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Here's a fun and clean riddle for the New Year! I found the material for it on Krabbe's site. I will provide the link when it is solved.

How many times can this position be repeated, with both sides helping, before either side can claim a draw repetition? No cheating, please!

enter image description here

Addendum 1/7/20201: With the acceptance of the answer, here is the link to Krabbe's AD magazine article "TWEEëNTWINTIG FOTO'S, ", which translates to "Twenty Two Photos."

Rewan Demontay
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  • With my understanding of your question, the answer is trivially 3. Perhaps I am misinterpreting? – Noah Caplinger Jan 02 '21 at 03:47
  • I think it will be after 9 repetitions; i.e. after the 10th occurrence of that position, since the first time it occurs it's not a repetition. The first time, White has the possibility to capture e.p. The next two times, there is no e.p. and both players have the right to castle on either side. The next two times White has lost the ability to castle long, etc. – bof Jan 02 '21 at 04:26
  • I don't see how it can be any more. I don't believe swapping the two white rooks or the two white knights makes it a different position. – bof Jan 02 '21 at 04:32
  • Oh, I counted wrong. Fencepost error. the number of castling possibilities goes 4,3,2,1,0. So a draw by repetition may be claimed after that setup occurs the 12th time. – bof Jan 02 '21 at 04:34
  • First time that setup occurs, there are 4 ways to castle and White has an e.p.; next 2 times, no e.p., still 4 ways to castle; next 2 times, 3 ways to castle; next 2 times, 2 ways to castle; next 2 times, 1 way to castle; last 3 times, no way to castle. – bof Jan 02 '21 at 04:39
  • @RewanDemontay I see now what the puzzle is trying to say (which I certainly didn't notice!), but after reading the relevant rule, I do think the answer three is correct. The word "position" is explicitly defined in the FIDE rule as containing information about castling and en passant. I'm going to stand on a technicality by saying the puzzle should not mention the repitition of "position," and instead "board layout" or something similar. Great puzzle though :) – Noah Caplinger Jan 02 '21 at 05:04
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    The puzzle would improve if White would have a provable e.p. right. – Hauke Reddmann Jan 08 '21 at 12:11

1 Answers1

5

The board layout can be repeated at most 11 times with White having the move, the 12th will be a draw.

[FEN "rnbqkbnr/1pp2pp1/4p3/p2pP2p/P6P/8/1PPP1PP1/RNBQKBNR w KQkq d6 0 1"]

1. Be2 {White can no longer capture en passant.} Be7 2. Bf1 Bf8 {2nd time} 3. Be2 Be7 4. Bf1 Bf8 {3rd time}
5. Rh2 {White can no longer 0-0} Be7 6. Rh1 Bf8 {4th time} 7. Rh2 Be7 8. Rh1 Bf8 {5th time}
9. Ra2 {White can no longer 0-0-0} Be7 10. Ra1 Bf8 {6th time} 11. Ra2 Be7 12. Ra1 Bf8 {7th time}
13. Be2 Rh7 {Black can no longer 0-0} 14. Bf1 Rh8 {8th time} 15. Be2 Rh7 16. Bf1 Rh8 {9th time}
17. Be2 Ra7 {Black can no longer 0-0-0} 18. Bf1 Ra8 {10th time} 19. Be2 Ra7 20. Bf1 Ra8 {11th time}

However (emphasis mine):

9.2.2 Positions are considered the same if and only if the same player has the move, pieces of the same kind and colour occupy the same squares and the possible moves of all the pieces of both players are the same.

So:

[FEN "rnbqkbnr/1pp2pp1/4p3/p2pP2p/P6P/8/1PPP1PP1/RNBQKBNR w KQkq d6 0 1"]

1. Bd3 Bc5 2. Be2 Bf8 3. Bf1

is not a repetition of the original position (it's Black's turn now), although it is the same board layout.

Therefore: the position can be repeated 10 times (excluding the initial position), multiply this by 2 (2 more times the layout can be repeated with Black's turn) and add 1 (the initial position) + the final repetition on which the draw is claimed = 22 times.

Rewan Demontay
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double-beep
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