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No. 116, Jan. 1996

3459 - N. Plaksin and A. Zolotarev

Authors' intention is to correct their problem #3241. So that the same analysis leads to (1) Add bNb5 & bBb6, and (2) Retract -1. ... f7xRe6 -2. Re5-e6 Na3-b5 -3. Rb5-e5+ Nb1-a3 -4. f3-f4 Na3-b1 -5. Ne3-c2 Nc2-a3+ -6. Nd5-e3 Bf2-b6 -7. Nb6-d5+ Bh4-f2 -8. f2-f3 Be7xPh4 -9. h3-h4 Bf8-e7 -10. h2-h3 e7xNd6.

Unfortunately, the problem has a new cook. Retract -1. f3-f4 f7xNe6 -2. Nf4-e6 h6-h5 -3. Nd5-f4 Bf2-b6 -4. Nb6-d5+ Bh4-f2 -5. f2-f3 Be7xPh4 -6. h3-h4 Bf8-e7 -7. h2-h3 e7xNd6 and everything unlocks.


3460 - Thomas Volet

Wh. caps are g2xh3, fxg, and c2xd3xPe4xf5. Bl. caps are a7xb6 and hxBg. wPa2 was promoted on a8 for balance. The wK can only be extracted from the f6 square (thus unlocking the NE cage) by retracting Kg5-f6. This requires that the bK is extracted first, so that g2xh3 must be retracted, after wRh1 and wBf1 are reconducted home.

Here is the unlocking maneuver: (1) retract wPf3xNg4, (2) retract wPf2-f3 & bPg4-g3-g2, (3) put bNg4 in g7 instead of the actual wN so that we may (4) set bK in h5, allowing to (5) remove wNg1 and drive wR back to h1, (6) unpromote wQ or wN & reconduct wP to a6, (7) retract bPh5xwBg4 with bK on h4, (8) retract -1. wBf3-g4 a7xb6 -2. wBg2-f3 c7-c6 -3. wBf1-g2 Kg4-h4 -4. g2xR/Nh3+ and everything unlocks.

It is important to observe that Bl. absolutely needs the two tempis a7xb6 & c7-c6 for step (8). He cannot retract h6-h5?? (or else the wK will be stuck on f6) and he has no free units because the uncaptured bN is stuck on g7. Wh. cannot uncapture more Bl. units through e4xf5 because, once the wPf5 is on e4, the bK can only reach e4 if the wP is reconducted on c2 and then there will be no way to reconduct the wK to his e1 home square.

Finally, because these two tempis must be kept in store for (8), and because e4xf5?? cannot be retracted before unlocking, the last move must have been f3xNg4.


3461 - Jean-Michel Trillon

1. Kb2 Qh1 2. Kb1 Bg1 3. Kb2 Rh2 4. Rh3 Nh4 5. Kb1 Qg2 6. Kb2 Rh1 7. Rh2 Qh3 8. Rg2 Bh2 9. Rg1 Bg2 10. Rf1 Bg1 11. Kb1 Rh2 12. Kb2 Bh1 13. Kb1 Ng2 14. Kb2 Qh4 15. Kb1 Rh3 16. Kb2 Bh2 17. Rg1 Rf1 18. Kb1 Ne1 19. Rg2 Rg1 20. Kb2 Nd3 21. Kb1 Re1! 22. Rg1 Rf1 23. Rg2 Ne1 24. Rg1 Ng2 25. Kb2 Re1 26. Rf1 Bg1 27. Kb1 Rh2 28. Kb2 Qh3 29. Kb1 Nh4 30. Kb2 Bg2 31. Kb1

Rh1 32. Kb2 Bh2 33. Rg1 Bf1 34. Rg2 Bg1 35. Rh2 Qg2 36. Rh3 Rh2 37. Kb1 Qh1 38. Kb2 Ng2 39. Rh4 Rh3 40. Kb1 Bh2 41. Kb2 Qg1 42. Kb1


3462 - Andrey Frolkin

1. h4 a5 2. Rh3 a4 3. Rb3 axb3 4. a3 Ra4 5. Ra2 bxa2 6. Nf3 axb1=B 7. Ne5 Ba2 8. Nxd7 Bd5 9. Nxf8 Bf3 10. Nxh7 Kd7 11. Ng5 Rh6 12. Nh3 Rd6 13. Ng1 Ke6 14. Nh3 Bd7 15. Ng1 Be8 16. Nh3 Nd7 17. Ng1 Qa8 18. Nh3


3463 - Thierry Le Gleuher

bNg1 and b1 have been promoted, using 8 captures. Last bl. capture is d/fxe6.
The wh. captures are g4xh5+ (last move), bRa8 and bNg8 captured on home squares, 4 captures of bl. units (bQ, bKB, dark-squared bP, light-squared bP en passant) for promotion into a wR now on f4.

bPd7 went d7-d5 x Pc4 x Pb3ep x Pa2 x Nb1=N.
bPc7 went c7-c5 x Pd4 x e3 x Pf2 x Ng1=N. bPf7 captured wKBe6.
wPh2 went h2-h4 x g5 x h6ep x g7 x f/h8=R.

bQB was captured on h5. Once g4xBh5+ is retracted, the Bl. position unlocks by retracting g2-g4. But we have to drive the wBe6 back home to f1 before. A few tempi can be given to save Bl. from retro-stalemate by driving the wQR back home, so that bPa2xb1 can be retracted. The quickest (and only possible) way is:

-1. g4xBh5+ f7xBe6 -2. Ke3-f2 f2xNg1=N -3. Kd4-e3 e3xPf2 -4. Ke5-d4 d4xPe3ep -5. e2-e4 c5xPd4 -6. Ra3-g3 c7-c5 -7. Ra1-a3 a2xNb1=N -8. Bd5-e6 b3xPa2 -9. Bg2-d5 c4xPb3ep -10. b2-b4 d5xPc4 -11. Qd1-f1 d7-d5 -12. Bf1-g2 Kf5-g6 -13. g2-g4+ and everything unlocks (Bl. did O-O, bPg7 was captured on g5).

This problem is an amelioration of problem #3300 by same author. It has a longer precise sequence and one more en-passant capture.