Stalemated only if facing a Kc3 and a Knight on one of the squaresī2, c5, b6 that control a4 (the first of these arises at the start of Namely, Kb2 to move can be stalemated by force only from this position Īnd its reflection about the a1-h8 diagonal and Ka3 to move is In the diagram) stalemate can only be forced in a few special positions. Shown in the following diagram Įven on the long edge of each triangle (the twelve White Kings One of the six-square triangular neighborhoods of the corners In fact, the defending King can avoid stalemate as long as it's not on But this idea would be less convincing than the tablebase.Īn exhaustive computer search shows that as expected K+N cannot in general Then make it calculate King+Knight vs King in a few positions where the lone King begins at an edge of the board (but not too close to a corner). There would only need to be 64圆3圆2 = 249984 positions.Ī second idea would be to get a basic engine and modify its code so that it takes into account stalemate as a win, and you can probably also throw away most of the code of the engine to make it calculate faster. One idea is to build an endgame tablebase which takes into account stalemate as a win, which is equivalent to say that White wins when he captures Black's King. I would like either an irrefutable proof or at least some very strong evidence. I obviously don't want a "yes/no" answer without any evidence to back it up. This doesn't prove that a King and a Knight can always force stalemate against a lone King, but it at least shows that it's not completely inconceivable that King+Knight could force the stalemate. The result: White can force a stalemate! The trick is the move 3. So I set up a random position with King+Knight vs King where the lone King was at the edge of the board, and I tried to analyze it. I first thought that forcing stalemate would be impossible. but what about a King and one Knight against a lone King? I believe that a King and two Knights are able to force stalemate against a lone King (though obviously not checkmate).
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