1

Usually in selfmate problems white has material advantage to be able to force black to checkmate white, so white likely has a conventionally better position or has a forced checkmate. But not always, as this example shows: Position where neither player can force a win and neither player can force a draw .

But in general, does having a checkmate imply having a selfmate or vice versa in most cases, and in what conditions is it not the case? If both sides play with selfmate as the goal, then either one side can force a selfmate or both sides can force at least a draw/not lose(here "loss" means checkmating the opponent), analogously to normal chess. So any position is potentially a selfmate for one side or "drawn". How are the set of positions that have selfmates and the set of positions that have checkmates related?

Improve this question

1 Answer 1

Reset to default
0

There are certainly exceptions. Most easily, when the opponent has only the King left, there's usually a forced checkmate but there cannot be a selfmate. Likewise with Q vs. N and Q vs. B. In fact there's no forced selfmate of Q vs. P and Q vs. R, even though in most such positions the side with the Queen has a forced win, i.e. an eventual forced checkmate. Conversely, there are "no-brainer" selfmates where any legal sequence will end with you being checkmated, so in particular you cannot give checkmate even with the opponent's assistance, let alone force checkmate. Another example:

[FEN "8/8/8/8/pp2B3/kp6/8/K7 w - - 0 1"]

1.Bb1?? (1.Bd5 b2+ 2.Kb1 b3 3.Bxb3) b2#

White can selfmate in one but cannot win by force (though there's an easy forced draw).

Improve this answer
1
  • Thanks! It's interesting that merely reversing the victory condition can change the set of winning positions in such complex ways.
    – alices_and_bobs
    Commented May 29 at 5:24

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