2

Suppose, in the endgame queen + pawn vs queen (KQPKQ) , the stronger side has a bishop-pawn (c-pawn or f-pawn). The weaker side cannot win the pawn, nor has a perpetual check, nor can play on stalemate.

Is there any drawing position left with all these conditions?

1

According to the tablebase here, this position is a draw. Of course, with the open position, Black threatens perpetual check with any free move.


[FEN "3k1q2/8/8/8/8/8/5P2/3KQ3 w - - 0 1"]

1

The weaker side cannot win the pawn, nor has a perpetual check nor can play on stalemate. Is there any drawing position left with all these conditions?

I will answer this with a no, and for a reason that's more general than the KQPKQ scenario you're specifically interested in. But ultimately, whether the answer to your question is yes or no comes technically down to definitions. First, consider this KQPKQ position, with White to move:

[FEN "8/5k2/5Pq1/4K3/8/8/8/5Q2 w - - 0 32"]

This is a drawn position, and perhaps one might be tempted to say it meets your criteria: White never needs to lose the pawn, nor allow a perpetual check from Black, nor play into a stalemate. So why don't I think this gives a "yes" answer to your question? Because even so, the only reason it's a draw is because White can only avoid those fates by playing into a different repetition himself.

My more general point is this: there are really only two essential reasons a game is ever drawn (ignoring mutual agreement, since we're talking theory).

  1. Stalemate
  2. Repetition

Perpetual check is a particular case where one side forces a repetition directly, and the 50-move rule is theoretically redundant: if mate can't be forced, players will unavoidably fall into a repetition once there are no more captures or pawn moves available (as there are only finitely many chess positions in total). The 50-move rule is merely a convenience for practice. Even insufficient mating material as a reason is theoretically redundant, as again, if a mate can't be forced and stalemate doesn't arise, eventually repetition will crop up.

What I'm getting at is, even beyond your KQPKQ endgame, the only base reason a position is ever theoretically drawn is because either stalemate or repetition is ultimately unavoidable; that's it. You could take my example position as giving your question a "yes" answer, since White can choose to enter a repetition that isn't Black perpetually checking the white king; my preference is to say the answer is still "no" as I consider that functionally equivalent to allowing a perpetual from Black, but it's all semantics at that point.


A preemptive point of clarification. It is of course the case that the set of positions which are drawn with the 50-move rule in effect is different than the set of positions which are drawn without it in effect. So the 50-move rule has a definite impact on "practical endgame theory." The sense in which I'm calling it "theoretically redundant" above is only that, in an ideal setting chess never needed such a rule in the first place, as draw by repetition would eventually arise for any position in which progress can't be made.

  • I think your answer confuses perpetual check and repetition. While a perpetual check eventually forces a repetition a repetition can be reached even without checks. – Jester May 14 '15 at 12:04
  • @Ignaz, I agree that a repetition can be reached even without checks, but where in my answer is it suggested otherwise? I explicitly point out that perpetual check can ultimately be seen as a special case of repetition, not the other way around. – ETD May 15 '15 at 2:41
  • then the correct answer to the question should be a clear yes, right? Your position does not allow to win the pawn nor can balck play for stalemate, nor does black have a perpetual check unless white allows it. Soo it seems to me it satisfies all the conditions asked for. – Jester May 15 '15 at 6:27
  • @Ignaz, as I say explicitly in the answer, whether the answer is Yes or No depends on how one chooses to interpret things. I go on to explain the sense in which I would want to say the answer is No, in the sense that I am of the opinion that distinguishing perpetual check as some different sort of case is misleading. But I tried to acknowledge that my position does indeed provide a Yes answer in a certain light. – ETD May 15 '15 at 13:44
  • I think our comments here bear that out: you lean toward a view in which this position offers a Yes answer, while I instead take the position rather to show how it isn't all that meaningful to distinguish perpetual check as a separate case in the first place. Both are perfectly reasonable viewpoints IMO, and I meant my answer to acknowledge that. – ETD May 15 '15 at 13:45

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.