Akari / Light Up
Akari / Light Up · Can the whole board be lit?
NP-completeBulb placements can encode an entire Boolean circuit.
NP: completeness
- Exact task
- Can lamps illuminate every white cell without shining on one another, while meeting every numbered adjacency clue?
- Scale & input
- Variable rectangle with black walls and numbered clues.
- Move rules
- Bulbs shine along rows/columns until walls; bulbs cannot see one another.
- Goal
- Light every white cell and satisfy adjacent-bulb counts.
- Repetition
- Static placement.
- Players & information
- One solver; full puzzle visible.
- Randomness
- None.
Counts are orthogonal neighbors, not all illuminated cells.
Result in cited thesis
McPhail §3–5; Pulles §2 rules.
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Akari / Light Up · Remove all numbered clues
Polynomial-time solvableEvery unnumbered layout can be completed by adding a bulb to an unlit cell until none remain.
P: polynomial-time algorithm
- Exact task
- Can lamps illuminate every white cell without shining on one another when there are no numbered clues?
- Scale & input
- Growing grid; unnumbered black walls only.
- Move rules
- Same lighting and nonvisibility rules.
- Goal
- Illuminate all white cells.
- Repetition
- At most one new bulb per step.
- Players & information
- All cells known.
- Randomness
- None needed.
A primary undergraduate thesis; the short greedy proof is explicit.
Result in cited thesis
§3.1.1, algorithm 1.
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Only the clue 4
PForce bulbs around each 4, reject conflicts, then greedily light dark cells.
P: polynomial-time algorithm
- Exact task
- Decide whether the puzzle can be completed.
- Scale & input
- Growing grid. Walls are unnumbered or labeled 4.
- Move rules
- Ordinary Akari lighting and bulb-count rules.
- Goal
- Light every white cell without bulbs seeing each other.
- Repetition
- Static completion, no play-history restriction.
- Players & information
- One solver. All clues and constraints are known.
- Randomness
- None.
Primary undergraduate thesis. Forced-bulb conflicts must be checked.
restriction: Akari / Light Up · Can the whole board be lit?. This comparison is not itself a reduction.
Result in cited thesis
§3.1.2, Algorithms 2–3.
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Only 1s, only 2s, or only 3s
NP-completeEach of these three fixed clue sets still encodes circuits.
NP: completeness
- Exact task
- Decide whether the puzzle can be completed.
- Scale & input
- Growing grid. Choose one allowed number: 1, 2, or 3. Unnumbered walls also allowed.
- Move rules
- Ordinary Akari lighting and bulb-count rules.
- Goal
- Light every white cell and meet each clue.
- Repetition
- Static completion, no play-history restriction.
- Players & information
- One solver. All clues and constraints are known.
- Randomness
- None.
Primary undergraduate thesis with mechanically checked gadgets.
restriction: Akari / Light Up · Can the whole board be lit?. This comparison is not itself a reduction.
Result in cited thesis
§3.1, NP membership; §§3.1.3–3.1.5, circuit constructions.
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Backgammon
Backgammon · Known rolls & opponent
NP-hardEven with those future choices disclosed, finding a winning play is NP-hard.
NP: hardness / lower bound
- Exact task
- Can you win against this prescribed opponent and roll sequence?
- Scale & input
- Explicit scalable line of points and polynomially many checkers. Six home points per side and two six-sided dice suffice. Input includes the prescribed future rolls and opponent policy.
- Move rules
- Move according to both dice, doubles four times. Two opposing checkers block a point. A hit checker must re-enter before others move.
- Goal
- Bear off every checker first.
- Repetition
- Captures permit arbitrarily long play. No fixed turn limit is assumed.
- Players & information
- Two players, with the opponent’s complete policy revealed in this task.
- Randomness
- Both players’ future rolls are supplied, using the paper’s compact descriptions.
Hardness only. This is neither a fixed 24-point board nor expected win probability under fair random dice.
Published result
Definition 2 and Theorem 1.
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Known rolls, strategic opponent
PSPACE-hardThe opponent now chooses variable assignments in the reduction.
PSPACE: hardness / lower bound
- Exact task
- Can you force a win against an opponent whose policy is not disclosed?
- Scale & rules
- Same scalable checker/point model and known roll schedule. Both players choose strategically.
- Result scope
- Theorem 2 is a hardness result. It does not prove PSPACE membership.
rule change: Replace the prescribed opponent by an adversarial player · Backgammon · Known rolls & opponent. This comparison is not itself a reduction.
Published result
Definition 4, Theorem 2.
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Adaptive rolls and strategic opponent
EXPTIME-hardThe proof uses roll choices that adapt to play to simulate a revisitable formula game.
EXPTIME: hardness / lower bound (Adversarially adaptive rolls in the source model. Hardness only, not fair-dice probability or completeness.)
- Exact task
- What is the hardness of the paper’s unknown-strategy, unknown-roll winner model?
- Scale & rules
- Same generalized board. Opponent and future rolls are unknown, with adaptively chosen rolls in the reduction.
- Randomness
- No expected-score or specified probability threshold is analyzed.
The theorem concerns adaptive rolls, not fair-dice win probabilities. It proves EXPTIME-hardness without a matching upper bound.
rule change: Allow an adversarial strategy and adaptive future rolls · Backgammon · Known rolls & opponent. This comparison is not itself a reduction.
Published result
Definition 6, Theorem 3, especially p.10 on adaptive dice.
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Matching upper bounds
Open in the 2021 sourceThe author explicitly leaves membership questions open.
Open question in 2021 source
- Exact task
- Which upper bounds match the three backgammon hardness models?
- Scope
- KSKR, USKR and USUR exactly as defined in the paper. Future-policy and roll encodings must be specified.
- Caution
- An unbounded game length alone does not rule out EXPTIME algorithms on finitely represented game graphs.
This records the question in the cited 2021 source, not a lower bound or a claim that every variant of this game is unresolved.
task change: Ask for membership, not only hardness · Backgammon · Known rolls & opponent. This comparison is not itself a reduction.
Open in the 2021 source
Conclusion, p.12.
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Bloxorz
Bloxorz · Reach the goal with switches and bridges
PSPACE-completeChanging a 1×1×2 rolling block into a unit cube makes this reachability problem polynomial-time.
PSPACE: completeness
- Exact task
- Can the block legally reach the goal standing upright?
- Scale & input
- Explicit list of board tiles, start/goal, bridge states and one-to-one switch-to-bridge links.
- Move rules
- Roll/tilt a 1×1×2 block. Entering a switch toggles its bridge. All occupied cells must be supported.
- Goal
- Reach the goal upright.
- Repetition
- Revisits and repeated switch activation allowed.
- Players & information
- One player. Complete board state known.
- Randomness
- None.
Scalable levels with switches and bridges only. No splitting portals or weak tiles required.
Published result
§1.1 and Theorem 2.
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Remove switches and bridges
POnly polynomially many block positions and orientations remain.
P: polynomial-time algorithm
- Exact task
- Decide whether the rolling block can reach the goal.
- Scale & input
- Explicit finite board with a 1×1×2 block, start and goal.
- Move rules
- Roll and tilt, always fully supported. Static tiles only.
- Goal
- Reach goal upright.
- Repetition
- Revisits allowed.
- Players & information
- One player. Complete board known.
- Randomness
- None.
restriction: Bloxorz · Reach the goal with switches and bridges. This comparison is not itself a reduction.
Published result
§1.2 and Table 2.
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Replace the long block with a cube
PA unit cube can expand its reachable area systematically.
P: polynomial-time algorithm
- Exact task
- Decide whether the cube can reach the goal using switches and bridges.
- Scale & input
- Same explicit switch/bridge board, with a 1×1×1 cube.
- Move rules
- Step to an adjacent supported square, toggling switches.
- Goal
- Reach goal.
- Repetition
- Revisits and repeated toggles allowed.
- Players & information
- One player. Full state known.
- Randomness
- None.
rule change: Bloxorz · Reach the goal with switches and bridges. This comparison is not itself a reduction.
Published result
Theorem 8.
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Fragile one-use tiles instead of switches
NP-completeA tile that disappears after use can enforce Boolean choices.
NP: completeness
- Exact task
- Decide whether the 1×1×2 block can reach the goal.
- Scale & input
- Explicit board. Normal and single-use tiles, no switches or bridges.
- Move rules
- Roll/tilt long block. Fragile tiles disappear after traversal.
- Goal
- Reach goal upright.
- Repetition
- Fragile tiles cannot be reused.
- Players & information
- One player. Full board known.
- Randomness
- None.
Single-use tiles are a theoretical variant, not a feature of the original Bloxorz.
rule change: Bloxorz · Reach the goal with switches and bridges. This comparison is not itself a reduction.
Published result
§5, Theorem 7.
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Checkers
Checkers · Every turn must capture
PSPACE-completeRequiring a capture every turn bounds the game’s length, but optimal play remains PSPACE-complete.
PSPACE: completeness
- Exact task
- Decision: can the first player force a win?
- Scale & input
- Supplied n×n position.
- Move rules
- Mandatory jumps; continue the capture sequence while possible. Kings jump in either direction.
- Goal
- Opponent cannot jump.
- Repetition
- Impossible: play permanently consumes resources.
- Players & information
- Two players; all information visible.
- Randomness
- None.
Noncapturing moves are forbidden, even when no capture exists.
Published result
§§2, 4.
- Losing at Checkers is HardJeffrey Bosboom; Spencer Congero; Erik D. Demaine; Martin L. Demaine; Jayson Lynch (2018).Published book chapter · The Mathematics of Various Entertaining Subjects, vol. 3 (2019), 103–118; cited 2018 manuscriptRead full text
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Checkers · Work together
NP-completeCooperation changes the always-jumping game into a solution-existence puzzle.
NP: completeness
- Exact task
- Decision: can cooperating players eliminate one color?
- Scale & input
- Supplied n×n position.
- Move rules
- Alternate mandatory capturing turns; continue available jumps.
- Goal
- Remove every piece of either color.
- Repetition
- Impossible: play permanently consumes resources.
- Players & information
- Two sides cooperate; complete information.
- Randomness
- None.
This claim concerns the always-jumping variant.
Published result
§5.
- Losing at Checkers is HardJeffrey Bosboom; Spencer Congero; Erik D. Demaine; Martin L. Demaine; Jayson Lynch (2018).Published book chapter · The Mathematics of Various Entertaining Subjects, vol. 3 (2019), 103–118; cited 2018 manuscriptRead full text
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Force your opponent to capture everything
NP-completeLosing immediately is computationally hard even though the horizon is only two turns.
NP: completeness
- Exact task
- Can you make a move forcing the opponent to capture all your remaining pieces next turn?
- Model & rules
- Arbitrary position on a growing n×n board. American checkers captures are compulsory and must continue while possible. Kings jump in both directions. Two players, full information, no randomness. Only the next two turns are relevant.
task change: Ask for an immediate forced loss · Checkers · Every turn must capture. This comparison is not itself a reduction.
Published result
§2; Theorems 3.2–3.3
- Losing at Checkers is HardJeffrey Bosboom; Spencer Congero; Erik D. Demaine; Martin L. Demaine; Jayson Lynch (2018).Published book chapter · The Mathematics of Various Entertaining Subjects, vol. 3 (2019), 103–118; cited 2018 manuscriptRead full text
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Cooperate for just two turns
NP-completeReplacing forced capture with a cooperative choice remains NP-complete.
NP: completeness
- Exact task
- Can your move let the opponent capture all your pieces on their next turn?
- Model & rules
- Arbitrary position on a growing n×n board. American checkers captures are compulsory and must continue while possible. Kings jump in both directions. Two players, full information, no randomness. Only the next two turns are relevant.
task change: The opponent helps instead of resisting · Force your opponent to capture everything. This comparison is not itself a reduction.
Published result
§5, first paragraph
- Losing at Checkers is HardJeffrey Bosboom; Spencer Congero; Erik D. Demaine; Martin L. Demaine; Jayson Lynch (2018).Published book chapter · The Mathematics of Various Entertaining Subjects, vol. 3 (2019), 103–118; cited 2018 manuscriptRead full text
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Chomp
Chomp · Winner on a full rectangle?
Polynomial timeEvery rectangle with more than one square is a first-player win by strategy stealing.
P: polynomial-time algorithm
- Exact task
- Who wins from a complete rectangular chocolate bar?
- Scale & input
- Positive dimensions m,n given in binary or unary, describing a full rectangle.
- Move rules
- Choose a remaining square and remove it and every square up and to its right. The lower-left square is poisoned.
- Goal
- Make the opponent take the poison.
- Repetition
- Each move removes at least one square.
- Players & information
- Two players, perfect information.
- Randomness
- None.
This decides the winner. The proof does not provide an efficient winning first move. The 1×1 board is losing.
Published result
§1.1, Theorem 2.
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Find the first winning bite
Open in the 2020 sourceThis search question is explicitly open even though deciding who wins is immediate.
PSPACE: upper bound / membership (Polynomial-space search for a winning opening move on unary rectangles.) · Open question in 2020 source
- Exact task
- Can a winning first move be found in polynomial time in unary dimensions?
- Input & rules
- A full m×n rectangle with m,n written in unary, at least two cells. Same poison and bite rules.
- Scope
- The authors place the search in polynomial space and explain why the sparse input is an obstacle to usual hardness proofs.
This records the question in the cited 2020 source, not a lower bound or a claim that every variant of this game is unresolved.
task change: Find a move rather than name the winner · Chomp · Winner on a full rectangle?. This comparison is not itself a reduction.
Open in the 2020 source
§3, “Hardness for Specific Games”, p.21:11.
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Start from a partially eaten bar
Open in the 2013 sourceHardness of unrestricted poset games does not settle their two-dimensional Chomp restriction.
PSPACE: upper bound / membership (Explicit partially eaten position, at most one move per remaining cell.) · Open question in 2013 source
- Exact task
- What is the winner-decision complexity for arbitrary Chomp positions?
- Input & rules
- Explicit remaining cells of a downward-closed chocolate shape with the poison retained, not a succinct exponentially large bar. Same bite rule.
- Upper bound
- In PSPACE by finite board storage and at most one move per cell.
This records the question in the cited 2013 source, not a lower bound or a claim that every variant of this game is unresolved.
rule change: Allow a partially eaten starting position · Chomp · Winner on a full rectangle?. This comparison is not itself a reduction.
Open in the 2013 source
Introduction, p.2.
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Domineering / Cram
Domineering · Winner from a position?
In PSPACEPolynomial-space search follows from the move bound. Exact complexity is open in the cited source.
PSPACE: upper bound / membership
- Exact task
- Can the player to move force a win?
- Scale & input
- A finite grid-cell board supplied explicitly, possibly with unavailable cells, plus the player to move.
- Move rules
- One player places vertical dominoes and the other horizontal dominoes. Occupied cells cannot be reused.
- Goal
- Make the last legal move.
- Repetition
- Every move consumes two empty cells, so play has at most N/2 moves.
- Players & information
- Two players, deterministic perfect information.
- Randomness
- None.
The 2026 graph-generalization results do not automatically settle domino placement on a grid.
Published result
Introduction §§1.1–1.2. PSPACE membership is the direct bounded-depth-search consequence.
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Winner complexity on general grid boards
Open in the 2024 sourceThe authors explicitly identify general-board winner complexity as open.
PSPACE: upper bound / membership (Explicit finite grid, at most one domino placement per two cells, as recorded in the assumptions.) · Open question in 2024 source
- Exact task
- Is deciding the Domineering winner efficiently solvable, or can hardness be proved?
- Rules
- Explicit remaining grid cells. Vertical versus horizontal placement, normal play.
- Known upper bound
- In PSPACE by at most N/2 moves.
This records the question in the cited 2024 source, not a lower bound or a claim that every variant of this game is unresolved.
task change: Close the upper/lower-bound gap · Domineering · Winner from a position?. This comparison is not itself a reduction.
Open in the 2024 source
§1.2, “Cram and Domineering”.
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Cram: either player may use either orientation
Open in the 2024 sourceThe general-board problem is explicitly open. The simple parity strategies for some empty rectangles do not settle it.
PSPACE: upper bound / membership (Explicit finite grid, at most one domino placement per two cells, as recorded in the assumptions.) · Open question in 2024 source
- Exact task
- What is the complexity of deciding the Cram winner on a general grid board?
- Rules
- Explicit remaining grid cells. Both players may place a horizontal or vertical domino. No overlap, last move wins.
- Bound
- At most N/2 moves, hence polynomial-space search.
This records the question in the cited 2024 source, not a lower bound or a claim that every variant of this game is unresolved.
rule change: Allow both orientations for both players · Domineering · Winner from a position?. This comparison is not itself a reduction.
Open in the 2024 source
Introduction and §1.2.
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Cram on an arbitrary graph: Arc Kayles
PSPACE-completeA September 2026 proof settles the unrestricted graph game.
PSPACE: completeness
- Exact task
- Who wins when domino placements are graph edges?
- Scale & rules
- Explicit finite undirected graph. A move selects an edge and removes its two endpoints. The last move wins.
- Players & information
- Two players, perfect information, no randomness, at most |V|/2 moves.
This does not prove PSPACE-hardness of grid Cram.
rule change: Remove orientation ownership and replace the grid by a graph · Domineering · Winner from a position?. This comparison is not itself a reduction.
Result in cited preprint
Bonnet, Theorem 1 and §3; Hanaka et al. identify Cram as the grid-graph restriction.
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Geography
Geography · Arrows · no vertex reuse
PSPACE-completeMove along arrows without returning to a visited vertex. Deciding who can force the last legal move is PSPACE-complete.
PSPACE: completeness
- Exact task
- Can the player to move force a win from the supplied graph and token position?
- Scale & input
- An explicitly given directed graph of arbitrary size and a starting vertex.
- Move rules
- Players alternate moving one shared token along an outgoing arrow to an unvisited vertex.
- Goal
- A player with no legal move loses.
- Repetition
- Vertices cannot be revisited; at most |V|−1 token moves.
- Players & information
- Two players; Perfect information
- Randomness
- None
The source counts placing the initial marker as a turn; specifying who moves next, or adding a forced prefix, translates conventions.
Published result
Definition 3 and Theorem 2; Theorem 3 for planarity
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Geography · Two-way · no vertex reuse
PMake the paths two-way and keep the ban on revisiting vertices. Maximum matching now gives a polynomial-time winner algorithm.
P: polynomial-time algorithm
- Exact task
- Can the player to move force a win from the supplied graph and token position?
- Scale & input
- An explicitly given undirected graph of arbitrary size and a starting vertex.
- Move rules
- Move the shared token to a neighbor, deleting the vertex just left.
- Goal
- A player with no legal move loses.
- Repetition
- No vertex revisits.
- Players & information
- Two players; Perfect information
- Randomness
- None
The player to move wins exactly when every maximum matching covers the token's vertex. Compare maximum-matching sizes in G and G−v.
Published result
Fraenkel et al., Theorem 1.1; Fox–Geissler, Section 5
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Geography · Two-way · no edge reuse
PSPACE-completeStill use two-way paths, but use up each edge rather than each vertex. Vertices can be revisited, and winner determination is PSPACE-complete again.
PSPACE: completeness
- Exact task
- Can the player to move force a win from the supplied graph and token position?
- Scale & input
- An explicitly given undirected graph of arbitrary size and a starting vertex.
- Move rules
- Move the shared token along an unused edge, then delete that edge.
- Goal
- A player with no legal move loses.
- Repetition
- Edges cannot be reused; vertices may be revisited. At most |E| moves.
- Players & information
- Two players; Perfect information
- Randomness
- None
Hardness persists on planar graphs of maximum degree 3. Bipartite graphs admit a polynomial algorithm.
Published result
Abstract; main hardness result
- Undirected edge geographyAviezri S. Fraenkel, Edward R. Scheinerman and Daniel Ullman (1993).Published journal article · Theoretical Computer Science (1993)Open manuscript not recorded. Evidence coverage is limited to the primary abstract; the full proof is not assessed in this record.
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Geography · Sparse planar directed graph
PSPACE-completeHardness survives even when edges can be drawn without crossings and every vertex has at most three neighbors.
PSPACE: completeness
- Exact task
- Decision: can the next player force a win?
- Scale & input
- Directed planar bipartite graph, maximum total degree three; start vertex supplied.
- Move rules
- Move to an unvisited out-neighbor.
- Goal
- Opponent has no move.
- Repetition
- Impossible: play permanently consumes resources.
- Players & information
- Two players; all information visible.
- Randomness
- None.
Directions still matter. This is not undirected vertex Geography.
Published result
§4 corollary.
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Compute the nimber, not just the winner
P search at degree 3; PSPACE-hard and polynomial-space computable at degree 4Degree three permits efficient nimber computation. Degree four already gives PSPACE-hardness, while recursive computation still needs only polynomial space.
P: polynomial-time algorithm (Compute the Grundy-value function on explicit graphs of maximum degree 3.) · PSPACE: hardness / lower bound (Grundy-value computation on explicit graphs of maximum degree 4. This is a function hardness result, not a PSPACE-complete decision-language claim.) · PSPACE: upper bound / membership (Polynomial-space computation of the Grundy-value function by depth-first mex evaluation.) [atlas-derived]
- Exact task
- Compute an undirected Geography position's Grundy value.
- Model & rules
- Explicit graph and token. Alternate to unvisited neighbors. The player with no legal move loses. Perfect information, no randomness.
Why the atlas-derived bound holds
At each recursion level store the remaining graph, token and a polynomial-size seen-nimber bitset. The nimber is at most the number of available moves, hence at most |V|. There are at most |V| levels because a vertex is removed each turn. This computes a function in polynomial space; it does not classify a decision language as PSPACE-complete.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
Burke, Ferland and Teng, Theorem 1.1. Polynomial-space computation follows from depth-first mex evaluation: at most |V| moves and at most |V| options per position.
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Play two boards together
PSPACE-completeEach board alone is easy. Their combined game is hard.
PSPACE: completeness
- Exact task
- Determine the winner of two undirected Geography positions combined.
- Model & rules
- Two explicit graphs and tokens. Each turn moves on exactly one board to an unvisited neighbor. A player unable to move on either board loses. Perfect information, no randomness.
Published result
Theorem 4.3.
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One shared pass changes the winner problem
PSPACE-completeAdding one shared pass makes winner determination PSPACE-complete.
PSPACE: completeness
- Exact task
- Determine the winner with one pass available to either player.
- Model & rules
- Explicit undirected graph and token. Move to an unvisited neighbor or consume the shared pass. Passing is allowed even when stuck. A player with no remaining action loses. Perfect information, no randomness.
Published result
Section 4, Theorem 4.3.
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Two shared passes restore polynomial time
PPass parity matters: one pass is PSPACE-complete, two passes give the original polynomial-time winner test.
P: polynomial-time algorithm
- Exact task
- Determine the winner with exactly two shared passes available.
- Model & rules
- Explicit undirected graph and token. Move to an unvisited neighbor or consume one of two shared passes, including when stuck. Two players, full information, no randomness. Every action removes a vertex or pass. The player with no legal action loses.
rule change: Increase the shared pass allowance from 1 to 2 · One shared pass changes the winner problem. This comparison is not itself a reduction.
Published result
Theorem 4.3, even total pass count.
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Hex
Hex · From a given position
PSPACE-completeFrom a supplied partly filled board, deciding who can force a connection is PSPACE-complete as the board grows.
PSPACE: completeness
- Exact task
- Does the player to move have a winning strategy from the supplied partly filled board?
- Scale & input
- An n×n rhombus of hexagonal cells, with n part of the input; arbitrary supplied positions.
- Move rules
- Alternate placing a stone of your color on an empty cell. No swap/pie rule.
- Goal
- Connect your assigned opposite sides with your stones.
- Repetition
- Stones never move or disappear; no cell is played twice.
- Players & information
- Two players; Perfect information
- Randomness
- None
This classification does not concern only a fixed 11×11 commercial board, or only empty starting boards.
Published result
Theorem II.2 and Corollary II.3
- Hex ist PSPACE-vollständigStefan Reisch (1981).Published journal article · Acta Informatica 15 (1981), 167–191Read English rendering The linked rendering is in English; the original article is in German. Translator/date are not specified in the copy.
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Hex · From an empty board
POn every empty square Hex board without a swap rule, the first player has a winning strategy. This settles the winner question, not how to play the winning moves.
P: polynomial-time algorithm
- Exact task
- On an empty n×n board without the swap rule, does the first player have a winning strategy?
- Scale & input
- Empty square rhombus boards of every size n≥1.
- Move rules
- The same placement and connection rules as Hex; no swap/pie rule.
- Goal
- Connect your assigned opposite sides with your stones.
- Repetition
- Stones remain fixed; a cell can be filled only once.
- Players & information
- Two players; Perfect information
- Randomness
- None
The decision answer is always yes. Strategy stealing establishes existence; this is not an efficient method for finding the winning moves.
Published result
Fact 0.1
- Hex ist PSPACE-vollständigStefan Reisch (1981).Published journal article · Acta Informatica 15 (1981), 167–191Read English rendering The linked rendering is in English; the original article is in German. Translator/date are not specified in the copy.
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Find the winning first move
Open in the 2020 sourceKnowing that the first player wins does not identify the move. The paper leaves polynomial-time search open for Hex itself.
Open question in 2020 source
- Exact task
- Given the empty board's dimensions in unary, output a winning first move.
- Model & rules
- Empty growing square Hex board, normal alternating placement and connection goals, no swap rule. Two players, perfect information, no randomness. Cells are never reused.
Its PSPACE-hardness theorem covers broader game classes, not this empty-Hex search problem. Open status is dated to the cited paper.
Open in the 2020 source
Section 3, Hardness for Specific Games.
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Require a win within k turns
FPT in kFor ordinary planar Hex, the dependence on board size can be polynomial with the expensive part confined to k.
FPT: parameterized result (Fixed-parameter tractable in input turn bound k. Not a polynomial-time claim when k varies freely.)
- Exact task
- Can the player force a connection within k total plies?
- Model & rules
- Growing explicit Hex board and input bound k on total alternating moves. Existing stones allowed, usual connection goals. Full information, no randomness. Each turn fills one cell.
FPT is not a claim of polynomial time when k varies freely. The proof’s dependence on k is not practical.
task change: Bound the winning horizon · Hex · From a given position. This comparison is not itself a reduction.
Published result
§5, Theorem 5, Corollaries 2–3
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Can optimal play have a small circuit?
Open in the 2020 sourceThe authors ask whether small strategies must exist even when finding them is hard.
Open question in 2020 source
- Exact task
- Does empty-board Hex necessarily admit a polynomial-size optimal-strategy circuit?
- Scope
- A scalable empty symmetric Hex board. A circuit takes a reached position and outputs the strategy’s next move.
- Distinction
- This is strategy representation size, not the problem of finding one opening move.
This records the question in the cited 2020 source, not a lower bound or a claim that every variant of this game is unresolved.
task change: Ask for a compact whole-game strategy · Hex · From an empty board. This comparison is not itself a reduction.
Open in the 2020 source
§3, “Other Notions of Constructiveness”, p.21:11.
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Jigsaw & packing
Jigsaw · Fit every piece?
NP-completeAmbiguous tab-and-pocket matches make scalable assembly NP-complete.
NP: completeness
- Exact task
- Can all supplied pieces assemble into the target square?
- Scale & input
- An explicit finite collection of n nearly square pieces and a target square. Shapes are encoded by finitely described matching types.
- Move rules
- Face-up orthogonal placement of tab-and-pocket pieces, matching shapes only. Use each supplied piece once.
- Goal
- Fill the target exactly, without overlap or gaps.
- Repetition
- Static completion problem. No move-history restriction.
- Players & information
- One solver, all input is visible.
- Randomness
- None.
This concerns scalable, potentially ambiguous piece sets, not the difficulty of a particular commercial picture puzzle.
Published result
§6, Theorem 5 and Corollary 2.
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Edge matching · Match every edge?
NP-completeExact edge matching is NP-complete and converts to jigsaw assembly.
NP: completeness
- Exact task
- Can all colored-edge tiles fill the target square?
- Scale & input
- n explicitly listed unit-square tiles, with four colors per tile and a square target of area n.
- Move rules
- Place tiles on the square grid. Adjacent edge colors must agree. Every tile is used exactly once.
- Goal
- Fill the target with compatible edges.
- Repetition
- Static completion problem. No move-history restriction.
- Players & information
- One solver, all input is visible.
- Randomness
- None.
The reduction forces tile orientations. Difficulty does not rely on image recognition.
Published result
§4, Theorem 3.
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Polyomino packing · Pack every piece?
NP-completeEven pieces fitting in O(log n) by O(log n) boxes can encode hard jigsaw puzzles.
NP: completeness
- Exact task
- Can the supplied polyominoes exactly tile the square?
- Scale & input
- n finite grid-cell shapes and a square target whose area equals their total area.
- Move rules
- Orthogonal placements of all supplied pieces, without overlap.
- Goal
- Cover every target cell exactly once.
- Repetition
- Static completion problem. No move-history restriction.
- Players & information
- One solver, all input is visible.
- Randomness
- None.
A fixed catalog of shapes is a different restriction from the variable piece sets here.
Published result
§7, Theorem 6 and Corollary 3.
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Every mate is locally identifiable
P (algorithmic consequence)Testing piece pairs gives a polynomial-time assembly method under this strong promise.
P: polynomial-time algorithm [atlas-derived]
- Exact task
- Can the puzzle be assembled when local fit identifies true neighbors?
- Scale & rules
- Explicit finite pieces. Every locally compatible pair is guaranteed to be adjacent in the final solution.
- Scope
- This is stronger than merely promising one global solution.
restriction: Promise that local compatibility identifies the true neighbor · Jigsaw · Fit every piece?. This comparison is not itself a reduction.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
Introduction, p.4, paragraph beginning “The shape-matching community”.
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Lemmings
Lemmings · Save one lemming with two skills
PSPACE-completeOne lemming is enough. Building and bashing let the player close and reopen passages.
PSPACE: completeness
- Exact task
- Can the single lemming be guided alive into an exit?
- Scale & input
- Explicit pixel landscape with steel and deadly zones. Skill budgets and allowed time can be exponentially large or unlimited.
- Move rules
- Only Builder and Basher skills; at most one skill assignment per time step.
- Goal
- Get the single lemming into an exit alive.
- Repetition
- Repeated paths and repeated construction/destruction allowed.
- Players & information
- One player; complete level and deterministic lemming behavior.
- Randomness
- None.
Not a polynomial-move-budget claim. The linked author manuscript is an earlier version whose main PSPACE theorem agrees with the 2015 final paper.
Published result
Section 4, Theorem 2; Figures 2–3.
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Lemmings · Polynomial building and digging budgets
NP-completePolynomially many terrain-changing commands give a short certificate. Even with only Climber skills, choosing which lemmings to save encodes a satisfiability problem.
NP: completeness [atlas-derived]
- Exact task
- Can at least the required number of lemmings be rescued?
- Scale & input
- Explicit landscape. Builder, Basher and Miner budgets polynomial in landscape size. Other skill budgets may be larger. The total number of lemmings is polynomial in the number of landscape cells.
- Move rules
- All eight classic skills permitted under the given budgets. Deadly zones/traps are permitted. Other unavailable skills have budget zero in the hardness construction.
- Goal
- Rescue at least a supplied number of lemmings.
- Repetition
- Revisits and long waits allowed.
- Players & information
- One player; full deterministic level.
- Randomness
- None.
This concerns saving at least a supplied threshold of lemmings. It does not claim hardness for rescuing every lemming or for one lemming under these bounded budgets.
Why the atlas-derived bound holds
Membership is Theorem 1 for polynomial Builder, Basher and Miner budgets. Theorem 3's construction uses only Climbers, so those three budgets are zero. For a 3CNF formula with m clauses, only the m clause lemmings can be rescued, and a strategy rescuing k corresponds to an assignment satisfying k clauses. Thus a strategy rescuing at least m exists exactly when the formula is satisfiable. Its threshold decision task is NP-hard and in NP in the same model.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
Viglietta, Theorems 1 and 3 and the value-preserving construction in Section 5. NP-completeness of the threshold decision task is a direct consequence, not the wording of Theorem 3.
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One lemming, unlimited skills, no traps or water
Open in the 2015 sourceThe paper asks whether removing deadly zones makes unlimited-skill rescue easier. Its PSPACE-hard construction uses such zones, and does not settle this restriction.
Open question in 2015 source
- Exact task
- Decide whether the single lemming can reach an exit.
- Scale & input
- Explicit scalable pixel landscape, one lemming, unlimited time and all skill types available without a usage bound.
- Move rules
- Original modeled skills and terrain rules. Remove deadly-zone objects, including traps and water. Long falls can still kill.
- Goal
- Save the one lemming.
- Repetition
- Revisits and terrain changes allowed.
- Players & information
- One solver, complete information.
- Randomness
- None.
Open in the cited 2015 source; later status is not established by this record. NP-hardness with limited skills does not classify levels where all skills are unlimited.
Open in the 2015 source
§6, especially Table 3's one-lemming, no-hazards, unlimited-skills column
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Magic: The Gathering
Magic: The Gathering · Forced endgame outcome
UndecidableA forced game encodes a Turing machine and wins exactly when the machine halts.
Undecidable: undecidability
- Exact task
- Will the designated player win this forced endgame?
- Scale & input
- A finite described reachable game state, using the paper’s card pool and rules. Standard-sized decks suffice, while counters and generated tokens have no fixed cap.
- Move rules
- All remaining actions are forced. Tokens, triggered effects and counters simulate a tape and transition rules. Numeric choices use binary notation.
- Goal
- Reach the designated player’s winning effect.
- Repetition
- Nonhalting simulation yields an unavoidable infinite loop, treated as a draw by the cited rule.
- Players & information
- Two players, no remaining strategic choices and no relevant hidden information.
- Randomness
- No stochasticity is used in the construction.
This does not model finite physical memory or tournament clocks, and it freezes the paper’s rules/card text instead of asserting compatibility with every later rules update.
Published result
Theorem 1, §§2, 4.6 and 5.2.
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Allow strategic choices before the forced endgame
Non-arithmetic lower boundThe 2020 preprint proves a stronger lower bound for strategic play.
Arithmetic truth: hardness / lower bound (Arithmetic-truth lower bound for strategic play in the frozen 2020 model. No matching arithmetic upper bound is claimed.)
- Exact task
- Does a player have a winning strategy?
- Scale & rules
- The paper’s Magic rules and unbounded tokens/counters. Binary numeric choices, full information, no randomness.
- Task
- Its mate-in-n target means reaching a forced winning endgame within n strategic moves, not resolving the full game within n turns.
rule change: Allow players to choose numbers before the simulation starts · Magic: The Gathering · Forced endgame outcome. This comparison is not itself a reduction.
Result in cited preprint
Theorems 2–3, §§2–3.
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Can every legal move be executed algorithmically?
Open in the 2021 sourceThe authors conjecture that the transition function is computable.
Open question in 2021 source
- Exact task
- Given a state and a legal move, is the next state computable?
- Scope
- All cards under the paper’s rules. Numerical choices must use binary notation. The input move is promised legal.
This records the question in the cited 2021 source, not a lower bound or a claim that every variant of this game is unresolved.
task change: Ask about one transition instead of the eventual outcome · Magic: The Gathering · Forced endgame outcome. This comparison is not itself a reduction.
Open in the 2021 source
§2, Conjecture 1.
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Is an arbitrary proposed move legal?
Open in the 2021 sourceThe authors conjecture that general move legality is undecidable.
Open question in 2021 source
- Exact task
- Is legality decidable for every proposed Magic move?
- Scope
- The paper’s rules/card pool and binary number convention. This is broader than the separately studied blocking step.
This records the question in the cited 2021 source, not a lower bound or a claim that every variant of this game is unresolved.
task change: Check legality instead of the eventual winner · Magic: The Gathering · Forced endgame outcome. This comparison is not itself a reduction.
Open in the 2021 source
§2, Conjecture 2.
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How far beyond halting-hard?
Open in the 2020 sourceThe lower bound is proved. The matching upper bound is Conjecture 3.
Open question in 2020 source
- Exact task
- Is winning-strategy existence Turing-equivalent to arithmetic truth?
- Scope
- Biderman’s strategic Magic model, full information and binary numeric choices, with the card pool/rules frozen as in the paper.
New card text could alter the conjecture. This is a dated research question, not an established upper bound.
task change: Classify strategic play more precisely · Magic: The Gathering · Forced endgame outcome. This comparison is not itself a reduction.
Open in the 2020 source
§5, Conjecture 3.
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Mancala
Kalah · Fixed 6×4 opening
Fixed board solvedThe first player wins by 10 stones under the cited rules. This is a finite-board solution, not an asymptotic hardness classification.
Fixed: fixed-instance result
- Exact task
- What is the optimal score from the Kalah(6,4) opening?
- Scale & input
- Exactly six pits per player, initially four stones each, and two empty stores.
- Move rules
- Sow counterclockwise from an own nonempty pit. Include your store, skip the opponent's store, and do not skip the starting pit on later laps. Ending in your store gives another move. Ending in an empty own pit captures that stone and the opposite pit, even when the opposite pit is empty.
- Goal
- When either side empties, the other player collects remaining stones. Maximize final score difference.
- Repetition
- Positions cannot repeat under these rules.
- Players & information
- Two players with fully known pit counts.
- Randomness
- None.
This paper does not classify the scalable winner problem. The empty-opposite capture convention matters.
Published result
§2.1 and Table 9.
- Solving KalahGeoffrey Irving, Jeroen Donkers, Jos Uiterwijk (2000).Published journal article · ICGA Journal 23(3) (2000), 139–147Read full text
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Kalah · Scalable board, unary counts
In PSPACEThis scalable winner problem is in PSPACE when the stone counts are unary. Every play has polynomially many sowing choices. This is an upper bound, not a hardness or completeness claim.
PSPACE: upper bound / membership (Atlas-derived membership for unary stone counts. Polynomial-length play follows from the displayed potential argument.) [atlas-derived]
- Exact task
- From an explicitly supplied position, can the designated player force a strictly larger final store count?
- Scale & input
- An explicit board with m pits per player and two stores, a player to move, and nonnegative stone counts encoded in unary. The initial position may be arbitrary. Let S be the total number of stones, including stores.
- Move rules
- Choose a nonempty own pit and sow counterclockwise. Include your store, skip the opponent's store, and do not skip the source pit on later laps. A final stone in your store gives another choice. A final stone in an empty own pit captures that stone and the opposite pit, including when the opposite pit is empty.
- Goal
- When either side empties, collect the remaining stones into the other player's store. The designated player wins exactly when their final store is larger. A tie is not a win.
- Repetition
- No extra repetition restriction is required. The decreasing rank proved below bounds every play by mS(S+1)+S sowings.
- Players & information
- Two players with fully visible counts and deterministic optimal play. Consecutive moves by one player are allowed when the sowing ends in their store.
- Randomness
- None.
The citation supplies the rules, not a published PSPACE theorem. This bound does not prove PSPACE-hardness. The unary length assumption matters to this particular proof.
Why the atlas-derived bound holds
Let C count stones in the two stores, and let Φ sum each non-store stone's distance (1 through m) along its own row to its own store. Then 0≤Φ≤mS. A move that stores no stone never passes its own store, so sowing k stones reduces Φ by 1+⋯+k. A move that stores a stone increases C by at least one. Consequently R=(S−C)(mS+1)+Φ is a nonnegative integer that decreases every move. Initially R≤mS(S+1)+S. A depth-first minimax traversal stores a polynomial number of polynomial-size positions when S is unary. Bonus turns only change which player selects the next child. The final collection of remaining stones cannot lengthen play.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
Rules: Irving, Donkers and Uiterwijk, §2.1. Atlas-derived upper bound: between moves that add to a store, the sum of each stone's distance to its own store decreases. There are at most S store-increasing moves and at most mS intervening sowings per phase. Depth-first minimax therefore uses polynomial space for unary counts.
- Solving KalahGeoffrey Irving, Jeroen Donkers, Jos Uiterwijk (2000).Published journal article · ICGA Journal 23(3) (2000), 139–147Read full text
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Awari: a fixed board is a draw
Fixed board solvedThe cited ruleset was strongly solved: perfect play draws. This does not classify growing-board Awari or other Oware rules.
Fixed: fixed-instance result
- Exact task
- Determine optimal play on the fixed Awari board.
- Scale & rules
- Two rows of six pits, four stones each. Sow from an own pit counterclockwise, skipping the starting pit on a lap. A last stone making an enemy pit hold two or three captures it and consecutive preceding enemy pits with two or three.
- Ending & repetition
- Choose a move leaving the opponent a stone whenever possible. Otherwise a grand slam captures every opposing stone, and the mover also collects all remaining stones because the opponent cannot move. Repetition ends play and divides uncollected stones equally, allowing half-stones. The companion rules note specifies threefold repetition.
- Players & goal
- Two players, known counts, no randomness. Capture more than 24 of the 48 stones to win.
Traditional Oware capture and ending conventions differ.
rule change: Change capture, feeding and repetition rules · Kalah · Fixed 6×4 opening. This comparison is not itself a reduction.
Published result
Romein–Bal §1 and Appendix; Donkers companion note, rules 4–8. The solution does not use the rule that cancels grand-slam captures.
- Awari Is SolvedJohn W. Romein, Henri E. Bal (2002).Published journal article · ICGA Journal 25(3) (2002)Open manuscript not recorded. The stored link is the journal issue contents; a direct open manuscript is not recorded.
- Comments on the Awari SolutionJeroen Donkers (2002).Published journal article · ICGA Journal 25(3) (2002)Open manuscript not recorded. The stored link is the journal issue contents; a direct open manuscript is not recorded.
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Tchoukaillon: solitaire sweep solvability
P (algorithmic consequence)A winning play must always use the nearest harvestable bin. This gives a polynomial-time solvability test.
P: polynomial-time algorithm [atlas-derived]
- Exact task
- Given a bin vector, can every stone reach the store?
- Scale & rules
- Input is an explicit vector of k nonnegative binary counts. Bin i is i steps from the store. Only a bin containing exactly i stones can be sown, depositing one in each earlier bin and finally the store.
- Players & goal
- One player, full information, no randomness. Empty all bins.
- Repetition
- Each move permanently stores one stone, so no repeated state.
P follows from the cited characterization: reject any count above its bin index, then simulate at most k(k+1)/2 forced moves. This is not a winner result for two-player Kalah.
rule change: Replace adversarial play with an all-sowings-end-in-store solitaire task · Kalah · Fixed 6×4 opening. This comparison is not itself a reduction.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
§1.1 and Theorem 2.1.
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Dakon: win before the opponent moves
Fixed openings solvedThe paper finds winning openings for Dakon-n at n=2 and 4 through 18, but none at n=3. These finite searches do not establish a complexity class.
Fixed: fixed-instance result
- Exact task
- Can the initial player capture enough stones during the first turn to prevent another round?
- Scale & rules
- Dakon-n has n pits per side, initially n stones per pit, and stores. Sow counterclockwise through your store, skipping the opponent's. If the last stone lands in a nonempty pit, pick that pit up and keep sowing. Ending in your store grants another choice. Ending in an empty own pit captures it and the opposite pit.
- Ending & information
- Two players, known counts, deterministic. Rounds refill pits with n stones each and skip unfillable pits. A player unable to fill one pit loses. This record concerns only the terminating first-turn construction, not general repetition behavior.
rule change: Add relay sowing and round-based refill rules · Kalah · Fixed 6×4 opening. This comparison is not itself a reduction.
Result in cited source
Rules pp. 79–80; winning openings pp. 81, 83–84.
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Captureless Kalah: still in PSPACE for unary counts
In PSPACE (upper bound)Removing capture does not break the polynomial play-length bound. Ordinary sowing still permanently adds stones to stores, and all other sowings move stones closer to their own store.
PSPACE: upper bound / membership (Atlas-derived membership for unary stone counts. Polynomial-length play follows from the displayed potential argument.) [atlas-derived]
- Exact task
- Can the designated player force a larger final store count after removing the opposite-pit capture rule?
- Scale & input
- Explicitly list m pits per side, two stores, whose turn it is and unary stone counts. Arbitrary initial positions are allowed.
- Rule change
- Use the Kalah rules in the linked unary winner record but disable the last-stone/opposite-pit capture. Keep sowing into your own store, skipping the opponent's store, bonus choices for a last stone in your store, and final collection when one side empties.
- Players & goal
- Two players, fully visible counts, no randomness. Finish with strictly more stored stones than the opponent. A tie is not a win.
- Repetition
- No repetition rule is needed: the same rank R=(S−C)(mS+1)+Φ strictly decreases.
The polynomial-space result establishes membership only, not PSPACE-hardness or completeness.
rule change: Disable the opposite-pit capture · Kalah · Scalable board, unary counts. This comparison is not itself a reduction.
Why the atlas-derived bound holds
The unary Kalah proof never relies on a capture taking place. Moves that deposit stones in stores decrease S−C. Every other move stays within the mover's own row and decreases Φ. Therefore the identical polynomial rank and minimax-space bound apply.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
Public rules are in §2.1. This is an explicitly defined captureless modification and a derived upper bound, not an attributed published result.
- Solving KalahGeoffrey Irving, Jeroen Donkers, Jos Uiterwijk (2000).Published journal article · ICGA Journal 23(3) (2000), 139–147Read full text
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Kalah with binary counts: an EXPTIME upper bound
In EXPTIME (upper bound)Binary notation can describe exponentially many stones. A finite-state search gives an EXPTIME upper bound. This weaker bound does not show that binary Kalah is harder or rule out PSPACE.
EXPTIME: upper bound / membership (Atlas-derived exponential-time membership for binary stone counts. Polynomial-size states alone do not prove PSPACE membership.) [atlas-derived]
- Exact task
- Decide the same winner problem when each pit count is written in binary.
- Scale & input
- The m-pit rows are explicitly listed. All pit and store counts are nonnegative binary integers, with their total S. The player to move and designated player are specified.
- Move rules
- Identical to the unary Kalah winner record, including captures, bonus choices, no source-pit skip and final collection. Only the encoding changes.
- Players & goal
- Two players, fully visible counts, deterministic play. The designated player must finish with a strictly larger store.
- Repetition
- The same decreasing rank guarantees termination, but its mS(S+1)+S bound is not polynomial in binary input length.
A polynomial-size position by itself does not imply a polynomial-space winner algorithm. No EXPTIME-hardness, separation from unary Kalah, or optimality of this upper bound is claimed.
encoding change: Write stone counts in binary · Kalah · Scalable board, unary counts. This comparison is not itself a reduction.
Why the atlas-derived bound holds
For input length L, m≤L and log(S+1)=O(L). At most 2(S+1)^(2m+2)=2^poly(L) states specify all counts and the player to move. There are at most m successors per state. A complete sowing can be computed without iterating over individual stones: divide its count by the 2m+1 visited positions, add the quotient to each, and distribute the remainder along the sowing order, then apply the last-stone rule. Arithmetic is polynomial in L. The decreasing rank makes the graph acyclic, so backward induction over its exponential number of states takes exponential time.
Atlas-derived consequence
The derivation is stated in this record, rather than quoted as a theorem from the cited work.
§2.1 supplies the rules. The upper bound is derived here by enumerating the finite game graph, with at most 2(S+1)^(2m+2) positions and a polynomial-time move calculation using quotient and remainder.
- Solving KalahGeoffrey Irving, Jeroen Donkers, Jos Uiterwijk (2000).Published journal article · ICGA Journal 23(3) (2000), 139–147Read full text
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Mastermind
Mastermind · Do the guesses fit any code?
NP-completeConsistency is hard even with only two colors.
NP: completeness
- Exact task
- Does at least one secret code produce every supplied guess rating?
- Scale & input
- Code length and explicit guess list grow. Two colors already suffice.
- Move rules
- Feedback gives correct positions and correct colors in wrong positions. Repeated colors allowed.
- Goal
- Find whether a compatible code exists.
- Repetition
- Static completion, no play-history restriction.
- Players & information
- Offline consistency problem. Guesses and ratings known, code hidden.
- Randomness
- None.
Not optimal adaptive codebreaking. In a truthful live game, consistency is already promised.
Published result
§2; Lemma 5 and Corollary 1.
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Count all compatible codes
#P-completeCounting remains hard with just two colors.
#P: completeness (Counting consistent secrets. A #P result, not winner determination or mere existence.)
- Exact task
- Count secret codes consistent with the supplied ratings.
- Scale & input
- Growing code length and explicit query list. Two colors suffice.
- Move rules
- Usual black and white feedback, repeated colors allowed.
- Goal
- Count compatible ordered codes.
- Repetition
- Static completion, no play-history restriction.
- Players & information
- Offline list known, hidden code not given.
- Randomness
- None.
task change: Mastermind · Do the guesses fit any code?. This comparison is not itself a reduction.
Published result
Theorem 1 and Lemma 5.
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Only exact-position feedback
NP-completeRemoving misplaced-color feedback leaves two-color consistency hard.
NP: completeness
- Exact task
- Decide whether supplied guesses and exact-position counts have a compatible code.
- Scale & input
- Growing length and query list. Two colors suffice.
- Move rules
- Feedback counts matching positions only.
- Goal
- A compatible ordered code exists.
- Repetition
- Static completion, no play-history restriction.
- Players & information
- Offline queries and ratings known.
- Randomness
- None.
rule change: Mastermind · Do the guesses fit any code?. This comparison is not itself a reduction.
Published result
Lemma 4 and Corollary 1.
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Ignore order and fix the color count
PA fixed palette gives only polynomially many color-count vectors to check.
P: polynomial-time algorithm
- Exact task
- Count compatible unordered codes.
- Scale & input
- Growing code length, constant number of colors, explicit query list.
- Move rules
- Feedback counts color matches ignoring position, including exact-position matches.
- Goal
- Count compatible color multisets.
- Repetition
- Static completion, no play-history restriction.
- Players & information
- Offline queries and ratings known.
- Randomness
- None.
The paper’s white-only score is total color overlap, not ordinary white feedback alone.
rule change: Mastermind · Do the guesses fit any code?. This comparison is not itself a reduction.
Published result
§2 definition of MSP-White; Lemma 1.
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Nim
Nim · Take the last counter
PCompute the bitwise XOR of the heap sizes. A nonzero result means a winning move exists. This takes polynomial time for binary or unary input.
P: polynomial-time algorithm
- Exact task
- Decision: does the player to move have a winning strategy?
- Scale & input
- A scalable explicit list of heap sizes, in binary or unary. A binary heap h uses O(log(h+1)) bits, but the input also includes every heap.
- Move rules
- Remove any positive number of counters from one heap.
- Goal
- Take the last counter.
- Repetition
- Impossible: play permanently consumes resources.
- Players & information
- Two players; all information visible.
- Randomness
- None.
The XOR algorithm makes Nim easy. Compact binary notation alone does not imply efficient solvability, and changing to unary does not make ordinary Nim hard.
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Nim · Avoid the last counter
PThe same XOR test works until every nonempty heap has size one. Then parity decides the outcome.
P: polynomial-time algorithm
- Exact task
- Decision: does the player to move have a winning strategy?
- Scale & input
- A scalable explicit list of heap sizes, in binary or unary.
- Move rules
- Remove any positive number from one heap.
- Goal
- Make the opponent take the last counter.
- Repetition
- Impossible: play permanently consumes resources.
- Players & information
- Two players; all information visible.
- Randomness
- None.
With only singleton heaps remaining, an even positive number of heaps is winning.
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Neighboring Nim: heaps of at most 1
PHeaps are 0/1. For direct equivalence with undirected vertex Geography, the previous-move vertex starts empty. Winner determination is polynomial-time.
P: polynomial-time algorithm
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input: explicit undirected graph, current vertex, heap at each vertex. Move to the same or an adjacent vertex and decrease that heap. Graph size scales.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every move strictly decreases one heap. No pass is permitted and no full state repeats.
- Randomness
- None.
equivalence: Geography · Two-way · no vertex reuse.
Published result
Definitions 2.1–2.2; discussion immediately after Definition 2.2
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Neighboring Nim: heaps of at most 2
PSPACE-completeHeaps are 0/1/2. PSPACE-complete by directed Geography reduction. Each arc becomes a seven-vertex internal gadget with one heap of 2, all others 1.
PSPACE: completeness
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input: explicit undirected graph, current vertex, heap at each vertex. Move to the same or an adjacent vertex and decrease that heap. Graph size scales.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every move strictly decreases one heap. No pass is permitted and no full state repeats.
- Randomness
- None.
rule change: Allow heaps of size 2 · Neighboring Nim: heaps of at most 1. This comparison is not itself a reduction.
Published result
Theorem 3.1 and Corollary 3.5
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Neighboring Nim: large binary heaps
PSPACE-hard; in EXPTIMEArbitrary binary heaps preserve PSPACE-hardness. An exponential-time upper bound follows by evaluating all positions. Whether this can be improved to polynomial space is Open Problem 6.1 in the 2017 source.
PSPACE: hardness / lower bound · EXPTIME: upper bound / membership (Atlas-derived finite-state exponential-time upper bound under this record’s explicit rules.) [atlas-derived] · Open question in 2017 source
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input: explicit undirected graph, current vertex, heap at each vertex. Move to the same or an adjacent vertex and decrease that heap. Graph size scales.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every move strictly decreases one heap. No pass is permitted and no full state repeats.
- Randomness
- None.
PSPACE membership was open in the 2017 source; later status is not established by this record. The EXPTIME upper bound is derived from its rules, not claimed as a named theorem in that paper.
restriction: Bound every heap by 2 · Neighboring Nim: heaps of at most 2. This comparison is not itself a reduction.
Open in the 2017 source · derived upper bound
The PSPACE-hardness is from the source. The EXPTIME upper bound is a finite-state consequence derived in the atlas.
Section 3.2 and Open Problem 6.1. Derived EXPTIME upper bound: there are at most |V| times the product of (heap size + 1) states. Each state has polynomially many bits, move enumeration is exponential at most, and every move decreases the total counters.
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Three-pile Nim with one shared pass
Open characterizationExactly three heaps and one shared pass, unavailable after all heaps empty. A May 2026 primary source still identifies the general losing-position formula as open. This does not establish PSPACE-hardness.
Open question in 2026 source
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input: binary heap sizes and whether a shared pass is unused. Make a normal Nim move or an allowed pass. A pass is consumed globally.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every move decreases a heap or uses the single pass. No repeated state.
- Randomness
- None.
This is an open characterization question in the May 2026 cited source. It is not itself a complexity lower bound.
rule change: Add one nonterminal pass · Nim · Take the last counter. This comparison is not itself a reduction.
Open in the 2026 source
2025 Definition 19 defines the rules. The 2026 preprint's abstract explicitly reaffirms the three-pile characterization question.
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Nim with a pass only while a pile has at least 2
PChanging pass availability restores a polynomial-time XOR test. XOR all heaps with a bit that is 1 exactly when a pass is unused and some heap is at least 2. Otherwise that bit is 0.
P: polynomial-time algorithm
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input: binary heap sizes and whether a shared pass is unused. Make a normal Nim move or an allowed pass. A pass is consumed globally.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every move decreases a heap or uses the single pass. No repeated state.
- Randomness
- None.
rule change: Pass threshold changes from 1 to 2 · Three-pile Nim with one shared pass. This comparison is not itself a reduction.
Published result
Corollary 2 with every threshold m_i=2
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Demi-quantum Boolean Nim
PSPACE-completeExplicit 0/1 matrix: rows are configurations, columns are heaps. Only single ordinary moves are permitted, with illegal rows discarded. PSPACE-complete and polynomially isomorphic to Avoid True: each row maps to a clause on its zero columns.
PSPACE: completeness
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Explicit scalable 0/1 matrix. Rows are alternative positions on named heaps. Choose one heap, remove its stick from each legal row, discard illegal rows. No superposed moves.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Each move removes one counter in every surviving row and discards rows where that move is illegal. No pass is allowed.
- Randomness
- None. Alternative configurations are deterministic, not random measurements.
equivalence: Avoid True.
Published result
Theorems 3–4
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Quantum Boolean Nim from a superposition
Σ₂^P-hard; in PSPACEExplicit polynomial-width list of 0/1 heap configurations. Proved Σ₂^P-hard, and in PSPACE by Theorem 7.1. Completeness is unresolved in the source. The rule change allows superposed moves.
Σ₂^P: hardness / lower bound · PSPACE: upper bound / membership
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input configurations share named heaps. Flavor D, width 2: select one ordinary move or two alternative moves, each legal somewhere. Retain every legal successor branch. This is deterministic, not probabilistic quantum mechanics.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every surviving branch removes one counter per turn. No pass is allowed.
- Randomness
- None. Superposed moves are deterministic branching rules.
rule change: Allow superpositions of moves · Demi-quantum Boolean Nim. This comparison is not itself a reduction.
Result in cited preprint
FUN Theorems 6–8; full version Theorem 7.1 and §8.1
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Quantum Nim: heaps of at most 2, ordinary start
In PSPACE; exact complexity open in 2020 sourceSingle ordinary starting configuration, scalable heaps of size 0/1/2. In PSPACE. Open Question 8.3 asks whether it is polynomial-time solvable or PSPACE-complete, without asserting either.
PSPACE: upper bound / membership · Open question in 2020 source
- Exact task
- Decide whether the next player wins under optimal play.
- Scale & input rules
- Input configurations share named heaps. Flavor D, width 2: select one ordinary move or two alternative moves, each legal somewhere. Retain every legal successor branch. This is deterministic, not probabilistic quantum mechanics.
- Players & information
- Two alternating players, complete information.
- Goal
- The player with no legal move loses.
- Repetition
- Every surviving branch strictly loses counters each turn. No pass is allowed.
- Randomness
- None. Superposed moves are deterministic branching rules.
Open question in the cited 2020 source. Later status is not established by this record.
rule change: Different starting-position promise, not a reduction · Quantum Boolean Nim from a superposition. This comparison is not itself a reduction.
Open in the 2020 source
Theorem 7.1; §8.1, Open Question 8.3
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Quantum Boolean Nim from an ordinary position
PAn ordinary starting position is easy by parity. A supplied superposition of such positions has a proved hardness lower bound.
P: polynomial-time algorithm
- Exact task
- Determine the winner from one ordinary 0/1-heap position.
- Scale & rules
- Scalable list of heaps, each 0 or 1. Flavor D, width 2: choose one ordinary move or two alternatives, each legal somewhere. Retain legal successor branches.
- Players & goal
- Two alternating players, perfect information. The player with no legal move loses.
- Repetition & randomness
- Each branch removes a stick per turn. No repeated positions or randomness. These are deterministic quantum-inspired rules.
restriction: Start with one ordinary position instead of a supplied superposition · Quantum Boolean Nim from a superposition. This comparison is not itself a reduction.
Result in cited preprint
Section 8.1, paragraph preceding Open Question 8.3.
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Long global Nim: explicit move set
In PSPACE (derived upper bound)A subtraction size can be used only once across all piles. With S explicitly listed, at most |S| moves remain. Depth-first game-tree evaluation uses polynomial space, for binary or unary integers.
PSPACE: upper bound / membership (Atlas-derived polynomial-space bound for an explicitly listed move set S, in unary or binary; excludes a succinct unexpanded interval.) [atlas-derived]
- Exact task
- Determine whether the player to move can win from supplied piles and available subtraction sizes.
- Scale & input
- An explicit list of nonnegative pile sizes and a finite explicit set S of positive integers. Pile count and |S| scale. Values may be binary or unary.
- Move rules
- Choose s in S and one pile containing at least s counters. Subtract s from that pile and delete s from S globally.
- Goal
- Normal play: the player with no legal move loses.
- Repetition
- No move size is reused. Every turn reduces |S| by one.
- Players & information
- Two alternating players, with all piles and S visible.
- Randomness
- None.
This record establishes PSPACE membership only, without a matching lower bound or polynomial-time algorithm. The bound does not cover a succinct interval S={1,...,M} when M is binary and the interval is not expanded.
rule change: Supply allowed move sizes and forbid their reuse globally · Nim · Take the last counter. This comparison is not itself a reduction.
Why the atlas-derived bound holds
Each move deletes one member of the explicitly listed set S, so a play has at most |S| turns. Pile sizes only decrease and never require more bits than the input. A state contains the piles, a used-size bitmask and the player to move. A depth-first minimax stack of at most |S| such states uses polynomial space. This argument applies with unary or binary numbers, but not to an unexpanded binary-sized interval of allowed moves.
Atlas-derived upper bound
The paper supplies the game rules. The explicit-move-set PSPACE argument is given by the atlas, not attributed as a theorem to the paper.
Definition 22 supplies the rules. Upper-bound derivation: depth is at most |S|. A state stores the piles and a used-size bitmask, each polynomial in input length. A depth-first minimax stack is therefore polynomial space.
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Quantum Unary Nim: ordinary start
In PSPACE; classification open in 2020 sourceTheorem 7.1 gives PSPACE membership. Open Question 8.3 asks whether this problem has a polynomial-time algorithm or is PSPACE-complete. Ordinary unary Nim still has the XOR solution.
PSPACE: upper bound / membership · Open question in 2020 source
- Exact task
- Determine the winner from one ordinary position with unary-encoded heaps under the specified quantum-inspired rules.
- Scale & input
- A scalable explicit list of heaps, with every heap size written in unary. The initial position is one ordinary configuration, not a supplied superposition.
- Move rules
- Flavor D, width 2: choose one ordinary Nim move or two alternative Nim moves, each legal in some current branch. Retain all legal successor branches.
- Goal
- Normal play: the player with no legal move loses.
- Repetition
- Every surviving branch decreases its counter total. The maximum play length is at most the initial unary counter total.
- Players & information
- Two alternating players. The initial position and all choices are public.
- Randomness
- None. The branching rules are deterministic and do not model random measurements.
The cited 2020 source poses P versus PSPACE-complete as an open question, not a proved dichotomy. Later status is not established by this record.
restriction: Restrict each unary heap to 0, 1 or 2 · Quantum Nim: heaps of at most 2, ordinary start. This comparison is not itself a reduction.
Open in the 2020 source
Theorem 7.1 and Section 8.1, Open Question 8.3. The membership theorem handles exponentially many possible branches without storing all of them at once.
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Portal
Portal · Cubes, floor buttons and doors
PSPACE-completePortals are not needed: a cube placed on either of two floor buttons stores a bit that controls doors elsewhere.
PSPACE: completeness
- Exact task
- Can the player reach the exit in the cube-and-door model?
- Scale & input
- Arbitrary-size discretized 3D level; fixed-point positions/velocities and polynomially described state.
- Move rules
- Pick up cubes, press weighted buttons, pass connected doors. No Portal Gun or long-fall mechanic required.
- Goal
- Reach the level exit.
- Repetition
- Backtracking and repeated switch changes allowed.
- Players & information
- One player; complete level.
- Randomness
- None in the construction.
The result concerns the specified generalized finite-state model, not the finite set of shipped test chambers.
Published result
Section 8, Theorem 13 and Metatheorem 11; Figure 9. NCL edge reversal reduction.
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Portal · Timed buttons and doors
NP-hard; in PSPACEThe exit opens only while every timer is active. Reaching it becomes a race to visit every switch without wasting a detour.
NP: hardness / lower bound · PSPACE: upper bound / membership (Membership applies to the source’s finite-state game model and stated encoding assumptions.)
- Exact task
- Can the player reach the exit while the required timed doors are open?
- Scale & input
- Explicit arbitrary-size level with configurable timer lengths. Polynomial-bit finite-precision state and polynomially bounded velocities, as in Theorem 12.
- Move rules
- Walk between timed buttons; buttons temporarily open connected doors. No Portal Gun needed.
- Goal
- Reach the exit before required doors close.
- Repetition
- Revisits allowed, but timers can make them fatal to a successful route.
- Players & information
- One player; full level and timer settings.
- Randomness
- None in the construction.
Configurable timers are required for hardness. PSPACE membership assumes the paper's finite-precision model, polynomially bounded velocities and polynomial-bit state. This does not assert NP-completeness.
Published result
Demaine, Lockhart and Lynch, Corollary 8 (hardness) and Theorem 12 (PSPACE membership).
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Lasers, relays and moving platforms
PSPACE-completeLaser-controlled platforms implement reusable memory latches.
PSPACE: completeness
- Exact task
- Decide whether the player can reach the exit.
- Scale & input
- Growing discretized 3D level, polynomially described finite state.
- Move rules
- Portals redirect lasers. Relays control moving platforms.
- Goal
- Reach the exit.
- Repetition
- Backtracking and repeated latch changes allowed.
- Players & information
- One player. Entire level known.
- Randomness
- None.
rule change: Portal · Cubes, floor buttons and doors. This comparison is not itself a reduction.
Published result
§9, Theorem 14.
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High-energy pellets and their catchers
NP-hard; in PSPACE (finite-precision model)Activating every catcher within the constructed timing constraint encodes a grid Hamiltonian cycle.
NP: hardness / lower bound · PSPACE: upper bound / membership (Atlas-derived membership for the stated finite-precision, polynomial-bit engine model.) [atlas-derived]
- Exact task
- Decide whether the player can reach the exit.
- Scale & input
- Explicit growing level with pellet launchers, catchers, portals and catcher-controlled doors. Positions, velocities and timer states have polynomial-bit finite-precision encodings. Velocities are polynomially bounded, as assumed by the paper's PSPACE argument.
- Move rules
- Redirect pellets through portals. Pellets reflect and can kill the player. A launcher replaces its pellet only after the prior pellet is destroyed, so the number of active pellets remains bounded by the input.
- Goal
- Reach exit.
- Repetition
- Revisits allowed, timing can prevent a successful detour.
- Players & information
- One player. Entire level known.
- Randomness
- None.
NP-hardness is a published theorem. The PSPACE upper bound applies to the specified finite-precision engine with polynomially bounded velocities and polynomial-bit state. It is not a claim about exact real-valued physics or NP-completeness.
rule change: Portal · Cubes, floor buttons and doors. This comparison is not itself a reduction.
Why the atlas-derived bound holds
The fixed level, player, doors and bounded number of active pellets have a polynomial-size state under the stated precision and velocity assumptions. A launcher creates a new pellet only when its previous pellet has disappeared. Thus the state graph has at most exponentially many vertices, and a successor can be computed in polynomial space. Nondeterministically search for an exit-reaching path while storing only the current state and a polynomial-bit step counter permitting exponentially many steps, then use NPSPACE=PSPACE. This is the argument given for the broader engine in the proof of Theorem 12, applied to the pellet-only mechanics.
Published hardness · atlas-derived upper bound
The source proves NP-hardness. The stated finite-state model gives the PSPACE upper bound.
Section 7, Theorem 9 proves NP-hardness. The finite-precision PSPACE upper bound is derived from Theorem 12's state-space argument, whose proof explicitly discusses pellets and their launchers, although its displayed mechanics list omits them.
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Only the portal gun
Open in the 2018 sourceThe authors explicitly leave the portals-only mechanics set unresolved.
Open question in 2018 source
- Exact task
- Reach an exit using only portal placement and ordinary movement.
- Scale & input
- Scalable polygonal level with fixed-point coordinates and discrete time.
- Move rules
- One portal pair, ordinary movement, no other puzzle devices.
- Goal
- Reach the target.
- Repetition
- Allowed.
- Players & information
- One player, complete information.
- Randomness
- None.
Open in the cited 2018 source. Later status is not established by this record.
Open in the 2018 source
§10.1; model in §2
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Only movable cubes
Open in the 2018 sourceCubes alone are an explicit open mechanics set. The proved cube result also uses buttons and doors.
Open question in 2018 source
- Exact task
- Reach an exit using movable cubes and ordinary movement.
- Scale & input
- Scalable polygonal level, discrete fixed-point model.
- Move rules
- Carry cubes and use them as platforms. No other puzzle devices.
- Goal
- Reach the target.
- Repetition
- Allowed.
- Players & information
- One player, complete information.
- Randomness
- None.
Open in the cited 2018 source. Later status is not established by this record.
restriction: Remove buttons and doors · Portal · Cubes, floor buttons and doors. This comparison is not itself a reduction.
Open in the 2018 source
§10.1; model in §2
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Rubik's Cube
Rubik's Cube · Find any solution
Polynomial-time searchA scalable, legally scrambled cube can be solved efficiently if the solution need not be shortest.
P: polynomial-time algorithm (Polynomial-time search under the exact input and move rules of this record.)
- Exact task
- Find a sequence of legal turns that solves the supplied reachable, generalized Rubik’s Cube configuration.
- Scale & input
- n × n × n cube, n grows; legal scramble promised.
- Move rules
- Rotate individual slices.
- Goal
- Restore solved face colors.
- Repetition
- Turns may be undone; no repetition ban.
- Players & information
- One player; all stickers known.
- Randomness
- None.
This is a search guarantee on valid scrambles, not a decision about arbitrary sticker colorings.
Published result
Constructive §4.1, theorem 3.
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Rubik's Cube · Can it be solved within k turns?
NP-completeRequiring the shortest route makes a different computational problem.
NP: completeness
- Exact task
- Can the supplied generalized Rubik’s Cube configuration be solved in at most the given number k of turns?
- Scale & input
- Variable n × n × n cube plus k.
- Move rules
- One slice per move; slice-turn or slice-quarter-turn metric.
- Goal
- Reach solved colors using at most k moves.
- Repetition
- Revisits allowed.
- Players & information
- One player; complete configuration known.
- Randomness
- None.
Not the fixed 3 × 3 × 3 cube, and not whole-cube rotations as free slice moves.
Published result
Problem definitions §2; theorem 5.3.
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Long thin cuboid: shortest solution
P searchFixing two dimensions permits optimal solving in polynomial time.
P: polynomial-time algorithm (Polynomial-time search under the exact input and move rules of this record.)
- Exact task
- Find a shortest solving sequence.
- Scale & input
- c1×c2×n cuboid. c1 and c2 are constants, n grows. Reachable configuration.
- Move rules
- Legal slice turns preserving the cuboid, under the paper’s move model.
- Goal
- Minimize number of turns to solve.
- Repetition
- Reversible turns, revisits allowed.
- Players & information
- One player. All sticker colors known.
- Randomness
- None.
restriction: Rubik's Cube · Can it be solved within k turns?. This comparison is not itself a reduction.
Published result
§6, Theorem 6.
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Turn several layers together
Open in the 2018 sourceThe shortest-solution theorem covers one slice at a time. The authors ask what happens when one move rotates a contiguous bundle of layers touching a face.
Open question in 2018 source
- Exact task
- Decide whether a scalable n × n × n cube can be solved within k wide turns.
- Scale & input
- Growing cube, explicit sticker configuration and move budget k.
- Move rules
- WTM permits quarter or half turns of a face-adjacent bundle. WQTM counts only quarter turns.
- Goal
- Make faces monochromatic within k moves.
- Repetition
- Repeated states allowed, subject to the budget.
- Players & information
- One solver, complete information.
- Randomness
- None.
Open in the cited 2018 source. Later status is not established by this record.
rule change: Count a wide turn as one move · Rubik's Cube · Can it be solved within k turns?. This comparison is not itself a reduction.
Open in the 2018 source
§6 Future work
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Sokoban
Sokoban · Crates onto targets?
PSPACE-completePushing crates is powerful enough to simulate a constraint graph. Hardness even survives removing fixed barrier pieces from the scalable puzzle.
PSPACE: completeness
- Exact task
- Decision: can every target square be occupied by a crate?
- Scale & input
- Growing finite grid with a walker, unit crates and an equal number of target cells. This record uses the no-fixed-barrier construction.
- Move rules
- Walker moves one orthogonal step, pushing at most one crate if the next cell is empty; pulling prohibited.
- Goal
- Crates occupy all targets.
- Repetition
- Repeated positions allowed; no repetition ban
- Players & information
- 1 player; Complete: entire input is known
- Randomness
- None
No fixed barriers does not mean no obstacles: packed crates can be immovable. Reduction starts from NCL configuration-to-configuration, not directly the edge task.
Published result
Robert A. Hearn: §4.3, Theorem 22
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Just get the player to an exit
PSPACE-completeReplacing crate-storage goals with one player destination still leaves a PSPACE-complete problem.
PSPACE: completeness
- Exact task
- Push-1F: can the player reach a designated cell?
- Model & rules
- Growing explicit rectangular grid with unit crates and fixed walls. Orthogonal steps may push one crate by one cell into empty space. No pulling or gravity. One player, full information, no randomness. Positions may repeat.
A related rule-and-goal variant called Push-1F, not the ordinary Sokoban storage task.
rule change: Remove internal fixed walls · Reach an exit without fixed walls. This comparison is not itself a reduction.
Published result
Sections 1 and 6.
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Reach an exit without fixed walls
PSPACE-completeThe no-fixed-wall reachability question was settled by a 2025 proof, revised in 2026.
PSPACE: completeness
- Exact task
- Push-1: can the player reach the specified destination?
- Model & rules
- Growing explicit rectangular grid, unit movable crates, start and destination. Orthogonal moves push at most one crate one cell. No pulling, gravity or internal fixed walls. Repeated positions allowed. One player, complete information, no randomness.
Clusters of movable crates can still be immobile. Older sources that list this task as open are outdated.
Published result
Sections 1 and 6 of arXiv v2.
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Sliding blocks with a unique route to the exit
Hardness conjectured in the 2025 sourceThe paper proves hardness for a storage-goal uniqueness promise, then conjectures analogous hardness when the objective is reaching an exit. Ordinary walks between pushes do not distinguish solutions.
Open question in 2025 source
- Exact task
- Decide solvability of PushPush pathfinding with at most one successful ordered sequence of pushes.
- Scale & input
- Explicit scalable bounded square grid, block positions, player start and exit.
- Move rules
- Push one adjacent block until it hits another block or the boundary. No momentum transfer.
- Goal
- Move the player to the exit, rather than cover storage sites.
- Repetition
- At most one successful push sequence is promised.
- Players & information
- One solver, complete information.
- Randomness
- None in gameplay. The proposed hardness comparison permits randomized reductions.
The cited 2025 source conjectures no polynomial algorithm unless NP = RP. This is not a proved PSPACE-hardness result for the promised version. Later status is not established by this record.
rule change: Blocks slide to obstacles; require a unique push sequence to an exit · Sokoban · Crates onto targets?. This comparison is not itself a reduction.
Open in the 2025 source
§4 Concluding Remarks
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Tetris
Tetris · Know every upcoming piece
NP-completeSeeing the whole piece sequence still leaves hard packing choices.
NP: completeness
- Exact task
- Can the known sequence of pieces be placed to clear at least k rows without losing?
- Scale & input
- Growing rectangular board, initial filled cells, explicit finite piece sequence, row target k.
- Move rules
- Translate/rotate falling tetrominoes; complete rows clear.
- Goal
- Clear at least k rows without losing.
- Repetition
- Finite supplied sequence; no repeating history rule.
- Players & information
- One player; all future pieces visible.
- Randomness
- None.
Uses the paper's rotation models, not an assertion about every modern implementation or a fixed 10 × 20 board.
Published result
Model §2, theorem 2 and §3 reduction.
- Tetris is Hard, Even to ApproximateRon Breukelaar, Erik D. Demaine, Susan Hohenberger, Hendrik Jan Hoogeboom, Walter A. Kosters, David Liben-Nowell (2004).Published journal article · International Journal of Computational Geometry & Applications (2004)Read full text
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Only two columns
PTwo columns admit a polynomial-time algorithm.
P: polynomial-time algorithm
- Exact task
- Decide whether the known piece sequence can survive, optionally clearing the board.
- Model & rules
- Explicit initial board of variable height, fixed width two, explicit tetromino sequence. One informed player, no randomness. Pieces fall, rotate and lock. Full rows disappear. Section 2 uses partial lock-out and specified rotation restrictions.
restriction: Restrict the width from eight columns to two · Eight columns still suffice. This comparison is not itself a reduction.
Published result
Sections 2–3, Theorem 3.2. The theorem also permits larger polyominoes.
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Eight columns still suffice
NP-completeA fixed width of eight preserves NP-completeness.
NP: completeness
- Exact task
- Decide survival or complete board clearing for the supplied sequence.
- Model & rules
- Height, initial occupied cells and finite tetromino sequence are input. One informed player, no randomness. Section 2 rotation and partial lock-out rules apply. This is not a fixed 10 × 20 board.
Published result
Sections 2 and 4, Theorem 4.1.
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Only one tetromino shape
NP-complete (explicit sequence)A 2026 result gives hardness for each single shape I, T, S, Z, J and L.
NP: completeness (Published NP-hardness plus polynomial certificate verification for an explicit sequence or unary piece count.) [atlas-derived]
- Exact task
- Decide clearing or survival for the supplied finite run of one piece type.
- Scale & input
- Explicit growing board, initially filled cells and prescribed number of identical pieces.
- Move rules
- Super Rotation System (SRS), including wall kicks.
- Goal
- Clear the board, or place all pieces without losing.
- Repetition
- Pieces consumed in order.
- Players & information
- One player. Entire input known.
- Randomness
- None.
- Encoding
- The finite sequence is explicitly listed, equivalently a unary count of identical pieces. A compact binary count is a different input model.
O (square) is excluded. The rotation model is essential. Source states NP-hardness.
rule change: Use SRS wall kicks and one piece shape, and change the goal from a row target to clearing the board or surviving the sequence · Tetris · Know every upcoming piece. This comparison is not itself a reduction.
Why the atlas-derived bound holds
For each of K explicitly given pieces, store its placement and a path through the polynomial-size orientation/position graph for the current fixed board. A simple path suffices. Verify legal SRS moves, locking and line clears. The total certificate is polynomial in K plus the explicit board size. The hardness construction uses polynomially many identical pieces.
Published hardness · atlas-derived membership
The source proves NP-hardness. Membership in NP uses the explicitly listed finite piece sequence (or unary count).
§1.1 and §§5–9. NP membership for this explicit-sequence model follows from polynomial-size legal-placement certificates.
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I-pieces with three empty top rows
PWith room at the top, indefinite survival is guaranteed and eventual clearing is polynomial-time.
P: polynomial-time algorithm
- Exact task
- Decide whether the board can eventually be completely cleared with enough I-pieces.
- Scale & input
- Growing board of height at least 7. Top three rows empty. Unlimited identical I-pieces.
- Move rules
- Translate and rotate straight tetrominoes under the paper’s model.
- Goal
- Eventually clear the board.
- Repetition
- Unlimited supply.
- Players & information
- One player. Initial board known.
- Randomness
- None.
Both initial conditions and the supply/goal differ from finite-count hardness.
task change: Only one tetromino shape. This comparison is not itself a reduction.
Published result
§3.1, Theorem 1 with k=4.
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Dominoes that can rotate downward
PThe falling rotation rule allows a polynomial-time algorithm.
P: polynomial-time algorithm
- Exact task
- Decide survival for a specified count, or eventual clearing with unlimited dominoes.
- Scale & input
- Growing board. Only 1×2 pieces. Explicit finite count for survival or unlimited supply for eventual clearing.
- Move rules
- Rotation keeps a lowest cell fixed and moves the other to an unoccupied cell no higher.
- Goal
- Survive the specified count, or eventually clear.
- Repetition
- Pieces may continue indefinitely for the latter task.
- Players & information
- One player. Initial board and supply known.
- Randomness
- None.
Does not classify clearing with an arbitrarily short prescribed domino count.
rule change: Tetris · Know every upcoming piece. This comparison is not itself a reduction.
Published result
§2.2; §3.2, Theorem 6 and Corollary 7.
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UNO
UNO · Order all the cards
NP-completeMatching colors or numbers is enough for NP-completeness in the scalable solitaire model.
NP: completeness
- Exact task
- Decision: is there a legal ordering of every card?
- Scale & input
- An explicit multiset of cards with color and number labels. Card count and both label sets may grow.
- Move rules
- Discard one card matching the previous color or number. The first card is unrestricted. No action cards, draw pile or voluntary passes.
- Goal
- Play all cards.
- Repetition
- Discarded cards cannot return.
- Players & information
- One player, all cards known.
- Randomness
- None.
The result concerns a growing explicit input family, not a fixed commercial set.
Published result
§2; Theorem 4
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UNO · Two players, visible hands
PThe corrected two-player model reduces to undirected vertex Geography.
P: polynomial-time algorithm
- Exact task
- Decision: can the first player avoid being the first unable to play?
- Scale & input
- An explicit multiset of cards with color and number labels. Card count and both label sets may grow.
- Move rules
- Discard one card matching the previous color or number. The first card is unrestricted. No action cards, draw pile or voluntary passes.
- Goal
- Determine the first player unable to play, under optimal opposition, in the paper’s uncooperative model.
- Repetition
- Discarded cards cannot return.
- Players & information
- Two alternating players, both hands visible.
- Randomness
- None.
Simplified model, not commercial hidden-hand UNO. The authors explicitly withdrew their 2010 PSPACE-completeness claim.
Published result
§2.3; Theorems 6–7
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Keep the number of colors fixed
PFixing either colors or numbers makes solitaire UNO polynomial-time.
P: polynomial-time algorithm
- Exact task
- Decide whether every card can be played.
- Model & rules
- Explicit growing card multiset, same matching rule, no action cards or draws. One player, all information known, no randomness. Each card is discarded once.
restriction: Bound colors or numbers · UNO · Order all the cards. This comparison is not itself a reduction.
Published result
§3.3, Theorem 5
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Two players cooperate
NP-completeCooperation is NP-complete in the scalable visible-hand model.
NP: completeness
- Exact task
- Can both players cooperate to empty player 1’s hand first?
- Model & rules
- Two visible hands, unbounded card colors and numbers, no draw pile or action cards. Alternate discards. A stuck player skips rather than loses. No voluntary pass. Cards never return.
rule change: Cooperate, and skip if stuck · UNO · Two players, visible hands. This comparison is not itself a reduction.
Published result
§§2.3, 3.1, Theorem 1
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