Chess is unsolved, so there is no known optimal strategy for one side to force a win or a draw.

However, is there an optimal strategy to lose? Consider a variant where checkmate results in a loss, but all other rules are the same as regular chess. Is there a known optimal strategy to win or force a draw?

For example, can one side force the opponent to checkmate (assuming the opponent tries to avoid checkmate)? If not, can one side force a draw by rules like stalemate, threefold repetition, or the fifty-move rule? Or is this as hard to solve as regular chess?

  • If both players avoid winning, it's going to be a long game no matter what. If just 1 side is trying to lose, it's fairly simple to do so. Fool's mate, Scholar's mate, etc. – Mast May 28 at 11:48
  • 1
    nitpick: I think you mean "Can any side force a loss...?". Because the answer to "Can either side..." is an obvious no, from game theory. At most only one side can. – Jeffrey May 29 at 13:55
  • you could lose by simply running out of time – BenTaylor May 29 at 15:00
  • @Jeffrey Not necessarily, because the two sides are not quite equal. I could imagine a possibility where white might be able to force it but black not, or vice versa. – WBT May 29 at 15:51
  • @WBT Maybe it's my english, but I read "Can either side" as "Does there exist strategies for both black and white that allow each to force a loss, if they want". – Jeffrey May 29 at 15:54

As far as I know, there is no chess variant that has been investigated that is exactly as you describe. However, antichess is a quite similar concept. The objective is to lose all your pieces/be stalemated, and it is required for you to take a piece if you can. It was weakly solved to be a win, starting with 1. e3. It is still interesting, however, as many of the lines are extremely long (over 100 moves).

In order to answer your exact question, it would be fairly difficult, but might be slightly easier to solve than regular chess. A powerful computer would likely be needed in order to figure it out.

| improve this answer | |
  • 2
    The antichess variant is different because it requires capturing when possible, but without this constraint, I guess it becomes a question of whether one side can get a selfmate position while the other side tries to move a piece back and forth until it's a draw or to get a stalemate position. – Victor May 28 at 2:06
  • 31
    Note that weakly solved is specific game theory terminology, which means that the solution gives a full strategy starting from the beginning of the game against all possible opponent choices (but not from positions which you never would reach if you followed that strategy, that's the "weakness"). Not quite as weak as someone would think if they don't know the term! – JiK May 28 at 8:46
  • 1

There is likely not a way to force a loss. Selfmates are relatively rare, and one can probably not be forced from the opening position.

The best plan, paradoxically, is likely to try to capture as many of the opponent's pieces (but not pawns) as possible, while keeping your own pieces to be better able to force your opponent. The opponent's pawns are kept because they have very little choice as to where they can move, and can also be easily blocked to avoid the opponent moving them before you want them to. For example, in the following position, the f-pawn is blocked and the king cannot move, so Black has no choice but to advance the g-pawn and will have to checkmate White in two moves:

[FEN "3Q4/8/8/4N3/6p1/7k/5p2/5N1K b - - 0 1"]

If not, can one side force a draw

Well, if checkmate indeed cannot be forced, then almost by definition a draw could be forced. The game can't go on forever, after all. If nothing else, the players will eventually run out of pawns to move or pieces to capture, and the 50 move rule will come into play.

| improve this answer | |
  • 2
    The question of whether one would be able to force stalemate in the absence of the 50-move and repetition rules would still be relevant. Obviously even without the 50-move rule, any game which doesn't result in a checkmate or stalemate would necessarily result in an eventual draw by repetition, since there are only a finite number of possible positions. – supercat May 28 at 14:47

Finding the optimal strategy requires the investigation of pretty much all (pretty much, in the grand scale of things) the positions that are possible to get to from the start.

While developing a computer program that plays extremely well in chess (or other similar games) is a solvable task—actually, it is a solved task somewhere in the beginning of the XXI century—finding and proving an optimal strategy is a totally different problem. Chess is hard to fully analyze because of the large number of possible positions.

To go more formal, consider Shannon number 10^120 which is a lower-bound on the game-tree complexity of chess. It is true that the sensible games would go much lower to around 10^40, but it will not matter too much.

For the game variant you describe, one has to find the Shannon-like number to say about the feasibility of finding the optimal strategy. However, simply by looking at the rules, I doubt it would result in something that is feasible.

Thus, I don't expect an optimal playing strategy for this variant to be found any time soon. While it might be slightly easier than for regular chess, the complexity will still be enormous.

NB: Antichess example is different because of the introduction of the compulsory capture, which severely reduces the number of possible variations.

| improve this answer | |

Forcing self-mate can be done from certain positions, the one forcing their opponent to checkmate them is usually far ahead in material. When I was a teenager (a long time ago) I actually studied the art of self-mate.

There is a game in Chess Companion where one player gave the other queen odds and the one with the extra queen had to force self-mate (and achieved it!). From that game I learned the technique.

Obviously it is not forced from the opening position.

To achieve it you have to be far ahead in material to restrict your opponent's legal moves. Their final move will be to capture one of yours that is giving check. (I can't unfortunately remember how it's done).

| improve this answer | |
  • But is there any proof that White can't force a selfmate from the opening position and there is a first move that prevents Black from forcing a selfmate? A proof for that would imply that neither side can force a loss. – Victor May 29 at 18:32
  • It's unlikely white can force a self-mate but if so white can also force a win. Because to force a self-mate your own side has to be totally in control of the position. Whether or not it is forced given queen odds (we can assume a win is forced given queen odds) is something I do not know, it would need a serious solver. – CashCow Jun 1 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.