asynch permits cannot affect the decision of rule 0, and no atomicity assumption is used below. (ii) When rule 0 fires, D = d and the destination belongs to P . By (i) the peer is at its initial position, and the observing robot is itself not moving at its own Look, so by induction it stands at a landing point of P . The possible cases and landing points are the four listed in part (i) of the proof of Lemma D.28, all of which belong to P . In particular each firing of rule 0 toggles the robot between its own initial position and B, respectively between its own initial position and A. The destination is computed from the snapshot taken at the Look, so delaying the Move (Corollary E.6) does not change the landing point.

46

(iii) The first phase. Let t be the earliest completion of a nonzero Move, say by a. By (ii), a(t) = B. If b has not begun a nonzero Move, then C1 holds at t. Otherwise, b is executing its first nonzero Move towards A (or completes it at t), and its color is already 1. Hence a remains at B until b completes this Move, when C3 holds with cumulative counts (1, 1). The case in which b completes first is symmetric. (iv) Permanent stopping. A robot applies rule 0 only if its Look took place before the first Comp of its peer. That Comp occurs at some finite time T , and each cycle is finite, so each robot performs only finitely many Look operations before T and therefore fires rule 0 finitely often. Under either rule the activated robot sets its color to 1, and after the last of these finitely many nonzero Move operations has been completed, only rule 1 fires and neither robot moves again. Hence Stopped(a) and Stopped(b) hold. (v) The terminal configuration. By (ii), a robot that fired rule 0 an even number of times is back at its initial position, and one that fired it an odd number of times stands at B, respectively at A. The four combinations are exactly C0 , C1 , C2 , C3 , and Definition D.29 admits each of them as a terminal configuration. Lemma D.30, Lemma D.31 and Lemma D.32 together prove Lemma 3.4 (5).

E

Auxiliary Execution Constructions

E.1

Normalization and Distance Updates

Throughout this appendix we fix two robots a, b at distinct positions and write d = ∥a − b∥ for their distance. We call the color-dependent part of a robot’s observation its visible state and denote it by γ. Thus OBLOT has no visible state, the visible state of an FSTA robot is its own color, that of an FCOM robot is the color of its peer, and that of a LUMI robot is the pair of both colors. Each lemma below carries applicability conditions on the starting state, namely the positions, the colors, and the pending operations, and on the scheduler class. When a lemma is applied inside a construction, these conditions are verified against the state actually reached by the preceding prefix. Lemma E.1 (Normalization). Fix a reference vector e ̸= 0 of the plane. For every algorithm A there are constants λγ ≥ 0 and θγ , determined by A alone, and a map g on visible states, such that the following hold. (1) Suppose that at an activation of a robot r the peer is at global vector w ̸= 0 and the visible state of r is γ. Then the adversary may choose the local coordinate system of r so that the global displacement of r is v = λγ Rθγ (w), where Rθ denotes the rotation by angle θ, and the color set by r is g(γ). (2) If w = 0, the local output is determined by γ alone, but the adversary may choose the direction and the magnitude of the resulting global displacement. In particular, two robots with equal visible states can be given equal global displacements. Unless stated otherwise, we use this coordinate system, called the normalized frame. Proof. (1) Place r at the origin. By chirality there is a unique orientation-preserving similarity ψ of the plane that maps w to e, and variable disorientation admits ψ as a local coordinate system. The local input is then (e, γ), so by determinism the output is a fixed pair (qγ , g(γ)). Writing qγ = λγ Rθγ (e) and observing that ψ −1 is a composition of a rotation and a positive scaling, hence commutes with rotations, the global displacement is
ψ −1 (qγ ) = λγ Rθγ ψ −1 (e) = λγ Rθγ (w). (2) If w = 0, the geometric part of the input is the origin in every local coordinate system, so the output q depends only on γ. The global displacement ψ −1 (q) changes direction and magnitude with the choice 47

of ψ, and choosing the same ψ for two robots with equal visible states makes the two displacements equal. Lemma E.2 (Contraction ratios). Put q c1 (γ) := 1 − 2λγ cos θγ + λ2γ ,

c2 (γ) :=

q

1 − 4λγ cos θγ + 4λ2γ .

Under normalized frames the following hold. (i) If exactly one robot with visible state γ completes a cycle while its peer is stationary, then d 7→ c1 (γ) d. (ii) If two robots, both with visible state γ, complete cycles simultaneously from the same snapshot, then d 7→ c2 (γ) d and the midpoint is invariant. Moreover c1 (γ) = 0 ⇐⇒ (λγ , θγ ) = (1, 0),

c2 (γ) = 0 ⇐⇒ (λγ , θγ ) = (1/2, 0).

(iii) Identify the plane with the complex plane and Rθ with multiplication by eiθ . The transformations of the difference vector in (i) and (ii) are multiplication by σ1 (γ) := 1 − λγ eiθγ ,

σ2 (γ) := 1 − 2λγ eiθγ ,

and cj (γ) = |σj (γ)|. When cj (γ) ̸= 0, put ψ(γ) := arg σ1 (γ) and φ(γ) := arg σ2 (γ). If ψ ̸≡ 0 (mod 2π), then there is k ≥ 1 with cos(kψ) ≤ − 21 , and for every θ0 there is k ≥ 1 with cos(θ0 + kψ) ≤ 0. The same statements hold for φ. Proof. Let w := b − a and û := w/d. (i) By Lemma E.1(1) the new difference vector is d(û−λRθ û), whose squared norm is d2 (1−2λ cos θ + 2 λ ). (ii) The displacement of b is the negative of that of a, so the new difference vector is d(û − 2λRθ û) and a + b is unchanged. The conditions for vanishing follow immediately. (iii) The first assertion is the computation in (i) and (ii) read as multiplication of complex numbers, together with |1 − λeiθ |2 = 1 − 2λ cos θ + λ2 and |1 − 2λeiθ |2 = 1 − 4λ cos θ + 4λ2 . For the second assertion, suppose first that ψ/2π is irrational. Then {kψ mod 2π}k≥1 is dense in [0, 2π): by the pigeonhole principle some integer multiple of ψ is arbitrarily close to 0, and its own multiples then fill the circle. Both assertions follow. Suppose next that ψ/2π = p/q in lowest terms with q ≥ 2. Then {kψ mod 2π}k≥1 = {2πj/q : 0 ≤ j < q}. If q is even, the value j = q/2 belongs to this set and cos π = −1. If q is odd, and therefore q ≥ 3, the points are spaced by 2π/q, so one of them lies within π/q of π, where cos ≤ − cos(π/q) ≤ − cos(π/3) = − 12 . For the version with an offset, the set {θ0 + kψ mod 2π}k≥1 consists of q equally spaced points with gap 2π/q ≤ π, so the closed arc [π/2, 3π/2], which has length π, contains at least one of them, and there cos ≤ 0. The argument for φ is identical. From now on we fix the reference vector of Lemma E.1 to be e = (1, 0). The choice is only for concreteness: the arguments below use nothing about e beyond the existence, for each visible state, of the pair (λγ , θγ ) supplied by that lemma, and the displacement of an activation is λγ Rθγ (w) for the actual peer vector w. Remark E.3 (The rule of EPG∗ in a normalized frame). Write (λ⋆ , θ⋆ ) := ( 12 , π), the displacement that moves a robot away from its observed peer by half the observed distance; this is the rule of Algorithm 5. In a round in which both robots are activated from the same configuration and apply the same rule (λ, θ), the midpoint is preserved and the difference vector is multiplied by σ2 = 1 − 2λeiθ (Lemma E.2(ii),(iii)). The expansion condition di+1 = 2di of Definition D.8 therefore holds at such a step if and only if σ2 = 2, ⋆ that is if and only if (λ, θ) = (λ⋆ , θ⋆ ). In that case σ1 = 1 − λ⋆ eiθ = 32 , so a solo activation multiplies the distance by c⋆1 = 32 and ψ ⋆ ≡ 0. 48

E.2

Admissible Execution Segments

The constructions of Appendices F and G are concatenations of a small number of execution segments. This subsection records each segment together with the scheduler classes that admit it and the state it requires at its beginning. Recall from Section 2.2 that we write F, R, S, A, ALC , AM for the scheduler classes, and that the phase times of the i-th cycle of a robot r are denoted tL (r, i) < tC (r, i) < tB (r, i) < tE (r, i). Lemma E.4 (Solo activation interval). Let K ∈ {S, ALC , AM , A}, and suppose that at the end of a finite admissible prefix the robot b has no pending operation. For every k ≥ 1 there is an execution that extends this prefix by an interval in which a completes k consecutive LCM cycles while b is never activated, followed by a fair tail. Throughout this interval the position and the color of b are unchanged, and Stopb (t) holds at every time t of the interval. Proof. Under Ssynch it suffices to take {a} as the activation set of k consecutive rounds. Under Asynch and its subclasses, delay the next Look of b and place the k cycles of a one after another inside that delay. The atomicity constraints only forbid a Look of one robot during prescribed phases of the other robot’s cycle, and b executes no cycle during the interval, so no constraint is violated. Appending a tail that activates both robots infinitely often makes the execution fair. The class K = R is excluded. Indeed, in an Rsynch execution the same robot cannot be activated twice in a row after the fully synchronous prefix, by Lemma E.8 below. Lemma E.5 (Delay before a phase). Under Asynch and its subclasses, the adversary may insert a finite delay before any of the first three phase times of a cycle of a robot r, that is, before tL (r, i), inside (tL (r, i), tC (r, i)), or inside (tC (r, i), tB (r, i)), and place events of the other robot inside that delay, within the limits imposed by the atomicity constraints. No delay is inserted inside a Move operation: once the Move of r begins at tB (r, i), the robot proceeds to the destination computed at tC (r, i) and the Move ends at tE (r, i) without interruption. The values used by r in the later phase are those obtained in the earlier one. Proof. The times tL , tC , tB , tE are chosen by the adversary subject only to the ordering constraint and to finiteness of each phase (Section 2.2); in particular the adversary controls the duration tE (r, i) − tB (r, i) of a Move operation, but not an interruption of it, which is why every delay used here is placed before tB (r, i). The second assertion is the determinism of the algorithm: the destination and the new color computed at tC (r, i) do not depend on anything occurring afterwards. Corollary E.6 (Delayed move with a stale destination). Let K ∈ {ALC , AM , A}. The adversary can construct a fair execution in which a computes its destination at tC (a, i), is then delayed, the robot b completes at least one nonzero Move during the delay, and only afterwards does a move to the destination it had already computed. The delay is placed before the Move of a begins, and the Move itself, once begun, is carried out in full. Writing wold and wnew for the vector from a to b at tC (a, i) and at tB (a, i) respectively, the difference vector after the move of a is wnew − λγ Rθγ (wold ). Proof. Apply Lemma E.5 with the delay placed between tC (a, i) and tB (a, i), and place the whole cycle of b inside it. What ALC forbids is a Look in (tL (a, i), tC (a, i)], and what AM forbids is a Look in [tB (a, i), tE (a, i)]; the delay interval (tC (a, i), tB (a, i)) meets neither, so the segment is admissible under ALC , AM and A alike. The displacement formula is Lemma E.1(1) applied to the snapshot that a obtained at its Look, which is released unchanged when the Move begins. Lemma E.7 (Observation during movement). Let K ∈ {ALC , A}, and suppose that b performs a nonzero Move from pB to pE ̸= pB while a has no pending operation. For every µ ∈ (0, 1) there is a fair execution in which a performs a Look at a time when b is at pµ = (1 − µ)pB + µpE . 49

Proof. The movement of b is an uninterrupted continuous motion along the segment [pB , pE ], whose duration the adversary may choose, so by the intermediate value theorem there is τµ ∈ (tB (b, i), tE (b, i)) with b(τµ ) = pµ . Setting tL (a, j) = τµ is unconstrained under A, and it is admissible under ALC as well, because τµ > tC (b, i). Lemma E.8 (Alternating singleton activation). Let K = R. For every p ≥ 0 there is a fair execution whose first p rounds are fully synchronous and whose tail activates {a}, {b}, {a}, . . . or {b}, {a}, . . .. In this tail, (i) the distance d is updated by the factor c1 at every round; (ii) the robot that is not activated remains stationary during that round; (iii) if the displacement of the moving robot is v, then a + b is translated by v; (iv) the two colors may evolve independently. Once the alternating tail has begun, no fully synchronous round is admissible. Proof. This is the definition of Rsynch in Section 2.2, and (i)–(iv) follow from the fact that exactly one displacement is applied in each round of the tail; for (i) this is Lemma E.2(i).

E.3

Invisibility and State Orbits

Several constructions insert a segment that the other robot cannot detect, so that its next activation reproduces the behavior it would have had without the insertion. The following notion isolates exactly what has to be preserved. Definition E.9 (r-invisible segment). A finite execution segment σ is r-invisible if the normalized local input that r receives at its next activation is the same before and after σ. Lemma E.10 (Invisible transitions). Let M ∈ {FSTA, OBLOT }. If r is not activated during σ and the peer r occupies a position distinct from that of r both before and after σ, then σ is r-invisible. Consequently, however the internal state of r changes during σ, the robot r computes the same output at its next activation. Proof. Neither ObsF STA (r) = (pr , cr ) nor ObsOBLOT (r) = pr contains the color of the peer, so only two components have to be preserved. The own color cr is unchanged because r is not activated. The relative position of the peer may change by a similarity during σ, but as long as it is nonzero it is mapped to e in the normalized frame by Lemma E.1(1). The last assertion is then the determinism of the algorithm. Lemma E.11 (Period-wise pumping in FSTA). Let M = FSTA and K ∈ {S, ALC , AM , A}. Consider a solo activation interval of a (Lemma E.4) in which every Look of a observes its peer at a position different from its own. Along such an interval, the color of a evolves as c0 , g(c0 ), g 2 (c0 ), . . . , which is eventually periodic, say with preperiod p and period q. For every n ≥ 1 such that the first p + nq activations of a satisfy this condition, there is a fair execution in which a is activated alone p + nq times; at the end of that interval the color of a is γp := g p (c0 ), and b, never being activated, stays at its position throughout the interval. If moreover a never reaches the position of b during the interval, then the inserted interval is b-invisible. Proof. The visible state of an FSTA robot is its own color. Every Look of a in the interval observes a nonzero relative position of its peer, so by Lemma E.1(1) each activation updates the color by g. The color sequence is therefore the orbit of g and is eventually periodic. After p + nq steps the orbit is back at γp . The robot b is not activated, so its own color is unchanged, and the color of a is not part of the input of b. The relative position of a changes by a similarity, but as long as it is nonzero the adversary can present the same normalized input to the next activation of b; hence the interval is b-invisible by Lemma E.10. 50

E.4

Synchronous Prefixes and Repetition

This subsection collects the segments in which the same rule is applied repeatedly, either by both robots simultaneously or by one robot alone. Lemma E.12 (Maintaining coincidence). Suppose that a(t0 ) = b(t0 ), that neither robot has a pending operation at t0 , and that the scheduler class permits activating both robots in every round from t0 on. The last condition always holds under Fsynch; under Ssynch and the subclasses of Asynch it holds for the full activation set with aligned phase times; under Rsynch it holds only while the execution is still in its fully synchronous prefix. Suppose moreover that in each such round the two local outputs are either both zero or both nonzero; this is in particular the case when the two visible states are equal. Then the adversary can construct a fair execution with a(t) = b(t) for every t ≥ t0 . Proof. At a coincidence the geometric part of both local inputs is the origin (Section 2.1). By Lemma E.1(2) the direction and the magnitude of the global displacement obtained from a nonzero local output are chosen by the adversary, so two nonzero outputs can be mapped to the same global displacement, while two zero outputs produce no movement. Activating both robots in every round therefore preserves the coincidence. If one output is zero and the other is not, the coincidence is destroyed, so the condition on the outputs cannot be omitted. In OBLOT the output at a coincidence is unique, hence that condition always holds. Lemma E.13 (Synchronous rigidity in OBLOT F ). Let M = OBLOT and K = F , so that the displacement rule is a single pair (λ, θ) by Lemma E.16(a), and put c2 := c2 (λ, θ). The adversary can construct a fair execution in which, in every round k with d(k) > 0, the midpoint is fixed at m and d(k) = c2k d(0), and in which exactly one of the following occurs. (a) λ = 0: neither robot moves, and moves(a) = moves(b) = 0. (b) c2 = 0: in round 1 the two robots coincide at m, and the coincidence is maintained by Lemma E.12. (c) 0 < c2 < 1: the sequence d(k) decreases strictly to 0 but remains positive, and moves(a) = moves(b) = ∞. (d) c2 ≥ 1: d(k) ≥ d(0) for every k, and moves(a) = moves(b) = ∞. Moreover the adversary lets the two robots start and complete their Moves simultaneously, with the same fractional progress along their segments. The two robots are then simultaneously stationary only before and after the common Move interval of a round, so every joint stop of the execution carries one of the round-boundary distances d(k). In particular the restriction of d to Σab is monotone. Proof. Give the normalized frame in every round, so that Lemma E.2(ii) applies to each round. In case (b) the execution is fully synchronous and the model is OBLOT , so the output condition of Lemma E.12 is satisfied. In cases (c) and (d) a nonzero Move occurs in every round, and since tE (r, i) < tL (r, i + 1) these are counted as distinct moves. Activating both robots in the same round with a common speed profile is admissible under Fsynch. In a round containing a nonzero Move the two Move intervals then coincide, so the instants at which both robots are stationary are those preceding the common Move, where the distance is d(k), and those following it, where it is d(k + 1). In a round containing no nonzero Move the distance is constant and equal to both. Hence d takes only the values d(k) = c2k d(0) on Σab , and that sequence is monotone in each of the four cases: constant in (a), constant at 0 from round 1 on in (b), strictly decreasing in (c), and non-decreasing in (d). Lemma E.14 (Synchronous rigidity in the colored models). Let M ∈ {LUMI, FCOM, FSTA} and let K be any scheduler class that permits activating both robots in every round; under Rsynch this means during the fully synchronous prefix. The adversary can construct a fair execution in which, in every round k, the two robots carry the same color γk , where γ0 = c0 , and the distance satisfies d(k +1) = c2 (γk ) d(k); 51

moreover, in every round k with d(k) > 0 the midpoint is fixed at m. While the robots occupy distinct positions, their common color satisfies γk+1 = g(γk ). After a coincidence, the common color instead follows the transition map for the zero geometric observation. In either case the two colors remain equal. Since coincidence is thereafter maintained, the common color sequence (γk ) is eventually periodic. Proof. Activate both robots in every round and give them the normalized frames at every Look at which they are distinct. The designated initial colors are equal, and equal colors give the two robots equal visible states, so by Lemma E.1 their next colors are equal as well; while the robots are distinct, this common next color is g(γk ). While the robots are distinct, the distance update and the invariance of the midpoint are Lemma E.2(ii). If c2 (γk ) = 0, the robots coincide at m at the end of round k; from then on their visible states are still equal, so the coincidence is preserved by Lemma E.12 and d(k+1) = 0 = c2 (γk ) d(k) holds trivially, whereas the common position may move, since by Lemma E.1(2) the adversary chooses the displacement at a coincidence. At every such round both robots observe the zero geometric configuration with equal visible states, so they apply the same transition map for that observation and their colors remain equal. Before a coincidence the common color sequence is an orbit of g; after it, the sequence is an orbit of the transition map for the zero observation. Either is the orbit of a map on a finite set, so (γk ) is eventually periodic. Lemma E.15 (Solo pumping in the models without private memory). Let M ∈ {F COM, OBLOT } and K ∈ {S, ALC , AM , A}. Let γ⋆ be the visible state of a at the beginning of a solo activation interval of Lemma E.4, and put λ := λγ⋆ , θ := θγ⋆ and c := c1 (γ⋆ ). Assume that the robots occupy distinct positions at the beginning of the interval. Throughout the interval Stopb (t) holds, so the peer remains stationary. At every cycle whose Look observes distinct positions, the vector from a to b is transformed by I − λRθ . Hence d(j) = c j d(0) up to and including the first landing at the peer, if such a landing occurs. Moreover: (a) if λ = 0, then a does not move during the interval; in particular, if M = OBLOT then b obeys the same rule, so the configuration stays at ι along every fair infinite execution; (b) if c = 0, the first activation makes a reach its peer. The normalized rule does not determine subsequent outputs at coincidence; those must be treated separately; (c) if λ > 0 and c > 0, every activation in the solo interval is a nonzero Move, and any prescribed finite number of such Moves can be scheduled. Proof. While b is not activated, the visible state of a is constant: the model OBLOT has no visible state, and in FCOM it is the color of b, which does not change, since a cannot observe its own color. Hence, at every Look that observes distinct positions, the same pair (λ, θ) is applied, and the transformation of the vector from a to b and the distance formula follow from Lemma E.1(1) and Lemma E.2(i). In case (c) the distance stays positive, so every Look observes distinct positions and every displacement has length λd(j) > 0. Concerning (a), when M = F COM the color of a may change during the interval, so the subsequent behavior of b is not determined by λ alone; this is why immobility of the whole configuration is asserted only for M = OBLOT . Lemma E.16 (A single displacement rule). In each of the following two cases, the displacement rule of the activated robot is a single pair (λ, θ): (a) any algorithm with M = OBLOT ; (b) an algorithm with M = FCOM, together with an interval during which the peer is not activated. When these activations are performed one at a time, with each cycle completed before the next begins, and every Look observes distinct positions, successive landing distances are multiplied by the same factor c1 (λ, θ), so that d(k) = c1k d(0) after k activations. Their sequence is therefore monotone and cannot alternate between strict decreases and strict increases. A landing at the peer ends this normalized argument and must be handled separately. If λ = 0, the robot does not move at all.

52

Proof. By Lemma E.1 the displacement is determined by the pair (λγ , θγ ) attached to the visible state γ. In case (a) the model OBLOT has a unique visible state, so the family of constants degenerates to a single pair. In case (b) the visible state of an F COM robot is the color of its peer alone, since it cannot observe its own color; as long as the peer is not activated that color is unchanged, hence the visible state is constant (Lemma E.15). For such singleton cycles, the landing distance is multiplied by c1 (λ, θ) by Lemma E.2(i), so the landing distances form a geometric sequence and are monotone. If λ = 0 the displacement is identically zero. Corollary E.17 (The first nonzero rule along a synchronous prefix). Let M ∈ {LUMI, FCOM, FSTA} with the designated common initial color c0 , and let the execution begin with rounds activating both robots, as in Lemma E.14, so that the common color evolves by γ0 = c0 , γk+1 = g(γk ) and is eventually periodic. Exactly one of the following holds. (a) λγk = 0 for every k. Then this fair execution contains no nonzero Move at all, the configuration remains ι forever, and both robots are permanently stationary. (b) There is a least t with λγt ̸= 0. Then the configuration at the beginning of round t is still ι, both robots are stationary with no pending operation, and their common visible state is γt . Hence any segment of Appendix E.2 may be appended there. In particular, replacing round t by a solo activation interval of a (Lemma E.4) leaves the local input of a unchanged, so the rule applied at its first activation is (λγt , θγt ) with λγt ̸= 0; the visible state then stays equal to γt throughout the interval when M = FCOM (Lemma E.15), whereas for M = FSTA it follows the orbit of g (Lemma E.11). In case (a) the joint-stop sequence of Definition D.2 has the single term C0 , so the perpetuity condition of EPG∗ fails. In case (b) the configuration is unchanged until round t, so the first activation of a in the solo activation interval is the first nonzero Move of the execution and it is performed while b is not activated; Lemma 5.1(1) then makes the expansion condition fail at i = 0. Since the color sequence is eventually periodic, the dichotomy is decided by a finite check. For M = OBLOT the same statement holds with the single rule (λ, θ) of Lemma E.16(a), case (a) being λ = 0. Proof. By Lemma E.14 the two robots carry the same color γk in round k, so by Lemma E.1(1) both apply (λγk , θγk ), and the displacement is nonzero exactly when λγk ̸= 0. In case (a) no displacement is ever nonzero, so the configuration never changes. In case (b) every round before t applies a zero displacement, hence the configuration at the beginning of round t is still ι; the robot a then has the same peer position and the same visible state as it would under simultaneous activation, so its rule there is (λγt , θγt ). The finite check is the eventual periodicity of the orbit of g on a finite color set.

E.5

Preservation and Continuation of Violations

Every common-failure execution in Appendices F and G is a concatenation of the segments collected above. The following lemma states when such a concatenation is admissible and which violations survive it. Lemma E.18 (Concatenation of segments). The violations exhibited by a common-failure construction split into two kinds. A safety violation, namely landing at a forbidden point, a forbidden overtaking, a forbidden coincidence, a forbidden transition, or an expansion that moves the center of gravity, occurs at a finite time and cannot be repaired by any continuation. A liveness violation, namely the failure of infinitely many alternations, of perpetuity, of reaching, of convergence, or of stable gathering or stopping, is decided only on an infinite execution. The following hold. 1. A finite segment that forces a safety violation may be followed by any continuation permitted by the scheduler class, in particular by a fair tail, and the violation is preserved.

53

2. Segments are concatenated starting from the state actually reached by the preceding prefix. Positions, colors, internal states, and pending operations are carried over and are not reset. A segment may be appended only if its applicability conditions hold in that state and if the concatenation is an execution of the scheduler class under consideration. In particular, no fully synchronous round may be placed after an alternating singleton tail under Rsynch (Lemma E.8); a solo activation interval (Lemma E.4) requires the robot that is not activated to have no pending operation; and a delayed move (Corollary E.6) requires K ∈ {ALC , AM , A}. 3. A liveness violation requires an explicit fair infinite continuation along which the violation persists; an arbitrary fair tail does not suffice. When a safety violation of one predicate is combined with a liveness violation of another, the segment forcing the safety violation is placed first, and the infinite continuation witnessing the liveness violation is built from the state that this segment leaves. Since the predicate of each task is evaluated relative to the fixed initial instance ι, the argument for a later segment remains valid even when earlier segments have already changed the configuration. Proof. A safety violation is witnessed by a finite prefix, and every extension of that prefix contains it; this gives (1). Claim (2) records that a concatenation is an execution of the algorithm in the given scheduler class exactly when each stage is admissible from the state already reached. The conditions listed are precisely those required by the corresponding segment lemmas. For (3), a liveness property is not decided by any finite prefix, so an infinite continuation violating it has to be exhibited. Finally, the task predicates refer to ι and to the trajectory, not to the configuration at which a segment happens to begin; hence a change of configuration by an earlier segment does not invalidate the argument for a later one.

E.6

The Solo-Expansion Obstruction

Proof of Lemma 5.1. (1) Say r̄ completes no nonzero Move, so r̄i+1 = r̄i . Consecutive terms of the sequence are distinct, hence ri+1 ̸= ri and ai+1 + bi+1 = (ai + bi ) + (ri+1 − ri ) ̸= ai + bi . (2) For K = R take the execution of Lemma E.8 with an empty fully synchronous prefix; every round activates exactly one robot. For K ∈ {S, ALC , AM , A} the adversary places the whole cycle of one robot after the whole cycle of the other, alternately: while one robot runs its cycle the other performs no operation, so no Look falls inside a Move interval of the peer and every atomicity constraint of Section 2.2 is met; both robots are activated infinitely often, so the execution is fair. In such an execution the boundaries between consecutive cycles are times of Σab . If no nonzero Move ever occurs, the configuration never changes and (Ci ) has the single term C0 ; then the first condition of EPG (exactly two terms) and the first condition of EPG∗ (an infinite sequence) both fail. Otherwise consider the first cycle containing a nonzero Move, performed by r. Every earlier cycle left the configuration unchanged, so the joint stop preceding that cycle realizes C0 and the joint stop following it realizes C1 ̸= C0 , while r̄ ran no cycle in between. By (1), 21 (a1 + b1 ) ̸= c0 , so the second condition of both tasks fails at i = 0. This is a safety violation and is preserved by any fair continuation.

E.7

The Two-Nonzero-Move Obstruction for SRO

We use the notation of Definition C.1: (Ci )i≥0 is the sequence of configurations at the times when neither robot is inside a nonzero Move, with consecutive repetitions removed, di = bi − ai , and Qi is the square with diagonal ai bi , interior included, with Q−1 = R2 . Condition C1 is di+1 = R−90◦ di , condition C2 is di+1 = 2−1/2 R−45◦ di , and condition C3 is ai+1 , bi+1 ∈ Qi−1 . Lemma E.19 (Two solo nonzero moves defeat SRO). Consider a situation in which only one robot moves while its peer remains stationary. The displacement that can satisfy C1 is unique, namely the one rotating the segment by 90◦ clockwise about the peer, and the displacement that can satisfy C2 is also unique, namely the one rotating the moving robot by 90◦ clockwise about the midpoint. No other 54

displacement satisfies C1 or C2. If the same robot performs two consecutive nonzero Move operations while its peer stays still, the rotation condition and C3 cannot both be satisfied. Consequently, two solo nonzero moves of one robot violate ΦSRO , irrespective of the internal states. Zero Move operations do not advance the sequence (Ci ). Proof. After each nonzero Move of the moving robot, both robots are stationary until the next Look, so the configurations reached after the two moves are consecutive terms Ci+1 , Ci+2 of the sequence (Ci ). Condition C1 requires ∥di+1 ∥ = ∥di ∥ together with di+1 = R−90◦ di , so with the peer held fixed the new position of the moving robot is uniquely determined. Condition C2 requires di+1 = 2−1/2 R−45◦ di and determines it uniquely in the same way. Both conditions are purely geometric and do not depend on the internal states. Exchanging the names of the robots replaces each dj by −dj , which leaves the conditions unchanged, and they are also invariant under orientation-preserving similarities. We may therefore assume that a is the moving robot and that bi = (0, 0), ai = (−1, 0), so that di = (1, 0) and o n Qi = (x, y) : x + 21 + |y| ≤ 21 . If the first move satisfies C1, then ai+1 = (0, 1), and the second move gives ai+2 = (1, 0) under C1 and ai+2 = ( 12 , 12 ) under C2. If the first move satisfies C2, then ai+1 = (− 21 , 12 ), and the second move gives ai+2 = ( 12 , 21 ) under C1 and ai+2 = (0, 12 ) under C2. In each of the four cases ai+2 ∈ / Qi , so C3 fails at Ci+2 . This is the computation of [FSW20, Lemma 23, Fig. 5]. The lemma is applied in two ways. When the two solo nonzero moves are available, it refutes ΦSRO outright. When only one solo move is available, its first part still shows that a displacement which is neither the 90◦ clockwise rotation about the peer nor the 90◦ clockwise rotation about the midpoint already violates the rotation condition. The degenerate case in which the moving robot lands on its peer is covered by this observation, since di+1 = 0 satisfies neither C1, which requires ∥di+1 ∥ = ∥di ∥ ̸= 0, nor C2, whose right-hand side is nonzero.

E.8

The Overtaking Construction

The delayed move of Corollary E.6 becomes an obstruction when the destination computed before the delay is long compared with the distance that remains when the move is released. The following lemma records the two forms in which this is used, one for MCv and one for ZCC. Lemma E.20 (Overtaking). Let K ∈ {ALC , AM , A}. Suppose that b computes at its Comp a destination from its saved Look snapshot, with global displacement v ̸= 0, that its Move is then delayed as in Corollary E.6, and that only a is activated during the delay. Let w be the vector from b to a at the moment at which the Move of b is released. Keep a stationary while b executes this pending Move. 1. If ∥w∥ < ∥v∥/2, then the distance after the Move of b is larger than ∥w∥. Hence the second condition of ΦMCv , monotonicity, is violated. 2. If b(0), the current position of b, and a lie on a line in this order, if v points towards a, and if ∥w∥ < ∥v∥, then b lands beyond a, and at its landing time t one has dis(b(0), b(t)) > dis(b(0), a(t)). Hence the second condition of ΦZCC , collision freedom, is violated. The delay is placed between tC (b, i) and tB (b, i). By Lemma E.5 and Corollary E.6, it is admissible under ALC , AM and A, and any prescribed finite number of complete cycles of a may be placed inside it. Throughout the delay b stays at its position, and its pending Move uses the destination computed at tC (b, i), independently of these cycles. When M ∈ {F COM, OBLOT } the visible state of a is constant during the delay (for FCOM it is the color that b adopted at tC (b, i)); when M = FSTA the color of a follows the orbit of g as long as every Look of a observes distinct positions. Hence the distance formulas of Lemmas E.15 and E.11 apply to these cycles, although b has a pending operation. 55

Proof. (1) After the Move of b, the vector from b to a is w − v, and ∥w − v∥ ≥ ∥v∥ − ∥w∥ > ∥w∥. (2) On the line we have dis(b(0), b + v) = dis(b(0), b) + ∥v∥ > dis(b(0), b) + ∥w∥ = dis(b(0), a), which is the stated inequality at the landing time. The admissibility of the delay and of the cycles of a placed inside it is Lemma E.5 together with Corollary E.6; the last assertion of Lemma E.5 shows that the pending Move of b is unaffected by them. The distance formulas follow from Lemma E.1(1) and Lemma E.2(i), exactly as in the proofs of Lemmas E.15 and E.11, since b does not move during the delay. Remark E.21 (The reachability hypothesis is not automatic). Neither ∥w∥ < ∥v∥/2 nor ∥w∥ < ∥v∥ holds by construction: they require the solo activations of a during the delay to bring a close enough to b. When the solo contraction ratio of a satisfies c < 1, we have d(j) = c j d(0) → 0 by QLemma E.15, and the hypothesis can be met. For M = FSTA the relevant quantity is the product Γ = j c1 (γj ) over one color period, by Lemma E.11, and the hypothesis can be met when Γ < 1. The reachability condition and, for part (2), the required collinearity and order must be verified in each application. The continuation arguments for MCv and ZCC are given directly in Lemmas C.7 and C.17.

E.9

Avoiding a Countable Set

Lemma E.7 lets the adversary choose, by a continuous parameter, the point of the peer’s movement at which an observation is made. Several impossibility arguments combine this freedom with the fact that the set of outcomes admitted by a task predicate is countable, or even finite. Corollary E.22 (Avoiding a countable set). Let K ∈ {ALC , A}, suppose that b performs a nonzero Move, and suppose that a observes it with parameter µ ∈ (0, 1) as in Lemma E.7. Let h : (0, 1) → R assign to µ some quantity of the resulting execution. If h is continuous and nonconstant, then for every countable set G ⊂ R the adversary can choose µ with h(µ) ∈ / G. Proof. The image of a nonconstant continuous function on an interval contains a nondegenerate interval, hence is uncountable, and therefore is not contained in G. The corollary is used with the following two observation parameters. Impossibility of VEC. In the proof that VEC ∈ / Sol(FSTAA ) (Lemma C.22), h is a coordinate of the segment determined by the two landing points, which varies continuously and nonconstantly with µ. The set G consists of the finitely many successor configurations admitted by the transition condition of ΦVEC , so a single µ avoids all of them. In Lemma F.11 the same h is used to avoid one further value, namely the unique successor admitted by the expansion condition of EPG∗ , and the countably many µ at which the two robots coincide. Impossibility of DMSD. In the proof that DMSD ∈ / Sol(FSTAA ) (Lemma C.25), the normalization a(0) = 0, b(0) = e1 , D0 = 1 gives h(µ) = ∥c + µw∥,

c := e1 − 2λRe1 ,

w := λ2 R2 e1 ̸= 0.

Here ∥c + µw∥2 is a quadratic polynomial in µ whose leading coefficient ∥w∥2 is positive, so h is nonconstant on (0, 1). The set G is the countable set D0 Z[1/2], and a value h(µ) ∈ / G witnesses a joint stop at which the ratio is not dyadic. In both applications the exceptional set is countable, so several such requirements can be imposed simultaneously: a finite or countable union of countable sets is still countable, and a single µ avoids all of them at once. This is what allows one execution to refute several task predicates, as in Appendices F and G. 56

F

Common-Failure Executions for Two-Task Disjunction

Each row lists one two-element antichain A, given by the corresponding disjunction BA of principal tasks from Table 2, together with NA = Max(Q \ UA ) and, for each Y ∈ NA , the statement exhibiting one admissible fair execution of an arbitrary algorithm in Y that violates both task predicates. The four obligations carrying a reference to Proposition 5.2 or Proposition 5.3 are proved in the main text; each of them refutes three predicates on a single execution, which is what lets one statement discharge two obligations at once, namely those of EPG∗ ∨ S-RDV and EPG∗ ∨ VSS for the first and those of SRO ∨ S-RDV and SRO ∨ VSS for the second. The column sum is the number 47 of two-task boundary obligations certified in Table 9 (Appendix H.3) and recorded in Table 3.

57

Table 4: The twenty-four two-element antichains, their maximal unsolvable classes NA , and the commonfailure execution that discharges each obligation (A, Y ). The rows are grouped by the construction families of Section 5 and Table 3. Four obligations are proved in the main text and are listed here only as references; all the others are proved in Appendix F. Disjunction BA NA and proof reference

|NA |

Expansion family, EPG∗ ∨ P with P ∈ Cexp , Appendix F.1 EPG∗ ∨ HET FCOMS Lem. F.1; FSTAR Lem. F.2 ∗ EPG ∨ OSPO FCOMS Lem. F.3; FSTAS Lem. F.4; OBLOT R Lem. F.5 ∗ EPG ∨ S-RDV FCOMA Lem. F.6; FSTAR Prop. 5.2 ∗ EPG ∨ ZCC FCOMS Lem. F.7; FSTAAM Lem. F.8; OBLOT R Lem. F.9 ∗ EPG ∨ VSS FSTAR Prop. 5.2 EPG∗ ∨ VEC FCOMS Lem. F.10; FSTAA Lem. F.11; OBLOT R Lem. F.12 ∗ EPG ∨ CFSM FCOMS Lem. F.13; OBLOT R Lem. F.14

2 3 2 3 1 3 2

Shrinking-rotation family, Appendix F.2 SRO ∨ S-RDV FCOMA Lem. F.15; FSTAS Prop. 5.3 SRO ∨ ZCC FCOMS Lem. F.16; FSTAAM Lem. F.17 SRO ∨ VSS FSTAS Prop. 5.3 SRO ∨ VEC FCOMS Lem. F.18; FSTAA Lem. F.19 SRO ∨ CFSM FCOMS Lem. F.20

2 2 1 2 1

Exchange family, Appendix F.3 OSPO ∨ S-RDV FCOMA Lem. F.21; FSTAS Lem. F.22; OBLOT F Lem. F.23 OSPO ∨ VSS FSTAS Lem. F.24; OBLOT F Lem. F.25 S-RDV ∨ ZCC FCOMA Lem. F.26; FSTAAM Lem. F.27; OBLOT F Lem. F.28 S-RDV ∨ VEC FCOMA Lem. F.29; FSTAA Lem. F.30; OBLOT F Lem. F.31 S-RDV ∨ CFSM FCOMA Lem. F.32; OBLOT F Lem. F.33 ZCC ∨ VSS FSTAAM Lem. F.34; OBLOT F Lem. F.35 VSS ∨ VEC FSTAA Lem. F.36; OBLOT F Lem. F.37 VSS ∨ CFSM OBLOT F Lem. F.38

3 2 3 3 2 2 2 1

Monotone-convergence family, Appendix F.4 MCv ∨ VSS FSTAAM Lem. F.39 MCv ∨ VEC FCOMA Lem. F.40; FSTAA Lem. F.41 MCv ∨ CFSM FCOMA Lem. F.42

1 2 1

Dyadic family, Appendix F.5 DMSD ∨ CFSM OBLOT A Lem. F.43

1

Total

47

58

F.1

Disjunctions Containing EPG∗

This subsection treats the seven antichains {EPG∗ , P } with P ∈ Cexp = {HET, OSPO, S-RDV, ZCC, VSS, VEC, CFSM}, which carry sixteen boundary obligations (A, Y ) in total. The two F STAR obligations, for EPG∗ ∨ S-RDV and EPG∗ ∨ VSS, are proved in the main text by Proposition 5.2 and are not repeated here. Every construction below obtains its EPG∗ violation from Lemma 5.1 in one of two ways. If the execution contains no nonzero Move at all, the joint-stop sequence has the single term C0 and the perpetuity condition fails, which is a liveness violation. If it does contain one, and the first one is performed by a robot while its peer performs none, then the configuration has not changed before it, so the joint stops surrounding it realize C0 and C1 , and Lemma 5.1(1) gives 12 (a1 + b1 ) ̸= c0 : the expansion condition fails at i = 0, which is a safety violation. Each proof only has to record which of the two cases applies and to refute the companion task on the same execution. Throughout this subsection and the next, we use the normalized frames of Lemma E.1 and the contraction ratios of Lemma E.2, and we write (λ⋆ , θ⋆ ) = ( 12 , π) for the rule of Algorithm 5 in a normalized frame, as recorded in Remark E.3: a round in which both robots are activated from the same configuration and apply the same rule (λ, θ) satisfies the expansion condition di+1 = 2di if and only if (λ, θ) = (λ⋆ , θ⋆ ), and a solo activation under that rule multiplies the distance by c⋆1 = 32 with ψ ⋆ ≡ 0. F.1.1

EPG∗ ∨ HET (EH)

Lemma F.1. EH ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of FCOM under Ssynch. Normalize the edge length of the hexagon of ι to 1, with a(0) = v0 and b(0) = v3 ; the initial colors are common. The visible state of an FCOM robot is the color of its peer, so while both robots are activated in every round their colors stay equal and follow a common sequence c0 , g(c0 ), . . . (Lemma E.14), which is eventually periodic since the palette is finite. During this prefix the configuration stays at ι. If λcj = 0 for every j, this fair execution contains no nonzero Move, so moves(a) = 0 and the perpetuity condition of HET fails, while the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails as well (Lemma 5.1(2)). Both are liveness violations of the same fair infinite execution, so Lemma E.18(3) is satisfied. Otherwise let t be the least j with λcj ̸= 0, replace round t by an interval in which a alone is activated k ≥ 2 times in a row (Lemma E.4), and append a fair tail. Since b is not activated, the visible state of a is constantly ct , so by Lemma E.16(b) its rule is the single pair (λ, θ) = (λct , θct ), and by Lemma E.15 the distance satisfies d(j) = c1j d(0). All rounds before the interval left the configuration at ι, so the first activation of a in the interval is the first nonzero Move of the execution and it is performed while b is not activated. By Lemma 5.1(1) the center of gravity of the next term of the joint-stop sequence is not c0 , so the expansion condition of EPG∗ fails at i = 0; this is a safety violation and is preserved by the tail. For HET, as b stands still at v√3 , the second condition requires −1 , v0 after 0, 1, 2 nonzero √ a to be at v0 , v√ Move operations, that is d = 2, 3, 2; this would force c1 = 3/2 and c1 = 2/ 3 simultaneously, which is impossible. (If c1 = 0 then a reaches v3 ̸= v−1 in one activation.) Hence the second condition fails, and being a safety violation it is preserved by the tail (Lemma E.18(1)). In both cases a single execution violates both predicates, so the common-failure property holds at FCOMS and Lemma 6.1 gives the claim. Lemma F.2. EH ∈ / Sol(F STAR ). Proof. Let A be any deterministic algorithm of FSTA under Rsynch. Take the alternating singleton activation with an empty fully synchronous prefix (Lemma E.8) followed by an arbitrary fair tail, with edge length 1, a(0) = v0 , b(0) = v3 , common initial colors, and normalized frames up to and including the round following the first nonzero Move. Every round of this execution activates exactly one robot, so Lemma 5.1(2) applies to it. The initial configuration is symmetric under the rotation by π about the 59

midpoint combined with exchanging the robots, and in F STA the state of the peer does not appear in the local input (Lemma E.10), so until the first nonzero Move both robots follow the same color sequence γ0 , g(γ0 ), . . . (Lemma E.11). If λgj (γ0 ) = 0 for every j, no nonzero Move ever occurs, so moves(a) = 0 and the perpetuity condition of HET fails, while the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. Otherwise let j be least with λgj (γ0 ) ̸= 0. The first nonzero Move is the (j + 1)-st activation of a, performed while b is not activated, so by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0. Write (λ, θ) for the displacement of that activation, with λ ̸= 0, and a∗ for its landing point. If a∗ = b(0), then a∗ = v3 ̸= v−1 violates the second condition of HET. If a∗ ̸= b(0) and a∗ ̸= v−1 , the second condition fails at once. If a∗ = v−1 , the robot b activated next has the same visible state as a had, so it applies the same (λ, θ); since λRθ maps v3 − v0 , of √ length 2, to v−1 − v0 , of length 1, we have λ = 1/2, and the displacement of b has length 12 ∥v−1 − v3 ∥ = 3/2, which differs from the ∥v2 − v3 ∥ = 1 required by the second condition (Lemma C.11). In each case a single execution violates both predicates, so Lemma 6.1 gives the claim. F.1.2

EPG∗ ∨ OSPO (EO)

Lemma F.3. EO ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of FCOM under Ssynch. Run fully synchronous rounds with normalized frames. By Lemma E.14 the colors follow a common sequence c0 , g(c0 ), . . . , and the configuration stays at ι while every Move is zero. If λcj = 0 for every j, continue forever. Then d is constant, so ΦOSPO has no strictly decreasing phase. The joint-stop sequence is (C0 ), so the perpetuity condition of EPG∗ fails. This fair execution witnesses both liveness violations. Otherwise let t be least with λct ̸= 0. Replace round t by two consecutive solo activations of a (Lemma E.4), and append a fair tail. Since b is inactive, the visible state of a stays ct , so by Lemma E.16(b) the robot a applies (λct , θct ) with λct ̸= 0 both times. The first of these activations is the first nonzero Move of the execution, and it is solo. Hence the expansion condition of EPG∗ fails at i = 0 (Lemma 5.1(1)). For OSPO: if d vanishes during the interval, the collision-freedom condition of ΦOSPO fails. Otherwise the second activation is also a nonzero Move. At its completion ma = 2 and mb = 0, which no execution satisfying ΦOSPO admits (as in Lemma D.14). In the latter case, all these violations are safety violations, preserved by the tail (Lemma E.18(1)). Lemma 6.1 gives the claim. Lemma F.4. EO ∈ / Sol(F STAS ). Proof. Let A be any deterministic algorithm of FSTA under Ssynch. Give normalized frames at every Look at which the robots are distinct. Then each robot advances along c0 , g(c0 ), . . . , one step per activation and independently of the schedule, and its displacement depends only on its color. Let J := {i ≥ 0 : λgi (c0 ) ̸= 0}. A collision violates the collision-freedom condition of ΦOSPO as a safety violation, so we assume none below. Any witness (ti ) of ΦOSPO has ma (ti ) = mb (ti ) = i. Hence |ma − mb | ≤ 1 at all times, and ma , mb are unbounded.

(∗)

Case J = ∅. Run fully synchronously forever. No nonzero Move occurs. So d is constant, and the joint-stop sequence is (C0 ). Both ΦOSPO and the perpetuity condition of EPG∗ fail, and this execution witnesses both liveness violations. Case J ̸= ∅. Let j := min J, and activate a alone j + 1 times (Lemma E.4). The last of these activations is the first nonzero Move, and it is solo. So the expansion condition of EPG∗ fails at i = 0 (Lemma 5.1(1)); this is a safety violation. 60

If |J| ≥ 2, let j ′ be the second element of J. Continue activating a alone up to its (j ′ + 1)-st activation, then append a fair tail. The robot a performs a second nonzero Move while b performs none. This contradicts (∗), a safety violation. If |J| = 1, append a fully synchronous tail. Each robot performs at most one nonzero Move in the whole execution, contradicting the unboundedness in (∗). This tail witnesses the liveness violation, and the safety violation of EPG∗ precedes it. Lemma E.18(1)(3) is satisfied in every case, and Lemma 6.1 gives the claim. Lemma F.5. EO ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch. An OBLOT robot has a unique visible state, so in a normalized frame the rule of A is a single pair (λ, θ) (Lemma E.16(a)). Case (λ, θ) ̸= (λ⋆ , θ⋆ ). Since Rsynch contains the fully synchronous executions, take the execution of Lemma E.13, with both robots activated in every round and a normalized frame at every Look. By Lemma D.13 it violates ΦOSPO , the degenerate cases λ = 0 and c2 = 0 being treated there as well. For EPG∗ : if λ = 0 the configuration never changes, the joint-stop sequence has the single term C0 and the perpetuity condition fails; if λ ̸= 0 then C1 ̸= C0 and, by Remark E.3, d1 = σ2 d0 with σ2 ̸= 2, so the expansion condition fails at i = 0. Case (λ, θ) = (λ⋆ , θ⋆ ). Take the alternating singleton activation with an empty fully synchronous prefix (Lemma E.8) and a normalized frame at every Look. Since λ⋆ ̸= 0, the first activation of a is the first nonzero Move of the execution and is performed while b is not activated, so by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0. For OSPO: each activation moves the active robot directly away from its peer by half the observed distance, so d is non-decreasing throughout the execution and ⋆ in fact d 7→ 23 d at every activation, by Lemma E.2(i) with σ1 = 1 − λ⋆ eiθ = 32 . Hence no interval on which d strictly decreases exists and ΦOSPO fails. In both cases a single execution violates both predicates, so Lemma 6.1 gives the claim. F.1.3

EPG∗ ∨ S-RDV (ES)

Lemma F.6. ES ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of F COM under Asynch. Fix distinct initial positions a0 , b0 , and put A = 2a0 − b0 , B = 2b0 − a0 , and P = {A, a0 , b0 , B}. Run fully synchronous cycles, choosing local frames that give the two robots identical normalized observations. Their colors remain equal. If every computed Move is zero, this fair, time-divergent execution keeps the robots at their distinct initial positions. Thus ΦS-RDV fails, and the joint-stop sequence consists only of C0 , violating the perpetuity condition of EPG∗ . Otherwise, consider the first round with a nonzero output. Let both robots perform their Looks before either performs its Compute, and then let both complete their Computes. They obtain the same normalized displacement (λ, θ), with λ > 0, and the same new color γ ′ . Delay the Move of b until after a completes its Move. At the landing of a, both robots are stationary, although b still has a pending Move. Writing p ̸= a0 for the landing point of a, this joint stop has midpoint p + b0 a0 + b0 ̸= . 2 2 Hence Lemma 5.1(1) gives a finite violation of ΦEPG∗ , which persists under every extension. Suppose first that (λ, θ) ̸= (2, 0). Since the Look of a observed the initial configuration, its landing point satisfies p ̸= B. No valid exchange time T1 precedes this first completion, because both cumulative nonzero-move counts were zero. If p ∈ / P , the stationary-position requirement of S-RDV is already violated. Otherwise p ∈ {A, b0 }. The exchange configuration requiring zero Moves by a is now impossible. The other two require a to reach B with exactly one completed nonzero Move. However, any subsequent arrival at B while respecting the segment constraint requires a second completed nonzero Move: the exterior endpoint B cannot lie in the interior of a straight movement contained in [A, B]. Thus no 61

continuation satisfies the exchange phase. Complete the pending cycle of b and append an arbitrary admissible fair, time-divergent continuation. Both predicates fail on the resulting execution. It remains to consider (λ, θ) = (2, 0). Then a lands at B. Without reactivating a, let b complete its already computed Move, whose destination is A. The robots are now at the distinct positions (B, A), both have color γ ′ , and neither has a pending cycle. Let Aγ ′ have exactly the same transition rule as A, but designated initial color γ ′ . Apply the general impossibility of stable RDV in FCOMA used in the proof of Lemma D.25 to Aγ ′ . It supplies an admissible fair, time-divergent execution from some distinct initial positions that fails stable rendezvous. An orientation-preserving similarity maps those positions to (B, A). Transforming the local frames accordingly preserves every local observation and color transition, so variable disorientation allows this counterexecution to be reproduced from the reached configuration. Append that execution to the finite prefix. The resulting execution is admissible, fair, and timedivergent. Its failure of stable rendezvous violates the terminal condition of S-RDV, while the earlier violation of ΦEPG∗ persists. Thus a single execution violates both predicates, and Lemma 6.1 gives the claim. The remaining obligation of this pair, the class FSTAR , is Proposition 5.2. F.1.4

EPG∗ ∨ ZCC (EZ)

Lemma F.7. EZ ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of F COM under Ssynch. The initial colors being common, while both robots are activated simultaneously their colors follow a common eventually periodic sequence c0 , g(c0 ), . . . (Lemma E.14) and the configuration stays at ι. If λcj = 0 for every j, the two robots never move and remain at distinct positions, so the convergence condition of ZCC fails, while the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. This fair execution witnesses both liveness violations. Otherwise let t be the least j with λcj ̸= 0 and replace round t by an interval in which a alone is activated repeatedly (Lemma E.4); the interval may be made arbitrarily long, the activations of b being placed far enough in the future to keep the execution fair. The configuration was unchanged before the interval, so the first activation of a in it is the first nonzero Move of the execution and it is solo; by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation. The visible state of a is constantly ct , so by Lemma E.16(b) its rule is a single pair (λ, θ) with λ ̸= 0, and b is at rest throughout. Under Ssynch every activation of a is a complete cycle, so no operation is pending at the end of any of them. Put σ1 := 1 − λeiθ and c1 := |σ1 |, and write a0 := a(0), b0 := b(0) and d := ∥b0 − a0 ∥. If c1 = 0, the robot a reaches the position of b at the first activation, and no operation is pending at that time. Use the coincidence continuation described in the proof of Lemma C.16, with a0 , b0 kept fixed. It either keeps both positions fixed forever, so that the zig-zag condition fails, or produces a finite violation of the collision-freedom condition; in either case ΦZCC fails. Assume c1 ̸= 0, and put ψ := arg σ1 (Lemma E.2(iii)). The vector wj := b0 − a(j) satisfies wj = σ1j w0 ̸= 0, so every Look of a in the interval observes distinct positions, and fa (j) = ∥a(j) − a0 ∥ = ∥w0 − σ1j w0 ∥ = d 1 − c1j eijψ . As b stands still, ∥b(j) − a0 ∥ = d, so the collision-freedom condition requires fa (j) ≤ d, that is c1j ≤ 2 cos(jψ). Note that σ1 = 1 would force λ = 0, so c1 ̸= 1 whenever ψ ≡ 0. • If ψ ̸≡ 0 (mod 2π), Lemma E.2(iii) provides k ≥ 1 with cos(kψ) ≤ − 12 < 0; making the interval long enough to reach the k-th activation, the collision-freedom condition fails there, a safety violation. Append an arbitrary fair, time-divergent continuation.

62

• If ψ ≡ 0 and c1 > 1, that is θ = π, then fa (j) = (c1j − 1)d exceeds d as soon as c1j > 2, and the collision-freedom condition again fails after finitely many activations. Append an arbitrary fair, time-divergent continuation. • In the remaining contraction case ψ ≡ 0 and 0 < c1 < 1, that is σ1 ∈ (0, 1), end the solo interval after its first activation. Then a(1) = b0 − σ1 (b0 − a0 ) lies in the open segment (a0 , b0 ) and b is still at b0 , so the robots still lie in the original initial segment in their original order, and neither has a pending operation. Apply the continuation property proved in Lemma C.16, keeping the original initial positions fixed. It supplies a fair, time-divergent continuation violating ΦZCC . In every case the safety violation of ΦEPG∗ occurs at the first activation of the solo interval, which belongs to the prefix of each construction above, and every construction is a fair, time-divergent continuation of that prefix. The violation therefore persists (Lemma E.18(1)), one fair infinite execution violates both predicates, and Lemma 6.1 gives the claim. Lemma F.8. EZ ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM , which admits synchronous activations of one robot at a time (Lemma E.4). Start with alternating singleton activations, each cycle being completed before the next one begins, with normalized frames at every Look that observes distinct positions. This interval activates one robot at a time, so Lemma 5.1 applies to it. If no nonzero Move ever occurs, the two robots stay at ι at distinct positions, so the convergence condition of ZCC fails, and the joint-stop sequence has the single term C0 , so the perpetuity condition of EPG∗ fails. This fair execution witnesses both liveness violations. Assume therefore that a nonzero Move occurs. The first one is performed while the peer is not activated, so by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation. End the prefix at the completion of the cycle containing this Move. This is a joint stop, with no pending operation. Every Look in the prefix observes distinct positions: before the first nonzero Move the configuration is ι, and that Move is the last operation of the prefix. Since the visible state of an F STA robot is its own color, Lemma E.1(1) shows that each robot that has been activated k times carries g k (c0 ). Hence the two colors lie on the same normalized color orbit, the orbit of c0 under g. If the robots coincide, use the coincidence continuation from Lemma C.16. Otherwise their colors lie on the same normalized color orbit, so the continuation statement proved in Lemma C.17 applies. In either case it supplies a fair, time-divergent continuation violating ΦZCC , with the original initial positions kept fixed. The safety violation established in the prefix persists (Lemma E.18(1)). A single fair infinite execution therefore violates both predicates, so Lemma 6.1 gives the claim. Lemma F.9. EZ ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch. An OBLOT robot has a unique visible state, so in a normalized frame the rule is a single pair (λ, θ) (Lemma E.16(a)). Case (λ, θ) ̸= (λ⋆ , θ⋆ ). Since Rsynch contains the fully synchronous executions, take the execution of Lemma E.13, with both robots activated in every round and a normalized frame at every Look. By Lemma C.15 that execution violates ΦZCC , the degenerate cases λ = 0 and c2 = 0 being treated there as well. For EPG∗ : if λ = 0 the joint-stop sequence has the single term C0 and the perpetuity condition fails; if λ ̸= 0 then C1 ̸= C0 and, by Remark E.3, d1 = σ2 d0 with σ2 ̸= 2, so the expansion condition fails at i = 0. Case (λ, θ) = (λ⋆ , θ⋆ ). Take the alternating singleton activation with an empty fully synchronous prefix (Lemma E.8) and a normalized frame at every Look. Since λ⋆ ̸= 0, the first activation of a is the first nonzero Move of the execution and is performed while b is not activated, so by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation. For ZCC: by Remark E.3, c1 = 23 , so by Lemma E.2(i) each activation maps d 7→ 32 d; hence d → ∞, the two robots do not converge to a single point, and the convergence condition fails. 63

In both cases a single execution violates both predicates, the safety violation being placed first and a fair infinite continuation of the liveness violation being exhibited, so Lemma E.18(1)(3) is satisfied and Lemma 6.1 gives the claim. F.1.5

EPG∗ ∨ VSS (EV)

This pair has the single boundary class FSTAR , and that obligation is discharged in the main text by Proposition 5.2, which exhibits one admissible fair execution violating EPG∗ , S-RDV and VSS simultaneously. F.1.6

EPG∗ ∨ VEC (EX)

Lemma F.10. EX ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of F COM under Ssynch. The initial colors being common, while both robots are activated simultaneously their colors follow a common eventually periodic sequence γ0 , g(γ0 ), . . . (Lemma E.14) and the configuration stays at ι. If λcj = 0 for every j, the two robots never move, the joint-stop sequence reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition of VEC fails, while the perpetuity condition of EPG∗ fails as well (Lemma 5.1(2)). Otherwise let t be the least j with λcj ̸= 0, replace round t by an interval in which a alone is activated k ≥ 3 times in a row (Lemma E.4), and append a fair tail. The configuration was unchanged before the interval, so the first activation of a in it is the first nonzero Move of the execution and it is performed while b is not activated; by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation. The visible state of a is constantly ct , so by Lemma E.16(b) it applies the same displacement rule at every activation. With the peer at rest, the only displacements that satisfy the transition condition of VEC are those rotating the segment clockwise by 45◦ or by 90◦ ; in particular if c1 = 0 then di = 0, while all the permitted targets are nonzero, so the transition condition fails. If a departs from those two displacements the transition condition fails at that point, and otherwise a reaches a configuration with di ⊥ d0 in at most two moves. At that moment both robots are at rest, but a repeats the same displacement at its next activation, so the stopping condition fails (Lemma C.21, [NNW26, Lemma 21]). All of these are safety violations and may be concatenated by Lemma E.18(1)(2). In each case a single execution violates both predicates, so Lemma 6.1 gives the claim. Lemma F.11. EX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch. Begin with a fully synchronous prefix, which is admissible under Asynch; by Lemma E.14 the two robots always carry the same state s0 , s1 , . . . . If λsk = 0 for every k, this fair execution contains no nonzero Move, the joint-stop sequence reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition of VEC fails and the perpetuity condition of EPG∗ fails as well (Lemma 5.1(2)). Otherwise take the least k with λsk > 0. All earlier rounds left the configuration at ι. Activate a alone in round k, so that it performs the first nonzero Move of the execution, and use the freedom of Asynch in the choice of Look times: by Lemma E.7 the robot b observes a at a point pµ , µ ∈ (0, 1), of its movement segment. Let b begin its Move before the Move of a ends, so that no time of Σab lies between the two landings and the configuration after both landings is the term C1 of the joint-stop sequence following C0 . An F STA robot cannot consult the internal state of its peer and retains no previously observed angle or distance, so the displacement of b is determined by pµ alone, and C1 varies continuously and non-constantly with µ. The transition condition of VEC admits only finitely many successors of C0 , and the expansion condition of EPG∗ admits exactly one, namely the configuration with midpoint c0 and difference vector 2d0 . Adding the countably many values of µ at which the two robots coincide, 64

Corollary E.22 provides a µ avoiding all of them, and both the transition condition of VEC (Lemma C.22, [NNW26, Lemma 22]) and the expansion condition of EPG∗ fail at i = 0. Both are safety violations witnessed at a finite time, hence preserved by any fair tail (Lemma E.18(1)), and Lemma 6.1 gives the claim. Lemma F.12. EX ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch. Take the alternating singleton activation of a and b with an empty fully synchronous prefix (Lemma E.8) followed by an arbitrary fair tail, and supply normalized frames. By Lemma E.16(a) the rule of A is a single pair (λ, θ). Every round activates exactly one robot, so Lemma 5.1 applies to this execution. If λ = 0, then since OBLOT has no colors the two robots do not move under any schedule, the jointstop sequence reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition of VEC fails and the perpetuity condition of EPG∗ fails. Assume λ ̸= 0. The activation of a in round 1 is then the first nonzero Move of the execution and b is not activated in that round, so by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation. For VEC: if c1 = 0, then a lands on the position of b in round 1 and d1 = 0, while all the targets permitted by the transition condition are nonzero, so that condition fails. Otherwise the only single displacements satisfying the transition condition are those rotating the segment clockwise by 45◦ or by 90◦ . If a departs from them the transition condition fails at that point; otherwise a reaches a configuration with di ⊥ d0 in at most two moves, and since the same displacement is applied at every activation the robots move again immediately afterwards, so the stopping condition fails. All of these are safety violations. In every case a single execution violates both predicates, so Lemma 6.1 gives the claim. F.1.7

EPG∗ ∨ CFSM (EC)

Lemma F.13. EC ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of F COM under Ssynch. The initial colors being common, while both robots are activated simultaneously their colors follow a common eventually periodic sequence γ0 , g(γ0 ), . . . (Lemma E.14) and the configuration stays at ι. If λcj = 0 for every j, this fair execution contains no nonzero Move, so moves(a) = 0 ̸= 1 and the single-move condition of CFSM fails, while the joint-stop sequence reduces to C0 and the perpetuity condition of EPG∗ fails (Lemma 5.1(2)). Otherwise let t be the least j with λcj ̸= 0, replace round t by an interval in which a alone is activated twice in a row (Lemma E.4), and append a fair tail. The configuration was unchanged before the interval, so the first activation of a in it is the first nonzero Move of the execution and it is performed while b is not activated; by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0. Since b is not activated, the visible state of a is constantly ct , so by Lemma E.16(b) it applies the same pair (λ, θ), with λ ̸= 0, at both activations. If c1 = 0 then a lands on the position of b at the first activation, so a(t) = b(t) at that time and the collision-freedom condition fails. If c1 ̸= 0 the distance stays positive, so both activations are nonzero Move operations and moves(a) ≥ 2 ̸= 1, which violates the single-move condition. All of these are safety violations witnessed at a finite time, hence preserved by the fair tail (Lemma E.18(1)(2)). In each case a single execution violates both predicates, so Lemma 6.1 gives the claim. Lemma F.14. EC ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch. Take the alternating singleton activation of a and b with an empty fully synchronous prefix (Lemma E.8) followed by an arbitrary fair tail, and supply normalized frames. The visible state of an OBLOT robot is unique, so the rule is a single pair (λ, θ) (Lemma E.16(a)). Every round activates exactly one robot, so Lemma 5.1 applies. 65

If λ = 0, then since OBLOT has no colors the two robots do not move under any schedule, so moves(a) = 0 ̸= 1 and the single-move condition of CFSM fails, while the joint-stop sequence reduces to C0 and the perpetuity condition of EPG∗ fails. Assume λ ̸= 0. The activation of a in round 1 is the first nonzero Move of the execution and b is not activated in that round, so by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation. For CFSM: if c1 = 0, then a lands on the position of b in round 1, so a(t) = b(t) at that time and the collision-freedom condition fails. Otherwise each activation maps d 7→ c1 d (Lemma E.2(i)) and d stays positive, so the activations of a in rounds 1 and 3 are both nonzero Move operations and moves(a) ≥ 2 ̸= 1, which violates the single-move condition. In every case a single execution violates both predicates, so Lemma 6.1 gives the claim.

F.2

Disjunctions Containing SRO

This subsection treats the five antichains {SRO, P } with P ∈ E ∪ P3 = {S-RDV, VSS} ∪ {ZCC, VEC, CFSM}, which carry eight boundary obligations in total. Every construction below uses the geometric fact of Lemma E.19: with the peer at rest, two nonzero Move operations of the same robot cannot both satisfy the rotation condition and C3, and a landing on the peer satisfies neither C1, which requires ∥di+1 ∥ = ∥di ∥ ̸= 0, nor C2, whose right-hand side is nonzero. Zero Move operations do not advance the sequence (Ci ). Two of the eight obligations are discharged by Proposition 5.3, stated in Section 5.2: for every algorithm in F STAS it exhibits one admissible fair execution violating SRO and both exchange tasks simultaneously, which settles the whole maximal negative boundary of SRO ∨ VSS and the F STAS obligation of SRO ∨ S-RDV. We give its proof first; the remaining six obligations, including the F COMA obligation of SRO ∨ S-RDV, are proved afterwards. Proof of Proposition 5.3. Fix an algorithm and initial positions a0 = d, b0 = 2d, with outer points 0 and 3d. Under normalized observations, write the state transition as s 7→ g(s) and the global displacement as Λs w, where w points to the peer and Λs = λs Rθs . If the state orbit from the designated initial state contains no nonzero output, fair alternating activations leave both robots stationary. Neither the exchange phase nor the perpetual-motion requirement of SRO is satisfied. Otherwise, let s be the first state in that orbit with a nonzero output. Activate only a until it completes that first nonzero Move; all preceding Move operations are zero. We distinguish two cases. Case 1: Λs = 2I. Robot a moves to 3d. The relative segment reverses direction without shrinking, violating both permitted rotation types of SRO. Keep a inactive and activate b until it reaches state s; its preceding normalized outputs are zero, so the configuration is unchanged and b sees a peer at distance d. By Lemma D.23 the robot b then lands at 4d, outside the interval allowed by S-RDV and outside the endpoint set allowed by VSS. All three violations are irreversible, so append a fair suffix. Case 2: Λs ̸= 2I. The first nonzero Move makes the common exchange phase unattainable, so both exchange tasks have already failed. If it also violates SRO, append a fair suffix. Otherwise the robots remain distinct. If the normalized state orbit of a after this Move contains another nonzero output, keep b inactive until that output is executed; Lemma E.19 then violates SRO, and a fair suffix again completes the construction. If no further nonzero output occurs in that orbit, keep a inactive and let b traverse the same initial state sequence and perform its first nonzero Move. If this violates SRO, we are done. Otherwise the robots are still distinct and both are now in state g(s), from which every subsequent normalized output is zero. Fair alternating activations therefore leave the configuration stationary forever, violating the perpetual-motion requirement of SRO. This specifies a fair infinite continuation rather than an infinite solo execution.

66

F.2.1

SRO ∨ S-RDV (RS)

Lemma F.15. RS ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch. Run fully synchronous cycles with identical normalized observations until the first nonzero output. Throughout this zero-Move prefix, the positions remain unchanged and the colors remain equal. If no nonzero output occurs, the resulting fair, time-divergent execution violates both the perpetuity condition of SRO and the terminal rendezvous condition of S-RDV. Otherwise, identify the plane with C using orientation-preserving coordinates in which a0 = 0 and b0 = 1. The outer exchange points are then A = −1 and B = 2. In the first round with a nonzero output, let both robots perform their Looks before either performs its Compute, and then let both complete their Computes. Write z = λeiθ ̸= 0 for their common normalized displacement coefficient and γ ′ for their common updated color. Their computed destinations are z for a and 1 − z for b. Delay the Move of b until strictly after a has landed, and do not reactivate a in the meantime. There is therefore a joint stop with positions (z, 1), whose relative-position vector is d1 = 1 − z,

d0 = 1.

Case z ̸= 2. The first nonzero Move of a does not land at B. Since its Look observed the initial configuration, the first-move obstruction of Lemma D.22 applies, so no extension can satisfy the exchange phase of S-RDV. If   1−i , d1 ∈ / −i, 2 then neither C1 nor C2 holds at the first transition, and SRO already fails. Otherwise   1+i z ∈ 1 + i, . 2 Let b complete its stored Move while a remains at rest. The next joint-stop vector is d2 = 1 − 2z, and ( 2 − i, z = 1 + i, d2 = d1 1 − i, z = (1 + i)/2. √ √ Its modulus is respectively 5 or 2. Thus the distance increases, whereas C1 and C2 permit only √ factors 1 and 1/ 2. Hence SRO fails at this second transition. Once the pending cycle of b is complete, append an arbitrary admissible fair, time-divergent continuation. Both finite violations persist. Case z = 2. The first joint-stop vector is d1 = −1, which satisfies neither C1 nor C2. Thus SRO has already failed. Without reactivating a, let b complete its stored Move. The robots are now at (B, A) = (2, −1), both have color γ ′ , and neither has a pending cycle. Let Aγ ′ have the same transition rule as A, but designated initial color γ ′ . Apply the general impossibility of stable RDV in F COMA used in the proof of Lemma D.25. It gives a fair, time-divergent counterexecution of Aγ ′ from distinct initial positions. As in the proof of Lemma F.6, an orientationpreserving similarity and corresponding choices of local frames reproduce this counterexecution from (B, A). The reached colors and absence of pending cycles make it a valid continuation of our prefix. This continuation violates the terminal rendezvous condition of S-RDV, while preserving the earlier violation of SRO. In every case, a single admissible fair, time-divergent execution violates both predicates. Lemma 6.1 gives the claim. The remaining obligation of this pair, the class FSTAS , is Proposition 5.3.

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F.2.2

SRO ∨ ZCC (RZ)

Lemma F.16. RZ ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of F COM under Ssynch. As above the colors follow a common eventually periodic sequence c0 , g(c0 ), . . . (Lemma E.14) while both robots are activated simultaneously, and the configuration stays at ι. If λcj = 0 for every j, no nonzero Move ever occurs, so the sequence of configurations reduces to C0 and the perpetuity condition of SRO fails, while the robots remain at distinct positions and the convergence condition of ZCC fails. This fair execution witnesses both liveness violations. Otherwise let t be the least j with λcj ̸= 0 and replace round t by an interval in which a alone is activated k times in a row (Lemma E.4, with k large enough for what follows). Since b is not activated, the visible state of a is constantly ct , so by Lemma E.16(b) the rule is the single pair (λ, θ) with λ ̸= 0, and b is at rest throughout. For SRO: if c1 = 0 the robot a lands on the position of b at the first activation, and di+1 = 0 satisfies neither C1 nor C2, so the second condition of SRO fails. If c1 ̸= 0, taking k ≥ 2 makes a perform two nonzero Move operations with its peer at rest, so by Lemma E.19 the second condition of SRO fails. Either way this is a safety violation in the prefix. For ZCC: put σ1 := 1 − λeiθ and c1 := |σ1 |, as in the proof of Lemma F.7. Under Ssynch every activation of a is a complete cycle, so neither robot has a pending operation at the end of any of them. If c1 = 0, end the solo interval after its first activation; the robots then coincide. Use the coincidence continuation described in the proof of Lemma C.16. This either keeps both positions fixed forever, so that the zig-zag condition fails, or produces a finite violation of the collision-freedom condition. If c1 ̸= 0 and σ1 ∈ / (0, 1), the cases ψ ̸≡ 0 and ψ ≡ 0 with c1 > 1 in the proof of Lemma F.7 apply verbatim: for k large enough the collision-freedom condition fails at a finite time, and an arbitrary fair, time-divergent continuation is appended. In
the remaining contraction case σ1 ∈ (0, 1), after the k activations we have a(k) = b(0) − σ1k b(0) − a(0) and b is still at b(0). Thus the robots still lie in the original initial segment in their original order, and neither has a pending operation. Apply the continuation property proved in Lemma C.16, keeping the original initial positions fixed. It supplies a fair, time-divergent continuation violating ΦZCC , while preserving the safety violation of ΦSRO already obtained in the prefix. The safety violation of ΦSRO occurs in the prefix and precedes every continuation used above, so Lemma E.18(1)(3) is satisfied and Lemma 6.1 gives the claim. Lemma F.17. RZ ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM , which admits synchronous activations of one robot at a time (Lemma E.4). Begin with an interval in which b is never activated and a alone is activated, one complete cycle at a time, with normalized frames at every Look at which the robots are distinct. Since in FSTA activations that only update the color may be interleaved, the length of this interval is measured in nonzero Move operations and not in activations. Continue it until a lands on the position of b or completes its second nonzero Move, whichever comes first. As long as a has not landed on b, every Look of a observes distinct positions, so its color follows c0 , g(c0 ), g 2 (c0 ), . . . (Lemma E.11). If in this interval a lands on the position of b, then di+1 = 0 and neither C1 nor C2 can hold, so the second condition of SRO fails at that point, a safety violation. Suppose next that a never lands on b and performs at most one nonzero Move along its whole solo orbit, which includes the case in which no state produces a nonzero Move. Then the periodic part of the orbit of c0 contains no nonzero rule. Stop activating a once its color has entered that periodic part, and activate b alone the same number of times. The color of b follows the same finite orbit as long as its Looks observe distinct positions. If b lands on the position of a, the second condition of SRO fails as in the previous case, and the prefix is ended there as described below. Otherwise both robots now carry colors in the periodic part, at distinct positions, and neither has a pending operation. Continue with alternating singleton activations forever. Every Look observes distinct positions and every output is zero, so the positions remain fixed. This fair, time-divergent execution contains only finitely many

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nonzero Move operations, so the sequence (Ci ) is finite and the perpetuity condition of SRO fails, while fr is eventually constant and the zig-zag condition of ZCC fails. In the remaining case a does not land on b and performs two nonzero Move operations with its peer at rest, so by Lemma E.19 the rotation condition and C3 cannot both hold, whatever the internal states, and the second condition of SRO fails. In each case with a safety violation of ΦSRO , the violation is produced by the last Move of the prefix constructed so far, and every Look in that prefix observed distinct positions. By Lemma E.1(1), each robot that has been activated k times carries g k (c0 ). End the prefix at this joint stop, with no pending operation. If the robots coincide, use the coincidence continuation from Lemma C.16. Otherwise their colors lie on the same normalized color orbit, so the continuation statement proved in Lemma C.17 applies. In either case it supplies a fair, time-divergent continuation violating ΦZCC , with the original initial positions kept fixed. The safety violation established in the prefix persists. All the blocks may be concatenated by Lemma E.18(1)(2)(3), and a single execution violates both predicates, so Lemma 6.1 gives the claim. F.2.3

SRO ∨ VSS (RV)

The whole maximal negative boundary of this pair is the single class F STAS , and that obligation is discharged in the main text by Proposition 5.3. F.2.4

SRO ∨ VEC (RX)

Lemma F.18. RX ∈ / Sol(F COMS ). Proof. Let A be any deterministic algorithm of FCOM under Ssynch. The colors follow a common sequence γ0 , g(γ0 ), . . . (Lemma E.14) while both robots are activated simultaneously, and the configuration stays at ι. If λcj = 0 for every j, no nonzero Move ever occurs, so the sequence of configurations reduces to C0 and both the perpetuity condition of SRO and the attainment condition of VEC, which would require d0 ⊥ d0 , fail. Otherwise let t be the least j with λcj ̸= 0, replace round t by an interval in which a alone is activated k ≥ 3 times in a row (Lemma E.4), and append a fair tail. The visible state of a is constantly ct , so by Lemma E.16(b) it applies the same pair (λ, θ), with λ ̸= 0, at every activation. If c1 = 0, the robot a lands on the position of b at the first activation and di+1 = 0; both C1, C2 and all the targets permitted by the transition condition of VEC are nonzero, so the second conditions of both tasks fail simultaneously. If c1 ̸= 0, the same robot performs two nonzero Move operations with its peer at rest, so by Lemma E.19 the second condition of SRO fails. For VEC the argument of Lemma F.10 applies: the only single displacements satisfying the transition condition rotate the segment clockwise by 45◦ or by 90◦ , so if a departs from them the transition condition fails, and otherwise a reaches a configuration with di ⊥ d0 in at most two moves and then repeats the same displacement, so the stopping condition fails. All of these are safety violations. In each case a single execution violates both predicates, so Lemma 6.1 gives the claim. Lemma F.19. RX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch. Define the execution by starting, as in the first stage of Lemma C.25, with a finite fully synchronous prefix that freezes the common state of the two robots, which Asynch admits (Lemma E.14), then placing an observation-during-movement interval (Lemma E.7), and finally appending an arbitrary fair tail. If λs = 0 for every state s, the configuration never changes, so the sequence (Ci ) is finite and d0 ̸⊥ d0 ; both the perpetuity condition of SRO and the attainment condition of VEC fail. Otherwise, in the round of the first common state s with λs > 0, activate a alone so that it performs a nonzero Move, and let the robot b, in state s, observe a point pµ , µ ∈ (0, 1), of that movement segment 69

(Lemma E.7, which is admissible for K = A). Delay the Move of b until after a has landed, so that the landing configuration of a is a joint stop. Do not reactivate a before b completes this Move. If this Move of a is neither the clockwise 90◦ rotation about the peer (C1) nor the clockwise 90◦ rotation about the midpoint (C2), then by Lemma E.19 the second condition of SRO already fails. If it is one of them, the robot b performs the displacement it computes from pµ , and the distance at the time both robots are at rest is a non-constant continuous function of µ. The distances permitted by SRO belong to the countable set generated from d by the ratios of C1 and C2, so by Corollary E.22 some µ avoids that set and ΦSRO fails. In the same observation, an FSTA robot b cannot consult the internal state of its peer and retains no previously observed angle or distance, so its displacement is determined by pµ alone. The segment obtained after both landings varies continuously and non-constantly with µ, whereas the transition condition of VEC admits only finitely many targets, so by Corollary E.22 some µ avoids them. Both excluded sets being countable, a single µ ∈ (0, 1) avoids both, and the transition condition of VEC fails in the same execution. Both are safety violations, so they may be concatenated by Lemma E.18(1)(2), and Lemma 6.1 gives the claim. F.2.5

SRO ∨ CFSM (RC)

Lemma F.20. RC ∈ / Sol(FCOMS ). Proof. Let A be any deterministic algorithm of FCOM under Ssynch. The colors follow a common sequence γ0 , g(γ0 ), . . . (Lemma E.14) while both robots are activated simultaneously, and the configuration stays at ι. If λcj = 0 for every j, no nonzero Move ever occurs, so moves(a) = 0 ̸= 1 and the single-move condition of CFSM fails, while the sequence of configurations reduces to C0 and the perpetuity condition of SRO fails. Otherwise let t be the least j with λcj ̸= 0, replace round t by an interval in which a alone is activated twice in a row (Lemma E.4), and append a fair tail. The visible state of a is constantly ct , so by Lemma E.16(b) it applies the same pair (λ, θ), with λ ̸= 0, at both activations. If c1 = 0, the robot a lands on the position of b at the first activation, so a(t) = b(t) at that time and the collision-freedom condition of CFSM fails, while di+1 = 0 satisfies neither C1 nor C2 and the second condition of SRO fails. If c1 ̸= 0, both activations are nonzero Move operations, so moves(a) ≥ 2 ̸= 1 and the single-move condition fails, while Lemma E.19 makes the second condition of SRO fail as well. All of these are safety violations witnessed at a finite time, hence preserved by the fair tail (Lemma E.18(1)). In each case a single execution violates both predicates, so Lemma 6.1 gives the claim.

F.3

Disjunctions of the Remaining Exchange Tasks

This subsection treats the eight antichains {E, P } with E ∈ E = {S-RDV, VSS} and P ∈ P4 = {OSPO, ZCC, VEC, CFSM}, which carry eighteen boundary obligations in total. The exchange component is always supplied by the common first phase of Appendix D.5: either no nonzero Move ever occurs, so that Lemma D.21 is contradicted, or the first nonzero Move does not land where Lemma D.22 requires, or it does and Lemma D.23 carries the peer out of the admissible region. The companion component is obtained from a solo activation interval, a delayed-move interval, an observation-duringmovement interval, or a violation of a move-count or stopping requirement, according to the companion task and the boundary class. F.3.1

OSPO ∨ S-RDV (OS)

Lemma F.21. OS ∈ / Sol(F COMA ).

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Proof. Let A be any deterministic algorithm of F COM under Asynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line. Run fully synchronous cycles with identical normalized observations (Lemma E.14). While every Move is zero, the positions stay fixed and the colors stay equal. No nonzero output. If no nonzero output occurs, continue forever. The robots stay distinct at constant distance. So ΦS-RDV fails (no rendezvous), and ΦOSPO fails (no decreasing phase). This execution witnesses both liveness violations. A nonzero output occurs. Otherwise let γ be the common color at the start of the first round with a nonzero output. Let (λ, θ), with λ > 0, be the normalized displacement for the initial configuration with peer color γ. Since A is deterministic, (λ, θ) is known before that round is scheduled. Case (λ, θ) ̸= (2, 0). Replace the round by two solo activations of a (Lemma E.4), followed by a fair tail. The peer color stays γ, so a applies (λ, θ) twice (Lemma E.16(b)). The first of these Move operations observes ι and does not land at B. So the exchange phase of S-RDV is unsatisfiable (Lemma D.22). As in Lemma F.3, either a collision occurs or ma = 2, mb = 0; either way ΦOSPO is violated. All these violations are safety violations. Case (λ, θ) = (2, 0). Let both robots Look, and then both Compute; their colors become a common γ ′ . Delay the Move of b so that it begins strictly after the Move of a ends (Corollary E.6). The robot a moves from d to 3d while b rests at 2d, so they collide, and ΦOSPO fails. Then let b complete its Move to A. The robots are now at (B, A) with common color γ ′ , and neither has a pending operation. Let Aγ ′ be A with initial color γ ′ . The impossibility of stable RDV in FCOMA used in Lemma D.25 gives a counterexecution of Aγ ′ . Transport it to (B, A) by an orientation-preserving similarity, as in Lemma F.6, and append it. This continuation violates the terminal rendezvous condition of S-RDV and witnesses that liveness violation. The safety violation of ΦOSPO precedes it. Lemma E.18(1)(2)(3) is satisfied in every case, and Lemma 6.1 gives the claim. Lemma F.22. OS ∈ / Sol(F STAS ). Proof. Let A be any deterministic algorithm of FSTA under Ssynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line, and normalized frames at every Look at which the robots are distinct. The next color and the displacement of a robot then depend only on its own color. First part. Activate {a}, {b}, {a}, . . . , which Ssynch admits. By the π-symmetry of ι and Lemma E.10, both robots follow c0 , g(c0 ), . . . until the first nonzero Move. If there is no nonzero Move, continue forever. The robots stay at ι, so they never meet; ΦS-RDV fails by Lemma D.21, and ΦOSPO fails because d is constant. This execution witnesses both liveness violations. Otherwise let k be least with λgk (c0 ) ̸= 0. The robot a performs the first nonzero Move at its (k +1)-st activation; let that Move complete, whether or not the two robots meet during it. • If its landing point is not B, ΦS-RDV is unsatisfiable (Lemma D.22). • If it is B, let b be activated next; it lands at 4d ∈ / [A, B] (Lemma D.23), so ΦS-RDV fails. Either way ΦS-RDV suffers a safety violation at a finite time. This argument uses only the landing points, so it does not assume that the two robots stay distinct. If a collision has occurred in the prefix constructed so far, the collision-freedom condition of OSPO has already failed. Both predicates have then suffered safety violations, so append an arbitrary fair, time-divergent continuation. Otherwise the first move of a did not land at B, since a move from d to 3d would pass through the stationary robot at 2d. The safety violation of ΦS-RDV has therefore already occurred, and b has not yet performed its (k + 1)-st activation. Activate b once. If a collision occurs during this cycle, both predicates again have safety violations and an arbitrary fair tail completes the construction. Otherwise let T be the end of this cycle. Both robots have observed distinct positions, so they now carry γ = g k+1 (c0 ), neither has a pending operation, and ma (T ) = mb (T ) = 1.

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Every execution satisfying ΦOSPO has ma (tj ) = mb (tj ) = j at its phase boundaries. Consequently |ma − mb | ≤ 1 at all times, and both counts are unbounded. Let Jγ = {i ≥ 0 : λgi (γ) ̸= 0}. If |Jγ | ≥ 2, let i′ be its second element and activate a alone for ′ i + 1 cycles, stopping this construction at the first collision if one occurs. A collision violates ΦOSPO ; otherwise the resulting counts are ma = 3, mb = 1, which also violate it. Append an arbitrary fair tail. If |Jγ | ≤ 1, use a fully synchronous fair tail. A collision again violates ΦOSPO . If no collision occurs, each robot performs at most one further nonzero move, contradicting the required unboundedness of the counts. Thus one fair infinite execution violates both predicates, and Lemma 6.1 gives the claim. Lemma F.23. OS ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line and take the infinite fully synchronous execution of Lemma E.13 from this instance, with a normalized frame at every Look. The visible state of an OBLOT robot is unique, so the rule is a single pair (λ, θ) (Lemma E.16(a)), and by Lemma E.2(ii) the distance at the end of round k is d(k) = c2k d(0). Violation of ΦOSPO . If the two robots ever occupy the same point, the collision-freedom condition fails at a finite time; this is a safety violation. It covers c2 = 0, where they coincide at the end of round 1 and the coincidence is maintained by Lemma E.12, and also the case in which they meet while travelling inside a round. Otherwise d(t) > 0 throughout. By Lemma E.13 every joint stop of the execution carries one of the round-boundary distances d(k), so the restriction of d to Σab is monotone. The turning times required by OSPO are joint stops, so by the criterion stated after Definition D.12 no strictly decreasing phase can be followed by a strictly increasing one, and ΦOSPO fails. This is a liveness violation, and the fully synchronous execution is itself the fair infinite continuation along which it persists. Violation of ΦS-RDV . If λ = 0 no nonzero Move ever occurs, so the first nonzero Move demanded by Lemma D.21 does not exist. This is a progress violation, witnessed by the same infinite fully synchronous execution. If λ ̸= 0, both robots perform a nonzero Move in round 1, so an admissible exchange configuration must be (B, A) = (3d, 0) with move counts (1, 1). When (λ, θ) ̸= (2, 0) the landing of round 1 is not this configuration, so the requirement of Lemma D.22 is unmet. When (λ, θ) = (2, 0) round 1 does land at (3d, 0), but round 2 applies the same rule and sends a to 3d + 2(0 − 3d) = −3d and b to 0 + 2(3d − 0) = 6d, both outside [A, B] = [0, 3d] (Lemma D.27). These are safety violations witnessed at a finite time. In each case a single execution violates both predicates, so Lemma E.18(1)(3) is satisfied and Lemma 6.1 gives the claim. F.3.2

OSPO ∨ VSS (OV)

Lemma F.24. OV ∈ / Sol(F STAS ). Proof. Let A be any deterministic algorithm of FSTA under Ssynch. Use the execution of Lemma F.22. It violates ΦOSPO : in that proof the violation comes from a collision up to T , from the move-count requirement (∗) in the second part, or, if no nonzero Move occurs, from the constancy of d, and none of these refers to the exchange task. We show that the same execution also violates ΦVSS , where P = {0, d, 2d, 3d}. If no nonzero Move occurs, (P1) fails by Lemma D.21; the execution of Lemma F.22 itself is the fair continuation witnessing this liveness violation. Otherwise, as in that lemma, the two robots traverse the same states until the first nonzero Move and a is activated first, so that Move is performed by a and its Look observes ι. • If it lands outside P , (P1-b) fails. • If it lands in P \ {B}, (P1-c) is unsatisfiable (Lemma D.22). 72

• If it lands at B, then b lands at 4d ∈ / P (Lemma D.23), and (P1-b) fails. In each case this is a safety violation at a finite time, whether or not a collision occurs, since VSS permits passing. In either case a single execution violates both predicates; Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. Lemma F.25. OV ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch. Take the fully synchronous execution of Lemma E.13 with normalized frames. By Lemma E.16(a), both robots apply one and the same pair (λ, θ) in every round. Use A = 0, a0 = d, b0 = 2d, B = 3d and P = {0, d, 2d, 3d}. OSPO. If a collision occurs, ΦOSPO fails. Otherwise the joint stops lie between rounds when λ > 0, and the distance there is c2k d(0) (Lemma E.2(ii)). This is monotone, so no decreasing phase can be followed by an increasing one. The execution itself witnesses this liveness violation. VSS. We distinguish three cases. • If λ = 0, (P1) fails by Lemma D.21. This is a liveness violation of the same execution. • If λ > 0 and (λ, θ) ̸= (2, 0), the round-1 Move of a completes earliest and does not land at B. So ΦVSS is unsatisfiable (Lemma D.22). • If (λ, θ) = (2, 0), round 1 yields (3d, 0) and round 2 yields (−3d, 6d). Neither point lies in P , so (P1-b) fails. The last two violations are safety violations. Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. F.3.3

S-RDV ∨ ZCC (SZ)

Lemma F.26. SZ ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of F COM under Asynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line. Run fully synchronous cycles with identical normalized observations (Lemma E.14). While every Move is zero, the positions stay fixed and the colors stay equal. No nonzero output. Continue forever. The robots stay at ι, so ΦS-RDV fails, while fr ≡ 0 for both robots, so the zig-zag condition admits no strictly increasing pair and ΦZCC fails. This execution witnesses both liveness violations. A nonzero output occurs. Let γ be the common color at the start of the first round with a nonzero output, and let (λ, θ), with λ > 0, be the normalized displacement for the initial configuration with peer color γ. Since A is deterministic, (λ, θ) is known before that round is scheduled. Case (λ, θ) = (2, 0). Let both robots Look, and then both Compute; their colors become a common γ ′ . Delay the Move of b so that it begins strictly after the Move of a ends (Corollary E.6). The robot a moves from d to 3d while b rests at 2d, so at the landing ∥a − a0 ∥ = 2d > d = ∥b − a0 ∥, and the collision-freedom condition of ZCC fails at a finite time, a safety violation. Then let b complete its Move to A. The robots are at (B, A) with common color γ ′ , and neither has a pending operation. As in Lemma F.21, the impossibility of stable RDV in FCOMA used in Lemma D.25, applied to A with initial color γ ′ , supplies a counterexecution that an orientation-preserving similarity transports to (B, A). Appending it violates the terminal rendezvous condition of S-RDV. The safety violation of ZCC precedes that liveness violation. Case (λ, θ) ̸= (2, 0). Replace the round by a solo activation interval of a (Lemma E.4), during which b is never activated and keeps the color γ; by Lemma E.16(b) the robot a applies (λ, θ) at every activation and b is at rest throughout. The interval may be made arbitrarily long, the activations of b being placed 73

far enough in the future to keep the execution fair. Every activation of a in it is a complete cycle, and b has no pending operation. The first of these Move operations observes ι and does not land at B, so the exchange phase of S-RDV is unsatisfiable (Lemma D.22), a safety violation at a finite time. For ΦZCC , put σ1 := 1 − λeiθ and c1 := |σ1 |. If c1 = 0, the robot a reaches the position of b at the first activation, and neither robot has a pending operation. The robots coincide; use the coincidence continuation described in the proof of Lemma C.16; it activates one robot at a time with complete cycles, which Asynch admits. This either keeps both positions fixed forever, so that the zig-zag condition fails, or produces a finite violation of the collisionfreedom condition. Assume c1 > 0, and put ψ := arg σ1 (Lemma E.2(iii)). Writing w0 for the vector from a to b at the beginning of the interval, the vector after j activations is σ1j w0 ̸= 0, so every Look of a in the interval observes distinct positions, and fa (j) = ∥a(j) − a(0)∥ = d 1 − c1j eijψ . Since b stands still, ∥b − a0 ∥ = d, and the collision-freedom condition requires fa (j) ≤ d, that is c1j ≤ 2 cos(jψ). Note that σ1 = 1 would force λ = 0, so c1 ̸= 1 whenever ψ ≡ 0. • If ψ ̸≡ 0 (mod 2π), Lemma E.2(iii) provides k ≥ 1 with cos(kψ) ≤ − 12 < 0. Making the interval long enough to reach the k-th activation, the collision-freedom condition fails there, a safety violation. • If ψ ≡ 0 and c1 > 1, that is θ = π, then fa (j) = (c1j − 1)d exceeds d as soon as c1j > 2, so the collision-freedom condition again fails after finitely many activations. • In the remaining contraction case ψ ≡ 0 and 0 < c1 < 1, that is σ1 ∈ (0, 1), end the solo interval after its first activation. Then a lies in the open segment (a0 , b0 ) and b is still at b0 , so the robots still lie in the original initial segment in their original order, and neither has a pending operation. Apply the continuation property proved in Lemma C.16, keeping the original initial positions fixed; its schedule activates one robot at a time with complete cycles, which Asynch admits. It supplies a fair, time-divergent continuation violating ΦZCC , while preserving the safety violation already obtained in the prefix. In every case a single execution violates both predicates, the safety violations preceding the liveness ones, so Lemma E.18(1)(3) is satisfied and Lemma 6.1 gives the claim. Lemma F.27. SZ ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM . Synchronous activations of one robot at a time are admissible under AM (Lemma E.4), so a and b may be activated alternately, each cycle being completed before the next one begins, and, under normalized frames, the argument of Lemma D.23 applies unchanged. If no nonzero Move ever occurs, which in FSTA is equivalent to λgj (c0 ) = 0 for every j since the color sequence is eventually periodic (Lemma E.11), the two robots stay at ι at distinct positions, so ΦS-RDV fails and the convergence condition of ZCC fails as well. Otherwise the first nonzero Move is performed by a. Either it does not land at B = 3d, so that the first-phase clauses become unsatisfiable by the argument of Lemma D.22, or it does and Lemma D.23 makes the robot b, activated next, land at 4d ∈ / [A, B]; in either case ΦS-RDV fails, a safety violation. Let the prefix end with the cycle that establishes this violation, that of a in the first case and that of b in the second. Every Look in this prefix observes distinct positions: before the first nonzero Move the configuration is ι, in the second case b observes a at B ̸= 2d, and a landing on the peer can only be the last Move of the prefix. Since the visible state of an F STA robot is its own color, Lemma E.1(1) shows that each robot that has been activated k times carries g k (c0 ).

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End the prefix at this joint stop, with no pending operation. If the robots coincide, use the coincidence continuation from Lemma C.16. Otherwise their colors lie on the same normalized color orbit, so the continuation statement proved in Lemma C.17 applies. In either case it supplies a fair, time-divergent continuation violating ΦZCC , with the original initial positions kept fixed. The safety violation established in the prefix persists (Lemma E.18(1)). A single fair infinite execution therefore violates both predicates, and Lemma 6.1 gives the claim. Lemma F.28. SZ ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch and take the infinite fully synchronous execution of Lemma E.13 with a normalized frame at every Look. The visible state of an OBLOT robot is unique, so the rule is a single pair (λ, θ) (Lemma E.16(a)). For this execution ΦZCC fails by Lemma C.15, which exhausts the degenerate cases λ = 0 and c2 = 0, the violation of the collision-freedom condition when φ ̸≡ 0, and the violation of the zig-zag condition when φ ≡ 0. For ΦS-RDV the argument is that of Lemma F.23: if λ = 0 no nonzero Move occurs and the first nonzero Move demanded by Lemma D.21 does not exist; if λ ̸= 0 and (λ, θ) ̸= (2, 0), the nonzero Move of round 1 does not meet the requirement of Lemma D.22; and if (λ, θ) = (2, 0), then in round 2 the robots land at −3d and 6d, leaving [A, B] = [0, 3d] (Lemma D.27). A single execution violates both predicates, so Lemma 6.1 gives the claim. F.3.4

S-RDV ∨ VEC (SX)

Lemma F.29. SX ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of F COM under Asynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line. The initial colors being common, while both robots are activated simultaneously their colors follow a common eventually periodic sequence c0 , g(c0 ), . . . (Lemma E.14) and the configuration stays at ι. If λcj = 0 for every j, continue forever. This fair execution contains no nonzero Move, so the two robots never occupy one and the same point permanently and ΦS-RDV fails, while the sequence of jointstop configurations reduces to C0 and d0 ̸⊥ d0 , so the attainment condition of VEC fails as well. This execution witnesses both liveness violations. Otherwise let γ be the common color at the start of the first round with a nonzero output, and let (λ, θ), with λ ̸= 0, be the normalized displacement for the initial configuration with peer color γ. Since A is deterministic, (λ, θ) is known before that round is scheduled. Case (λ, θ) = (2, 0). Let both robots Look, and then both Compute; their colors become a common γ ′ . Delay the Move of b so that it begins strictly after the Move of a ends (Corollary E.6). At the landing of a neither robot is inside a Move, although b still has a pending one, so that instant is a joint stop and realizes C1 = (3d, 2d), whence d1 = −d0 . This is neither the clockwise rotation by 45◦ nor the one by 90◦ , so the transition condition of VEC fails at i = 0, a safety violation at a finite time. Then let b complete its Move to A. The robots are at (B, A) with common color γ ′ , and neither has a pending operation. As in Lemma F.6, the impossibility of stable RDV in FCOMA used in Lemma D.25, applied to A with initial color γ ′ , supplies a counterexecution that an orientation-preserving similarity transports to (B, A). Appending it violates the terminal rendezvous condition of S-RDV. The safety violation of VEC precedes that liveness violation. Case (λ, θ) ̸= (2, 0). Replace the round by an interval in which a alone is activated k ≥ 3 times in a row (Lemma E.4), followed by an arbitrary admissible fair, time-divergent tail. The visible state of a is constantly γ, so by Lemma E.16(b) it applies the same pair (λ, θ) at every activation. The first of these Move operations observes ι, and the cumulative nonzero-move counts being (1, 0), an admissible exchange configuration would require a to land at B. It does not, so the exchange phase of S-RDV is unsatisfiable (Lemma D.22), a safety violation at a finite time. For VEC: if c1 = 0, the robot a lands on the position of b at the first activation and d1 = 0, while all the targets permitted by the transition condition are nonzero, so that condition fails. If c1 ̸= 0, then 75

as in Lemma F.10 the only single displacements satisfying the transition condition rotate the segment clockwise by 45◦ or by 90◦ ; if a departs from them the transition condition fails, and otherwise a reaches a configuration with di ⊥ d0 in at most two moves and then repeats the same displacement, so the stopping condition fails. Either way this is a safety violation in the prefix, preserved by the tail. In every case a single execution violates both predicates, so Lemma E.18(1)(3) is satisfied and Lemma 6.1 gives the claim. Lemma F.30. SX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of F STA under Asynch, and let E0 be the execution of alternating singleton activations {a}, {b}, . . . under normalized frames (Lemma E.4); E0 is fair and is determined by A. If E0 contains no nonzero Move, the two robots stay at ι, so ΦS-RDV fails, while the sequence of configurations reduces to C0 and d0 ̸⊥ d0 , so the attainment condition of VEC fails as well; Lemma 6.1 gives the claim. Otherwise, as long as no nonzero Move has occurred the two robots traverse the same sequence of states, so the first nonzero Move is performed by a, which is activated first, from some state s with λs > 0, while b is in the same state s and has no pending operation. Let pE be its landing point. Case pE = B. Follow E0 up to the round following this Move. When a has landed, d1 = b0 −B = −d0 , which is none of the successors admitted by the transition condition of VEC; and Lemma D.23 makes b, activated next, land at 4d ∈ / [A, B], so ΦS-RDV fails. Case pE ̸= B. Follow E0 up to the start of this Move, and by Lemma E.7 let b perform its Look while a is at pµ = (1 − µ)a0 + µpE , µ ∈ (0, 1), delaying the Move of b until a has landed. The Move of a is then the nonzero Move with earliest completion time and it does not land at B, so ΦS-RDV fails by Lemma D.22. By Lemma E.1(1) the robot b lands at b0 + λs Rθs (pµ − b0 ), an affine function of µ with nonzero slope λs Rθs (pE − a0 ), so a suitable coordinate of the segment after both landings is continuous and non-constant in µ. The transition condition of VEC admits only finitely many successors, so by Corollary E.22 some µ avoids them and the transition condition fails (cf. Lemma C.22). In both cases the two violations are safety violations forced by a finite prefix, which extends to a fair execution by Lemma E.18(1)(2); Lemma 6.1 gives the claim. Lemma F.31. SX ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch and take the fully synchronous execution in which both robots are activated in every round (Lemmas E.13 and E.14), with a normalized frame at every Look. The violation of ΦS-RDV is that of Lemma F.23. For ΦVEC the argument of Lemma C.20 applies to the same execution: if λ = 0 the attainment condition fails, if c2 = 0 the transition condition fails, and otherwise the robots move again immediately after any orthogonal configuration, so the stopping condition fails, or no such configuration is ever reached and the attainment condition fails. A single execution violates both predicates, so Lemma 6.1 gives the claim. F.3.5

S-RDV ∨ CFSM (SC)

Lemma F.32. SC ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of F COM under Asynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line. The colors follow a common sequence c0 , g(c0 ), . . . (Lemma E.14) while both robots are activated simultaneously, and the configuration stays at ι. If λcj = 0 for every j, continue forever. This fair execution contains no nonzero Move, so moves(a) = 0 ̸= 1 and the single-move condition of CFSM fails, while the two robots never occupy one and the same point permanently and ΦS-RDV fails as well. This execution witnesses both liveness violations. Otherwise let γ be the common color at the start of the first round with a nonzero output, and let (λ, θ), with λ ̸= 0, be the normalized displacement for the initial configuration with peer color γ. Case (λ, θ) = (2, 0). Let both robots Look, and then both Compute; their colors become a common γ ′ . Delay the Move of b so that it begins strictly after the Move of a ends (Corollary E.6). The robot a 76

moves from d to 3d while b rests at 2d, so the two robots occupy the same point while a travels and the collision-freedom condition of CFSM fails, a safety violation at a finite time. Then let b complete its Move to A. The robots are at (B, A) with common color γ ′ , and neither has a pending operation. As in Lemma F.6, the impossibility of stable RDV in FCOMA used in Lemma D.25, applied to A with initial color γ ′ , supplies a counterexecution that an orientation-preserving similarity transports to (B, A). Appending it violates the terminal rendezvous condition of S-RDV. The safety violation of CFSM precedes that liveness violation. Case (λ, θ) ̸= (2, 0). Replace the round by an interval in which a alone is activated twice in a row (Lemma E.4), followed by an arbitrary admissible fair, time-divergent tail. The visible state of a is constantly γ, so by Lemma E.16(b) it applies the same pair (λ, θ) at both activations. The first of these Move operations observes ι, and the cumulative nonzero-move counts being (1, 0), an admissible exchange configuration would require a to land at B. It does not, so the exchange phase of S-RDV is unsatisfiable (Lemma D.22), a safety violation at a finite time. For CFSM: if c1 = 0, the robot a lands on the position of b at the first activation, so a(t) = b(t) there and the collision-freedom condition fails. If c1 ̸= 0, both activations are nonzero Move operations, so moves(a) ≥ 2 ̸= 1 and the single-move condition fails. Either way this is a safety violation in the prefix, preserved by the tail. In every case a single execution violates both predicates, so Lemma E.18(1)(3) is satisfied and Lemma 6.1 gives the claim. Lemma F.33. SC ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch and take the fully synchronous execution in which both robots are activated in every round (Lemmas E.13 and E.14), with a normalized frame at every Look. The violation of ΦS-RDV is that of Lemma F.23. For ΦCFSM , Lemma E.13 leaves only three possibilities, namely moves(r) = 0, a permanent coincidence of the two robots, and moves(r) = ∞, each of which violates the single-move condition or the collision-freedom condition of CFSM. A single execution violates both predicates, so Lemma 6.1 gives the claim. F.3.6

ZCC ∨ VSS (ZV)

Lemma F.34. ZV ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM . Synchronous activations of one robot at a time are admissible (Lemma E.4), so begin with alternating singleton activations, each cycle being completed before the next one begins, under normalized frames. If no nonzero Move ever occurs, condition (P1) of VSS fails by Lemma D.21, and the two robots remain at distinct positions, so the convergence condition of ZCC fails as well. Otherwise the first nonzero Move is performed by a. Either it does not land at B = 3d, so that the first-phase clauses become unsatisfiable by the argument of Lemma D.22, or it does and Lemma D.23 makes the robot b, activated next, land at 4d ∈ / P ; in either case the endpoint condition of VSS fails, a safety violation. Let the prefix end with the cycle that establishes this violation, that of a in the first case and that of b in the second. As in the proof of Lemma F.27, every Look in this prefix observes distinct positions, so by Lemma E.1(1) each robot that has been activated k times carries g k (c0 ). End the prefix at this joint stop, with no pending operation. If the robots coincide, use the coincidence continuation from Lemma C.16. Otherwise their colors lie on the same normalized color orbit, so the continuation statement proved in Lemma C.17 applies. In either case it supplies a fair, time-divergent continuation violating ΦZCC , with the original initial positions kept fixed. The safety violation established in the prefix persists (Lemma E.18(1)). A single fair infinite execution therefore violates both predicates, and Lemma 6.1 gives the claim. Lemma F.35. ZV ∈ / Sol(OBLOT F ).

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Proof. Let A be any deterministic algorithm of OBLOT under Fsynch and take the fully synchronous execution in which both robots are activated in every round (Lemmas E.13 and E.14), with a normalized frame at every Look. The violation of ΦZCC is that of Lemma F.28, and the violation of ΦVSS is that of Lemma F.25, which rests on Lemma D.31. A single execution violates both predicates, so Lemma 6.1 gives the claim. F.3.7

VSS ∨ VEC (VX)

Lemma F.36. VX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of F STA under Asynch, and let E0 be the execution of alternating singleton activations {a}, {b}, . . . under normalized frames (Lemma E.4); E0 is fair and is determined by A. If E0 contains no nonzero Move, condition (P1) of VSS fails by Lemma D.21, while the sequence of configurations reduces to C0 and d0 ̸⊥ d0 , so the attainment condition of VEC fails as well; Lemma 6.1 gives the claim. Otherwise, as in the proof of Lemma F.30, the first nonzero Move is performed by a from some state s with λs > 0, while b is in the same state s and has no pending operation. Let pE be its landing point. Case pE = B. Follow E0 up to the round following this Move. When a has landed, d1 = b0 −B = −d0 , which is none of the successors admitted by the transition condition of VEC; and Lemma D.23 makes b, activated next, land at 4d ∈ / P , so the endpoint condition of VSS fails. Case pE ̸= B. Construct the execution as in the corresponding case of Lemma F.30: b performs its Look while a is at pµ , µ ∈ (0, 1) (Lemma E.7), and its Move is delayed until a has landed. The Move of a is the nonzero Move with earliest completion time and does not land at B, so by Lemma D.22 no valid T1 exists and ΦVSS fails. The landing point of b is affine in µ with nonzero slope, while the successors admitted by the transition condition of VEC are finitely many and independent of µ, whether or not the configuration (pE , b0 ) is counted; so by Corollary E.22 some µ avoids them and the transition condition fails. In both cases the two violations are safety violations forced by a finite prefix, which extends to a fair execution by Lemma E.18(1)(2); Lemma 6.1 gives the claim. Lemma F.37. VX ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch and take the fully synchronous execution in which both robots are activated in every round (Lemmas E.13 and E.14), with a normalized frame at every Look. The violation of ΦVSS is that of Lemma F.25 and the violation of ΦVEC is that of Lemma F.31. A single execution violates both predicates, so Lemma 6.1 gives the claim. F.3.8

VSS ∨ CFSM (VC)

Lemma F.38. VC ∈ / Sol(OBLOT F ). Proof. Let A be any deterministic algorithm of OBLOT under Fsynch and take the fully synchronous execution in which both robots are activated in every round (Lemmas E.13 and E.14), with a normalized frame at every Look. By Lemma E.13, if λ = 0 then moves(r) = 0 and the single-move condition of CFSM fails, while by Lemma D.21 the first phase of VSS cannot be achieved. If c2 = 0 the two robots coincide in round 1, so the collision-freedom condition of CFSM fails, and their landing point is the midpoint 1.5d ∈ / P , so ΦVSS fails as well. Otherwise both robots perform a nonzero Move in every round, so moves(r) = ∞ and the single-move condition fails, while as in Lemma D.31 the landings of round 2 leave the set P of designated points. In each case a single execution violates both predicates, so Lemma 6.1 gives the claim.

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F.4

Disjunctions Containing MCv

This subsection treats the three antichains {MCv, P } with P ∈ {VSS, VEC, CFSM}, which carry four boundary obligations in total. For FCOM, the constructions use the delayed-Comp argument of Lemma C.6. For F STA, after establishing the companion safety violation, we apply the continuation statement of Lemma C.7 from a configuration with distinct positions, a common internal color, and no pending operation. F.4.1

MCv ∨ VSS (MV)

Lemma F.39. MV ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of F STA under AM , which admits both the simultaneous activation of the two robots and synchronous activations of one robot at a time (Lemma E.4). Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line, with P = {0, d, 2d, 3d}, and normalized frames at every Look at which the robots are distinct. Activate both robots simultaneously while every output is zero. By Lemma E.14 they carry the same color at every round boundary and the configuration stays at ι. If every output along this execution is zero, continue forever. Condition (P1) of VSS fails by Lemma D.21, and the two robots remain at distinct positions, so the convergence condition of MCv fails as well. This execution witnesses both liveness violations. Otherwise let γ be the common color at the start of the first round with a nonzero output, let (λ, θ), with λ ̸= 0, be the corresponding normalized displacement, and put c1 := c1 (γ). Since A is deterministic, these are known before that round is scheduled. Case c1 = 0, that is (λ, θ) = (1, 0). Activate both robots in that round. Each moves to the position of its peer, so they exchange positions and the distance returns to d after passing through 0; the monotonicity condition of MCv therefore fails, a safety violation at a finite time. The cumulative nonzero-move counts are (1, 1), so an admissible exchange configuration would have to be (B, A) = (3d, 0), whereas the landing is (2d, d); no valid T1 exists and VSS fails by Lemma D.22. Append an arbitrary admissible fair, time-divergent tail, which preserves both violations. Case c1 ̸= 0. Define the execution as an interval of alternating singleton activations carried up to the round following the first nonzero Move, each cycle being completed before the next one begins, followed by the continuation described below. The first nonzero Move is performed by the robot a, which is activated first. Either it does not land at B, so that the first-phase clauses become unsatisfiable by the argument of Lemma D.22, or it does and Lemma D.23 makes the robot b, activated next, land at 4d ∈ / P ; in either case the endpoint condition of VSS fails, a safety violation. Since c1 ̸= 0, neither of these two Move operations lands on the peer, so the two robots are at distinct positions at the end of this interval. Both robots have now been activated equally often, and every Look so far has observed distinct positions, so by Lemma E.1(1) they carry the same color. At this point the robots occupy distinct positions, carry the same internal color, and have no pending operation. Apply the continuation statement proved in Lemma C.7. It supplies a fair, time-divergent continuation violating ΦMCv , while all safety violations established in the prefix persist. In every case a single execution violates both predicates, the violations being finite except where stated, so Lemma E.18(1)(2)(3) is satisfied and Lemma 6.1 gives the claim. F.4.2

MCv ∨ VEC (MX)

Lemma F.40. MX ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch. Run fully synchronous cycles with normalized frames; by Lemma E.14 the two robots carry the same color at every round boundary, and the configuration stays at ι while every output is zero.

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If every output along this execution is zero, continue forever. No nonzero Move occurs and the two robots remain at distinct positions, so the convergence condition of MCv fails, while the sequence of joint-stop configurations reduces to C0 and d0 ̸⊥ d0 , so the attainment condition of VEC fails as well. This execution witnesses both liveness violations. Otherwise stop the zero-Move prefix just before the first round with a nonzero output, let γ be the common color at this point, so that λγ > 0, and put c := c1 (γ). Build from here the delayed-Comp interval of Lemma C.6: let b perform its Look and delay its Comp, the saved snapshot determining the nonzero displacement v = λγ Rθγ (wold ), where wold is the vector from b to a at that Look. Throughout the delay the position of b and its visible color γ are unchanged, so by Lemma E.16(b) the robot a may be activated alone for any prescribed finite number of complete cycles, applying the same pair (λγ , θγ ), with λγ ̸= 0, at each of them. Activate a alone k times, where k = 1 if c = 0 and otherwise k ≥ 3 is large enough for what follows, and append an arbitrary admissible fair, time-divergent tail. Violation of ΦVEC . If c = 0, the robot a lands on the position of b at the first activation and d1 = 0, while all the targets permitted by the transition condition are nonzero, so that condition fails. If c ̸= 0, then as in Lemma F.10 the only single displacements satisfying the transition condition rotate the segment clockwise by 45◦ or by 90◦ ; if a departs from them the transition condition fails, and otherwise a reaches a configuration with di ⊥ d0 in at most two moves and then repeats the same displacement, so the stopping condition fails. Either way this is a safety violation inside the interval. Violation of ΦMCv . If c > 1, the first of these Move operations already increases the distance between its starting and ending times. If c = 1, its starting and ending positions are distinct points at the same distance from the stationary robot b, so the midpoint of the movement segment is strictly closer to b than either endpoint and the distance increases between the time a passes that midpoint and the completion of the Move. In both cases the monotonicity condition fails. If c < 1, choose k above so large that the vector wk from b to a after k complete cycles satisfies ∥wk ∥ = c k ∥wold ∥ < ∥v∥/2; for c = 0 this already holds with k = 1. Keep a stationary and let b complete its Comp using the saved snapshot, update its color, and perform the resulting Move with displacement v. The distance immediately before that Move is ∥wk ∥, whereas at its completion it satisfies ∥wk − v∥ ≥ ∥v∥ − ∥wk ∥ > ∥wk ∥, so the monotonicity condition fails again. Both violations are finite and occur inside the interval, so they may be concatenated by Lemma E.18(1) and are preserved by the fair tail. Lemma 6.1 gives the claim. Lemma F.41. MX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch. Run a fully synchronous prefix, which Asynch admits; by Lemma E.14 the two robots carry the same state at every round boundary. If λs = 0 for every state s of that common sequence, this fair execution contains no nonzero Move: the robots stay at their distinct initial positions, so the convergence condition of MCv fails, and the sequence of configurations reduces to C0 with d0 ̸⊥ d0 , so the attainment condition of VEC fails as well. Otherwise truncate the prefix at the round in which both robots carry the first state s with λs > 0; the configuration is still ι. Activate a alone, so that it performs a nonzero Move from a0 to some pE ̸= a0 , and by Lemma E.7 let b, in state s, perform its Look while a is at pµ = (1 − µ)a0 + µpE , µ ∈ (0, 1), delaying the Move of b until a has landed. By Lemma E.1(1) the displacement of b is v = λs Rθs (pµ − b0 ), so its landing point is affine in µ with nonzero slope λs Rθs (pE − a0 ). We exclude the at most one µ with pµ = b0 , so that v ̸= 0, and the at most one µ for which b lands at pE . VEC. The successors admitted by the transition condition of VEC are finitely many and independent of µ, whether or not the configuration (pE , b0 ) is counted, so by Corollary E.22 some µ avoids them as well, and the transition condition fails at a finite time (cf. Lemma C.22). MCv. If pE = b0 , then a lands on b, and the Move of b raises the distance from 0 to ∥v∥ > 0, so the monotonicity condition fails. Otherwise the robots are at distinct positions after both landings, and we use the following dichotomy, applied to every nonzero Move whose peer is at rest. Along such a Move the 80

distance to the stationary peer is non-increasing and then non-decreasing. If the landing point is not the point of the segment closest to the peer, the distance strictly increases at the end and the monotonicity condition fails; we may therefore assume that every such Move ends at its closest approach. Both robots have performed the same number of activations, and every Look so far has observed distinct positions, so by Lemma E.1(1) they carry the same color. At this point the robots occupy distinct positions, carry the same internal color, and have no pending operation. Apply the continuation statement proved in Lemma C.7. It supplies a fair, time-divergent continuation violating ΦMCv , while all safety violations established in the prefix persist. In every case a single execution violates both predicates. Whenever a nonzero output occurs, the violation of ΦVEC is a safety violation at a finite time. In the case without nonzero Move, both liveness violations are witnessed by the explicit fair execution given there. Otherwise ΦVEC has already suffered a safety violation. The continuation from Lemma C.7 witnesses either a finite monotonicity violation or a failure of convergence. Thus all predicates fail on the same fair infinite execution. Hence Lemma E.18(1)(2)(3) is satisfied, and Lemma 6.1 gives the claim. F.4.3

MCv ∨ CFSM (MC)

Lemma F.42. MC ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch. Run fully synchronous cycles with normalized frames; by Lemma E.14 the two robots carry the same color at every round boundary, and the configuration stays at ι while every output is zero. If every output along this execution is zero, continue forever. No nonzero Move occurs, so moves(a) = 0 ̸= 1 and the single-move condition of CFSM fails, while the two robots remain at distinct positions and the convergence condition of MCv fails as well. This execution witnesses both liveness violations. Otherwise stop the zero-Move prefix just before the first round with a nonzero output, let γ be the common color at this point, so that λγ > 0, and put c := c1 (γ). Build from here the delayed-Comp interval of Lemma C.6, exactly as in Lemma F.40: the robot b performs its Look and delays its Comp, the saved snapshot determining the nonzero displacement v = λγ Rθγ (wold ), and during the delay the robot a is activated alone k times, applying the same pair (λγ , θγ ) at each of them (Lemma E.16(b)). Take k = 1 if c = 0 and otherwise k ≥ 2 large enough for what follows, and append an arbitrary admissible fair, time-divergent tail. Violation of ΦCFSM . If c = 0, the robot a lands on the position of b at the first activation, so a(t) = b(t) at that time and the collision-freedom condition fails. If c ̸= 0, the k ≥ 2 activations are all nonzero Move operations, so moves(a) ≥ 2 ̸= 1 and the single-move condition fails. Either way this is a safety violation inside the interval. Violation of ΦMCv . The three cases c > 1, c = 1 and c < 1 are those of Lemma F.40 and the arguments there apply verbatim: a single Move of a already breaks monotonicity when c ≥ 1, and when c < 1 the choice ∥wk ∥ < ∥v∥/2 makes the distance grow from ∥wk ∥ to ∥wk − v∥ ≥ ∥v∥ − ∥wk ∥ > ∥wk ∥ when b releases its Move. Both violations are finite and occur inside the interval, so they may be concatenated by Lemma E.18(1) and are preserved by the fair tail. Lemma 6.1 gives the claim.

F.5

DMSD ∨ CFSM (DC)

This subsection treats the single antichain {DMSD, CFSM}, whose maximal negative boundary is the one class OBLOT A . The construction uses an observation made during the movement of the peer to produce a non-dyadic distance at a joint stop, and then one further nonzero Move of a robot that has already moved. Lemma F.43. (DMSD ∨ CFSM) ∈ / Sol(OBLOT A ). Proof. Let A be any deterministic algorithm of OBLOT under Asynch. Define the execution as a finite fully synchronous prefix freezing the common state (Lemma E.13), then an observation-during-movement 81

interval (Lemma E.7), then an interval consisting of one solo activation of a, and finally an arbitrary fair tail. The first prefix is the first stage of Lemma C.25; an OBLOT algorithm is an FSTA algorithm with |C| = 1, so that lemma applies here and yields Corollary C.26. Normalize a(0) = 0, b(0) = e1 and D0 = 1, and write λ := λs and R := Rθs for the frozen common state s. If λ = 0 for every state, no nonzero Move ever occurs, so moves(a) = 0 ̸= 1 and the single-move condition of CFSM fails, while the distance stays equal to D0 at every joint stop and the progress condition (D2) of DMSD fails as well. Otherwise, in the observation interval the robot b performs a nonzero Move from b(0) to pE = e1 − λRe1 , and by Lemma E.7 the robot a performs a Look at a time when b is at pµ = (1 − µ)b(0) + µpE for a prescribed µ ∈ (0, 1). If pµ ever coincides with the position of a, the two robots occupy the same point and the collision-freedom condition of CFSM fails at once, so we may assume that this does not happen; then λ > 0 makes the resulting Move of a nonzero, and a lands at λRpµ . Let both Move operations complete and delay the next Look of each robot; throughout that delay neither robot is inside a nonzero Move, so the whole interval is contained in Σab . The distance there is the non-constant continuous function h(µ) = ∥c + µw∥,

c := e1 − 2λRe1 ,

w := λ2 R2 e1 ̸= 0,

of µ, as in the third stage of Lemma C.25. Since Z[1/2] is countable, Corollary E.22 provides a µ ∈ (0, 1) with h(µ)/D0 ∈ / Z[1/2], so the dyadic condition (D1) of DMSD fails. As 0 ∈ Z[1/2], this µ also keeps the two robots apart, so the distance stays positive. At that point a has already performed one nonzero Move. In the last interval a is activated alone; the distance being positive, in the normalized frame it receives a local input isomorphic to the earlier one, and λ > 0 makes this Move nonzero as well, so moves(a) ≥ 2 ̸= 1 and the single-move condition of CFSM fails. Both are safety violations, so they may be concatenated by Lemma E.18(1)(2) and are preserved by the fair tail, and Lemma 6.1 gives the claim. This completes the forty-seven boundary obligations of the two-element antichains: fourteen in Appendix F.1, six in Appendix F.2, eighteen in Appendix F.3, four in Appendix F.4, one here, and four in the main text, namely the two FSTAR obligations of Proposition 5.2 and the two FSTAS obligations of Proposition 5.3. Together they establish the case |A| = 2 of Proposition 6.2.

G

Common-Failure Executions for Three-Task Disjunctions

Each row lists one three-element antichain A, given by the corresponding disjunction BA of principal tasks from Table 2, together with NA = Max(Q \ UA ) and, for each Y ∈ NA , the statement exhibiting one admissible fair execution of an arbitrary algorithm in Y that violates all three task predicates. Every Y ∈ NA also satisfies Y ∈ / UA′ for each two-element antichain A′ ⊂ A, so with two exceptions the execution is the one already constructed for A′ in Table 4, and the proof only adds the failure of the third predicate on that execution. The exceptions are the boundary classes OBLOT S and OBLOT AM , which occur in no row of Table 4: for them the execution is constructed from scratch, once in Lemma G.21 and once in Lemma G.33, and reused by the later rows of the same family. The column sum is the number 34 of three-task boundary obligations certified in Table 9 (Appendix H.3) and recorded in Table 3; together with Table 4 this discharges all 47 + 34 = 81 boundary obligations, and Lemma 6.1 then gives Prof(BA ) = UA for every A ∈ Ant(Q).

82

Table 5: The sixteen three-element antichains, their maximal unsolvable classes NA , and the commonfailure execution that discharges each obligation (A, Y ). The rows are grouped by the construction families of Section 5 and Table 3, and within each group they follow the order of Table 4. All thirty-four obligations are proved in Appendix G. Disjunction BA

NA and proof reference

|NA |

Expansion family, EPG∗ ∨ E ∨ P with E ∈ E, P ∈ P4 , Appendix G.1 EPG∗ ∨ OSPO ∨ S-RDV FCOMA Lem. G.1; FSTAS Lem. G.2; OBLOT R Lem. G.3 EPG∗ ∨ OSPO ∨ VSS FSTAS Lem. G.4; OBLOT R Lem. G.5 ∗ EPG ∨ S-RDV ∨ ZCC FCOMA Lem. G.6; FSTAAM Lem. G.7; OBLOT R Lem. G.8 ∗ EPG ∨ S-RDV ∨ VEC FCOMA Lem. G.9; FSTAA Lem. G.10; OBLOT R Lem. G.11 ∗ EPG ∨ S-RDV ∨ CFSM FCOMA Lem. G.12; OBLOT R Lem. G.13 EPG∗ ∨ ZCC ∨ VSS FSTAAM Lem. G.14; OBLOT R Lem. G.15 ∗ EPG ∨ VSS ∨ VEC FSTAA Lem. G.16; OBLOT R Lem. G.17 ∗ EPG ∨ VSS ∨ CFSM OBLOT R Lem. G.18

3 2 3 3 2 2 2 1

Shrinking-rotation family, SRO ∨ E ∨ P with E ∈ E, P ∈ P3 , Appendix G.2 SRO ∨ S-RDV ∨ ZCC FCOMA Lem. G.19; FSTAAM Lem. G.20; OBLOT S Lem. G.21 SRO ∨ S-RDV ∨ VEC FCOMA Lem. G.22; FSTAA Lem. G.23; OBLOT S Lem. G.24 SRO ∨ S-RDV ∨ CFSM FCOMA Lem. G.25; OBLOT S Lem. G.26 SRO ∨ ZCC ∨ VSS FSTAAM Lem. G.27; OBLOT S Lem. G.28 SRO ∨ VSS ∨ VEC FSTAA Lem. G.29; OBLOT S Lem. G.30 SRO ∨ VSS ∨ CFSM OBLOT S Lem. G.31

3 3 2 2 2 1

Monotone-convergence family, MCv ∨ VSS ∨ P , Appendix G.3 MCv ∨ VSS ∨ VEC FSTAA Lem. G.32; OBLOT AM Lem. G.33 MCv ∨ VSS ∨ CFSM OBLOT AM Lem. G.34

2 1

Total

34

83

G.1

Three-Task Disjunctions Containing EPG∗

Throughout Appendix G we use the following principle. If A = {X1 , X2 , X3 } is a three-element antichain and Y ∈ NA , then for every two-element antichain A′ ⊂ A we have UA′ ⊆ UA and hence Y ∈ / UA′ . In S AM , some such A′ every triple treated below, except for the boundary classes OBLOT and OBLOT ′ ′ satisfies Y ∈ NA , so the execution constructed for A at Y in Appendix F applies to the algorithms of Y and we only have to show that the same execution violates the third predicate as well. For OBLOT S and OBLOT AM a new execution is constructed; those cases occur in Appendices G.2 and G.3. This subsection treats the eight antichains {EPG∗ , E, P } with E ∈ E = {S-RDV, VSS} and P ∈ P4 = {OSPO, ZCC, VEC, CFSM}, which carry eighteen boundary obligations in total. When the reused pair is one of Cexp , the EPG∗ component is the violation already recorded in Appendix F.1. Otherwise no nonzero Move other than the first one begins in the reused execution before the first nonzero Move has been completed. Hence either that execution contains no nonzero Move at all, in which case the joint-stop sequence is (C0 ) and the perpetuity condition fails, or the completion of the first nonzero Move is a joint stop realizing C1 at which only one robot has moved, so by Lemma 5.1(1) the centre of gravity has shifted and the expansion condition fails at i = 0. The peer may still have a pending Move at that instant; this does not affect the argument, since a robot that has not begun to move is stationary. We keep the notation (λ⋆ , θ⋆ ) = ( 21 , π) of Remark E.3. G.1.1

EPG∗ ∨ OSPO ∨ S-RDV (EOS)

Lemma G.1. EOS ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch. Use the execution of Lemma F.21, which violates ΦOSPO and ΦS-RDV . We show that it also violates ΦEPG∗ . Recall c0 = 23 d. No nonzero output. The joint-stop sequence is (C0 ), so the perpetuity condition fails. The same infinite execution witnesses this. Case (λ, θ) ̸= (2, 0). The first solo activation of a is the first nonzero Move. The joint stops before and after it realize C0 and C1 , and b does not move in between. So the expansion condition fails at i = 0 (Lemma 5.1(1)). Case (λ, θ) = (2, 0). The Move of b begins strictly after that of a ends. So the landing time of a is a joint stop, with configuration (3d, 2d) and midpoint 52 d ̸= c0 . The expansion condition fails at i = 0, as in Lemma F.6. In each case a single execution violates all three predicates; in the last two cases the violation of ΦEPG∗ is a safety violation in the prefix, so Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. Lemma G.2. EOS ∈ / Sol(F STAS ). Proof. Let A be any deterministic algorithm of FSTA under Ssynch. Use the execution of Lemma F.22, which violates ΦOSPO and ΦS-RDV . To verify the failure of EPG∗ , only its first part, the alternating singleton activation, is needed. If no nonzero Move occurs, the joint-stop sequence is (C0 ) and the perpetuity condition of EPG∗ fails; the execution of Lemma F.22 itself is the fair continuation witnessing this. Otherwise, as in that lemma, the two robots traverse the same states until the first nonzero Move, and a is activated first, so that Move is performed by a in a round that activates a alone. The surrounding round boundaries are joint stops realizing C0 and C1 , and b does not move in between, so the expansion condition fails at i = 0 (Lemma 5.1(1)). This is a safety violation at a finite time, and it precedes the continuation witnessing any liveness violation of ΦOSPO . In either case a single execution violates all three predicates; Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. Lemma G.3. EOS ∈ / Sol(OBLOT R ).

84

Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.5, which violates ΦEPG∗ and ΦOSPO . The visible state of an OBLOT robot is unique, so in a normalized frame the rule is a single pair (λ, θ) (Lemma E.16(a)). If λ = 0 no nonzero Move occurs, so the first nonzero Move demanded by Lemma D.21 does not exist and ΦS-RDV fails. If λ ̸= 0 and (λ, θ) ̸= (λ⋆ , θ⋆ ), the execution of Lemma F.5 is the fully synchronous one. When (λ, θ) ̸= (2, 0), the nonzero Move of round 1 does not meet the requirement of Lemma D.22, namely a landing of a at B = 3d or of b at A = 0, so ΦS-RDV fails at that landing. When (λ, θ) = (2, 0), round 1 takes a to 3d and b to 0, and round 2 takes a to −3d and b to 6d, both outside [A, B] = [0, 3d] (Lemma D.27), so ΦS-RDV fails. If (λ, θ) = (λ⋆ , θ⋆ ), the execution of Lemma F.5 is the alternating singleton one. Since ( 21 , π) ̸= (2, 0), the robot a moves from d to d/2 at its first activation, which is not the landing at B = 3d required by Lemma D.22, so ΦS-RDV fails. In each case, a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.2

EPG∗ ∨ OSPO ∨ VSS (EOV)

Lemma G.4. EOV ∈ / Sol(F STAS ). Proof. The execution of Lemma F.24 is that of Lemma F.22 and violates ΦOSPO and ΦVSS . The argument of Lemma G.2 also gives a violation of ΦEPG∗ on that same execution. Hence Lemma 6.1 gives the claim. Lemma G.5. EOV ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.5, which violates ΦEPG∗ and ΦOSPO . By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs and condition (P1) of VSS fails by Lemma D.21. If λ ̸= 0 and (λ, θ) ̸= (λ⋆ , θ⋆ ), the execution is the fully synchronous one. When (λ, θ) ̸= (2, 0), the nonzero Move of round 1 does not meet the requirement of Lemma D.22, so no valid T1 exists and ΦVSS fails. When (λ, θ) = (2, 0), round 2 takes a to −3d and b to 6d, neither of which lies in P = {0, d, 2d, 3d}, so the endpoint condition (P1-b) fails. If (λ, θ) = (λ⋆ , θ⋆ ), the execution is the alternating singleton one and a moves from d to d/2 ∈ / P at its first activation, so (P1-b) fails there. In each case, a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.3

EPG∗ ∨ S-RDV ∨ ZCC (ESZ)

Lemma G.6. ESZ ∈ / Sol(F COMA ). Proof. Use the execution of Lemma F.26, which violates ΦS-RDV and ΦZCC . If it contains no nonzero move, the perpetuity condition of EPG∗ fails. Otherwise consider the landing of the first nonzero move of a. In both branches of that construction, b remains at its initial position until after this landing, even when it already has a pending move. Thus the landing time is a joint stop, and the center of gravity has changed by half the nonzero displacement of a. The expansion condition of EPG∗ therefore fails at this first transition. This safety violation persists along the continuation constructed in Lemma F.26. Hence the same fair execution violates all three predicates, and Lemma 6.1 gives the claim. Lemma G.7. ESZ ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM and use the execution of Lemma F.27, where ΦS-RDV and ΦZCC were shown to fail. Its first phase is an interval of alternating singleton activations, which activates one robot at a time, and the configuration is unchanged before the first nonzero Move. 85

In the branch in which no nonzero Move ever occurs the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. Otherwise the first nonzero Move occurs in that first phase and is performed while the peer is not activated, so by Lemma 5.1(1) the expansion condition fails at i = 0, a safety violation. The prefix of Lemma F.27 is then followed by the continuation used there, namely the coincidence continuation from Lemma C.16 or the continuation from Lemma C.17, and the violation persists along it (Lemma E.18(1)). In each branch a single fair execution violates all three predicates, so Lemma 6.1 gives the claim. Lemma G.8. ESZ ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.9, which violates ΦEPG∗ and ΦZCC . By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs, so the first nonzero Move demanded by Lemma D.21 does not exist and ΦS-RDV fails. If λ ̸= 0 and (λ, θ) ̸= (λ⋆ , θ⋆ ), the execution of Lemma F.9 is the fully synchronous one. When (λ, θ) ̸= (2, 0), the nonzero Move of round 1 does not meet the requirement of Lemma D.22 and ΦS-RDV fails at that landing. When (λ, θ) = (2, 0), the robots land at 3d and 0 in round 1 and at −3d and 6d in round 2, leaving [A, B] = [0, 3d] (Lemma F.23, Lemma D.27), so ΦS-RDV fails. If (λ, θ) = (λ⋆ , θ⋆ ), the execution is the alternating singleton one and a moves from d to d/2 at its first activation, which is not the landing at B = 3d required by Lemma D.22, so ΦS-RDV fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.4

EPG∗ ∨ S-RDV ∨ VEC (ESX)

Lemma G.9. ESX ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch and use the execution of Lemma F.29, which violates ΦS-RDV and ΦVEC . We show that it also violates ΦEPG∗ . The initial centre of gravity is 12 (a0 + b0 ) = 32 d. No nonzero output. The execution contains no nonzero Move, so the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. The same infinite execution witnesses this, together with the two liveness violations of Lemma F.29. Case (λ, θ) = (2, 0). The Move of b begins strictly after that of a ends, so at the landing of a neither robot is inside a Move, although b still has a pending one. That instant is a joint stop and realizes C1 = (3d, 2d), of centre of gravity 25 d ̸= 32 d; equivalently, b completes no nonzero Move between C0 and C1 , so Lemma 5.1(1) applies. The expansion condition of EPG∗ fails at i = 0, a safety violation in the prefix. Case (λ, θ) ̸= (2, 0). The configuration is unchanged before the solo activation interval of a, so the first activation of a in it is the first nonzero Move of the execution and b is not activated during it. The maximal joint-stop intervals before and after it realize C0 and C1 , so by Lemma 5.1(1) the expansion condition fails at i = 0, a safety violation in the prefix. In the first case all three violations are liveness violations of one infinite execution. In the other two the violation of ΦEPG∗ is finite and precedes the continuations supplied by Lemma F.29. In either case Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. Lemma G.10. ESX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch and use the execution of Lemma F.30, which violates ΦS-RDV and ΦVEC . If no nonzero Move occurs, the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. Otherwise, in both cases of that proof the first nonzero Move is performed by a from the initial configuration, and b completes no nonzero Move before a has landed: for pE = B it is not activated, and for pE ̸= B its Move is delayed until after the landing. The landing of a is therefore a joint stop realizing C1 , and by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation preserved by the rest of the execution (Lemma E.18(1)(2)). 86

In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. Lemma G.11. ESX ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.12, which violates ΦEPG∗ and ΦVEC and which is in every case the alternating singleton one with an empty fully synchronous prefix and normalized frames. By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs, so the first nonzero Move demanded by Lemma D.21 does not exist and ΦS-RDV fails. If λ ̸= 0 and (λ, θ) ̸= (2, 0), the landing of round 1 is not the B = 3d required by Lemma D.22, so ΦS-RDV fails at that landing. If (λ, θ) = (2, 0), then c1 = |1 − 2| = 1 ̸= 0 and Lemma D.23 applies: the robot b, activated immediately after a lands at 3d, receives an isomorphic local input and lands at 4d ∈ / [A, B], so ΦS-RDV fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.5

EPG∗ ∨ S-RDV ∨ CFSM (ESC)

Lemma G.12. ESC ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch and use the execution of Lemma F.32, which violates ΦCFSM and ΦS-RDV . That execution has the same three cases as the one of Lemma F.29, and the argument of Lemma G.9 applies verbatim to each of them: with no nonzero output the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails; when (λ, θ) = (2, 0) the landing of a is a joint stop realizing C1 = (3d, 2d) and Lemma 5.1(1) makes the expansion condition fail at i = 0; and otherwise the first nonzero Move is the first activation of a in its solo interval, with b not activated, so Lemma 5.1(1) applies again. In the first case all three violations are liveness violations of one infinite execution; in the other two the violation of ΦEPG∗ is finite and precedes the continuations supplied by Lemma F.32. In either case Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. Lemma G.13. ESC ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.14, which violates ΦEPG∗ and ΦCFSM and which is the alternating singleton one with an empty fully synchronous prefix and normalized frames. By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs, so the first nonzero Move demanded by Lemma D.21 does not exist. If λ ̸= 0 and (λ, θ) ̸= (2, 0), the landing of round 1 is not the B = 3d required by Lemma D.22. If (λ, θ) = (2, 0), then c1 = 1 ̸= 0, which is the third case of Lemma F.14, and Lemma D.23 makes b land at 4d ∈ / [A, B]. In every case ΦS-RDV fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.6

EPG∗ ∨ ZCC ∨ VSS (EZV)

Lemma G.14. EZV ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM and use the execution of Lemma F.34, which violates ΦVSS and ΦZCC . Its first phase is an interval of alternating singleton activations, and the configuration is unchanged before the first nonzero Move. In the branch in which no nonzero Move ever occurs, the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. Otherwise the first nonzero Move occurs in that first phase and is performed while the peer is not activated, so by Lemma 5.1(1) the expansion condition fails at i = 0, a safety violation. It persists along the continuation from Lemma C.16 or Lemma C.17 used in Lemma F.34 (Lemma E.18(1)). In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. 87

Lemma G.15. EZV ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.9, which violates ΦEPG∗ and ΦZCC . By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs and condition (P1) of VSS fails by Lemma D.21. If λ ̸= 0 and (λ, θ) ̸= (λ⋆ , θ⋆ ), the execution of Lemma F.9 is the fully synchronous one. When (λ, θ) ̸= (2, 0), the nonzero Move of round 1 does not meet the requirement of Lemma D.22, so no valid T1 exists and ΦVSS fails. When (λ, θ) = (2, 0), the landings 3d and 0 of round 1 both belong to P = {0, d, 2d, 3d}, but in round 2 the robots land at −3d and 6d, leaving P , so the endpoint condition (P1-b) fails. If (λ, θ) = (λ⋆ , θ⋆ ), the execution is the alternating singleton one and a moves from d to d/2 ∈ / P at its first activation, so (P1-b) fails there. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.7

EPG∗ ∨ VSS ∨ VEC (EVX)

Lemma G.16. EVX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch and use the execution of Lemma F.36, which violates ΦVSS and ΦVEC . If no nonzero Move occurs, the joint-stop sequence has the single term C0 and the perpetuity condition of EPG∗ fails. Otherwise, in both cases of that proof the first nonzero Move is performed by a from the initial configuration, and b completes no nonzero Move before a has landed: for pE = B it is not activated, and for pE ̸= B its Move is delayed until after the landing. The landing of a is therefore a joint stop realizing C1 , and by Lemma 5.1(1) the expansion condition of EPG∗ fails at i = 0, a safety violation preserved by the rest of the execution (Lemma E.18(1)(2)). In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. Lemma G.17. EVX ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.12, which violates ΦEPG∗ and ΦVEC and which is the alternating singleton one with normalized frames. By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs and (P1) fails by Lemma D.21. If λ ̸= 0 and (λ, θ) ̸= (2, 0), the landing of round 1 is not the B = 3d required by Lemma D.22 and no valid T1 exists. If (λ, θ) = (2, 0), then c1 = |1 − 2| = 1 ̸= 0 and Lemma D.23 applies: the robot b, activated immediately after a lands at 3d, lands at 4d ∈ / P = {0, d, 2d, 3d}. In every case ΦVSS fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.1.8

EPG∗ ∨ VSS ∨ CFSM (EVC)

Lemma G.18. EVC ∈ / Sol(OBLOT R ). Proof. Let A be any deterministic algorithm of OBLOT under Rsynch and use the execution of Lemma F.14, which violates ΦEPG∗ and ΦCFSM and which is the alternating singleton one with normalized frames. By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 no nonzero Move occurs and (P1) fails by Lemma D.21. If λ ̸= 0 and (λ, θ) ̸= (2, 0), the landing of round 1 is not the B = 3d required by Lemma D.22 and no valid T1 exists. If (λ, θ) = (2, 0), then c1 = 1 ̸= 0 and Lemma D.23 makes b land at 4d ∈ / P = {0, d, 2d, 3d}. In every case ΦVSS fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim.

88

G.2

Three-Task Disjunctions Containing SRO

This subsection treats the six antichains {SRO, E, P } with E ∈ E and P ∈ P3 = {ZCC, VEC, CFSM}, which carry thirteen boundary obligations in total. The class OBLOT S occurs here as a boundary class for five of these antichains, and no corresponding lemma exists in Appendix F; the execution for it is constructed from scratch in Lemma G.21 and reused by the four later OBLOT S lemmas. G.2.1

SRO ∨ S-RDV ∨ ZCC (RSZ)

Lemma G.19. RSZ ∈ / Sol(F COMA ). Proof. Use the construction of Lemma F.26. If no nonzero output occurs, the perpetuity condition of SRO fails as well. If the first nonzero rule is (2, 0), the first solo move of a reverses the segment and satisfies neither C1 nor C2. Thus ΦSRO has a safety violation before the continuation used in Lemma F.26. Otherwise keep the peer inactive during the initial solo interval of a. If the first move lands on the peer, the resulting zero segment violates the rotation condition of SRO. If it does not, choose the interval to contain at least two nonzero moves of a. Lemma E.19 then gives a safety violation of ΦSRO . For the ZCC component, use the finite safety violation or the continuation property of Lemma C.16, as in the proof of Lemma F.26. In the contraction case σ1 ∈ (0, 1) the robots remain in the original initial segment in their original order after any number of solo activations of a, so the continuation property applies at the end of the interval chosen above. The earlier safety violations persist. Thus one fair infinite execution violates all three predicates, and Lemma 6.1 gives the claim. Lemma G.20. RSZ ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of FSTA under AM and modify the execution of Lemma F.17 as follows. If no nonzero Move ever occurs, ΦS-RDV fails by Lemma D.21, and the first condition of SRO and the third condition of ZCC fail as shown in Lemma F.17. If the first nonzero Move of a does not land at the B = 3d required by Lemma D.22, then ΦS-RDV fails at that point, and the execution of Lemma F.17 may be used unchanged, so ΦSRO and ΦZCC fail as shown there. If it does land at B, then (λs , θs ) = (2, 0), and this displacement is neither the 90◦ clockwise rotation about the peer required by C1 nor the one about the midpoint required by C2, so by Lemma E.19 the second condition of SRO fails already at this move and no second nonzero Move of a is needed. Insert, immediately after this move, a solo activation interval of b: activating b alone until it reaches the state s leaves the configuration unchanged, since in all earlier states λ = 0 and the state transition of an F STA robot depends only on its own state; then b in state s observes a peer at distance d and, in the normalized frame, receives a local input isomorphic to that of a, so it applies the same (2, 0) and, by Lemma D.23, lands at 4d ∈ / [A, B], which violates ΦS-RDV . Moreover, the first move of a alone takes it from d to 3d while b rests at b0 = 2d, so ∥a − a0 ∥ = 2d > d = ∥b − a0 ∥ and the collision-freedom condition of ZCC has already failed. No additional delayed Move is needed; append an arbitrary fair, time-divergent tail. In every case a single fair execution violates all three predicates, and the safety violations established in its prefix persist (Lemma E.18(1)), so Lemma 6.1 gives the claim. Lemma G.21. RSZ ∈ / Sol(OBLOT S ). Proof. Let A be any deterministic algorithm of OBLOT under Ssynch. No lemma of Appendix F has OBLOT S as a boundary class, so we construct the execution directly. The visible state of an OBLOT robot is unique, so in a normalized frame the rule is a single pair (λ, θ) (Lemma E.16(a)). Case λ = 0. Since OBLOT has no colors, the two robots do not move under any schedule. Hence the sequence of configurations reduces to C0 and the perpetuity condition of SRO fails, the robots remain 89

at distinct positions and the convergence condition of ZCC fails, and no nonzero Move exists, so ΦS-RDV fails by Lemma D.21. All three are independent of the schedule, hence persist along any fair infinite continuation. Case (λ, θ) = (2, 0). Define the execution by one solo activation of a, then one solo activation of b, then alternating singleton activations and an arbitrary fair tail, all admissible under Ssynch. The displacement of a is neither the 90◦ clockwise rotation about the peer (C1) nor the one about the midpoint (C2), so by Lemma E.19 the second condition of SRO fails at this move. Moreover σ1 = 1 − 2 = −1, so c1 = 1 and ψ ≡ π (Lemma E.2(iii)), whence fa (1) = d |1 − eiπ | = 2d > d = ∥b(0) − a(0)∥ and the collision-freedom condition of ZCC fails. Finally a has landed at B = 3d, so the robot b, activated immediately afterwards, receives in the normalized frame an isomorphic local input, applies the same (2, 0) and, by Lemma D.23, lands at 4d ∈ / [A, B], so ΦS-RDV fails. Case λ ̸= 0 and (λ, θ) ̸= (2, 0). Define the execution by an interval in which a alone is activated k times in a row (Lemma E.4, with k large enough for what follows), then alternating singleton activations and an arbitrary fair tail. The first nonzero Move of a is not the landing at B required by Lemma D.22, so ΦS-RDV fails at that point. For ΦSRO : if that move is neither of type C1 nor of type C2, the second condition fails at once, which includes the case c1 = 0, since then di+1 = 0 while both types require a nonzero value; if it is of one of the two types, the second solo nonzero Move of a makes the second condition fail by Lemma E.19. For ΦZCC , put σ1 := 1 − λeiθ and regard A as an F COM algorithm with a single color, so that the constructions in the proof of Lemma C.16 apply. Every activation is a complete cycle under Ssynch, so no operation is pending at the end of the solo interval. If c1 = 0, the robots coincide after the first activation; end the solo interval there and use the coincidence continuation described in the proof of Lemma C.16. This either keeps both positions fixed forever or produces a finite violation of the collision-freedom condition. If c1 ̸= 0 and σ1 ∈ / (0, 1), the collision-freedom condition fails at a finite time for k large enough, as in the proof of Lemma F.26. In the remaining contraction case σ1 ∈ (0, 1), the robots still lie in the original initial segment in their original order after the k activations, and neither has a pending operation; the continuation property proved in Lemma C.16 supplies a fair, time-divergent continuation violating ΦZCC , while preserving the safety violations already obtained in the prefix. In every case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.2.2

SRO ∨ S-RDV ∨ VEC (RSX)

Lemma G.22. RSX ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch and use the execution of Lemma F.29, which violates ΦS-RDV and ΦVEC . We show that it also violates ΦSRO , under the same case distinction. No nonzero output. No nonzero Move ever occurs, so the sequence of joint-stop configurations reduces to C0 and the perpetuity condition of SRO fails. The same infinite execution witnesses this, together with the two liveness violations of Lemma F.29. Case (λ, θ) = (2, 0). The landing of a is a joint stop realizing C1 = (3d, 2d), so d1 = −d0 . Neither C1, which requires d1 = R−90◦ d0 , nor C2, which requires d1 = 2−1/2 R−45◦ d0 , is satisfied by a rotation by 180◦ of the same norm. The second condition of SRO therefore fails, a safety violation in the prefix. Case (λ, θ) ̸= (2, 0). In the solo activation interval of a the robot applies the same pair (λ, θ), with λ ̸= 0, at every activation (Lemma E.16(b)). If c1 = 0 then a lands on the position of b at the first activation and d1 = 0, while C1 preserves the norm and C2 scales it by 2−1/2 , so neither can hold with d0 ̸= 0; the second condition of SRO fails. If c1 ̸= 0, the interval has length k ≥ 3, so a performs two nonzero Move operations with its peer at rest, and by Lemma E.19 the rotation condition and C3 cannot both hold, so the second condition fails again. Either way this is a safety violation in the prefix. In the first case all three violations are liveness violations of one infinite execution. In the other two the violation of ΦSRO is finite and precedes the continuations supplied by Lemma F.29. In either case Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim.

90

Lemma G.23. RSX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch and use the execution of Lemma F.19, which violates ΦSRO and ΦVEC . If λs = 0 for every state s, no nonzero Move occurs and ΦS-RDV fails by Lemma D.21, while ΦSRO and ΦVEC fail in the first case of Lemma F.19. Otherwise the robot a performs the first nonzero Move in the round of the first common state s with λs > 0. If its landing is not the B = 3d required by Lemma D.22, then ΦS-RDV fails at that point. If it is, then (λ, θ) = (2, 0), which is neither the C1 rotation about the peer nor the C2 rotation about the midpoint, so by Lemma E.19 ΦSRO fails at once. In this case the points of the movement segment of a are pµ = (1 + 2µ)d with µ ∈ (0, 1), and the robot b, observing pµ in state s, applies the same (2, 0) in
the normalized frame and lands at 2d + 2 (1 + 2µ)d − 2d = 4µd. Taking µ ∈ (3/4, 1) gives 4µd > 3d, so the robots leave [A, B] = [0, 3d] and ΦS-RDV fails. The set of parameters to be excluded for the VEC violation is countable, so such a µ can be chosen inside (3/4, 1) and all three violations occur in the same execution. All three are safety violations, so they may be concatenated by Lemma E.18(1)(2), and Lemma 6.1 gives the claim. Lemma G.24. RSX ∈ / Sol(OBLOT S ). Proof. Let A be any deterministic algorithm of OBLOT under Ssynch and use the execution of Lemma G.21, in which a alone is activated and then a and b are activated alternately. The violations of ΦSRO and ΦS-RDV were shown there. By Lemma E.16(a) the rule is a single pair (λ, θ). If λ = 0 the robots do not move, the sequence of configurations reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition of VEC fails. If c1 = 0, the robot a lands on the position of b at its first activation and di+1 = 0, while all the targets permitted by the transition condition of VEC are nonzero, so that condition fails. Otherwise the only single displacements satisfying the transition condition rotate the segment clockwise by 45◦ or by 90◦ ; if a departs from them the transition condition fails at that point, and otherwise a reaches a configuration with di ⊥ d0 within its two activations, at which moment both robots are at rest, but the same displacement is repeated in the alternating singleton activations that follow, so the robots move again and the stopping condition fails. All of these are safety violations, and a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.2.3

SRO ∨ S-RDV ∨ CFSM (RSC)

Lemma G.25. RSC ∈ / Sol(F COMA ). Proof. Let A be any deterministic algorithm of FCOM under Asynch and use the execution of Lemma F.32, which violates ΦCFSM and ΦS-RDV . The three cases are those of Lemma G.22, and the argument there applies verbatim, the solo activation interval of a now having length two, which still supplies the two nonzero Move operations required by Lemma E.19. In the first case all three violations are liveness violations of one infinite execution; in the other two the violation of ΦSRO is finite and precedes the continuations supplied by Lemma F.32. In either case Lemma E.18(1)(3) is satisfied, and Lemma 6.1 gives the claim. Lemma G.26. RSC ∈ / Sol(OBLOT S ). Proof. Let A be any deterministic algorithm of OBLOT under Ssynch and use the execution of Lemma G.21, where the violations of ΦSRO and ΦS-RDV were shown; ΦCFSM fails under the same case distinction. If λ = 0, the two robots do not move under any schedule, so moves(a) = 0 ̸= 1 and the single-move condition fails.

91

√ If (λ, θ) = (2, 0), then c1 = 1 − 4 + 4 = 1 ̸= 0, so the distance remains d and a performs a nonzero Move both in round 1 and at its second activation in the alternating singleton part; hence moves(a) ≥ 2 ̸= 1 and the single-move condition fails. If λ ̸= 0 and (λ, θ) ̸= (2, 0), then for c1 = 0 the robot a lands on the position of b at its first activation, so a(t) = b(t) and the collision-freedom condition fails, while for c1 ̸= 0 both of its two consecutive solo activations are nonzero Move operations and moves(a) ≥ 2 violates the single-move condition. All of these are safety violations. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.2.4

SRO ∨ ZCC ∨ VSS (RZV)

Lemma G.27. RZV ∈ / Sol(F STAAM ). Proof. Let A be any deterministic algorithm of F STA under AM and use the execution of Lemma G.20 with ΦS-RDV replaced by ΦVSS . If no nonzero Move ever occurs, condition (P1) of VSS fails, and the first condition of SRO and the third condition of ZCC fail as in Lemma F.17. If the first nonzero Move of a does not land at the B = 3d required by Lemma D.22, no valid T1 exists and ΦVSS fails at that point; the execution of Lemma F.17 may then be used unchanged. If it does land at B, then (λs , θs ) = (2, 0), which is neither the C1 rotation about the peer nor the C2 rotation about the midpoint, so by Lemma E.19 the second condition of SRO fails at this move. Insert immediately afterwards a solo activation interval of b: activating b alone until it reaches the state s leaves the configuration unchanged, since λ = 0 in all earlier states, and then b in state s receives, in the normalized frame, a local input isomorphic to that of a, applies the same (2, 0) and by Lemma D.23 lands at 4d ∈ / P = {0, d, 2d, 3d}, so ΦVSS fails. Moreover, the first move of a alone takes it from d to 3d while b rests at b0 = 2d, so ∥a − a0 ∥ = 2d > d = ∥b − a0 ∥ and the collision-freedom condition of ZCC has already failed. No additional delayed Move is needed; append an arbitrary fair, time-divergent tail. In every case a single fair execution violates all three predicates, and the safety violations established in its prefix persist (Lemma E.18(1)), so Lemma 6.1 gives the claim. Lemma G.28. RZV ∈ / Sol(OBLOT S ). Proof. Let A be any deterministic algorithm of OBLOT under Ssynch and use the execution of Lemma G.21, where the violations of ΦSRO and ΦZCC were shown; ΦVSS fails under the same case distinction. If λ = 0, the two robots do not move, so no nonzero Move exists and condition (P1) of VSS fails. If (λ, θ) = (2, 0), then in that execution a is activated alone in round 1 and b in round 2, both under normalized frames. The robot a lands at B = 3d, and c1 = 1 ̸= 0, so Lemma D.23 applies and the robot b, activated immediately afterwards, receives an isomorphic local input and lands at 4d ∈ / P = {0, d, 2d, 3d}, so ΦVSS fails. If λ ̸= 0 and (λ, θ) ̸= (2, 0), the first nonzero Move of a is not the landing at B required by Lemma D.22, so no valid T1 exists and ΦVSS fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.2.5

SRO ∨ VSS ∨ VEC (RVX)

Lemma G.29. RVX ∈ / Sol(F STAA ). Proof. Let A be any deterministic algorithm of FSTA under Asynch and use the execution of Lemma F.19, which violates ΦSRO and ΦVEC . If λs = 0 for every state s, no nonzero Move occurs and condition (P1) of VSS fails, while ΦSRO and ΦVEC fail in the first case of Lemma F.19. 92

Otherwise the robot a performs the first nonzero Move in the round of the first common state s with λs > 0. If its landing is not the B = 3d required by Lemma D.22, no valid T1 exists and ΦVSS fails at that point. If it is, then (λ, θ) = (2, 0) and, as in Lemma G.23, the points of the movement segment of a are pµ = (1 + 2µ)d with µ ∈ (0, 1), and the robot b observing pµ in state s applies the same (2, 0) in the normalized frame and lands at 4µd. The set P = {0, d, 2d, 3d} being finite, only finitely many µ give 4µd ∈ P , so together with the countable set to be excluded for the VEC violation a µ ∈ (0, 1) avoiding all of them can be chosen. For that µ we have 4µd ∈ / P , so ΦVSS fails, and all three violations occur in the same execution. All three are safety violations, so they may be concatenated by Lemma E.18(1)(2), and Lemma 6.1 gives the claim. Lemma G.30. RVX ∈ / Sol(OBLOT S ). Proof. Let A be any deterministic algorithm of OBLOT under Ssynch and use the execution of Lemma G.21 with the same case distinction λ = 0 / (λ, θ) = (2, 0) / otherwise. The violation of ΦSRO is that of Lemma G.21 and the violation of ΦVSS is that of Lemma G.28. For ΦVEC : if λ = 0 the robots do not move, the sequence of configurations reduces to C0 , and d0 ̸⊥ d0 , so the attainment condition fails. If (λ, θ) = (2, 0) then di+1 = −di , which is none of di , R−90◦ di and 2∓1/2 R−45◦ di , so the transition condition fails. Otherwise, if c1 = 0 then di+1 = 0 while all the permitted targets are nonzero and the transition condition fails, and if c1 ̸= 0 then, as in Lemma G.24, the only single displacements satisfying the transition condition rotate the segment clockwise by 45◦ or by 90◦ ; departing from them makes the transition condition fail, and otherwise an orthogonal configuration is reached within the two activations of a and the same displacement is repeated afterwards, so the stopping condition fails. In each case a single execution violates all three predicates, so Lemma 6.1 gives the claim. G.2.6

SRO ∨ VSS ∨ CFSM (RVC)

Lemma G.31. RVC ∈ / Sol(OBLOT S ). Proof. Let A be any deterministic algorithm of OBLOT under Ssynch and use the execution of Lemma G.21. All three violations follow the same case distinction λ = 0 / (λ, θ) = (2, 0) / otherwise: the violation of ΦSRO is that of Lemma G.21, the violation of ΦVSS is that of Lemma G.28, and the violation of ΦCFSM is that of Lemma G.26. A single execution violates all three predicates, so Lemma 6.1 gives the claim.

G.3

Three-Task Disjunctions Containing MCv

This subsection treats the two antichains {MCv, VSS, P } with P ∈ {VEC, CFSM}, which carry three boundary obligations in total. The class OBLOT AM occurs here as a boundary class for both antichains, and no lemma of Appendix F has it as a boundary class; the execution for it is constructed from scratch in Lemma G.33 and reused in Lemma G.34. We record once the two activation patterns used there. Under AM the adversary may activate one robot at a time (Lemma E.4, whose statement includes K = AM ), and it may also activate both robots in lockstep: giving the two robots identical phase times tL < tC < tB < tE in each round places every Look before both Move intervals, so no Look falls inside the Move interval of the peer and M -atomicity is respected. In such a round both robots compute from the same snapshot, so Lemma E.2(ii) applies and gives d 7→ c2 d with the midpoint fixed. We refer to the latter pattern as a fully synchronous round; note that Lemma E.13 is stated for K = F and is not invoked below. G.3.1

MCv ∨ VSS ∨ VEC (MVX)

Lemma G.32. MVX ∈ / Sol(F STAA ).

93

Proof. Let A be any deterministic algorithm of FSTA under Asynch. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line, with P = {0, d, 2d, 3d}, and normalized frames at every Look at which the robots are distinct. Activate both robots simultaneously while every output is zero; they then carry the same state at every round boundary and the configuration stays at ι. No nonzero output. Continue forever. Condition (P1) fails by Lemma D.21, so ΦVSS fails; the sequence of joint-stop configurations reduces to C0 and d0 ̸⊥ d0 , so the attainment condition of VEC fails; and the two robots remain at distinct positions, so the convergence condition of MCv fails as well. This execution witnesses all three liveness violations. Otherwise let σ be the common state at the start of the first round with a nonzero output, let (λ, θ), with λ ̸= 0, be the corresponding normalized displacement, and put c1 := c1 (σ). Case c1 = 0, that is (λ, θ) = (1, 0). Activate both robots in that round. Each moves to the position of its peer, so they exchange positions and the distance returns to d after passing through 0; the monotonicity condition of MCv fails. The cumulative nonzero-move counts are (1, 1), so an admissible exchange configuration would have to be (B, A) = (3d, 0), whereas the landing is (2d, d); no valid T1 exists and ΦVSS fails by Lemma D.22. Finally C1 = (2d, d) gives d1 = −d0 , which is neither the clockwise rotation by 45◦ nor the one by 90◦ , so the transition condition of VEC fails at i = 0. All three are safety violations at a finite time; append an arbitrary admissible fair, time-divergent tail. Case c1 ̸= 0. At the round boundary at which both robots carry σ the configuration is still ι; from there continue as in Lemma F.36, activating a first, so that a performs the first nonzero Move from σ while b is in σ with no pending operation. Let pE be its landing point. If pE = B, then as in that lemma d1 = −d0 violates the transition condition of VEC, and b lands at 4d ∈ / P by Lemma D.23, so ΦVSS fails. Moreover the Move of a from d to 3d passes through b0 = 2d, so the distance to the stationary peer drops to 0 and returns to d, and the monotonicity condition of MCv fails. All three are safety violations; append an arbitrary fair tail. If pE ̸= B, then as in that lemma b observes pµ and its Move is delayed until a has landed; ΦVSS fails by Lemma D.22, and the transition condition of VEC fails for all µ outside a finite set. The landing point of b is affine in µ with nonzero slope, so it equals pE for at most one µ; excluding it as well, and since c1 ̸= 0 gives pE ̸= b0 , the robots are at distinct positions after both landings. It remains to violate MCv in this case. For MCv we use the following dichotomy, applied to every nonzero Move of the execution whose peer is at rest. Along such a Move the distance to the stationary peer is the distance from a fixed point to a point travelling along a segment, hence non-increasing and then non-decreasing. If the landing point is not the point of that segment closest to the peer, the distance strictly increases on a nonempty final part of the Move, so the monotonicity condition of MCv fails at a finite time and we are done. We may therefore assume that every such Move ends exactly at its closest approach, so that the distance is non-increasing along the execution so far. Both robots have now performed the same number of activations, and every Look so far has observed distinct positions (excluding also the at most one µ with pµ = b0 ), so by Lemma E.1(1) they carry the same color. At this point the robots occupy distinct positions, carry the same internal color, and have no pending operation. Apply the continuation statement proved in Lemma C.7; its schedules are admissible under AM , hence under Asynch. It supplies a fair, time-divergent continuation violating ΦMCv , while all safety violations established in the prefix persist. In the case without nonzero output, all three liveness violations are witnessed by the explicit fair execution given there. In the cases c1 = 0 and pE = B, and whenever the dichotomy above yields a monotonicity violation, all three violations are safety violations at a finite time. In the remaining case, the companion predicates have already suffered safety violations. The continuation from Lemma C.7 witnesses either a finite monotonicity violation or a failure of convergence. Thus all predicates fail on the same fair infinite execution. Hence Lemma E.18(1)(2)(3) is satisfied and {MCv, VSS, VEC} has the common failure property on FSTAA . Lemma 6.1 gives the claim. Lemma G.33. MVX ∈ / Sol(OBLOT AM ). Proof. Let A be any deterministic algorithm of OBLOT under AM . No lemma of Appendix F has 94

OBLOT AM as a boundary class, so we construct the execution directly, using the two activation patterns recorded above. Use coordinates A = 0, a0 = d, b0 = 2d, B = 3d on the initial line, with P = {0, d, 2d, 3d}. The visible state of an OBLOT robot is unique, so in a normalized frame the rule is a single pair (λ, θ) (Lemma E.16(a)). Put c1 := c1 (λ, θ) and c2 := c2 (λ, θ) as in Lemma E.2. Case λ = 0. Neither robot ever moves, under any schedule. Hence (P1) fails by Lemma D.21 and ΦVSS fails; the sequence of joint-stop configurations reduces to C0 and d0 ̸⊥ d0 , so the attainment condition of VEC fails; and the robots remain at distinct positions, so the convergence condition of MCv fails. All three are independent of the schedule and persist along any fair infinite continuation. Case c1 = 0, that is (λ, θ) = (1, 0). Activate both robots simultaneously in round 1, which AM permits. Each moves to the position of its peer, so they exchange positions and the distance returns to d after passing through 0; the monotonicity condition of MCv therefore fails. The cumulative nonzero-move counts are (1, 1), so an admissible exchange configuration would have to be (B, A) = (3d, 0), whereas the landing is (2d, d); no valid T1 exists and ΦVSS fails by Lemma D.22. Finally C1 = (2d, d) gives d1 = −d0 , which is neither the clockwise rotation by 45◦ nor the one by 90◦ , so the transition condition of VEC fails at i = 0. All three are safety violations at a finite time; append an arbitrary admissible fair, time-divergent tail, which preserves them. Case 0 < c1 < 1. Define the execution by an interval of alternating singleton activations of a and b, carried up to the round following the first nonzero Move and given normalized frames (Lemma E.4), followed by a delayed-Move interval of b (Corollary E.6; the suspension is placed between tC (b, i) and tB (b, i) and is therefore admissible under AM ), followed by an arbitrary fair tail. In the first interval, if the landing point of the first nonzero Move is not the point required by Lemma D.22, no valid T1 exists; and if it is that point, the robot activated in the next round receives an isomorphic local input and lands at 4d ∈ / P by Lemma D.23. Either way ΦVSS fails. For ΦVEC the argument of Lemma F.12 applies verbatim: the only single displacements that the transition condition admits while the peer stands still are the 45◦ and 90◦ clockwise rotations of the segment, so if a departs from them that condition fails at that move, and if it does not, a perpendicular configuration is reached within at most two moves, after which the same displacement is applied again and the stopping condition fails. Since c1 ̸= 0, neither of the two Move operations of this interval lands on the peer, so the robots are at distinct positions at its end. In the second interval, if no robot performs a nonzero Move from that point on, the two robots stay at these distinct positions and the convergence condition of MCv fails. Otherwise b computes a destination with displacement v ̸= 0 approaching a, suspends its Move, and in the meantime a alone is activated repeatedly. Since OBLOT has no colors, every such activation applies the same (λ, θ), so d(j) = c1j d → 0 by Lemma E.2(i) and the vector w from b to a at the release can be made to satisfy ∥w∥ < ∥v∥/2. The distance immediately before the Move of b is ∥w∥, whereas at its completion it satisfies ∥w − v∥ ≥ ∥v∥ − ∥w∥ > ∥w∥, so the monotonicity condition of MCv fails [CP05, Lemma 4]. Case c1 ≥ 1. The solo activations of a no longer bring the distance below ∥v∥/2, so define the execution to be an infinite sequence of fully synchronous rounds. For λ > 0 the inequality c1 ≥ 1 is equivalent to λ ≥ 2 cos θ, which implies λ ≥ cos θ, that is c2 ≥ 1. Hence Lemma E.2(ii) gives d(k) = c2k d(0) ≥ d(0) > 0 for every k, so the two robots do not converge to a single point and the convergence condition of MCv fails. For ΦVEC the argument of Lemma C.20 applies verbatim to this execution, since it uses only that both robots are activated simultaneously in every round and that the displacement rule is the single pair (λ, θ). For ΦVSS : if (λ, θ) ̸= (2, 0), the nonzero Move of round 1 does not meet the requirement of Lemma D.22; and if (λ, θ) = (2, 0), then round 1 takes a to 3d and b to 0, so in round 2 the robot a lands at −3d and b at 6d, both outside P , which violates (P1-b). In every case a single execution violates all three predicates, so {MCv, VSS, VEC} has the common failure property on OBLOT AM and Lemma 6.1 gives the claim. G.3.2

MCv ∨ VSS ∨ CFSM (MVC)

Lemma G.34. MVC ∈ / Sol(OBLOT AM ). 95

Table 6: The eighteen covering pairs of the signature-inclusion order. Each pair µ, ν in a row satisfies µ ≥ ν by previously established inclusions and equivalences. Representative µ

Cσ (µ)

LUMI F OBLOT F LUMI R FSTAR OBLOT R FCOMS FCOMA FSTAS FSTAAM FSTAA OBLOT S OBLOT AM OBLOT A

{OBLOT F , LUMI R } {OBLOT R } {FSTAR , FCOMS } {OBLOT R , FSTAS } {OBLOT S } {FCOMA , OBLOT S } {OBLOT AM } {FSTAAM , OBLOT S } {FSTAA , OBLOT AM } {OBLOT A } {OBLOT AM } {OBLOT A } ∅

Proof. Let A be any deterministic algorithm of OBLOT under AM and use the execution of Lemma G.33, where the violations of ΦMCv and ΦVSS were established. We show that ΦCFSM fails under the same case distinction. Case λ = 0. Neither robot ever moves, so moves(a) = 0 ̸= 1 and the single-move condition fails. Case c1 = 0. The two robots exchange positions in round 1 and therefore occupy the same point while they travel, so the collision-freedom condition fails. Case 0 < c1 < 1. The robot a performs its first nonzero Move in the interval of alternating singleton activations, and in the delayed-Move interval of b the distance stays positive, so each of the repeated solo activations of a is again a nonzero Move; hence moves(a) ≥ 2 ̸= 1 and the single-move condition fails. Case c1 ≥ 1. The execution consists of fully synchronous rounds, and c2 ≥ 1 gives d(k) ≥ d(0) > 0 for every k, as shown in Lemma G.33. Since λ ̸= 0, the displacement of each robot in round k has norm λ d(k) > 0, so both robots perform a nonzero Move in every round, and these are counted as distinct Move operations because tE (r, i) < tL (r, i + 1). Hence moves(a) = ∞ ̸= 1 and the single-move condition fails. In every case a single execution violates all three predicates, so {MCv, VSS, CFSM} has the common failure property on OBLOT AM and Lemma 6.1 gives the claim.

H

Order-Theoretic Certificates and Enumeration

H.1

A Certificate for the Signature Order

Let R be the thirteen representative settings in Table 1. Define Cσ (µ) = {ν ∈ R : σ(ν) ⊊ σ(µ), ∄ξ ∈ R : σ(ν) ⊊ σ(ξ) ⊊ σ(µ)}. Thus ν ∈ Cσ (µ) means that the signature of ν is immediately below the signature of µ among the thirteen signatures. This definition uses only the six-column table, not an assumed computational comparison between its rows. Reading the thirteen binary rows gives Table 6. Every displayed comparison follows from Proposition 3.1 and transitivity, except for the following two comparisons, which also use the previously established equivalences: LUMI R ≡ FCOMR ≥ FCOMS ,

FCOMA ≡ FCOMAM ≥ OBLOT AM . 96

No strictness or principality is assumed in these derivations. For any proper inclusion σ(µ) ⊋ σ(ν), finiteness supplies a chain from σ(µ) to σ(ν) whose successive inclusions are covering relations among the thirteen signatures. The table certifies the corresponding computational inclusion at each step. Transitivity therefore gives µ ≥ ν. Equality of two signatures occurs only for the same row. This proves the reverse implication used in Theorem 3.5 without relying on a previously established complete comparison table.

H.2

Meet, Join, and Join-Irreducibility

This subsection certifies the lattice claims made after the computational order has been established by Theorem 3.5. It is not used to prove that theorem: the earlier signature-order certificate relies only on the covering comparisons and the task-specific boundary results. Lemma H.1 (The computational lattice and its join-irreducibles). The poset Q is a finite lattice with least element OBLOT A and greatest element LU MI F . Its nonleast join-irreducible elements are exactly J(Q) = {OBLOT F , OBLOT R , OBLOT S , FSTAA , FCOMA , OBLOT AM }. Let Sσ = {σ(X) : X ∈ Q}. The family contains ∅ = σ(OBLOT A ) and B6 = σ(LUMI F ). For every incomparable pair, the intersection of its signatures is the signature of the class in the meet column of Table 7. For a comparable pair X ≤ Y , the intersection is σ(X). These checks exhaust all pairs, so Sσ is closed under intersection. For any T ⊆ B6 , define \ cl(T ) = {S ∈ Sσ : T ⊆ S}. The family being intersected is nonempty because it contains B6 . Finiteness and intersection closure give cl(T ) ∈ Sσ . It is the least signature containing T . Therefore the lattice operations are characterized by σ(X ∧Q Y ) = σ(X) ∩ σ(Y ),
σ(X ∨Q Y ) = cl σ(X) ∪ σ(Y ) .

(2)

The join column of the table is obtained by this closure operation. For comparable X ≤ Y , one has X ∧Q Y = X and X ∨Q Y = Y , so those pairs need not be listed. Identifying the join-irreducibles. In any finite lattice, a nonleast element j with a unique lower cover j∗ is join-irreducible. Indeed, every x < j lies below j∗ , and hence the join of two elements strictly below j is also at most j∗ . Conversely, if j has two distinct lower covers u, v, they are incomparable. Their join lies below j, but cannot lie strictly below j without contradicting that u and v are lower covers. Thus j = u ∨ v is not join-irreducible. The covering relations in Table 6 give the six elements and lower covers in Table 8. For explicit certificates excluding every other nonleast element, Table 7 gives FSTAAM = FSTAA ∨Q OBLOT AM , FSTAS = FSTAA ∨Q OBLOT S , FSTAR = FSTAA ∨Q OBLOT R , FCOMS = FCOMA ∨Q OBLOT S , LUMI R = FSTAR ∨Q FCOMS , LUMI F = OBLOT F ∨Q LUMI R . In each equality, both joinands are strictly smaller than the left-hand side. The least element OBLOT A is excluded from J(Q) by definition. This proves the complete list in Lemma H.1. 97

Table 7: Meet and join for all twenty-four incomparable pairs. The order and signatures are those already established in the main text. X

X ∧Q Y

Y F

OBLOT OBLOT F OBLOT F OBLOT F OBLOT F OBLOT F OBLOT F FSTAR FSTAR OBLOT R OBLOT R OBLOT R OBLOT R OBLOT R FCOMS FCOMS FCOMS FCOMA FCOMA FCOMA FCOMA FSTAAM FSTAA FSTAA

R

X ∨Q Y R

LUMI FSTAR FCOMS FCOMA FSTAS FSTAAM FSTAA FCOMS FCOMA FCOMS FCOMA FSTAS FSTAAM FSTAA FSTAS FSTAAM FSTAA FSTAS FSTAAM FSTAA OBLOT S OBLOT S OBLOT S OBLOT AM

OBLOT OBLOT R OBLOT S OBLOT AM OBLOT S OBLOT AM OBLOT A OBLOT S OBLOT AM OBLOT S OBLOT AM OBLOT S OBLOT AM OBLOT A OBLOT S OBLOT AM OBLOT A OBLOT AM OBLOT AM OBLOT A OBLOT AM OBLOT AM OBLOT A OBLOT A

LUMI F LUMI F LUMI F LUMI F LUMI F LUMI F LUMI F LUMI R LUMI R LUMI R LUMI R FSTAR FSTAR FSTAR LUMI R LUMI R LUMI R LUMI R LUMI R LUMI R FCOMS FSTAS FSTAS FSTAAM

Table 8: Join-irreducible model classes, their unique lower covers, and the principal tasks already established at those classes. j OBLOT OBLOT R OBLOT S FSTAA FCOMA OBLOT AM

Join-irreducible generation.

Task with profile ↑j

j∗ F

R

OBLOT OBLOT S OBLOT AM OBLOT A OBLOT AM OBLOT A

EPG∗ SRO MCv CFSM VSS DMSD

For completeness, in any finite lattice L with least element 0, _ x = {j ∈ J(L) : j ≤ x},

(3)

where the empty join is 0. To prove this, induct on the number of elements at most x. For x = 0 it is immediate, and for join-irreducible x the right-hand side includes x. Otherwise x = u ∨ v for two strictly smaller elements. The induction hypothesis expresses each of them as a join of join-irreducibles at most x, proving the claim. Consequently, if x ≰ y, some j ∈ J(L) satisfies j ≤ x and j ≰ y. This is the sufficiency argument in Proposition 3.8. For necessity, let j ∈ J(Q) and let j∗ be its unique lower cover. A principal upset ↑ x distinguishes j from j∗ only if x ≤ j and x ̸≤ j∗ . Since every x < j lies below j∗ , this forces x = j. Thus every principal separating family contains ↑ j for every j ∈ J(Q), proving uniqueness. 98

Order embedding versus lattice embedding. joins are not in general set unions. For example,

The six-task signatures preserve meet, but their

FSTAA ∨Q FCOMA = LUMI R , whose signature properly contains σ(FSTAA ) ∪ σ(FCOMA ). No distributivity of the model lattice is asserted. In fact, OBLOT R ∧Q (FSTAA ∨Q FCOMA ) = OBLOT R , (OBLOT R ∧Q FSTAA ) ∨Q (OBLOT R ∧Q FCOMA ) = OBLOT AM . This does not affect distributivity of the upset lattice Up(Q), whose operations are union and intersection.

H.3

Antichain and Maximal-Complement Enumeration

All counts in this subsection are determined solely by the finite order of the thirteen rows. They certify the number of abstract profiles and boundary obligations, not the existence of algorithms or of commonfailure executions. Proposition H.2 (Antichains and maximal-complement counts). The antichain polynomial of Q is X AQ (z) = z |A| = 1 + 13z + 24z 2 + 16z 3 , A∈Ant(Q)

so Up(Q) has 54 elements, of which 53 are nonempty. For an antichain A put UA = NA = Max(Q \ UA ). Then X X |NA | = 47, |NA | = 34. A∈Ant(Q) |A|=2

S

X∈A ↑X

and

A∈Ant(Q) |A|=3

Lemma H.3 (Width of the computational lattice). The width of Q is three. Proof. The thirteen classes are partitioned into the three chains C1 :

OBLOT A < OBLOT AM < OBLOT S < OBLOT R < OBLOT F < LUMI F ,

C2 :

FSTAA < FSTAAM < FSTAS < FSTAR ,

C3 :

FCOMA < FCOMS < LUMI R .

Each Ci is a chain in the order established by Theorem 3.5, and the three cover all thirteen classes. An antichain meets each chain at most once, so its size is at most three. Conversely {OBLOT F , FSTAR , FCOMS } is a three-element antichain, since no two of its signatures are comparable. Hence width(Q) = 3. Lemma P H.4 (Deletion recurrence for the antichain polynomial). For a finite poset P and x ∈ P , let AP (z) = A∈Ant(P ) z |A| . Then AP (z) = AP \{x} (z) + z AP \(↓P x∪↑P x) (z),

A∅ (z) = 1,

where deletions carry the induced order. Proof. Antichains not containing x are precisely those of P \ {x}. If an antichain contains x, its other elements must be incomparable with x, and form an antichain of P \ (↓P x ∪ ↑P x). Conversely, adjoining x to any such antichain is valid and increases its size by one, accounting for the factor z.

99

Table 9: Finite enumeration certificate, grouped by the smallest row index. The omitted row indices have zero in all four columns. i

qi

a2 (i)

a3 (i)

b2 (i)

b3 (i)

2 4 5 6 7 9 10

OBLOT F FSTAR OBLOT R FCOMS FCOMA FSTAAM FSTAA

7 2 5 3 4 1 2

8 0 6 0 2 0 0

16 5 8 8 6 2 2

18 0 13 0 3 0 0

Total

24

16

47

34

Number the classes q1 , . . . , q13 in the row order of Table 1, and let Si = σ(qi ). Thus (q1 , . . . , q13 ) = (LUMI F , OBLOT F , LUMI R , FSTAR , OBLOT R , FCOMS , FCOMA , FSTAS , FSTAAM , FSTAA , OBLOT S , OBLOT AM , OBLOT A ). For k ∈ {2, 3}, let ak (i) be the number of k-element antichains whose smallest row index is i, and let bk (i) be the sum of |NA | over those antichains. The term “smallest” here refers to the numerical row index, not to a least element in the computational order. Here is an explicit counting rule using the signature table. Put Ii = {j > i : Si ⊈ Sj and Sj ⊈ Si }. Then

a2 (i) = |Ii |,

a3 (i) = {(j, k) : j, k ∈ Ii , j < k, qj ∥ qk } .

Thus the incomparable-pair list in Table 7 also lists exactly the possible pairs used by this counting rule. Alternatively, the polynomial follows by applying Lemma H.4 to the same order. For the boundary counts, Y belongs to NA exactly when σ(X) ⊈ σ(Y )

for every X ∈ A,

for every Z > Y there is X ∈ A with σ(X) ⊆ σ(Z). The first condition puts Y outside UA , and the second makes it maximal there. In the second condition it suffices to test the upper covers of Y , because UA is an upset and Q is finite. Applying these rules gives Table 9. Proof of Proposition H.2. There is one empty antichain and there are thirteen singleton antichains. Lemma H.3 excludes antichains of size four or more, and the columns a2 and a3 of Table 9 give 24 and 16. Thus AQ (z) = 1 + 13z + 24z 2 + 16z 3 , and by the antichain–upset bijection of Proposition 2.2 the number of upsets is 1 + 13 + 24 + 16 = 54. The columns b2 and b3 give the totals 47 and 34 for the maximal-complement obligations.

100

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