Answers.Body: <p>In fact K+R+R can force checkmate against K on an open board.</p>
  <p>First move both Rooks away from the King and to different rows and columns.</p>
  <p>Then check the King with one Rook, say on a row.
  Whichever side the King goes, move the other Rook to limit it to a single row.</p>
  <p>Now the Rooks, moving only horizontally,
  can limit the defending King to just four squares on that row,
  with enough spare moves for the attacking King to approach:</p>
  <pre><code>[Title &quot;RR keep K boxed in&quot;]
  [FEN &quot;R7/5k2/2R5/8/8/8/8/K7 w - - 0 0&quot;]
  
  1. Rh8 Kg7 2. Rch6 Kf7 3. Kb2 Ke7 4. Rc8 Kd7 5. Rhc6 Ke7 6. Kc3 Kf7 7. Rh8
  </code></pre>
  <p>Once the King has joined the fray, mate can soon follow.  For example:</p>
  <pre><code>[Title &quot;KRR mate K on infinite board&quot;]
  [FEN &quot;7R/5k2/2R5/8/3K4/8/8/8 w - - 0 0&quot;]
  
  1. Ke5 Kg7 (1... Ke7 2. Rc7#) 2. Rch6 Kf7 3. Kd6 Kg7 4. Kd5 Kf7 
  5. Ke5 Kg7 (5... Ke7 6. R6h7#) 6. Kf5 Kf7 7. R6h7#
  </code></pre>
  
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Answers.Text: Can you prove that 3 bishops are insufficient? I think it is, but am unable to prove it. 2 white-sq bishops plus 2 black-sq bishops are trivially enough, but even 4 white-sq bishops plus 1 black-sq bishop seems hard if at all possible.
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Answers.Text: With only one dark-square Bishop there is no mating position, let alone a forced mate. The attacking King and dark-squared Bishop would have to control either a diamond of four or an X-shape of five squares around the bare king, and neither of these is possible.
Answers.UserDisplayName: Noam D. Elkies
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Answers.Text: Oh lol that was trivial; I was so focused on trying to trap the king that I didn't try to find the final position needed.
Answers.UserDisplayName: user21820
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Answers.UserDisplayName: Noam D. Elkies
Answers.Body: <p>An <em>infinite</em> number of knights would be insufficient to force checkmate, assuming the enemy king starts some distance away and not surrounded. The knights are too slow and not enough of them can chase an enemy king that simply runs away.</p>
  <p>A rook and two bishops, on the other hand, could force checkmate. It's rather easy to force a position like this:</p>
  <pre><code>[FEN &quot;3BB3/8/4k3/R7/8/8/8/4K3 w - - 0 1&quot;]
  </code></pre>
  <p>The Black king can only shuffle between two squares while the White king gets into position on d4 or e4 to set up Ra6#.</p>
  <p>Four bishops could also force a checkmate:</p>
  <pre><code>[FEN &quot;3BB3/8/4k3/8/3BB3/3K4/8/8 w - - 0 1&quot;]  
  
  1. Bc5 Ke5 2. Bd7 Kf4 3. Bd6#
  </code></pre>
  <p>I'm guessing that three bishops and a knight, or two bishops and two knights, could also force checkmate. The bishop pair is very effective at corralling the enemy king.</p>
  
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Answers.Text: An infinite number of Knights on an infinite board may force mate if the local density of Knights near the opposing king is high.
Answers.UserDisplayName: Michael West
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Answers.Text: They'd have to be very near the king (like, within two ranks or files) or on both sides of it.
Answers.UserDisplayName: D M
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Answers.Text: @MichaelWest If you can add constraints to the initial position, then a finite number of knights can suffice. In effect --- that is what you are talking about, since only finitely many of those knights in the vicinity of the king will play a role. It could be of interest to characterize starting positions for which knights can checkmate on an infinite board.
Answers.UserDisplayName: John Coleman
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Answers.Text: The original statement added constraints to the initial position, namely: "assuming the enemy king starts some distance away and not surrounded."  This made me start to think about other initial conditions and if there was an unstated assumption about knight distribution. The initial position could be mate, but that wouldn't be very interesting :)
Answers.UserDisplayName: Michael West
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Answers.Text: @MichaelWest The implicit logic of the original statement is that infinitely many knights are not always able to force mate *because* the king might be too far away. Two rooks, for example, can always force mate no matter where the enemy king is. even Infinitely many knights lack that property. I agree that there are some interesting questions involving the distribution of those infinitely many knights relative to the king. When the king is partially but not wholly surrounded it could be an interesting king hunt. Sort of like the king is a blockade runner.
Answers.UserDisplayName: John Coleman
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Answers.Text: @MichaelWest 5 knights can do it. 4 knights to cover the 8 adjacent squared, and a 5th solely to target the center and cause the mate
Answers.UserDisplayName: Alexander
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Answers.Text: @Alexander 4 knights and a king (or, indeed, any combination of 4 pieces - including pawns! - and a king) could, *if* the initial setup allows it and/or the lone king cooperates. But it seemed like OP wanted to know what pieces could *force* checkmate, and knights (or knights and a king) can't do that unless you make assumptions about their starting position relative to the lone king.
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Answers.UserDisplayName: D M
Answers.Body: <p>Since this post was made, people have begun to play infinite chess somewhat seriously. The following piece combinations (together with the king) are known to force mate:</p>
  <ul>
  <li>2 Queens</li>
  <li>Queen + Rook</li>
  <li>Queen + Bishop</li>
  <li>Queen + Knight</li>
  <li>2 Rooks</li>
  <li>Rook + 2 Bishops of different colors</li>
  <li>Rook + 2 Bishops of the same color</li>
  <li>Rook + Knight + Bishop</li>
  <li>Rook + 2 Knights</li>
  <li>2 Bishops of one color + 2 Bishops of the other color</li>
  <li>2 Bishops of one color + 1 Bishop of the other color + 1 Knight</li>
  <li>1 Bishop of one color + 1 Bishop of the other color + 2 Knights</li>
  </ul>
  <p>No subset of any of the above can force mate for the trivial reason that no mates exist. No amount of knights is enough if they all start to one side of the king. No amount of bishops is enough if you do not have at least 2 of each color.</p>
  <p>The remaining open questions are in the situation where there are only bishops of a single color and knights. There is a quite lazy argument that 10 or so light squared bishops + 2 knights are able to force mate, but we don't even know if 1 bishop + 3 knights can force mate or not. I will probably ask this as a separate question.</p>
  <p>I will now talk briefly about the methods for each checkmate in the above list. Since there are a lot I won't be as thorough as I'd like, but please ask about any endgame you find suspicious.</p>
  <ul>
  <li><p>The endgames involving a Queen are easy enough, on its own the Queen can confine the king to a 5x5 region by moving between the corners of a 7x7 square as needed. While doing this there are always moves where the Queen needn't budge to maintain the prison, which the attacker can use to approach with their other pieces. Forcing the final mate is simple.</p>
  </li>
  <li><p>Noam Elkies has already described in this thread a technique with 2 Rooks</p>
  </li>
  <li><p>The endgames with a Rook and two minors can all be done in essentially the same way (though for the ones with a bishop there are easier ways):</p>
  <ol>
  <li>Using a distant rook to prevent the Black king from running forever in one direction, get your king onto the same row as Black's king.</li>
  <li>Use your rook to confine Black's king to 3 rows (by always moving in time to prevent its escape), centered on the row your king is on. By doing this you gain a free move every other move, and prevent the kings passage beyond the column your king is on.</li>
  <li>Run one of your minor pieces to the opposite side of the Black king, so that it is sandwiched between your king and this minor piece. If your minor piece is a knight you may need to switch to confining the Black king to 4 rows instead of 3 once it is suitably far from your king, otherwise the knight won't have enough free time to catch up.</li>
  <li>Move your rook to be guarded by the minor piece, so that the king cannot gain access to the columns next to your minor piece. The king is now trapped in a finite area - the three rows we have been keeping it on, and the columns between your king and minor piece.</li>
  <li>Squeeze this area as small as possible, then approach with the second minor piece. Eventually you can force a trap of the form</li>
  </ol>
  </li>
  </ul>
  <p>For example</p>
  <pre><code>[FEN &quot;8/2R5/8/4kNK1/8/8/8/8 w - - 0 1&quot;] 
  </code></pre>
  <p>or</p>
  <pre><code>[FEN &quot;8/2R5/8/B2k1K2/8/8/8/8 w - - 0 1&quot;]
  </code></pre>
  <p>, from which mating is not so bad.</p>
  <ul>
  <li>The endgames with two bishops of opposite colors are not so bad. Two bishops of different colors can confine the king to a finite area, which lets you position the pieces as needed. Then you can switch to confining the king to 7 half diagonals and use the king and other minor pieces to squeeze the king into one of several small trap. I used a variant of my <a href="https://gist.github.com/mjtb49/51020875338082e7369e6b72d7cd3fb0" rel="nofollow noreferrer">tablebase script</a> to verify that the positions I can force like this are indeed forced mates, though allegedly other people have cleaner arguments for the finish here.</li>
  </ul>
  <p>As a bonus, it is known that a Queen and a Pawn (which we say is unable to promote or move two squares at once) is enough, but there is a unique (up to a reflection) mate in 4 in that endgame which you must pass through.</p>
  <pre><code>[FEN &quot;8/3K4/8/2k5/8/3P4/1Q6/8 w - - 0 0&quot;]
  </code></pre>
  <p>I am unsure of the status of other endgames with a pawn on the board.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2024-08-02T05:13:27.473Z
Answers.Text: "It is known that a Queen and a Pawn is enough" -- do you have a reference
  for this?  I found this a few months ago but thought it was new.
  I used this concluding sequence (1 Kc7 Kd5 2 Qf6 Kc5 3 Qb6+ Kd5 4 Qd6#)
  but there are other possible finishes: put Q on a1 instead and it can end
  1 Qa4 Kb6 2 Kc8 Kc5 3 Kc7 Kd5 4 Qe4+ Kc5 5 Qc4#.  (If 1 Qa4 Kd5 then
  2 Qa7 Ke5 3 Qf2 Kd5 4 Qb2 Kc5 and we reach your diagram.)
Answers.UserDisplayName: Noam D. Elkies
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Answers.CreationDate: 2024-08-02T05:19:06.533Z
Answers.Text: Also, in your KRB/K diagram, if White also has a Pe3 then it's mate in 3
  (1 Bc3 Kd6 2 Be5+ Kd5 3 e4#), suggesting that RBP may be enough.
  There are similar RNP setups too.
Answers.UserDisplayName: Noam D. Elkies
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2024-08-02T13:45:02.987Z
Answers.Text: If you found the QKP/K a few months ago then probably you are the first person to. I was told that others had a method but I could never track them down, so we had to work out a new one.
  
  Your comment made me doubt the uniqueness result I claimed in the post, which came from an old version of my tablebase script. I modified the script to check the claim. The correct claim is that it is the unique (up to symmetry) mate in 4 which can be forced by White (so the previous position with the king on d5 and Black to move is the truly unique one).
Answers.UserDisplayName: Matthew Bolan
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Answers.CreationDate: 2024-08-02T13:51:50.770Z
Answers.Text: The updated script runs much faster than my old one, and it reveals that I got quite lucky with the method I found. The position 3K4/8/4k3/8/8/3P4/5Q2/8 w - - 0 1 , which  my method passes through, is one of only 4 (up to symmetry) mates in 22 ply (meaning after Black's move, White has at worst a mate in 11). There also is apparently only 1 mate in 10 ply,  2 mates in 18 ply, and 2 mates in 20 ply.
Answers.UserDisplayName: Matthew Bolan
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Answers.CreationDate: 2024-08-02T14:37:53.410Z
Answers.Text: Though I should mention that my script only checks for moves of some bounded length (though it won't let the king move into some distant check). The above results are true provided that there isn't some other short mate where White needs to make a move of more than 200 squares.
Answers.UserDisplayName: Matthew Bolan
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Answers.LastEditDate: 2024-08-01T21:03:10.710Z
Answers.UserDisplayName: Matthew Bolan
Answers.Body: <p>Here are some mates I found:</p>
  <ol>
  <li><p>Queen and rook (limit the black king until the white arrives to help)</p>
  <pre><code>[Title &quot;KQR vs k mate&quot;]
  [FEN &quot;1R6/5k2/8/8/8/8/6Q1/K7 w - - 0 1&quot;]
  
  1. Qg8+ Ke7 2. Rd8 Kf6 3. Re8 Kf5 4. Kb2 Kf4
  5. Qg2 Kf5 6. Kc3 Kf6 7. Kd4 Kf7 8. Qg8+ Kf6
  9. Ke3 Kf5 10. Kf3 Kf6 11. Re6+ Kf5 12. Qg6#
  </code></pre>
  </li>
  <li><p>Queen and pawn (not sure if the position can be reached with optimal play)</p>
  <pre><code>[FEN &quot;8/8/4Q3/3Pk3/8/4K3/8/8 w - - 0 1&quot;]
  </code></pre>
  </li>
  <li><p>Queen and bishop (again, limit the king with the queen and the bishop, so the attacking king finds the time to approach)</p>
  <pre><code>[Title &quot;KQB vs k mate&quot;]
  [FEN &quot;Q7/2k5/8/8/B7/8/8/7K w - - 0 1&quot;]
  
  1. Kg2 Kb6 2. Kf3 Kc5 3. Qb7 Kc4 4. Ke3 Kc5
  5. Ke4 Kd6 6. Qd7+ Kc5 7. Qd4#
  </code></pre>
  </li>
  <li><p>Queen and knight (limit the black king with the queen and the knight as much as possible)</p>
  <pre><code>[Title &quot;KQN vs k mate&quot;]
  [FEN &quot;Q7/2k5/8/8/8/1N6/8/7K w - - 0 1&quot;]
  
  1. Kg2 Kd7 2. Kf3 Ke7 3. Ke4 Kf6 4. Qg8 Ke7
  5. Ke5 Kd7 6. Kd5 Ke7 7. Nc5 Kf6 8. Ne4+ Ke7
  9. Ke5 Kd7 10. Nd6 Kc6 11. Qb8 Kc5 12. Qb5#
  </code></pre>
  </li>
  <li><p>Two rooks (<a href="https://chess.stackexchange.com/a/33385">Noam's answer</a> mentions how the mate can be forced)</p>
  <pre><code>[FEN &quot;8/8/R7/1R2k3/8/4K3/8/8 w - - 0 1&quot;]
  </code></pre>
  </li>
  </ol>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-12-18T18:17:57.820Z
Answers.Text: Are all of them possible with black's optimal play?
Answers.UserDisplayName: Saat
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Answers.Text: Two rooks seems pretty easy.
Answers.UserDisplayName: xehpuk
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Answers.CreationDate: 2020-12-19T16:18:09.823Z
Answers.Text: Now that I think about it, I don't see how that KQP/K mate can be forced.
  What was Black's last move?  There are a few other mating positions:
  Ke5 mated also by Kc5,Pe4(g4),Qf5 or Ke7,Pd3,Qe4 or even 
  Kf7,Pd3,Qe4 or Ke3,Pe4,Qe7.  A few of those can be traced back a move or two, 
  e.g. Ke7,Qg1,Pd3 mate in three against either Kd5 or Ke5;
  but even there it's not clear that the bK can be 
  corralled into such a mate from further out.
Answers.UserDisplayName: Noam D. Elkies
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Answers.UserDisplayName: double-beep
Body: <p>Most of the winnable endgames require cornering the opposing king to checkmate, often using zugzwang. However, if you have an infinite board (infinite in every direction), you won't be able to corner the opposing king as easily, as they will be able to retreat unless there is a threat behind them. For example, queen vs king will be a draw in this scenario, as the king and queen will never cover all escape routes. I also believe that two rooks will be unable to deliver mate, as you cannot force opposition. However, queen and rook will be able to checkmate, by using the queen to create a &quot;corner&quot; and checkmating using rook and king. In general, if a combination mates in a regular board, that combination plus a queen will be able to mate here; but I'm looking for more interesting combinations.</p>
  <p>So, which piece combinations will be able to mate in this scenario?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-12-18T20:40:18.320Z
Comments.Text: Presumably three rooks doesn't count, as on an infinite board there's no way to promote :)
Comments.UserDisplayName: Especially Lime
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Comments.CreationDate: 2020-12-18T20:53:41.267Z
Comments.Text: Well, this whole question is a theoretical exercise, so any combination of pieces is allowed :)
Comments.UserDisplayName: Saat
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-12-20T20:31:25.137Z
Comments.Text: This gets even more interesting if you bring in transfinite mathematics.  I'm intrigued by mates that require that the white king is a finite distance away from the black king, or to counter the "unbounded number of knights can't mate" answer with ω² knights or something abusrd.
Comments.UserDisplayName: Cort Ammon
Tags: <checkmate><chess-variants>
Title: On an infinite board, which pieces are needed to checkmate?
ThreadText: # On an infinite board, which pieces are needed to checkmate?
  
  (Asked by: username_0 on 2020-12-18)
  
  Most of the winnable endgames require cornering the opposing king to checkmate, often using zugzwang. However, if you have an infinite board (infinite in every direction), you won't be able to corner the opposing king as easily, as they will be able to retreat unless there is a threat behind them. For example, queen vs king will be a draw in this scenario, as the king and queen will never cover all escape routes. I also believe that two rooks will be unable to deliver mate, as you cannot force opposition. However, queen and rook will be able to checkmate, by using the queen to create a "corner" and checkmating using rook and king. In general, if a combination mates in a regular board, that combination plus a queen will be able to mate here; but I'm looking for more interesting combinations.
  
  So, which piece combinations will be able to mate in this scenario?
  
  **Tags:** checkmate, chess-variants
  
  > **Comments:**
  > **username_1**: Presumably three rooks doesn't count, as on an infinite board there's no way to promote :)  
  > **username_0**: Well, this whole question is a theoretical exercise, so any combination of pieces is allowed :)  
  > **username_2**: This gets even more interesting if you bring in transfinite mathematics.  I'm intrigued by mates that require that the white king is a finite distance away from the black king, or to counter the "unbounded number of knights can't mate" answer with ω² knights or something abusrd.  
  
  ---
  
  ## 4 answers
  
  ### Answer by username_10 on 2020-12-18. Score: 15 | Accepted answer
  
  Here are some mates I found:
  
  1. Queen and rook (limit the black king until the white arrives to help)
  
     ```
     [Title "KQR vs k mate"]
     [FEN "1R6/5k2/8/8/8/8/6Q1/K7 w - - 0 1"]
  
     1. Qg8+ Ke7 2. Rd8 Kf6 3. Re8 Kf5 4. Kb2 Kf4
     5. Qg2 Kf5 6. Kc3 Kf6 7. Kd4 Kf7 8. Qg8+ Kf6
     9. Ke3 Kf5 10. Kf3 Kf6 11. Re6+ Kf5 12. Qg6#
  
     ```
  2. Queen and pawn (not sure if the position can be reached with optimal play)
  
     ```
     [FEN "8/8/4Q3/3Pk3/8/4K3/8/8 w - - 0 1"]
  
     ```
  3. Queen and bishop (again, limit the king with the queen and the bishop, so the attacking king finds the time to approach)
  
     ```
     [Title "KQB vs k mate"]
     [FEN "Q7/2k5/8/8/B7/8/8/7K w - - 0 1"]
  
     1. Kg2 Kb6 2. Kf3 Kc5 3. Qb7 Kc4 4. Ke3 Kc5
     5. Ke4 Kd6 6. Qd7+ Kc5 7. Qd4#
  
     ```
  4. Queen and knight (limit the black king with the queen and the knight as much as possible)
  
     ```
     [Title "KQN vs k mate"]
     [FEN "Q7/2k5/8/8/8/1N6/8/7K w - - 0 1"]
  
     1. Kg2 Kd7 2. Kf3 Ke7 3. Ke4 Kf6 4. Qg8 Ke7
     5. Ke5 Kd7 6. Kd5 Ke7 7. Nc5 Kf6 8. Ne4+ Ke7
     9. Ke5 Kd7 10. Nd6 Kc6 11. Qb8 Kc5 12. Qb5#
  
     ```
  5. Two rooks ([username_3's answer](https://chess.stackexchange.com/a/33385) mentions how the mate can be forced)
  
     ```
     [FEN "8/8/R7/1R2k3/8/4K3/8/8 w - - 0 1"]
  
     ```
  
  > **Comments:**
  > **username_0**: Are all of them possible with black's optimal play?  
  > **username_11**: Two rooks seems pretty easy.  
  > **username_3**: Now that I think about it, I don't see how that KQP/K mate can be forced. What was Black's last move?  There are a few other mating positions: Ke5 mated also by Kc5,Pe4(g4),Qf5 or Ke7,Pd3,Qe4 or even  Kf7,Pd3,Qe4 or Ke3,Pe4,Qe7.  A few of those can be traced back a move or two,  e.g. Ke7,Qg1,Pd3 mate in three against either Kd5 or Ke5; but even there it's not clear that the bK can be  corralled into such a mate from further out.  
  
  ---
  
  ### Answer by username_3 on 2020-12-19. Score: 27
  
  In fact K+R+R can force checkmate against K on an open board.
  
  First move both Rooks away from the King and to different rows and columns.
  
  Then check the King with one Rook, say on a row.
  Whichever side the King goes, move the other Rook to limit it to a single row.
  
  Now the Rooks, moving only horizontally,
  can limit the defending King to just four squares on that row,
  with enough spare moves for the attacking King to approach:
  
  ```
  [Title "RR keep K boxed in"]
  [FEN "R7/5k2/2R5/8/8/8/8/K7 w - - 0 0"]
  
  1. Rh8 Kg7 2. Rch6 Kf7 3. Kb2 Ke7 4. Rc8 Kd7 5. Rhc6 Ke7 6. Kc3 Kf7 7. Rh8
  
  ```
  
  Once the King has joined the fray, mate can soon follow. For example:
  
  ```
  [Title "KRR mate K on infinite board"]
  [FEN "7R/5k2/2R5/8/3K4/8/8/8 w - - 0 0"]
  
  1. Ke5 Kg7 (1... Ke7 2. Rc7#) 2. Rch6 Kf7 3. Kd6 Kg7 4. Kd5 Kf7 
  5. Ke5 Kg7 (5... Ke7 6. R6h7#) 6. Kf5 Kf7 7. R6h7#
  
  ```
  
  > **Comments:**
  > **username_4**: Can you prove that 3 bishops are insufficient? I think it is, but am unable to prove it. 2 white-sq bishops plus 2 black-sq bishops are trivially enough, but even 4 white-sq bishops plus 1 black-sq bishop seems hard if at all possible.  
  > **username_3**: With only one dark-square Bishop there is no mating position, let alone a forced mate. The attacking King and dark-squared Bishop would have to control either a diamond of four or an X-shape of five squares around the bare king, and neither of these is possible.  
  > **username_4**: Oh lol that was trivial; I was so focused on trying to trap the king that I didn't try to find the final position needed.  
  
  ---
  
  ### Answer by username_5 on 2020-12-19. Score: 13
  
  An *infinite* number of knights would be insufficient to force checkmate, assuming the enemy king starts some distance away and not surrounded. The knights are too slow and not enough of them can chase an enemy king that simply runs away.
  
  A rook and two bishops, on the other hand, could force checkmate. It's rather easy to force a position like this:
  
  ```
  [FEN "3BB3/8/4k3/R7/8/8/8/4K3 w - - 0 1"]
  
  ```
  
  The Black king can only shuffle between two squares while the White king gets into position on d4 or e4 to set up Ra6#.
  
  Four bishops could also force a checkmate:
  
  ```
  [FEN "3BB3/8/4k3/8/3BB3/3K4/8/8 w - - 0 1"]  
  
  1. Bc5 Ke5 2. Bd7 Kf4 3. Bd6#
  
  ```
  
  I'm guessing that three bishops and a knight, or two bishops and two knights, could also force checkmate. The bishop pair is very effective at corralling the enemy king.
  
  > **Comments:**
  > **username_6**: An infinite number of Knights on an infinite board may force mate if the local density of Knights near the opposing king is high.  
  > **username_5**: They'd have to be very near the king (like, within two ranks or files) or on both sides of it.  
  > **username_7**: @username_6 If you can add constraints to the initial position, then a finite number of knights can suffice. In effect --- that is what you are talking about, since only finitely many of those knights in the vicinity of the king will play a role. It could be of interest to characterize starting positions for which knights can checkmate on an infinite board.  
  > **username_6**: The original statement added constraints to the initial position, namely: "assuming the enemy king starts some distance away and not surrounded."  This made me start to think about other initial conditions and if there was an unstated assumption about knight distribution. The initial position could be mate, but that wouldn't be very interesting :)  
  > **username_7**: @username_6 The implicit logic of the original statement is that infinitely many knights are not always able to force mate *because* the king might be too far away. Two rooks, for example, can always force mate no matter where the enemy king is. even Infinitely many knights lack that property. I agree that there are some interesting questions involving the distribution of those infinitely many knights relative to the king. When the king is partially but not wholly surrounded it could be an interesting king hunt. Sort of like the king is a blockade runner.  
  > **username_8**: @username_6 5 knights can do it. 4 knights to cover the 8 adjacent squared, and a 5th solely to target the center and cause the mate  
  > **username_5**: @username_8 4 knights and a king (or, indeed, any combination of 4 pieces - including pawns! - and a king) could, *if* the initial setup allows it and/or the lone king cooperates. But it seemed like OP wanted to know what pieces could *force* checkmate, and knights (or knights and a king) can't do that unless you make assumptions about their starting position relative to the lone king.  
  
  ---
  
  ### Answer by username_9 on 2024-08-01. Score: 3
  
  Since this post was made, people have begun to play infinite chess somewhat seriously. The following piece combinations (together with the king) are known to force mate:
  
  * 2 Queens
  * Queen + Rook
  * Queen + Bishop
  * Queen + Knight
  * 2 Rooks
  * Rook + 2 Bishops of different colors
  * Rook + 2 Bishops of the same color
  * Rook + Knight + Bishop
  * Rook + 2 Knights
  * 2 Bishops of one color + 2 Bishops of the other color
  * 2 Bishops of one color + 1 Bishop of the other color + 1 Knight
  * 1 Bishop of one color + 1 Bishop of the other color + 2 Knights
  
  No subset of any of the above can force mate for the trivial reason that no mates exist. No amount of knights is enough if they all start to one side of the king. No amount of bishops is enough if you do not have at least 2 of each color.
  
  The remaining open questions are in the situation where there are only bishops of a single color and knights. There is a quite lazy argument that 10 or so light squared bishops + 2 knights are able to force mate, but we don't even know if 1 bishop + 3 knights can force mate or not. I will probably ask this as a separate question.
  
  I will now talk briefly about the methods for each checkmate in the above list. Since there are a lot I won't be as thorough as I'd like, but please ask about any endgame you find suspicious.
  
  * The endgames involving a Queen are easy enough, on its own the Queen can confine the king to a 5x5 region by moving between the corners of a 7x7 square as needed. While doing this there are always moves where the Queen needn't budge to maintain the prison, which the attacker can use to approach with their other pieces. Forcing the final mate is simple.
  * username_3 username_3 has already described in this thread a technique with 2 Rooks
  * The endgames with a Rook and two minors can all be done in essentially the same way (though for the ones with a bishop there are easier ways):
  
    1. Using a distant rook to prevent the Black king from running forever in one direction, get your king onto the same row as Black's king.
    2. Use your rook to confine Black's king to 3 rows (by always moving in time to prevent its escape), centered on the row your king is on. By doing this you gain a free move every other move, and prevent the kings passage beyond the column your king is on.
    3. Run one of your minor pieces to the opposite side of the Black king, so that it is sandwiched between your king and this minor piece. If your minor piece is a knight you may need to switch to confining the Black king to 4 rows instead of 3 once it is suitably far from your king, otherwise the knight won't have enough free time to catch up.
    4. Move your rook to be guarded by the minor piece, so that the king cannot gain access to the columns next to your minor piece. The king is now trapped in a finite area - the three rows we have been keeping it on, and the columns between your king and minor piece.
    5. Squeeze this area as small as possible, then approach with the second minor piece. Eventually you can force a trap of the form
  
  For example
  
  ```
  [FEN "8/2R5/8/4kNK1/8/8/8/8 w - - 0 1"] 
  
  ```
  
  or
  
  ```
  [FEN "8/2R5/8/B2k1K2/8/8/8/8 w - - 0 1"]
  
  ```
  
  , from which mating is not so bad.
  
  * The endgames with two bishops of opposite colors are not so bad. Two bishops of different colors can confine the king to a finite area, which lets you position the pieces as needed. Then you can switch to confining the king to 7 half diagonals and use the king and other minor pieces to squeeze the king into one of several small trap. I used a variant of my [tablebase script](https://gist.github.com/mjtb49/51020875338082e7369e6b72d7cd3fb0) to verify that the positions I can force like this are indeed forced mates, though allegedly other people have cleaner arguments for the finish here.
  
  As a bonus, it is known that a Queen and a Pawn (which we say is unable to promote or move two squares at once) is enough, but there is a unique (up to a reflection) mate in 4 in that endgame which you must pass through.
  
  ```
  [FEN "8/3K4/8/2k5/8/3P4/1Q6/8 w - - 0 0"]
  
  ```
  
  I am unsure of the status of other endgames with a pawn on the board.
  
  > **Comments:**
  > **username_3**: "It is known that a Queen and a Pawn is enough" -- do you have a reference for this?  I found this a few months ago but thought it was new. I used this concluding sequence (1 Kc7 Kd5 2 Qf6 Kc5 3 Qb6+ Kd5 4 Qd6#) but there are other possible finishes: put Q on a1 instead and it can end 1 Qa4 Kb6 2 Kc8 Kc5 3 Kc7 Kd5 4 Qe4+ Kc5 5 Qc4#.  (If 1 Qa4 Kd5 then 2 Qa7 Ke5 3 Qf2 Kd5 4 Qb2 Kc5 and we reach your diagram.)  
  > **username_3**: Also, in your KRB/K diagram, if White also has a Pe3 then it's mate in 3 (1 Bc3 Kd6 2 Be5+ Kd5 3 e4#), suggesting that RBP may be enough. There are similar RNP setups too.  
  > **username_9**: If you found the QKP/K a few months ago then probably you are the first person to. I was told that others had a method but I could never track them down, so we had to work out a new one.  Your comment made me doubt the uniqueness result I claimed in the post, which came from an old version of my tablebase script. I modified the script to check the claim. The correct claim is that it is the unique (up to symmetry) mate in 4 which can be forced by White (so the previous position with the king on d5 and Black to move is the truly unique one).  
  > **username_9**: The updated script runs much faster than my old one, and it reveals that I got quite lucky with the method I found. The position 3K4/8/4k3/8/8/3P4/5Q2/8 w - - 0 1 , which  my method passes through, is one of only 4 (up to symmetry) mates in 22 ply (meaning after Black's move, White has at worst a mate in 11). There also is apparently only 1 mate in 10 ply,  2 mates in 18 ply, and 2 mates in 20 ply.  
  > **username_9**: Though I should mention that my script only checks for moves of some bounded length (though it won't let the king move into some distant check). The above results are true provided that there isn't some other short mate where White needs to make a move of more than 200 squares.  
  
  ---
  
  