Answers.Body: <p>I have made partial progress on the second part of this question, and now believe two light squared bishops suffice given enough knights. My main piece of evidence for this is the existence of a method by which two light squared bishops and a knight can confine the king to a diamond shaped region of side length 20, while periodically gaining a move to bring other pieces closer. I found this method using a <a href="https://github.com/mjtb49/InfiniteChessEndgameScripts" rel="nofollow noreferrer">script I wrote</a>. The kNBB_20_3_2.5_23.txt file in the linked repository is a set of 217208 positions with Black to move, encoded as tuples of four coordinates, where the first is the coordinate of the Black king, the second the knight, and the last two the bishops. The claim is that in all of these positions, no matter what Black plays, White has a response that both keeps Black in one of these positions and will eventually allow White to gain a tempo with which to approach with the other knights. One can test the trap against a computer opponent by running <code>play_vs_trap(load_trap(&quot;kNBB_20_3_2.5_23.txt&quot;), 22, box_size=21)</code> in trap_tester.py. The opponent is only programmed to eventually avoid repetition, it may allow you to repeat a position many times before it finds an answer which gains a tempo.</p>
  <p>While I do not claim to understand every last detail of the trap, the main themes are fairly understandable, and games against fellow humans suggest that it is not too difficult to force and maintain slightly larger versions of the trap (with side length say 30). <a href="https://i.sstatic.net/iYGfVaj8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iYGfVaj8.png" alt="A diagram of the main bishop positions" /></a>
  For the most part, the bishops stay on the green squares, and the knight tracks the king closely, preferring to be on the side of the king closest to the edge. The knight and bishops need to always be in time to catch the king as in the below diagram, but since knights are so much faster than the king and the bishops take at most 4 moves to reposition this tends to be easy. Sometimes near the corners and edges the knight will skirt around the outside of the trap to manage this while getting ahead of the king.</p>
  <p><a href="https://i.sstatic.net/Y1tIltx7m.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Y1tIltx7m.png" alt="A diagram of the knight catching the black king" /></a></p>
  <p>This method of catching the king has a slight problem, which is that the knight interferes with his own bishop. Luckily whenever this is a problem there is a bishop move available which can keep the king controlled and eventually gains a tempo.</p>
  <p><a href="https://i.sstatic.net/l752Gi9Fm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/l752Gi9Fm.png" alt="The key bishop move on the side" /></a></p>
  <p>Finally, the play near the corner is somewhat delicate, and depending how it is approached seems to need one of several setups from White. One of the more important configurations to know is</p>
  <pre><code>[FEN &quot;7K/8/8/8/3k4/3N4/8/3B1B2 b - - 0 1&quot;]
  
  1... Ke3 2. Kg8 Kd2 3. Bfe2 Kc3 4. Kh8 Kc4 5. Ba4
  </code></pre>
  <p>where the knight is well placed to guard most squares near the corner, but White must allow Black to slightly displace the bishops in order to gain tempi.</p>
  <p>Given the size of the trap and my current understanding of how it works I suspect that the trap can always be forced. Even if it cannot be, a trap in the same vein but with a larger side length should exist and would allow the knight to track the king even more loosely, meaning eventually one of the traps in this family should work. If this is the case, then some amount of knights will always work (for example, one could simply threaten every square on the perimeter of some box containing the trap, then treat the endgame like KBN v K). If I get more time to work on this I will attempt to find smaller traps and hopefully arguments they are forceable, I actually suspect even two knights could be enough given a king and two light squared bishops.</p>
  
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Answers.Body: <p>If the black king is surrounded by white units, then surely there is an n such that white king + bishop + n knights can deliver checkmate.</p>
  <p>However if bK can run away in some direction, and the white pieces have to catch up, then I think no value of n is sufficient. The white bishop can move arbitrarily far with a single move, so that can always keep pace with the fleeing king. But if bK stays on the opposite colour squares, it will be able to keep running away.</p>
  <p>To deliver checkmate, 9 squares must be covered. So that means at least 3 wN, or 2wN with wK. But if bK makes zigzag NW &amp; NE moves while trying to move in a N direction, the wK can never catch up, and only 2wN can keep up with the running bK. So checkmate can never be forced.</p>
  <p>Can bishop &amp; knight delay bK and slow his progress northwards? A position where delay might happen is like this, with black to move.</p>
  <pre><code>[FEN &quot;8/8/8/3N4/4B3/8/3k4/8 w - - 0 1&quot;]
  </code></pre>
  <p>Of course this is just a window onto an infinite board. There are no edges. Black King generally avoids light squares, but exceptionally can move to e2, then f2 then g3 then h4.</p>
  <p>Being charitable to White, we may say that White might manage to make another block at:</p>
  <pre><code>[FEN &quot;8/7N/6B1/8/7k/8/8/8 w - - 0 1&quot;]
  </code></pre>
  <p>Let's do some counting. Black has made 4 king moves and advanced 2 ranks. White has made 2 knight moves and 1 bishop move, and some other distant knight has moved 2 ranks closer. So overall the distant knight is no nearer to bK.</p>
  <p>If on the other hand, bK tried to run just in NE direction, then I think wN can move faster than bK. So that's not a good strategy for Black.</p>
  <p>This isn't a proof at the level of pure mathematics or chess solving engine, but it does convince me. Most human chess analysis is at the level of such approximate reasoning.</p>
  <p>A question remains, what is the smallest n for which a cage exists such that bK can always be mated? n=2 is too small to checkmate even a willing bK in the middle of the board, but with n=3 a position does exist so that whoever has the move, White can checkmate with their next move.</p>
  <pre><code>[FEN &quot;8/6N1/8/3B4/3k4/2NN4/2K5/8 w - - 0 1&quot;]
  
  1. Nf5#
  </code></pre>
  
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Answers.Text: While I agree the mate is likely not possible for roughly the reasons you describe, I am looking for a proof of impossibility "at the level of pure mathematics". In part this is because this question arose out of my attempts to extend a pre-existing table of endgames for fairy chess on an infinite board, where the previous maintainers of the table claimed several endgames impossible incorrectly with reasoning not much different from what you have written here. The common theme was that even if the pieces that can catch the king can't mate, they might be able to slow it down for others.
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Answers.Text: In this case, I think that any strategy that amounts to just running in a direction cannot be the whole story. A single bishop and a knight can always slow progress in any direction, and perhaps this can be efficient enough to allow a second knight to catch up, which in turn might allow a third, etc. Running diagonally is actually particularly scary to me, as a single knight on its own can slow such a king enough for a second knight to keep pace, and the bishop and the knight together can coordinate to form an efficient diagonal barrier.
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Answers.Text: I have actually done quite a bit of playing around on the board. My current opinion is that naively running diagonally (Say if you declare an order of preference of moves NE, E, N, etc. and stick with it) probably allows a second knight to catch up. Running orthogonally is better, but you actually still need to be quite careful about how you dodge out of the various blockades. The blockades can be forced infinitely often, but I think if the bK always dodges to the side away from the lead knight, they cannot be reestablished "efficiently enough".
Answers.UserDisplayName: Matthew Bolan
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Answers.Text: Have added an exploration of whether wBN can sufficiently delay bK moving North in a zigzag way. I think the answer is no
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Answers.Body: <p>You just said that their King has to be 'somewhere' which means that it could be a lot of squares away which would mean that it would be a draw due to 50 move rule.</p>
  <p>But, I am assuming that you mean that the 50 move rule doesn't exist, then that means that if you place 1 bishop and an arbitrary amount of Knights, then you cannot checkmate black because:</p>
  <p>A) You cannot bring the king in, because the other king will keep running in the opposite direction, preventing progress</p>
  <p>B) A bishop and a knight cannot bring checkmate without a king and in the  center of the board.</p>
  <p>C) It will require something similar to this position:</p>
  <pre><code>[Variant &quot;From Position&quot;]
  [FEN &quot;8/7k/8/8/8/8/BNN5/KNN5 w - - 0 1&quot;]
  
  1. Nd4 Kh8 2. Ncd3 Kg7 3. Nc3 Kf6 4. Bd5 Ke7 5. Nc4 Kf6 6. Ne4+ Kg6 7. Nf4+ Kg7 8. Ncd6 Kf8 9. Nc6 Kg7
  </code></pre>
  <p>where the knights and bishop create a blockade from behind. So, theoretically, a lot of knights (could be 10, 100, 1000) could be placed extremely, nearly infinitely, far away from the 'center'. Then, wherever the opposition places the black King, you can 'push' the king from behind towards your other forces, which will assist checkmate in the center.</p>
  
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Answers.Text: I do not know how to included the position in the answer. Could someone please show me how?
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Answers.Text: You almost had it - you just had to indent it 4 spaces.
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Answers.Text: Good point about the 50 move rule, I will add that to the main question. Unfortunately I don't think you have addressed what I view as the main possibility for a mating construction, that a bishop and a knight might be capable of slowing down a king enough to allow a second knight to catch up, which in turn might allow one to bring in more knights. Your final claim also does not make sense to me, as there are only finitely many knights Black can always ensure that their king is far to the right of all of the knights. I will add an answer with what I know about the problem later today.
Answers.UserDisplayName: Matthew Bolan
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Answers.Text: Good point Matthew, I will edit it into the answer.
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Body: <p>Two players agree to play chess on an infinite board with no edges. White begins by placing down a bishop, a king, and some finite number of knights of their choosing, in whatever configuration they so choose. Black then places down their king somewhere. Ignoring the 50 move rule, can White always checkmate Black? I suspect not, but cannot prove it.</p>
  <p>If White did not have a bishop, then Black can avoid mate forever by placing their king at least two squares to the right of all the White pieces, and then moving to the right each turn. One can check that White's pieces are not fast enough to guard all three squares to the right of Black's king, at best White can catch up with a single knight. With the bishop, this argument breaks down as a bishop and a knight together can easily catch the king and coordinate to prevent a move to the right. The worry is that they can do this often enough to allow the other White pieces to approach or drive the king into some trap. Perhaps the argument can be salvaged by modifying the underlying monovariant to consider vertical motion as well, or by careful tracking of the inefficient moves (bishop moves or knight moves not in the ideal direction) needed to reestablish the blockading positions.</p>
  <p>Parity based arguments like running on the opposite color of the bishop are difficult to make work. I think this is because several similar (fairy) endgames are wins:</p>
  <ul>
  <li>If Black promises to only ever move on squares the opposite color of the bishop (So White is trying to checkmate a <a href="https://en.wikipedia.org/wiki/Ferz" rel="nofollow noreferrer">ferz</a> instead of a king), White can easily catch them by eventually moving their knight adjacent to the king, after which it is trapped.</li>
  <li>If White places 10 or so light squared bishops instead of just one (and at least two knights), then White can always force mate by placing a knight adjacent to the Black king while controlling appropriate light squares. (EDIT: See my answer for evidence that two light squared bishops and enough knights can force checkmate. I would be interested if anyone can get a more understandable method or better bounds on the number of knights required)</li>
  </ul>
  <p>These are not great evidence for the single bishop case, as both rely on a knight adjacent to the king being excellent at confining the king.</p>
  
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Comments.Text: Does this help? https://chess.stackexchange.com/questions/33381/on-an-infinite-board-which-pieces-are-needed-to-checkmate/45997#45997
Comments.UserDisplayName: Hauke Reddmann
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Comments.Text: No, as per my answer to that question I believe the endgames of the sort described in this question are the only pawnless endgames where Black has no material that are not understood on the infinite board.
Comments.UserDisplayName: Matthew Bolan
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Comments.CreationDate: 2024-08-07T22:13:08.840Z
Comments.Text: Specifically, to completely answer that question needs an answer to which endgames of the form m light squared bishops + n knights + king vs king are winning. I think I have an argument that m = 3 occurs for some large enough n (though many details are unchecked so please take this with a grain of salt), and I have some evidence that m = 2 might. Infuriatingly though, m=1 has resisted my attempts to prove impossible.
Comments.UserDisplayName: Matthew Bolan
Tags: <endgame><checkmate><chess-variants>
Title: Can a Bishop and some Knights force mate on an infinite board?
ThreadText: # Can a Bishop and some Knights force mate on an infinite board?
  
  (Asked by: username_0 on 2024-08-01)
  
  Two players agree to play chess on an infinite board with no edges. White begins by placing down a bishop, a king, and some finite number of knights of their choosing, in whatever configuration they so choose. Black then places down their king somewhere. Ignoring the 50 move rule, can White always checkmate Black? I suspect not, but cannot prove it.
  
  If White did not have a bishop, then Black can avoid mate forever by placing their king at least two squares to the right of all the White pieces, and then moving to the right each turn. One can check that White's pieces are not fast enough to guard all three squares to the right of Black's king, at best White can catch up with a single knight. With the bishop, this argument breaks down as a bishop and a knight together can easily catch the king and coordinate to prevent a move to the right. The worry is that they can do this often enough to allow the other White pieces to approach or drive the king into some trap. Perhaps the argument can be salvaged by modifying the underlying monovariant to consider vertical motion as well, or by careful tracking of the inefficient moves (bishop moves or knight moves not in the ideal direction) needed to reestablish the blockading positions.
  
  Parity based arguments like running on the opposite color of the bishop are difficult to make work. I think this is because several similar (fairy) endgames are wins:
  
  * If Black promises to only ever move on squares the opposite color of the bishop (So White is trying to checkmate a [ferz](https://en.wikipedia.org/wiki/Ferz) instead of a king), White can easily catch them by eventually moving their knight adjacent to the king, after which it is trapped.
  * If White places 10 or so light squared bishops instead of just one (and at least two knights), then White can always force mate by placing a knight adjacent to the Black king while controlling appropriate light squares. (EDIT: See my answer for evidence that two light squared bishops and enough knights can force checkmate. I would be interested if anyone can get a more understandable method or better bounds on the number of knights required)
  
  These are not great evidence for the single bishop case, as both rely on a knight adjacent to the king being excellent at confining the king.
  
  **Tags:** endgame, checkmate, chess-variants
  
  > **Comments:**
  > **username_1**: Does this help? https://chess.stackexchange.com/questions/33381/on-an-infinite-board-which-pieces-are-needed-to-checkmate/45997#45997  
  > **username_0**: No, as per my answer to that question I believe the endgames of the sort described in this question are the only pawnless endgames where Black has no material that are not understood on the infinite board.  
  > **username_0**: Specifically, to completely answer that question needs an answer to which endgames of the form m light squared bishops + n knights + king vs king are winning. I think I have an argument that m = 3 occurs for some large enough n (though many details are unchecked so please take this with a grain of salt), and I have some evidence that m = 2 might. Infuriatingly though, m=1 has resisted my attempts to prove impossible.  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_0 on 2024-08-11. Score: 5
  
  I have made partial progress on the second part of this question, and now believe two light squared bishops suffice given enough knights. My main piece of evidence for this is the existence of a method by which two light squared bishops and a knight can confine the king to a diamond shaped region of side length 20, while periodically gaining a move to bring other pieces closer. I found this method using a script I wrote. The kNBB_20_3_2.5_23.txt file in the linked repository is a set of 217208 positions with Black to move, encoded as tuples of four coordinates, where the first is the coordinate of the Black king, the second the knight, and the last two the bishops. The claim is that in all of these positions, no matter what Black plays, White has a response that both keeps Black in one of these positions and will eventually allow White to gain a tempo with which to approach with the other knights. One can test the trap against a computer opponent by running play_vs_trap(load_trap("kNBB_20_3_2.5_23.txt"), 22, box_size=21) in trap_tester.py. The opponent is only programmed to eventually avoid repetition, it may allow you to repeat a position many times before it finds an answer which gains a tempo.
  While I do not claim to understand every last detail of the trap, the main themes are fairly understandable, and games against fellow humans suggest that it is not too difficult to force and maintain slightly larger versions of the trap (with side length say 30). 
  For the most part, the bishops stay on the green squares, and the knight tracks the king closely, preferring to be on the side of the king closest to the edge. The knight and bishops need to always be in time to catch the king as in the below diagram, but since knights are so much faster than the king and the bishops take at most 4 moves to reposition this tends to be easy. Sometimes near the corners and edges the knight will skirt around the outside of the trap to manage this while getting ahead of the king.
  
  This method of catching the king has a slight problem, which is that the knight interferes with his own bishop. Luckily whenever this is a problem there is a bishop move available which can keep the king controlled and eventually gains a tempo.
  
  Finally, the play near the corner is somewhat delicate, and depending how it is approached seems to need one of several setups from White. One of the more important configurations to know is
  [FEN "7K/8/8/8/3k4/3N4/8/3B1B2 b - - 0 1"]
  
  1... Ke3 2. Kg8 Kd2 3. Bfe2 Kc3 4. Kh8 Kc4 5. Ba4
  
  where the knight is well placed to guard most squares near the corner, but White must allow Black to slightly displace the bishops in order to gain tempi.
  Given the size of the trap and my current understanding of how it works I suspect that the trap can always be forced. Even if it cannot be, a trap in the same vein but with a larger side length should exist and would allow the knight to track the king even more loosely, meaning eventually one of the traps in this family should work. If this is the case, then some amount of knights will always work (for example, one could simply threaten every square on the perimeter of some box containing the trap, then treat the endgame like KBN v K). If I get more time to work on this I will attempt to find smaller traps and hopefully arguments they are forceable, I actually suspect even two knights could be enough given a king and two light squared bishops.
  
  
  ---
  
  ### Answer by username_2 on 2024-09-29. Score: 1
  
  If the black king is surrounded by white units, then surely there is an n such that white king + bishop + n knights can deliver checkmate.
  However if bK can run away in some direction, and the white pieces have to catch up, then I think no value of n is sufficient. The white bishop can move arbitrarily far with a single move, so that can always keep pace with the fleeing king. But if bK stays on the opposite colour squares, it will be able to keep running away.
  To deliver checkmate, 9 squares must be covered. So that means at least 3 wN, or 2wN with wK. But if bK makes zigzag NW & NE moves while trying to move in a N direction, the wK can never catch up, and only 2wN can keep up with the running bK. So checkmate can never be forced.
  Can bishop & knight delay bK and slow his progress northwards? A position where delay might happen is like this, with black to move.
  [FEN "8/8/8/3N4/4B3/8/3k4/8 w - - 0 1"]
  
  Of course this is just a window onto an infinite board. There are no edges. Black King generally avoids light squares, but exceptionally can move to e2, then f2 then g3 then h4.
  Being charitable to White, we may say that White might manage to make another block at:
  [FEN "8/7N/6B1/8/7k/8/8/8 w - - 0 1"]
  
  Let's do some counting. Black has made 4 king moves and advanced 2 ranks. White has made 2 knight moves and 1 bishop move, and some other distant knight has moved 2 ranks closer. So overall the distant knight is no nearer to bK.
  If on the other hand, bK tried to run just in NE direction, then I think wN can move faster than bK. So that's not a good strategy for Black.
  This isn't a proof at the level of pure mathematics or chess solving engine, but it does convince me. Most human chess analysis is at the level of such approximate reasoning.
  A question remains, what is the smallest n for which a cage exists such that bK can always be mated? n=2 is too small to checkmate even a willing bK in the middle of the board, but with n=3 a position does exist so that whoever has the move, White can checkmate with their next move.
  [FEN "8/6N1/8/3B4/3k4/2NN4/2K5/8 w - - 0 1"]
  
  1. Nf5#
  
  > **Comments:**
  > **username_0**: While I agree the mate is likely not possible for roughly the reasons you describe, I am looking for a proof of impossibility "at the level of pure mathematics". In part this is because this question arose out of my attempts to extend a pre-existing table of endgames for fairy chess on an infinite board, where the previous maintainers of the table claimed several endgames impossible incorrectly with reasoning not much different from what you have written here. The common theme was that even if the pieces that can catch the king can't mate, they might be able to slow it down for others.  
  > **username_0**: In this case, I think that any strategy that amounts to just running in a direction cannot be the whole story. A single bishop and a knight can always slow progress in any direction, and perhaps this can be efficient enough to allow a second knight to catch up, which in turn might allow a third, etc. Running diagonally is actually particularly scary to me, as a single knight on its own can slow such a king enough for a second knight to keep pace, and the bishop and the knight together can coordinate to form an efficient diagonal barrier.  
  > **username_0**: I have actually done quite a bit of playing around on the board. My current opinion is that naively running diagonally (Say if you declare an order of preference of moves NE, E, N, etc. and stick with it) probably allows a second knight to catch up. Running orthogonally is better, but you actually still need to be quite careful about how you dodge out of the various blockades. The blockades can be forced infinitely often, but I think if the bK always dodges to the side away from the lead knight, they cannot be reestablished "efficiently enough".  
  > **username_2**: Have added an exploration of whether wBN can sufficiently delay bK moving North in a zigzag way. I think the answer is no  
  
  ---
  
  ### Answer by username_3 on 2024-09-27. Score: 0
  
  You just said that their King has to be 'somewhere' which means that it could be a lot of squares away which would mean that it would be a draw due to 50 move rule.
  But, I am assuming that you mean that the 50 move rule doesn't exist, then that means that if you place 1 bishop and an arbitrary amount of Knights, then you cannot checkmate black because:
  A) You cannot bring the king in, because the other king will keep running in the opposite direction, preventing progress
  B) A bishop and a knight cannot bring checkmate without a king and in the  center of the board.
  C) It will require something similar to this position:
  [Variant "From Position"]
  [FEN "8/7k/8/8/8/8/BNN5/KNN5 w - - 0 1"]
  
  1. Nd4 Kh8 2. Ncd3 Kg7 3. Nc3 Kf6 4. Bd5 Ke7 5. Nc4 Kf6 6. Ne4+ Kg6 7. Nf4+ Kg7 8. Ncd6 Kf8 9. Nc6 Kg7
  
  where the knights and bishop create a blockade from behind. So, theoretically, a lot of knights (could be 10, 100, 1000) could be placed extremely, nearly infinitely, far away from the 'center'. Then, wherever the opposition places the black King, you can 'push' the king from behind towards your other forces, which will assist checkmate in the center.
  
  > **Comments:**
  > **username_3**: I do not know how to included the position in the answer. Could someone please show me how?  
  > **username_4**: You almost had it - you just had to indent it 4 spaces.  
  > **username_0**: Good point about the 50 move rule, I will add that to the main question. Unfortunately I don't think you have addressed what I view as the main possibility for a mating construction, that a bishop and a knight might be capable of slowing down a king enough to allow a second knight to catch up, which in turn might allow one to bring in more knights. Your final claim also does not make sense to me, as there are only finitely many knights Black can always ensure that their king is far to the right of all of the knights. I will add an answer with what I know about the problem later today.  
  > **username_3**: Good point Matthew, I will edit it into the answer.  
  
  ---
  
  