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Mate with two knights

Mate with two knights

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We've discussed this topic several times, and this morning I came across this position from Znosko Borovsky - Seitz, Nice 1931

Black to move, mate in six

  • 8
  • a
  • 7
  • b
  • 6
  • c
  • 5
  • d
  • 4
  • e
  • 3
  • f
  • 2
  • g
  • 1
  • h

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I didn't think mate could be *forced* with just two knights. I'm guessing the white pawn is part of the reason it's forced in this puzzle?

1 edit
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1... Kg4! 2. Kg1 (A) Kg3 3. Kh1 Nh4 4. f5 (B) Nf3 5. f6 Nd1/g4 6. f7 Nf2#

(A) 2. Kh1 Kg3 3. Kg1 Nh4 4. Kh1(C) Nf3 5. f5 Nd1/g4 6. f6 Nf2#

(B) 4. Kg1 Nf3+ 5. Kh1 Nd1/g4 6. f5 Nf2#

(C) 4. f5 Nf3+ 5. Kh1 Nd1/g4 6. f6 Nf2#

In words: After 1... Kg4! White's K gets trapped in the corner, gets stalemated and is forced to move his pawn while black proceeds to deliver mate.

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Originally posted by magnublm
I didn't think mate could be *forced* with just two knights. I'm guessing the white pawn is part of the reason it's forced in this puzzle?
Thats correct, mate can be forced with 2 knights when the weaker side has a pawn behind a certain place. This is because with the pawn stalemate can be avoided but the mate is horrendously complex and if anyone succeeds in doing it here they are probably using a tablebase.

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Originally posted by Dragon Fire
Thats correct, mate can be forced with 2 knights when the weaker side has a pawn behind a certain place. This is because with the pawn stalemate can be avoided but the mate is horrendously complex and if anyone succeeds in doing it here they are probably using a tablebase.
Mate can be forced with only one knight with the aid of the opponents judas pawn.

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Originally posted by Dragon Fire
Thats correct, mate can be forced with 2 knights when the weaker side has a pawn behind a certain place. This is because with the pawn stalemate can be avoided but the mate is horrendously complex and if anyone succeeds in doing it here they are probably using a tablebase.
I did most certainly not use a tablebase to arrive at the solution, just shuffling around some pieces. It's not that hard really.

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Originally posted by Dragon Fire
Thats correct, mate can be forced with 2 knights when the weaker side has a pawn behind a certain place. This is because with the pawn stalemate can be avoided but the mate is horrendously complex and if anyone succeeds in doing it here they are probably using a tablebase.
I created these tablebases when I was in eleventh grade three decades ago. Geez, there are only a few dozen possibilities in this position.

These are old-style Ruy Lopez type tablebases--using a quill pen, I list every possible white response to black's choice moves.

After 1...Kg4, white has two legal moves. In both cases 2...Kg3 forces the other, etc.

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http://www.chess-poster.com/english/mail/mail_2006/mate_with_two_knights.htm

This site will sum it up nicely. The confusion about a mate with two knights is that it cannot be forced but it can happen with some help from your opponent.

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Originally posted by Karl Childers
...a mate with two knights...cannot be forced, but it can happen with some help from your opponent.
A mate with two knights alone cannot be forced. As this thread has mentioned, K+N+N can force mate against K+P.

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Originally posted by zzyw
I did most certainly not use a tablebase to arrive at the solution, just shuffling around some pieces. It's not that hard really.
The position given here is not difficult but most K & 2N vs K & P would require a tablebase to solve and if achieved in a nornal game here would be highly suspect.

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Originally posted by Wulebgr
I created these tablebases when I was in eleventh grade three decades ago. Geez, there are only a few dozen possibilities in this position.

These are old-style Ruy Lopez type tablebases--using a quill pen, I list every possible white response to black's choice moves.

After 1...Kg4, white has two legal moves. In both cases 2...Kg3 forces the other, etc.
Sorry for confusion I did not mean this particular position which is indeed solvable but rather the sort of positions likely to occur in actual play which may have a forced mate is 40+ moves. It is these positions that I refer to as (probably) requiring a tablebase.

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Originally posted by Dragon Fire
Sorry for confusion I did not mean this particular position which is indeed solvable but rather the sort of positions likely to occur in actual play which may have a forced mate is 40+ moves. It is these positions that I refer to as (probably) requiring a tablebase.
Sorry I went after you as a rattlesnake, then.

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