# US Problem Bulletin

Solutions

nnn?? - Unto Heinonen
1.d4 h5 2.Bh6 Nf6 3.f4 Ne4 4.Nf3 Nd2 5.e4 a5 6.Ba6 e5 7.c4 Bb4 8.Qa4 c5 9.Qc6 b5 10.Bb7 Ra6 11.a4 Rb6 12.Na3 Na6 13.0-0-0 Nf1 14.Rd3 Be1 15.b4 Qh4 16.Rb3 Qf2 17.h4 Kd8 18.Rh3 d5 19.Rg3 Bh3 20.Rg6 f5 21.Rf6 g5 22.g4

Incredibly there is no dual in this SPG where all 16 Ps are on the 4th rank.

2544v - A. Frolkin
Retract -1 ... Rg3-h3# -2. f5-f4 Ke6-f7 -3. f6-f5 Kd5-e6 -4. f7-f6 Bf6-e7 -5. e2-e1=B Re8-e3 -6. e3-e2 Rc8-e8 -7. e4-e3 c7-c8=R -8. e5-e4 Ke4-d5 -9. e6-e5 c6-c7 -10. e7-e6 Ne6-d4 -11. d4-d3 Ke3-e4 -12. d5-d4+ c5-c6 -13. d6-d5 c4-c5 -14. d7-d6 c2-c4 -15. c3xRb2 Qd1-b1 -16. c4-c3 Rb1-b2 -17. c5-c4 Bb2-f6 -18. c6-c5 Bc1-b2 -19. c7-c6 b2-b3 -20. b3xBa2 Ke2-e3 -21. b4-b3 Bb3-a2 -22. b5-b4 Ba4-b3 -23. b6-b5 Bb5xPa4 -24. a5-a4 Ke1-e2 -25. a6-a5 Bf1-b5 -26. a7-a6 and then the position unlocks with the retraction of e2xN/Bf3.

Unfortunately, this problem is cooked as Gerd Wilts found out. Retract -1 ... Rg3-h3# -2. e2-e1=B Bf6-e7 -3. f5-f4 Re8-e3 -4. e3-e2 Nb5-d4 -5. e4-e3 Rc8-e8 -6. e5-e4 c7-c8=R -7. e6-e5 c6-c7 -8. e7-e6 c5-c6 -9. d4-d3 c4-c5 -10. d5-d4 c2-c4 -11. c3xRb2 Qd1-b1 -12. c4-c3 Rb1-b2 -13. c5-c4 Bb2-f6 -14. d6-d5 Bc1-b2 -15. f6-f5 b2-b3 -16. b3xBa2 Ke6-f7 -17. b4-b3 Bc4-a2 -18. c6-c5 a2-a3 -19. c7-c6 Na3-b5 -20. b5-b4 Kd5-e6 -21. b6-b5 Nb5xPa3 -22. a4-a3 Bf1-c4 -23. a5-a4 Ke4-d5 -24. a6-a5 Ke3-e4 -25. a7-a6 Ke2-e3 -26. d7-d6 Ke1-e2 -27. f7-f6 e2xBf3

2819 - Thomas Volet
Retract: 1.c3(+Rd4) Rd1 2.c2 Rg1 3.Qf1 Rg2 4.Qa1 b7 5.Qa8 b6 6.a7(-Qa8) c6 7.a6 a7(+Nb6) 8.Nd5 c7 9.Nf4 Qd7 10.Nh3 Bd6 11.a5 Bf8 12.a4 e7(+Bf6) 13.Bc3 Qd1 14.~ d2(-Qd1) 15.~ d3 16.Bd2 d4 17.Bc1 d5 18.d2(+N/Qe3) Nf5 (or Qb6) 19.Ke3 Nd6 (or Qa6) etc. {uncapture Rh8 on e4 before the wQ on g5} Try: 1.c3(+Nd4)? Ne6 2.c2 Nf4 3.Qf1 Ng2 .... 19.K?? as Ng2 controls e3.

3034 - Mark Kirtley
1. h4 a5 2. Rh3 a4 3. Rc3 a3 4. Rxc7 axb2 5. a4 bxc1=Q 6. a5 Qxb1 7. a6 Qb6 8. Qb1 Nc6 9. Qb3 Rb8 10. O-O-O Qxf2 11. a7 Qxf1 12. a8=R Qxg1 13. Ra1 Ra8 14. Rxc8 Ra6 15. Ra8 Qa7 16. Rh1

A beautiful cyclic rotation of 3 wh. Rooks.

3228 - Peter Wong
1. e3 g6 2. Ba6 Bg7 3. Ne2 Bc3 4. bxc3 Nh6 5. Ba3 O-O 6. Bxe7 Nxa6 7. Na3 Nc5 8. Rb1 Ne4 9. Rxb7 Ng3 10. Qb1 Nf1!! 11. Ng3 Qe8 12. Nxf1

10 ... Nf1 loses a tempo !

3229 - Noam D. Elkies
1. Nc3 c6 2. Ne4 Qc7 3. Ng5 Qg3 4. hxg3 Kd8 5. Rh4 Kc7 6. Rb4 Kd6 7. e4 Ke5 8. Bb5 Kf6 9. d3 Kg6 10. Be3 Kh6 11. Qd2 Kh5 12. Kf1 Kg6 13. Re1 Kf6 14. Qd1 Ke5 15. Bd4 Kd6 16. Re3 Kc7 17. Rf3 Kd8 18. Rf6 Ke8 19. Rxc6 d6 20. Rxc8#

15 moves long King tour for loss of a tempo.

3230 - Unto Heinonen
1. g4 a5 2. g5 a4 3. g6 a3 4. gxh7 axb2 5. a4 bxa1=R 6. Bb2 g5 7. Bg7 b5 8. Bh6 Bg7 9. a5 Bc3 10. a6 Nf6 11. a7 Rg8 12. h8=N Bb7 13. Ng6 Bxh1 14. Nf4 Nc6 15. Nh5 Qb8 16. f4 Qb6 17. Nf3 O-O-O 18. a8=B Ne5 19. Be4 Ra8 20. Bh7 Ba5 21. Nc3 g4 22. Qa1 g3 23. Nd1 g2 24. Qd4 Rg3 25. Qg1 Rh8 26. Nd4 Ra3 27. c3 Ra1

Cyclic permutation of Rooks a1, a8 & h8.

3231 - Alexander Zolotarev and Nikita M. Plaksin
Wh. caps are Qh6xg5+, e2xf3 & dxe. Bl. caps are a7xBb6 & d7xBc6. We can only add one bl. Pawn and one bl. officer. One of the wNs is wPa2 promoted on a8.

Add bPe2 & bBg4. Then retract -1. Qh6xNg5+ e3-e2 -2. d4xRe5 Re4-e5 -3. Rd7-e7 Re7-e4+ -4. Rd5-d7 Ne6-g5 -5. Rg5-d5+ e4-e3 -6. Nf5-g7 Ng7-e6+ -7. Ne3-f5 Bc8-g4 -8. Ng4-e3+ d7xBc6 -9. Bd5-c6 e5-e4 -10. Bc4-d5 e6-e5 -11. Bf1-c4 Ke4-f4 -12. e3xRf3 and everything unlocks.

Impressive precision of unlocking moves, with tricky retro-tactics.

3237 - Gerd Wilts and Norbert Geissler
1. Nc3 d5 2. Nxd5 Bf5 3. Nxe7 Bxc2 4. Nxg8 Rxg8 5. a3 Bxa3 6. Qxc2 Bxb2 7. Qxc7 Bxc1 8. Qxb8 Qxb8 9. Rxa7 Qxh2 10. Rxb7 Qxg1 11. Rxh7 Qxg2 12. Rxg7 Rxg7 13. Rxf7 Rxf7 14. Bxg2 Rxf2 15. Bxa8 Rxe2 16. Kxe2 Bxd2 17. Kxd2

3308 - Andrey 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 Bf3 11. exf3 c6 12. Bc4 Qc7 13. Bg8 Nf7 14. Qe2 Rh4 15. Qe6 dxe6 16. b7 Nd7 17. b8=Q Kh5 18. Qb5 Rb8 19. Qe2 Rbb4 20. Qd1 Rbg4 21. fxg4

Phoenix theme with the Qd1 & anti-phoenix theme with the Bg8.

3309 - René J. Millour
Let's write +U, +R, -U, -R for Up, Right, Down & Left. The Imitator did not start above the 5th rank so that its total moves amount to at least +U.

The missing wR did not move beyond the wPs barrier so that it could only contribute +U. Adding +3U for wPa3, b3, g3 and -4U for bPc3 does not explain how the Imitator could move up. Only explanation is that the missing bP did contribute -U: this is possible if it promoted and then reached the 8th rank (where it was captured, or where it replaced an original bl. unit).

Thus the bPb7 did play b7-b3xRa2-a1=X. (Btw, the captured wR was the original wQR because b2-b3 could only be played after the capture.) Thus the wh. units totaled -7R, while the Ig6 can only account for -R. Only explanation is that the missing bPb7, after promotion on a1, reached h8, yielding a total of +6R for bl. units.

We now see that the bPb7 promoted into a bNa1, and that a bN was captured on a8, necessarily by a wN. The Imitator started on h5. The captures were bPb3xQRa2 and wNg6xNh8.

3369 - Michel Caillaud
1. f3 h5 2. Kf2 Rh6 3. Ke3 Rg6 4. Kd4! Rg3 5. hxg3 b6 6. Rh4 Bb7! 7. Re4 Ba6 8. Re6 dxe6 9. Ke4 Qd6 10. Ke3 Qa3 11. bxa3 Bc8 12. Bb2 a6 13. Bf6 gxf6 14. Kf2 Bh6 15. Ke1 Bf4 16. gxf4 f5

Two intertwined tempo walks.

3370 - Noam D. Elkies
1. a4 b6 2. a5 Bb7 3. a6 Qc8 4. axb7 Na6 5. b8=B Qb7 6. f3! Qc6 7. f4 Qc3 8. dxc3 Kd8 9. Be3 Kc8 10. Bc5 Kb7 11. Bb4 Kc6 12. Ba5 bxa5 13. e3 Kb6! 14. Bb5 Kb7 15. Ne2 Kc8 16. Rf1 Kd8 17. Rf3 Ke8 18. Qxd7#

3371 - Nikita M. Plaksin and Alexander Zolotarev
Wh. all five caps are c2xd3+ (last move), b2xc3, f2xe3 & exdxc. This required promotions on b1 and f1 of bl. Pa7, b7, g7, h7 after they captured axb, gxf & hxgxf.

Shortest trip of wQe1 can be d1-e1-e2 as in the following unlocking maneuver:
-1. c2xQd3+ Qf5-d3
-2. Qe1-e2+ Qf1-f5
-3. Qd1-e1 f2-f1=Q
-4. d4xNc5 g3xR/B/Nf2
and wh. retropat is avoided.
Further unlocking requires (1) unpromoting bNc5 on f1, (2) retracting wPf2xRe3 & wPe3xRd4, (3) unpromoting bRe3 & d4 on b1, (4) reconducting wKR and wQB home, after they have been resurrected by hxg or gxf, (5) retracting wPb2xRc3.

This displays a four-fold Ceriani theme with Q + N + R + R.

3372 - Thomas Volet
Wh. caps are b2xBc3 and g2xf3xe4. Bl. caps are d7xNe6 and a3xBb2 for promotion on b1. Position unlocks by reconducting bPh4 to h5 and replacing bNg2 by some wh. unit. Because f7-f6 must be retracted in the final phase, we cannot retract c7-c6 and lock the bl. royalities out.

Retract
-1. Na3-b1 c5-c4
-2. Nc4-a3 f6-f5
-3. Na5-c4 f7-f6
-4. Nb3-a5 c6-c5
-5. Nc1-b3 Nd4-e2
-6. Ne2-c1+ Nb5-d4
-7. Rg5-g4 Na3-b5
-8. Rb5-g5 Nb1-a3
-9. Rb6-b5 b2-b1=N
-10. Rb5-b6 a3xBb2
-11. Bc1-b2 a4-a3
-12. Rb1-b5 a5-a4
-13. b2xBc3
and bl. retropat is avoided. Further unlocking requires
(1) reconducting bBc3 to home square f8,
(2) retracting g7-g6,
(3) reconducting bBh5 to home square c8,
(4) retracting -n. d7xNe6 h5-h4 -(n+1). Nf4-e6 Ng2-h4 -(n+2). Ng2-f4+,
(5) extracting the wKh3 through g5, this requires retracting f7-f6.

The pattern of Bl. pawn retractions cannot be modified. bPf4 must be retracted at least once for tempo reason, but then at least twice so that the wRg4 can leave, and then at least three times so that the wK can cross g5. Hence c7-c6 must be kept in store for after bl. Q and K can be driven home. First move of the Black King was Ke8-d8 !

White moved last so that mate in 1 is with 1 ... Bxg4 and not 1. Rxf4??

3406 - Andrey Frolkin
1. Nf3 a5 2. Nd4 a4 3. Nb3 axb3 4. a3 Ra4 5. Ra2 bxa2 6. Rg1 axb1=N 7. Rh1 Nc3 8. Rg1 Ne4 9. c3 Ng3 10. Qc2 Nh1 11. Qg6 hxg6 12. Kd1 Rh5 13. Kc2 Nh6 14. Kd3 Rd5 15. Ke3 e5 16. Kf3 Qh4 17. Ke3 Bc5 18. Kf3 Ba7 19. h3 b6 20. Ke3 Ba6 21. Kf3 Bd3 22. Ke3 Bf5 23. Kf3 e4 24. Ke3 Ng4 25. Kf4 Qh8 26. Kg5 f6 27. Kf4 g5#

3407 - Noam D. Elkies
1. h4 f6 2. Rh3 Kf7 3. Re3 Kg6 4. Re6 Kf5 5. e4 Kf4 6. Qh5 Nc6 7. Qe5 Kg4 8. Ne2 Nd4 9. Nec3 Nf5 10. Ke2 Kh5 11. Kf3 Kg6 12. Kg4 Kf7 13. Kh5 Ke8 14. g4 g6#

3408 - Noam D. Elkies
1. c3 g5 2. Qa4 g4 3. Qf4 d5 4. h3 d4 5. Qh2 g3 6. Kd1 Bg4 7. Kc2 e6 8. Kd3 Bb4 9. Ke4 c5 10. Ke5 Nc6 11. Kf4 Qf6 12. Ke4 Nge7 13. Kd3 O-O 14. Kc2 Kh8 15. Kd1 Rg8 16. Ke1 Raf8

3460 - Thomas Volet
Retract -1. Kf7xNf8 Ne6-f8
-2. Kf8-f7+ Nd4-e6+
-3. a4-a5 Nb3-d4
-4. a3-a4 Nc1-b3
-5. a2-a3 c2-c1=N
-6. h4-h5 b3xBc2
-7. Bd3-c2 b4-b3
-8. Bf1-d3 a5xNb4
-9. e2xRf3 Rg3-f3+
and Bl. retropat is avoided just in time. Unlocking is completed with
(1) reconduct wNb4 to g8,
(2) extract both wRs through the f-file, wK through c7, b6,
(3) reconduct wNg8 to b1,
(4) retract bPf7xNg6, bPc7xQd6, wPc2xRd3xNe4xNf5xQg6xPh7.

All this needs at least 19 Knight moves.

3461 - Alexander A. Kislyak
The try:
-1. Rc8xc7 Rd7-c7
-2. Rc7-c8+ Rd6-d7
-3. d5-d4 Rh6-d6
-4. d6-d5 Rh3-h6
-5. d7-d6 Rf3xPh3!!
-6. h4-h3 Rf2-f3
-7. h5-h4 Rc2-f2
-8. h6-h5 Rc3-c2
-9. h7-h6 Bd2-c1
-10. Rf1-b1 etc.
fails because any attempt to reconduct the wR to b1, the wB to c1 and retract c2-c4 Kc4-b4 d2-d3+ needs one more bl. tempo when the bR is pinned.

Really the position unlocks with the following:
-0 ... c2-c4!!
-1. Rc8xc7 Rd7-c7
-2. Rc7-c8+ Rd6-d7
-3. d5-d4 Rh6-d6
-4. d6-d5 Rh3-h6
-5. d7-d6 Rh5xPh3!!
-6. h4-h3 Rc5-h5
-7. h5-h4 Rc3-c5
-8. h6-h5 Bd2-c1
-9. Rf1-b1 etc.
and now we have the required bl. tempo in store.

Finally, Bl. has the move and mates in one with 1 ... Na7-c8#

3462 - Andrey Frolkin
The last moves were: -1. Rg3-h3 f6-f5 -2. Kd5-e6 f7-f6 -3. Bf6-e7 e2-e1=B -4. Re8-e3 e3-e2 -5. Rc8-e8 e4-e3 -6. c7-c8=R e5-e4 -7. Ke4-d5 e6-e5 -8. c6-c7 e7-e6 -9. Ne6-d4 d4-d3 -10. Ke3-e4 d5-d4 -11. c5-c6 d6-d5 -12. c4-c5 d7-d6 -13. c2-c4 c3xRb2 -14. Qd1-b1 c4-c3 -15. Rb1-b2 c5-c4 -16. Bb2-f6 c6-c5 -17. Bc1-b2 c7-c6 -18. b2-b3 b3xBa2 -19. Ke2-e3 b4-b3 -20. Bb3-a2 b5-b4 -21. Ba4-b3 b6-b5 -22. Bb5xPa4 a5-a4 -23. Ke1-e2 a6-a5 -24. Bf1-b5 a7-a6 -25. e2x[B/N]f3 and the position unlocks.

3519 - Gerd Wilts
(a) 1. d4 Nc6 2. d5 Nd4 3. f4 Nxe2 4. Kf2 Ng3 5. hxg3 Nf6 6. Rh6 Nh5 7. Re6 f6 8. Qg4 Kf7 9. Ne2
(b) 1. d4 Nc6 2. d5 Nd4 3. Kd2 Nxe2 4. Ke3 Ng3 5. hxg3 Nf6 6. Rh6 Nh5 7. Re6 f6 8. Qg4 Kf7 9. Qh3 Kg6 10. Ne2 Kg5 11. f4+ Kh6 12. Kf2 Kg6 13. Qg4+ Kf7

Incredible twins in the proof game genre !

3520 - Gianni Donati
1. Nf3 b5 2. Nd4 Bb7 3. Nxb5 Bf3 4. gxf3 f6 5. Bh3 Kf7 6. Be6+ Kg6 7. Bxg8 e6 8. O-O Bd6 9. Kg2 Bxh2 10. Nd6 Qe7 11. Nb7 d6 12. Rh1 Nd7 13. Kf1 Rf8 14. Ke1 Qd8

Here castling-uncastling is motivated by the need of losing one white tempo !!

3521 - Peter Wong
Author's intention: 1. h4 g6 2. h5 g5 3. h6 Nxh6 4. Rh3 Ng4 5. Re3 h6 6. Re6 dxe6 7. a3 Qd3 8. exd3 Nd7 9. Qf3 Nb6 10. Qf5 exf5 11. Kd1 Be6 12. Ke2 Ba2 13. b3 Kd7 14. Bb2 Kc6 15. Bf6 exf6 16. Nc3 Be7 17. Re1 Rhd8 18. Kd1 Rd5 19. Kc1 Ra5 20. Rd1 Kc5

Wh. castle by hand for tempo-related reasons. Additionally, there are two very cute tempo moves.

Unfortunately, there are duals (mentioned by Joost de Heer): 11. Ke2 Be6 12. Kd1 Ba2 13. b3 Kd7 14. Bb2 Kc6 15. Bf6 exf6 16. Kc1 Be7 17. Kb2 Rhd8 18. Nc3 Rd5 19. Rd1 Ra5 20. Kc1 Kc5

3522 - Thomas Volet
Wh. 8 caps are the bBf8 at home and b2xc3, c2xBd3 and a2xb3xc4xd5xe6xPf7. Bl. 3 caps are the wBf1 at home, axb and d7xe6. It is not possible to retract d7xe6 or b2xc3 without first reconducting the corresponding QB. But we also have to reconduct the corresponding KR and, if we further retract c2xd3, the wQ must have been reconducted.

Retract
-1. Ra2-a1 b3-b2
-2. Rd1-a1 b4-b3
-3. Bc1-a3 a5xQb4
-4. b2xRc3 and bl. retropat is avoided.
Further unlocking requires
(1) reconducting the bR in f8, the wK in g1, the wR in f1,
(2) retracting c3-c2, reconducting the wQ in d1 and retracting c2xBd3,
(3) reconducting the bB in c8, retracting d7xRe6.
Now it seems the least number of Knight moves is obtained with decapturing the bN on c4 and d5, thus giving wNb1->a4 (2 moves), wNg1xf8->a8 (7), bNg8->d5 (2), bNb8xf1->c4 (7) for a total of 18 Knight moves. But there is one difficulty: Wh. cannot play the shortest route NxBf8-e6-c7-a8 because Bl. has to play O-O in the proof game. Hence two more Knight moves are required, for a total of 20 Knight moves at least.

Here is a proof game 1. Nf3 c5 2. Ne5 Qb6 3. Ng6 Qb3 4. axb3 Nc6 5. Nxf8 Ne5 6. Ne6 Ng4 7. Ra4 Ne3 8. Re4 Nxf1 9. Nd4 Nf6 10. Nc3 O-O 11. Ne6 a5 12. Nc7 Ra6 13. Na8 Ne3 14. Na4 Nc4 15. bxc4 Nd5 16. cxd5 Re6 17. dxe6 Kh8 18. exf7 c4 19. Re6 dxe6 20. O-O Bd7 21. Re1 Bb5 22. Kf1 c3 23. Kg1 Bd3 24. cxd3 Rc8 25. Qb3 c2 26. Qb4 Rc3 27. Kf1 Rc4 28. Rd1 Rc5 29. Ke1 Rc3 30. bxc3 axb4 31. Ba3 b3 32. Ra1 b2 33. Ra2

3528 - Philippe Schnoebelen
May White castle? The answer is NO! as spelled out by the bl. units. Hence White cannot avoids the Qf1# threat by 1. O-O-O.

Analysis: Black's promoted B is from h2xg1=B if we assume Wh. may castle. Hence the Bl. pawns require the captures (e.g. with bxa, cxd, exfxgxhxg1, fxg) of the seven missing Wh. units (not counting the two Wh. Bishops captured at home). So that Black's last move could not have been a capture and White last move just before must have been with his K or R: the O-O-O is not legal.