# Phénix

Solutions

349 - A. Frolkin
The problem is to figure out how we can avoid exposing the black king to illegal checks. Black made the last move, so his king just came from d5 (there is no other possible move). We must retract -1...Kd5xBd6 -2. Be5-d6+ Kd6-d5 -3. Bb2-e5+ Kd5xBd6 -4. Be5-d6+ Kd6-d5 -5. Bc3-e5+ Kd5xBd6 -6. Be5-d6+ Kd6-d5 -7. Bd4-e5+ Kd5xBd6 -8. Be5-d6+ Kc4-d5 -9. Bc6-b5+ Kb3-c4 -10. Q~-c2+, and now Black's king can shuffle between b3 and c4 while the white rooks move out of the way and unlock the position.

However, the big question is whether the resulting position (after these retractions) is actually possible. Each side has 4 promoted pieces on the board: Wh. has 3B+Q and Bl. has 3R+B. Also, White has not captured any pieces, and 3 of his 4 promotions must take place on dark squares; since d8 is not possible, they must be b8, f8, h8. Therefore White's b,f,h pawns have promoted. Proceeding from here, we can deduce exactly what happened to all of the pawns:
Bl h-pawn captured Wh g-pawn on g4,
Wh h-pawn promoted to B on h8,
Bl f-pawn captured Wh e-pawn and promoted to R on e1,
Wh f-pawn promoted to B on f8,
Bl c-pawn captured Wh d-pawn and promoted to B on d1,
Wh c-pawn promoted to Q on c8,
Bl b-pawn captured Wh a-pawn and promoted to R on a1,
Wh b-pawn promoted to B on b8,
Bl a-pawn promoted to R on a1.

This accounts for all pawns and all captures. So the position is indeed possible. We have determined precisely which pawns captured which other ones as well as the squares on which each promotion took place. An absolutely remarkable composition!!! [ThHwa]

## Phénix 28, Dec. 1994

1936 - G. Wilts and N. Geissler
1. e4 d5 2. exd5 Bg4 3. Qxg4 Qxd5 4. Qxg7 Qxa2 5. Qxh7 Qxb1 6. Rxa7 Qxc2 7. Qxc2 Rxh2 8. Qxc7 Rxg2 9. Qxb7 Rxg1 10. Qxb8 Rxb8 11. Rxg1 Rxb2 12. Rxg8 Rxd2 13. Rxf8 Kxf8 14. Rxe7 Rxf2 15. Rxf7 Rxf7 16. Bg5 Rxf1 17. Kxf1

1937 - L. Jury
Solution: add wKc4. Then Bl. has the move and play goes 0 ... a2 1. Nb2#.

Adding wKc5?? fails because then Wh. has the move (-1. Kb3-a4 legal).

1938 - N. Plaksin and A. Zolotarev
Bl. caps are h7xg6xf5xe4, d7xe6 and c7xBd6. Wh. caps are gxf, a6xRb7 and bxRc so that his a, b & d Pawns promoted on b8, c8 & d8. Wh. has no last move and Bl. must uncapture a wh. promotee who must quickly uncapture a bl. unit, avoiding Bl. retropat.

Retract e.g. -1. f5xQe4 Qb7-e4 -2. g6xBf5 Qb8-b7 -3. h7xNg6 b7-b8=Q -4. a7-a6 a6xRb7 etc. wBf5 is unpromoted on c8 (later wNg6 on d8) so that c7xd6 (later d7xe6) can be retracted.

Last move is -1. Pf5xQe4. (Other moves are not precisely determined.)

1939 - M. Caillaud
1. g4 g6 2. g5 h5 3. gxh6 ep g5 4. e4 g4 5. h4 gxh3 ep 6. Qg4 e6 7. Qg7 Qg5 8. e5 f5 9. exf6 ep e5 10. c4 e4 11. f4 exf3 ep 12. Kf2 c6 13. Kg3 Bd6 14. Kh4 Bh2 15. c5 d5 16. cxd6 ep c5 17. Ba6!! c4 18. d4 cxd3 ep 19. Bc4 a6 20. a4 Ra7 21. a5 b5 22. axb6 ep a5 23. Ne2 a4 24. b4 axb3 ep 25. Bb2 Na6 26. Be5 Rc7

1940v - T. Le Gleuher
1. Nc3 b6 2. Nd5 Bb7 3. Nxe7 Bf3 4. Nxg8 Bb4 5. e3 d6 6. Ne2 Kd7 7. Nc3 Kc6 8. Nb1 Nd7 9. Bb5 Kxb5 10. Nf6 c6 11. Nxh7 Qc7 12. Ng5 Rae8 13. Nh3 Qb8 14. Ng1

Wh. Knights are swapped.

1941 - M. Caillaud
(a) 1. a4 a5 2. Ra3 Ra6 3. Rh3 Rc6 4. c3 Rc5 5. Qc2 c6 6. Qg6 hxg6 7. Kd1 Rh4 8. Kc2 Rxa4 9. Rh5 gxh5
(b) 1. c3 h5 2. Qb3 Rh6 3. Qb6 axb6 4. Kd1 Rxa2 5. Kc2 Ra4 6. Ra3 Rc6 7. Rb3 Rc5 8. Rb5 c6 9. Ra5 bxa5

1942 - T. Le Gleuher
1. g3 c5 2. Bh3 Qb6 3. Bf5 Qb3 4. axb3 g5 5. Ra6 g4 6. Rc6 a5 7. c4 a4 8. Qc2 a3 9. Qe4 a2 10. d3 a1=Q 11. Bf4 Qa5 12. Nd2 Qd8 13. Bc7 e5 14. f4 Qh4 15. gxh4 g3 16. Ngf3 g2 17. Rf1 g1=Q 18. Bg6 Qg5 19. Kf2 Qd8

Two promoted Queens go back to (same) home square !!

## Phénix 31, Jul. 1995

2023c - T. Le Gleuher
1. Nc3 h5 2. Ne4 h4 3. Ng3 hxg3 4. h4 Rh5 5. Rh2 gxh2 6. Rb1 hxg1=N 7. Ra1 Nh3! 8. Rb1 Nf4 9. Ra1 Nd5 10. Rb1 Nb4 11. Ra1 Rd5 12. Rb1 g5 13. Ra1 Bg7 14. Rb1 Bd4 15. Ra1 Bb6 16. Rb1 c5 17. Ra1 Qc7 18. Rb1 Qe5 19. Ra1 d6 20. Rb1 Kd7 21. Ra1 Ke6 22. Rb1 Kf5 23. Ra1 Kf4 24. Rb1 Bf5 25. Ra1 Nd7 26. Rb1 Rc8 27. Ra1 Rc6 28. Rb1 Bc7 29. Ra1 Ra6 30. Rb1 Ra3 31. Ra1 Rh3 32. Rb1 Rh1 33. Ra1 Rg1 34. Rb1 Nd3+

29 pendulum moves by wRa1 in a dual-free SPG.

2024 - P. Wong
1. Nf3 a6 2. Nd4 Ra7! 3. Nb5 axb5 4. f3! Ra3 5. bxa3 Nc6 6. Bb2 Na5 7. Be5 Nc4 8. Bd6 cxd6 9. c3 Qa5 10. Qc2 Qa4 11. Qg6 Qc2 12. Qh6 gxh6 13. f4 Bg7 14. f5 Bd4 15. f6 Nxf6 16. h3! Rg8 17. h4 Rg3 18. h5 Rh3 19. g3! Ba7 20. Bg2 b6 21. Ba8 Bb7 22. O-O Bg2 23. Bf3 Ne4 24. g4 f6 25. g5 Kf7 26. g6 Ke6 27. g7 Kd5 28. g8=N

Four losses of a tempo.

2025 - A. Frolkin
1. h4 f5 2. h5 Kf7 3. h6 Kg6 4. hxg7 h5 5. b4 Nh6 6. g8=B h4 7. Bc4 h3 8. Ba6 bxa6 9. b5 Bb7 10. b6 Be4 11. b7 Nc6 12. b8=Q Bd3 13. Qb2 Rb8 14. Qf6 exf6 15. exd3 Bb4 16. d4 Ba5 17. Bc4 Rb4 18. Bg8 Nf7 19. Qe2 Rh4 20. Qb5 Kh5 21. Qb8 Qe7

Anti-phoenix theme with wQ + wB.

2026 - T. Le Gleuher
1. a4 c6 2. a5 Qb6 3. axb6 Na6 4. Ra5 Nc7 5. bxc7 Rb8 6. cxb8=Q Nf6 7. Qf4 Nd5 8. Qa4 Nc3 9. dxc3 Kd8 10. Kd2 Kc7 11. Kd3 Kd6 12. Kc4 Ke6 13. Kb3 Kf6 14. Bg5 Kg6 15. Nd2 Kh5 16. Bh4 Kh6!! 17. e3 Kg6 18. Rg5 Kf6 19. Be2 Ke6 20. Bg4 Kd6 21. Ngf3 Kc7 22. Re1 Kd8 23. Re2 Ke8 24. Qh1

Long King walk (15 moves) for loss of a tempo.

## Phénix 34, Oct. 1995

2104 - J. Lörinc
(a) -1. d4xe3 ep+ ! e2-e4 -2. d5-d4+

(b) Here -1. d4xe3 ep?? is not legal because the bPd4 was not covered and thus could not capture. The only way to explain the check assumes that the bBa8 was not covered previously: -1. O-O+ !

(c) Here -1. d4xe3 ep e2-e4?? is not legal because the wPe4 and bPf5 are paralyzed. The only way to explain the check assumes that the bBa8 could not capture on f3 previously because f3 was covered by a wB. Unfortunately, -1. Nb2xBd1?? or -1. Nf2xBd1 are illegal because of the paralyzed Ns. Thus -1. c2xBd1=N+ ! (not -1. e2xBd1=N??)

Here, the last move is an en-passant capture, a castling move, or a promotion depending on which fairy rule is considered.

2105 - T. Le Gleuher
1. b4 b5 2. Bb2 Bb7 3. Bd4 Qc8 4. Bb6 axb6 5. c3 Ra3 6. Qc2 Rb3 7. Qg6 hxg6 8. a4 Rxh2 9. a5 Rxg2 10. Rh8 Rh2 11. axb6 Rh1 12. Ra8 Ra3 13. Bh3 Ra1

Belfort Theme with 4 Rooks !

2106c - T. Le Gleuher
Another Belfort Theme with 4 Royalties !!

1. d4 c5 2. Kd2 Qa5+ 3. Kd3 Qd2+ 4. Kc4 a5 5. Kb5 c4 6. Kb6 c3 7. Kc7 cxb2 8. c3 Ra6 9. Qb3 Rg6 10. Qb6 Rg3 11. hxg3 f6 12. Rh5 Kf7 13. Rb5 Kg6 14. Rb3 Kf5 15. Kd8 Ke4 16. f3+ Ke3 17. Ke8 Kf2 18. Qd8 Qd1 19. Be3+ Ke1

2107 - T. Le Gleuher
1. d4 h5 2. Bh6 g5 3. e3 Bg7 4. Ba6 b5 5. Nc3 Bb7 6. Rb1 Bxg2 7. Nd5 Bf1 8. Nxe7 Kxe7 9. d5 Bxb2 10. Bc8 Bc1 11. Bf8

Belfort Theme with 4 Bishops ! PvdH suggests comparison with

F. Christiaans and H. Boumeester

16+14. Proof game in 7.5 moves

2108 - A. Frolkin
1. a4 Nf6 2. a5 Nd5 3. a6 Nb6 4. axb7 a5 5. f4 a4 6. f5 Ra5 7. f6 Na6 8. b8=N a3 9. Nc6 dxc6 10. fxe7 Kd7 11. e8=R Kd6 12. Re6 Kd5 13. Rg6 hxg6 14. Nf3 Rh3 15. Ne5 Rc3 16. h4 Bd6 17. Rh3 Qf8 18. Re3 Bh3 19. Nd7 f5 20. Re8 Be5 21. Nb8

Anti-phoenix theme with wN + wR. Unfortunately cooked (as indicated by G. Wicklund on the retro-list) e.g. with 1. Nf3 Nf6 2. Ne5 Nd5 3. Ng6 hxg6 4. a4 Rh3 5. a5 Rc3 6. a6 Nb6 7. axb7 a5 8. h4 a4 9. Rh3 a3 10. Rd3 Ra5 11. Rd6 Na6 12. Rc6 dxc6 13. f4 Bh3 14. f5 Kd7 15. f6 Kd6 16. fxe7 Kd5 17. e8=R Bd6 18. Re6 Qf8 19. Re7 f5 20. Re8 Be5 21. b8=N

2109 - T. Le Gleuher
1. Nc3 b6 2. Nd5 Bb7 3. Nxe7 Bf3 4. Nxg8 Bb4 5. e3 d6 6. Ne2 Kd7 7. Nc3 Kc6 8. Nb1 Nd7 9. Bb5 Kxb5 10. Nf6 c6 11. Nxh7 Qc7 12. Ng5 Rae8 13. Nh3 Qb8 14. Ng1

Wh. Knights are swapped.

2110 - T. Le Gleuher
1. Nf3 Nc6 2. Nd4 Ne5 3. Nc6 Nf6 4. Nb8 c6 5. Nc3 Qa5 6. Nd5 Qc3 7. dxc3 Ne4 8. Bg5 Nd2 9. Bf6 exf6 10. Ne7 Nb1 11. Ng8 Bc5 12. Qd5 Be3 13. fxe3 Nf3 14. Kf2 Ng1

Belfort Theme with 4 Knights and very nice maneuvers avoiding dual paths for the Knights !

## Phénix 36, Dec. 1995

2181 - M. McDowell and J. Quah
Authors' intention was 1. e3 f6 2. Bd3 Kf7 3. Bxh7 Rxh7 4. Ke2 Rxh2 5. Kf3 Rxg2 6. Rh5 Rg4 7. Ra5 Ra4 8. Rxa7 Rxa2 9. Rxa8 Rxa1 10. Rxb8 Rxb1 11. Rxc8 Rxc1 12. Rxd8 Rxd1 13. Re8 Re1 but there are many cooks, e.g. 1. e3 h5 2. Qg4 hxg4 3. h4 Rxh4 4. Rh3 gxh3 5. Ke2 Ra4 6. Kf3 Rxa2 7. Rxa2 hxg2 8. Rxa7 gxf1=R 9. Rxa8 Rxc1 10. Rxb8 Rxb1 11. Rxc8 f6 12. Rxd8 Kf7 13. Re8 Re1 where permutations are possible after 6.5 moves.

2182 - N. Plaksin and A. Zolotarev
Bl. caps are g7xf6, axb, bxc, cxd & -1. Bh5xg4+. One bB is bPh7 promoted on h1. Wh. caps are exf/d, hxg allowing the wPs in the balance. Wh. promoted two Ps on a8, two on g8. The position unlocks by retracting g7xf6. This requires unpromoting first the wPh2 & g2. We also have to reconduct the bBf8, which can only be resurrected by e5xd6 (hxg must wait for bPh7). Thus the wPe2 did promote on d8.

Retract: -1. Bh5xRg4+ Re4-g4 -2. Rg4-g3+ Re6-e4 -3. c3xQd2 Rd6-e6 -4. c4-c3 Rd8-d6 -5. c5-c4 d7-d8=R -6. c6-c5 d6-d7 -7. d5-d4 e6xBd6

Notice that only a wRg4 could reach d8 before Bl. retro-stalemate. Only a wQd2 can reach a8. Also observe that only 2wQs can reach g8 after they are uncaptured on the left, and they have to pass through d2 to go there. So that it is not possible to retract d2-d3 now.
The unlocking maneuver is as follows (with some latitude in sequencing): (1) unpromote wQd2 through a6-a7-a8=Q, (2) retract bPa7xQb6, (3) unpromote Qb6 through g7-g8=Q (via d2), (4) retract wPb6-b7 & bPb7xQc6, (5) unpromote Qc6 through g7-g8=Q (via d2), (6) reconduct bBd6 to f8, (7) retract d2-d3, avoiding Bl. retro-stalemate, (8) retract g7xNf6 (with wPs on g5 & g6).

This displays 4 completely determined Ceriani-Frolkin promotions.

2184c - P. Raican
1. e4 Nf6(=P) 2. Bd3(=N) f5 3. ef5(=N) h6 4. Nxe7(=B) Qxe7+ 5. Ne2(=P) Qf6(=R) 6. Rf1(=B) Rxf2(=Q)+ 7. Nxf2(=B) Rh7(=B) 8. Bg1(=N) Bg8(=N)

2185 - J. de Heer
1. h4 g5 2. Rh3 gxh4 3. Re3 h3 4. Re6 dxe6 5. Nf3 Bd7 6. Ne5 Ba4 7. f3 Bb3 8. axb3 h2 9. Ra4 h1=B 10. Rh4 Qd4 11. g4 Bg2 12. Rh1 Bh3 13. Bg2

2186 - M. Ott and D. Borst
1. d4 Nc6 2. Be3 Ne5 3. dxe5 d6! 4. Bc5 d5 5. e3 d4 6. Ke2 d3 7. Kf3 d2 8. Qe1 d1=R 9. Bb5 Rd7 10. c4 Nf6 11. Nc3 Nd5 12. Rd1 Nb4 13. Rd6 Na6 14. Rc6 Rd6 15. exd6 Kd7 16. Kg4 Nb8 17. Ra6

Impressive trip of the bPd7, aiming at loss of a tempo.

## Phénix 40, Apr. 1996

2262 - T. Volet
Bl. caps are f7xNg6 & e7xBd6. Wh. caps are g2xf3, h2xg3xf4xe5xd6xPc7 and cross-caps a4xb5 & bxa. Position unlocks by reconducting bBf8, but this bB must be uncaptured by bxa, so that we must first unpromote bPa7, retract a4xb5 and reconduct bPb4 to b7. This further requires driving back home the bBc8 and the bQd8.

Here is the unlocking maneuver:
(1) retract -1. Rg1-g2 f7xNg6 -2. g2xQf3 (and not g2xBf3) escaping bl. retropat,
(2) drive a wN to b7, releasing the bN,
(3) unpromote the bN on a1, reconduct the bP to a5,
(4) retract a4xBb5, releasing the wBa6, and then the wNb7 and the bK,
(5) drive bBb5 to its home square c8,
(6) reconduct the bPb4 to b7,
(7) retract b6xBa7,
(8) drive bK, bQ and bBa7 to their home d8, e8 & f8,
(9) retract e7xBd6, allowing d6xPc7 etc.

Last 6 captures and their succession are completely determined.

2263 - A. Minost
1. d4 h6 2. Bxh6 e5 3. Bc1 exd4 4. Qxd4 Rxh2 5. Qxa7 Rh8 6. Qxb7 Rxa2 7. Qd5 Ra8 8. Qd1

Switchbacks of four different units.

2264 - A. Frolkin and A. Shvichenko
1. h4 d5 2. Rh3 Bg4 3. Ra3 Bf3 4. gf3 (=bP) Nc6 5. Bh3 Rc8 6. Bxc8 (=bB) d6 7. Rxa7 (=bR) Bb4 8. a4 Bc3 9. bc3 (=bP) Nf6 10. Ba3 Rf8 11. Bxf8 (=bB) Ra8 12. Na3 Na7 13. Qb1 c5 14. Qb6 c4 15. Qxd8 (=bQ)

Four bl. units on origin squares really are transmuted wh. units.

2265 - T. Le Gleuher
1. f4 Nc6 2. f5 Rb8 3. f6 exf6 4. d4 Ba3 5. d5 Nge7 6. d6 O-O 7. dxe7 Bxb2 8. e8=Q Ba3 9. Qe4 Ne7 10. Qg6 fxg6 11. e4 Kf7 12. Bc4 Ke8 13. Ne2 Rh8 14. O-O Ng8 15. Rf5 Bf8

Bl. did O-O though all four K-side units are on their home squares !! (The Ng8 really is from b8).

Unfortunately, this is cooked e.g. by 1. f4 Nh6 2. f5 Nc6 3. f6 exf6 4. d4 Ne5 5. dxe5 Ba3 6. e6 O-O 7. e7 Bxb2 8. e8=Q Ba3 9. Qe4 Rb8 10. Qg6 fxg6 11. e4 Kf7 12. Bc4 Ke8 13. Ne2 Rh8 14. O-O Ng8 15. Rf5 Bf8

2266 - A. Frolkin
Intended solution is

1. e3 d5 2. Bd3 d4 3. Be4 d3 4. Ne2 dxe2 5. d3 f5 6. Bd2 f4 7. Bb4 f3 8. Kd2 e1=R 9. Nc3 Rg1 10. Ne2 fxe2 11. f4 e1=R 12. f5 Ref1 13. f6 Bf5 14. a4 e6 15. a5 Bc5 16. a6 Ne7 17. axb7 a5 18. Qe1 Na6 19. b8=N O-O 20. Nc6 Rb8 21. Nd4 Rf2 22. Ne2 Rxe2

but unfortunately the problem is cooked, e.g. with

1. a4 b5 2. Ra3 bxa4 3. Rd3 a3 4. Rxd7 a2 5. Rd3 a1=R 6. e3 f5 7. Ra3 f4 8. Bd3 f3 9. Ne2 fxe2 10. f4 a5 11. f5 Na6 12. f6 Bf5 13. Be4 e6 14. d3 Bc5 15. Bd2 Ne7 16. Bb4 O-O 17. Kd2 e1=R 18. Nc3 Rg1 19. Qe2 Raf1 20. Qe1 Rf2 21. Ne2 Rb8 22. Ra1 Rxe2

## Phénix 43, Jul.-Aug. 1996

2340 - T. Le Gleuher
Bl. caps include f7xe6 and a7xb6xc5xb4 (because a7-a5xb4 would not let the wh. King reach a5). Wh. caps include g2xf3.

The bPe5 could not capture, so that it went straight from e7. The bl. King had to escape his home square before the lock g2xf3 is played, so that e7-e5 and f7xe6 were played even earlier. But the wKR cannot reach h5 once gxf3 and e7-e5 are played. So that it had to escape before g2xf3, which means that the wKB was captured on its home square f1, before g2xf3, by a bl. promotee issued from the bPb7 who had to capture 3 wPs. With these observations, it becomes simple to complete a proof game where, necessarily, the bNb8 was captured by a promoted wQ issued from the promotion on h8 of the wPf2.

Here is a proof game:
1. f4 g5 2. Kf2 Bg7 3. Ke3 Bh6 4. Kd4 e5 5. Kc3 Qf6 6. fxg5 Qf2 7. gxh6 Qxg1 8. Kb4 Qf2 9. Kc3 Qg3 10. Kb4 Qg7 11. hxg7 b5 12. gxh8=Q Bb7 13. Qf6 Bc8 14. Qb6 Ba6 15. Qxb8 Bc8 16. Qb6 axb6 17. Kc3 Bb7 18. Kb4 Bd5 19. Kc3 Bb3 20. axb3 Ra4 21. Ra3 Rc4 22. bxc4 bxc4 23. d4 cxd3 24. Kb4 dxc2 25. Rc3 cxb1=Q 26. Qd3 Qc2 27. Rc5 bxc5 28. Ka5 Qd1 29. Bh6 Qxf1 30. Qg6 Qf3 31. Bf8 Qe4 32. Qxg8 Qf3 33. Qg4 Qe4 34. Qe6 fxe6 35. Rd1 Kf7 36. Rd3 Kg6 37. Rh3 Kf5 38. Bh6 Qf3 39. Bd2 Ke4 40. gxf3 Kd5 41. Bb4 cxb4 42. Rh5

2341 - F. Laroussinie and P. Schnoebelen
1. a4 c5 2. a5 c4 3. a6 c3 4. axb7 a5 5. f4 Ra6 6. f5 Rg6 7. f6 Rxg2 8. fxe7 g6 9. exd8=B Bc5 10. bxc8=N Bxg1 11. d4 d6 12. Bh6 Kd7 13. e3 Kc6 14. Bb5 Kxb5 15. Na3 Ka4 16. Qb1 Ne7 17. Kd1 Rxd8 18. Bf8 Nxc8 19. b3

The Frolkin promotions on d8 and c8 are completely determined, even though the promotees do not move at all.

2342 - P. Schnoebelen
1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 a5 5. f4 Ra6 6. f5 Rg6 7. f6 Rxg2 8. fxe7 g5 9. exd8=B Bc5 10. bxc8=R Bxg1 11. d4 g4 12. Bh6 d6 13. e3 Kd7 14. Be2 Kc6 15. Bf3 Kb6 16. Qe2 Ne7 17. Kd1 Rxd8 18. Bf8 Nxc8

The Frolkin promotions on d8 and c8 are completely determined, even though the promotees do not move at all.

2343 - T. Volet
Wh. caps include a-b & c-d cross-captures. Bl. caps include e7xf6. The position unlocks by resurrecting a bl. Knight which can be interposed on b2.

Retract
-1. Ra8-a7 e6-e7
-2. Rd8-a8 e5-e6
-3. Rd4-d8 h5-h6
-4. Rh4-d4 e4-e5
-5. Rh2-h4 h4-h5
-6. Rg2-h2 h3-h4
-7. Rg1-g2 h2-h3
-8. g2-g1=R e3-e4
-9. h3xBg2 Bf3-g2
-10. h4-h3 Bd1-f3
-11. h5-h4 Bf3xNd1
-12. Nb2-d1
and the bl. K can escape through e2.

2344 - J. Rotenberg
Visible caps include gxh, d2xc3xb4 for wh., d7xc6 for bl. It is possible to include the bPa7 in the balance if he did capture axb, and the wPe2 if he did promote on d8 after exd.

(a) It it possible to assume Wh. has the move. Retract
-1. Qa2-a1 g3xPh4
-2. Qa1-a2 Qa2-b1
-3. Rb1-b2 h2-h3
-4. b2-b1=R Qb1-a2
-5. h5-h4 a2-a3
-6. a3xNb2 Nd3-b2
-7. Qb2-a1 Nf4-d3
-8. Qa1-b2 Ne6-f4
-9. Qb2-a1 Nd8-e6
-10. Qa1-b2 d7-d8=N
-11. Qb2-a1 e6xRd7
-12. R~ Qa1-b1
-13. a4-a3 Rb1-c1
-14. Qa3-b2 Rb2-b1+
-15. R~ Qc1-a1
-16. R~ Rb1-b2
-17. Qb2-a3 Ra1-b1
-18. Qb1-b2 Qa3-c1
-19. Qb2-b1 e5-e6
-20. Rb1-d1 e4-e5
-21. Qc1-b2 e3-e4
-22. Qd1-c1 Qc1-a3 and everything unlocks.

So that mate in 1 is with 1. Ne4#!! and not 1 ... Bb6#??

(b) The solution for (a) can be used again because we have enough tempi. So that it is tempting to assume none of this is the author's intention...

2345 - M. Caillaud
(a) 1. c3 g5 2. Qa4 g4 3. Qd4 g3 4. Qxh8 gxh2 5. Qe5 hxg1=R 6. Rh6 Rxg2 7. Qh2 Rxf2 8. Bg2 Nf6 9. Bf3 Ne4 10. Rf6 Ng5
(b) 1. c3 g5 2. Qc2 g4 3. Qg6 g3 4. Qxg8 gxh2 5. Qg3 hxg1=N 6. Rh6 Rg8 7. Qh2 Rxg2 8. Rf6 Rxf2 9. Bg2 Nh3 10. Bf3 Ng5

## Phénix 46, Nov. 1996

2428 - T. Le Gleuher
1. Nf3 Nf6 2. Ne5 Ne4 3. Nxd7 Nxd2 4. Nxb8 Nxb1 5. Bf4! h5 6. e3 h4 7. Ke2 h3 8. Kf3 hxg2 9. h4 g1=N 10. Kg2 Bg4 11. h5 e6 12. h6 Ke7 13. hxg7 Kf6 14. g8=N Kg7 15. Bg5

2429 - T. Le Gleuher
1. Nc3 h5 2. Ne4 h4 3. Ng3 hxg3 4. h4 b5 5. h5 b4 6. h6 b3 7. h7 bxc2 8. b4 g5 9. Bb2 c1=B 10. Bg7 f6 11. Bxf8 Kf7 12. Qb3 Kg6 13. Kd1 gxf2 14. g3 Kf5 15. Bg2 f1=B 16. Bb7 c6 17. Bxc8

2430 - T. Le Gleuher
1. c4 f5 2. c5 f4 3. c6 f3 4. cxb7 c5 5. h4 c4 6. Rh3 c3 7. Rg3 c2 8. Rxg7 cxb1=R 9. Qc2 Rxa1 10. Qg6 hxg6 11. bxa8=R Rxh4 12. Rh7 Rh1 13. Rh8

2431 - T. Le Gleuher
1. f4 a5 2. f5 a4 3. f6 a3 4. fxe7 f6 5. h4 Kf7 6. Rh3 Kg6 7. Rb3 Kh5 8. e3 Kxh4 9. Ke2 Kg3 10. Kd3 Kf2 11. Qf3 Ke1 12. Qc6 dxc6 13. Ke4 Bf5 14. Kxf5 Qxd2 15. Ke6 Nd7 16. Kf7 Rd8 17. exd8=Q Qd1 18. Ke8

2432 - M. Caillaud
1. h4 Na6 2. h5 Nc5 3. h6 Ne4 4. hxg7 h5 5. g4 Rh6 6. g5 Ra6 7. g6 Ngf6 8. g8=N e6 9. Ne7 Kxe7 10. g7 Kd6 11. g8=N Ke5 12. Ne7 Qxe7 13. c4 Qa3 14. c5 d6 15. c6 Bd7 16. cxb7 Rd8 17. b8=N Be8 18. Nd7 Rxd7

2433 - G. Donati
1. Na3 a5 2. Nc4 a4 3. Na5 b5 4. c4 b4 5. Qc2 b3 6. Qg6 hxg6 7. Nf3 Rh3 8. Nh4 Rc3 9. bxc3 g5 10. Ba3 b2 11. Bd6 b1=N 12. Bf4 d6 13. Kd1 Qd7 14. Kc2 Kd8 15. Kd3 Qh3 16. Ke4 g4 17. Rxb1 Bf5 18. Kxf5 Nd7 19. Kg5 Rc8 20. Rb8 Ngf6 21. Ra8 Nh7 22. Kh5 g6 mate.

## Phénix 51, Apr. 1997

2526 - P. Wong
1. g3 b6 2. Bg2 Bb7 3. Bh3 Bh1 4. Bf1 Bd5 5. h3 Bb3 6. ab3 h6 7. Qc2 Rh7 8. Qh7 Qc8 9. Qg6 hg6 10. Kd1 Kf7 11. Kc2 Qe8 12. Kd3 Qd8 13. Ke3

2527 - T. Le Gleuher
1. d4 Nc6 2. Bg5 Na5 3. Bf6 exf6 4. a4 Ba3 5. d5 Ne7 6. d6 O-O 7. dxe7 d6 8. e8=Q Bd7 9. Qe4 Re8 10. Qg6 Kf8 11. e4 Ke7 12. Ke2 Rh8 13. Kf3 Be8 14. Kg4 Kd7 15. Kh5 Qe7 16. g4 Rd8 17. Bh3 Kc8 18. Qf1 hxg6#

Castling, uncastling, and false castling.

2528 - P. Schnoebelen
1. e4 a6 2. Bb5 axb5 3. Nf3 Ra6 4. O-O Rg6 5. Re1 b6 6. Re3 Bb7 7. Rd3 Bc6 8. Rd6 exd6 9. e5 Qh4 10. e6 f6 11. e7 Kf7 12. e8=R Ne7 13. Kf1 Nc8 14. Re1 Be7 15. Ke2 Ke6 16. Rh1 Qc4+ 17. Ke1

Castling and uncastling with promoted Rook.

## Phénix 54, Jul.-Aug. 1997

2611 - M. Velucchi
Black may not castle: his last move was his K or R. This holds even assuming White just captured a Bl. unit. Hence e.g. 1. O-O?? a4 2. Rf1 Ra3 3. Rd1 Rb3 4. Kh8 Rb8# does not work.

Sol: 1. Kf7!! a4 2. Rb8 Ra3 3. Kg8 Rb3 4. Kh8 Rxb8#

2612 - A. Kisliak
Last moves were -1. .. Qc1-a1 -2. g6xRh7 Rh8-h7 -3. g5-g6 h7-h6 -4. h4xBg5 Bh6-g5 -5. h3-h4 Bf8-h6 -6. h2-h3 and now 2 cases:
(1): -6. .. g7xRf6 -7. Rb6-f6 Rg8-h8 -8. Rb1-b6 Rh8-g8 -9. Ka1-a2 Rg8-h8 -10. a2-a3 Qa3-c1 -11. b3xPa4 and forward 1. Qd1#
(2): -6. .. g7xNf6 -7. Ne4-f6 Nb1-e2 -8. Ne2-f4 Rg8-h8 -9. Ka1-a2 Rh8-g8 -10. a2-a3 Qa3-b1 -11. b3xPa4 and there's no mate in 1.
But white could also retract -7. Nd5-f6 and give mate in 1 with Nd5-b4#.

2613 - A. Zolotarev
Mate in 1 is with 1. cd5 Na3#.

Wh. 5 caps are axbxc & fxexdxc. For balance, Bl. had to promote his a, f, g, and h Pawns, so that his 3 caps are bxc, gxf for promotion on f1 and hxg for promotion on g1. The position unlocks by retracting a2xb3 but we must first unpromote on a1.

Retract
-1. Rd3-d5+ d5xRc6
-2. Rg3-d3 e4xBd5
-3. Rg1-g3 e3-e4
-4. g2-g1=R g3-g4
-5. h3xR/Ng2 and Wh. retropat is avoided.
Further unlocking requires:
(1) unpromoting bBd5 on f1,
(2) unpromoting bRc6 on f1,
(3) retracting wPf2xbQe3,
(4) unpromoting bQe3 on a1,
(5) retracting wPa2xNb3.

2614 - C. Poisson
1. Na3 a5 2. Nc4 Ra6 3. Na5 Ra8 [+bPa7] 4. Nc6 Nc6 5. Nh3 Nb8 6. Ng1

2615 - T. Le Gleuher
The missing wBc1 was necessarily captured on h2 as in the following proof game:
1. f4 e5 2. fxe5 Bd6 3. exd6 b5 4. dxc7 Bb7 5. cxd8=R Bd5 6. h4 Bb3 7. axb3 a5 8. Ra3 Ra6 9. Ra1 Rc6 10. Ra3 Rc4 11. bxc4 bxc4 12. d4 cxd3 13. Bf4 dxe2 14. Bh2 g5 15. Rc3 gxh4 16. g4 hxg3 17. Bh3 gxh2 18. Be6 dxe6 19. Qd3 hxg1=Q 20. Qc4 Kd7 21. Qb3 Kc6 22. Qa4 Kd5 23. Qe8 Qc5 24. Qxg8 Qa3 25. bxa3 Ke4 26. Rd4 Kf3 27. Qg6 Rd8 28. Rb4 Rd4 29. Rc5 Rh4 30. Rd4 Rh6 31. Qh5 Ke4 32. Qh3 Rh4 33. Qf1 exf1=Q 34. Rg5 Kf3 35. Rg7 Ke2 36. Rg5 Kd1 37. Rg7 Rh2 38. Rg5 Qf3 39. O-O

2616 - R. J. Millour
The bBf8 was captured at home so that one bB is a bP promoted on g1 who later changed square color using the "Plus Chess" effect. This required at least 5 captures: d5/4-+>e5xf4xg3xh2xg1=B and one for changing Bg1's square-color. With 2 wBs captured at home, this accounts for all 5 bl. caps.

Therefore last move is -1. O-O+ and not -1. Be8xf7+. So that the wPh2 could not visit f7 or d7 on its way to promotion into the wBb3. With just bQ, bQR and 2 bNs as possible captures, the wPh2 had to use the Plus Chess effect: h2xg3xf4xe5-+>d4/5xc5/6. Then it was blocked by the bPc7 so that wPcxb was necessary and then wPbxa too, for promotion on white-squared a8, because a wBb8 could not change color by capturing the bPa anywhere between a3 and a7.

Finally, the wPh2 made h2xg3xf4xe5-+>d4xc5xPb6xPa7-a8=B. It had to capture the bQR on its way and this bQR could only get out through the d file. Because the bPd7 could not reach d4 before wPe5-+>d4xc5, then the bQR had to escape via d6, when the bP was on d5.

Necessarily, the bRa8 visited d6.

## Phénix 58, Dec. 1997

2699 - A. Zolotarev
The last move was f7xe6 (otherwise Bf1 would be a promotion piece, but there are 8 black pawns on the board). Black captures are f7xe6, g7xf6, c7xd6, dxexfxg. So the white a-, b- and c-pawn all promoted on c8, needing 3 captures. Because the black h-pawn never left its column, the white h-pawn must've captured once, and both the g- and the h-pawn promoted on g8. Last moves were: -1. .. f7xQe6 -2. Qc8-e6 f4xNg3 -3. c7-c8=Q e5xNf4 -4. b6xc7. Next, unpromote the knights on g3 and f4 on c8. Now black can uncapture c7xRd6 and d6xRe5. Both rooks unpromote on g8. Finally, black can uncapture g7xRf6. The position then unlocks. Bishop-moves needed are: Bf1-e2-d1, Bc8-h3-f1, and Bc1-b2-f6-h4.

2700 - G. Donati
1. g4 d6 2. Bh3 Qd7 3. Kf1 Qe6 4. Qe1 Qe3 5. fxe3 Kd7 6. Qh4 Kc6 7. Qh6 gxh6 8. Nf3 Bg7 9. Rg1 Bd4 10. Rg3 Nf6 11. Bg2 Rg8 12. Rh3 Rg5 13. Rh5 Re5 14. Rg5 Bf5 15. Rg8 Nbd7 16. Rh8 Rg8 17. Ke1 Rg5 18. Ra8 Rh5 19. g5 Ng8 20. g6 Rg5 21. Bf1 Rg1 22. g7 Rh1 23. Ng1

2701 - V. Liskovets
1. Qd7 2. Kb7 3. Kc6 4. Kd5 5. Ke4 6. Bd5 7. Be6. In this position, only one of the two castlings of white is legal.

• White may castle kingside: 8. Kf3 9. Bg4 OO#
• White may castle queenside: 8. Kd3 9. Bc4 OOO#

2702 - P. Wong
1. Nf3 Nc6 2. Ne5 Nd4 3. Nd7 Kd7 4. h4 Kc6 5. h5 Kb5 6. h6 c6 7. hg7 h5 8. Rh3 h4 9. Rh2 h3 10. Rh1 h2 11. Rg1 Rh3 12. Rh1 Rd3 13. Rg1 Nf3 14. ef3 h1=Q 15. Be2 Qh8 16. Rf1 Bh3 17. Rh1 e6 18. Bf1 Qg5 19. Ke2 Rd8 20. Rg1 Bd6 21. Ke1
But unfortunately this problem is cooked: 15. Qe2 Qh8 16. Qe3 Bh3 17. Kd1 e6 18. Qe1 Qg5 19. Ke2 Rdd8 20. Qd1 Bd6 21. Ke1

2703 - M. Caillaud
1. a4 c5 2. Ra3 c4 3. Rd3 cd3 4. b4 de2 5. b5 ef1=N 6. b6 Ng3 7. ba7 b5 8. c4 Bb7 9. c5 Bd5 10. c6 b4 11. c7 Nc6 12. c8=B Qc7 13. Ba6 Rd8 14. a8=Q Na7 15. Qc6 Nc8 16. Qe6 fe6 17. Bf1 Kf7 18. d3 Kg6 19. Bh6 gh6 20. a5 Bg7 21. a6 Bd4 22. a7 Nf6 23. a8=R Rhe8 24. Ra1 Ba7 25. hg3

2704 - T. le Gleuher
1. a4 e5 2. Ra3 Qg5 3. Rc3 Qe3 4. fe3 Be7 5. Kf2 Bg5 6. Kg3 Bf4 7. Kh4 Bg3 8. Kg5 Bf2 9. Rc7 Bg1 10. c4 Bf2 11. Qc2 Bg3 12. Qh7 Bf4 13. Kh4 Bg5 14. Kg3 Be7 15. Qg8 Bf8 16. Qh7 Bb4 17. Qg6 Bc3 18. Qa6 ba6 19. bc3 Bb7 20. Ba3 Be4 21. Bd6 Bb1 22. Rc5 Be4 23. Bb8 Bf3 24. Ba7 Rc8 25. ef3 Rc6 26. Bd3 Rg6 27. Rf1 Rg4 28. fg4 OO 29. Rf3 Rb8 30. Kf2 Rb2 31. Rh3 Rd2

## Phénix 62, Apr. 1998

2796 - A. Zolotarev
Intention:
(a): The last moves were -1. d3xNe4 Nc3-d4 -2. f5xNg6 Nb1-c3 -3.e4xNf5 b2-b1=N -4. d2-d3 a3xQb2. Further retracting involves: Both uncaptured knights are unpromoted on c1, and a bishop is uncaptured by one of these pawns by b5xBc4. The white c-pawn uncaptures 2 queens on d3 and e4. These queens unpromote on g1. The bishop on c4 goes back to f1, and the position is unlocked with g2xNf3. Rook moves are: Ra8-d8, Rh8-g8-g5, Rh1-h4-g4-g7-h7, so 7 rook moves.
(b): The last moves were -1. d2xQe3 Qd3-e3 -2. Ra1-b1 Qb1-d3 -3. f5xNg6 b2-b1=Q -4. e4xNf5 a3xQb2 and the same retractions as in (a). So in total 7+1=8 rook moves

However, there is an easy cook that gives in both (a) and (b) one rook move less: Last moves were -1. d3xQe4/d2xQe3 Ra8-d8 -2. Ke7-f6 Qb8-e8 and only 6 rook moves are needed in (a) (Ra8-d8, Rh8-g8-g5, Rh1-g1-g7-h7), and one extra in (b) (Ra1-b1).

2797 - G. Donati
1. h4 g5 2. hg5 Bh6 3. Rh6 Kf8 4. Rc6 bc6 5. g4 Ba6 6. Bg2 Bd3 7. Bd5 Be4 8. f4 Bf3 9. ef3 Na6 10. Ke2 Rb8 11. Kd3 Rb3 12. Kc4 Nb4 13. Ne2 cd5 14. Kb5 Nc6 15. Ka6 Ra3 16. Kb7 Nb8 17. Ka8 Ke8

2798 - M. Caillaud
1. Nf3 Nc6 2. Ne5 Nd4 3. Nc6 Ne2 4. Nb8 Ng1 5. Ba6 d5 6. d3 Bh3 7. Bh6 g5 8. f4 g4 9. Qg4 b5 10. Qc8 f5 11. g4 Kf7 12. g5 Qd6 13. Qd8 Bf1 14. Bc8 a5 15. h3 Ra6 16. Rh2 Rc6 17. Re2 Rc2 18. Re6 Rh2 19. Nc3 Rh1 20. Kf2 Bg7 21. Rae1 Bc3 22. Bf8 h5 23. Rh6 b4 24. Rh8 Qh6 25. Re6 b3 26. Ra6 ba2 27. Ra8 a1=R 28. g6 Kf6 29. g7
Unfortunately cooked: 1. g4 h5 2. Bg2 Rh6 3. Bb7 d5 4. g5 Bf5 5. Bc8 Bc2 6. Nf3 Bb1 7. Ne5 Ba2 8. Nd7 Bc4 9. Ra6 Be2 10. Rg6 f5 11. Rg7 Ra6 12. Rh7 Bg7 13. d3 Ra1 14. g6 a5 15. Bh6 Ra6 16. f4 Bc3 17. Rf2 Nc6 18. Bf8 Nd4 19. g7 Rg6 20. Re1 Bf1 21. Re6 Rg1 22. Ra6 Rh1 23. Nb8 Qd6 24. Ra8 Rf7 25. Qa4 Ne2 26. Qd7 Qh6 27. Qd8 Ng1 28. Rh8 Rf6 29. h3

2799 - T. le Gleuher
1. g3 h5 2. Bh3 Rh6 3. Kf1 Re6 4. Kg2 f6 5. Kf3 Re3 6. Kf4 e5 7. Kf5 Qe7 8. Kg6 Qb4 9. Kh7 Bc5 10. Kg8 d6 11. Kg7 Bg4 12. Kg6 Bf3 13. ef3 Re1 14. Ne2 Rg1 15. Kf5 Rg2 16. Re1 a5 17. Ng1 a4 18. Re4 a3 19. Rc4 Ra4 20. Ke4 Na6 21. Kd3 e4 22. Ke2 e3 23. Ke1
But this is cooked, since the position can be reached in 22.0 moves too: 1. g3 e5 2. Bh3 Qh4 3. Be6 f6 4. Nh3 Bc5 5. OO d6 7. Bg8 Bg4 7. Re1 Bf3 8. ef3 Qb4 9. Re4 a5 10. Be6 a4 11. Rg4 a3 12. Rg7 Ra4 13. Rg4 Rg8 14. Rc4 Rg4 15. Kf1 Na6 16. Ke2 Re4 17. Kd3 Re1 18. Bf5 Rg1 19. Ke2 Rg2 20. Ng1 e4 21. Ke1 e3 22. Bh3 h5

2800 - V. Liskovets
Only four rooks could still be volage (they resulted from promotion by the pawns from a2, c2, e2, and g2). The h1-rook must've visited a black square to reach any of the current positions of the rooks. Dependant on which rook is no longer volage, this results in the following variants:
I: Rg4 is no longer volage: 1. Rg1? d1=wB!; 1. Rb4! d1=wN/wB Ra2#
II: Re6 is no longer volage: 1. Rb6 d1=wN/wB 2. Ra2#
III: Re2 is no longer volage: 1. Red2 Kb1 2. Rd1#
IV: Rc8 is no longer volage: 1. Rb8! d1=wN/wB 2. Ra2#
V: Rc2 is no longer volage: 1. Rcd2 Kb1 2. Rd1#

2801 - T. Le Gleuher
The both kings are C and c. There are four possible orientations.
"B" on b2: Then C is the black king, and c is the white king. if A was a black rook, Ac4 and Ag2 are both original rooks, and both are on the same colour. So A isn't a black rook. A can't be a bishop either, since aa4 would deliver an impossible check to the black king. If A is a pawn, then the white pawns on c7 and g7 captured 6 times on a black square. Maximal one of these captures could be a capture on a white square (with an en passant capture), since there are 3 black pawns on white squares present. So there are left for captures: Rh8, Bf8, Qd8, 2 pawns on a black square, and 1 pawn on a white square. But that leaves no black piece for b2. So this is impossible.
"B" on b7: c is the black king, C is the white king. A isn't a pawn (ad8 can't be a pawn), A isn't a bishop (ad8 would give an impossible check to the white king), A isn't a rook (ag2 and ag6 would be both original rooks then), so this orientation is impossible
"B" on g7: C is the black king, c is the white king. A can't be a rook: Af6 would give check to the white king, and for B the only other possibilities are B, Q or P, both giving check too. A can't be a bishop either, ah5 would give an impossible check to the black king. Suppose A is a pawn. The white king went to h6 via g7xf6 Kg5-h6. So the bishop on f8 was captured at home. This could only have been done by a white promotion piece. This requires 4 captures on black squares (or 3 captures on black squares and 1 en passant capture on a white square). There are 4 black pieces on a black square missing, apart from the bishop from f8). What is B? A rook or a bishop would imply a black promotion. This implies at least 4 captures on black squares for black: Ra1, Bc1, the promoted piece, and Ph2. But Ph2 needs 2 captures to reach a square where it could get captured. Together with the 3 captures needed for the white promotion, this is too much. If B is a black queen, the last moves must've been Bh8xXg7 Qf8xBg7, or Rg5xXg7 Qf8xBg7, so a promotion piece is needed after all, but this was impossible. So this orientation isn't possible either.
"B" on g2: C is the white king, c is the black king. A isn't a pawn (a pawn on e1 is impossible). A isn't a rook either, because all three black rooks would be promotion pieces, needing 12 captures, which is too much. So A is a bishop (and the last move was f2xe1=B). Now what is B? It can't be a queen, bishop (illegal check) or a rook (this would require an extra capture), so it's a pawn.

So the coloured and orientated position is

Black had 2 promotions, and white had 3 promotions. Black's pawns didn't capture the knight on g1, so the other 8 missing white pieces were captured by the black pawns. All missing black pieces were captured by the white pawns. The bishop on e1 must've captured c5(e5)xd4xe3(ep)xf2xRe1=B. So the promotion on g8 was the white c-pawn, after capturing c4xd5xRe6xf7xNg8=B. Other promotions were b2-b4xc5xQb6xa7xNb8, b7-b5xa4xRb3xQa2xNb1 and h2-h4xg5xh6(ep)xBg7xRf8. So the black c-pawn didn't promote, and the promotion on e1 was by the black e-pawn.
So the history of Pe7 is e7-e5xd4xe3(ep)xf2xRe1.

## Phénix 65, July 1998

• 1. h4 Nc6 2. h5 Nd4 3. h6 Nb5 4. hg7 h5 5. Nc3 Nh6 6. g8=N Bg7 7. Ne7 Rf8 8. Nc6 dc6 9. Ne4 Qd5 10. c3 Bd7 11. Qb3 OOO 12. Nf6 Bh8
• 1. Nc3 Nc6 2. Nd5 Nd4 3. Ne7 Nb5 4. Nc6 dc6 5. h4 Qd5 6. h5 Bd7 7. h6 OOO 8. hg7 h5 9. c3 Nh6 10. g8=N Bg7 11. Nf6 Rf8 12. Qb3 Bh8

2873 - A. Frolkin
1. h4 f5 2. h5 f4 3. h6 f3 4. hg7 h5 5. g4 h4 6. g5 Rh5 7. g6 Nh6 8. g8=B Bg7 9. Bd5 Bh8 10. Bc6 dc6 11. g7 Qd3 12. g8=B Qh7 13. Bc4 Bf5 14. Bb5 cb5 15. d4 Nc6 16. Bf4 OOO 17. e3 Rg8 18. Bc4 Rg7 19. Bg8

2874 - M. Velucchi
1. e4 f5 2. Bc4 fe4 3. Bg8 d5 4. Bd5 Be6 5. Bc4 Kf7 6. Bf1 Bb3 7. d4 ed3 8. Bf4 Bc2 9. Bc7 Bb1 10. Bb8 Bc2 11. Bf4
However, this is cooked: 1. e4 f5 2. Bc4 d5 3. Bd5 Bd7 4. Bg8 Ba4 5. Bc4 fe4 6. Bf1 Kf7 8. Bf4 Bc2 9. Bc7 Bb1 10. Bb8 Bc2 11. Bf4

2875 - M. Velucchi
1. c4 d5 2. cd5 c5 3. dc6 Bh3 4. cb7 Bg2 5. ca8=B Bf1 6. Bg2 Kd7 7. Bf1

2876 - M. Caillaud
The last move in the diagram position was f2-f4. But g4xf3(ep) isn't allowed, since white has no legal last move in this position. So a pre-plan is needed: 1. Rb3 2. Nb4 3. Rf3 (last move was d2-d4) 4. cd3 ep (last move was Kb3-a4) 5. d2 6. Rb3 7. d1=B (last move now was f2-f4, since g2-g3 would, after retracting h2-h1=B, result in 7 pawns in the h2-h7-c7 triangle, which is impossible) 8. gf3 ep (last move now could've been g2-g3, since there are only 6 pawns in the aforementioned triangle left) 9. Kf5 10. Kg6 11. Kh5 12. g6 g4#.

## Phénix 69, December 1998

2947 - B. Kampmann
Solution:

Last move was c7xb8=Q. But black has no legal last move before that. Removing any queen in the northwest corner gives black, after uncapturing the proper piece, a retro-move, removing the queen on f1 gives black Kh1-h2 or Kh3-h2 as retromove.

2948 - T. le Gleuher
1. Kd7 2. Kc8 3. Kb7 4. Ka6 5. Ka5 6. Kb4 7. Kc3 8. Kd4 9. Kc5 10. Kd4 11. c5 12. c4 13. c3 14. c2 15. c1=B 16. Bd2 17. Be1 18. Bf2 19. Bg1 20. Rf1 21. Rd1 22. Rd3 23. Ke4 24. Rd4 Now white's last move was f2-f4. 25. gf3 26. Bf2 27. Be1 28. Bb4 29. Bd6 30. Be5 31. Bf4 32. e5 gf3#

2949 - T. Petrovic
Suppose white may castle. White's captures are axbxc, bxcxd, cxd and h2xg3. The only two possible last moves are f2-f3 and e2-e4. If the last move was f2-f4, then the bishop on g1 is a promoted one, and the promotion occurred on g1. But this would require 7 captures by black, one too many. So if white may castle, his last move was e2-e4.
1. fe3 d8=N 2. Kd5 OOO!# (Rd1?)

2950 - S. Hashimoto
1. h4 g5 2. hg5 e5 3. Rh7 Qf6 4. Rf7 Rh3 5. g4 Re3 6. de3 Bb4 7. Bd2 d6 8. Bc3 Be6 9. Qd4 Ba2 10. Nd2 Bd5 11. Ra7 Bf3 12. ef3 Ne7 13. Bd3 Kd7 14. Kf1 Kc6 15. Kg2 Nd7 16. Kg3 Rh8 17. Ra8 Rh1 18. Nh3 Ra1 19. Rc8 Ra8

2951 - A. Zolotarev
The position unlocks when white can get a piece on b8. Retroplay is: -1. Ra1xBa2 b5-b4 -2. h6-h7 Rb4-b2 -3. Nb2-d3 Rh4xPb4 -4. b3-b4 Rh1-h4 -5. h5-h6 h2-h1=R -6. h4-h5 g3xNh2 -7. Nf3-h2 g4-g3 -8. Ne4-f3 g5-g4 -9. Nc6-e4 g6-g5 -10. Nb8-c6 b6-b5 -11. Ka8-a7 a7-a6 -12. Nc6-b8 Qa6-c8 and the white bishop on e8 gets freed. It goes back to f1, and white releases the whole position by retracting e2-e3.

2952 - M. Caillaud
1. e3 b5 2. Bc4 b4 3. d3 b3 4. Nd2 ba2 5. b4 g6 6. Bb2 Bg7 7. Qc1 Be5 8. Be5 d6 9. Qa3 Nd7 10. OOO a1=B 11. Be6 Bc3 12. Nc4 Be1 13. f3 Bg3 14. h4 Bf4 15. h5 Bh6 16. hg6 Bf8 17. g7 d5 18. gf8=N dc4

2953 - M. Kozulya
1. a4 d5 2. a5 d4 3. a6 d3 4. ab7 de2 5. Ra6 ed1=B 6. Rc6 a5 7. h4 a4 8. h5 a3 9. h6 a2 10. hg7 Ra3 11. Rhh6 Na6 12. Rhf6 h5 13. b8=B h4 14. Ba7 h3 15. Be3 h2 16. Bg5 Rhh3 17. f4 Nh6 18. g8=B Bdg4 19. Bh7 Bgd7 20. Bhd3 e6 21. Bb5 Bb4 22. c4 Bc3 23. dc3 Ke7 24. Nd2 Qh8 25. Ndf3 Be8 26. Nh4 Bcd7
1. h4 d5 2. h5 d4 3. h6 d3 4. hg7 de2 5. Rh6 ed1=B 6. Rf6 h5 7. a4 h4 8. a5 h3 9. a6 h2 10. ab7 Rh3 11. Raa6 Nh6 12. Rac6 a5 13. g8=B a4 14. Bh7 a3 15. Bhd3 a2 16. Bb5 Raa3 17. c4 Na6 18. b8=B Bdg4 19. Ba7 Bgd7 20. Be3 e6 21. Bg5 Bb4 22. f4 Bc3 23. dc3 Ke7 24. Nd2 Qh8 25. Ndf3 Be8 26. Nh4 Bcd7

2954 - A. Frolkin & A. Vasilenko
1. d4 e5 2. Nd2 e4 3. Ndf3 ef3 4. Kd2 fe2 5. Kc3 e1=R 6. Be2 Qf6 7. Bh5 Ne7 8. Kb4 Qf3 9. d5 Ng6 10. d6 Re7 11. dc7 f5 12. cb8=B f4 13. Be5 d6 14. Qd5 Rf7 15. Bd2 Nh4 16. Qc6

2955 - P. van den Heuvel
1. d3 Nc6 2. Kd2 Ne5 3. Ke3 c6 4. Kf4 Qb6 5. Kg5 Qb3 6. ab3 Nf6 7. Ra4 Rg8 8. Rf4 Ne4 9. Kh5 h6 10. Rf6 gf6 11. d4 Rg3 12. hg3 Bg7 13. Rh4 Kf8 14. Rg4 Kg8 15. Kh4 h5 16. Kh3 h4 17. Kh2 h3 18. Kh1 h2 19. d5

## Phénix 73, Apr. 1999

3044 - Paul Raican
Add the white kin on d5, the black king on b4, and the 6 pawns on a4, a6, a7, c3, d2, and d6. There are 13 captures by the white pawns. Retracting b2xc3 would result in two extra captures, so the only legal last move is c2-c3. But black has no move before that. Removing Nc8 allows b2xc3.

3045 - Nikita Plaksin & Alexander Zolotarev
White captures are d2xe3xf4xg5, h2xg3, and b3xBc4. Black's capture is cxRb or cxRd. The black h-, c- and b-pawn promoted. To avoid retro-stalemate, the last moves must've been: -1. f4xRg5 Rb5-g5 -2. e3xXf4 Rb1-b5 -3. d2xXe3 b2-b1=R -4. f2-f3 c3xRb2. One of the uncaptured pieces was an original rook, the other must be the promoted b-pawn. The only piece that can go from b1 to e3 or f4 is a rook. So the white d-pawn uncaptured three rooks, of which two promoted.

3046 - Michel Caillaud
1. a4 b5 2. ab5 Bb7 3. Ra6 Bf3 4. Rb6 Bh5 5. g4 Nc6 6. Bg2 Ne5 7. Bb7 a5 8. Ba6 c6 9. Rb7 Qb6 10. h4 Qe3 11. de3 000 12. Rc7 Kb8 13. Qd6 Ka8 14. Qf6 gf6 15. e4 Bh6 16. Be3 Bf4 17. Ba7 Nh6 18. Bb8 Rdg8 19. Rc8 Rg5 20. Rg8 Rf5 21. gf5 Bh2 22. Rg3 Rg8 23. Ra3 Rg2 24. Ra1 Neg4 25. Bf4 Ng8 25. Bc1
The white rook makes a long trip to allow the black castling, and the white bishop makes a circuit to screen the black king.

3047 - Mikhail Kozulya
1. Nc3 f5 2. Ne4 f4 3. Ng3 fg3 4. f4 gh2 5. Kf2 hg1=Q 6. Kg3 Qd4 7. Kh3 Qd3 8. cd3 h5 9. Qb3 h4 10. Qb6 cb6 11. b3 Qc7 12. Bb2 Qc5 13. Rc1 Qg1 14. Rc6 dc6 15. g4 hg3

3048 - Alexander Zolotarev
The black bishop in the southwest corner escapes after retracting c3-c4, but before this move can be retracted, the black king must leave via b4 first, so the white king has to leave a5.
(a): Last moves were: -1. .. g4xNh3 -2. Nf2-h3 g5-g4 -3. Nd1-f2 Nd3-b2 -4. Nb2-d1 Ne1-d3 -5. h5-h6 f2xNe1=N (uncapturing a queen and unpromoting it on g8 fails because this queen needs to many moves to reach g8) -6. Nd3-e1 f3-f2 -7. Nb4-d3 Nc3-a2 -8. Na2-b4 Ne4-c3 -9. h4-h5 Nc5-b4 -10. Qb5-a6 Na6-c5 -11. Qd5-b5 f4-f3 -12. Qg8-d5 f5-f4 -13. g7-g8=Q f6-f5 -14. g6-g7 f7-f6 -15. f5xBg6 Bh5-g6 -16. f4-f5 Bg4-h5 -17. f(2/3)-f4 Bc8-g4 -18. Kb5-a5 d7xc6 and everything unlocks.
(b): Now the last moves were: -1. .. c7xNb6 -2. Nd5-b6 g4-g3 -3. Nc3-d5 Nb4-a2 -4. Na2-c3 Nd3-b4 -5. h5-h6 Ne1-d3 -6. h4-h5 f2xNe1=N -7. Nd3-e1 f3-f2 -8. Nf2-d3 f4-f3 -9. Nd1-f2 Nd3-b2 -10. Nb2-d1 Nc5-d3 -11. Qb5-a6 Na6-c5 -12. Qd5-b5 and further as in (a).

A minor change in the position changes the role of the black knights, and thus their last move too.

3049 - Andrei Frolkin
1. h4 c5 2. h5 c4 3. h6 c3 4. hg7 cb2 5. c4 ba1=B 6. c5 Bf6 7. c6 Bg5 8. c7 Nf6 9. g8=N b5 10. Nh6 b4 11. Ng4 b3 12. Nh2 ba2 13. g4 a1=N 14. Bg2 Nb3 15. Be4 Nc5 16. Qb3 a5 17. Kd1 a4 18. Kc2 a3 19. Kc3 a2 20. Kd4 a1=Q 21. Nc3 Qa6 22. Bc2 Qe6 23. Bd1 Ra1 24. Ba3 Rc1

## Phénix 76, July-Aug. 1999

3146 - Paul Raican
The author's intention was: 1. d4 Nc6 2. Qd3 Nd4 3. Qh7 Ne2 4. Qg7 Rh2 5. Bf4 Rg2 6. Bc7 Qc7 7. Bg2 Qc2 8. Bb7 Qb2 9. Bc8 Qa2 10. Bd7 Kd7 11. Qf7 Qb1 12. Ke2 Qg1 13. Qf8 Qa1 14. Ra1 Rf8 15. Ra7 Kd8 16. Re7 Rf2 17. Kf2 Ne7
But Vlaicu Crisan found the following cook: 1. e4 a6 2. Ba6 d5 3. Bb7 de4 4. Be4 Ra2 5. Bh7 Rb2 6. Ra7 Rh7 7. Rc7 Rh2 8. Rc8 Rg2 9. Rb8 Rg1 10. Rg1 Rb1 11. Rg7 Rc1 12. Rf7 Rc2 13. Qe2 Rd2 14. Qe7 Be7 15. Rd8 Kd8 16. Re7 Rf2 17. Kf2 Ne7

3147 - Vilimantas Satkus
1. d4 h6 2. Bh6 d6 3. Bg2 Rh2 4. Bf8 Rh1 5. Be7 Rg1 6. Bd6 Rg2 7. Bc7 Qd4 8. Bg2 Qb2 9. Bb7 Qa2 10. Bc8 Qb1 11. Ra7 Ra7 12. Qb1 Rc7 13. Qb8 Rc8

3148 - Nikita Plaksin & Alexander Zolotarev
Black captures are cxd, dxexfxg3 and g7xf6. So the white captures are a6xb7-b8, bxc-c8, and hxg. The captured piece on g3 can't be the missing g-pawn, because the cage would never unlock. So a promotion occurred on g8. The position unlocks after retracting g7xf6. For this, a white piece must be unpromoted on g8, and the black bishop should be brought back to f8. The piece that must be unpromoted on g8 can't be a knight or a bishop (they can't reach a square where they could get captured), or a queen (because of the stipulation). So it must be a rook. Last moves were: -1. Ne2-f4 f4xNg3 -2. e4-e5 e5xNf4 -3. d5-d6 d6xNe5 -4. Nc6-e5 a5-a4 -5. Nb8-c6 a6-a5 -6. b7-b8=N a7-a6 -7. a6xNb7. The two knights on f4 and g3 unpromote now on c8, one of the pawns uncaptures a black-squared bishop. Black uncaptures c5xRd4, and this rook is brought to g8, where it unpromotes. The bishop goes back to f8, and black uncaptures g7xNf6, after which the position unlocks. Only 1 queen move is needed: Qd8-d7.
Four Frolkin promotions, to NNNR.

3149 - Michel Caillaud
White captures are f4xe5 and exf. Black captures are a3xb2, bxc, cxd and dxe. The black h-pawn was captured on its own file. To open the cage, b2-b3 must be retracted. But before this happens, a black piece must unpromote and take back a3xb2. The only piece which can unpromote on b1 is the black queen. So a screening piece on the b1-e1 line is needed. The only piece that can do the screening is the bishop on g1. It must leave the southeast corner via g3 and h4. So a screening piece on f2 is needed. The only piece which can do this is the white rook on g4. But this rook must go to f2 via g3, thus a screening piece on f3 is needed, and only the black queen can do this.
Schematic retroplay looks like: wRg4-h5; bPg3-g7; bQb2-f3; wRh5-h1; bBg1-h2; wRh1-f2; wPh3-h4; bBh2-c1; bQf3-b1; a3xb2-b1=Q, and b3-b2.

3150 - Andrei Frolkin
White captures are axbxc and bxc. Black captures are axb, gxf and fxg. To release the position, a white bishop must be brought to a2, so the white queen can escape via a1. The white king is then brought to a1, so the white bishop can screen the king, allowing the rook on d2 to escape. Last moves were: -1. Na4-b2 g7xPf6 -2. f5-f6 g4-g3 -3. f4-f5 f5xBg4 -4. f3-f4 f6-f5 -5. Be6-g4 d6-d5 -6. Ba2-e6 f7-f6 -7. Qa1-b1 h7-h6 -8. Qf6-a1 e5-e4 -9. Kb1-c1 e6-e5 -10. Ka1-b1 d7-d6 -11. Bb1-a2 Rd1-d2. White has no time to bring the queen to b1, to be the screen. So b1 was necessarily visited by the white king, white queen, and white bishop (and the white knight who started on that square of course).
However, as Henrik Juel pointed out, white does have time to bring the queen to b1 after all: -9. Be6-a2 Ka2-a3 -10. Bh3-e6 and now bK spends time between a2 and a3 while wQ retracts to the first row, then (with bK on a3) -15. Kb1-c1 e6-e5 -16. Ka1-b1 d7-d6 -17. Qb1-f1 Rd1-d2. So the answer to the question should ke 'King, queen and knight'.

3151 - Andrei Frolkin
The position unlocks when black can retract h6xBg5. The last moves were: -1. .. c6xNd5 -2. Nc3-d5 c7-c6 -3. Nb5-c3 a6-a5 -4. Nd6-b5 a7-a6 -5. Nf5-d6 Nf3-h4 -6. Nh4-f5 Nd4-f3 -7. Nf5xRh4 Rh3-h4 -8. Rh4-h5 Rh5-h6 -9. N~ h6xBg5. This also shows that white couldn't have made the last move: the black bishop on white squares was captured on a4, so white's last move can't have been a3-a4. Black's b-pawn promoted, so white's last move can't have been b3xBa4. f3 is needed by the black knight so f3-f4 can't have been the last move.
So mate in 1 with 1. Rh1# (and not with 1. .. Qg3?)

3152 - Michel Caillaud, Thomas Volet & Pascal Wassong
Black captured dxe, and promoted his a- and h-pawn without captures. White captured axb, bxc, exf, fxe and h5xg6. To release the position, the white knight must be brought to b3, so black can retract Qc4-d3. But the knight must perform as screen for the white king too. Last moves were: 1. Nf7-h6 Bg8-h7 2. Nd8-f7 Bh7-g8 3. Ne6-d8 Bg8-h7 4. Nc7-e6 Bh7-g8 5. Nd5-c7 Bg8-h7 6. Nb6-d5 Bh7-g8 7. Nc4-b6 Bg8-h7 8. Na5-c4 Bh7-g8 9. Nb3-a5 Dc4-d3 10. d5-d6 Nd3-e1 11. Rh1-d1, with a nice stairway by the white knight.

## Phénix 80, Dec. 1999

3251 - Michel Caillaud
1. Nf3 g6 2. Ne5 Bg7 3. Nd7 Bc3 4. dc3 h6 5. Bh6 a6 6. Bf8 Rh2 7. Kd2 Rh5 8. Ke3 Re5 9. Kf4 Re2 10. Kg5 Rf2 11. Ba6 Nf6 12. Bb7 Ra2 13. Rh8 Rb2 14. Ra8 Na6 15. Nb8 Bh3 16. Nd2 Ng4 17. Ne4 f5 18. Nf6 Kf7 19. Ng8 Ke6 20. Kg6 Ke5 21. Kf7 Qd5 22. Ke8 Qa2 23. Qd8 Rd2 24. Bc8

3252 - Alexander Zolotarev
The last moves were: -1. Qb1xNb2+ e7xNd6 -2. Ne4-d6 g5-g4 -3. Nc3-e4 Qb3-a2 -4. Na2-c3 Qe3-b3 -5. b4-b5 Qe1-e3 -6. c3-c4 e2-e1=Q -7. f6-f7 e3-e2 -8. f5-f6 e4-e3 -9. f4-f5 f5xNe4 -10. Nf2-e4 f6-f5 -11. Nd3-f2 Na4-b2 -12. Nb2-d3 Nc5-a4 -13. Bf1-a6 Na6-c5 -14. d3-d4 f7-f6 -15. e2xBd3 Bf5-d3 -16. f3-f4 Bc8-f5 -17. f2-f3 d7xRc6, and the whole position unlocks.

3253 - A. Kislyak
The last moves were: -1. .. Qh4xNh5 -2. Nf4-h5 Rh5-h7 -3. Nd5-f4 Rh7-h5, and now there is not enough time for -4. Ne3-d5? Rh5-h7 -5. Nf5-e3 b4-b3 -6. Nh6-f5 b5-b4 -7. Kh7-g6 b6-b5 -8. Ng6-f8??, so we must continue -4. Nb6-d5 Rh5-h7 -5. Nc8-b6 Rh7-h5 -6. c7-c8=N Rh5-h7 -7. c6-c7 Rh7-h5 -8. c5-c6 Rh5-h7 -9. c4-c5 Rh7-h5 -10. c3-c4! Rh5-h7 -11. c2-c3 c4xNb3 -12. Nd4-b3 Rh7-h5 -13. Nf5-d4 Rh5-h7 -14. Nh6-f5 c5-c4 -15. Kh7-g6 c6-c5 -16. Ng6-f8 c7-c6 -17. Rf8-f7 Bf7-e8, and the position is released fairly easily: White captured hxNg and axPb and promoted on b8, while Black captured hxQg and promoted on a1.

3254 - Michel Caillaud
1. d4 d5 2. Bh6 gh6 3. Kd2 Bg7 4. Ke3 Bd4 5. Kf3 Be3 6. Nc3 d4 7. Na4 d3 9. Qb3 d1=Q 10. Rc1 Qd6 11. Rc2 Qg3 12. fg3 Bd2 13. Kf2 Bg4 14. Qe6 Kf8 15. Qc8
1. Nc3 d5 2. Na4 d4 3. c3 d3 4. Qc2 dc2 5. d4 Bg4 6. Bh6 gh6 7. d5 Bg7 8. d6 Kf8 9. d7 Qe8 10. d8=Q Be5 11. Qc8 Bg3 12. fg3 c1=B 13. Kf2 Bd2 14. Rc1 Qd8 15. Rc2

3255 - Pascal Wassong
1. f3 b6 2. Kf2 Ba6 3. Qe1 Bd3 4. ed3 Qc8 5. Qe6 de6 6. Ke2 Qd7 7. Ke1 Qd8

3256 - Jean-Marc Lousteau
1. e4 Nf6 2. Bc4 Nd5 3. Ne2 Nb6 4. Bd5 Nc6 5. c4 Rb8 6. Qb3 Na8 7. Qb6 Nd4 8. b4 Nxe2 9. Bb2 Nf4 10. Be5 Ng6 11. Bg3 Rg8 12. e5 Nh8

3256b - Jean-Marc Lousteau
13. Ke2 Ng6 14. Rc1 Nh4 15. Rc3 Nf5 16. Rf3 Nh6 17. Rf6 Rh8 18. Rg6 Ng8 19. Qf6 Nb6 20. c5 Nc4 21. a4 Na5 22. Bc4 Nc6 23. a5 Ra8 24. Ra4 Nb8

3256c - Jean-Marc Lousteau
25. Be6 Nc6 26. Bh3 Rb8 27. e6 Ne5 28. Rg4 Ng6 29. Qa1 Nf6 30. Ra3 Nd5 31. Rd3 Nb6 32. Rd6 Na8 33. Rb6 Rg8 34. Bd6 Nh8

3256d - Jean-Marc Lousteau
35. Rf4 Ng6 36. Rf6 Ne5 37. Bf5 Ng4 38. h3 Nh6 39. g4 Rh8 40. Bg3 Ng8 41. Rd6 Nb6 42. Rd3 Nc4 43. Rc3 Ne5 44. Rc1 Nc6 45. Qe5 Ra8 46. Nc3 Nb8

## Phénix 84, Apr. 2000

3345 - Nicolai Zinovyev
Not 1. Ra7?, because black has no last move. So mate in 7 with 0. .. Ka5 1. Ka7 Ka4 2. Ka6 Ka3 3. Ka5 Ka2 4. Ka4 Ka1 5. Kb3 Kb1 6. Rc7 Ka1 7. Rc1#

3346 - Thierry Le Gleuher
1. h5 2. h4 3. h3 4. h2 5. h1=R 6. Rh8 7. OO 8. Ra8 9. Kf8 10. Ke8 11. OOO 12. Kc7 13. Ka6 14. Ka5 15. Ka4 16. a4 Nc3#

3347 - Alexander Zolotarev
There are at least three queen-moves necessary (Qd1-h5, Qa8-a5, and Qd8-e8). It's possible to release the position without any further queen moves: -1. .. Nd7-f5 -2. e3xNd4 Ne2-d4 -3. b6xBa7 Ng1-e2 -4. c5xNb6 g2-g1=N -5. d4xNc5 h3xRg2. Further retractions are (Black moves between parentheses): (f7-f1=N-c5); (g5xBf4-f1=N-c6); Ba3-f4; (e2-e1=B-a7); f2xRe3; (b4-b1=R-e3); (e7-e2); e3xRd4; (b3-b1=R-d4); Bc1-a3; b2xBc3; g3-g4; Qd1-h5; e2-e3; and the whole position unlocks. Note that white can't retract g3-g4 too early, since both black rooks had to go over g3. Before the black king can escape white needs to retract e2-e3, so black must uncapture a white rook on g2, otherwise this rook can't get back behind the pawn barrier.
Sixfold Ceriani-Frolkin promotion (RRBNNN).

3349 - Pascal Wassong
The last moves were for instance: -1. Rb5-c5 a7-a6 -2. Ba3-d6 b7-b6 -3. Bc1-a3 c7-c6 -4. b2xBc3 Be5-c3 -5. Rc5-b5 Bf4-e5 -6. Rc4-c5 f6-f5 -7. Re4-c4 Bd6-f4 -8. Re5-e4 Bf8-e6 -9. Qh1-g1 e7-e6 -10. e6xNf7 Ng5-f7 -11. Qh6-h1 -12. Nf3-g5 -13. Rh5-e5 Ng1-f3 -14. Rh1-h5 h2xNg1 -15. Ne2-g1 h3-h2 -16. h2xRg3. Now the rook goes back to the 8th line, and black can retract d6-d5, allowing the white pawn on f6 to uncapture all missing black officers.
So the missing white knight was captured on g1.

3350 - Gianni Donati & Michel Caillaud
1. h4 e5 2. Rh3 Qf6 3. Rc3 Kd8 4. Rc6 bc6 5. Nf3 Ba6 6. Nd4 Be2 7. f3 Bd1 8. Bc4 Bc5 9. Bd5 cd5 10. Kf1 Qa6 11. Kf2 c6 12. Ke1 Kc7 13. Ne2 Kb6 14. Ng1 Ka5

3351 - Satoshi Hashimoto
1. c4 Na6 2. c5 Rb8 3. c6 bc6 4. b4 Rb5 5. Ba3 Re5 6. b5 Re3 7. de3 Nh6 8. Qd6 ed6 9. b6 Qg5 10. b7 Be7 11. b8=Q OO 12. Qb2 Bb7 13. Kd2 Ra8 14. Kc3 Nb8 15. Kb4 Kf8 16. Qg7 Ke8 17. Nc3 Bf8 18. Rb1 Qd8 19. Ka5 Bc8 20. Rb7 Ng8

Six black pieces leave their homesquare and return to it again to let the black rook from h8 go to a8!

3352 - Gianni Donati
1. a4 e5 2. Ra3 Qe7 3. Rg3 Qa3 4. Nh3 Qa1 5. Na3 Qc1 6. Nc4 Qa1 7. Qb1 Qa3 8. Qa2 Qf3 9. gf3 Ba3 10. Bg2 d6 11. OO Bg4 12. Ra1 Nd7 13. Kf1 OOO 14. Ke1 Re8 15. Bf1 Re6 16. Rg1 Rh6 17. Rh1 Ndf6 18. Ng1 Rh3 19. Qb1 Bh5 20. Qd1 Ng4

Four white pieces leave their homesquare, and go back, and the two white rooks swap places, just to let the black queen capture the bishop on c1!

3353c - Michel Caillaud
1. g3 c6 2. Bg2 Qb6 3. Bd5 cd5 4. f4 Nc6 5. Nf3 Nd8 6. Rf1 Qg1 7. h4 b6 8. h5 Ba6 9. h6 Nh6 10. f5 Nf5 11. a4 h5 12. a5 Rh6 13. Ra4 Rc6 14. Rh4 e6 15. Rh1 Bb4 16. Nh2 Bc3 17. dc3 Nd6 18. Kd2 N6b7 19. Rf4 d6 20. Ra4 Bc4 21. Ra1 Ba2 22. Qf1 Rc4 23. Ke1

## Phénix 87, July/Aug. 2000

3438 - E. Minerva
Retract -1. Ke6xNf7 Bc5xPe7, and h#1 with 1. Nd8 ed8=N#.

3439 - T. le Gleuher
1. e4 d5 2. ed5 Nc6 3. dc6 Qd5 4. cb7 Qa2 5. ba8=R Qb1 6. R8a7 Qa1 7. Ra1

3440 - T. le Gleuher
1. e4 d5 2. ed5 Sc6 3. dc6 Qd5 4. cb7 Qg2 5. ba8=B Qg1 7. Bag2 Qf1 8. Bf1

3441 - A. Frolkin & A. Kornilov
The last moves were: -1. Qg8xQg6 Bb7-c8 -2. c5-c6 Be4-b7 -3. c4-c5 Bb1-e4 -4. c3-c4 a2xNb1=B -5. Na3-b1 b6-b5 -6. Nc4/b5-a3 a3-a2 -7. Nd6-b5/c4 a4-a3 -8. Ne8-d6 a5-a4 -9. Ng7-e8 a6-a5 -10. Ne8xRg7 Rh7-g7 -11. Nd6-e8 Rg7-h7 -12. Nb5-d6 Rh7-g7 -13. Na7-b5 Rg7-h7 -14. Nc6-a7 a7-a6 -15. Nb8-c6 Rh7-g7 -16. b7-b8=N Rg7-h7 -17. a6xNb7 Rh7-g7 -18. Qh8-g8 Nd6-b7 -19. Qg8-h8 Ne8-d6 -20. Qh8-g8 Ng7-e8 -21. Kg8xNf8 Re8-e7 -22. c2-c3 Be7-d8 -23. a5-a6 Bb4-e7 -24. a4-a5 e7-e6 -25. Rc6-f6 and the whole position unlocks.
But Henrik Juel found the following cook: -1. Qg8xRg6 Bb7 -2. c5 Be4 -3. c4 Bb1 -4. c3 a2xNb1=B -5. Na3 b6 -6. Nc4 a3 -7. Ne5 Rg7 -8. Ng6 Rh7 -9. Qh8 a4 -10. Kg8xNf8 Re8 -11. Nf4xNg6 Be7 -12. Nh3 Ba3 -13. Ng1 e7 -14. Re6 Nf4 -15. Re3 g6xQf5 etc.

3442 - D. Innocenti
1. a3 e6 2. Ra2 Ba3 [+wPa2] 3. Na3 [+bBf8] Ba3 [+wNg1] 4. b4 Ne7 5. Bb2 Nec6 (OO is forbidden because square f8 is controlled) 6. Qc1 Bb2 [+wBc1] 7. Bb2 [+bBf8] Bb4 [+wPb2] 8. e4 OO 9. Bb5 Kh8 10. Ne2 Re8 11. OO Bf8 12. Bc6 [+bNg8] Nc6 [+wBf1] 13. e5 Ne5 [+wPe2]
However, this seems to be cooked: 1. a3 e6 2. Ra2 Ba3[+wPa2] 3. b4 Ne7 4. Bb2 Nec6 5. Qc1 Bb2 [+wBc1] 6. Na3 Ba3 [+wNg1] 7. Bb2 Bb4 [+wPb2] and further as in the author's intention.

3443 - S. Tkachenko & A. Frolkin
The last moves were: -1. .. a2xSb1=N -2. Ka4-a3 a3-a2 -3. Kb5-a4 a4-a3 -4. Kc5-b5 b5xNa4 -5. Kd4xPc5 c6-c5 -6. Nc5-a4 and now c7-c6 -7. a4-a5 b6-b5 -8. a2-a4 b7-b6 -9. Ra3-b3 Rb3-c3, but the pawn moves by the b- and c-pawns can be interchanged.

3444 - M. Caillaud
1. c4 Nc6 2. c5 Nd4 3. c6 dc6 4. e3 Bf5 5. Bc4 Bc2 6. d3 Ba4 7. Bd2 Bb5 8. a4 Qd7 9. ab5 Rd8 10. b6 Qc8 11. ba7 Rd5 12. a8=Q Qd8 13. Qc8 Ra5 14. Qe6 Ra8 15. Ba5 fe6

3445 - M. Kirtley
1. c3 f5 2. Qb3 f4 3. Qe5 f3 4. ef3 a5 5. Ba6 e5 6. d3 e4 7. Bg5 e3 8. f4 e2 9. Kd2 e1=R 10. Nf3 Re6 11. Rc1 Kf7 12. Rc2 Kg6 13. Kc1 Re7 14. Be7