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Econ-661 · economics · computer science · Computational economics

Encoding and complexity of token-rich Battle Sheep

LLM status Solved unverified

Statement

ID: GT-091. Category: Complexity of combinatorial games. Status: R; source-stated question retained with a rule and encoding audit, not certified as an unresolved classification.

Definitions and mathematical statement

Use the finite-board, two-player, normal-play version of generalized Battle Sheep. The input explicitly lists a finite set $B$ of hexagonal-grid cells, stack owners, positive integer stack sizes written in binary, and the player to move. Missing cells are board boundaries. A legal move chooses a stack of the moving player's sheep of size $s\geq2$, chooses an integer $q$ with $1\leq q<s$, and chooses one of the six straight grid directions having an adjacent empty board cell. The $q$ sheep travel along that direction through consecutive empty cells to the last empty cell before a boundary or occupied cell. They form a new stack there, and $s-q\geq1$ sheep remain at the source. Stacks cannot pass occupied cells. A player without a legal move loses.

Let $\mathcal W_{\mathrm{BS}}$ be the language of such explicitly encoded positions from which the player to move can force a win. The catalog question asks how its complexity changes when stack sizes, and hence the total number of sheep, can be exponential in $|B|$.

An essential scope observation is that every legal move occupies one previously empty cell and empties no cell. If $e_0$ cells are initially empty, every play has at most $e_0$ moves, independently of the number of sheep. Moreover every stack size has at most the bit length of the conserved total. Thus the explicit-board, binary-stack model admits a polynomial-space game-tree evaluation. The correct classification target must take this observation into account; merely increasing stack sizes does not supply exponential play depth.

Short English statement

Check whether large binary-encoded sheep stacks actually change winner-determination complexity once the splitting and movement rules are enforced.

Variants and scope boundaries

The source's Open Problem 1 asks about EXPTIME-hardness with exponential stack sizes. In the explicit-board model above, EXPTIME-hardness under polynomial-time many-one reductions would imply $\mathsf{PSPACE}=\mathsf{EXPTIME}$. A succinctly encoded exponentially large board, moves that empty the source, or moves onto occupied cells would require a different analysis. These variants are not inferred from the original rules. The polynomial depth observation does not bound the number of legal splits by a polynomial.

Source relationship and status

The source proves PSPACE-completeness for its restricted family and explicitly states Open Problem 1 for exponentially large stacks. The occupancy observation is an editorial audit of Section 1.1's rules, not a claimed published resolution. Accordingly this entry preserves the historical question while declining to repeat the catalog's claim that token magnitude defeats polynomial-depth membership. Combining the audit with the source hardness result gives PSPACE-completeness for the explicit-board extension, provided the restricted instances use the same winner convention.

References

[S1] Kyle Burke and Hirotaka Ono, Battle Sheep is PSPACE-complete (2025), arXiv:2505.06414v1. Inspected locators: Section 1.1 (bottom sheep stays, empty-cell movement), Theorem 1 and Corollary 1, Section 3, Open Problem 1. https://arxiv.org/html/2505.06414v1

Common conventions: Source mathematical conventions

Finite games. For a finite set $A$, $\Delta(A)$ is its probability simplex. Players choose independent mixed strategies in Nash equilibrium; correlation is allowed only when explicitly part of the solution concept. In a game with payoff functions $u_i$ and strategy profile $x$, an additive $\varepsilon$ Nash equilibrium satisfies \[ u_i(x)\geq u_i(x_i',x_{-i})-\varepsilon \quad\text{for every player $i$ and every admissible deviation $x_i'$}. \] For numerical approximation, payoffs are normalized as locally specified. An $\varepsilon$ payoff residual is distinct from distance at most $\varepsilon$ to the set of exact equilibria. Point convergence, convergence of empirical play, and convergence in distance to an equilibrium set are also distinct.

Computational representations. Unless stated otherwise, a complexity question takes rational data with binary encoding; polynomial time is measured in total bit length. A fixed-accuracy polynomial algorithm is not necessarily polynomial in $1/\varepsilon$, and neither is necessarily polynomial in $\log(1/\varepsilon)$. Succinct games and value, demand, or sampling oracles must declare their interfaces. Exact real solutions require an explicit representation; an unspecified list of arbitrary real numbers is not a finite computational output. A claimed algorithmic barrier is meaningful only for its specified model and reductions.

Dynamic games. Histories, information, observation, and allowable randomization are part of the model. A stationary strategy depends only on the current state; a positional strategy in a deterministic graph game chooses one action at each controlled vertex. Nash equilibrium at an initial state and equilibrium after every history are different requirements. An expectation of a pathwise $\liminf$ payoff, a $\liminf$ of expected averages, a limiting expected average, and a uniform finite-horizon equilibrium are different criteria. None is silently substituted for another.

Allocation and mechanisms. A complete allocation assigns every item exactly once unless otherwise stated. Zero-valued goods, negative values, capacity constraints, feasibility, lotteries, and copies of goods affect fairness definitions and are specified locally. Dominant-strategy incentive compatibility, Bayesian incentive compatibility, universal truthfulness, and truthfulness in expectation impose different inequalities. Whenever an objective ratio has zero denominator, its convention is stated or those instances are excluded.

Infinite-dimensional models. Expectations, integrals, stochastic equations, and optimization objectives require their local regularity, integrability, filtrations, and admissible domains. A backward--forward mean-field system is a boundary-value problem; it is not an initial-value semiflow without an additional construction. Long-time, large-population, and vanishing-noise limits require an order of limits and a convergence topology. If a minimum need not be attained, the objective is its infimum or supremum and the approximation requirement is stated separately.

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Econ-661-A2 · Full solution claimed · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-661

Encoding and complexity of token-rich Battle Sheep

Result and scope

For the explicit-board, binary-stack, normal-play model in the statement, this attempt gives a complete polynomial-space upper bound and an exact stack-capping lemma. Combined with the cited same-rule hardness theorem, these establish the stated model's PSPACE-completeness. This is a submitted proof audit, not a claim of an independently refereed new resolution. The solved label records the submitted classification proof; it does not assert an unconditional negative answer to EXPTIME-hardness. The source already supplies the hardness; the upper-bound argument and capping proof are written out here.

Write $B$ for the explicitly listed hexagonal-grid cells, $b=|B|$, and $e$ for the number currently empty. Each occupied cell has an owner and a positive binary integer size. A move splits a size $s\geq2$ into $q$ and $s-q$, both positive. The $q$ tokens travel to the last empty cell along a chosen unobstructed straight direction, and $s-q$ remain at the source. The player with no legal move loses. This rule is essential throughout.

Polynomial depth is independent of token count

Lemma: Every legal move decreases the number of empty cells by exactly one. Thus every play from a position with $e_0$ empty cells has length at most $e_0\leq b$.
Proof: The destination was empty and becomes occupied because $q\geq1$. The source remains occupied because $s-q\geq1$. Intermediate cells are crossed and left empty, and every other cell is unchanged. These are the only changes to occupancy. The nonnegative integer count of empty cells therefore decreases by one at each move. ∎

There can nevertheless be exponentially many choices of $q$ at one node. Polynomial depth does not imply polynomial game-tree size. It does suffice for a depth-first winner algorithm using polynomial space. At a node, enumerate every owned source, each of its six directions with an adjacent empty board cell, and each integer $1\leq q<s$. Recursively test the successor with the opposite player to move. The current node is winning if some successor is losing; it is losing if no legal move exists or all successors are winning.

Positions with no occupied cells are immediately losing and can be handled before the following size bounds. Let $L$ be the maximum input bit length of a stack size and let $K$ be the maximum bit length of a coordinate. The conserved total is at most $b2^L$, so every stack uses at most $L+\lceil\log_2 b\rceil+1$ bits. A split counter requires the same order of space, even though incrementing through its full range may take exponential time. A board state, a source index, a direction, and a counter occupy polynomial space in $b,L,K$. Storing one such frame at each of at most $b+1$ recursion levels is still polynomial. A ray endpoint can be found by repeated coordinate addition and membership/occupancy checks; at most $b$ consecutive listed cells can be traversed. No assumption that the coordinate bounding rectangle has polynomial area is needed.

An exact bound on strategically relevant stack sizes

The space argument already proves membership. The following stronger lemma shows that most of an enormous stack's binary representation is strategically irrelevant on a fixed explicit board.

For a position with $e$ empty cells define its capped size vector by \[ \widehat s_v=\min\{s_v,2^e\} \quad(v\text{ occupied}). \tag{1} \]

Theorem: Suppose two positions have the same board, occupied cells, owners, and player to move, and the same vector (1). They have the same winner. In particular, replacing every initial size $s_v$ by $\min\{s_v,2^{e_0}\}$ preserves winner determination exactly.
Proof: Induct on $e$. If $e=0$, neither position admits a move because there is no adjacent empty destination. Both are losing for the player to move.

For $e\geq1$, put $h=2^{e-1}$, so the current cap is $2h$. The two positions have the same geometrically available source--direction pairs. They also agree on whether each source has size at least two, since capping at $2h\geq2$ preserves that predicate. Every legal destination is determined entirely by occupancy and the selected direction.

Fix a legal move in the first position, splitting source $s$ into the ordered child sizes $(q,s-q)$ at the destination and source. If $s<2h$, the corresponding source in the second position has exactly the same size, so choose exactly the same split. The new child caps at level $h$ then agree.

If $s\geq2h$, its counterpart $s'$ also satisfies $s'\geq2h$. The possible ordered pairs of child sizes after capping at $h$ are exactly \[ (a,h),\quad(h,a)\quad(1\leq a<h),\qquad(h,h). \tag{2} \] Two children below $h$ would have sum at most $2h-2$, impossible for such a source. Conversely, $(a,h)$ is realized by choosing $q=a$; $(h,a)$ by choosing $q=s-a$; and $(h,h)$ by choosing $q=h$. Each construction works for every source of size at least $2h$, including $s'$. Thus match the first move by any split of $s'$ giving the same pair in (2). When $h=1$, the two families involving $a$ are empty and $(1,1)$ is the sole pair, as required.

Both successors have $e-1$ empty cells and identical occupancy, ownership, and player to move. Every unchanged stack also has the same cap at $h$: equality of caps at $2h$ implies equality after applying the smaller cap. By induction the matched successors have the same winner. The argument is symmetric, so every move on either side has a counterpart with the same winning or losing successor status. A position has a winning move exactly when its counterpart does. This completes the induction. ∎

Capping uses a threshold whose binary representation has just $e_0+1$ bits. Comparing a long input integer with that threshold and replacing it is polynomial in its input length. This produces an equivalent position with at most $b+1$ bits per stack. The result is a compression of large numbers, not a polynomial-time winner algorithm: the capped number itself can still be exponential in $b$, and the game tree can remain exponential. No claim that a cap of $b$ or $e_0+1$ tokens suffices follows from the proof.

Classification and the source's question

Burke and Ono's Theorem 1 proves PSPACE-hardness for a restricted family of explicitly represented Battle Sheep positions under the same normal-play convention. Their rules retain the bottom token and require movement to empty cells. Such instances are a subclass of the binary-stack model here, so the identity encoding is a polynomial-time reduction. Together with the depth-first upper bound, this gives \[ \mathcal W_{\mathrm{BS}} \text{ is PSPACE-complete}. \tag{3} \] The hardness theorem is an external input to (3); the reduction gadgets are not claimed as this attempt's work.

The source's Open Problem 1 asks whether exponentially large stacks yield EXPTIME-hardness. Under the exact explicit-board rules above, that hardness would imply $\mathsf{EXPTIME}\subseteq\mathsf{PSPACE}$, by composing the polynomial reduction with the proved polynomial-space algorithm. Since the reverse containment is standard, it would imply $\mathsf{PSPACE}=\mathsf{EXPTIME}$. Conversely, if those classes coincide, PSPACE-completeness already gives EXPTIME-hardness. Thus this formulation cannot establish a larger complexity class merely by increasing token magnitudes. It would be incorrect to assert an unconditional separation between PSPACE and EXPTIME.

Rule checks and limitations

If a move were allowed to empty the source, the occupancy potential would no longer decrease strictly. If moves could land on occupied cells, the split-pair and geometry argument would need revision. If the board were succinctly described and could contain exponentially many cells, its number of empty cells would no longer be bounded by the input length. Scoring by occupied area at the end instead of normal play would also change the terminal condition. None of these variants is included in (3). The capping proof uses the stated winner convention and explicitly listed finite board, and preserves ownership rather than identifying the players.

Completion Estimate

100%

This is a post hoc, heuristic estimate for the full catalogue target. The attempt proves a polynomial-space winner algorithm independent of token magnitude and an exact stack-capping equivalence under the canonical occupancy-preserving moves. Combined with the cited same-rule hardness theorem, this completes the explicit-board classification and correctly relates exponential-stack hardness to the unresolved equality of standard complexity classes.

References

The reference is Kyle Burke and Hirotaka Ono, Battle Sheep is PSPACE-complete, arXiv:2505.06414v1 (2025), Section 1.1, Theorem 1, and Section 3. https://arxiv.org/html/2505.06414v1. The contribution of this submitted attempt is the complete encoding audit and proved cap equivalence, with the source's hardness clearly attributed.

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Econ-661-A1 · Full solution claimed · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-661, version 2

Encoding and complexity of token-rich Battle Sheep

Problem ID: C79-657 Model: GPT 6 Astra Pro Version: 2

Result and scope

For the explicit-board, binary-stack, normal-play model in the statement, this attempt gives a complete polynomial-space upper bound and an exact stack-capping lemma. Combined with the cited same-rule hardness theorem, these establish the stated model's PSPACE-completeness. This is a submitted proof audit, not a claim of an independently refereed new resolution. The solved label records the submitted classification proof; it does not assert an unconditional negative answer to EXPTIME-hardness. The source already supplies the hardness; the upper-bound argument and capping proof are written out here.

Version 2 status and provenance. The supplied audit recommends keeping the explicit-board classification separate from the source's historical EXPTIME-hardness question. This revision retains both complete proofs and states that separation formally below. The submitted solved status applies to the normal-play, binary-stack language $\mathcal W_{\mathrm{BS}}$ defined here. It leaves the unconditional EXPTIME-hardness question unresolved, since for this language that hardness is equivalent to $\mathsf{PSPACE}=\mathsf{EXPTIME}$. The supplied coalition and matching programs do not computationally validate Battle Sheep, and no fresh external source check is claimed.

Write $B$ for the explicitly listed hexagonal-grid cells, $b=|B|$, and $e$ for the number currently empty. Each occupied cell has an owner and a positive binary integer size. A move splits a size $s\geq2$ into $q$ and $s-q$, both positive. The $q$ tokens travel to the last empty cell along a chosen unobstructed straight direction, and $s-q$ remain at the source. The player with no legal move loses. This rule is essential throughout.

Polynomial depth is independent of token count

Lemma: Every legal move decreases the number of empty cells by exactly one. Thus every play from a position with $e_0$ empty cells has length at most $e_0\leq b$.
Proof: The destination was empty and becomes occupied because $q\geq1$. The source remains occupied because $s-q\geq1$. Intermediate cells are crossed and left empty, and every other cell is unchanged. These are the only changes to occupancy. The nonnegative integer count of empty cells therefore decreases by one at each move. ∎

There can nevertheless be exponentially many choices of $q$ at one node. Polynomial depth does not imply polynomial game-tree size. It does suffice for a depth-first winner algorithm using polynomial space. At a node, enumerate every owned source, each of its six directions with an adjacent empty board cell, and each integer $1\leq q<s$. Recursively test the successor with the opposite player to move. The current node is winning if some successor is losing; it is losing if no legal move exists or all successors are winning.

Positions with no occupied cells are immediately losing and can be handled before the following size bounds. Let $L$ be the maximum input bit length of a stack size and let $K$ be the maximum bit length of a coordinate. The conserved total is at most $b2^L$, so every stack uses at most $L+\lceil\log_2 b\rceil+1$ bits. A split counter requires the same order of space, even though incrementing through its full range may take exponential time. A board state, a source index, a direction, and a counter occupy polynomial space in $b,L,K$. Storing one such frame at each of at most $b+1$ recursion levels is still polynomial. A ray endpoint can be found by repeated coordinate addition and membership/occupancy checks; at most $b$ consecutive listed cells can be traversed. No assumption that the coordinate bounding rectangle has polynomial area is needed.

An exact bound on strategically relevant stack sizes

The space argument already proves membership. The following stronger lemma shows that most of an enormous stack's binary representation is strategically irrelevant on a fixed explicit board.

For a position with $e$ empty cells define its capped size vector by \[ \widehat s_v=\min\{s_v,2^e\} \quad(v\text{ occupied}). \tag{1} \]

Theorem: Suppose two positions have the same board, occupied cells, owners, and player to move, and the same vector (1). They have the same winner. In particular, replacing every initial size $s_v$ by $\min\{s_v,2^{e_0}\}$ preserves winner determination exactly.
Proof: Induct on $e$. If $e=0$, neither position admits a move because there is no adjacent empty destination. Both are losing for the player to move.

For $e\geq1$, put $h=2^{e-1}$, so the current cap is $2h$. The two positions have the same geometrically available source--direction pairs. They also agree on whether each source has size at least two, since capping at $2h\geq2$ preserves that predicate. Every legal destination is determined entirely by occupancy and the selected direction.

Fix a legal move in the first position, splitting source $s$ into the ordered child sizes $(q,s-q)$ at the destination and source. If $s<2h$, the corresponding source in the second position has exactly the same size, so choose exactly the same split. The new child caps at level $h$ then agree.

If $s\geq2h$, its counterpart $s'$ also satisfies $s'\geq2h$. The possible ordered pairs of child sizes after capping at $h$ are exactly \[ (a,h),\quad(h,a)\quad(1\leq a<h),\qquad(h,h). \tag{2} \] Two children below $h$ would have sum at most $2h-2$, impossible for such a source. Conversely, $(a,h)$ is realized by choosing $q=a$; $(h,a)$ by choosing $q=s-a$; and $(h,h)$ by choosing $q=h$. Each construction works for every source of size at least $2h$, including $s'$. Thus match the first move by any split of $s'$ giving the same pair in (2). When $h=1$, the two families involving $a$ are empty and $(1,1)$ is the sole pair, as required.

Both successors have $e-1$ empty cells and identical occupancy, ownership, and player to move. Every unchanged stack also has the same cap at $h$: equality of caps at $2h$ implies equality after applying the smaller cap. By induction the matched successors have the same winner. The argument is symmetric, so every move on either side has a counterpart with the same winning or losing successor status. A position has a winning move exactly when its counterpart does. This completes the induction. ∎

Capping uses a threshold whose binary representation has just $e_0+1$ bits. Comparing a long input integer with that threshold and replacing it is polynomial in its input length. This produces an equivalent position with at most $b+1$ bits per stack. The result is a compression of large numbers, not a polynomial-time winner algorithm: the capped number itself can still be exponential in $b$, and the game tree can remain exponential. No claim that a cap of $b$ or $e_0+1$ tokens suffices follows from the proof.

Classification and the source's question

Burke and Ono's Theorem 1 proves PSPACE-hardness for a restricted family of explicitly represented Battle Sheep positions under the same normal-play convention. Their rules retain the bottom token and require movement to empty cells. Such instances are a subclass of the binary-stack model here, so the identity encoding is a polynomial-time reduction. Together with the depth-first upper bound, this gives \[ \mathcal W_{\mathrm{BS}} \text{ is PSPACE-complete}. \tag{3} \] The hardness theorem is an external input to (3); the reduction gadgets are not claimed as this attempt's work.
Theorem (Explicit-board classification): For finite explicitly listed hexagonal-grid boards with binary-encoded positive stack sizes, normal-play winner determination is $\mathsf{PSPACE}$-complete under polynomial-time many-one reductions, provided the legal split leaves at least one token at the source and places the moved tokens at the last consecutive empty cell in a chosen direction. The hardness part uses Burke and Ono's cited same-rule theorem. No succinct-board or alternative winner convention is included.
Proof: The empty-cell lemma bounds play length by the explicit number of cells, and the depth-first recursion stores only polynomially many bits even when it enumerates exponentially many binary split choices. This proves membership. The cited restricted same-rule family embeds identically into the language defined here and gives the polynomial-time hardness reduction. The stack-capping theorem is an additional winner-preserving compression result; it is not needed to justify the space bound. ∎

The source's Open Problem 1 asks whether exponentially large stacks yield EXPTIME-hardness. Under the exact explicit-board rules above, that hardness would imply $\mathsf{EXPTIME}\subseteq\mathsf{PSPACE}$, by composing the polynomial reduction with the proved polynomial-space algorithm. Since the reverse containment is standard, it would imply $\mathsf{PSPACE}=\mathsf{EXPTIME}$. Conversely, if those classes coincide, PSPACE-completeness already gives EXPTIME-hardness. Thus this formulation cannot establish a larger complexity class merely by increasing token magnitudes. It would be incorrect to assert an unconditional separation between PSPACE and EXPTIME.

Corollary (Exact status of the historical hardness question): For this explicit-board, binary-stack, normal-play language, \[ \mathcal W_{\mathrm{BS}}\text{ is }\mathsf{EXPTIME}\text{-hard} \quad\Longleftrightarrow\quad \mathsf{PSPACE}=\mathsf{EXPTIME}, \] where hardness uses polynomial-time many-one reductions. Consequently the explicit-board classification gives no unconditional negative answer to the historical EXPTIME-hardness question.
Proof: If the language is $\mathsf{EXPTIME}$-hard, its polynomial-space algorithm and the reductions decide every $\mathsf{EXPTIME}$ language in polynomial space. Combine this with the standard containment $\mathsf{PSPACE}\subseteq\mathsf{EXPTIME}$. Conversely, equality of the classes turns the proved $\mathsf{PSPACE}$-hardness into $\mathsf{EXPTIME}$-hardness under the same reductions. ∎

Rule checks and limitations

If a move were allowed to empty the source, the occupancy potential would no longer decrease strictly. If moves could land on occupied cells, the split-pair and geometry argument would need revision. If the board were succinctly described and could contain exponentially many cells, its number of empty cells would no longer be bounded by the input length. Scoring by occupied area at the end instead of normal play would also change the terminal condition. None of these variants is included in (3). The capping proof uses the stated winner convention and explicitly listed finite board, and preserves ownership rather than identifying the players.

Verification and final scope. The submitted reasoning checks the strict occupancy potential, polynomial storage of binary splits, and the exact attainable child-cap pairs (2). Burke and Ono retain credit for the imported hardness theorem; this revision does not present their gadgets as newly verified. The precise classification and conditional complexity-class equivalence are the conclusions. Neither an unconditional failure of EXPTIME-hardness nor a separation of PSPACE from EXPTIME has been proved, and the supplied finite audit results do not settle either claim.

Completion Estimate

100%

This is a post hoc, heuristic estimate for the full catalogue target. The estimate covers the audited canonical explicit-board classification: the polynomial-space algorithm, exact cap equivalence, and cited same-rule hardness account for it. It does not assign completion to an unconditional answer to the source's separate EXPTIME-hardness question. For this language that question is exactly equivalent to $\mathsf{PSPACE}=\mathsf{EXPTIME}$, which remains undecided here.

References

The reference is Kyle Burke and Hirotaka Ono, Battle Sheep is PSPACE-complete, arXiv:2505.06414v1 (2025), Section 1.1, Theorem 1, and Section 3. https://arxiv.org/html/2505.06414v1. The contribution of this submitted attempt is the complete encoding audit and proved cap equivalence, with the source's hardness clearly attributed.

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