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?

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    Presumably three rooks doesn't count, as on an infinite board there's no way to promote :) – Especially Lime Dec 18 '20 at 20:40
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    Well, this whole question is a theoretical exercise, so any combination of pieces is allowed :) – Saat Dec 18 '20 at 20:53
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    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. – Cort Ammon Dec 20 '20 at 20:31

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 (Noam's answer mentions how the mate can be forced)

    [FEN "8/8/R7/1R2k3/8/4K3/8/8 w - - 0 1"]
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    Are all of them possible with black's optimal play? – Saat Dec 18 '20 at 18:17
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    Two rooks seems pretty easy. – xehpuk Dec 19 '20 at 13:51
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    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. – Noam D. Elkies Dec 19 '20 at 16:18

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#
  • 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. – user21820 Dec 19 '20 at 15:47
  • 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. – Noam D. Elkies Dec 19 '20 at 18:00
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    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. – user21820 Dec 20 '20 at 2:00

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.

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    An infinite number of Knights on an infinite board may force mate if the local density of Knights near the opposing king is high. – Michael West Dec 19 '20 at 14:51
  • They'd have to be very near the king (like, within two ranks or files) or on both sides of it. – D M Dec 19 '20 at 20:48
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    @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. – John Coleman Dec 19 '20 at 22:37
  • 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 :) – Michael West Dec 19 '20 at 22:56
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    @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. – John Coleman Dec 20 '20 at 0:03

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