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Europe Echecs

Solutions

424/16 - E. Angelini

Bl. needs 8 moves to promote his bPa7 into a bNa7. This requires capturing the wKN on b3, the wPc2 and the wQd1. Then, the actual wNg1 really is from b1 !!! Here is a proof game:

1. Nf3 a5 2. Nd4 a4 3. Nb3 axNb3 4. Nc3 bxPc2 5. Nd5 cxQd1=N 6. Nf4 Nc3 7. Nh3 Nb5 8. Ng1 Na7


427/16 - E. Angelini

Author's intention is "wNb1 captured on a7 by bRa8". As in the following proof game: 1. f4 h6 2. f5 Rh7 3. f6 exf6 4. d4 c6 5. d5 Na6 6. d6 Bxd6 7. g4 Bc7 8. h4 d6 9. h5 Bxg4 10. e4 Bxh5 11. e5 Ne7 12. e6 fxe6 13. c4 Bf7 14. c5 Nxc5 15. a4 g6 16. a5 Nf5 17. a6 bxa6 18. b4 Ng7 19. b5 Ke7 20. b6 Qd7 21. Na3 Nb7 22. Nb5 axb6 23. Na7 Rxa7

Unfortunately, it is possible to capture the wN on e7: 1. f4 Na6 2. f5 c6 3. f6 exf6 4. d4 h6 5. d5 Rh7 6. d6 Bxd6 7. c4 Bc7 8. c5 Nxc5 9. g4 d6 10. h4 Bxg4 11. h5 Bxh5 12. e4 Qd7 13. e5 Ne7 14. e6 fxe6 15. a4 Bf7 16. a5 g6 17. a6 Nf5 18. b4 bxa6 19. b5 Ng7 20. b6 axb6 21. Nc3 Nb7 22. Nd5 Ra7 23. Ne7 Kxe7 (found by C. Wermelinger).


429/16 - E. Angelini

The wh. royalties swapped place, requiring tight maneuvers during which the bl. units had to get captured somehow. Precisely, the bl. Pg7 has been captured by 5. Kxg2, while the bl. Rh8 has been captured by 12. Nxg1. Here is a proof game:
1. g4 h5 2. Nf3 hxg4 3. Bh3 g3 4. Kf1 g2 5. Kxg2 Rh6 6. Qg1 Rg6 7. Kf1 Na6 8. Ke1 Nb8 9. Kd1 Na6 10. Qe1 Nb8 11. Bf1 Rg1 12. Nxg1.

Observe that the following try: 1. g4 h5 2. Bh3 Rh6 3. Kf1 Rf6 4. Kg2 Rf3 5. Nxf3 hxg4 6. Qg1 g3 7. Kf1 g2 8. Ke1 Na6 9. Kd1 Nb8 10. Qe1 g1=Q 11. Bf1 ?? 12. Nxg1 fails because Bl. lacks a waiting 11th move.


431/16 - E. Angelini

Wh. Pawns have been captured at home, Bl. Pawns have been captured on their file, respectively on b7, d4, e4, f3, g6. Here is one proof game:

1. Nf3 e5 2. Nc3 Qg5 3. Rb1 Qxg2 4. Bxg2 d5 5. Kf1 d4 6. Nxd4 f5 7. Bxb7 f4 8. Kg2 f3 9. Kxf3 Ba3 10. Qg1 g6 11. Qxg6 Ke7 12. Qd3 Nf6 13. Rg1 Nh5 14. Rg3 Bh3 15. Rxh3 Na6 16. Ke3 Rab8 17. Bf3 Rxb2 18. Bxb2 e4 19. Nxe4 Ng3 20. Nxg3 Rb8 21. Bc3 Rb3 22. Rxb3 Nc5 23. Rxa3 Nb3 24. Nxb3.


433/16 - Eric Angelini
Europe Echecs 433, Apr. 1995

White did just move his King to e5, but from which square? Whatever square we try, the King was in a seemingly impossible double check.

As a matter of fact, one such double check is possible. If we assume White did just play n. Kf5-e5, then the double check on f5 could have been given by an en-passant capture n-1 ... f4xg3 e.p.

But then the capture was only legal if White did play n-1. g2-g4. This move answered a check given by the bBh3, again seemingly impossible. The trick is that this must have been a discover check, given by some bNg4 that has been captured by the wK.

Last moves are:
n-2 ... Ng4-(/x)e5 +
n-1. g2-g4 f4xg3 ep +
n. Kf5xNe5.

In retro-parlance, this cute problem is a "Last move (Type B)" problem with 3 precise last single moves. The 4th retro-move is not precise because it could have been any capture Ng4xe5.


439/16 - B. Rothmann

Author's intention was 1. b4 c5 2. bxc5 Qb6 3. cxb6 g5 4. bxa7 Bg7 5. axb8=B Bxa1 6. Be5 Bb2 7. Bxh8 Bxc1 8. Bb2 e5 9. f4 exf4 10. g4 fxg3 11. Bxc1 gxh2 12. Kf2 hxg1=B 13. Kg3 Bb6

However, Peter van den Heuvel posted the following demolition to the retros list:

"This problem has some duals in the solution, at least in the solution that I have found. Here is just one proof game:
1. b4 c5 2. f4 e5 3. bxc5 Qb6 4. cxb6 exf4 5. g4 fxg3 e.p. 6. bxa7 g5 7. axb8B gxh2 8. Bd6 hxg1B 9. Bxf8 Bd4 10. Bg7 Bxa1 11. Bxh8 Bxh8 12. Kf2 Bd4+ 13. Kg3 Bb6.

The first two moves can be interchanged. And black can also play g5 some moves later, e.g. on move 7, 8 or 9.

Looking for the author's intention I found the following:
1. b4 c5 2. bxc5 Qb6 3. cxb6 g5 4. bxa7 Bg7 5. axb8B Bxa1 6. Be5 Bd4 7. Bxh8 e5 8. f4 exf4 9. g4 fxg3 e.p. 10. Bxd4 gxh2 11. Kf2 hxg1B + 12. Kg3 Be3 13. Bb6 Bxb6

This is dualfree and with black tempo on move 12, so this looks like the author's intention. But here white can also play 5. axb8Q and proceed as above, or with 6... Bb2, 10. Qxb2 ..., 12... Be3/d4/c5/a7 13. Qb6 Bxb6."


441/16 - E. Angelini

Author's intention: the Black Queen was captured on g6, as in 1. a4 Nc6 2. Nc3 Nd4 3. Na2 Nb3 4. h4 Nxc1 5. Nf3 Nb3 6. Nc1 Nd4 7. Nh2 Nb5 8. axb5 Nf6 9. b6 Ne4 10. bxc7 Rb8 11. cxb8=B Qb6 12. h5 Qg6 13. hxg6 Rg8 14. gxf7 Kd8 15. fxg8=B Ng3 16. Ba2 Nxf1 17. Bb1 Ke8 18. Nxf1 Kf7 19. Bh2 Kg8 20. Bg1

But it is possible to capture it on g5, as in 1. h4 Nf6 2. a4 Ne4 3. a5 Ng3 4. Nf3 Nxf1 5. Nh2 Ne3 6. Nf1 Nd5 7. Nc3 Nb6 8. axb6 Nc6 9. bxc7 Rb8 10. cxb8=B Nd4 11. Bh2 Qa5 12. Bg1 Qg5 13. hxg5 Nb3 14. g6 Rg8 15. gxf7 Kd8 16. fxg8=B Nxc1 17. Ba2 Ke8 18. Bb1 Kf7 19. Na2 Kg8 20. Nxc1


445/16 - Eric Angelini
Europe Echecs 445, May. 1996

Here is the solution posted by Richard.Sabey@swindon.gpsemi.COM to the retros-list.

Solution: The last check is of course b7xc8=B+.

White has made 7 captures. The promoted pawn came from g2, and captured on f3, e4, d5, c6, b7 and c8. There is only one other white capture, that of the h8R, the only missing black man which couldn't have got in the way of the promoting pawn. So that pawn captured Black's a8R, b8N, c8B, Q, g8N, b7P. (Black's b-pawn couldn't have promoted because Black had only one man, the f1B, to capture, but the b-pawn would have had to capture 5 times and get to g2.)

White's g-pawn might have played gxNf3 and fxNe4, but couldn't get past e4 without capturing something which couldn't have got captured before Black's b-pawn moved. (So White played Bc6, Black played bxBc6, and that freed the a8R, c8B, Q and other knight to get to e4 d5 b7 c8.) Therefore, before any white pawn moved, the only black men which could have got to f3 are the knights, so White must have played gxNf3.

When the black knight arrived at f3, no white pawn or bishop had moved, so the white king hadn't moved and no white man was missing.

So one check of the game was ... Nf3+. This was followed by White's first capture, and therefore Black couldn't have checked earlier, for then White would still be unable to move his king, and would have had to capture the checking piece (which would have to be a knight). White couldn't have checked earlier either, because no black pawn or bishop had yet moved, so black would have been unable to move his king and would have had to get out of check by capturing the checking man (which would have to be a knight), but I've already accounted for all captures, so no more are allowed. So ... Nf3+ was the first check of the game.

The White King was first under check !!


451/16 - E. Angelini
Europe Echecs 451, Dec. 1996


281 - Janko Furman

First and last moves are identical for both wBs.

Retract -1. Nd3-f2 b6-b5 -2. Kf2-f1 b7-b6 -3. Nf4-d3 d6-d5 -4. Nh5xPf4 f5-f4 -5. Rf1-e1 f6-f5 -6. Ke1-f2 f7-f6 -7. Nf6xPh5 h6-h5 -8. Rf2-f1 h7-h6 -9. Bf1-g2 Qg2-h1 -10. Nd3-c1 Qh1-g2 -11. Nf4-d3 Qg2-h1 -12. Bc1-d2 Qh1-g2 -13. Kd2-e1 Qg2-h1 114. Kd3-d2 Qh1-g2 -15. Ke4-d3 Qg2-h1 -16. Bd2-c1 Kc1-b1 -17. Be1-d2+ Qh1-g2 -18. Rg2-f2 Kd2-c1 -19. Bf2-e1+ a3xQb2.

Here Bl. did play g3xRh2 before White's g2-g3 , so that the 1st move of the wKB was Bf1-g2 to let the wQR in.

Regarding the wQB, we have to find out a way to let the bK in, keeping in mind the necessity of cross-captures b2xc3 and c2xb3. The bK could have come through b4-a3-b2, but then the sequence was bKb4; b2xc3+; Kb4-a3 so that Bc1-d2 was played first. If the bK came through g3-f2, then it is necessay to retract b2xc3 to let bBd1 and then wQR out. Again, the 1st move of wQB was Bc1-d2.


404 - N. Plaksine

Solution: 1. Bb2! forcing the reflex 1 ... Qa2xb2 mate.

The interesting problem is to find how the position can be reached by a legal game without any earlier mate in one opportunity. The last few moves are problematic because we have to explain how Black had no earlier checkmating possibility.

The only way to avoid the mate in one situations is to retract -1. f7xNg6 Nf8(e5)xBg6+ -2. Bb1xNg6 Ne5(f8)xBg6+ -3. Bc2-g6. Note that it is important to uncapture wNs so that the bK was in check and Black could not give mate.

After this sequence is retracted, there is no difficulty with explaining the position: two bl. Bs are promoted, ...


447 - M. Caillaud

1. f4 h5 2. f5 h4 3. f6 h3 4. fxe7 hxg2 5. h4 g5 6. h5 g4 7. Rh4 g3 8. Nh3 g1=B 9. Bg2 Be3 10. Bc6 g2 11. dxe3 g1=R 12. Kf2 Rg3 13. e4 Rb3 14. axb3 f5 15. Ra6 f4 16. Ba4 f3 17. Kg3 f2 18. Rh6 f1=Q 19. Qd6 Qf7 20. Bf4 Qc4 21. bxc4 a5 22. Bb3 a4 23. Qa6 a3 24. Bd6 a2 25. Nf4 axb1=N 26. Ba2 Nc3 27. bxc3

All four Frolkin promotions in a very economical SPG.


498 - A. Frolkine

1. Nc3 c5 2. Nd5 c4 3. Nxe7 c3 4. Ng6 hxg6 5. Nf3 Rh5 6. Nd4 Ra5 7. Nb5 f5 8. Na3 Qf6 9. Nb1 Ba3 10. Rg1 d6 11. Rh1 Bd7 12. Rg1 Ba4 13. Rh1 Kd7 14. Rg1 Kc6 15. Rh1 Kb5 16. Rg1 Nc6 17. Rh1 Re8 18. Rg1 Re3 19. Rh1 Rd3 20. cxd3 c2 21. Rg1 cxd1=N 22. Rh1 Nc3 23. Rg1 Ne4 24. Rh1 Qc3 25. Rg1 Ngf6 26. Rh1 Ne8 27. Rg1

wNg1 replaces wNb1. Incredibly, it is not possible to economize one wh. move and there is no solution in 26 moves only. wRh1 has to make 16 waiting moves.


503v - D. Pronkine

1. b3 h5 2. Ba3 h4 3. Qc1 h3 4. Kd1 hxg2 5. h4 e5 6. h5 e4 7. h6 e3 8. h7 exf2 9. e4 f5 10. Ne2 g1=B 11. Bg2 f1=B 12. e5 Bb6 13. d4 Bfc5 14. dxc5 f4 15. cxb6 f3 16. Qf4 f2 17. Nc1 Ba6 18. Bc5 f1=B 19. a3 Bfb5 20. c4 Ne7 21. cxb5 Rg8 22. hxg8=B g5 23. Bc4 d5 24. bxa6 Bf5 25. e6 Nc8 26. e7 Kf7 27. e8=B Kg8 28. Beb5 c6 29. Ra2 cxb5 30. Rc2 Nc6 31. Na2 dxc4

Quintuple Frolkin B-promotion.


507 - M. Caillaud

1. f3 c6 2. Kf2 Qb6 3. Kg3 Qxg1 4. h4 Qxf1 5. h5 Qxh1 6. h6 Qh5 7. hxg7 Nh6 8. g8=N Bg7 9. a4 Bc3 10. a5 f6 11. a6 Kf7 12. axb7 Kg6 13. Rxa7 Nf7 14. Ra1 Ra4 15. Nh6 Na6 16. b8=R Rf8 17. Rb4 Nh8 18. Rh4 Rf7 19. Rh1 Rh4 20. Ng4 Nb4 21. Nf2 Ba6 22. Nh3 Bd3 23. Ng1 Kg5 24. Kf2 Bg6 25. Ke1

Phoenix theme: wNg1 & wRh1 are promotees.


511 - N. Plaksine and A. Kisliak

Bl. cap is f7xe6. For balance, there must have been 5 wh. promotions on a8 (or c8), f8 & g8 so that no uncapture by Wh. is possible immediately. Last moves must have been -1. Nf6-g8+ Kf7-f8 -2. g7-g8=N+ Kf6-f7 -3. g6-g7+ Kf5-f6 -4. g5-g6+ Ke4-f5 -5. g4-g5+ Kd4-e4 -6. d5xc6ep+ c7-c5.

A far-fetched en-passant !


520 - L. Borodatov

Bl. caps are g4xRf3. Wh. seven caps have been made by his Ps.

Last moves have been a tricky maneuver : -1. Na3-b5+ b7-b6 -2. b5xc6 ep+ c7-c5+ -3. Rb6-e6+ Ke4-d4 -4. g5xf6 ep+ f7-f5 -5. Rg6-b6+.

Observe that -3 ... Ke5-d4?? requires illegal -4. g3xf4+, while -4. f5-f6+?? requires -4 ... Kd4-e4 -5. Nf6-e8+ Ne4-d4 -6. Ne8-f6+ and what now??.


521 - P. Wassong

1. Nc3 h5 2. Ne4 h4 3. Ng3 hxg3 4. h3 Rh4 5. Rh2 gxh2 6. Rb1 hxg1=B 7. Ra1 Bh2 8. Rb1 Be5 9. Ra1 Bxb2 10. c3 e5 11. Qb3 Be7 12. Qe6 dxe6 13. Kd1 Qd5 14. Kc2 Qa5 15. Kd3 Nd7 16. Ke3 Ndf6 17. Kf3 Bd7 18. Kg3 Rd8 19. Kh2 Bc8 20. Kg1 (or 20. Kh1 & 21. Kg1) Rdd4 21. Kh1 Ra4 22. Kh2 Ra3 23. Kg3 Nh5 24. Kf3 f6 25. Ke3 Kf7 26. Kd3 Kg6 27. Kc2 Kh7 28. Kd1 Kh8 29. Ke1

The Wh. King makes a long 17 moves journey just to lose a tempo (with a minor dual in the triangulation).


528 - A. Frolkine and L. Liubachevsky

1. e4 c5 2. Bd3 c4 3. Ne2 cxd3 4. O-O dxe2 5. e5 e1=R 6. e6 Re5 7. a4 Rb5 8. Ra3 d5 9. Rg3 Bd7 10. Rg6 fxg6 11. exd7 Kf7 12. Qg4 e5 13. f3 Qg5 14. Kf2 Bc5 15. Ke1 Ne7 16. d8=B Rf8 17. Ba5 Kg8 18. Bc3 a5 19. Rh1 Na6 20. Qe6

A devilish problem where Wh. apparently possible O-O has in fact already been done, while Bl. apparently has played O-O, but in fact never did so.


530 - G. Wilts

Wh. caps are e2xd3, fxe & h6xg7. Bl. caps are bxa & dxe/c or hxg for promotion into bB.

Position unlocks by retracting e2xd3. This requires the reconduction of wBf1, to be uncaptured through b3xBa2. This further requires the retraction of b2-b3, after wBc7 has been reconducted to c1. The bRh1 needs much time to reach c7 and free the wB.

Retract -1. Bd7-f5+ b5-b6 -2. f4-f3 b4-b5 -3. f5-f4 b3-b4 -4. f7-f5 f6xNe7 -5. Rh3-h1 f5-f6 -6. Rg3-h3 f4-f5 -7. Rg5-g3 f3-f4 -8. Rb5-g5 f2-f3 -9. Rb7-b5 Ba5-c7 -10. Rc7-b7+ Bb4-a5 -11. e4-e3 Ba3-b4 -12. e5-e4 Bc1-a3 -13. e6-e5 b2-b3 -14. b3xBa2 Bb1-a2 -15. b4-b3 Bc2-b1 -16. b5-b4 Bd1-c2 -17. b6-b5 Be2-d1 -18. b7-b6 Bf1-e2 -19. Ke4-d4 e2xPd3+

Last 38 single moves are precisely determined !!


537 - M. Caillaud

Wh. caps are a2xb3xc4, hxg & bQB at home. bBh3 is bPa7 promoted through a2xNb1=B. Other bl. caps are c7xBd6 and e3xQf2 for promotion f2-f1=R justifying wh. captures.

The K-side mess unlocks with the retraction of f2-f3. When f2-f3 is made, the position is something like:

[r3k1n1/1ppp1p1p/8/4p1P1/2P3pb/6Rb/1PPPPPPN/R1BQKB2]

where the free units may be on different squares. But, under the reflex condition, f2-f3 would have been immediately followed by the reflex Bh4xRg3#. That is, unless the bK was in e4 so that f2-f3 gave him check. As this is the only way to explain the position, we conclude that, under the reflex condition, the bl. King had to visit e4 and Bl. O-O-O is not legal in the final position.

Solution: 1. Rxd7!! Kxd7# and not 1. Rxb7?? O-O-O# illegal.


540 - N. Plaksine

Bl. caps are h7xg6, cxdxe & exfxg. To fit the balance, wPa2, b2, and c2 had to promote (on b8 and c8) after a6xb7 and bxc. Third wh. cap. is fxe so that wPg2 got captured on its file and wPh2 promoted on h8.

Position unlocks by retracting: -1. d6xNe5+ Nc6-e5+ -2. f4xNg3 Nb8-c6 -3. f5-f4 b7-b8=N -4. e6xNf5 a6xRb7 and bl. retropat is avoided.

Unlocking goes on with unpromoting uncaptured wNg3 and wNf5 on c8. Then c7xQd6 can be retracted and the Qd6 unpromoted on h8. Then h7xPg6 unlocks everything.

Quadruple Ceriani-Frolkin theme with Q+3N.


541 - G. Wilts

To balance wh. caps axb & b2xc3 and explain bPh2, we need to assume that Bl. promoted his Pf7 after f3xg2 so that bP e & h have been captured on their column. The cage unlocks by retracting b2xc3, but this requires reconducting the wQBc1, thus uncapturing it through g3xBh2, thus retracting g2-g3, thus reconducting wKBf1 and the royalties. All this while Bl. is cruelly lacking retro-tempi...

Retract -1. Na3-c4+ g2-g1=N -2. Qg1-d4 d4-d3 -3. Kf6-e7 d5-d4 -4. Kg5-f6 d6-d5 -5. Kh4-g5 d7-d6 -6. Kg5xPh4 h5-h4 -7. Kf4-g5 h6-h5 -8. Ke3-f4 h7-h6 -9. Kf2xPe3 e4-e3+ -10. Ra1-d1 e5-e4 -11. Qd1-g1 e6-e5 -12. Ke1-f2 e7-e6 -13. f2-f3 f3xRg2 -14. Rg1-g2 f4-f3 -15. Bf3-g4 f5-f4 -16. Bg2-f3 f6-f5 -17. Bf1-g2 f7-f6 -18. g2-g3 g3xBh2 -19. Nb1-a3 g4-g3 -20. Bd6-h2 g5-g4 -21. Ba3-d6 g6-g5 -22. Bc1-a3 Kb4-a5 -23. b2xNc3+ and everything unlocks. First move of the bKR was Rh8-g8 to let the wN in.

Last 45 single moves are precisely determined. A fantastic task.


546 - G. Wilts

Bl. caps are bxPc, g7xf6, f7xBe6 & hxg for promotion g1=B. Thus wh. caps include h2xg3xh4 allowing the bBg1 out, and requiring h2xg1=B. 3rd capture is dxc. wPb2 promoted to Bb8.

Retracting a wQ move or a bN move would break bl. O-O or wh. O-O-O. Otherwise, the position unlocks by reconducting a bBf8 and retracting g7xf6. For this, the wBf4 has to open the way to the bBe3. This can only be done by disrupting bRh8, or wRa1 or wKe1. So that one castling is broken but we cannot tell which.

Castlings are mutually exclusive.

550 - M. Caillaud and J. Rotenberg

(a) 1. Nf3 f5 2. Ne5 f4 3. Nxd7 Kf7 4. Ne5 Kf6 5. Nf3 Bg4 6. Ng1 Bxe2 7. f3 Qe8

(b) 1. e4 f5 2. e5 Kf7 3. e6 Kf6 4. exd7 Qe8 5. d8=Q f4 6. Qd3 Bg4 7. Q3e2 Bxe2 8. f3

Even though (b) is longer, one cannot take (a) and adapt it. Instead an Excelsior combined with a switchback is required.


551 - P. Wassong

Wh. caps are d6xc7, g5xf6, h3xg4 & bBc8 at home; Bl. caps are e7xd6 and axb. Bl. promoted two B's, on b1 and on h1. Wh. promoted his wPe2 on e8.

It is not possible to retract e.g. -1. Bf7-a2+ a4-a5 -2. h2-h1=B a3-a4 -3. h3-h2 and what??, or -1. Bf7-a2+ a4-a5 -2. Bd5-h1 a3-a4 -3. Ba2-d5 b3-b4 -4. Bb1-a2 a2-a3 -5. b2-b1=B and what??. Some tricky retro-tactic is needed to save on tempis.

Retract -1. Bf7-a2+ a4-a5 -2. Bd5-h1 Kf8-e7!! -3. Be8-f7+ Ke7-f8!! -4. Bf7-d5 Kf8-e7!! -5. Ba2-f7 a3-a4 -6. Bb1-a2 a2-a3 -7. b2-b1=B b3-b4 -8. a3xQb2 Ke7-f8 -9. Bf7-e8+ Qe5-b2 -10. Rf8-d8 Qe6-e5 -11. Rd8-f8 Kf8-e7 -12. Fe8-f7+ Qf7-e6 -13. a4-a3 Ke7-f8 -14. a5-a4 Qf8-f7 -15. Bf7-e8+ Qe8-f8 -16. a6-a5 Kf8-e7 -17. a7-a7 e7-e8=Q -18. Be8-f7+ e6-e7 -19. e7xNd6 Nc4-d6+ -20. Kb8-c8 d6xNc7+ and everything unlocks.


545v - A. A. Kisliak

Wh. caps are bBc8 at home, a2xb3 & e3xf4, so that we must add 2 wh. units. Bl. caps are cxd and gxf/h for promotion into bB. For balance, wPg2 promoted to g8 and we can precise bPg3xh2 and then bPc6xBd5. The position unlocks by reconducting wBc1 so that d2-d3 can be retracted. This wQB is resurrected through g3xBh2 so that we must add +wNa8 & +wNb8.

Retract -1. Rg2-h2 c5-c6 -2. Rg5-g2 c4-c5 -3. Rc5-g5 Ne8-c7 -4. Rc7-c5+ Nf6-e8 -5. Bd1-c2 Ng8-f6 -6. Be2-d1 g7-g8=N -7. Bf1-e2 g6-g7 -8. Bg2-f1 g5-g6 -9. Bh1-g2 g4-g5 -10. h2-h1=B g2-g4 -11. g3xBh2 c2-c4!! -12. g4-g3 Bg3-h2 -13. h5-h4 Bh4-g3 -14. h6-h5 Bf6-h4 -15. d5-d4 Bd4-f6 -16. h7-h6 Be3-d4 -17. g5-g4 Bc1-e3 -18. g6-g5 d2-d3 -19. c6xBd5 Qf1-a6 -20. Ra5-a7 and everything unlocks. Mate in 1 is through 1. Qc4#.

Observe that it not possible to lose a tempo when we have a free wN and a free bB, so that g2-g4 and c2-c4 as retracted are indeed forced moves: there has been two opportunities for en-passant captures.


557 - T. Le Gleuher

Bl. caps are a7xb6, d7xc6 & f3xe2. Wh. caps are a2xb3, c2xd3 . The wPf2 promoted on f8. b3-b4 & c2xd3 can only be retracted once the bK is extracted from his corridor.

Retract -1. h5-h4 Ke1-f2 -2. h6-h5 Nf3-g1 -3. Kg1-h1 Ne5-f3+ -4. Kh1-g1 Ng6-e5 -5. Kg1-h1 Nf8-g6 -6. Kh1-g1 f7-f8=N -7. Kg1-h1 f6-f7 -8. Kh1-g1 f5-f6 -9. Kg1-h1 f4-f5 -10. Kh1-g1 f2-f4!! -11. f3xNe2 Nd4-e2 -12. Kg1-h1 Nc2-d4 -13. Kh1-g1 Kf1-e1 -14. f4-f3 Ne1-c2 -15. f5-f4 Ke2-f1 -16. Kg1-h1 Kf3-e2 -17. Kf1-g1 Kg3-f3 -18. Ke2-f1 Nf3-e1 -19. h7-h6 Rh1-d1 -20. Kd1-e2 Ne1-f3+ -21. Kc2-d1 Nf3-e1+ -22. Kb3-c2 ~ -23. Kc4-b3 c2xd3+ -24. Kc5-c4 b3-b4+ and everything unlocks.

Observe that retracting -10. f3-f4?? would have required one additional bl. tempo: f6-f5 so that the bKR cannot be reconducted to h8 after we unlock it by retracting c7xd6. First move of the wPf2 was f2-f4.


567 - A. Kornilov

One K is under double check by Re4 and Qg4. This can only be explained by an en-passant capture, so that -1. e5xf6 ep++ f7-f5 -2. Qh3xg4 are the last moves. We deduce wKf4, bKe6, wPf6, wPg5, bBg8, bNf8, wQg4, wRe4 & wRf3.

There has been no promotion (all Ps are accounted for) so that bRh4, wNa4, wNd7, wBb1. Furthermore wPh7 allowing bBg8 in, and we can precise -2. Qh3xNg4.

All missing units are accounted for by dxc, fxg, gxh, Qh3xNg4 and e5xf6 ep, so that wPa2, bPa5, wPb3, bPb7, bPe7, bPg3, wPh5, bPh6 are forced. Then wPc7 (because of Bd8), so that wPc4, bPc6, bPd6. All caps have been by Wh., so that wBd8.

Mate in 1 is through 0 ... Kf7 1. Rxe7#. Last 6 single moves are -1. e6xf5 ep f7-f5 -2. Qh3xNg4+ Kd5-e6 -3. c3-c4+ Ke6-d5.

Order of captures has been wPd6xQc7, then g6xRh7 after the Bc8 reached g8, then f4xBg5 after the bPg7 advanced, then Qh3xNf4, and then e5xf6 ep.