Solutions for Drawn Proof Games

{A}-{I}

The general approach for most of these problems is:

(1) Find the unique draw cycle (from diagram position to diagram position), usually in 2.0 moves. This "flipping" will repeat twice.
(2) Find the unique Shortest Proof Game which reaches the diagram in n-4.0 moves.
(3) Take care not to change castling status or e.p. status during the flipping, because this will prevent the position from repeating.[1]

{A} 1. e3 f5 2. Bd3 f4 3. Bh7 f3 4. Bf5 Rh5 5. Bg4 Rf5 6. Bh5 g6 7. Bg6 (7. ... Rf7 8. Bh7 etc. =)
The White bishop executes a round trip of length 5.

{B} 1. f4 e5 2. Kf2 Bb4 3. Ke1 Bd2+ (4. Kf2 Be3+ etc. =)
This is the shortest known position of this class. Note that wK must execute a switchback in moves 2 & 3 so that castling rights are already disrupted before the flipping begins.

{C} 1. f4 e6 2. Kf2 e5 3. Kg3 e4 4. Kh3 d5+ (5. Kg3 Nd7 etc. =)
Which Black piece will block on d7 the bB attack on h3? It can't be bK, because this would break the castling rights which are certainly intact at the 4.0 move mark.

{D} 1. Nc3 Nf6 2. Nd4 Nd4 3. f3 Ng3 4. hg3 a6 (5. Nh3 Nc6 etc. =)
This is the first example with no check in the final position. Control of castling rights is used to render unique the "flipping".

{E} 1. Na3 Nc6 2. Nc4 Ne5 3. Ne5 h5 4. Ng4 hg4 5. Nh3 Rh3 6. Rb1 Re3 7. h4 f6 8. fe3 (8. ... Nh6 9. Ra1 etc. =)
This extends the ideas further, now using control of en passant rights to render unique the last White moves: 7. fe3? f6 8. h4? would be cooked as en passant would be live at 7.5 mark.

{F} 1. e4 f6 2. Be2 f5 3. Bg4 f4 4. Bf5 Nh6 5. Bg6+ (5. ... Nf7 6. Bf5 etc. =) But {F} highlights an issue, however, first pointed out by Guus Rol. Suppose the game repeated for the third time *prior* to the diagram position repeated. Then did the game not end at that point? In this example, this interpretation would cook the problem, but in other instances this principle can be used constructively to render a solution unique.

Article 18 of the Chess Problem Codex:

A position is considered as a draw if it can be proved that an identical position [i.e. the same kinds of pieces on the same squares with the same move rights] has occurred three times in the proof game combined with the solution.

There are two self-consistent opinions on this.

(1) The first approach would rely on an over-arching principle, one that says that the compositions try to remove all human decision-making. Just as players in chess problems don't resign, and just as we codify the meaning of "direct play" and "help play", so we should not be trying to reason whether a player claimed a draw they were entitled to. We should simply specify that when a trigger happens, the draw will be claimed.

This approach is basically simpler than any other, particularly in cases where we know a position did repeat 3 times historically, by one of many different ways. Or do we look at different points in history and alternative paths at each point?

Moreover, some retro compositions from the past depend upon the pre-eminence of Article 18, and it would be churlish to render those problems cooked after the fact.

(2) The second approach would apply a different over-arching principle: that conventions should apply only where there is otherwise irresolvable uncertainty about the solution. They should never reverse a logical implication from the diagram and stipulation. The e.p. & castling conventions are well-behaved from this point of view.

Now apply this principle to draw by repetition. If a game is known to have lasted past the third repetition (because we are told that the game was e.g. a Draw at SPG 14.0, and there is no other route to the final position), then the logical conclusion is that no player chose to claim a draw through repetition. This is simple retro reasoning, and no recourse to any convention is necessary.

If there are two possible histories on the other hand, and exactly one would have traveled via a third repetition, then Article 18 would apply to ensure that the solution is unique.

See the solution to {J} below.

Prior Art

{G} Position after Black's next move will look like position after 11th. In diagram, 12 Black moves are visible. So next move must take some piece nearer home to reduce to 11. This could be by Rc8, Kd7, Bb6, Nh6, Na4. Other Black pieces never moved. In particular g-pawn & h-pawns were captured on their starting squares.

Some White unit was captured on c6: this can't have been unpromoted missing h-pawn, so that unit promoted, capturing on g7. Since h-pawn was also captured, not enough moves for another White piece to be captured on c6. So Ceriani-Frolkin situation with promotion to Q/B: 10 White moves accounted for. Q/Bc6 earliest move 8, then followed 4-1 Black moves. So moves are tight. White's last three moves to lose parity must be slow walk with e-pawn. Bb6 only released after 5 h8, then has 3 moves to make, so N on h6 already, so path is unique. Bf8 moved while promoted piece still on g8, so it was B not Q. N could only move to h6 after 4 hg7. So Na4 was first 3 move for Black.

Black's next move is Ke8 or Ra1. But if Ke8, then Black lost kside castling rights during the last 4 double moves, and so no draw by repetition. So Ra1. If 11...f5, then en passant status changed during last 4 moves, so again no draw. That wraps it up: 1 h4 Na6 2 h5 Nc5 3 h6 Na4 4 hg7 Nh6 5 h8B Bg7 6 B7 Bd4 7 Be4 Bb6 8 Bc6 dc6 9 e3 Bh3 10 e4 f5 11 e5 Kd7. Following that came 12 ??? Rc8 13 ??? Ra8 14 ??? Rc8 15 ??? (diagram) and 15 Ra8=. Rather nice.

{H} is rather different. The only escape from perpetual retrogression is b4, but that is not available until 19.

The Black QR was captured in cage, and bishop & knight by pawns. Black captured one unit on b file, but two other captures unclear. Last move 23...Kh6Mg6 Prior 23 Rg5-h5+. So M=0 and 22...Kg6M'h6 Prior 22 Rh5-g5+ So M'=0 again. Same for previous series, since the only escape from retrogression is b4, not available until 19.

So the last moves are: 20 Rhg5+ KM"h6 21 Rh5+ Kg6 22 Rhg5+ Kh6 23 Rh5+ Kg6. If we can conclude that M"=0, then Black can draw (as long as it is still the end of Black's turn, and not the beginning of White's [1]).

Count to position after b4. White has made at least 18 moves with the visible pieces. So others (a/c-pawn, e-pawn & KB) made at most 1 move. In fact, it must be KB which moved to b5 to be captured there. So in fact M"=0. Natch gives:

1.f3 Nc6 2.Kf2 Nd4 3.Kg3 Ne2 4.Kg4 Ng3 5.hg3 c6 6.Rh5 Qa5 7.Nh3 Qa2 8.Na3 Qb3 9.cb3 Kd8 10.Nc2 Kc7 11.Ra5 Kd6 12.Na1 Ke6 13.Qe1 Kf6 14.Rf5 Kg6 15.Bb5 Rb8! 16.d3 Ra8! 17.Bf4 Rb8 18.Bb8 cb5 19.Qe5 b4 20.Rg5 Kh6 21.Rh5 Kg6 22.Rg5 Kh6 23.Rh5 Kg6.

Note the bonus switchback in Black's 15/16 moves.

My own approach to Olin-style stipulation

{I} Some move by White must allow a claim of draw by repetition. It must be a rook or king move, leading to a position which is a proof game in 9.0 + 0.5 - 4.0 = 5.5 moves. The only possibility that works is 9.Rg1, with the proof game 1.Nc3 Nf6 2.Ne4 Ne4 3.a4 Nf2 4.a5 Nh3 5.a6 Ng1 6.Rg1. Now there is a unique continuation for the draw 6...Nc6 7.Rh1 etc, to avoid disrupting other castling rights. Note moreover that the diagram position cannot be reached in 5.0 moves, so at 9.0 is not yet a draw by repetition, and White must therefore move before claiming.

In fact some of the earlier problems could also be retracted by half a move, and given a similar stipulation to {P}. For example in {K} if 8. Nc6-b8 is retracted, then that nevertheless is the unique continuation which could lead to the draw by repetition. On the other hand, {L} does not admit such a step: retracting 12. Ra1-b1 would allow the alternative 12. Rh1-g1.

The diagram for {I} is very clean, and in fact inspired Nicolas Dupont to turn to the "homebase" theme, with which we have both since had some success.

Just fifteen years after this problem was composed, I have finally managed to generate a fully-homebase version, which has been entered for a competition so I am not free yet to publish it.

Approach to Article 18.

{J} There are apparently two solutions: 1. Sc3 Sf6 2. Se4 Se4 3. Sh6 Sd2 4. Kd2 g6 5. Ke3 Bh6+/g7 6. Kf3 0-0 7. B()h6 Kh8 8. B()g7+ and then repeating by 8 ... Kg8 9. Bh6 Kh8 10. Bg7+ Kg8 11. Bh6 Kh8 12. Bg7+. But after 6 ... Bh6+, the game might end prematurely by repetition at 9.5.

Under the strict interpretation of Article 18, the game after 6. ... Bh6+ ends prematurely upon 11. Bh6.

Using the principle of taking conventions only to resolve uncertainty, we have genuine uncertainty, irresolvable by logic so the convention *does* trigger, and the again the game would terminate upon 11. Bh6.

So either way, 6 ... Bg7 is the only possibility, and the solution is unique.

Open Challenges

(1) Design a "Drawn. Proof game n" problem in which there are three earlier points at which a draw might have been claimed.

(2) Design a "Drawn. Proof game n" problem with Ceriani-Frolkin, but where the whole game is determined, including the repetition. (C.f. composition {I}.]

[1] Article 9.2 of FIDE Laws of Chess

The game is drawn, upon a correct claim by the player having the move, when the same position, for at least the third time (not necessarily by sequential repetition of moves):

  1. is about to appear, if he first writes his move on his scoresheet and declares to the arbiter his intention to make this move, or
  2. has just appeared, and the player claiming the draw has the move.

Positions as in a. and b. are considered the same, 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.

Positions are not the same if a pawn that could have been captured en passant can no longer be captured or if the right to castle has been changed temporarily or permanently.

[2] In the stipulation for problems {G} & {I}, I discuss claiming a draw "with" the next move. This choice of preposition is intended to glide over an issue with clause a in Article 9.2. Namely: if the draw comes before the move, does the move still count towards the total number of moves in the solution? One advantage of the convention, is that it simplifies the procedure.

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