# Longest sequence of mutually forced moves

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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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 '14 at 19:18
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 '14 at 20:01
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 '14 at 20:08
@Remco To be fair, you can have infinite loops as long as both players play along, because the 50 moves rule only says may be drawn - technicality obviously. – Voo Mar 9 '14 at 21:26
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 '14 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.

P.S. Thanks to Rosie F. for sourcing the problem. Quoting Rosie F.'s comment:

This problem is by Vilhelm Röpke. PDB cites it as Skakbladet, 1942.

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I'm not sure c1=Q can be considered forced, you had other promotions – Alan May 21 '14 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 '14 at 19:52
This problem is by Vilhelm Röpke. PDB cites it as Skakbladet, 1942. – Rosie F May 26 at 5:37
@RosieF Thank you. I edited your comment into my answer. – bof May 26 at 5:57

A simpler infinite-loop setting:

``````[FEN "1kb5/1p1p4/1P1P4/8/8/4p1p1/4P1P1/5BK1 w - - 0 1"]
``````
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One can [put white pawns on a7 and f3, put black pawns on c6 and h2, and move black's king to a8] to get 9 consecutive only-legal moves. ​ (without an immediate draw by dead reckoning; see Remco's comments to the other answer) ​ ​ ​ ​ – Ricky Demer Feb 22 at 4:39
True, and then wBf1 and the e2/e3 pawns are not needed. – Noam D. Elkies Feb 22 at 6:29
It would be the Bc8 and the d pawns that aren't needed, rather than the Bf1 and e pawns. ​ ​ – Ricky Demer Feb 22 at 7:46
Sorry, you're right. – Noam D. Elkies Feb 22 at 14:30