As title says. My question is about legal positions (positions which could be reached in actual games), but if there is a known solution with an illegal position to start the sequence, I'd be interested in knowing that too. Just to be clear, I'm not interested in the longest forced checkmate in actual games, but the longest possible. I do not necessarily mean "doubly forced", and I'm not interested in tournament rules like 50 move and other similar.

  • Does this answer your question? What's the longest checkmate in a 7-piece tablebase?
    – Allure
    Oct 27 at 9:13
  • 6
    Please clarify: How do you play chess with illegal moves? :-) Do you mean "doubly forced" (only one legal move for both players at any time) or White can force a mate against best defense of Black (in that case the table bases give the answer, AFAIK #549, see Allure's comment)? Do you play under tournament conditions (50 move rule interferes)? The table bases can also answer that but some more research is needed then. Oct 27 at 9:18
  • infinite considering running away from a checkmate as illegal
    – cmgchess
    Oct 27 at 15:42
  • @HaukeReddmann what's table base? Oct 29 at 8:14
  • 1
    @MauroGiliberti: Computers were used to find the absolutely best move (by exhaustively looking at the game tree) in positions with few pieces. For example, we know that KQ/K can mate in at most ten moves from any position (with White to move and legal position, of course). Thus you now can look up any (legal) position with at most 7 pieces and know for sure how long the mate takes. The link Allure gave might or might not answer your question, since with 8 pieces the longest possible mate may still be longer - we don't know yet. (Computer time and space needed is immense.) Oct 29 at 8:23

The answer is "yet unknown, but we have a known current record holder".

There are two main sources for records.

  • Chess problems. The longest "mate in n" which is also testedly correct (in the pre-computer age there were even longer attempts) is around 300 moves. This is also probably the maximum - there are only so many Black pawn tempo moves and long zugzwang triangulations to give the move to Black, so you won't get much more than 300 moves that way (this even holds with positions which are illegal due to impossible chains of pawn captures). See also here on CSE.
  • Table bases. By listing all mate positions, then working backwards (finding all mate in 1, then mate in 2 etc.), a computer can in principle know the value of any position without a doubt. For example KQ/K is at most #10 for any position. For time and space reasons, though, the table bases are currently known only for 7 pieces at most. This means:
  • The current record holder for the longest mate with ignoring the 50 move rule was implicitely given by Allure in his link: 549 moves. I give another with some additional info: see here. EDIT: Newsflash! Make that 584 (contains a promoted B).
  • Snag 1: If we now play by the 50 move rule, that game would have been drawn long ago. To my best knowledge, the record holder for this variant is unknown, and it would be quite a feature to repeat all the calculations. This would be best asked in a chess programmers community. We have a contender, though: see here on CSE, found accidentally.
  • Snag 2: The 8-piece table base are fervently in the works, don't hold your breath. If you now expect any additional piece will break the record exponentially: Nope. An email with preliminary results by Marc Bourzutschky argues possible exchanges lower the record lengths, and for this reason the author expected we already are close to the natural maximum.

OP asks for the longest win under basic rules, so Snag 1 in @HaukeReddmann's answer wouldn't apply.

Records in chess problems have already been far exceeded in tablebases so far generated, so I think these can be discounted as a pointer to the answer.

The most convincing pointer I have come across so far is "Haworth's Law". (Guy Haworth sadly died very recently.) This is available via the following link: http://centaur.reading.ac.uk/36276/3/HaworthLaw.pdf

I reproduce it here.

enter image description here

I should note that the points shown on the graph for 3 to 7 men (8 to 10 are only predictions) form a better straight line than I ever managed in my physics pracs at school. Also the conjecture has so far proved itself as a predictive method. I believe it was first proposed on the the basis of the 3-4-5 man Nalimov tables and the values for 6 and 7 man tables have morphed from predictions into verified measurements.

The author wisely doesn't predict more than a few increments ahead, but I would say that the best guess for the longest mate with 32 men would be arrived at by extending the straight line already accurately measured by the EGTBs (as you would in verifying Boyle's law or whatever).

Because the y axis is a logarithmic scale this would give a longest forced mate with 32 men on the board of around 1.2 x 10¹¹ moves. (Not a footnote; twelve hundred trillion moves in American trillions.)

One possible snag here is that the EGTBs don't exclude all illegal positions and I imagine OP is looking for the longest forced mate from a legal position.

For the EGTBs produced so far this is not a problem because the maximal mates are all from legal positions. However starting from the 12 man EGTBs some of the endgame classifications become illegal (e.g. KQQQQQQQQQQK). This effect progresses as the number of men increases until by 32 men, although the number of classifications is enormous, the only one that contains any legal positions is KQRRBBNNPPPPPPPPKQRRBBNNPPPPPPPP.

It's not clear if that would affect the rate of increase. The straight line graph for legal positions could conceivably flatten beyond 12 men (though there's no reason I can think of to suggest it would). At any rate it's hard to imagine that the author's first conjecture could fail, so on the basis of tablebase results to date, it would be reasonable to assume the answer would at least exceed the intersection of Haworth's line with the 12 man ordinate, viz. about 25,000 moves.

A second possible snag may be "Snag 2" mentioned in @HaukeReddmann's answer.

I haven't seen the paper he refers to. I would be very grateful if he could actually dig it out.

Without further details this snag is less than convincing for two reasons.

The first is that the possibility of exchanges is already built into the tablebase results from the off. Obviously there are more exchanges possible in all 4 man endgames than there are in any three man endgames. Any trend apparent from the tablebase results should already take the possibility of exchanges into account.

The second is that in practice it becomes apparent that one of the major factors adding to the length of a mate is actually the necessity of avoiding exchanges. Anyone who has practiced queen v rook endings will probably have come to the conclusion that he could do the things faster if he didn't have to worry about the possibility of a queen for rook exchange.

OK, in that case he's swapping down. Here is another example where a maximal length mate in bishop & knight v knight is played by Nalimov. You will notice he has any number of opportunites to make an equal swap while Black's knight is on the board, but all such swaps would immediately draw.

[Event "?"]
[Site "?"]
[Date "2021.3.11"]
[Round "?"]
[White "Nalimov"]
[Black "Nalimov"]
[Result "?"]
[SetUp "1"]
[FEN "8/8/8/8/8/8/1n5B/2K1N2k w - - 0 1"]

1.Bh2-g3  Nb2-c4  2.Kc1-d1  Nc4-e3+  3.Kd1-e2  Ne3-f5  4.Bg3-f2  Nf5-e7  5.Ke2-f3  
Ne7-c6  6.Kf3-g3  Nc6-d4  7.Kg3-g4  Nd4-b3  8.Kg4-h3  Nb3-c1  9.Ne1-c2  Nc1-d3  10.Bf2-b6  
Nd3-f2+  11.Kh3-g3  Nf2-e4+  12.Kg3-f4  Ne4-d2  13.Nc2-e3  Kh1-h2  14.Kf4-g4  Nd2-b3  15.Bb6-a7  
Nb3-c1  16.Ba7-d4  Nc1-d3  17.Kg4-h4  Nd3-f2  18.Bd4-e5+  Kh2-g1  19.Be5-c3  Nf2-e4  20.Bc3-e1  
Ne4-f2  21.Kh4-g3  Nf2-e4+  22.Kg3-f4  Ne4-c5  23.Kf4-f3  Nc5-e6  24.Be1-c3  Kg1-h2  25.Kf3-g4  
Ne6-c5  26.Bc3-e5+  Kh2-g1  27.Kg4-f3  Nc5-e6  28.Kf3-g3  Ne6-c5  29.Be5-d4  Nc5-b3  30.Bd4-a7  
Nb3-d2  31.Ne3-c4+  Kg1-h1  32.Nc4-e5  Nd2-c4  33.Ne5-d3  Nc4-d2  34.Kg3-h3  Nd2-f3  35.Ba7-e3  
Nf3-e5  36.Nd3-f4  Ne5-c4  37.Be3-d4  Nc4-d2  38.Nf4-h5  Nd2-e4  39.Bd4-e3  Ne4-f2+  40.Kh3-g3  
Nf2-d3  41.Nh5-g7  Nd3-e5  42.Ng7-e6  Ne5-d3  43.Ne6-g5  Nd3-e5  44.Ng5-h3  Ne5-c4  45.Be3-f4  
Nc4-b2  46.Bf4-d6  Nb2-d3  47.Bd6-e7  Nd3-e1  48.Be7-c5  Ne1-d3  49.Bc5-d4  Nd3-c1 50.Kg3-f3  
Nc1-d3  51.Nh3-g5  Kh1-h2  52.Bd4-c3  Nd3-c5  53.Bc3-e5+  Kh2-h1  54.Be5-d4  Nc5-a6  55.Ng5-e4  
Na6-c7  56.Ne4-f2+  Kh1-g1  57.Nf2-d3+  Kg1-f1  58.Nd3-f4  Kf1-e1  59.Kf3-e3  Nc7-a6  60.Bd4-c3+  
Ke1-d1  61.Ke3-d4  Kd1-c2  62.Nf4-d5  Na6-b8  63.Kd4-c4  Nb8-c6  64.Nd5-e3+  Kc2-c1  65.Kc4-d5  
Nc6-b8  66.Kd5-d6  Nb8-a6  67.Kd6-c6  Na6-b8+  68.Kc6-c7  Nb8-a6+  69.Kc7-b6  Na6-b8  70.Bc3-g7  
Nb8-d7+  71.Kb6-c6  Nd7-b8+  72.Kc6-b5  Nb8-d7  73.Ne3-d5  Kc1-d1  74.Nd5-b4  Nd7-b8  75.Kb5-b6  
Nb8-d7+  76.Kb6-c6  Nd7-b8+  77.Kc6-c7  Kd1-e2  78.Kc7xb8  Ke2-e1  79.Kb8-c7  Ke1-f2  80.Kc7-d6  
Kf2-f3  81.Kd6-e5  Kf3-e3  82.Nb4-d5+  Ke3-f3  83.Nd5-f4  Kf3-e3  84.Ke5-f5  Ke3-f2  85.Kf5-e4  
Kf2-g1  86.Ke4-f3  Kg1-h1  87.Nf4-h3  Kh1-h2 88.Nh3-f2  Kh2-g1 89.Bg7-e5  Kg1-f1 90.Be5-h2  
Kf1-e1 91.Nf2-e4  Ke1-d1  92.Bh2-e5  Kd1-e1  93.Be5-d4  Ke1-f1  94.Ne4-d2+  Kf1-e1 95.Kf3-e3  
Ke1-d1 96.Ke3-d3  Kd1-e1  97.Bd4-c5  Ke1-d1 98.Bc5-f2  Kd1-c1 99.Nd2-c4  Kc1-d1  100.Nc4-e3+  
Kd1-c1 101.Kd3-c3  Kc1-b1 102.Kc3-b3  Kb1-c1  103.Bf2-e1  Kc1-b1 104.Be1-d2  Kb1-a1 105.Ne3-c4  
Ka1-b1 106.Nc4-a3+  Kb1-a1 107.Bd2-c3# 

It may be that more cogent reason's are given in the actual paper. There is obviously a discrepancy between the paper @HaukeReddmann mentions and the predictions given in Haworth's paper but at least we don't have to wait for the end of the universe for a resolution. Haworth predicts that the 8 man tablebases should throw up a mate that needs at least 1200 moves, so the resolution should arrive with the completion of those tablebases (assuming they're DTM tablebases).

  • 1
    I know that paper of Haworth (RIP). And now I found the paper (which is no paper, but a preliminary email by Marc Bourzutschky) and edited the link into my answer. Nov 3 at 20:24
  • Many thanks for that. I'm still evaluating the contents of the e-mail. I have some queries already, which I'll post shortly as an edit to the above answer (shortly as in how long is a piece of string btw.). Nov 4 at 14:07

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