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In this position, the black move 1. ... Qc3 results in two consecutive forced moves, Kxc3 and Ke7:

2rkr3/2p5/2q5/8/8/8/3KP3/3R4 b - - 0 1

1... Qc3 2. Kxc3 Ke7

What is the longest sequence of mutually forced moves that can be constructed on a standard chessboard?

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2 Answers 2

up vote 6 down vote accepted

Inspired by Ed Dean's answer, here is another "infinite loop":

[FEN "8/6p1/1p3pPk/1P3Pp1/1Pp3p1/KpP3P1/1P6/8 - - - 0 0 "]
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There we go! (And I deleted my answer, since it was flawed, as its infinite loop wasn't fully forced.) –  ETD Mar 9 at 19:18
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Although technically, since neither side can possibly checkmate the other here, it's an immediate draw (just like stalemate). So there aren't any legal moves at all here, the game is over. –  RemcoGerlich Mar 9 at 20:01
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@RemcoGerlich That is a necessary property of a forced infinite loop, isn't it, that neither player can possibly checkmate the other. –  bof Mar 9 at 20:05
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Yes. It's in Article 1 of the rules, even ( fide.com/fide/handbook.html?id=124&view=article ). Infinite loops aren't really compatible with the laws of chess. –  RemcoGerlich Mar 9 at 20:08
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No, this has nothing to do with the 50 moves rule. See rules 1.3 or 5.2.b in the rules (my link). Or even rule 9.6 -- the rules describe this situation three times! It's an immediate draw, no claims. –  RemcoGerlich Mar 9 at 21:44

You mean like this?

[FEN "K1k5/P1Pp4/p2P4/Pp6/P1p5/2P5/8/8 - - - 0 0 "]

1. axb5 axb5 a6 b4 cxb4 c3 b5 c2 b6 c1=Q b7# 1-0 

White mates in 6.

I guess that's 9 consecutive forced moves. It would be eleven except for black's choice of promotion piece on his fifth move. I don't know if it's a record, and I don't know who composed this classic chess problem.

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I'm not sure c1=Q can be considered forced, you had other promotions –  Alan May 21 at 18:33
    
@Alan That's why I said "I guess that's 9 consecutive forced moves. It would be eleven except for black's choice of promotion piece on his fifth move." –  bof May 21 at 19:52

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