Find a chess position that can occur twice but not thrice in a single game

Question

For purposes of this puzzle, a chess position consists of an arrangement of pieces on the board, as well as the side to move. No other information about the game state is included.

The objective of this puzzle is to identify a position that can occur twice in a single game but cannot occur three times.

More generally, what numbers $n$ are such that some position can occur $n$ times but not $n+1$ times?

Note that we're using the FIDE ruleset, so 3-fold repetition and 50 move rules are optional for the players to call and basically irrelevant to this puzzle (but the 5-fold and 75 move rules are mandatory and may or may not be useful to a solution).

Answer 1

Here's my go. The final key idea is lifted from loopy walt's solution, but the rest of the setup feels different enough to warrant posting separately:

White to play:

Here, the first time around, white can castle, which is the only non-checkmate move that avoids both a stalemate and a capture. To get back to the same position later, white can bring a rook to e4, which makes the king position safe again, so white has time to get all their ducks in a row while black wiggles the knight around elsewhere.

White will then play the discovered check by moving the rook from e4 to e8, and after the black knight blocks, the same situation is reached again. This time castling is not allowed anymore, so either a checkmate, a stalemate, or a capture (Qg2+ Kxg2) is unavoidable.

For double checking purposes, here's a sample game that reaches the position after moves 28 and 38, and then never again:

1. e4 d5 
2. exd5 Qxd5 
3. Qh5 Qxa2 
4. Qxh7 Qxb1 
5. Qxh8 Qxb2 
6. Qxg8 Qxc2 
7. Qxf7+ Kd7 
8. Qxf8 Ke6 
9. Qxe7+ Kf5 
10. Qxc7 Nc6 
11. Qxb7 Ne5 
12. Bb2 Qxb2 
13. Qxc8+ Kf4 
14. Nf3 g5 
15. g3+ Kxf3 
16. Qxa8+ Kg4 
17. Qxa7 Qb6 
18. Bc4 Qh6 
19. Bd5 Qxh2 
20. Rxh2 Nc4 
21. Rh8 Nxd2 
22. Re8 Nc4 
23. Re4+ Kf3 
24. Ba8 g4 
25. Qh7 Nd6 
26. Qh2 Nf7 
27. Qg1 Ng5 
28. Re8+ Ne4 
29. O-O-O Ke2 
30. Bb7 Kf3 
31. Ba6 Nf6 
32. Re4 Nh5 
33. Bb7 Nf6 
34. Kd2 Nh5 
35. Ke1 Nf6 
36. Ra1 Nh5 
37. Ba8 Nf6 
38. Re8+ Ne4

---

EDIT: In the comments below, @Retudin proposes an outrageously beautiful modification that not only does away with an unnecessary pawn, but two of them!

Answer 2

White to move:

Accounting only for non-pawn non-capturing moves no matter what white does black will have to move the Bg1 either checkmating or forcing white to capture on f2 unless white's move was

0-0-0

which is available only once.

Proof game:

1. h4 h5 2. Rh3 Nf6 3. Rg3 Ne4 4. Rg6 Ng3 5. Rh6 Nxf1 6. Rxh8 d5 7. Nf3 Bf5 8. Ng5 Bh7 9. f4 Bg8 10. Nh7 e6 11. f5 Bc5 12. f6 Bg1 13. c4 a5 14. cxd5 Kd7 15. d6 Kc6 16. d7 Kd5 17. Qb3+ Ke4 18. Qc3 Kf5 19. Qf3+ Kg6 20. Qxf1 Kh6 21. Qf5 g5 22. Qf4 a4 23. Qe4 a3 24. Qe5 axb2 25. Na3 Rxa3 26. Qxc7 Rh3 27. Qxd8 Rh1 28. Qc7 b1=B 29. Bb2 g4 30. Be5 b5 31. Bh2 g3 32. Qxb8 gxh2 33. Qxb5 Bxa2 34. Qb1 Bxb1 35. Nf8+ Bbh7 36. O-O-O Bf2 37. Ng6 Be3 38. Nf4 Bd4 39. Nd5 Bc5 40. Kb2 Kg6 41. Ra1 Bg1 42. Kc1 Kf5 43. Nf4 Ke5 44. Nh3 Bb1 45. Kd1 Ke4 46. Ng5+ Kf5 47. Nh7 Kg6 48. Ke1 Kh6 49. Nf8+ Bbh7

The shown position occurs after 35 and 49 moves.

Answer 3

Here's a full solution. The set of possible $n$ is

$n \le 22.$

We first note that $n \le 22$ is ensured by the 5-fold repetition rule. A position can occur once with an en passant right, then 4 more times each with 4, then 3, then 2, then 1 castle right, and then 5 times with 0 castle rights. A draw will necessarily occur before any position reaches a 23rd occurrence.

Now we will verify every $n \le 22$ is achieved by some position. We will avoid positions constrained by the 75 move rule when possible.

First some remarks about the $n=0$ case. Of course, any position where the number of kings is not 2 is impossible, but also these are arguably not chess positions at all. If we demand that a chess position be reachable in game, then trivially $n=0$ is impossible. If we merely demand that a chess position be reachable by some legal sequence of moves, the following arrangement has this property yet cannot be achieved in a game (with either to move), since the previous position is necessarily dead:

Next, we quickly list the easy cases. We have $n=1$ for any position in a checkmate (or a stalemate, or a king in check to a pawn, etc.). The starting position has $n=21,$ and after a few pawn moves, we can reach a position with $n=22$ involving an en passant right. We can get $n=20$ with a position where the last move is uniquely determined, e.g.:

Any value of $n \in [4, 22]$ which is not congruent to 3 mod 4 can be achieved by moving/removing rooks from the positions mentioned in the preceding paragraph.

Other answers have provided solutions for $n=2.$ Here is the one I originally found.

White to move:

White must immediately castle to avoid an irrevocable move occurring in the next couple turns. The mutually pinned bishops make entrance into the position "one-way."

Next we handle the cases $n=7$ and $n=11.$ For $n=11,$ here's a white-to-move position that can occur once with three castling rights, once with two castling rights, 4 times with 1 castling right, and 5 times with 0 castling rights:

Then $n=7$ is achieved by removing a black rook from this position.

The remaining cases are $n=3,15,19,$ for which we will use positions constrained by the 75 move rule, which I suspect is necessary. First, our example for $n=3:$

It takes 26 turns to return to this position, with one white knight having to reach g2 to unpin the other white knight. This sequence of moves can only occur twice before running afoul of the 75 move rule.

For $n=15,$ we make a position which requires 5 turns to cycle, in addition to 2 extra turns each cycle in which a white rook moves back and forth to spend a castling right (the black rook can spend its castling right without lengthening the cycle). This position can occur 4 times with 3 castling rights, 4 times with 2 rights, 4 times with 1 right, and 3 times with 0 rights, requiring $3*5+2+4*5+2+4*5+3*5=74$ turns.

Finally, here's our example for $n=19.$ This one takes 7 turns to cycle if the white king does not move, but there's a 2 turn cycle with movement of the white king. This position can occur 4 times with 4 castling rights, 3 times with 3 castling rights, 4 times with 2 rights, 4 times with 1 right, and 4 times with 0 rights (it can't appear a 5th time with 0 rights, due to the previous position being determined). Including 2*3 turns spent on using castling rights (the kingside white rook never needs to move, since the white king moving spends that castling right), this takes $3*7 + 2*3 + 3*7+ 4*2 + 4*2 + 4*2 = 72$ turns. It would be impossible to achieve this position with 3 or more castling rights any more times, for that would require 79 turns.

Answer 4

I think with retudin's answer we are near the economy minimum. But there's a small improvement possible, to the challenge of finding a minimal diagram+move that can be achieved twice but not thrice.

White to play:

I used Popeye with the option "half-duplex" to find all "h~1" from this position. This asks, with White to move, for all the ways that White then Black can move. There are 24 sequences possible, but in 23 of them, Black must capture. The only exception is

1. 0-0-0 Ke2. Following this the position can be reset with e.g. 2. Kc2 Sf6 3. Re4+ Kf3 4. Rb1 Sg8 5. Ra1 Sf6 6. Kd2 Sg8 7. Ke1 Sf6 8. Rg4+ Se4.

There's a broader question of what are valid n for how many times can a diagram+move be repeated. This is discussed thoroughly in another response, but I agree that the maximum is 22, see: https://chess.stackexchange.com/questions/33707/extended-new-years-math-riddle-5-fold-repetition-75-move-rules/34091#34091.

EDIT: Pulled out from loopy walt's comment, even better!

Again, Popeye proves this is sound.

Answer 5

Minimalistic approach for n=2: (which incorrectly ends with the other color to move)

White to move must 0-0-0 to prevent a pawn move, capture or mate.
It can then give the black King some space with a3b3 and eventually get into the same position with e.g. Rba1 Kb8 Rba3. (and then 0-0-0 is not available anymore)
[FEN "1k6/1ppP4/8/8/8/R4p2/2PPPrp1/R3K3 w Q - 0 1"]
1. O-O-O Rf1 2. Rb3 Ka7 3. Kb2 Rf2 4. Rb1 Kb8 5. Kc1 Ka7 6. Kd1 Ka8 7. Ke1 Ka7 8. Ra1+ Kb8 9. Rba3

Answer 6

Partial Answer

$n=0$:

$n=1$:

$n=2$:

I don't really know, the best I have come up with is

setting the 50-move counter to 18, although for this puzzle 47 is required.

$n=3$: