Does there exist such an algorithm where, if given infinite processing power, a computer could play chess perfectly, i.e., it can generate perfect moves from any position?

If so, where can I find the pseudo-code for it?

  • 9
    What do you mean by perfect chess?
    – Herb
    Dec 19, 2017 at 19:53
  • 5
    @HerbWolfe I assume he means that it never makes a move that permits its opponent to force it to lose and resigns if, and only if, every possible move permits its opponent to force it to lose. Dec 19, 2017 at 20:36
  • 5
    @DavidSchwartz - "perfect chess", of course, can't be defined. Neither can "infinite processing power". Does this mean "executes all instruction sequences in 0 time"? "Has an infinite number of processors available"? FWIW - my definition of "perfect chess" is "never loses a game". Dec 19, 2017 at 23:27
  • 25
    Yes, it's called brute force. With infinite processing power you don't need to do alpha-beta pruning, although you may also need a rather large amount of storage to hold your search tree.
    – Michael
    Dec 19, 2017 at 23:45
  • 4
    The concept of an "algorithm" and the concept of infinite processing power don't really mix. The theory of algorithms and of computability is all based on an assumption of achieving a result in a finite number of steps. If you're allowed an infinite number of steps, the distinction between what is computable and what isn't disappears. Dec 20, 2017 at 23:48

8 Answers 8


Does an algorithm exist? Yes. According to Zermelo's Theorem, there are three possibilities for a finite deterministic perfect-information two-player game such as chess: either the first player has a winning strategy, or the second player has a winning strategy, or either player can force a draw. We don't (yet) know which it is for chess. (Checkers, on the other hand, has been solved: either player can force a draw.)

Conceptually, the algorithm is quite simple: construct a complete game tree, analyze the leaf nodes (the game-ending positions), and either make the winning initial move, resign (if the opponent has a forced win), or offer a draw (if the position is a draw).

The problem lies in the details: there are approximately 1043 possible positions, and an even larger number of moves (most positions can be reached in more than one way). You really need your infinitely-powerful computer to take advantage of this, since a computer that can take advantage of this algorithm either can't fit in the known universe, or won't finish computation until sometime after the universe ends.

  • 13
    @Wildcard No, it doesn't assume anything: it just contains all possibile legal games of chess and it will pick all those ones where the player at hand does not lose.
    – gented
    Dec 19, 2017 at 23:12
  • 11
    @gented, I was referring to the "resign" step of the algorithm. That's not a necessary step at all.
    – Wildcard
    Dec 19, 2017 at 23:19
  • 39
    The three-repetition rule bounds the search space, so the computer does not have to be infinitely powerful, merely astronomically powerful. Dec 20, 2017 at 9:25
  • 9
    For reference, compare a lower bound for the number of possible games (10^120) to the number of atoms in the observable universe (on the order of 10^80). The simplest algorithm would have to find all those games and store their data. Storing one game per atom would take 10^40 times as many atoms as we estimate in the observable universe. Dec 20, 2017 at 13:49
  • 6
    This answer is great until the very end when you refer to an "infinitely powerful computer". That's not what you mean, and that phrase doesn't belong in the question nor the discussion.
    – Don Hatch
    Dec 21, 2017 at 22:52

See https://en.wikipedia.org/wiki/Endgame_tablebase.

With infinite computer power, one could build such a table for the starting position and solve chess.

In practice, only positions with up to seven "men" (pawns and pieces, counting the kings) have been solved using current supercomputers, so we are very far from solving chess. The complexity of the problem increases exponentially with the increase in the number of pieces.

  • 9
    As a side note, if you actually produced such a table, no matter what you stored the information on, it would weigh roughly 10^43 times as much as the observable universe; considering there are ~10^123 possible chess positions and only ~10^80 baryons in the observable universe. Dec 19, 2017 at 22:49
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    @Shufflepants who said i was storing it using baryons?
    – Michael
    Dec 19, 2017 at 23:46
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    @Christoph And assuming conservation of information, and assuming you had a detector and your super computer with infinite processing power, you could slowly over the course of something like a googolplex years read out the tablebase as hawking radiation. Dec 20, 2017 at 14:58
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    @Shufflepants Note that an actual winning strategy might require much less space than a full tablebase. For instance, Nim has a winning strategy that is simple to describe, there is no need to build a huge table of all possible states. Dec 20, 2017 at 17:50
  • 2
    This solution as stated is not viable. The mass of such a table would form a black hole and it would be impossible to exfiltrate data from it.
    – emory
    Dec 23, 2017 at 9:57

If you really had infinite processing power, such an algorithm would be actually trivial to write. As chess has a finite number of possible states, you could in theory just iterate through them all until you find a path of perfect play. It would be horribly inefficient, but if you have infinite processing power, it wouldn't matter.

  • That's not true. He said you have infinite processing power, but not did not say anything about infinite space.
    – ubadub
    Dec 23, 2017 at 20:05
  • @ubadub : We wouldn't need infinite space. The length of a game is limited due to the 50-move rule, and a rule can be made up to sort all possible moves from a position. As they can be sorted, they can be stored as an integer. This is all the memory required to walk the whole tree. And if you have infinite time, you can walk the tree as often as you want, so you don't have to store every possible chess game.
    – vsz
    Dec 23, 2017 at 21:35
  • The length of the game is limited, but it is extremely large; as someone else pointed out, if you produced a table to store all such games, "no matter what you stored the information on, it would weigh roughly 10^43 times as much as the observable universe; considering there are ~10^123 possible chess positions and only ~10^80 baryons in the observable universe
    – ubadub
    Dec 23, 2017 at 23:43
  • 2
    @ubadub : That is true, but I was not talking about "a table to store all such games". There are many tree-related algorithms which don't have to hold all the nodes of the whole tree in memory.
    – vsz
    Dec 24, 2017 at 9:57
  • @ vsz good point
    – ubadub
    Dec 25, 2017 at 2:45

To directly address the question: yes there is such an algorithm. It is called minimax. (The endgame tablebases are generated by using this algorithm (backwards!), but the plain old simple minimax algorithm is all you need). This algorithm can play any two player zero sum game perfectly. Find pseudocode here:


note that variants of this algorithm are used by modern computer chess programs.

This algorithm basically generates the move which will minimize the opponent's chances, i.e., it would choose a move for White that is +2 instead of +1. Then, it generates the opponent's move which will maximize his chances, i.e., it would choose a move for Black that is -2 instead of -1.


Not only is there an algorithm to play perfect chess, it is possible to write a short program that will (given infinite resources) play any deterministic perfect-knowledge finite-duration two-player game perfectly.

The game engine does not even need to know the rules of the game it is playing. All it needs is an opaque representation of a "game state" and functions that (a) given any game state, provide a list of legal next game states and (b) given a game state, decide if it is a win for player 1, a win for player 2, a draw, or it is not an end state.

Given those functions a simple recursive algorithm "solves" the game.

This fact has been alluded to in previous answers by chessprogrammer (minimax) and by Acccumulation (who provides a version of the program in python).

I wrote such a program over 20 years ago. I tested it by playing noughts-and-crosses (tic-tac-toe if you are American). Sure enough it played a perfect game.

Of course this will fall over quickly on any imaginable computer for any serious game. Because it is recursive it is effectively building the entire game tree on the stack, so you will get a "stack overflow" (pun very much intended) before you get anywhere near analysing the 10^123 states of chess referred to in other answers. But it is fun to know that in principle this small program would do the job.

For me this also says something interesting about AI: however much "intelligence" you think is exhibited by Deep Blue, or Go Zero, or indeed by a human playing Chess or Go there is a sense in which these games have trivial, exactly computable optimal solutions. The challenge is how to get a good though not optimal solution in a reasonable time.

  • Your algorithm only works for perfect-knowledge two-player games. It will fall over for hidden-information games such as Stratego, because any implementation of function (a) violates the game rules. It also fails for games of potentially infinite duration: for example, drop the 50-move rule from chess, and it can't tell that two kings chasing each other around the board isn't a winnable state. All it can tell is that it's not an end state.
    – Mark
    Dec 22, 2017 at 18:59
  • Valid points. I will edit my answer.
    – gareth
    Dec 23, 2017 at 19:36

I will ignore the possibilities of draws or infinite sequences of moves for simplicity. Once the algorithm is understood, it is not particularly difficult to extend it to those cases.

First, some definitions:

  1. Any move that wins the game for the player who makes that move is a winning move.

  2. Any move that loses the game for the player who makes that move is a losing move.

  3. Any move that leaves the other player with at least one winning move is also a losing move. (Since the opponent can take that move and force a loss.)

  4. Any move that leaves the other player with only losing moves is also a winning move. (No matter what move your opponent makes, you will win.)

  5. A perfect strategy means always making winning moves if any remain and resigning when one has only losing moves remaining.

Now, it's trivial to write a perfect strategy. Simply explode all possible move sequences and identify winning/losing moves. Ignoring stalemate, this will eventually identify every move as either a winning move or a losing move.

Now, the strategy is trivial. Look at all your possible moves. If any winning moves remain, take one and win. If only losing moves remain, resign, since your opponent can force you to lose.

It is not difficult to adjust the strategy to include the possibility of a stalemate.

Update: Just in case it's not clear how this identifies every move as a winning more or a losing move, consider:

  1. Every move that results in a win is a winning move.
  2. Every move that results in a loss is a losing move.
  3. Every move that results in the opponent having only winning or losing moves is either a winning or a losing move.
  4. Call n the number of moves in the longest possible chess game. (We are ignoring unbounded sequences for now, though including them is not difficult.)
  5. There are no moves with n prior moves we need to consider.
  6. Every move with n-1 prior moves is either a winning move or a losing move since n moves ends the longest game.
  7. Thus every move at depth n-2 is followed by only winning moves or losing moves and thus is itself a winning move or losing move.
  8. And so on back to the first move.
  • 1
    Your definitions of winning and losing moves are not comprehensive enough. The first move, for instance, neither wins the game (#1), nor leaves the opponent with only losing moves (#4), so it isn't a "winning move". Neither does it lose the game (#2), nor leave the opponent with any winning move (#3), so it isn't a "losing move". Your strategy requires that every move is defined either as a "winning move" or a "losing move", which simply isn't the case as you've defined it. Dec 19, 2017 at 20:47
  • 2
    @NuclearWang It does define every move as either a winning move or a losing move. What do you think the third alternative is? Visualize the tree of all possible chess games (and remember, we're excluding ties or infinite sequences for now). Every chain ends in either a win or a loss. This percolates up through the tree eventually identifying every move as either a winning move or a losing move. Dec 19, 2017 at 20:59
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    @NuclearWang either the first move is a winning move for one player, or else chess is (like tic-tac-toe) a drawn game with perfect play. We don't know which because no one has ever had the computing power to run this algorithm to completion, and no one has found a more direct proof.
    – hobbs
    Dec 19, 2017 at 21:04
  • 8
    There is no randomness and no hidden information in chess, which leaves no room for "maybe". Every position is won, lost, or drawn (even if we haven't managed to identify them as such). And this explanation is leaving out the "drawn" option for simplicity, but it mostly amounts to 1) a position is drawn if it's drawn according to the rules, and 2) a position is drawn if it has no winning moves, but has at least one move that leaves the opponent with no winning moves.
    – hobbs
    Dec 19, 2017 at 21:22
  • 2
    @DavidSchwartz: Unless someone is in a losing position, every move that isn't perfect is bad. In a losing position, there would generally be no single "perfect" move [except in a forced-move situation] since any legal move could have some probability of being the only winning or drawing move in some conceivable (possibly highly contrived) circumstances. Resigning, however, would seem the unambiguous worst "move". Suppose the game is proven solved as a win for White with d4. Would you want to play a chess program which responded to 1. d4 with ...resigns?
    – supercat
    Dec 20, 2017 at 18:27

Suppose you have three functions: win_state , get_player, and next_states. The input for win_state is a game state, and the output is -1 if white is in checkmate, 0 if it’s a draw, 1 if black is in checkmate, and None otherwise. The input for get_player is a game state, and the output is -1 if it’s black’s turn and 1 if it’s white’s turn. The input for next_states is a list of possible next game states that can result from a legal move. Then the following function, when given a game state and a player, should tell you what game state to move to for that player to win.

def best_state(game_state,player)
  def best_result(game_state):
     if win_state(game_state):
         player = get_player(game_state)
         return max([best_result(move)*player for move in next_states(game_state)])*player
  cur_best_move = next_states(games_state)[0]
  cur_best_outcome = -1
  for state in next_states(game_state):
     if best_result(state)*player > cur_best_outcome:
           cur_best_outcome = best_result(state)*player
           cur_best_move = state

Use a look-up table

Yes. It's easy. You don't even need infinite processing power. All you need is a look-up table that contains, for each possible board position, the best move to play in that position. Here is the pseudo-code:

def play-move(my-color, board-position):
    return table-of-best-moves[my-color, board-position]

The catch

The only catch is that this look-up table would have to be very, very large—perhaps larger than the Milky Way galaxy—and it would take a long time to construct it—perhaps longer than the current age of the universe, unless there's some undiscovered regularity in chess that makes it much simpler than we can see right now. But if you had this look-up table, the subroutine to choose a perfect move every time could be implemented in as little as one CPU instruction.

Also, given our current knowledge of chess, there's no way to be sure that perfect play guarantees that you won't lose. For example, if perfect play guarantees a win for White, then Black would lose even if Black plays perfectly.

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