Answers.Body: <p>It is trivially a win for white even on an 8x8 board</p>
  <pre><code>[fen &quot;8/8/8/8/8/8/8/rPPPPPPP w - - 0 1&quot;] 
  
  1. h2 Rxb1 2. g2 null 3. h3 Rxc1 4. f2 null 5. g3 null 6. h4 Rxd1 7. e2 null 8. f3 null 9. g4 null 10. h5 Re1 11. e3 null 12. f4 null 13. g5 null 14. h6 Rxe3 15. f5 null 16. g6 null 17. h7 Rf3 18. f6 null 19. g7 null 20. h8=Q Rxf6
  </code></pre>
  
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Answers.CreationDate: 2021-11-04T14:42:38.713Z
Answers.Text: At the risk of being naive - even with the looming threat of g8=Q, is it guaranteed that with perfect play, two queens *always* can capture a rook? I see the logic here that lead to an inevitable loss for Black on the infinite board, but I'm not sure if it's true for an 8x8.
Answers.UserDisplayName: AKemats
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Answers.Text: I agree, it's still interesting whether two queens are enough to trap black's rook on the 8-by-8 board. I suspect it is, but not exactly sure yet. It took me a while to figure out why the win is trivial in the infinite case, but it has to do with white being able to guarantee a promotion of at least 8 pawns, after which the queens take away one row each while protecting one another.
Answers.UserDisplayName: Scounged
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Answers.Text: Thanks for clarifying, I got tripped up by the formulation of the question. Actually, it's possible to prove that white cannot win given optimal play in the example above, since white only gets two queens. We see it as follows: a white queen can cover at most 3 squares on the f-file at the same time, so 2 white queens can cover at most 6 squares on the f-file. Therefore, the black rook will always be able to find a safe square on the f-file, no matter what white does.
Answers.UserDisplayName: Scounged
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Answers.CreationDate: 2021-11-05T05:56:36.490Z
Answers.Text: It's a trivial win on the infinite board after white gets one queen, as the queen can protect one pawn to queen each time, until white has 8 queens. At that point the rook is mated.
Answers.UserDisplayName: justhalf
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Answers.Text: @Scounged: Two queens can't take all rook flights unless they are both attacking the rook (one on the rank, one at the file) and protect each other. But in that case, why didn't one queen take the rook in the move before? Three should suffice, methinks. (For controlling *all* squares of an 8*8 board, five are needed)
Answers.UserDisplayName: Hauke Reddmann
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Answers.CreationDate: 2021-11-05T10:51:30.783Z
Answers.Text: @HaukeReddmann Indeed, two queens aren't enough. Interestingly, three queens also seem to be insufficient, but I think that four queens should be enough to trap the rook on a standard chessboard. This raises a new question: given an n-by-8 board, how many queens (or maybe allow some knights as well) are needed to capture the rook?
Answers.UserDisplayName: Scounged
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Answers.CreationDate: 2021-11-05T11:07:11.187Z
Answers.Text: @Scounged: Dunno, but five queens protect even all squares of a 13x13 board, except their own squares. (Questions like these call for fairy tablebases, which are poorly researched.)
Answers.UserDisplayName: Hauke Reddmann
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Answers.CreationDate: 2024-09-21T00:50:19.417Z
Answers.Text: 4Q vs. R on an 8x8 board is a win at least if "check" is not required: Qc2,g3,f7,b6 limit the Rook to the corners, and then if Rh8 then Qc2-c6 and the Rh8 has no moves.   If the Rook is allowed to pass then the further Qb6-d4 traps it.
Answers.UserDisplayName: Noam D. Elkies
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Answers.CreationDate: 2024-09-22T01:38:50.807Z
Answers.Text: @Scounged 7 queens suffice to kill a Rook on an 8-by-anything board.  Start with Q's on a1,a2,...,a7 so the R must be on the 8th.  Then Qa8, Qa6-b7, Qa5-b6, Qa4-b5, Qa3-b4, Qa2-b3, Qa1-b2, Qb1, Qb3-b2, Qb4-b3, Qb5-b4, Qb6-b5, Qb7-b6, Qa8-b7.  Each move attacks the Rook with a protected Queen, so the Rook must keep moving vertically.  Now repeat the procedure: Qb8, Qb6-c7, Qb5-c6, etc., inching one column to the right every 14 moves.  Eventually the wall of Queens engulfs the Rook.
Answers.UserDisplayName: Noam D. Elkies
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Answers.CreationDate: 2021-11-04T14:32:34.370Z
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Answers.UserDisplayName: Brian Towers
Answers.Body: <p>The &quot;trivial&quot; white wins solution has been refuted by the fact that the &quot;win&quot; is actually a loss. You can't construct a mate even without figuring out a forcing sequence of moves.
  The infinte board is another story. Black will need to figure out a different way to stop white when she has to go through an infinte number of pawns. I am not 100% convinced that there isnt some way to stop white with the rook other than brute force pawn gobble, and on an infinte board that ain't gonna work.
  It feels like white should win on an infinte board - but before coming to this conclusion, there are possibly other options which dont begin with ...Rxb1. I couldn't find any way to stop the pawns without going for the &quot;let him only get 2 queens &quot; strategy, but I certainly didn't exhaust all the possibilities.</p>
  
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Answers.CreationDate: 2021-11-05T05:01:18.417Z
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Answers.UserDisplayName: potmo
Answers.Body: <p>White has an infinite number of pawns of which black can capture one in his first move. Since black can always capture only one pawn per move at some point white has all his remaining pawns on the second rank.</p>
  <p>&quot;All his reamining pawns&quot; are infintely many divided by two - which is still infinitely many. You could, instead of naming the lines &quot;a, b, c, d, ...&quot; also number then and then &quot;every second pawn&quot; means &quot;any pawn on a line with an even number&quot;. Since there are infinitely many even numbers (equally many as there are integers) there are also (countably) infintely many pawns left.</p>
  <p>From there on white will advance every pawn in a move and black will take exactly one. Since infinity minus 7 is still infinity, after 7 more moves white will have infinitely many queens and black still has one rook. From there on the solution seems trivial.</p>
  
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Answers.CreationDate: 2024-09-20T15:50:52.157Z
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Answers.UserDisplayName: bakunin
Body: <p>I've had this puzzle rattling around in my brain for the past few days and I want to know what you all make of it.</p>
  <p>Imagine chess played on a board with eight ranks and an infinite number of files. Black has a single rook on a1. White has an infinite number of pawns, one on every square in the first rank besides a1.</p>
  <p>The rules here are slightly different from normal chess.</p>
  <blockquote>
  <ol>
  <li>White's goal is to capture the rook. Black's goal is to never get captured.</li>
  <li>White's pawns can only move one space forward and never have the option to move forward two spaces. They still capture diagonally.</li>
  <li>White may move one pawn off of the first rank, and any number of pawns already off the first rank simultaneously (notated with an ampersand).</li>
  </ol>
  </blockquote>
  <p>Is White guaranteed to capture Black's rook at some point in time, or does Black have a strategy to survive? I <em>think</em> White has a guaranteed victory, but I can't say for sure. Consider the following string of moves.</p>
  <pre><code>1.b2 Rxc1
  2.b3&amp;d2 Rxe1
  3.b4&amp;d3&amp;f2 Rxg1
  4.b5&amp;d4&amp;f3&amp;h2 Rxi1
  5.b6&amp;d5&amp;f4&amp;h3&amp;j2 Rb1
  6.b7&amp;d6&amp;f5&amp;h4&amp;j3&amp;k2 Rxb7
  7.d7&amp;f6&amp;h5&amp;j4&amp;k3&amp;l2 Rxd7
  8.f7&amp;h6&amp;j5&amp;k4&amp;l3&amp;m2 Rxf7
  9.h7&amp;j6&amp;k5&amp;l4&amp;m3&amp;n2 Rxh7
  10.j7&amp;k6&amp;l5&amp;m4&amp;n3&amp;o2
  </code></pre>
  <p>Now Black is stuck. If they play 10... Rxj7, then White follows with 11. kxj7 and wins. If they do anything else, then 11. j8=Q is sure to follow, which spells out doom for Black. Does Black have a more optimal strategy that leads to their survival, or is this it?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-04T13:41:58.400Z
Comments.Text: What is the starting position of the rook? And can white's pawns promote? Otherwise the question is rather dull, since black's rook can just hide behind white's pawns then.
Comments.UserDisplayName: Scounged
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-04T14:02:43.193Z
Comments.Text: 1. Is the ∞ the length or the width. 2. Do the pawns promote?
Comments.UserDisplayName: Shambhav
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-04T14:31:49.763Z
Comments.Text: @ShambhavGautam There's an infinite number of files (a, b, c, d...), but in any file, pawns *can* promote on the 8th rank.
Comments.UserDisplayName: AKemats
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-04T16:04:02.910Z
Comments.Text: I think with rule 3, the game is trivially won for White, but without it... it's at least interesting.
Comments.UserDisplayName: Annatar
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Comments.CreationDate: 2021-11-04T16:09:59.893Z
Comments.Text: Even without rule 3 it should still be an easy win for White no? If White just plays some pawns far away, a defense from behind would not be possible. So the only defense would be from the side (or from in front but then White protects those advancing pawns then advances elsewhere) when White can slowly advance with a mass of pawns.
Comments.UserDisplayName: koedem
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-04T18:36:36.667Z
Comments.Text: Commenters on the answer by Brian Towers seem to be reading your "every other square" as meaning something different than "every square on the first rank other than the one the rook is on". Could you make that more explicit in your formulation?
Comments.UserDisplayName: L. Scott Johnson
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-05T07:53:23.010Z
Comments.Text: @koedem Yes, that was my first thought as well. The rook can only properly attack from in front, but needs 3 moves to go from n1 to x1, so the pawns gain free moves and win, right? However, what if the rook only captures bases of those pawn chains where the head pawn has advanced to the third (or fourth?) rank? That would work out tempo wise and we would have to show whether the resulting pawn island can still make it to the finish line as it does not have infinite cover anymore.
Comments.UserDisplayName: Annatar
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-05T08:24:47.617Z
Comments.Text: On a *third* thought, infinity still works in White's favor, doesn't it. He can have an infinite number of pawns on the second rank, cluttering it just like the first, while the rook still needs more than one move to capture the new second-rank-bases, so then White has time for an infinite number of pawns on the third rank and so on...
Comments.UserDisplayName: Annatar
Tags: <chess-variants><puzzles>
Title: On an 8-by-∞ board, who wins between an infinite number of pawns and a rook?
ThreadText: # On an 8-by-∞ board, who wins between an infinite number of pawns and a rook?
  
  (Asked by: username_0 on 2021-11-04)
  
  I've had this puzzle rattling around in my brain for the past few days and I want to know what you all make of it.
  Imagine chess played on a board with eight ranks and an infinite number of files. Black has a single rook on a1. White has an infinite number of pawns, one on every square in the first rank besides a1.
  The rules here are slightly different from normal chess.
  
  
  White's goal is to capture the rook. Black's goal is to never get captured.
  White's pawns can only move one space forward and never have the option to move forward two spaces. They still capture diagonally.
  White may move one pawn off of the first rank, and any number of pawns already off the first rank simultaneously (notated with an ampersand).
  
  
  Is White guaranteed to capture Black's rook at some point in time, or does Black have a strategy to survive? I think White has a guaranteed victory, but I can't say for sure. Consider the following string of moves.
  1.b2 Rxc1
  2.b3&d2 Rxe1
  3.b4&d3&f2 Rxg1
  4.b5&d4&f3&h2 Rxi1
  5.b6&d5&f4&h3&j2 Rb1
  6.b7&d6&f5&h4&j3&k2 Rxb7
  7.d7&f6&h5&j4&k3&l2 Rxd7
  8.f7&h6&j5&k4&l3&m2 Rxf7
  9.h7&j6&k5&l4&m3&n2 Rxh7
  10.j7&k6&l5&m4&n3&o2
  
  Now Black is stuck. If they play 10... Rxj7, then White follows with 11. kxj7 and wins. If they do anything else, then 11. j8=Q is sure to follow, which spells out doom for Black. Does Black have a more optimal strategy that leads to their survival, or is this it?
  
  **Tags:** chess-variants, puzzles
  
  > **Comments:**
  > **username_1**: What is the starting position of the rook? And can white's pawns promote? Otherwise the question is rather dull, since black's rook can just hide behind white's pawns then.  
  > **username_2**: 1. Is the ∞ the length or the width. 2. Do the pawns promote?  
  > **username_0**: @username_2Gautam There's an infinite number of files (a, b, c, d...), but in any file, pawns *can* promote on the 8th rank.  
  > **username_3**: I think with rule 3, the game is trivially won for White, but without it... it's at least interesting.  
  > **username_4**: Even without rule 3 it should still be an easy win for White no? If White just plays some pawns far away, a defense from behind would not be possible. So the only defense would be from the side (or from in front but then White protects those advancing pawns then advances elsewhere) when White can slowly advance with a mass of pawns.  
  > **username_5**: Commenters on the answer by username_6 seem to be reading your "every other square" as meaning something different than "every square on the first rank other than the one the rook is on". Could you make that more explicit in your formulation?  
  > **username_3**: @username_4 Yes, that was my first thought as well. The rook can only properly attack from in front, but needs 3 moves to go from n1 to x1, so the pawns gain free moves and win, right? However, what if the rook only captures bases of those pawn chains where the head pawn has advanced to the third (or fourth?) rank? That would work out tempo wise and we would have to show whether the resulting pawn island can still make it to the finish line as it does not have infinite cover anymore.  
  > **username_3**: On a *third* thought, infinity still works in White's favor, doesn't it. He can have an infinite number of pawns on the second rank, cluttering it just like the first, while the rook still needs more than one move to capture the new second-rank-bases, so then White has time for an infinite number of pawns on the third rank and so on...  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_6 on 2021-11-04. Score: 2
  
  It is trivially a win for white even on an 8x8 board
  [fen "8/8/8/8/8/8/8/rPPPPPPP w - - 0 1"] 
  
  1. h2 Rxb1 2. g2 null 3. h3 Rxc1 4. f2 null 5. g3 null 6. h4 Rxd1 7. e2 null 8. f3 null 9. g4 null 10. h5 Re1 11. e3 null 12. f4 null 13. g5 null 14. h6 Rxe3 15. f5 null 16. g6 null 17. h7 Rf3 18. f6 null 19. g7 null 20. h8=Q Rxf6
  
  > **Comments:**
  > **username_0**: At the risk of being naive - even with the looming threat of g8=Q, is it guaranteed that with perfect play, two queens *always* can capture a rook? I see the logic here that lead to an inevitable loss for Black on the infinite board, but I'm not sure if it's true for an 8x8.  
  > **username_1**: I agree, it's still interesting whether two queens are enough to trap black's rook on the 8-by-8 board. I suspect it is, but not exactly sure yet. It took me a while to figure out why the win is trivial in the infinite case, but it has to do with white being able to guarantee a promotion of at least 8 pawns, after which the queens take away one row each while protecting one another.  
  > **username_1**: Thanks for clarifying, I got tripped up by the formulation of the question. Actually, it's possible to prove that white cannot win given optimal play in the example above, since white only gets two queens. We see it as follows: a white queen can cover at most 3 squares on the f-file at the same time, so 2 white queens can cover at most 6 squares on the f-file. Therefore, the black rook will always be able to find a safe square on the f-file, no matter what white does.  
  > **username_7**: It's a trivial win on the infinite board after white gets one queen, as the queen can protect one pawn to queen each time, until white has 8 queens. At that point the rook is mated.  
  > **username_8**: @username_1: Two queens can't take all rook flights unless they are both attacking the rook (one on the rank, one at the file) and protect each other. But in that case, why didn't one queen take the rook in the move before? Three should suffice, methinks. (For controlling *all* squares of an 8*8 board, five are needed)  
  > **username_1**: @HaukeReddmann Indeed, two queens aren't enough. Interestingly, three queens also seem to be insufficient, but I think that four queens should be enough to trap the rook on a standard chessboard. This raises a new question: given an n-by-8 board, how many queens (or maybe allow some knights as well) are needed to capture the rook?  
  > **username_8**: @username_1: Dunno, but five queens protect even all squares of a 13x13 board, except their own squares. (Questions like these call for fairy tablebases, which are poorly researched.)  
  > **username_9**: 4Q vs. R on an 8x8 board is a win at least if "check" is not required: Qc2,g3,f7,b6 limit the Rook to the corners, and then if Rh8 then Qc2-c6 and the Rh8 has no moves.   If the Rook is allowed to pass then the further Qb6-d4 traps it.  
  > **username_9**: @username_1 7 queens suffice to kill a Rook on an 8-by-anything board.  Start with Q's on a1,a2,...,a7 so the R must be on the 8th.  Then Qa8, Qa6-b7, Qa5-b6, Qa4-b5, Qa3-b4, Qa2-b3, Qa1-b2, Qb1, Qb3-b2, Qb4-b3, Qb5-b4, Qb6-b5, Qb7-b6, Qa8-b7.  Each move attacks the Rook with a protected Queen, so the Rook must keep moving vertically.  Now repeat the procedure: Qb8, Qb6-c7, Qb5-c6, etc., inching one column to the right every 14 moves.  Eventually the wall of Queens engulfs the Rook.  
  
  ---
  
  ### Answer by username_10 on 2021-11-05. Score: 0
  
  The "trivial" white wins solution has been refuted by the fact that the "win" is actually a loss. You can't construct a mate even without figuring out a forcing sequence of moves.
  The infinte board is another story. Black will need to figure out a different way to stop white when she has to go through an infinte number of pawns. I am not 100% convinced that there isnt some way to stop white with the rook other than brute force pawn gobble, and on an infinte board that ain't gonna work.
  It feels like white should win on an infinte board - but before coming to this conclusion, there are possibly other options which dont begin with ...Rxb1. I couldn't find any way to stop the pawns without going for the "let him only get 2 queens " strategy, but I certainly didn't exhaust all the possibilities.
  
  
  ---
  
  ### Answer by username_11 on 2024-09-20. Score: 0
  
  White has an infinite number of pawns of which black can capture one in his first move. Since black can always capture only one pawn per move at some point white has all his remaining pawns on the second rank.
  
  "All his reamining pawns" are infintely many divided by two - which is still infinitely many. You could, instead of naming the lines "a, b, c, d, ..." also number then and then "every second pawn" means "any pawn on a line with an even number". Since there are infinitely many even numbers (equally many as there are integers) there are also (countably) infintely many pawns left.
  
  From there on white will advance every pawn in a move and black will take exactly one. Since infinity minus 7 is still infinity, after 7 more moves white will have infinitely many queens and black still has one rook. From there on the solution seems trivial.
  
  
  ---
  
  