Battle Sheep is PSPACE-complete

TL;DR

This work establishes that Battle Sheep is PSPACE-complete, even when each pile contains at most 3 tokens, by reducing from Bounded Two-Player Constraint Logic (B2CL) using a network of gadgets that simulate logical gates and wires. The authors design Primitive gadgets (Variable, Goal, And, Or, Choice, Fanout) plus a Makeup gadget to balance move counts and enforce the target game's structure, ensuring Blue can win if and only if the B2CL instance is a win. They further show the reduction yields a polynomial-depth game tree, placing Battle Sheep in PSPACE, and discuss several open problems including hardness for smaller piles and EXPTIME hardness with exponential token counts.

Abstract

Battle Sheep is a board game published by Blue Orange Games. With two players, it is a combinatorial game that uses normal play rules. We show that it is PSPACE-complete, even when each stack has only up to 3 tokens.

Figures (11)

• Figure 1: Legend: (a) is a blank space with no sheep tokens. (b) is a space with 1 token. Since the color doesn't matter, as there are no excess sheep to move, we leave them black. (c) is a space with $k$ Blue sheep tokens. (d) is a space with $k$ Red sheep tokens.
• Figure 2: A position in Battle Sheep. Blue's options are to move 1 to 4 tokens from the pile with 5 to space $a$, $b$, or $c$. Blue cannot move straight up because there are tokens directly in the way. They cannot move to the space between their stack and $a$ because they must move as far as possible in the chosen direction. The same is true of the space between the 5-stack and $c$. They cannot move from the stack with 1 sheep token because they must leave at least one token behind and there are not any excess tokens to move.
• Figure 3: A Wire between two gadgets. The output of the bottom gadget is connected to the input of the top gadget. The ellipsis denotes that there could be many hexagons along that path. (a) is how the Wire begins as a connection between two gadgets. (b) is the result of the case where Blue was able to play the token below the output to another location, meaning the extra token on the input side could come down and block it. This is the active case, where a positive signal is sent from the lower gadget to the upper. (c) is the result of the case where Blue was not able to play a token onto the lower gadget, and instead had to send it up, propagating an inactive signal. Blue will have to find another place to play the top token of the upper stack.
• Figure 4: Wire turns of 30 and 60 degrees. In both cases, an active input allows the output to be activated.
• Figure 5: Goal gadget. If the input is active (empty), then the Blue player is able to move one token there to win the game. The black sheep stacks can belong to either player since there is only one sheep in each stack.

Theorems & Definitions (4)

• Theorem 1: main
• proof
• Corollary 1: Completeness
• proof