The Problemist


R197 - Ladislav Packa
1.d4 h5 2.Bh6 c5 3.Kd2 Qa5+ 4.Kc1 Qe1 5.e3 a5 6.Bb5 Na6 7.Ba4 b5 8.Nf3 d5 9.Ne5 Bh3 10.Ng4 e5 11.b3 Be7 12.Kb2 Bh4 13.Qd3 g5 14.Qh7 f5

Eight double bP moves in a 14 moves SPG.

R200 - Nikita Plaksin
The Bl. pawns captured all 6 missing wh. units, which implies that wh. Pawns a2, b2, c2 & h2 promoted on b8, c8 and h8 (after capturing the 2 missing bl. units with a6xb7 and bxc). It is possible to find a proof game where bl. O-O-O is still legal.

-1. d6xNe5 Nc6-e5
-2. e6xBf5 Nb8-c6
-3. d7xBe6 b7-b8=N
-4. a7-a6 a6xBb7
and bl. retropat is avoided. Further unlocking requires (1) unpromoting wBs e6 & f5 on c8, (2) retracting c7xBd6, (3) unpromoting wBd6 on h8, (4) retracting h7xBg6. Then everything unlocks.

This displays four completely determined Ceriani promotions.

R201 - Peter Wong
(a) 1.c3 Nh6 2.Qa4 Nf5 3.Qxd7+ Qxd7 4.Nf3 Kd8 5.Nd4 Qxd4 6.e3 Nd7 7.Bd3 Qxd3
(b) 1.Nf3 Nh6 2.Ne5 Nf5 3.Nxd7 Qxd7 4.e3 Kd8 5.Bb5 Qxb5 6.c3 Nd7 7.Qb3 Qxb3
(c) 1.e3 Nh6 2.Bb5 Nf5 3.Bxd7+ Qxd7 4.c3 Kd8 5.Qa4 Qxa4 6.Nf3 Nd7 7.Nh4 Qxh4
The sequence of captures by bQ changes cyclically: QNB, NBQ, BQN.

R202v - Unto Heinonen
1.Nf3 a5 2.Nd4 a4 3.Nb3 axb3 4.Nc3 bxc2 5.Ne4 cxd1=B 6.Ng5 Bb3 7.Nxh7 Bc4 8.Ng5 Rh5 9.Nh7 Ra5 10.Kd1 d5 11.Kc2 Kd7 12.Kc3 Ke6 13.Kd4 Kf5 14.Ke3 Be6 15.Kf3 Nd7 16.Kg3 Rc8 17.Kh4 g5+ 18.Kh3 Bg7 19.Kg3 Be5+ 20.Kf3 Bd6 21.Ke3 Kg4 22.Kd4 c5+ 23.Kc3 Qb6 24.Kc2 Qb4 25.Kd1 b5 26.Ke1

The wh. King makes a tempo-correcting seventeen move round-trip to add one single move to the shortest proof game (which lasts 25.0 moves).

R203 - Yury Lebedev and Andrei Frolkin
All 4 missing bl. Ps promoted so that last move is Bh3xRg4+ and no wh. P ever captured. Hence (1) bPf7 promoted to bBg1 after fxg, (2) bPc7, d7 & e7 promoted to bQs on d1 after xd and xwQB, and (3) wPc2 promoted to wQc8. Last moves are:
-1. Bh3xRg4+ Rg3-g4+
-2. Ng4-f2+ Ne3-c4+
-3. Nd5-b4+ Nc4-d6+
-4. Nb6-d5+ Kd6-e6+
-5. e4-e5+ Be7-f6+
-6. Nf6-g4+ etc.

11 consecutive checks.

R204 - Andrei Frolkin
Black has the move and mates by Qg7#. 12.5 moves ago the position must have been the one shown below which allows the white rooks and bishops to have come in play:


The problem position is reached by playing:
1.h5 h6 2.hxg6 h5 3.a3 h4 4.a4 h3 5.a5 h2 6.a6 h1=N 7.f3 Nf2 8.c3 Nd3 9.c4 Nb4 10.c5 Nxa6 11.Nc6+ Nb8 12.Nxb8 Nh5 13.c6

Trying to retract Bl. first fails because if we begin by -1 ... Ng7-h5 -2. c5-c6 Nh5-g7 -3. Nc6xNb8 then Wh. will have to retract wPd3-d4 eventually and then wRa1 cannot return home.

R205 - Valery Liskovets
1.Ke4 2.Bxe6 and now if White had just moved to reach this position, he cannot castle both ways. Depending on which castling is still deemed possible, we continue: 3.Kd3 4.Bc4 O-O-O# or 3.Kf3 4.Bg4 O-O#

The Problemist, Jan. 1994

R221 - Aleksandr Shuryakov
Author's intention: Add WPs e3 and f2. Then the position is drawn. Black has no last move, so that White moved last and Black is stalemate. (Same idea adding WQs d2 and f2 were it not for the restriction in the stipulation).
Unfortunately cooked because we can also add WPs e3 and d2.

R222 - Michel Cailaud
Author's intention: 1. h4 a5 2. h5 a4 3. h6 a3 4. hxg7 axb2 5. Rh6 bxa1=R 6. Rc6 h5 7. Nh3 h4 8. Nf4 h3 9. Nd5 h2 10. f4 h1=R 11. Kf2 Rxf1 12. Kg3 Rhh1 13. a4 Nh6 14. g8=B dxc6 15. Bh7 Nd7 16. Bf5 Nb6 17. Bh3 Bf5 18. a5 e6 19. a6 Bc5 20. a7 Be3 21. dxe3 Qd6 22. Qd3 Kd7 23. Bd2 Rh8 24. a8=Q Bh7 25. Qg8 Ra8 26. Qgg6 Kc8 27. Ba5 Kb8 28. Nbc3 Ra1 29. Kh4 Qd8 30. Kh5, which would be a fantastic cyclic exchange of 4 BRs !!

Unfortunately has duals: ... 18. Kh4 e6 19. Kh5 Bc5 20. a5 Be3 21. a6 Qe7/Qf6 22. a7 Kd7 23. dxe3 Rh8 24. a8=Q Bh7 25. Qg8 Qd8 26. Bd2 Ra8 27. Ba5 Kc8 28. Nbc3 Kb8 29. Qd3 Ra1 30. Qgg6

The Problemist, Mar. 1994

R223v - Frank Christiaans
Black has the move: 1... Kb2# and not 1. Bd4=.

Analysis: Bl. K could not reach b1 via the Q-side because this would require that Wh. played b2-b3 first and only then a2-a3, in contradiction with bPa2. Thus Bl. K entered through h3 and g2, when Wh. had already played g2-g3 and not yet h2-h3, with a wN already on h1.

Possible last moves by Bl. can be ruled out as follows: (a) 0 ... Kb2xRb1: the 2 wRs won't allow the bK out. (b) 0 ...Kb2xBb1: wBb1 requires a forward sequence a2-a3, Bf1-..-a2, b2-b3, Bc1-b2-.. so that bRa1 is promoted. It will not be possible to extract the bPa2 and bRa1.

(The original R223 problem had no wRg2, and it was possible to retract bKb2xRb1, .. bRa1-g1, bPf3xg2-g1=R, wRa1, wPh2-h3, bKc1-..-g2 etc.)

R224 - Andrei Frolkin
Retract -1. Rb7-c7 Qc7-b8 -2. Rb8xNb7 Na5-b7 -3. b7-b8=R+ Qb8-c7 -4. e4-e5 Bc7-d8 -5. Rd8xNd7 Ne5-d7 -6. d7-d8=R+ Bd8-c7 -7. e3-e4 Kc7-c6 -8. Kb5-a6 etc.

The Problemist, May 1994

R225 - Sergei Tkachenko
1. Nc3 d5 2. Ne4 dxe4 3. Nf3 exf3 4. Rb1!! fxg2 5. Ra1 gxf1=N 6. Rb1 Ne3 7. Ra1 Nxd1 8. O-O Kd7 9. Rxd1

R226 - Aleksandr Kislyak
1. c4 b6 2. c5 Ba6 3. c6 Nxc6 4. Qb3 Qb8 5. Qxf7 Kd8 6. Qxg7 Kc8 7. Qxe7 Kb7 8. Qe4 Ba3 9. Nxa3 Ne7 10. Nc4 Rg8 11. a4 Rxg2 12. Ra3 Rxg1 13. Rb3 Rxh1 14. Qxh1 Nc8 15. Na5 #

The Problemist, Jul. 1994

R227 - Ian Gent
(a) If only Black's O-O is legal, then 1. Nf5! (threat 2. Rxa8; Rxh8) O-O 2. Ne7#.
(b) If only Black's O-O-O is legal, then 1. d6! O-O-O 2. Ra8#.
(c) If both are legal, then the last move was 0 ... b7-b5, and then 1. cxb6 ep! O-O-O 2. Ra8#.

R228 - Viktory Kievsky
All 4 present Bs from promotion. Inventory shows BPa7 could not capture on its way to a1=B. So that WPa2 or WPc2 required two captures before promoting on a8. Second Bl. promotion was on d1 or f1, so that WBf1 was captured by BPe7. Inventory shows WPh2 could not capture on its way to h8=B, that WPg6 is from f2, that WPe5 is from e2.

Last move was 0 ... f7-f5, not h5xg4??, not Kf7-e6 (h5xg6??), not Kd6-e6 (d4xe5??).

But 1. exf6 ep? (2. Qxg4) fails on 1 ... g3! (a vicious trap ...). Working solution is 1. Qb3+! d5 (or Kxe7) 2. e5xd6 ep (or Qf7) #

The Problemist, Sep. 1994

F1489R - Juraj Lörinc
1. d3 e5 2. Kd2 Be7 3. Ke3 Bh4 4. Kf4 Nf6 5. Kg5 g6 6. Kh6 O-O 7. Bg5 Re8 8. e3 Re7 9. Qf3 Ne8 10. Kxh7 d6 11. Kh8 Kh7 12. Qxf7 Bf5 13. f4 Re6 14. Qe7+ Ng7.

8 ... Re7 cannot be escaped, so that no proof game ends with White's 14th.

(b) Wh. mates with 1. Qxg7 # (not 1 ... Qg8).

R229 - Yury Lubkin
Sol 1: Add WKg5 and 1. Bd5#
Sol 2: Add WKg4. Then 1. Bd5? illegal because Wh. does not have the move ! Instead, 0 ... c6/e6/e2 1. Qe5/Nf6/Rxe2 #.

R229v - Yury Lubkin
Add WKh5. Then Wh. does not have the move ! Mate in 1 with 0 ... d6/f3/h3/h6 1. Be6/Rxf3/g4/Bg6 #.

R230 - Leonid Borodatov
All missing Wh. men (but WBf1) required for Bl. Ps captures. Retract -1. Nh3-f4+ Kf5-g6 -2. e7-e6+ Kg6xBf5! (-2 ... d3-d2?? leads to retropat) -3. Be6-f5+ Kf5-g6 -4. Bf7-e6 Kg6xSf5!! -5. Be6-f7+ d2-d3 and Bl. is now free to retract Pc6xd5 or Pe3xf2.

The Problemist, Nov. 1994

R231 - Maurice Jago
Author's intention: Black has the move and mates with 1 ... Qa6. White cannot mate by 1. Rxd7? because Black has no last move. Retracting -1. Kc7-c8 Ne8-d6+ -2. Kd8-c8 Nd6-b7+ -3. Pb7-b6+ leads to illegal position. -1 ... Ne8xN/R/Pd6+ leads to retropat. -1 ... Ne8xQd6+ to illegal position.

"Unfortunately" it is possible to retract -1 ... Ne8xBd6+ -2. Bc7-d6 leading to a perfectly legal position, so that really White may mate with 1. Rxd7.

R232 - Jasper van Atten
Author's intention: 1. e4 h6 2. e5 Rh7 3. e6 fxe6 4. Bd3 Kf7 5. Bg6 Kf6 6. d3 Ke5 7. Qh5 Kd6 8. Ke2 Qe8 9. Kf3 Qf7 10. Kg4 Qf6 11. Nf3 Qe5 12. Rf1 Qxh2 13. Bf4 Kc6 14. Nbd2 Qg1 15. Bh2 Qh1 16. Bg1 Qh2 17. Rae1 Qe5 18. Qh2 Qf6 with 10 moves by the BQ, including a tempo loss by triangulation.

Unfortunately cooked by 1. e4 Nf6 2. e5 Ng4 3. e6 fxe6 4. Bd3 Kf7 5. Bg6 Kf6 6. d3 Ke5 7. Bf4 Kd5 8. Nd2 Nxh2 9. Qh5 Kc6 10. Ngf3 Ng4 11. O-O Nh6 12. Kh2 Qe8 13. Kh3 Qf7 14. Bh2 Qf4 15. Rae1 Ng8 16. Bg1 Qf6 17. Kg4 h6 18. Qh2 Rh7.

The Problemist, Jan. 1995

R233 - Peter Wong
1. h3! Nc6 2. h4 Ne5 3. h5 Ng6 4. hxg6 f6! 5. gxh7 f5 6. hxg8=B Rh3 7. Bd5 Rb3 8. axb3 f4 9. Ra6 f3 10. Re6 a6! 11. exf3 a5 12. Bb5 a4 13. Bbc6 bxc6 14. Ke2 Bb7! 15. Kd3 Ba6 16. Kc3 Be2 17. Nxe2 Qb8 18. Rg1 Qb4 19. Kxb4 O-O-O 20. Ka3 Kb7! 21. Ka2 Kb8 22. Ka1

Wonderful five tempo-losing maneuvers in a dual-free SPG.

R234 - Unto Heinonen
1. b4 e5 2. b5 Qh4 3. b6 Nf6 4. bxc7 b5 5. Nc3 b4 6. Nd5 b3 7. c3 b2 8. Qa4 bxc1=B 9. Qxh4 e4 10. d4 Bh6 11. e3 g5 12. O-O-O Ba3 13. Kb1 Bc1 14. a3 d6 15. Ka2 Bf5 16. c8=B Bg6 17. Bh3 Nbd7 18. g4 O-O-O 19. Ba6 Kb8 20. Ne2 Rc8 21. Rhe1 Rc6 22. Bf1 Rhc8 23. h3 R8c7 24. Bc8 Bf8

2 interchanges between original and promoted Bs.

The Problemist, Mar. 1995

R235 - Frank Christiaans
NW corner: Add wBb7, bRc7, bB a7, bBa8, bBc8, bPb6, bPc6, bPd7.
Alas cooked by adding wBb5, bRa7, bBa4, bBc6, bBc8, bPa5, bPb6, bPc4

SE corner: Add wBb3, bRc6, bBa2, BBc2, bBc7, bPa4, bPb6, bPd6.
Alas cooked e.g. by adding wBa4, bRd4, bBb3, bBb4, bBe8, bPb6, bPb7, bPd6. Other cooks exist.

R236 - Nikita Plaksin and A. Zolotarev
Wh. caps are axb, bxc, bxc, cxd, e2xd3, f2xe3 and last move -1. Qa6xb7#. Bl. caps are bxa and gxf & hxg for promotion of bPf7, g7 & h7.

Retract -1. Qa6xNb7 Nd8/6-b7 -2. Bb7-c8+ N~ -3. c4xQd5 Qg2/h1-d5 -4. b3xBc4 Qg1-g2/h1 -5. a2xBb3 g2-g1=Q -6. g3-g4 h3xNg2 and everything unlocks by unpromoting the 2 bBs on f1, the bN on e1, so that f2xBe3 and e2xd3 can be retracted. Indeed, the opening could have been 1. b2-b4.

4 Ceriani-Frolkin promotions with Q, B, B, N. Unfortunately, it is also possible to retract -3. c4xRd5 Rf5-d5 -4. e2xRd3 Rd6-d3+ -5. Kd3-c3 Nd5-b4++ with only three promotions.

The Problemist, May 1995

R237 - Sergei Tkachenko
The dualistic mates 1. Rce3/Rf4/Nf6/Ng5 #?? are illegal: Black has no possible last move.

Really, White played last and Black has the move. 0 ... Rxd4/Rxc5/Rxb4/Rxc3 is followed by 1. Rce3/Rf4/Nf6/Ng5 #.

R238v - Andrei Frolkin
1. f4 Nc6 2. f5 Ne5 3. f6 Ng4 4. fxe7 f5 5. Nh3 Kf7 6. e8=Q Kf6 7. Qe3 Nf2 8. Qxa7 b6 9. Qb7 Ra3 10. Rg1 Rf3 11. a4 Bc5 12. a5 d6 13. a6 Bd7 14. a7 Bb5 15. a8=Q Qd7 16. Qe8 Nxd1 17. Ra8 Bxg1 18. Rc8 c5 19. Qa8 with place exchange between the 2 promoted wQs.

The Problemist, Jul. 1995

F1557R - Peter Fayers and Cedric Lytton
The bEb5 is a promotee (Equihoppers only reach 1/4 of the board). Last move is -1. bRd7xd8+. Other Bl. caps are axbxc, f7xe6 & g3xRh2 for promotion of bPg7 into bEh1, now on b5. Assume -1. Rd7x(Q/R)d8+?? What can we retract now? Not h2-h3, not any E move, not d3xBe4 and not d3xEe4 because this bEe4 gives check and can only retract Ea4-e4+ without capture so that Wh. is then retropat. Thus -1. bRd7xEd8+ allowing the retraction of Eb6/f6/f8xd8 now or one move before.

But this wEd8 is another promotee, through a2...a7xb8=E, so that all Wh. caps are accounted for. Finally, we can precise all capture squares: wKRh1 on h2, wKEg1 on c5, wQBc1 on b6, wQd1 on e6, bKBf8 on b8, bKEg8 on e4, bQEb8 on d8, and a promoted wPa2 (=wEb8) on d8.

R239 - Yuri Lubkin
(a) + wKa6 & 1. Qf8#
+ wKa4 & 1. Qg1#
+ wKa5 & Bl. has the move: 0 ... c3 1. Qc8#

(b) + wKa6 & 1. Qa3#
+ wKa4 & 1. Qe3#
+ wKa5 & Bl. has the move: 0 ... c3 1. Qxc3#

R240 - Maurice Jago
One bR is a promotee. Wh. caps are cxd, dxe and exf. For balance, the bPh7 had to promote on g1 after h3xg2. Other Bl. cap is axb. Then the wPa2 promoted on a8. Only a bR could have been promoted on g1, so that bPf7 was captured on f3 or f4.

The cage unlocks by retracting e3-e4. E.g. -1. Rb1-b2 Bb2-c1+ -2. Re1-b1 Nb1-d2 -3. Qc1-e3 e3-e4 -4. Re4-d4 d2x(B/N)e3. White has the move and mate with 1. Rxa4#.

It is interesting to observe that Bl. cannot be given the move. Try -1. Nb1-d2 Ka3-a2 -2. Nd2-b1+ Ra2-b2 -3. Rb2-c2+ Nc2-a1 -4.?? Another try is -1. Nb1-d2 Ka3-a2 -2. Nd2-b1+ Rb1-b2 -3. Rb2-c2+ Nc2-a1 -4.??


The Problemist, Sep. 1995

R241 - Frank Christiaans
Suppose black can castle. The white bishop from f1 was captured on c6. Before this capture, the rook from h8 was captured on f3. In order to let this rook out of the northeast corner, f5 had to be played. Also e6 or g6 had to be played, either to let out the bishop from f8, or to let the rook out via Rh8-h7-f7-f5. But this means that there's a 'screen' from pawns on e2, f3, f5, e6/g6, one the bishop can't get past unless he gives check. (Actually, g6 has to be free in order to let the bishop pass, so e6 is occupied). So black can't castle.
1. Qg7! (2. Qg8#) Nh7/Ne7/Kd8 2. Qf7/Qe7/Qd7#

R242 - Frank Christiaans
(a)A rook was captured on h6. That rook couldn't have escaped the white pawn-barrier until g2xf3 was played. So the rook had to leave h8 via f8, and hence e6 was already played to let out the bishop from f8. So black can't castle anymore, since the bishop from f1 had to go to c6 via e8. So #3 with 1. Qg7!
(b)Black did visibly 23 moves (Nb8: 1, Bc8: 1, Qd8: 2, Ke8: 2, Bf8: 2, Ng8: 3, Rh8: 5, a7: 1, b7: 2, d7: 1, e7: 1, f7: 1, g7: 1), and the only piece that can visit c1 in an extra 4 moves is the black queen (d8-h4-e1-c1-c2-a4-a7). So Qd8 captured the bishop on c1. A possible proof game is: 1. Nc3 e6 2. Nd5 Qh4 3. c3 Bc5 4. Qa4 Nh6 5. Qd4 f5 6. Nf3 Rf8 7. Ne7 Rf6 8. a3 Rg6 9. Qf6 Rg3 10. Ne5 Rf3 11. gf3 Be3 12. Bh3 d6 13. OO Kd8 14. fe3 Qe1 15. Bg4 Qc1 16. Bh5 Qc2 17. Be8 Nf7 18. Bc6 bc6 19. Kh1 c5 20. Rg1 Ke8 21. Rg4 Qa4 22. Rb4 Nc6 23. Rb6 ab6 24. Rg1 Qa7 25. Rg6 Ng5 26. Rh6 gh6 27. Ng8 Ba6

The Problemist, Nov. 1995

R243 - Andrey Frolkin
Bl. caps: hxg, gxf, requiring the promotion of wPa2 and wPb2. Wh. caps: hxg, gxf, fxe and axb for promotions on b8. The N-E cage unlocks with the retraction of g6xf5, but this requires some more room. Getting free room will temporarily freeze the bBe8 so that we need five black tempis. Here is the way to obtain them:

Retract -1. Bg8-f7 Bf7-e8 -2-6. Nb8-f1 ~ -7. b7-b8=N ~ -8-12. Nb8-h1 ~ -13 b6-b7 ~ -14 b5-b6 ~ -15. b2-b4! ~ -16-21. b3-b8=N Bf7-e8 -22. a2xPb3 Be8-f7 -23. Bf7-g8 a5-a4 -24. Qg8-h8 a6-a5 -25. Kh8-g7 a7-a6 -26. Bg7-f8 b4-b3 -27. Qf8-f8 b5-b4 -28. Bg8-f7 Bf7-e8 -29. Qa8-f8 and everything unlocks.

(a) Bl. has the move 0 ... Bxg5/Rxg5/~ 1. Rgxh6/Rxg5/Rhxh6 #
(b) First move of wPb2 was b2-b4 because parity arguments (and time pressure) forbid b2-b3.

R244v - Unto Heinonen
1. h4 a5 2. h5 a4 3. h6 a3 4. hxg7 axb2 5. Rh6 bxa1=R 6. Rc6 d6 7. a4 Bg4 8. a5 e6 9. a6 Qf6 10. axb7 Qf3 11. exf3 h5 12. Bd3 h4 13. Bf5 h3 14. d3 h2 15. Be3 h1=R 16. Nd2 R1h7 17. Ne2 Nh6 18. g8=N Bg7 19. Ne7 Bb2 20. Nd4 Kf8 21. Ke2 Kg7 22. Qh1 Kf6 23. Qh3 Rh1 24. Nf1 Ra1 25. Nh2 Na6 26. b8=B Bc1 27. Ba7 Ra8 28. Bc5 Rhh8

Wonderful rotation of 4 bl. Rooks on their home squares, without castling.

The Problemist, Jan. 1996

F1596R - Nikita Plaksin
Author's intention was -1. Pc7xQd8=N+ [Nd8->g1] Pg2xNh1=Q [Qh1->d8] but, unfortunately, there are duals e.g. -1 ... Qd8xNd1 [Qd1->d8] -2. Qc1/2/3-c6+ Kd7-c8 -3. Qd2-c1/2/3+ etc.

R245 - Yury Arefiev
Bl. only capture is g7xQf6. Wh. caps are b6xRa7, axb, bxc & cxd. Possibly, bPc7 promoted to c1. We can't retract g7xf6 or b7-b6 without first reconducting bKR or bQB.

(a) Add bQh2 and unlock with -1. Rd2-d1+ Qg1-h2 -2. h2-h3 Qc1-g1 -3. Rd1-d2 etc. One bQ is bPc7 promoted.
(b) Add bRg4 and unlock with -1. Rd2-d1+ Rg8-g4 -2. h2-h3 g7xQf6 etc. (Qf6 go to a3).
(c) Add bNf5 and unlock with -1. Rd2-d1+ Nd6-f5 -2. h2-h3 Nb5-d6 -3. Ka4-b4 Na3-b5+ etc.
(d) Add bBd5 and unlock with -1. Rd2-d1+ Bb7-d5 -2. h2-h3 Bc8-b7 -3. Kc5-b4 b7-b6+ etc.

R246 - Tim Sparrow
A shortest proof game is: 1. e4 d5 2. Qg4 Bxg4 3. a4 e6 4. Na3 Bxa3 5. h4 Qd6 6. Rh3 Nd7 7. Rhxa3 O-O-O 8. d3 g5 9. Bxg5 b6 10. Be2 Kb7 11. O-O-O Bxe2 12. Bxd8 Bxd1 13. Bxc7 Ne7 14. Ra1 Ra8.

There are some possible changes in move ordering etc., but the bBd1 did Bc8xQg4xBe2xQRd1.

The Problemist, Mar. 1996

R247 - Yury Lubkin
(1) Add wKc3 for 1. cb7#

(2) Add wKc4. Now Bl. has the move, so that mate in 1 is with 0 ... ba6/bc6/b6/b5+ 1. c7/Bxc6/ab6/ab6ep#

R248 - Per Olin
Authors' intention: 1. d4 e6 2. Bf4 Ne7 3. Bxc7 Nec6 4. Bxb8 Nxb8 5. Nf3 Qa5+ 6. Nfd2 Qxa2 7. Rg1 Qxa1 8. Rh1 Qxb1 9. Nxb1. This is the position (with Wh having lost the right to castle) which is repeated after 9 ... move 10. move return 11. return, and again after 11 ... move 12. Kd2 return (diagram position) and now 13. Ke1.

The idea is that 7. Rg1 ... 8. Rh1 is mandatory because if Wh. still have castling rights after 8.5 moves, then the repeated positions will not be exactly the same and he will not be able to claim draw by repetition.

However there are cooks, e.g. 1. d4 Nh6 2. Bxh6 e6 3. Bf4 Qg5 4. Kd2 Qa5 5. Kc1 Qxa2 6. Bxc7 Qxa1 7. Ba5 Qxa5 8. Na3 Qd2 9. Kxd2 so that play may now proceed e.g. 9 ... Be7 10. Nb1 Bf8 11. Na3 Be7 12. Nb1 Bf8 (diagram position) 13. Na3 and draw.

The Problemist, May 1996

R249 - N. Nagnibida
Bl. captures have been d7xc6xb5xa4 so that he has no last move: a promotion into a Nd1 would be one capture too many & the bK is retro-pat. Hence Bl. has the move and mate in 1 by 1. Qb4/Qf4/Nf5/Ne2?? is illegal.

Rather 0 ... Nxc2/Nxd3/Nxe3/Nxf2 1. Ne2/Qb4/Qf4/Nf5#

R250 - Pavel Kamenik
1. c4 Nf6 2. c5 Nd5 3. c6 f6 4. cxb7 c6 5. bxc8=N Qa5 6. Nxe7 Kd8 7. Ng6 hxg6 8. Qb3 Rh4 9. Qb7 Ra4 10. Qxa8 Kc7 11. Kd1 Kb6 12. Kc2 Kb5 13. Kd3 Nb4 14. Ke4 N4a6

The Problemist, Jul. 1996

R251 - Frank Christiaans
Analysis: The SE cage required wPh2-h3, wPg3xh2xg1=R, bNf3-g1 (while wK is on d1) & wPf2-f3 before the wK could get out. Thus the wQ was captured at home by a bN. The other missing bP could not promote and the wPa2 had not enough captures available to promote on d8. (So that necessarily, the g3xh2xg1 capturees are the 2 wNs.) The wPa2 could reach c6 and get captured there if it captured bQ and bR, but then this was before d7xc6 let the bQB out, so that this would require capture of the bKR and entail illegality of Bl. O-O. Thus, if O-O is legal then the wQR was captured on c6.

(a) Retract -1. Rh5-h8 !! & forward 1. Re5#. The tries -1. Rh5xNh8?? or -1. Rh5xBh8?? are illegal.

(b) Retract -1. Na2xRb4!!. Now Bl. cannot O-O, so that forward 1. Rxb7 ~ 2. Rb8# cannot be defended by 1 ... O-O (nor by a check to the wK). The tries -1. Bf8xQg7?? or -1. Bh6xQg7?? are illegal.

R252 - A. Frolkin and S. Tkachenko
1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 hxg2 5. bxc8=B gxh1=B 6. Ba6 Bb7 7. Bg2 Bc8 8. Bgb7 Rh4 9. f3 Re4 10. h4 Nh6 11. h5 Nf5 12. h6 Nd4 13. h7 f5 14. h8=R Kf7 15. Rxf8 Ke6 16. Rh8 Qe8 17. Rh1

The Problemist, Sep. 1996

R254 - Yury Arefiev
Yes, the position is legal. Rhe retro-play is: 1. .. Rc3?#8212;Nc2 2. Ne3-c2 Bc2-d1 3. Qd1-e1 Kg1-h2 4. Nf1-e3 Kh1-g1 5. h2?#8212;Ng3 Kg1-h1 6. Qe1-d1 ... 9. Qd1-e1 Ne1-f3 10. Ne3-f1 Kg1-h1 ... 13. Nh3-f4 Nf3-e1 14. Ng1-h3 ... 17. Qe1-d1 Nf1-~ 18. Nh3-g1 Kg1-h1 19. Nf4-h3 ... 22. Na8-b6 Kg1-h1 23. a7-a8=N ... 26. a4-a5 a5?#8212;Bb4 27. Qd1-e1 a6-a5. .. and the position unlocks.

The Problemist, Nov. 1996

R256 - Cedric Lytton
White's only missing piece was captured on c1. So black has no last move. #1 with 0. .. Be5/Bd6/Bc7 1. Rf8/Rc8/Bf8#

The Problemist, Jan. 1997

R258 - Stanislav Vokal
(a)To get the king past the rook, black had to castle. So the queen got captured on d8, and the queen on h8 is a promoted one. If white still can castle, black would need 7 captures to promote the pawns to a bishop and queen. This includes the a-pawn, but that pawn can capture only once, and none of the seven captures towards promotion is on the a- or b-file. So white may not castle.
1. Kd2! (2. Qa7 Ka7 3. Ra1#) Bb1/Bc2 2.Bb7 Rb7 3. Qb7#
(b) The promotions now can take both place on g1, so white may castle.
1. OO! and variants are as in (a).

The Problemist, Mar. 1997

R260 - Philippe Schnoebelen
(a)Both the white bishop and one of the knights are promotees. In order to preserve as much castle-rights as possible, the white h- or f-pawn must've captured [h/f]xg-g6xh7xg8=B, and b6xa7-a8=N. So black may not castle queenside, a fact which is indicated too by the position of the white men!
(b)h#1.5 with 1. ... Nh5(Nd5?) 2. OO Qg7#.

The Problemist, May 1997

R262 - Denis Saunders
(a) Rotation of 90 degrees clockwise results in a #1 with 1. b8=N#. Rotation of 180 degrees clockwise results in a #1 with 1. Rf5#. Rotation of 270 degrees clockwise results in an illegal position (Ba1 is promoted, and the pawn structure requires too many captures.
(b) Now only the 180 degrees clockwise rotation gives a legal position. #1 with 1. Rf5#.

The Problemist, July 1997

R265 - Peter Wong
1. c3 Na6 2. Qa4 Rb8 3. Qc6 dc6 4. h3 Bg4 5. hg4 Rc8 6. Rh6 Ra8 7. Rd6 ed6 8. f3 Be7 9. Kf2 Bf6 10. Ke3 Bd4 11. Ke4 Nf6 12. Kd3 OO 13. Kc2 Re8 14. Kd1 Re3 15. Ke1 Rd3 16. e4

The Problemist, Sep. 1997

R266 - Alexandr Kisliak
White's caps are axb, d2xc3 and e/gxf. Black's caps include axb and hxg. bPe7 was promoted. Retract
-1. Qd8-e7# f2-f3
-2. a4xNb3 Nd4-b3
-3. Na7-b5 Nb5-d4+
-4. Nc6-a7 f3-f4
-5. Ne5-c6 Re7-d7
-6. Nd7-e5+ Re8-e7
-7. Ne5xQd7 Re7-e8
-8. Nd3-e5 Re8-e7
-9. Ne1-d3 Re7-e8
-10. e2-e1=N Re8-e7
-11. e3-e2 e2xRf3
-12. Rf5-f3 Re7-e8
-13. Rd5-f5 Re8-e7
-14. Rc5-d5 Qc6-d7
-15. Rf7-d7+ Re7-e8
-16. Qe8-d8 Kf7-e6
-17. Qd8-e8+ Ne6-f8 etc.
and everything unlocks easily.

R267 - Unto Heinonen
1. b4 a6 2. b5 axb5 3. Bb2 Rxa2 4. Bc3 Rxc2 5. Ra3 Rxd2 6. Nxd2 c6 7. Nb3 Qb6 8. Qd3 Qxf2+ 9. Kd2 Qxg2 10. Ke3 Qxh2 11. Bg2 h5 12. Bf3 Qxe2+ 13. Nxe2 h4 14. Ng3 h3 15. Rxh3 c5

The Problemist, Nov. 1997

R268 - Miroslav Brada
Black has no last move in both cases.
(a) (1. Nb3 Ke4 2. Bg6#) 0. .. Bc5 1. Nf3 Ke4/Kc4 2. Nd2#; 0. .. Be5 1. Qd7 Kc4/Kc5 3. Qd5#
(b) (1. Nf3 Kc4 2. Qa6#) 0. .. Bc5 1. Qd7 Ke4/e5 2. Qd5#; 0. .. Be5 1. Nb3 Ke4/c4 2. Nd2#

R269 - Andrey Frolkin and Sergey Tkachenko
1. c4 d5 2. c5 Kd7 3. c6+ Ke6 4. cxb7 Nc6 5. b8=Q Ba6 6. Qa4 Bc4 7. Qbb3 Rb8 8. Kd1 Rb6 9. Kc2 Nd4+ 10. Kc3 Rc6 11. Qd1 Nb5+ 12. Kc2 d4 13. Qe1 d3+ 14. Kd1 Qd4 15. Qc2 Qc3 16. a4 dxc2#

The Problemist, Jan. 1998

R270 - Yuri Lubkin

The Problemist, Mar. 1998

R272 - N. Nagnibida
Black has no legal last move (f7xg8=B Kd5-c6 requires 11 captures, one too many). So the solution is:
1. ... e5/e6/ed6/ef6/gf6 2. Na7/Qf3/Qc3/Nd4/Nd4#

R273 - Stanislav Vokal
The white pawn structure takes 8 captures, and the rook originally from h8 was captured in his corner. So black's last move couldn't have been d7-d6. So black has no last move. Solution:
1. .. e5 2. e4 ~ 3. Ra2 ~ 4. Ra8 ~ 5. Rf8#
1. .. e6 2. e3 ~ 3. Ra2 ~ 4. Ra8 ~ 5. Rf8#

F1772R - Mario Velucchi
1. e4 f5 2. Ba6 b5 3. Bc8 fe4 4. f4 ef3 5. Ba6 fg2 6. Bb5 gh1=B 7. Ba4 Bc6 8. Qh5

C9206 - Denis Saunders
(a) Ba7 is clearly promoted. Suppose white can still castle. Then the promotion to bishop took place on g1, after black captured fxgxh-h2xg1. Together with d7xc6, these are all black captures. The white e-pawn is one of the captured pieces. However, this pawn could not capture to get past the black e7-pawn. This is a contradiction, since this pawn needs to promote to get captured. So white may not castle.
1. Rd1! [OOO?] (2. Qc4#) Ne3/Ne5/cb5/Rf4/Qc2/Qd3/Qe4 2. Qe3/Qe5/Rb5/Qe7/Rc2/Nd3/Ne4#.
(b) Now the e-pawn could have promoted, and then get captured by the black f-pawn. So white may castle.
1. Rd1? Re8! 1. OOO! (2. Qc4#) and the same variations as in the previous case.

The Problemist, May 1998

R274 - Kivanc Cefle
Suppose black can still castle.The white queen was captured on d1, so the other two missing pieces were captured on b6 and d6. One of the missing pieces is the pawn from h2. This pawn can't reach d6 or b6 without promotion, so it promoted on g8. Because it has to capture 3 pieces on its way to g8, the piece it promoted to must still be on the board, since both the captures on d6 and b6 are needed to free the pieces the pawn has to capture on its way to g8. The only pieces white can promote to are a knight or a rook, but both pieces can't escape g8 without breaking black's castle rights (h6 is already on its place, to let the pawn pass h7). So black may not castle. #2 with 1. ab8=Q.

R275 - ?#8364;rpàd Molnàr
The last moves were: -1. .. a4xNb3 -2. Qd8-b6 b6xNa5 -3. Rc8-c6 c7xNb6 -4. Bc6-b5 b5xNa4 -5. Na8-b6 b6-b5 -6. Bb7-c6 Kb5-b4 -7. c3-c4 Bb4-a3, and the position unlocks. White unpromotes four knights on e8.

PS755 - Gianni Donati
1. h4 h5 2. Rh3 Rh6 3. Rg3 Rc6 4. Rg5 Rc3 5. dc3 d5 6. Bf4 d4 7. e3 d3 8. Qh5 d2 9. Ke2 d1=R 10. g3 Rd6 11. Bg2 Rh6 12. Bh1 Rh8 13. Qh8

The Problemist, July 1998

R276 - Charles Frankiss
1. b4 g5 2. b5 Bg7 3. b6 Be5 4. ba7 Bh2 5. ab8=Q Bg1 6. Rh7 Ra2 7. Rh3 Rh4 8. Ra3 Ra4 9. f3 Ba7 10. Qa8 b6 11. Qd5 Ba6 12. Qa5 Qa8

R277 - Michel Caillaud & Graham Lee
1. f4 Nc6 2. Kf2 Nd4 3. Kg3 Ne2 4. Kg4 Ng3 5. Ne2 Nh5 6. Ng3 Rb8 7. Be2 Ra8 8. Qg1 Rb8 9. Qa7 Ra8 10. Qb8 Ra5 11. d4 Rg5 12. fg5 Nh6 13. gh6 Rg8 14. Bf4 Rh8 15. Nd2 Rg8 16. Re1 Rh8 17. Bd1 d6 18. Re6 Kd7 19. Rd6 Ke8 20. Rd7

The Problemist, Sept. 1998

R278 - Mikhail Kozulya
1. e4 Nf6 2. e5 Ne4 3. h4 Nd2 4. Kd2 a5 5. Kd3 a4 6. Bf4 a3 7. Bh2 ab2 8. a4 b5 9. a5 b4 10. a6 b3 11. a7 bc2 12. ab8=Q Ra7 13. Qb6 Rb7 14. Qh6 e6 15. Ra8 Qg5 16. Na3 Qc1 17. f4 b1=Q 18. Nf3 cd1=Q

The Problemist, Nov. 1998

R281 - Unto Heinonen
1. Nf3 e5 2. Rg1 Ba3 3. ba3 Qh4 4. Bb2 Qh3 5. gh3 f5 6. Rg5 f4 7. Rf5 g5 8. Bg2 g4 9. Ng5 h5 10. Bd5 e4 11. Be5 Nh6 12. Nc3 Rf8 13. Bg8 d5 14. Rb1 d4 15. Nd5 a5 16. c3 Ra6 17. Qa4 Rc6 18. Rb6 h4 19. Ra6 b5 20. Kd1 b4 21. Qb5 a4 22. Qb8 Rc3 23. Qa8 c5 24. Bc7 c4

The Problemist, Mar. 2000

R296 - Thierry le Gleuher
1. Nc3 Nf6 2. Nd5 Rg8 3. Ne7 Nd5 4. Ng6 Bc5 5. e4 d6 6. Ne2 Kd7 7. Nc3 Kc6 8. Nb1 Nd7 9. Bb5 Kb5 10. e5 c6 11. e6 Qa5 12. e7 b6 13. e8=N Ba6 14. Nf6 Rae8 15. Ne4 Re5 16. Nf8 Rh5 17. Ng3 g5 18. Ne2 Rg7 19. Ng1
The knight on f8 is the knight originally from b1, the knight on b1 is the one originally from g1, and the one on g1 is the promoted knight!

R297 - Alexandr Yaroch
White captures are the whitesquared bishop on c8, and h5xg6. The black captures are the pawn on the b-line, and e7xPd6. The black h-pawn promoted on h1 without capturing anything. The position unlocks when white can uncapture h5xg6. For this, a knight has to unpromote on h1, and the pawn needs to get retracted to h6.
The last moves were: -1. .. Rh1-h6 -2. a4-a5 Rb1-h1 -3. a3-a4 Rb6-b1 -4. a2-a3 Rb1xPb6 -5. b5-b6 Nh1-g3 -6. b4-b5 h2-h1=N -7. b3-b4 h3-h2 -8. b2-b3 h4-h3 -9. e4-e5 h5-h4 -10. e3-e4 h6-h5 -11. h5xNg6 and the position unlocks.
So white has the move. Mate in 1 with 1. Ne6!# and not with 1. .. Qe5?