18 problem(s) found in 1816 milliseconds (displaying 18 problem(s)). [PROBID IN 'P0008596; P0009007; P1003993; P1004352; P1004353; P1004354; P1011937; P1186643; P1281158; P1298075; P1298076; P1338392; P1376322; P1398780; P1398781; P1398782; P1398783; P1401677'] [download as LaTeX]

1 - P0008596

9508 Die Schwalbe 163 02/1997

(15+14)

Stellung nach dem 15. weißen Zug. Schwarz kann mit dem nächsten Zug durch Stellungswiederholung Remis erzwingen. Wie waren die ersten 22 Einzelzüge?

**Per Olin**9508 Die Schwalbe 163 02/1997

(15+14)

Stellung nach dem 15. weißen Zug. Schwarz kann mit dem nächsten Zug durch Stellungswiederholung Remis erzwingen. Wie waren die ersten 22 Einzelzüge?

1. h4 Sa6 2. h5 Sc5 3. h6 Sa4 4. hxg7 Sh6 5. g8=L Lg7 6. Lxh7 Ld4 7. Le4 Lb6 8. Lc6 dxc6 9. e3 Lh3 10. e4 f5 11. e5 Kd7 12. Sc3 Ke8 13. Sb1 Kd7 14. Sc3 Tc8 15. Sb1, dann 15. ... Ta8= draw by 3Rep

**Keywords:**Draw by repetition, Ceriani-Frolkin Theme, Unique Proof Game, Constrained problem

**Genre:**Retro

**FEN:**2rq3r/pppkp3/1bp4n/4Pp2/n7/7b/PPPP1PP1/RNBQKBNR

**Input:**Gerd Wilts, 1997-03-19

**Last update:**A.Buchanan, 2021-06-05 more...

2 - P0009007

9796 Die Schwalbe 167 10/1997

Spezielles Lob

(13+13) C+

BP in 23,0

Schwarz spielte 19. ... b4! Ergebnis?

**Andrey Frolkin**

Sergii I. TkachenkoSergii I. Tkachenko

9796 Die Schwalbe 167 10/1997

Spezielles Lob

(13+13) C+

BP in 23,0

Schwarz spielte 19. ... b4! Ergebnis?

1. f3 Sc6 2. Kf2 Sd4 3. Kg3 Sxe2+ 4. Kg4 Sg3 5. hxg3 c6 6. Th5 Da5 7. Sh3 Dxa2 8. Sa3 Db3 9. cxb3 Kd8 10. Sc2 Kc7 11. Taa5 Kd6 12. Sa1 Ke6 13. De1+ Kf6 14. Taf5+ Kg6 15. Lb5 Tb8 16. d3 Ta8 17. Lf4 Tb8 18. Lxb8 cxb5 19. De5 b4 19. De5 b4 20. Thg5+ Kh6 21. Th5+ Kg6 22. Thg5+ Kh6 23. Th5+ Kg6= draw by 3Rep

Result is draw by repetition.

Result is draw by repetition.

The theme would be impossible without the condition, since oscillating moves might otherwise have been taken earlier in the game, or not at all.

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**Keywords:**Unique Proof Game, Draw by repetition, Constrained problem, Switchback (t)

**Genre:**Retro

**Computer test:**Euclide 1.01 for first 18.5 moves, then move given in stipulation, and finally simple manual retro-logic for last 4.0 moves.

**FEN:**1Bb2bnr/pp1ppppp/6k1/4QR1R/1p4K1/1P1P1PPN/1P4P1/N7

**Reprints:**Die Schwalbe 193 02/2002

**Input:**Gerd Wilts, 1998-06-26

**Last update:**A.Buchanan, 2021-06-05 more...

1. Sa3 Sc6 2. Sc4 Se5 3. Sxe5 h5 4. Sg4 hxg4 5. Sh3 Txh3 6. Tb1 Te3 7. h4 f6 8. fxe3 Sh6 9. Ta1 Sg8 10. Tb1 Sh6 11. Ta1 Sg8 12. Tb1= draw by 3Rep

Clearly we look for PG in 7.5 moves, then see how to repeat. By Popeye 4.79, there are two apparent PGs in 7.5, the solution and the transposition of h3-h4 & fxe3. In the transposition, en passant is enabled so that position can never repeat. Moreover, white kside & black qside castling are enabled so cannot be disrupted in the repetition sequence: so only wRa1 & bSg8 are free to move.

Clearly we look for PG in 7.5 moves, then see how to repeat. By Popeye 4.79, there are two apparent PGs in 7.5, the solution and the transposition of h3-h4 & fxe3. In the transposition, en passant is enabled so that position can never repeat. Moreover, white kside & black qside castling are enabled so cannot be disrupted in the repetition sequence: so only wRa1 & bSg8 are free to move.

**Keywords:**Unique Proof Game, Draw by repetition, En passant, Castling, Constrained problem

**Genre:**Retro

**Computer test:**Popeye v4.79 + a little retro reflection

**FEN:**r1bqkbn1/ppppp1p1/5p2/8/6pP/4P3/PPPPP1P1/1RBQKB1R

**Input:**Gerd Wilts, 2002-04-05

**Last update:**A.Buchanan, 2020-05-25 more...

4 - P1004352

R186 Probleemblad 06/2002

1. Lob

PR Piet le Grand

(15+12) C+

The game is drawn.

PG in 17.0

**Andrew Buchanan**

Guus RolGuus Rol

R186 Probleemblad 06/2002

1. Lob

PR Piet le Grand

(15+12) C+

The game is drawn.

PG in 17.0

1. c4 b5 2. Dc2 Lb7 3. Kd1 Ld5 4. cxd5 g5 5. Dxc7 Sc6 6. Kc2 Dxc7 7. Kb3 Dg3+ 8. hxg3 e5 9. Txh7 Sce7 10. Tg7 Th1 11. Ka3 Sh6 12. Tg8 g4 13. f4 Sg6+ 14. Kb3 Se7 15. Ka3 Sg6+ 16. Kb3 Se7 17. Ka3 Sg6+= draw by 3Rep

The intention was that the position at 13.0 repeats at 15.0 & 17.0. Popeye finds 87 possible sequences of moves up to 13.0. Under the assumption [1] that an e.p. which can't be played due to check or pin still affects the definition of position, we could eliminate those 52 which end with e7-e5+, leaving us with 35.

The moves which follow can only be 14. Ka3-b3 Sg6-e7! (not 14. ... Kd8-e7? which disrupts Black castling rights) 15. Kb3-a3 Se7-g6+. And repeat the switchbacks. But all 35 candidate solutions have 13. ... Se7-g6+. Thus normally the game would have ended at at 17. Kb3-a3 (or possibly earlier). However, there is one exceptional solution with 13. f2-f4, which (again using the ep rule) would have prevented the game from ending prematurely. This part of the logic is not in question.

[1] But is this assumption justified? At the time of composition, the definition for position in the FIDE Laws was unclear on this point, and the composers had to take a view. Since then, the definition has been improved, and the problem has been retrospectively cooked.

The Codex does not opine on what happens to the soundness of a problem if the rules (or indeed the conventions) change, and materially impact a problem. The composers choose to take it that the problem remains historically sound. But if a historically unsound problem were *rendered* sound by changes to rules or conventions, we would want to consider it sound today. No reason why composers shouldn't have their cake and eat it here: clearly the relevant concept would be one of "peak soundness" - to take the problem to be as sound as its maximal soundness as rules and conventions swing wildly like open doors in a hurricane.

The intention was that the position at 13.0 repeats at 15.0 & 17.0. Popeye finds 87 possible sequences of moves up to 13.0. Under the assumption [1] that an e.p. which can't be played due to check or pin still affects the definition of position, we could eliminate those 52 which end with e7-e5+, leaving us with 35.

The moves which follow can only be 14. Ka3-b3 Sg6-e7! (not 14. ... Kd8-e7? which disrupts Black castling rights) 15. Kb3-a3 Se7-g6+. And repeat the switchbacks. But all 35 candidate solutions have 13. ... Se7-g6+. Thus normally the game would have ended at at 17. Kb3-a3 (or possibly earlier). However, there is one exceptional solution with 13. f2-f4, which (again using the ep rule) would have prevented the game from ending prematurely. This part of the logic is not in question.

[1] But is this assumption justified? At the time of composition, the definition for position in the FIDE Laws was unclear on this point, and the composers had to take a view. Since then, the definition has been improved, and the problem has been retrospectively cooked.

The Codex does not opine on what happens to the soundness of a problem if the rules (or indeed the conventions) change, and materially impact a problem. The composers choose to take it that the problem remains historically sound. But if a historically unsound problem were *rendered* sound by changes to rules or conventions, we would want to consider it sound today. No reason why composers shouldn't have their cake and eat it here: clearly the relevant concept would be one of "peak soundness" - to take the problem to be as sound as its maximal soundness as rules and conventions swing wildly like open doors in a hurricane.

2018 Laws 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. Thus positions are not the same if:

9.2.2.1 at the start of the sequence a pawn could have been captured en passant

9.2.2.2 a king had castling rights with a rook that has not been moved, but forfeited these after moving. The castling rights are lost only after the king or rook is moved.

2016 Codex Article 18: "A position is considered as a draw if it can be proven that an identical position has occurred three times in the proof game combined with the solution. The draw ends the game immediately."

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9.2.2.1 at the start of the sequence a pawn could have been captured en passant

9.2.2.2 a king had castling rights with a rook that has not been moved, but forfeited these after moving. The castling rights are lost only after the king or rook is moved.

2016 Codex Article 18: "A position is considered as a draw if it can be proven that an identical position has occurred three times in the proof game combined with the solution. The draw ends the game immediately."

**Henrik Juel**: 1st Commendation awarded by Piet le Grand in 12/2010, p.227-232 (2019-11-18)more ...

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**Keywords:**Unique Proof Game, Draw by repetition, Constrained problem, Golden Age (old e.p. rule), En passant

**Genre:**Retro

**Computer test:**Euclide, then manual elimination of all solutions which don't fit constraints.

**FEN:**r3kbR1/p2p1p2/6nn/1p1Pp3/5Pp1/K5P1/PP1PP1P1/RNB2BNr

**Input:**Gerd Wilts, 2002-12-26

**Last update:**A.Buchanan, 2021-06-05 more...

5 - P1004353

R187 Probleemblad 06/2002

Andrew Buchanan gewidmet

(14+13)

Die Stellung ist Remis

BP in 11.5

**Guus Rol**R187 Probleemblad 06/2002

Andrew Buchanan gewidmet

(14+13)

Die Stellung ist Remis

BP in 11.5

1. Sc3 Sf6 2. Se4 Sxe4 3. f3 Sg3 4. hxg3 a6 5. Txh7 Txh7 6. g4 Th1 7. Sh3 Tg1 8. Sxg1 Sc6 9. Sh3 Sb8 10. Sg1 Sc6 11. Sh3 Sb8 12. Sg1= draw by 3Rep

There are two tries:

4. ... h5,h6 5. Txh5,Txh6 Txh5,Txh6 6. g4 Th3 7. Sxh3 a6 8. Sg1 ... . But under either of these, the position repeats for the third time at 11.0. By the current convention, the game would have ended obligatorily at this point, prior to 11.5.

Black must lose a tempo: either by moving hP before its capture, or by saccing R on g1 rather than h3. In fact it has to be the latter, so that the repetition cycle begins as late as possible, avoiding premature draw.

There are two tries:

4. ... h5,h6 5. Txh5,Txh6 Txh5,Txh6 6. g4 Th3 7. Sxh3 a6 8. Sg1 ... . But under either of these, the position repeats for the third time at 11.0. By the current convention, the game would have ended obligatorily at this point, prior to 11.5.

Black must lose a tempo: either by moving hP before its capture, or by saccing R on g1 rather than h3. In fact it has to be the latter, so that the repetition cycle begins as late as possible, avoiding premature draw.

The proof games R186-R188 have ended in a draw in accordance with the Codex rules. Article 18 says: a position is considered as a draw if it can be proven that an identical position has occurred three times in the proof game combined with the solution. The draw ends the game immediately. This implies that no earlier triple repetition of position can occur.

(AB: One might think that the convention is too dogmatic, because we may *know* that the position has passed through an earlier triple repetition, and the fact that the game survived to a later point proves that there was no draw declared. On the other hand, if one has a choice between histories where an earlier triple repetition took place, and those where there was no such event, the we can clearly exclude the former. Under such a looser convention this problem would still be sound. However the current convention has the merit of simplicity. Until a bunch of problems emerge which need a looser convention, we should stick with the current convention.)

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(AB: One might think that the convention is too dogmatic, because we may *know* that the position has passed through an earlier triple repetition, and the fact that the game survived to a later point proves that there was no draw declared. On the other hand, if one has a choice between histories where an earlier triple repetition took place, and those where there was no such event, the we can clearly exclude the former. Under such a looser convention this problem would still be sound. However the current convention has the merit of simplicity. Until a bunch of problems emerge which need a looser convention, we should stick with the current convention.)

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**Keywords:**Unique Proof Game, Draw by repetition, Castling, Constrained problem

**Genre:**Retro

**Computer test:**For first 7.5 moves, Euclide 0.99, ©2000-2011 Étienne Dupuis.

**FEN:**rnbqkb2/1pppppp1/p7/8/6P1/5P2/PPPPP1P1/R1BQKBN1

**Input:**Gerd Wilts, 2002-12-26

**Last update:**A.Buchanan, 2021-12-19 more...

1. c4 g6 2. c5 Lh6 3. c6 Le3 4. cxb7 c5 5. Db3 Da5 6. Db6 axb6 7. f3 Da7 8. dxe3 e6 9. e4 Ke7 10. Lg5 Kd6 11. Ld8 Sc6 12. b8=D+ Dc7 13. Da7 Db8 14. Db7 Da7 15. Db8+ Dc7 16. Da7 Db8 17. Db7 Da7 18. Db8+= draw by 3Rep

There are a number of tries with no promotion, having instead 12. Db7-b8+.

For example: 1. c4 b6 2. c5 e6 3. Da4 Lxc5 4. Dxa7 Le3 5. Da6 c5 6. dxe3 Sc6 7. e4 Dc7 8. f3 Da7 9. Db7 Ke7 10. Lg5 Kd6 11. Ld8 g6 12. Db8+ etc.

However, in this case, the position at 17.0 is already a triple repetition, and by the 3R convention would already have been drawn and the game ended prior to 17.5.

There are a number of tries with no promotion, having instead 12. Db7-b8+.

For example: 1. c4 b6 2. c5 e6 3. Da4 Lxc5 4. Dxa7 Le3 5. Da6 c5 6. dxe3 Sc6 7. e4 Dc7 8. f3 Da7 9. Db7 Ke7 10. Lg5 Kd6 11. Ld8 g6 12. Db8+ etc.

However, in this case, the position at 17.0 is already a triple repetition, and by the 3R convention would already have been drawn and the game ended prior to 17.5.

The proof games R186-R188 have ended in a draw in accordance with the Codex rules. Article 18 says: a position is considered as a draw if it can be proven that an identical position has occurred three times in the proof game combined with the solution. The draw ends the game immediately. This implies that no earlier triple repetition of position can occur.

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**Keywords:**Unique Proof Game, Draw by repetition, Constrained problem

**Genre:**Retro

**Computer test:**HC+ For first 11,5 moves Jacobi v0.7.5: 80,267 of which all but one have 12. Qb7-b8+ leading to premature draw. Using a=>b feature, can show that the shortest loop from/to the final diagram is 3.0 moves.

**FEN:**rQbB2nr/q2p1p1p/1pnkp1p1/2p5/4P3/5P2/PP2P1PP/RN2KBNR

**Input:**Gerd Wilts, 2002-12-26

**Last update:**A.Buchanan, 2021-12-19 more...

7 - P1011937

R222 Probleemblad 12/2003

2nd Prize Probleemblad 2003

(13+13) C+

Die Stellung ist Remis.

BP in 41.5

**Guus Rol**R222 Probleemblad 12/2003

2nd Prize Probleemblad 2003

(13+13) C+

Die Stellung ist Remis.

BP in 41.5

1. h4 g5 2. hxg5 d5 3. Th6 d4 4. Tg6 Lh6 5. Tg7 d3 6. g6 dxc2 7. d4 Dd5 8. Lg5 Dxg2 9. Kd2 Dd5 10. e4 Dxa2 11. Df3 De6 12. d5 Sf6 13. dxe6 Sd5 14. Df6 Sc6 15. f3 Sd8 16. Lb5+ c6 17. Se2 Tg8 18. Txg8+ Lf8 19. Df4 Sf6 20. Th8 Sg8 21. Lf6 Lh6 22. Lg7 Lg5 23. Lf8 Sf6 24. Tg8 Sd7 25. Lg7+ Sf8 26. Lf6 Lh6 27. Tg7 Sd7 28. Lg5 Sb6 29. Df6 Sd5 30. Tg8+ Lf8 31. Df4 Sf6 32. Th8 Sg8 33. Lf6 Lh6 34. Lg7 Lg5 35. Lf8 Sf6 36. Tg8 Sd7 37. Lg7+ Sf8 38. Lf6 Lh6 39. Tg7 Sd7 40. Lg5 Sb6 41. Df6 Sd5 42. Tg8+= draw by 3Rep

**Keywords:**Unique Proof Game, Draw by repetition, Constrained problem

**Genre:**Retro

**Computer test:**Resolution time : 24 mn 0.88 s., Natch 2.3 Copyright (C) 1997,98,99,2001,2002,2003,2004 Pascal Wassong

**FEN:**r1bnk1R1/pp2pp1p/2p1PQPb/1B1n2B1/4P3/5P2/1PpKN3/RN6

**Input:**Gerd Wilts, 2004-01-04

**Last update:**James Malcom, 2021-01-24 more...

1. Sf3 Sc6 2. Se5 Sxe5 3. h3 Sg4 4. Th2 Sxh2 5. a3 Sg4 6. hxg4 h6 7. Sc3 Sf6 8. Sb1 Sg8 9. Sc3 Sf6 10. Sb1 Sg8= draw by 3Rep

**Henrik Juel**: The Draw (by repetition) is part of the stipulation

Without it, there would be cooks, e.g.

1.Sg1-f3 Sb8-a6 2.Sf3-g1 Sa6-b4 3.Sg1-f3 Sb4-d5

4.Sf3-g1 Sd5-e3 5.Bh2-h3 Se3-g4 6.Th1-h2 Sg4xh2

7.Sg1-f3 Sh2-g4 8.Sf3-h2 Sg4xh2 9.Ba2-a3 Sh2-g4

10.Bh3xg4 Bh7-h6

and short solutions, e.g.

1.Sg1-f3 Sb8-a6 2.Sf3-d4 Sa6-c5 3.Sd4-b3 Sc5xb3

4.Sb1-a3 Sb3-d4 5.Ta1-b1 Sd4-f5 6.Tb1-a1 Sf5-g3

7.Sa3-b1 Sg3xh1 8.Ba2-a3 Sh1-g3 9.Bh2xg3 Bh7-h6

10.Bg3-g4 (2018-12-25)

**Henrik Juel**: The shortest solution (without Draw) is unique

1.Sg1-f3 Sb8-c6 2.Sf3-e5 Sc6xe5 3.Bh2-h3 Se5-g4

4.Th1-h2 Sg4xh2 5.Ba2-a3 Sh2-g4 6.Bh3xg4 Bh7-h6

and identical to the first 6.0 moves of the intended solution

Why could the subsequent triple repetition not be achieved by rook moves?

Probably because the position after 7.Ta2 Tb8 8.Ta1 Ta8 is not considered to be a repeated position, as the castling rights have changed (2018-12-25)

**A.Buchanan**: Yes that's exactly it Henrik: the surviving rooks are nailed to their start squares (2021-06-05)

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**Keywords:**Unique Proof Game, Draw by repetition, Constrained problem, Castling

**Genre:**Retro

**FEN:**r1bqkbnr/ppppppp1/7p/8/6P1/P7/1PPPPPP1/RNBQKB2

**Input:**Gerd Wilts, 2011-02-05

**Last update:**Arnold Beine, 2021-06-05 more...

9 - P1281158

R404c Probleemblad 12/2012

3. Lob

Probleemblad Retros 2009-2012

(15+15) C+

Proof game in 7.5

Circe

**Andrew Buchanan**R404c Probleemblad 12/2012

3. Lob

Probleemblad Retros 2009-2012

(15+15) C+

Proof game in 7.5

Circe

1. f3 d5 2. Kf2 Kd7 3. Ke3 d4+ 4. Kxd4 Ke6+ 5. Kc5 Dxd2 6. Lxd2[+sDd8] a6! 7. Lc3 a5 8. Dd5+

Not 6. ... a5? 7. Lc3 Dd5+ 8. Dxd5[+sDd8]+ because the game is already dead after 7.0 moves.

Not 6. ... a5? 7. Lc3 Dd5+ 8. Dxd5[+sDd8]+ because the game is already dead after 7.0 moves.

Originally published as R404 with magazine's typo (wKc5 placed on c4). The diagram shown here is the correct diagram published in Probleemblad 71-2, apr-jun 2013 as R404c.

Threefold Repetition (or the 50 Move Rule) could only terminate the game at a later point, and require the conventions to trigger them. I think it's more accurate to use A1.3 to stop the game earlier at 7.5, including the pleasant try, and with no recourse to the conventions needed.

PDB & WinChloe contain no problems combining Circe with Threefold Repetition or the 50 move rule. Has anyone exploited the forced looping effect thematically in Circe before? (2013-12-16)

Forced looping interacts first with draw by repetition, so the 50M question is not so much of an issue. (2018-09-15)

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**Ladislav Packa**: I don`t know whether this is the right example for the application of Rule A.1.3. It is a very general rule and in the Laws of Chess can be found more accurately examples, for example A.5.2. or A.9.2. - Threefold repetition of position. (2013-12-16)**A.Buchanan**: Thanks very much for your comment Ladislav.Threefold Repetition (or the 50 Move Rule) could only terminate the game at a later point, and require the conventions to trigger them. I think it's more accurate to use A1.3 to stop the game earlier at 7.5, including the pleasant try, and with no recourse to the conventions needed.

PDB & WinChloe contain no problems combining Circe with Threefold Repetition or the 50 move rule. Has anyone exploited the forced looping effect thematically in Circe before? (2013-12-16)

**A.Buchanan**: This little problem has been reprinted in Probleemblad 74-2, apr-jun 2016 as 3e BV, but the typo from R404 has been *reintroduced*! Maniacal laughter! :-( (2016-06-16)**A.Buchanan**: To be clear, this problem would be unsound without DP, as the try would be a cook. Draw by Repetition & 50 Move are Roger Irrelevant here - even without recourse to these rules or conventions, there is no mate in the infinite loop of descending positions. Curiously, I have seen no other Circe problem which depends upon the forced looping that can so easily be set up, particularly in the end game, and which one might term "Circe Circuits". (2018-09-09)**Adrian Storisteanu**: (The 50-move rule may not really fit circe problems. A different number might be more appropriate for circe endgames -- though I'm not familiar with the theory!?) (2018-09-09)**A.Buchanan**: Circe endgames one might expect to be longer, because the stronger side cannot so easily eliminate the opposing pieces. In adapting 50M condition, one might interpret capturing to mean the genuine removal of a piece from the board, but pawn moves can be undone: the only pawn move which is a genuine progression is a promotion.Forced looping interacts first with draw by repetition, so the 50M question is not so much of an issue. (2018-09-15)

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**Keywords:**Circe, Dead Position, Unique Proof Game, Tempo Move, Draw by repetition

**Genre:**Retro, Fairies

**Computer test:**C+ Jacobi v0.1 1-Nov-2017

**FEN:**rnbq1bnr/1pp1pppp/4k3/p1KQ4/8/2B2P2/PPP1P1PP/RN3BNR

**Reprints:**535 Variant Proof Games 26/03/2015

Probleemblad 74-2 04-06/2016

**Input:**A.Buchanan, 2013-12-15

**Last update:**A.Buchanan, 2021-12-15 more...

1. f4 e5 2. Kf2 Lb4 3. Ke1 Lxd2+ 4. Kf2 Le3+ 5. Ke1 Ld2+ 6. Kf2 Le3+ 7. Ke1 Ld2+= draw by 3Rep

Shortest drawn PG. (Alternative diagram: could have sBe5 on e6 instead.)

According to Popeye 4.79, there are 12 ways to reach the diagram in 3.0 moves. After that, 4.0 moves are required to repeat the position twice. This must include moving wK so this unit must have moved by 3.0 point, to lose castling rights, in order for the position to repeat. Out of the 12 candidate solutions, only one involves wK switchback.

Shortest drawn PG. (Alternative diagram: could have sBe5 on e6 instead.)

According to Popeye 4.79, there are 12 ways to reach the diagram in 3.0 moves. After that, 4.0 moves are required to repeat the position twice. This must include moving wK so this unit must have moved by 3.0 point, to lose castling rights, in order for the position to repeat. Out of the 12 candidate solutions, only one involves wK switchback.

**Keywords:**Unique Proof Game, Draw by repetition, Economy record, Constrained problem, Castling

**Genre:**Retro

**Computer test:**Popeye 4.79 then a little bit of thinking

**FEN:**rnbqk1nr/pppp1ppp/8/4p3/5P2/8/PPPbP1PP/RNBQKBNR

**Input:**A.Buchanan, 2015-02-25

**Last update:**A.Buchanan, 2021-06-05 more...

11 - P1298076

www.anselan.com 2002

(13+15) C+

White can draw with his next move

PG in 9.0

**Andrew Buchanan**www.anselan.com 2002

(13+15) C+

White can draw with his next move

PG in 9.0

1. Sc3 Sf6 2. Se4 Sxe4 3. a4 Sxf2 4. a5 Sh3 5. a6 Sxg1 6. Txg1 Sc6 7. Th1 Sb8 8. Tg1 Sc6 9. Th1 Sb8 then 10. Tg1= draw by 3Rep

Trying each reversible White move in turn (7 of them) only 10. Rg1 yields any way to reach the new diagram in 5.5 moves. The remaining 4.0 moves must be executed in a way that leaves castling rights unchanged. The position cannot repeat at 5.0 mark or any shorter number of moves.

Precursor of P1338392.

Trying each reversible White move in turn (7 of them) only 10. Rg1 yields any way to reach the new diagram in 5.5 moves. The remaining 4.0 moves must be executed in a way that leaves castling rights unchanged. The position cannot repeat at 5.0 mark or any shorter number of moves.

Precursor of P1338392.

**Keywords:**Unique Proof Game, Draw by repetition, Homebase (S), Constrained problem, Castling

**Genre:**Retro

**Computer test:**Popeye 4.79 then a little bit of thinking

**FEN:**rnbqkb1r/pppppppp/P7/8/8/8/1PPPP1PP/R1BQKB1R

**Input:**A.Buchanan, 2015-02-25

**Last update:**A.Buchanan, 2021-06-05 more...

12 - P1338392

J.Lois-70 JT 2016-2017

Special Prize

(9+9) C+

Draw exists in 1.0 move.

PG in 13.0

**Andrew Buchanan**J.Lois-70 JT 2016-2017

Special Prize

(9+9) C+

Draw exists in 1.0 move.

PG in 13.0

1. Sf3 b5 2. Sd4 Lb7 3. Sxb5 Lxg2 4. Sxa7 Sc6 5. Sxc6 Lxf1 6. Sxd8 Lxe2 7. Sxf7 Lxd1 8. Sh6 Lxc2 9. Sxg8 Lxb1 10. Txb1 Txg8 11. Ta1 Th8 12. Tb1 Tg8 13. Ta1 Th8 then 14. Tb1 Tg8= draw by 3Rep

**Henrik Juel**: 1.Sf3 b5 2.Sd4 Lb7 3.Sxb5 Lxg2 4.Sxa7 Sc6 5.Sxc6 Lxf1 6.Sxd8 Lxe2 7.Sxf7 Lxd1 8.Sh6 Lxc2 9.Sxg8 Lxb1 10.Txb1 Txg8 11.Ta1 Th8 12.Tb1 Tg8 13.Ta1 Th8

14.Tb1 Tg8 draws by three repetitions

Judge J. Lois:

The problem is well explained by the author:

“The position at 10.0 is SPG in 10.0. Following this, only the two rooks which have moved can move, to avoid disrupting the remaining castling rights.

Validation cases:

(1) Show that E is indeed a unique SPG in 10.0. Observe that the unique solution does not repeat a position for the third time until 14.0.

(2) Show that E has no solutions in a smaller number of moves. (Checking 8.5, 9.0 & 9.5 is sufficient.)

(3) Show that D cannot be reached even non-uniquely in 9.0 moves or less.

(4) Show that none of the 6 positions after a single reversible White move from D can be reached even non-uniquely in 9.5 moves or less.

(5) Show that none of the other 6x11-1=65 positions after 1.0 reversible moves from D can be reached even non-uniquely in 10.0 moves or less. (In fact (5) implies (4) implies (3)) In the context of the condition "draw in 1.0", the problem is indeed SPG. "PG" would incorrectly suggest that a shorter solution exists”.

An excellent problem. (2017-08-30)

comment

**Keywords:**Unique Proof Game, Draw by repetition, Homebase (2), Constrained problem, Castling

**Genre:**Retro

**FEN:**r3kb1r/2ppp1pp/8/8/8/8/PP1P1P1P/R1B1K2R

**Reprints:**AA012 The Hopper Magazine 1 24/12/2021

**Input:**A.Buchanan, 2017-08-30

**Last update:**A.Buchanan, 2022-01-31 more...

1. Sc3 Sf6 2. Se4 Sxe4 3. f3 Sg3 4. hxg3 a6 5. Sh3 Sc6 6. Sg1 Sb8 7. Sh3 Sc6 8. Sg1 Sb8= draw by 3Rep

Shortest drawn PG without check at end. (Alternative diagram: could have wBf3 on h3 instead.)

Shortest drawn PG without check at end. (Alternative diagram: could have wBf3 on h3 instead.)

**Henrik Juel**: Pawn moves and captures ruin the repetitions, so let us do them at the beginning

1.Sc3 Sf6 2.Se4 Sxe4 3.f3 Sg3 4.hxg3 a6

There is probably some esoteric stuff about repetitions and castling rights, so let us pendulate with the surviving knights, leaving castling rights intact

5.Sh3 Sc6 6.Sg1 Sb8 7.Sh3 Sc6 8.Sg1 Sb8

The position of the men after 4... a6 just occurred for the third time, so it is draw by repetition (2020-05-25)

**Henrik Juel**: Afterwards I cheated and looked at the PDB definition of Draw by repetition, and sure enough:

FIDE 2005, article 9.2 (2020-05-25)

comment

**Keywords:**Unique Proof Game, Draw by repetition, Castling, Constrained problem, Economy record

**Genre:**Retro

**Computer test:**HC+ Popeye v4.87 + simple thinking about 3Rep

**FEN:**rnbqkb1r/1ppppppp/p7/8/8/5PP1/PPPPP1P1/R1BQKBNR

**Input:**A.Buchanan, 2020-05-25

**Last update:**A.Buchanan, 2021-11-29 more...

**Keywords:**Draw by repetition, Unique Proof Game, Constrained problem

**Genre:**Retro

**FEN:**rnb1kbnr/ppp3pp/5pq1/3pP3/8/7K/PPPPP1PP/RNBQ1BNR

**Input:**A.Buchanan, 2022-01-31

**Last update:**A.Buchanan, 2022-01-31 more...

Avoidance of premature draw

**Keywords:**Draw by repetition, Unique Proof Game, Constrained problem

**Genre:**Retro

**FEN:**rnbqkb1N/1ppppppr/p6p/8/8/5P2/PPPPP1PP/R1BQKB1R

**Input:**A.Buchanan, 2022-01-31

**Last update:**A.Buchanan, 2022-01-31 more...

16 - P1398782

AA009 The Hopper Magazine 1 24/12/2021

(15+15)

(15+15) Drawn game. PG in 12.0

**Andrew Buchanan**AA009 The Hopper Magazine 1 24/12/2021

(15+15)

(15+15) Drawn game. PG in 12.0

**Keywords:**Draw by repetition, Unique Proof Game, Constrained problem

**Genre:**Retro

**FEN:**rnb1k1nr/1ppp1ppp/4p3/8/q2N1P2/N7/PPPbP1PP/R1BQKB1R

**Input:**A.Buchanan, 2022-01-31

**Last update:**A.Buchanan, 2022-02-08 more...

17 - P1398783

AA010 The Hopper Magazine 1 24/12/2021

Motto: "No, no, no!"

(13+13)

(15+15) Drawn game. PG in 14.0

hint rot13.com: vtaber cerzngher qenjf

**Andrew Buchanan**AA010 The Hopper Magazine 1 24/12/2021

Motto: "No, no, no!"

(13+13)

(15+15) Drawn game. PG in 14.0

hint rot13.com: vtaber cerzngher qenjf

Constructed specifically to maximize the number of times 3Rep convention is ignored (3 times)

**Keywords:**Draw by repetition, Unique Proof Game, Constrained problem, Golden Age (3Rep convention)

**Genre:**Retro

**FEN:**rnbqkb1Q/pppppp2/7B/8/1K6/1P6/PP3PPP/nN3BNR

**Input:**A.Buchanan, 2022-01-31

**Last update:**A.Buchanan, 2022-01-31 more...

18 - P1401677

SS014 The Hopper Magazine 1 24/12/2021

(14+15)

Drawn game. PG 11.0

(2 tries)

**Andrew Buchanan**SS014 The Hopper Magazine 1 24/12/2021

(14+15)

Drawn game. PG 11.0

(2 tries)

**Keywords:**Unique Proof Game, Constrained problem, Draw by repetition

**Genre:**Retro

**FEN:**r2q1bnr/ppp1pppp/1k6/2n5/8/6P1/PPPPPPbP/RNBQ1R1K

**Input:**A.Buchanan, 2022-06-08

**Last update:**A.Buchanan, 2022-07-19 more...

Show statistic for complete result. Show search result faster by using ids.

https://pdb.dieschwalbe.de/search.jsp?expression=PROBID+IN+%27P0008596%3B+P0009007%3B+P1003993%3B+P1004352%3B+P1004353%3B+P1004354%3B+P1011937%3B+P1186643%3B+P1281158%3B+P1298075%3B+P1298076%3B+P1338392%3B+P1376322%3B+P1398780%3B+P1398781%3B+P1398782%3B+P1398783%3B+P1401677%27

The problems of this query have been registered by the following contributors:

Gerd Wilts (8)A.Buchanan (10)

hans: 1. h4 Sa6 2. h5 Sc5 3. h6 Sa4 4. hxg7 Sh6 5. g8=L Lg7 6. Lh7 Ld4 7. Le4 Lb6 8. Lc6 dxc6 9. e3 Lh3 10. e4 f5 11. e5 Kd7 12. Sc3 Ke8 13. Sb1 Kd7 14. Sc3 Tc8 15. Sb1 and now black can claim a draw after Ta8, because this position, with the same possibilities, occurred after the 11th and 13th move too. Errors possible: Switch f5 and Kd7: then white has lost the ep-right, so no 3-fold repetition. Tc8 instead of Kd7: Then black has still kingside castling rights, so no repetition there either. (2015-09-04)Per Olin: The solution ends after White's 11th move; the stipulation asks for 22 halvmoves (the repetition moves starting from White's 12th move are not unique). That there is a possibility to a threefold repetition in Black's 15th move dictates the exact move order 9. e3 Lh3 10. e4 f5 11. e5 Kd7. Tries: 1) Switch f5 and Kd7: then white has the ep-right, which is lost later when the position is repeated, so no threefold repetition. 2) 11. - Tc8 instead of 11. - Kd7: then black has still kingside castling rights, so no repetition there either. (2015-09-06)comment