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Probleemblad

Solutions

Probleemblad, Nov. 1993

1 - F. Christiaans and H. Boumeester
1. e4 d5 2. Be2 Bd7 3. Bg4 Bb5 4. Bc8 Bf1 5. d3 g5 6. Bxg5 Bh6 7. Bxe7 Bc1 8. Bf8

Probleemblad, Sep. 1994

1 - F. Christiaans
(1) add bRe7, bBe8, bPc7d6f7,
(2) add bRf7, bBf8, bPe7f6g7,
(3) add bRf4, bBe3, bPd2e4f2.

Three different kinds of illegality.

2 - U. Heinonen and O. Heimo
1. h3! Na6 2. h4 Nc5 3. h5 Ne4 4. h6 c5 5. hxg7 Nh6 6. g8=N Bg7 7. b3 Bxa1 8. Na3 Be5 9. Nb5 Bc7 10. Nxa7 e5 11. Nc6! Ra5 12. Nge7 Rb5 13. Nd5 O-O 14. Nce7+! Kh8 15. Ng8 Qe7 16. Nc3 Rd8 17. Nb1 Qf8

Nice tempo move, and white knights b1 and g8 paradoxically switch places.

Probleemblad, Nov. 1994

1 - D. Wissmann
1. d4 Nh6 2. Bxh6(bBh6) g5 3. d5 Bfg7 4. Qd4 O-O 5. Qxg7(bQg7) Qxb2(wQb2) 6. Nc3 Kh8 7. O-O-O Rg8 8. Rd3 Rg6 9. Rh3 Ra6 10. Nb1+ e5 11. dxe6ep(bPe6)+ Qf6 12. c4 g4+ 13. f4 gxf3ep(wPf3)+

Two castlings and two en passant captures.

Probleemblad, Jan. 1995

1 - S. Tkatsjenko
1. d3 h6 2. Bxh6 a6 3. Be3 Rh3 4. Ba7 Rf3 5. gxf3 b6 6. Bh3 Bb7 7. Bf5 Be4 8. Bh7 c6 9. Bxg8 Bh7 10. e4 g6 11. Ne2 Bh6 12. Nd4 Bd2+ 13. Ke2 Ba5 14. b4

Probleemblad, May 1995

1 - F. Christiaans
1. c4 d5 2. cxd5 Nc6 3. dxc6 Qxd2 4. Bxd2 Bf5 5. Bb4 Bxb1 6. Qxb1 O-O-O 7. Ba3 Re8 8. Qf5 Kd8 9. O-O-O #

Probleemblad, Jul. 1995

1 - Fernand Joseph
One wB, one bB captured at home. bPc5 came from c7 with no capture because we cannot account for 2 black-squared captures of wh. units. So that the wPs required five (i.e. all available) captures to reach their current position. Then the other caps are bPaxBb (on white square) and bPg3xNh2 (followed by bPh2-h1=something) to explain how the missing bPs can enter the balance count.

Tries: +bKf4?? requires -1. Qxe4#, one capture too many. +bKa5?? requires -1. Rxa4#, one capture too many. +bKa6?? requires -1. Rxa4#, one capture too many, or -1. wPa5xb6# but then Bl. played bPa7xb6 on a black square.

Solution: +bKa1!! with (completely determined) last move -1. Na2-c1#.

Unfortunately, Frank Christiaans found that +bKa6! is possible, and is an ugly dual solution. Then the bPa7 captured a6xBb5 (on a white square) so that it can later be captured by wPc2. Last move is -1. a5xb6 where the captured unit is not determined, it can even be the promoted bPg7.

Probleemblad, Sep. 1995

1 - H. J. Vleugels
If Wh. can O-O, then his last move can only be e2-e4, and then en-passant capture d4xe3 ep is legal. This is a case where A-Posteriori validation is necessary if we want to use en-passant capture.

Solution: 1. de3 ep O-O 2. Be4 Kh2 3. Bg3 fg3 mate.

2 - Michel Caillaud
1. Nc3 c6 2. Nd5 Qc7 3. Nb4 Qg3 4. hg3 (+bQd8) d6 5. Rh6 Bh3 6. gh3 (+bBc8) Be6 7. Bg2 Bb3 8. ab3 (+bBc8) Qb6 9. Ra6 Qe3 10. de3 (+bQd8) Qa5 11. Bh1 Qa3 12. ba3 (+bQd8) Qd7 13. Bb2 Qxh3 (+wPh2) 14. Ba1 Bf5 15. Qxd6 (+bPd7) Bd3 16. ed3 (+bBc8) Qxg3 (+wPg2) 17. Ke2 Qg6 18. Kf3 Qxd3 (+wPd2) 19. Ne2 Qxb3 (+wPb2) 20. Ng3 Qxe3+ (+wPe2) 21. Kg4 Qxa3 (+wPa2) 22. Rxc6 (+bPc7) Qa5

A triple place-exchange between Pawns, not possible with orthodox chess.

Probleemblad, Nov. 1995

1 - Mario Velucchi
Both Ks are in check so that we may only add a P on one of the main diagonals. Adding Pe4/f3/g2 leaves no explanation for check to bKf6 (not -1. b2-b3??, not -1. Kd4-d5??) so that wKd5 is in check, through g2xh1=B+. This promotion used the missing bP so that only a wP can be added. Add wPd4 (b2? c3? e5? trivially impossible).

Last move is -1. g2xBh1=B because five missing wh. men have been taken on bl. squares through a7xb6 and e7xd6xc5xb4 and wBg8 is a promotee.

+wPd4 leaves a position requiring six captures by wh.: bBc8, exd, fxexd, and gxf, fxg for promotion in g8 around bPg7-g2.

Probleemblad, Jan. 1996

1 - Thomas Brand
1. a4 e5 2. Ra3 e4 3. Rb3 e3 4. Rb6 axb6 5. Nh3 Ra5 6. Nf4 Rf5 7. Nd5 Rf3 8. exf3 Bc5 9. Ke2 Ne7 10. Kd3 e2 11. a5 e1=R 12. a6 Re4 13. a7 Be3 14. a8=R c5 15. Ra1 Ra4 16. Ke2 Ra8 17. Ke1

Probleemblad, Mar. 1996

2964 - Sergei Tkatsjenko
1. c3 g6 2. Qb3 Bg7 3. Qb6 cxb6 4. Kd1 Qc7 5. Kc2 Kd8 6. Kd3 Qc4 7. Ke3 Qxa2 8. b3 Kc7 9. Ba3 Kc6 10. Bd6 Kb5 11. c4 Ka6 12. Nc3 Nc6 13. Nxa2 Rb8 14. Nb4#

Probleemblad, May 1996

3000 - Henrik Juel
1. c3 h5 2. Qc2 Rh6 3. Qh7 f5 4. Kd1 Kf7 5. Kc2 Ke6 6. Kd3 Kd5 7. Ke3 Kc4 8. a3 Kb3 9. Nh3 Kc2 10. Nf4 Kd1 11. h3 Ke1 12. Rh2 Kxf1 13. Kf3 Ke1 14. Kg3 Kd1 15. Rh1 Kc2 16. Kh2 Rd6 17. Kg1 Rd5 18. Kf1 d6 19. Ke1 Be6

A very neat round-trip of the wh. King, together with a long trip of his bl. companion.

Probleemblad, Jul. 1996

3035 - Juri Arefjev
Bl. caps are a7xb6, bxc and e-f cross-captures, Wh. caps are d6xc7, h6xg7. For balance, Wh. promoted his wPa2 on a8 and Bl. promoted his bPh7 on h1. Clearly the Rd7 is white, the Rf6 is black, and there only remains to see which one is white and which one is black among the remaining two rooks. Then we can unpromote the Rc3 and free the position.

If we assume the Rc3 is white and unpromote it on a8, we fail on a bl. retropat when we retract a7-a8=R because it blocks the Bb8 oscillations. Hence the Rc3 is black and the Rd5 is white. We can unpromote the Rc3 on h1, and after we retract h7-h5/6, forbidding further oscillations from the Bg8, Wh. retropat is avoided because h6xNg7 is now possible.

Mate in 1 is with 1. f8=N#.

3036 - Thomas Volet
Wh. caps are axb, dxe & exf. Bl. caps are g7xh6, c7xd6xe5, and Bl. did promote his bPa7 on a1. Right now, it is not possible to unlock the SE cage without reconducting one wB home. The wK entered behind the bPs through the b6 hole, so that it is not possible to retract c7xd6 before we extract the wK.

Position unlocks by retracting:
-1. d6xQe5 c5-c6
-2. Rc6-c7 Q.
-3. Ra6-c6 Q.
-4. Ra4-a6 Q.
-5. Re4-a4 Q.
-6. Re5-e4 Q.
-7. Rg5-e5 Qg6-..
-8. Rg4-g5 Qg8-g6
-9. Rg5-g4 Qg7-g8
-10. Rd5-g5 Qf8-g7+
-11. Rd1-d5 b4-b5
-12. Ra1-d1 b3-b4
-13. a2-a1=R c4-c5
-14. a3-a2 a2xNb3
-15. Nc5-b3 c3-c4
-16. Ne6-c5 c2-c3
-17. Ng7-e6 Qg8-f8
-18. a4-a3 Kf8-f7
-19. a5-a4 Bg6-e8
-20. Ne8-g7+ Kf7-f8
-21. Ng7-e8 Bd3-g6
-22. Re8-d8 Bf1-d3
-23. Rd8-e8 e2xNf3
and everything unlocks.

It is now possible to count the number of Knights moves, knowing that the wNs have been captured on h6 and d6, the 2nd bN on f3, and we reach a total of at least 18 knight moves.

3039v - Frank Christiaans
(a) wRg7 wRh7 - bRh8 bBg8 bPg6 bPh6
(b) I: wRf7 wRh6 - bRg6 bBf8 bPg5 bPg7; II: wRh7 wRh8 - bRg8 bBf8 bPg7 bPh6
(c) I: wRa1 wRb1 - bRc1 bBa8 bPc6 bPc7; II: wRc7 wRc8 - bRd8 bBa7 bPb6 bPc6; III wRa7 wRc8 - bRb7 bBd8 bPb6 bPc6

Probleemblad, Sep. 1996

3074 - Michel Caillaud
1. d3 f6 2. Bh6 Nxh6 3. e4 Ng4 4. Be2 Nxh2 5. Bg4 Nxg4 6. Rh6 Nxh6 7. f3 Ng8
1. h4 f6 2. h5 Kf7 3. h6 Kg6 4. Rh5 Kxh5 5. e4 Kxh6 6. Bc4 Kg6 7. Bf7 Kxf7 8. d3 Ke8 9. Bh6 Nxh6 10. f3 Ng8

3075 - Philippe Schnoebelen
1. a4 g5 2. a5 g4 3. a6 g3 4. axb7 a5 5. d4 Ra6 6. d5 Rh6 7. d6 Rh3 8. dxe7 h5 9. exf8=N Qh4 10. bxc8=B Qc4 11. Qd4 Ke7 12. Qh4 Kd6 13. Bg5 Ne7 14. e3 Nxc8 15. Bd8 Rxf8

Hyper-Frolkin theme where Frolkin promotees do not move at all between promotion and capture.

Probleemblad, Nov. 1996

3110 - Gianni Donati
1. e4 Nh6 2. Qe2 Rg8 3. Qa6 bxa6 4. Nc3 Bb7 5. Nd5 Qc8 6. Nf6+ Kd8 7. Nxg8 Nc6 8. Nf6 Ng8 9. Nd5 Ke8 10. Nc3 Qd8 11. Nb5 Bc8

3111 - Unto Heinonen
1. h4 a5 2. h5 a4 3. h6 Ra5 4. hxg7 h5 5. Rh3 h4 6. Ra3 h3 7. g4 h2 8. g5 h1=N 9. g6 Rg5 10. f4 Rg2 11. f5 Ng3 12. f6 Rh1 13. fxe7 f5 14. Nh3 Nf6 15. g8=R Na6 16. g7 Kf7 17. e8=R Bc5 18. Re5 Bg1 19. Ra5 b5 20. e4 Bb7 21. e5 Qa8 22. Rb8 Be4 23. Rb6 cxb6 24. g8=R bxa5 25. Rg4 fxg4 26. e6+ Kg6 27. e7 Nf5 28. e8=R Ne7 29. Rc8 Ne8 30. Rc6+ dxc6

Four Frolkin promotions of Wh. Rooks in a very precise game.

Probleemblad, Jan. 1997

3144 - Henrik Juel
1. a4 c6 2. a5 Qb6 3. a6 Qb2 4. ab7 Qc2 5. ba8=R Qb1 6. R8a7 Qa1 7. Ra1
1. a4 c6 2. a5 Qa5 3. c3 Qc3 4. Ra7 Qb2 5. Rb7 Qb1 6. Rb1 Ra1 7. Ra1

3145 - Stanislav Vokal
Try: retract -1. Qa3-a1 and play forward 1. Re6 [2. Qxe7#] but 1 ... O-O !!

Solution: retract -1. Qa3xBa1 !! Now Black's O-O is broken.

Indeed, bBa1 is bPf7 promoted after f7xe6xd5xc4xb3xPa2-a1=B. These (and wBc1 at home) are all 6 captures by Bl. So that the 2nd missing wP had to be captured one way or the other. Thus, beside e2xd3xc4 & bBc8 at home, Wh. caps are gxf for a promotion on f8, or gxf & fxe for a capture on e6. A promotion on f8 required visiting f7, disturbing the bKe8 and breaking the bl. O-O. A double capture gxfxe required extracting the bQd8 from the N cage, again disturbing the bK.

3146 - Tom Volet
Bl. caps are exf, fxg & d7xc6xb5. Wh. caps are f2xe3, hxgxf & axb, allowing the wPa2 to be captured on the b file.

Bl. is under a pressing retropat threat, so that we have no time for retracting c6xPb5 allowing the resurrection of Black's KB through a3xBb4. Rather we must retract
-1. Ke4-d5 c6xPb5
-2. a4xRb5
and now we must
(1) reconduct the bRb5 to a8, the bBa6 to c8,
(2) retract d7xQc6,
(3) reconduct the wQc6 to d1, the wK to e1,
(4) retract f2xBe3, reconduct the bBe3 to f8,
(5) retract e7xNf6
and the position unlocks.

The last five captures (and their order) are precisely determined in a very attractive position.

3147c - Tom Volet
Wh. caps are d2xe3, g2xf3, h2xg3xf4xe5xd6xc7 & bBc8 at home. Bl. caps are e7xd6, f7xe6 & axb. Necessarily, the bPc7 was captured at home by d6xc7.

The position unlocks by
-1. Bg2-h1 a5xBb4
-2. Bf1-g2 a6-a5
-3. g2xRf3 Rf7-f3
-4. Rh4-h7 Re7-f7
-5. Rd4-h4 Re8-e7
-6. Kh7-h8 Ng6-f8
-7. Rd1-d4 Nh4-g6
-8. Kg6-h7 Nf5-h4
-9. Kg5-g6 f7xQe6
and now we go on with
(1) reconducting wRd1 to a1, wBb4 to c1, wQe6 to d1, wK to e1,
(2) retracting d2xBe3 and reconducting bBe3 to c8
(3) retracting e7xNd6
and the position unlocks with d6xPc7 etc.

The last six captures (and their order) are precisely determined in a very attractive position.

Probleemblad, Mar. 1997

3179 - Frank Christiaans
Due to tempo reasons, black must have the move. The fastest way towards this castling is:
1. ... d4 2. e3 d3 3. e4 d2 4. e5 d1=B 5. e6 Bb3 6. e7 Bg8 7. hg8=B h4 8. Bc4 h3 9. Bf1 gf1=N 10. Kh1 Ne3 11. Kg1 Nd5 12. Kh1 Ne7 13. Kg1 OO.

3180 - Valeri Surkov
(a)1. a4 b5 2. ab5 Na6 3. Ra6 Bb7 4. Ra1 Ba6 5. ba6
(b)1. a3 b5 2. Ra2 b4 3. ab4 Ba6 4. Ra6 Nc5 5. Ra1 Na5 6. ba5

3181 - Gianni Donati
1. e3 h6 2. Qf3 Rh7 3. Qf6 ef6 4. d4 Bd6 5. d5 Bg3 6. d6 Ne7 7. dc7 Ng6 8. cb8=Q Qb6 9. Qd6 Qa6 10. Qd1
1. d4 h6 2. Bg5 Rh7 3. Bf6 ef6 4. d5 Ne7 5. d6 Ng6 6. dc7 Bd6 7. cb8=B Bg3 8. Bf4 Qb6 9. Bc1 Qa6 10. e3

3182 - Gianni Donati
1. h4 a5 2. Rh3 a4 3. Rb3 ab3 4. cb3 Ra4 5. Qc2 Rg4 6. Qc6 bc6 7. h5 Ba6 8. h6 Bd3 9. hg7 h5 10. gh8=R Bh6 11. Rh7 Bg5 12. Rf7 Nh6 13. Rf3 Bf5 14. Rh3 Kf7 15. Rh1
1. Nf3 a5 2. Nd4 a4 3. Nb3 ab3 4. cb3 Ra4 5. Qc2 Rg4 6. Qc6 bc6 7. h4 Ba6 8. h5 Bd3 9. h6 Bf5 10. hg7 h5 11. gh8=N Bh6 12. Nf7 Bg5 13. Ne5 Nh6 14. Nf3 Kf7 15. Ng1

3183 - Gianni Donati
1. Nc3 a5 2. Ne4 a4 3. Ng5 a3 4. Nh7 ab2 5. Nf6 ef6 6. a4 and now:

Probleemblad, May 1997

3217 - Tivadar Kardos and ?#8364;rpàd Molnàr
Two white pawns promoted. To do this as economical as possible, the a-pawn captured 3 times towards d7, white captured e6xd7 before black played e7-e5, and he captured d5xe6 after black played e7-e5. Together with the capture on the g-file, this accounts for all 6 missing black pieces. So black can't retract d7-d6 as last move, which means that white moved last.
1. ... d5/dc5/Qg7/Qh5/Qg6/Qh7/ab6/ba6/bc6/cb6/Nb6/Ne7/Nf6 2. f7/Rd8/Ng7/Rg8/Bg6/gh7/Qa8/Rb8/Qc8/Nd6/Nc7/Re7/Nf6#

13 different mating moves!

3218 - Henrik Juel
White already has 10 bishops, so the bishop that's gonna be added is black. The only squares where black would give mate are d5, e4, g2 or h1. The last three require a capture as last move.
All pawns promoted. To let the white and black pawns pass each other, at least 8 captures are necessary, one on every line. After these mandatory captures, an equal number of white pawns and black pawns promote to a white-squared bishop. The problem has 7 white white-squared bishops, and 5 black white-squared bishops. So 2 more captures are needed for that. So all ten captures were done by the pawns, so the bishops didn't capture. If the mandatory captures were done by 5 white pawns, and 3 black ones, the extra captures have to be done by black pawns to a black promotion square, and the same story goes for 3 white captures and 5 black ones. So neither Bxe4, Bxg2, g2xf1=B, or g2xh1=B could be the last move.
So the only possibility to add a legally mating bishop is on d5.

3219 - Gianni Donati
1. Nf3 a5 2. Nd4 a4 3. Nb3 ab3 4. g3 Ra4 5. Bh3 Rf4 6. Bc6 dc6 7. Kf1 Bh3 8. Kg1 Kd7 9. Qf1 Ke6 10. Qg2 Kf5 11. Qe4 Kg4 12. Qe7 Nd7 13. Qe4 Be7 14. Qg2 Bg5 15. Kf1 Ngf6 16. Ke1 Re8 17. Qe4 Nf8 18. Qe7 Qe7

Probleemblad, Sept. 1997

R001 - Mario Velucchi
1. h3 Nf6 2. h4 Ne4 3. h5 Nd2 4. h6 Nb1 5. Qd6 Nd2 6. Qe6 d6 7. Kd1 d5 8. Kd2 d4 9. Kd3 Qd5 10. Ke3 d3 11. Kf3 d2 12. Kg3 d1=N 13. Kh3 Nc3 14. Kh2 Ne4 15. Nh3 Nf6 16. Rg1 Ng8 17. Kh1

R002 - Evgeni Markov
Last moves were: -1. ... c7-c8=N -2. Kc6xNd6 c5xb6 ep -3. b7-b5 Nb5-d6 -4. Kb6-c6 c4-c5 -5. Kc6-b6

R003 - Juha Saukkola
1. h4 b5 2. h5 b4 3. h6 b3 4. hg7 ba2 5. Rh6 ab1=B 6. Re6! h5 7. b4 h4 8. b5 h3 9. b6 h2 10. bc7 h1=B 11. cb8=B Qa5 12. Bh2 Rb8 13. g3 Bhb7 14. gh8=B Qd5 15. Bhb2 a5 16. c3 Bh7 17. Ra6
e6 is the only square on which the rook doesn't interfere with the bishops and the queen!

R004 - Henrik Juel
1. h4 c5 2. h5 c4 3. h6 c3 4. hg7 cb2 5. gf8=N ba1=B 6. Ng6 fg6 7. f4 Kf7 8. f5 Qf8 9. f6 Ke6 10. f7 Bf6 11. fg8=R Ke5 12. Rg7 Rg8 13. Rf7 Rg7

R005 - Gianni Donati
1. c3 g6 2. Qa4 Bh6 3. Qc6 Bd2 4. Kd2 Nf6 5. Ke3 Ne4 6. Kf4 Nd2 7. Kg5 Nb1 8. Kh6 Nd2 9. Kg7 Nf3 10. ef3 Rf8 11. Ba6 ba6 12. Ne2 Bb7 13. Rd1 Qc8 14. Rd4 Kd8 15. Rh4 Re8 16. Bf4 Rh8 17. Rd1 Ke8 18. Rd4 Qd8 19. Rb4 Bc8 20. Nd4
Four switchbacks because black needs to lose a tempo.

R006 - Unto Heinonen
1. a4 h5 2. a5 h4 3. a6 h3 4. Ra5 hg2 5. h4 Rh6 6. h5 Rc6 7. h6 g5 8. h7 g4 9. Rg5 f5 10. h8=B f4 11. Be5 f3 12. Rh8 fe2 13. f4 g3 14. Nf3 g1=B 15. Bg2 Bb6 16. Bh1 g2 17. f5 g1=B 18. f6 Bgc5 19. d4 d5 20. dc5 Bh3 21. cb6 Bf1 22. Rg2 Kf7 23. Bg3 Kg6 24. f7 Bh6 25. f8=B e5 26. Bd6 Qh4 27. Bg5 e4 28. Kd2 e1=B 29. Kc1 Ba5 30. Nc3 e3 31. Kb1 e2 32. Ka1 e1=B 33. Qe2 d4 34. Qe7 d3 35. Ne2 Beb4 36. c3 d2 37. cb4 d1=B 38. ba5 Ba4 39. b3 Kh5 40. ba4

Probleemblad, Sep. 1997

R007 - Gianni Donati
Try: +bKc1??. This assumes -1. 0-0# and then the bK cannot get out of the SW corner. Illegal position...

Try: +bKd4??. This assumes -1. d2xc3#. Then the wBh2 is wPh2 promoted on h8. Bl. caps are wBc1 at home, b7xa6, cxb, e7xf6 and g-h cross-captures. The bPd7 has no remaining capture for reaching the e-file where it could promote and/or get captured. Illegal position...

Sol: +bKc4!!. and there are plenty of non-problematic last moves.

But wait !! The black units are actually spelling out "c4" in some kind of billboard. I am stumped...

R008 - Stanislav Vokal
(a) Let's assume last move was -1. O-O, and that Black may castle, we deduce that both KBs and both Qs have been captured at home (by Knights). The wh. Knights have been captured by h7xg6xf5 and the bl. Knights have been captured by f2xe3xd4. We can use the parity argument: all in all, White made an odd number of moves, Black made an odd number of moves. This contradicts the Black to move assumption so that Black had to move his K or KR and cannot O-O anymore.

(b) Now Black made an even number of moves and 1... 0-0 is legal.

R009 - Gianni Donati and Peter van den Heuvel
1. Nh3 e5 2. Rg1 Ba3 3. bxa3 Nc6 4. Bb2 Nd4 5. Qc1 Nf3+ 6. Kd1 Nxg1 7. Nc3 Nf3 8. Rb1 Nd4 9. Ke1 a5 10. Qd1 Ra6 11. Bc1 Rf6 12. Rxb7 c6 13. Rxd7 Nh6 14. Rxd8+ Ke7 15. Ng1 Be6 16. Rb8 Bd5 17. Rb1 Rb8 18. Ra1 Ng8 19. Nb1

6 white units (and the bKN) make a switchback !! The b7, d7 & d8 captures could be done by the wKN but this would result into 18 or 20 wh. moves, not 19 as required.

R010 - Juha Saukkola
1. g4 a5 2. g5 a4 3. g6 a3 4. gxh7 axb2 5. hxg8=B Rh6 6. a4 Rha6 7. a5 R6a7 8. a6 c6 9. axb7 Na6 10. b8=B Bb7 11. h4 Qc8 12. Bh2 e5 13. h5 Be7 14. h6 Bh4 15. hxg7 f6 16. Ba2 d5 17. g8=R+ Kf7 18. Rg2

Probleemblad, Nov. 1997

R011 - Gianni Donati
1. a4 Na6 2. Ra3 Nc5 3. Rd3 Ne4 4. Rd6 exd6 5. Nf3 Qh4 6. Rg1 Qh3 7. gxh3 Be7 8. Bg2 Bh4 9. Rf1 Ke7 10. Rh1 Kf6 11. Ng1 Kg5 12. Bf1 Bxf2#

Subtle tempo losing maneuver.

R012 - Gianni Donati
1. Nf3 h5 2. Ne5 Nh6 3. Nxd7 Kxd7 4. g4 Kc6 5. Bg2+ Kb5 6. Be4 Bxg4 7. Bh7 g6 8. Kf1!! Bg7 9. Kg1 Bd4 10. Qf1 Bb6 11. Qg2 c5 12. Qxb7 Qc7 13. Qg2 Rd8 14. Qf1 Rd5 15. Qd1 Nd7 16. Kf1 Rc8 17. Ke1 Qb8

There is a beautiful try with the continuation 8. Rg1 Bg7 9. Rg3 Bd4 10. Rd3 Bb6 11. Rd7 c5 12. Rxb7 Qc7 13.?? and the white Rook can't make it.

R013 - Guus Rol
1. d4 d6! 2. Qd3 Kd7 3. Qh3 Kc6 4. d5 Kd5 5. a4 Kc6 6. Ra3 d5 7. Rf3 d4 8. Rf5 Kd7 9. e4 Ke8 10. Bc4 d3 11. e5 d2 12. Ke2 d1=R 13. e6 Rd7 14. ed7(=bP)

R014 - Tom Volet
Wh. caps are fxgxh, exfxg and bPa on its file. Bl. caps f7xe6, gxf, hxg. The position unlocks with the retraction of d7-d6 but it is necessary to provide a bl. screen inside the NE cage so that the wh. Rooks can be extracted.

-1. Bc5xPa3 a4-a3
-2. Bf2-c5 a5-a4
-3. Bh4-f2 Rh1-g1
-4. c2-c3 Rh3-h1
-5. B~-h4 Rg3-h3+
-6. B~ Rg4-g3
-7. B~ Be2-f1
-8. B~ Bf3-e2
-9. B~ Bc6-f3
-10. B~ Bd7-c6
-11. B~ Rc4-g4
-12. B~ Rc6-c4
-13. B~ Ra6-c6
-14. B~ Ra8-a6
-15. B~ Rd8-a8
-16. B~ Bc8-d7
-17. B~ d7-d6
-18. Nd6-e8 Rg8-d8
-19. Rf8-f7 and the position unlocks.

The Rg1 necessarily visited g8, d8, a8, a6, g4, h3, h1 & g1 (in that order). A beautiful problem displaying a rich content in an economic position.

Probleemblad, Jan. 1998

R015 - Gianni Donati
1. e4 d6 2. e5 Kd7 3. e6+ Kc6 4. exf7 e6 5. f4 Be7 6. f8=N Qe8 7. Nd7 Qg6 8. Nb6 axb6 9. f5 Ra5 10. f6 Rf5 11. f7 Bf6 12. f8=N Ne7 13. Nd7 Rf8 14. Nc5 Rf7 15. Na6 bxa6 16. Nf3 Kb7 17. Ne5 Ka8 18. Nd7 Bb7 19. Nf8

Two Frolkin and one Anti-Phoenix Knights on f8.

R016 - Gianni Donati
1. Nf3 Nc6 2. Ne5 Nd4 3. Nc6 bxc6 4. b4 Ba6 5. b5 Qc8 6. b6 Bd3 7. b7 Bg6 8. b8=N f5 9. Na6 Qxa6 10. d3 O-O-O 11. Bg5 Kb8 12. Bf6 exf6 13. e4 Bd6 14. e5 Rf8 15. e6 Bxh2 16. e7 Bg1 17. Rh6 gxh6 18. e8=B h5 19. Bf7 Rxf7

Two Frolkin promotees promote and leave their promotion square just so they can find a capture square ! Here the Frolkin promotions are not motivated because you have Pawns that must have captured.

R017 - Peter Wong
1. Na3 Nf6 2. Nc4 Ne4 3. Na5 Nxd2 4. Kxd2 h5 5. Kc3 Rh6 6. Kb4 Re6 7. c3 Rxe2 8. Nxe2 a6 9. Ng3 Ra7 10. Bb5 axb5 11. Re1 Ra6 12. Re6 Ra8 13. Ra6 e6+ 14. Qd6 Be7 15. Ra7 Bg5 16. Qa6 Bh6 17. Bg5 Qf6 18. Rd1 Qe7+ 19. Rd6 Qd8 20. Rb6 d6

Very pleasant tempo losing maneuvers by the bl. Rook and the bl. Queen.

Probleemblad, Mar. 1998

R018 - Gianni Donati
1. f4 Nf6 2. f5 Nd5 3. f6 Rg8 4. fxg7 f5 5. gxf8=B Rg6 6. b4 Rc6 7. b5 d6 8. b6 Kd7 9. bxc7 b6 10. h3 Ba6 11. c8=N Bd3 12. exd3 Kc7 13. Qe2 Nd7 14. Qxe7 Rxc8 15. Qf6 Nxf8

Two very tricky "aimless" promotions !

R019 - Unto Heinonen
1. Nf3 b5 2. Nd4 Bb7 3. Nxb5 Bxg2 4. Nxa7 c6 5. Nc8 Rxa2 6. Nc3 Rxb2 7. Ra8 Qa5 8. Na2 Qxd2+ 9. Kxd2 Na6 10. Kc3 Nc5 11. Qd4 Rxc2+ 12. Kb4 Rxe2 13. Ba3 Rxf2 14. Ba6 h5 15. Ra1 Bf1 16. Qxg7 Rxh2 17. Qxg8 Rh4+ 18. Qg4 Bh6 19. Ka5 Bc1 20. Qa4 Rh6 21. Na7#

R020 - T. Volet
All missing pieces are captured by pawns. The south-western corner is unlocked when the black king can go to a4. For this, the white king has to go to c5. But to be able to do that, black must retract d7xc6 and e7xd6, and this means that bBc8, bBf8 and bRh8 have to return first. The white queen-pawns however can't uncapture yet, and the white king-pawns can only uncapture 2 pieces. This means that the black rook on b2 has to go back to h8, and has to be replaced by a white knight on b2. White played e2xBf3 and h2xBg3 or h4xBg5. Therefore the white king bishop has to return as fast as possible to f1. Black will have to uncapture a bishop and a knight as fast as possible. We'll see that for the return of the black rook to h8, a white interference piece to protect the white king is needed, and that this piece has to be a white rook.
Both sides are under time-pressure. The black pawn on d3 may not go back to f6 or f7 too soon, because the black queen and king can't go back to their begin array square then. A possible last move sequence is 1. e4xBd3 Bf1-d3 2. f5xNe4 e2xBf3, then the black bishop to c8 and the white knight to d3 or d1, then Rb6-b2 Nb2-d3; d7xRc6 Rc5-c6; Rb6-c6 Rb5-c5; Rc5-c6 g4-g5 (or h4xBg5); Re5-c5 Rb6-b5; Re8-e5 etc. Here we see that white needs a free move (g5 or hg5) to let the black rook return. The rest of the retro-play is easy: Black rook to h8, black bishop to f8, white's king to c5, e7xRd6, and the position is unlocked. Notice that black can't uncapture 2. f5xe4 and d7xNc6, because the white knight can't reach b2 in time. So we have seen that white can have the move, and the last move was e4xBd3
Suppose white moved last. This can only have been h4xBg5 or g4-g5.

So white has the move, and the last moves were e4xBd3 Bf1-d3; f5xNe4 e2xBf3

H078 - Gianni Donati
Try: 1. c5? Kb6 2. Ne5 Ba6 3. Nc6 Bc4 4. OOO? Bd5 5. Nb8 Bb7#, because castling isn't allowed: To give white a last move, Ra8 must have moved. So: 1. d6 Be6 2. Ra5 Bf7 3. Kd7 Bh5 4. Rc5 Ka6 5. Kc6 Be8#.

Probleemblad, May 1998

R021 - G. Donati
Author's intention was: 1. e3 a6 2. Ba6 Nh6 3. Bb5 Ra6 4. d4 Rg6 5. Qd3 Rg3 6. hg3 Ng8 7. Rh6 Na6 8. Rb6 Nb8 9. Qg6 hg6 10. Ke2 Rh1 11. Nh3 Rc1 12. Kf3 Rb1 13. Kg4 Ra1 14. f3 Rh1 15. Ng1 Rh3 16. gh3
Olli Heimo however found a cook: 1. Na3 a6 2. Nb5 ab5 3. d4 Ra3 4. Be3 Re3 5. Qd3 Rg3 6. hg3 Na6 7. Rh6 Nc5 8. Rb6 Nb3 9. Qg6 Na1 10. e3 Nb3 11. Bb5 Nc5 12. Ke2 Na6 13. Kf3 Nb8 14. Kg4 hg6 15. f3 Rh3 16. gh3
A corrected version can be found at Probleemblad, Sep. 1998

R022 - Unto Heinonen
1. Nf3 h5 2. Ne5 Rh6 3. Ng4 e5 4. a3 Bxa3 5. e4 Be7 6. Ra6 c5 7. Rc6 a5 8. Bb5 Ra6 9. Ba4 b5 10. Qf3 Bb7 11. Rc8 Bxe4 12. Kd1 Bc6 13. Re1 d5 14. Re4 Nd7 15. Rf4 Nb6 16. Rf6 Kd7 17. Re6 f5 18. Qa3 Bf6 19. Re8 Ne7 20. d3 Ke6 21. Bg5 Qd6 22. Bh4 g5 23. Red8 Ng6

R023 - T. Volet
The position is a draw according to the 50 move rule. A possible begin position is:
[r1R5/1ppppppp/8/NKp1R3/2P5/1P3P2/B1PP1PPP/N4rk1]
Beginning with this position, one can play: 1. Bb1 c6 2. Kb6 Rb8 3. Kc7 Ra8 4. Kd8 Rb8 5. Ke8 Ra8 6. Rd8 Rc8 7. Re6 Rc7 8. Ra8 Rc1 9. Ra6 Rd1 10. Rb6 Rc1 11. Rb4 Rd1 12. Ra4 Rc1 13. Ra2 Rd1 14. Rb2 Kh1 15. Ba2 Re1 16. Rb1 Rf1 17. Rbe1 Kg1 18. R1e3 Re1 19. Rd3 Re4 20. Kf8 Kf1 21. Kg8 Ke1 22. Re3 Kd1 23. Kh8 Kc1 24. Re1 Kb2 25. Rh1 Re1 26. Rg1 Rd1 27. Ree1 Kc1 28. Ref1 Re1 29. Rh1 Kd1 30. Rfg1 Ke2 31. Rf1 Rb1 32. Rd1 Rb2 33. Bb1 Ra2 34. Rc1 Ra4 35. Ba2 Rb4 36. Rb1 Rb6 37. Rb2 Ra6 38. Bb1 Ra7 39. Ra2 Ra6 40. Ra4 Rb6 41. Ba2 Ra6 42. Rb1 Ra7 43. Rb2 Ke1 44. Bb1 Kd1 45. Rba2 Kc1 46. Rb4 Kd1 47. Raa4 Kc1 48. Rb6 Kb2 49. Rab4 Ka3 50. Ra6 Kb2 51. Rbb6 Ka3 52. Ba2
Starting from move 2, no capture or pawn move has occurred, so 50,5 moves have been played after the last pawn move.
However, Henrik Juel found out the diagram position can be reached after only 48.5 moves without capture!
1. Bb1 c6 2. Kb6 Kh1 3. Kc7 Kg1 4. Kd8 Kh1 5. Ke8 Kg1 6. Rd8 Rc8 7. Rd5 Rc7 8. Ra8 Kh1 9. Ra6 Kg1 10. Rb6 Kh1 11. Rb4 Kg1 12. Ra4 Kh1 13. Ra2 Kg1 14. Rb2 Kh1 15. Ba2 Kg1 16. Rb1 Kh1 17. Re1 Kg1 18. Re3 Rb1 19. Rc3 Kf1 20. Kf8 Ke1 21. Re3 Kd1 22. Rde5 Kc1 23. Re1 Kb2 24. Rh1 Rd1 25. Ree1 Kc1 26. Reg1 Re1 27. Kg8 Kd1 28. Kh8 Ke2 29. Rf1 Rb1 30. Rfg1 Rb2 31. Bb1 Ra2 32. Rd1 Ra4 33. Ba2 Rb4 34. Rb1 Rb6 35. Rb2 Ra6 36. Bb1 Ra7 37. Ra2 Ra6 38. Ra4 Ra7 39. Rb4 Ra6 40. Rb6 Ra7 41. Ba2 Ra6 42. Rb1 Ra7 43. Rb2 Ke1 44. Bb1 Kd1 45. Ra2 Kc1 46. Ra4 Kb2 47. Rab4 Ka3 48. Ra6 Kb2 49. Rbb6 Ka3 50. Ba2

Probleemblad, July 1998

R024 - Mario Velucchi
1. Nf3 Nc6 2. Ne5 Nd4 3. Nd7 Ne2 4. Nf8 Nc1 5. Ng6 Nb3 6. Ne7 Nd2 7. Nc8 Nf1 8. Kf1 Qc8 9. Ke1 Qd8

R025 - Michel Caillaud
1. c4 g6 2. c5 Bh6 3. c6 Be3 4. dxe3 f5 5. Qd6 cxd6 6. Bd2 Qb6 7. Ba5 Qd4 8. Bd8 b6 9. c7 Bb7 10. c8=Q f4 11. Qc6 f3 12. Bc7 fxg2 13. Qf3 d5 14. Bd6 Nh6 15. Ba3 Rf8 16. b4 Rxf3 17. Bc1 Rh3 18. f4 Nf7 19. Kf2 h5

Tricky round trip by the wh. Bc1 used as a mask for the Frolkin promotion.

R026 - Joost de Heer
Author's intention is 1. h4 a5 2. h5 a4 3. h6 a3 4. hg7 ab2 5. Rh6 Ra3 6. Ra6 h5 7. f4 h4 8. f5 h3 9. f6 h2 10. fe7 f6 11. d4 Kf7 12. e8=R Ne7 13. g8=N Ng6 14. R8e6 Qe7 15. Reb6 Qc5 16. Ne7 cb6 17. Nc6 dc6 18. d5 Bh3 19. d6 Bg2 20. d7 Bd5 21. d8=Q Bb3 22. Q8d3 Bd6 23. Qb5 cb5. Unfortunately this is cooked e.g. by 1. f4 d5 2. f5 d4 3. f6 Bh3 4. fxe7 Bxg2 5. h4 f6 6. h5 Kf7 7. h6 Bd5 8. hxg7 Bb3 9. Rh5 Qc8 10. e8=R Ne7 11. g8=R c6 12. Rb5 cxb5 13. Rg3 Ng6 14. Rc3 dxc3 15. d4 h5 16. d5 h4 17. d6 h3 18. d7 Bd6 19. d8=Q h2 20. Qb6 axb6 21. Re4 Ra3 22. Ra4 Qc5 23. Ra6 cxb2

R027 - Ben Donati & Gianni Donati
Both white rooks and one of the bishops are promoted. The pawns that promoted are the h-, d-, and a-pawn. These pawns all need to capture once: The h- and d-pawn to enter the cage, and the a-pawn to promote on a black square. So the b- and e-pawn got captured on their own file. The black pawn now on d4 captured only on the black squares, so it captured the b-pawn and both original rooks, and the e4-pawn captured the e-pawn and the bishop. Since those account for all captures, Rd5 didn't make a capture on its last move, so the last move was c5xRd4.

Probleemblad, September 1998

R021c - G. Donati
1. e3 a6 2. Ba6 Na6 3. Bb5 Ra6 4. d4 Rg6 5. Qd3 Rg3 6. hg3 Ng8 7. Rh6 Na6 8. Rb6 Nb8 9. Qg6 hg6 10. Kd2 Rh5 11. Kc3 Rc5 12. Kb4 Rc3 13. Bd2 Ra3 14. Bc3 Ra8

R026c - J. de Heer
1. h4 a5 2. h5 a4 3. h6 a3 4. hg7 Ra4 5. Rh6 Rb4 6. Ra6 h5 7. f4 h4 8. f5 h3 9. f6 h2 10. fe7 f6 11. d4 Kf7 12. e8=R Ne7 13. g8=N Ng6 14. Ree6 Qe7 15. Reb6 Qc5 16. Ne7 cb6 17. Nc6 dc6 18. d5 Bh3 19. d6 Bg2 20. d7 Bd5 21. d8=Q Bb3 22. Q8d3 Bd6 23. Qb5 cb5

R028 - T. le Gleuher
Intended solution was: 1. e4 d5 2. e5 Kd7 3. e6 Kc6 4. ef7 e5 5. Ke2 Bd6 6. f8=B Qf6 7. Be7 Kb5 8. Bd8 c5 9. Bc7 Be6 10. Bb8 Bf7 11. Bc7 Re8 12. Bb8 Ka5 13. Kd3 b5 14. Kc3 e4 15. d4 ed3 16. Kb3 d4 17. c4 dc3 18. Ka3 c4 19. b4 cb3
But this problem has been cooked: 1. d4 d5 2. Bf4 c5 3. Bb8 e5 4. e4 ed4 5. e5 Kd7 6. e6 Kc6 7. ef7 Bd6 8. f8=B Be6 9. Be7 Bf7 10. Bg5 Qf6 11. Bc7 Re8 12. Kd2 b5 13. Bb8 Kb6 14. c3 Ka5 15. Kc2 d3 16. Kb3 d4 17. Ka3 dc3 18. Bc1 c4 19. b4 cb3
A correction can be found at Probleemblad Jan. 1999

R029 - A. Frolkin
1. a4 f5 2. Ra3 f4 3. Rh3 f3 4. Na3 fe2 5. f4 Nf6 6. Kf2 e1=R 7. Kg3 Re3 8. Kh4 Rb3 9. cb3 e5 10. Nc2 e4 11. Na1 e3 12. Qc2 e2 13. Qh7 e1=R 14. f5 Re3 15. Be2 Rc3 16. dc3 Ba3 17. Be3 d6 18. Bb6 Qd7 19. Bh5 Kd8 20. Nf3 ab6 21. Re1 Ra5 22. Re7 Re5 23. g4 Nd5 24. f6 Re1
The Anti-Pronkin theme: two black pawns promote to a rook on e1, then they get captured, and finally the original rook from a8 ends up on e1.

R030 - M. Caillaud
White's last move was 1. g2-g4. But 1. hg3 isn't allowed, because in the resulting position, white doesn't have a last move. So a pre-plan is necessary: 1. e3 2. e2 3. e1=B 4. Bc3 5. Ba1 6. Qc3 7. Re4 8. a4 9. a3 10. a2 11. Ra3. Now the last moves can't have been Ka5-b5 a4xb3ep; b2-b4, because this results in one capture too many: The bishop on a1 is imprisoned, so it is a promoted one, and that takes 12 captures, while white only misses 11 pieces. So the last move was g2-g4: 12. hg3ep! Now the last move was Ka5-b5!! because now retracting a4xb3 ep b2-b4 results in a position in which black has captured 11 times, and white misses 12 pieces. 13. Nde5 14. Ng4 15. Qe5 hg3#

R031 - T. Volet
The white pawns have captured 6 times, and the black once, so all captures were done by pawns. The black a- and h-pawn promoted without capturing. The cage in the upper right corner can only be released by retracting h6xg7. This has many implications: The black h-pawn must have returned to h7, the promotion on h1 has to be undone, the knight on h1 must've left that square via g3 to make the promotion possible, the white pawn must go back to g2, so the white bishop must return to f1, the white bishop has to be uncaptured with b7xBc6, and for that the black queen bishop has to go back to c8. The black rook can't have been captured on g7 because black can't retract that check. Therefore the black rook has to return to d8 before the black bishop goes to c8, otherwise it can't return to h8. What did white capture on g7 then? A queen or rook give an illegal check, a second black-squared bishop could only arise on a1, and it can't have left that square, so a knight, which arose from a promotion on a1, was captured on g7. This means that white has played a2xb3xc4 to let the a-pawn promote to a knight, and to let the knight out of the left corner. So the three captures white has done after h6xg7 are c4xd5xe6 and d3xe4. To let the black bishop go past the white pawns to c8, white must uncapture d3xBe4 before d5xe6. Black has no last move. White must uncapture a black piece to give him a last move. So the last move was d3xBe4

Probleemblad, November 1998

R032 - Fernand Joseph
Black's last move could've been 1. a2xb1=N, so white mates in 2 with 1. Rc1 Ke2 2. Qe3#.

R033 - Dr. Tomislav Petrovic
Suppose white can still castle. Before black played f3xg2, the king and bishop should've been on their present suqare, and after that, white played f2-f[3/4]. So the rook couldn't have left the lower righthand corner. So c7xb6 couldn't have been black's last move. Neither can the last black move have been b7-b6, because this implies that the bishop on f1 is a promoted one, and the only pawn that could've promoted was the c-pawn, but that pawn needs a capture to promote on a white square. Because black captured twice in the lower righthand corner, both the a- and the b-pawn must have promoted, needing two captures to get past the black pawns on a7 and b6. So white's last move was not a capture. The only move giving black a last move was therefore f2-f4 (and black's move before that was f3xg2). So if white can castle, his last move was f2-f4. So h#2 with: 1. fg3ep! OOO 2. gh1=R gh4#

R034 - Unto Heinonen
1. h4 g5 2. hg5 c5 3. Rh6 Qb6 4. Rc6 Qb3 5. ab3 dc6 6. Ra6 Bh3 7. Rb6 ab6 8. gh3 Ra3 9. ba3 Nf6 10. Bb2 Nd5 11. Bf6 ef6 12. d4 cd4 13. Kd2 Bb4 14. Kc1 OO 15. Kb2 Re8 16. Ka2 Re2 17. Qe2 Nc7 18. Qe6 fe6 19. Ba6 ba6
Unfortunately, Andrei Frolkin found a cook 1. g4 b6 2. Bg2 Ba6 3. Ba8 Be2 4. b3 Bd1 5. Bb2 Bf3 6. Bg7 Bh1 7. Bh1 c5 8. Bc6 dc6 9. Bh8 Qd2 10. Nd2 e6 11. Bf6 Nf6 12. O-O-O Nd5 13. Nb1 Nc7 14. Rd4 cd4 15. a3 a6 16. Kb2 Bb4 17. Ka2 Kf8 18. g5 Kg8 19. h3 f6.

R035 - Michail Kozulia
1. f4 c6 2. f5 Qb6 3. f6 Qg1 4. fg7 Qg2 5. gh8=B Bg7 6. b4 Bb2 7. b5 Ba3 8. Bhb2 Qg7 9. Bg2 cb5 10. Bc6 dc6 11. c3 Bf5 12. Qb3 Bg6 13. Qe6 fe6 14. Kf2 Kf7 15. Rf1 Kf6 16. Kg1

R036 - Andrej Frolkin
1. h4 d5 2. h5 d4 3. h6 d3 4. hg7 dc2 5. d4 f5 6. Bd2 c1=B 7. Qa4 Kf7 8. Qa7 e5 9. Ba5 Bh6 10. Na3 f4 11. O-O-O e4 12. Kc2 e3 13. Kb3 ef2 14. Nc2 Ba3 15. Ka4 Ne7 16. b4 Re8 17. g8=R Bf8 18. Rh6 Bh3 19. gh3 Bc1 20. Bg2 f1=B 21. Bc6 Qd5 22. Bb5 Nbc6 23. h4 Rad8 24. Nh3 Rd6 25. Nf2 Bh3 26. a3 Bc8

R037 - Michel Caillaud
1. Nf3 b5 2. Ne5 b4 3. Nd7 b3 4. Nb8 Qd6 5. Na6 Kd7 6. a3 Kc6 7. Nb4 Kc5 8. Na2 ba2 9. a4 ab1=B 10. Ra3 Ba2 11. Rd3 Bc4 12. b3 Kb4 13. Bb2 c5 14. Bf6 gf6 15. Qa1 Bh6 16. Qd4 Be3 17. de3 Nh6 18. Rd1 Rg8 19. Ra1 Rg3 20. Qd1 Rh3 21. gh3 Bb5 22. Bg2 Be8 23. OO

R038 - Gianni Donati
1. h4 Nf6 2. h5 Nd5 3. h6 gh6 4. Nf3 Bg7 5. Rg1 Bc3 6. Ne5 Kf8 7. f3 Kg7 8. Kf2 Qg8 9. Kg3 Kf8 10. Kh4 Qg6 11. d4 Qb6 12. Kh5 Qb3 13. ab3 Ke8 14. Ra6 Kd8 15. Rf6 ef6 16. Na3 Ne7 17. d5 Re8 18. Qd4 Ng8 19. Qa4 Bb4 20. c4 Bf8 21. Nc2 Re7 22. Ne1 Ke8

Probleemblad, Januari 1999

R028c - T. le Gleuher
1. e4 d5 2. e5 Kd7 3. e6 Kc6 4. ef7 e5 5. Ba6 Qf6 6. Ke2 Be7 7. f8=Q Be6 8. Qb8 b5 9. Qa8 Kb6 10. Qe8 c5 11. Qg6 hg6 12. Kd3 Rh4 13. Kc3 e4 14. d4 ed3 15. Kb3 d4 16. c4 dc3 17. Ka3 c4 18. b4 cb3

R039 - J. de Heer
1. g4 h5 2. g5 Rh6 3. gh6 g5 4. h7 Bh6 5. h8=R Kf8 6. Rh6 Kg7 7. Rb6 ab6 8. Bg2 Ra3 9. Bc6 Re3 10. a4 dc6 11. Ra3 Bf5 12. Rd3 Nd7 13. Rd6 Bd3 14. Rh6 Ndf6 15. Rh7 Kg6 16. Rh8 Qd4

R040 - G. Donati
a) 1. g4 c6 2. g5 Qb6 3. g6 Qb3 4. ab3 a6 5. Ra4 a5 6. Rh4 a4 7. Rh7 a3 8. Rg7 a2 9. Rh7 a1=Q 10. Rh4 Qa5 11. Ra4 Qd8 12. Ra1
b) 1. g3 a5 2. g4 Ra6 3. g5 Rb6 4. g6 Rb3 5. ab3 Nc6 6. Ra4 Ne5 7. Rh4 a4 8. Rh7 a3 9. Rg7 a2 10. Rh7 a1=R 11. Rh4 Ra8 12. Ra4 c6 13. Ra1

R041 - U. Heinonen
1. e4 h5 2. Qh5 g6 3. Qd1 Rh2 4. g3 Rg2 5. Rh8 Bh6 6. a4 Kf8 7. Ra3 Kg7 8. Re3 Nf6 9. b3 Qg8 10. Ba3 Qh7 11. Rc8 Na6 12. Rh8 Qg8 13. Ba6 Qd8 14. Bf1 Ng8 15. d3 Kf8 16. Kd2 Ke8 17. Kc1 Bf8 18. Rh1 Rh2 19. f3 Rh8
Eight switchbacks (5 black, 3 white)!

R042 - M. Caillaud
1. g4 c5 2. Bg2 Qa5 3. Bc6 Qa2 4. Bb5 Qd5 5. Ra6 Nh6 6. Rc6 dc6 7. b3 Bf5 8. Bb2 Nd7 9. Bf6 ef6 10. g5 Bd6 11. g6 Ke7 12. gh7 Rhb8 13. h8=R a5 14. Rc8 a4 15. Rc7 Rh8 16. f3 Ke8 17. Kf2 Bf8 18. Kg3 a3 19. Kh4 Ra4 20. Kh5 Rg4 21. d4 Nb8 22. Qd3 Qd8 23. Nd2 Bc8 24. Qh7 Ng8
Seven black switchbacks, just because the white promoted rook has to go over the 8th rank to go to c7!

R043 - M. Caillaud & T. Volet
White still has all his pieces. The white pawns captured the missing 5 black pieces. The black c- and h-pawn promoted without capturing. White can only retract moves with his g7-pawn, so black must give white more retro moves. If white moved last, this doesn't work, with a black last move this will just go: the last moves are Bb8-a7 g6-g7; Bd6-b8 g5-g6; Bb4-d6 Qa4-a5. The following retro-moves must happen too, although there can be a different move order: Bc3-b4 Qc2-a4; Bd4-c3 Bc5-d4; Qc6-e4 and both sides have enough retro-moves now. Black now retracts a4-a3, Nb1 to d3, and the king-battery can finally be disarmed.

R044 - H. Juel
Solution:
[n3nnn1/4nknP/4nn2/P7/8/1PKPP1P1/2P2P2/nN1N3n]
After adding the knights we need to have a legal position, and both sides are missing 5 pieces. How many pawn captured happened with this given material and white pawn-structure? The black a- and b-pawn could have promoted with only 2 captures, by white or black. In both cases, the knight that appeared on a1 must still be there, since b3 was on its place before the promotion. To get the knight out of the corner, an extra capture is necessary. The same story goes for the g- and h-pawns. The black c-pawn must've captured once, as must the black f-pawn. The d- and e-pawn could have promoted with only 3 pawn captures, either 2 by white and 1 by black, or 3 by black. This shows that the extra capture, needed to free the knights on a1 or h1, isn't possible. So any legal position must have a knight on a1 and h1. The other knights are needed for the mate.
Guus Rol gives a version in which 10 knights, and the king must be added:
[8/7P/8/P7/6K1/1P1PP1P1/2P2P2/2NN4] [3nnnn1/3nkn1P/3nn3/P7/6K1/1P1PP1P1/2P2P2/n1NN3n]

Probleemblad, Mar. 1999

R045 - Mario Parinello
1. d4 b6 2. Qd3 Ba6 3. Qg6 hg6 4. d5 Rh5 5. d6 Ra5 6. Bg5 c5 7. e3 Qc7 8. dc7 d5 9. Bc4 d4 10. Ke2 d3 11. Kf3 d2 12. Kg3 d1=Q 13. c8=Q Qd8 14. Qg4 Nd7 15. Qd1

R046 - Gianni Donati
1. f4 b5 2. f5 b4 3. f6 b3 4. fe7 bc2 5. ed8=R Ke7 6. Rc8 cb1=Q 7. Rb8 Qa1 8. Rb6 Qb1 9. Rf6 Qf5 10. Qc2 Qf1 11. Rf1 h6 12. Kf2 Kf6 13. Qb1 Be7 14. Qa1 Kg6 15. Rd1 Kh7 16. Ke1

R047 - Thierry le Gleuher
1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 hg2 5. h4 g5 6. Rh3 g4 7. Rd3 g3 8. Nf3 g1=N 9. Rd7 Nh3 10. Bh3 g2 11. d3 g1=B 12. Kf1 a5 13. Ne1 a4 14. f3 Be3 15. Be3 a3 16. Nd2 a2 17. Rc1 a1=Q 18. Nb3 Qa5 19. Qd2 Qc3 20. Qc3
However, this is cooked: 1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 hg2 5. h4 g5 6. Rh3 g4 7. Rd3 g3 8. Nf3 g1=N 9. Rd7 Nh3 10. Bh3 g2 11. d3 g1=N 12. Kf1 a5 13. Ne1 Nf3 14. Be3 Nd2 15. Qd2 a4 16. Qc3 a3 17. Nd2 a2 18. Rc1 a1=N 19. f3 Nb3 20. Nb3

R048 - Satoshi Hashimoto
1. e4 Nf6 2. Qf3 Nh5 3. Qf6 ef6 4. e5 Bd6 5. e6 OO 6. e7 Nc6 7. e8=Q Rb8 8. Qe4 Re8 9. Bd3 Re5 10. Ne2 Rb5 11. OO Rb3 12. ab3 Kf8 13. Ra6 ba6 14. Qa4 Rb5 15. Be4 Re5 16. d3 Re8 17. Bf4 Ke7 18. Nd2 Rh8 19. Ra1 Ke8 20. Kf1 Bf8 21. Ke1 Ne7 22. Nf1 c6 23. Bb8 Ng8

R049 - Unto Heinonen
1. c4 h5 2. c5 h4 3. c6 h3 4. cb7 hg2 5. h4 Rh6 6. h5 Rc6 7. h6 Rc2 8. h7 c5 9. h8=B Nc6 10. b8=B g5 11. Bf4 Rb8 12. Bhe5 Rb3 13. Bb8 d6 14. Rh8 Bh3 15. Nc3 e6 16. Ne4 Be7 17. Ng3 Kf8 18. Nh1 Rg3 19. b3 Kg7 20. Bb2 Nf6 21. Rg8 Kh6 22. Rc1 Nd5 23. Bh8 Rb2 24. Rc4 Qb6 25. Re4 Bd8 26. d4 Nce7 27. Bc1 Nc8 28. Rh4
However, this is cooked: 1. c4 h5 2. c5 h4 3. c6 h3 4. cb7 hg2 5. d4 Rh3 6. Bf4 Rd3 7. Nd2 Na6 8. Rc1 Nb4 9. Rc6 Nd5 10. Rh6 c5 11. h4 d6 12. h5 g5 13. Rh4 Bh3 14. b8=B e6 15. Bc7 Rb8 16. Ne4 Rbb3 17. Ng3 Rd2 18. Nh1 Rg3 19. Rh8 Be7 20. h6 Kf8 21. h7 Kg7 22. Bb8 Qb6 23. Rg4 Bd8 24. b3 Rb2 25. Bc1 Nge7 26. Rg8 Kh6 27. h8=B Nc8 28. Rh4

R050 - Andrei Frolkin
The last moves were: -1. f2xRe1=R a2-a3 -2. Ra1-b1 a4-a5 -3. Bb1-c2 a3-a4 -4. c2-c1=B Bc1-d2 -5. d2-d1=N Rd1-e1 -6. a7-a6 Re1-e2 -7. f5-f4 Ne2-g1 -8. f6-f5 Qg1-h1. Now the position only unlocks when white can play his king to g2 (Re1 can go to f1 then, and the black king can open the cage by going to e2). This is only possible if Qh2 gets unpromoted on h1. So the promoted pieces are: Bc1, Nd1, Re1, Qh2. An AUW!

R051 - Michel Caillaud
White captures are d2xe3, and c6xd7. The white f-pawn didn't capture on its way to f7, so black captured fxe and e7xf6. The missing a-pawn must have promoted, so white captured a6xb7. This accounts for all captures.
White is almost out of retro moves. Black can't retract c7-c5 too soon, the reason for this will be clear during the solution.
Retroplay is: -1. .. Qg5-h4 -2. g3-g4 Qg6-g5 -3. Bh7-g8 Bd2-c1 -4. Qh8-f8 Bb4-d2 -5. Bg8-h7 Qg5-g6 -6. Qh6-h8 Qf4-g5 -7. Qg5-h6 Ba3-b4 -8. Qg4-g5 h4-h3 -9. Qh3-g4 Qf3-f4 -10. Qg2-h3 Bb4-a3 -11. Qf2-g2 Ba3-b4 -12. Qe1-f2 Qf2-f3 -13. Qa5-e1 Qe1-f2 -14. Qb5-a5 Qb4-e1 -15. Qa5-b5 Qb5-b4 -16. Qa6-a5 Qc6-b5 -17. Qb7-a6 B~ -18. Qb8-b7 Qb5-c6 -19. b7-b8=Q B~ -20. a6xBb7 Further retroplay is b7-b6; Nb6-c8 Rd8-c8-c6-b6; c7-c5; c6xRd7; Rd7->a8; bB->c8; d7-d6; bB-> f8; e7xf6; and the rest is easy.
There are 19 squares which must have been occupied by a queen: h4 (b), g5 (b+w), g6 (b), f8 (w), h8 (w), h6 (w), f4 (b), g4 (w), h3 (w), f3 (b), g2 (w), f2 (w+b), e1 (w+b), a6 (w), c6 (b), b7 (w), b8 (w), d1 (w), d8 (b). (a5 and b5 aren't necessary, there are ways to solve the position without visiting those squares!)
The main idea is: The same piece (the black queen) shields the king from four different sides.

Probleemblad, May 1999

R052 - Stanislav Vokal
(a): 1. Rd4 any 2. Qf5 any 3. Qf8#.
(b): 1. OO! and now black may not castle: The bishop from c1 got captured on g5 by the black pawn on h6. Before this capture, the rook on h8 was imprisoned in that corner. Because white may still castle, and because b3 was on that square before the capture of the bishop, Ra2 can not be the original rook from h8, so it's the original rook from a8. So black may not castle anymore.

R053 - Etienne Dupuis
1. d3 Nf6 2. Bh6 gh6 3. g4 Bg7 4. g5 OO 5. g6 Re8 6. gf7 Kf8 7. fe8=N c6 8. Nc7 Ng8 9. Na8 Ke8 10. Nb6 Bf8 11. Nd5 cd5

R054 - Jouwert Turkstra
1. g3 e6 2. Bg2 Ba3 3. b4 c5 4. Bb2 Nc6 5. Bf6 Nd4 6. Bc6 Qe7 7. f3 Qf8 8. Kf2 Ne7 9. Qf1 Kd8 10. Ke1 Qe8 11. Kd1 Rf8 12. Qe1

R055 - Michel Caillaud
1. h4 d6 2. h5 Nd7 3. h6 Nh6 4. e4 Ng4 5. Rh7 Ndf6 6. Rg7 Rh1 7. Bc4 Nh2 8. d3 Bh3 9. Rg4 Nh7 10. Bf7 Kf7 11. Bg5 Kg6 12. Be7 Kh5 13. Bg5 Bh6 14. Bc1 Qh4 15. Qf3 Rh8

R056 - Michel Caillaud
1. h4 b5 2. h5 b4 3. h6 Nh6 4. c3 Ng4 5. Qa4 b3 6. Qd7 Nd7 7. Rh7 Ndf6 8. Rg7 Rh1 9. Rf7 Nh7 10. Rf3 Bh6 11. Rg3 Kf7 12. ab3 Kg6 13. Ra7 Kh5 14. Rc7 Nh2 15. Re7 Bh3 16. Ree3 Qh4 17. Kd1 Rh8

R057 - Gianni Donati
1. g3 Nh6 2. Bg2 Nf5 3. Bc6 dc6 4. Kf1 Kd7 5. Kg2 Kd6 6. Kf3 Nh4 7. Ke3 Bh3 8. Qf1 e6 9. Qg2 Be7 10. Qe4 Bf6 11. Qg6 hg6 12. Nf3 Rh5 13. Rg1 Rd5 14. Rg2 Rd3 15. ed3 Kd5 16. Ke2 Qd6 17. Kf1 Nd7 18. Kg1 Rh8 19. Kh1 Rh5 20. Rg1 Re5 21. Rf1 Re1 22. Kg1 Rf1

Shorties I

1 - Tibor Orban
1. e4 e6 2. Bb5 Ke7 3. Bd7 c6 4. Be8 Ke8
The position can be reached in 3 moves on each side, with duals, and each time the Queen's Pawn captures the bishop. But in the unique 4-move solution, it's the other way round!

2 - Gianni Donati
1. Sa3 e6 2. Rb1 Ba3 3. ba3 Se7 4. Rb7 Sec6 5. Rb8 Rb8 6. a4 Rb3 7. ab3 Sb8
Cross-captures and a "sibling imposter" for a total of three units that are not what they seem to be!

3 - Henrik Juel
(a): 1. b4 e6 2. Ba3 Ke7 3. Qc1 Kd6 4. b5 c5 5. bc6 Kc6 6. Be7 d6
(b): 1. b4 c5 2. Ba3 cb4 3. Bb4 e6 4. Be7 d6 5. Qc1 Kd7 6. Kd1 Kc6
In the first solution the Pawn that captures en passant is alst itself captured -- this proof game is unusually mischievous!

4 - Markus Ott
1. c4 d6 2. Qb3 Bf5 3. Qb4 Bc2 4. d3 Qd7 5. Be3 Qb5 6. Bb6 c5 7. Bc7
Considering the symmetrical position, it is strange that white moves one more time than Black, and even stranger that no black move is a mirror of the immediately preceding white move.

5 - Gerd Wilts
1. d4 g5 2. Bf4 Bg7 3. Be5 Kf8 4. Bg7 Kg7
1. d3 g6 2. Bh6 g5 3. Bf8 Kf8 4. d4 Kg7
In the second solution, 1.d3 could be replaced by 1.d4 if white could pass its 4th move, and 1...g6 is unnecessary if this move could be passed. Can two tempo-manoeuvres on the same side be shown in a one-line shorty?

6 - Gerd Wilts
1. b4 e6 2. Bb2 Bc5 3. Be5 Be3 4. Bd6 e5
1. b3 e5 2. Ba3 Be7 3. Bd6 Bg5 4. b4 Be3

7 - Gerd Wilts
1. f4 b5 2. f5 b4 3. f6 b3 4. fg7 bc2 5. Qc2 Bg7 6. Qd1 Bf8
1. c4 g5 2. c5 g4 3. c6 g3 4. cb7 gf2 5. Kf2 Bb7 6. Ke1 Bc8
Note that a reversed position, with the white b- and g-pawns and the black c- and f-pawns omitted, would not work as a composition because of a too-early check.

8 - Gianni Donati
(a): 1. d4 Sh6 2. Sd2 Sf5 3. Sdf3 Sd4 4. Kd2 Sc2 5. Se1 Se1 6. Ke1
(b): 1. c3 Sf6 2. c4 Se4 3. c5 Sd2 4. c6 Sb1 5. Rb1 Sc6 6. Ra1 Sb8
Alternate white switchbacks for alternate places to capture the Knights.

9 - Gianni Donati
1. d3 h6 2. Bg5 hg5 3. e4 Rh3 4. Qh5 Rd3 5. cd3 g6 6. d4 gh5
1. d4 h5 2. Bh6 Rh6 3. c3 Ra6 4. Qa4 Ra4 5. e4 Rd4 6. cd4 g5
In the first solution the two black pawns make single-steps, and then cross-captures. In the second, they make double-steps.

10 - Gianni Donati
1. e4 h5 2. Qh5 g5 3. Qd1 Rh3 4. e5 Re3 5. de3 f6
1. d4 h6 2. Bh6 f6 3. Bc1 Rh3 4. e5 Re3 5. de3 g5

11 - Gianni Donati
1. e4 b6 2. Qh5 Bb7 3. Qd5 Ba6 4. Qa8 Bc8 5. Qb7
1. e3 b6 2. Qf3 Ba6 3. Qa8 Bb7 4. e4 Bc8 5. Qb7
If we think of the three arrival squares of the Bishop as arranged in a triangle, then one circuit is clockwise, the other counterclockwise.

12 - Mark Kirtley & Michel Caillaud
1. Sh3 Sc6 2. Sf4 Se5 3. Sd5 c6 4. Sb6 Qc7 5. Sc8 Qb8 6. Sb6 Qd8 7. Sd5

13 - Satoshi Hashimoto
1. e3 Sf6 2. e4 Se4 3. Sa3 Sc3 4. Rb1 Sb1 5. Sb5 Sa3 6. Sc3 Sc4 7. Sb1

14 - Dieter Müller
1. a4 d6 2. a5 Bg4 3. a6 Be2 4. ab7 Bd1 5. ba8=Q Bg4 6. Qf3 Bc8 7. Qd1

15 - Michel Caillaud
1. b4 h5 2. b5 Rh6 3. b6 Rc6 4. bc7 Rc2 5. cb8=Q Rd2 6. Qd6 Rd1 7. Qd1
3. .. Rd6 4. bc7 Rd2 5. cb8=B Rc2 6. Bbf4 Rc1 7. Bc1

16 - Michel Caillaud
1. e4 f5 2. e5 Sf6 3. ef6 e6 4. fg7 Qh4 5. gh8=Q Qc4 6. Qf6 Bh6 7. Qg6 hg6
1. e3 e6 2. e4 Qh4 3. e5 Sf6 4. ef6 Qc4 5. fg7 f5 6. gh8=S Bh6 7. Sg6 hg6
By the way, can anyone compose a shorty AUW?

Probleemblad, July 1999

R058 - Mario Parrinello
1. e3 b5 2. Ne2 b4 3. Nec3 b3 4. Ke2 ba2 5. b4 ab1=Q 6. b5 Qa1 7. Nb1 Qe5 8. c3 Qg3 9. hg3 f6 10. Rh4 Kf7 11. Ra4 Ke6 12. Ra1

R059 - Gianni Donati
1. Nc3 Nc6 2. Nd5 Nd4 3. Ne7 Ne2 4. Ke2 a5 5. Kf3 Ra6 6. Bd3 Rd6 7. Be4 Rd2 8. Kg4 Rd3 9. Bf4 Rg3 10. hg3 c6 11. Rh6 Qb6 12. Bc7 Qd4 13. Rd6 Qf6 14. Ng6 Qd8

R060 - Mark Kirtley
1. a4 Nf6 2. Ra3 Ne4 3. Rh3 Nd2 4. Rh7 Nf1 5. Rh8 Ng3 6. hg3 a5 7. R1h7 Ra6 8. Bh6 Rh6 9. Nc3 e6 10. Nd5 Qh4 11. Ne7 c5 12. Ng8 Bd6 13. Nf6 Ke7 14. Ne4 c4 15. Nd2 c3 16. Nb1

R061 - Satoshi Hashimoto
1. e3 f5 2. Bb5 f4 3. Bc6 dc6 4. Ne2 Qd5 5. Rg1 Qf3 6. gf3 Bh3 7. Rg6 Bf1 8. Rd6 g5 9. b4 g4 10. Bb2 g3 11. Bd4 g2 12. c3 g1=Q 13. Qc2 Qg5 14. Ng3 Ba6 15. Qf5 b5 16. Kf1 ed6 17. Kg1 Qd8 18. Nf1 Bc8

R062 - Michel Caillaud
1. d4 c6 2. Bf4 Qb6 3. Be5 Qc5 4. dc5 Na6 5. Qd6 Nb4 6. Qc7 d5 7. Nd2 Bg4 8. Rd1 Bf3 9. ef3 Rd8 10. Ba6 Rd6 11. Ne2 Rh6 12. OO Rh3 13. gh3 e6 14. Kg2 Be7 15. Kg3 Bh4 16. Kg4 Bg3 17. fg3 Nd3 18. Rf2 Nf4 19. Nf1 Ng6 20. Rd3 Nf8 21. Ra3 Nd7 22. b3 Nb8

R063 - Unto Heinonen
1. a4 b5 2. ab5 Na6 3. Ra6 d5 4. Rg6 hg6 5. b4 Rh3 6. Ba3 Rg3 7. hg3 Nh6 8. Rh6 Bh3 9. gh3 gh6 10. Bg2 Bg7 11. Be4 Bc3 12. dc3 a5 13. Kd2 de4 14. Ke3 Qd3 15. cd3 OOO 16. ba5 Rd5 17. Bd6 ed6
Unfortunately, this correction of R034 is cooked too, as Michael Kozulya found out: 1. a4 b5 2. ab5 Na6 3. Ra6 d5 4. b4 Qd6 5. d3 Qg3 6. hg3 Nh6 7. Rh3 Bh3 8. Bh6 g6 9. Bf8 OOO 10. gh3 Rhf8 11. Bg2 h6 12. Be4 de4 13. Kd2 Rd5 14. Ke3 Rfd8 15. c3 R8d6 16. Rd6 a5 17. ba5 ed6

R064 - Andrei Frolkin
The only letters that are only once in the diagram are the C and the D. So these letters are the two kings. Piece A gives either of those two kings check. The only legal check would be if A is a white rook, C is the white king, and D is the black king. This means that there are 10 white rooks, so no white pawns. H isn't white (there are 3 H's, and all the pawns were used for rook promotions). It can't be a black Q, R, B or P either (illegal check, or on the first rank), so H is a black knight. B isn't white (illegal check, or needs a promotion). It's not a black Q, R, or P either, so it's a black bishop. E isn't white either (again, illegal check or needs a promotion), it's not a black R or Q either (illegal check), so it's a black pawn. Now for F, G and I we have as candidates the black queen, the black rook, the white knight and the white bishop. Every black promotion implies an extra capture (all white pawns promote, so either the white pawn has to capture to let the black pawn pass it, or the black pawn has to capture to let the white pawn pass). Now suppose both the black queen and the black rook are on the board. This would require 7 captures (4 promotion pieces are still on the board, the captures on the c-line and on e8, and on e8 is a promoted piece captured too). But only 6 pieces are missing. So the white bishop and knight are both on the board. This is only possible if I is the white bishop, and G is the white knight.
The white c-pawn needed a capture to get passed the black c-pawns, and the last move was a capture too. So only 3 captures of black pieces are left to allow black pawns to promote, so there are only 3 black promotion pieces. 2 of them (bishop and knight) are already on the board, so only one of the three F's is promoted, so F must be a rook, and the piece captured on e8 must be the black queen.
This results in the following position:

[RRRRRRbR/1nKbk3/2p3R1/1N6/2p2n2/rR5B/7B/RNnrr3]

The last move was -1. f7xQe8.

R047c - Thierry le Gleuher
1. a4 h5 2. a5 h4 3. a6 h3 4. ab7 hg2 5. h4 g5 6. Rh3 g4 7. Rd3 g3 8. Nf3 g1=N 9. Rd7 Nh3 10. Bh3 g2 11. d4 g1=B 12. Kf1 a5 13. Ne1 a4 14. f3 Be3 15. Be3 a3 16. Nd2 a2 17. Rc1 a1=Q 18. c4 Qb1 19. c5 Qf5 20. Bf5

Probleemblad, September 1999

R065 - Satoshi Hashimoto
1. a4 d5 2. Ra3 Bg4 3. Rh3 e6 4. Rh7 Qf6 5. Rg7 Rh2 6. Rh7 Rg2 7. R1h3! Qb2 8. Ra3 f6 9. Ra1 Qa2 10. Rh1
Exchanging rooks. 7. R7h3 would lose the last tempo.

R066 - Gianni Donati & Michel Caillaud
1. c3 Nf6 2. Qa4 Ne4 3. Qc6 Nd2 4. Kd2 e5 5. Kd3 and now:

A theme that has been seen a lot of times lately, but here for the first time in 2-phases form

R067 - Gianni Donati
1. d4 c5 2. d5 Nc6 3. dc6 d5 4. c7 Bf5 5. c8=R Kd7 6. Rc6 Qe8 7. Rh6 gh6 8. g4 h5 9. g5 Bh6 10. g6 Be3 11. g7 Nh6 12. g8=R Rd8 13. Rg4 Bg6 14. Rb4 cb4 15. c4 Bb6 16. c5 d4 17. c6 Kd6 18. c7 Rd7 19. c8=R Bd8 20. Rc5 b6 21. Ra5 ba5
Three white Frolkin-rooks without capture of adversive pawns.

R068 - ?#8240;tienne Dupuis
1. a4 h5 2. a5 h4 3. a6 h3 4. Ra5 Rh4 5. Rh5 g6 6. Rh8 Bh6 7. g3 Kf8 8. Bh3 Kg7 9. Kf1 Nf6 10. Kg2 Qg8 11. Kf3 Qh7 12. Rc8 Nc6 13. Rh8 Qg8 14. Bf1 Qd8 15. Nh3 Ng8 16. Rg1 Kf8 17. Rg2 Ke8 18. Ng1 Bf8 19. Rh5 Rc8 20. Rg5 Rh8 21. Kg4 Nb8
The scheme for the switchbacks is the same as R041 by Heinonen, but in this version all the switchbacks are of the same kind (allowing another piece to travel through the homesquare)

R069 - Unto Heinonen
1. g4 c5 2. g5 Nc6 3. g6 Rb8 4. gh7 g5 5. Nf3 Bh6 6. Rg1 Kf8 7. Rg4 Qe8 8. Rc4 g4 9. e4 g3 10. e5 g2 11. e6 g1=R 12. ed7 Rg3 13. d8=Q Bh3 14. b4 Bf1 15. b5 Bh3 16. b6 Bc8 17. ba7 e6 18. a8=Q Nge7 19. Qa4 Rhg8 20. h8=B Ra8 21. Bd4 Rh8 22. Na3 Ng8 23. Qf6 Qd8 24. Rb1 Ke8 25. Rb5 Bf8 26. Ra5 b5 27. Ng1 b4 28. Qdf3 b3 29. Qb4 b2 30. Qfb3 b1=R 31. f3 Ra1 32. Nb1 Nb8
Ten switchbacks, including all black pieces! An amazing proofgame...

R070 - Thomas Volet
The white pawn on f7 is the pawn starting on a2, which means that all missing black pieces were captured by the white pawns, so the black h-pawn must have promoted. The position unlocks when the black king can move. For this, the white bishop on g8 must go back to f1. During this trip, many screens are needed to avoid illegal checks. The retro-play is:
bBh4 ->d4; wQg7 ->b3; bBb1-a2; wQb3-c4; bBa2-b3; wQc4~; bBb3-a4; bBa4 -> g6; wBg8-h7; bRe8-h8; wBh7-g8; bRh8 ->f5; bBg6~; wBg8 ->f1; bKe4-f4; e3-e2; and the rest of the retro-play isn't very difficult.

R056v - Michel Caillaud
1. Nc3 d5 2. Ne4 d4 3. Ng5 Qd5 4. Nh7 Qa2 5. Ng5 Rh2 6. N1f3 Rg2 7. Rh8 Rf2 8. Nh2 Rf6 9. Bh3 Rg6 10. Kf2 Qb2 11. Kg3 Qc2 12. Kh4 Qd2 13. Nh7 Qe2 14. Bh6 Qa6 15. Qh5 d3 16. Rh1

R049c - Unto Heinonen
1. c4 h5 2. c5 h4 3. c6 h3 4. cb7 hg2 5. h4 Rh6 6. h5 Rc6 7. h6 Rc2 8. h7 c5 9. h8=B Nc6 10. b8=B g5 11. Bf4 Rb8 12. Bhe5 Rb3 13. Bb8 d6 14. Rh8 Bh3 15. Nc3 e6 16. Ne4 Be7 17. Ng3 Kf8 18. Nh1 Rg3 19. b3 Kg7 20. Bb2 Nf6 21. Rg8 Kh6 22. Rc1 Nd5 23. Bh8 Rb2 24. d4 cd4 25. Rc5 Qf8 26. Bc1 Bd8

Probleemblad, November 1999

R071 - M. Caillaud
1. e4 h5 2. Qg4 hg4 3. Nf3 gf3 4. Be2 fe2 5. Rf1 ef1=B 6. b4 Ba6 7. b5 Rh3 8. ba6 b5 9. Ba3 Bb7 10. Bd6 ed6 11. Nc3 Be7 12. Ne2 Bf6 13. Ng1 Bc3 14. e5 Ba5 15. e6 Rc3 16. h4 Be4 17. h5 Nc6 18. h6 Rb8 19. h7 Rb7 20. h8=Q Qb8 21. Qh5 Nd8 22. Qd1 f6 23. e7 Kf7 24. e8=N c6 25. Nc7 b4 26. Nb5 b3 27. Na3 b2 28. Nb1

R072 - Satoshi Hashimoto
1. Nc3 h5 2. Nd4 h4 3. Ng3 hg3 4. h4 d5 5. Rh2 gh2 6. Rb1 hg1=B 7. Ra1 Bh2 8. Rb1 Be5 9. Ra1 Bb2 10. Rb1 Bc1 11. Rb3 Qd6 12. Rh3 Nd7 13. Rh1 Qh2 14. Rg1 e5 15. Rh1 Bfa3 16. Rg1 Ne7 17. Rh1 OO 18. Rg1 Kh8 19. Rh1 Rg8 20. Rg1 Nf8 21. Rh1 Bd7 22. Rg1 Re8 23. Rh1 Nc8 24. Rg1 Re6 25. Rh1 Rc6 26. Rg1 Rc3 27. Rh1 c5 28. Rg1 c4 29. Rh1
However, Etienne Dupuis showed this is cooked, for instance: 1. b4 d5 2. h4 c6 3. Nf3 h6 4. Ng5 Qc7 5. Bb2 Nd7 6. b5 hg5 7. Be5 cb5 8. Rg1 g4 9. Bh2 g3 10. Na3 gh2 11. Rb1 hg1=B 12. Nc4 Bd6 14. Rh3 Ba3 15. Rh1 Bc1 16. Nb6 Qh2 17. Rg1 e5 18. Rh1 Bfa3 19. Rg1 Ne7 20. Rh1 00 21. Rg1 Kh8 22. Rh1 Rg8 23. Rg1 Nf8 24. Rh1 Bd7 25. Rg1 Rc8 26. Rh1 Rc3 27. N4 Nc8 28. Rg1 bxc4 29. Rh1. In fact, playing 15. Rh2 and 16. Rh1, the position can be reached in 28,0 moves instead of 28,5.

R073 - Unto Heinonen
1. a4 Na6 2. a5 Nc5 3. a6 Nh6 4. ab7 Nf5 5. b8=N Ba6 6. Nc6 Bc4 7. Ra6 Rc8 8. Rb6 a5 9. g4 a4 10. Na5 c6 11. g5 Qc7 12. g6 Kd8 13. gf7 g5 14. Nc3 Bg7 15. Nd5 Bc3 16. Nf3 Bb4 17. c3 g4 18. d4 g3 19. Bh6 g2 20. Qb1 Qg3 21. Ne5 Qg8 22. fg8=N g1=N 23. Ngf6 Nh3 24. Nh5 Ng5 25. Bh3 a3 26. OO a2 27. Rd1 ab1=N 28. Rd3 Na3 29. Re3 Nb5
However, Satoshi Hashimoto found the following cook: 1. a4 Na6 2. a5 Nc5 3. a6 Nh6 4. ab7 Nf5 5. b8=N Ba6 6. Nc6 Bc4 7. Ra6 Qb8 8. Rb6 a5 9. g4 a4 10. Na5 c6 11. g5 Kd8 12. g6 a3 13. gf7 g5 14. Nc3 Bg7 15. Nd5 Bc3 16. Nf3 Bb4 17. c3 g4 18. d4 g3 19. Bh6 g2 20. Qb1 Qg3 21. Ne5 Qg8 22. fg8=N g1=N 23. Ngf6 Nh3 24. Nh5 Ng5 25. Bh3 a2 26. O-O ab1=N 27. Rd1 Na3 28. Rd3 Nb5 29. Re3 Rc8

R074 - G. Donati
White captured h6xg7, so a black pawn promoted on the queen side. The black pawns present in the diagram captured 5 times. So if white may castle, then either white played axb-b8 and black played axb1, or the black pawn was captured on the a-line. In any case, black can't have captured a white piece with his last name, so the only move that gives white a retromove, and which doesn't ruin white's castling rights, is h7-h5. White wants to mate with O-O-O after black played ef1=B, and the only move that doesn't ruin that mate is g5xh6 ep!. So the solution is:
1. ... gh6 ep 2. ef1=B O-O-O#. The Valladao-theme: a combination of en passant capture, minor promotion and castling.

Probleemblad, Januari 2000

R075 - Joost de Heer
(a) 1. b4 f6 2. Bb2 f5 3. Bf6 ef6 4. a3 Bb4 5. ab4 Ne7 6. Ra6 OO 7. Rf6
(b) 1. b4 f5 2. Bb2 Kf7 3. Bf6 ef6 4. a3 Bb4 5. ab4 Ne7 6. Ra6 Rf8 7. Rf6 Kg8
To castle, or not to castle, that's the question.

R076 - Michel Caillaud
(a) 1. e4 Nf6 2. Bb5 Ne4 3. c4 Nd2 4. Bd2 h5 5. Ba5 Rh6 6. Kd2 Rg6 7. Kc3 Rg2 8. Kb4 Rh2 9. Nc3 Rh3 10. Rh3
(b) 1. e3 h6 2. Bd3 Rh7 3. Bh7 Nf6 4. Bd3 Ne4 5. Bb5 Nd2 6. c4 Nf1 7. Bd2 Nh2 8. Ba5 Ng4 9. Kd2 Ne3 10. Kc3 Ng2 11. Kb4 Nf4 12. c4 Nh3 13. Rh3 h5
A new record in difference (3.5 moves) between the shortest proof game with white and the shortest proof game with black to move.

R077 - Gianni Donati
1. Nf3 e5 2. Nd4 ed4 3. c3 d3 4. Qc2 dc2 5. d3 Ne7 6. Bh6 g5 7. Na3 Ng6 8. Rd1 c1=Q 9. Rg1 Qe3 10. Ra1 Qh3 11. gh3 Be7 12. Rg4 Rf8 13. Rd4 g4 14. Bc1 Bg5 15. Nb1
Switchback of the white queenside figures.

R078 - Satoshi Hashimoto
1. d4 h6 2. Bg5 hg5 3. h4 Rh4 4. g4 Rg4 5. Rh6 b5 6. Rb6 ab6 7. a4 Ra4 8. Bg2 Rd4 9. Ra6 Na6 10. b4 Nb4 11. Bd5 Nd5 12. f4 Nf4 13. Qd3 d5 14. Qf5 Bf5 15. c4 Bb1 16. e4 de4 17. Nf3 Qd5 18. Ne5 Qc4 19. Nd7 Kd7
Black captures all white pieces. All white's pawns are captured after doing a double step.

R072c - Satoshi Hashimoto
1. Nc3 h5 2. Nd4 h4 3. Ng3 hg3 4. h4 d5 5. Rh2 gh2 6. Rb1 hg1=B 7. Ra1 Bh2 8. Rb1 Be5 9. Ra1 Bb2 10. Rb1 Bc1 11. Rb3 Qd6 12. Rh3 Nd7 13. Rh1 Qh2 14. Rg1 e5 15. Rh1 Bfa3 16. Rg1 Ne7 17. Rh1 OO 18. Rg1 Kh8 19. Rh1 Rg8 20. Rg1 Nf8 21. Rh1 Bd7 22. Rg1 Re8 23. Rh1 Nc8 24. Rg1 Re6 25. Rh1 Rf6 26. Rg1 Rf3 27. Rh1 f5 28. Rg1 f4 29. Rh1
The rook oscillates on two different places.

R073c - Unto Heinonen
1. a4 Na6 2. a5 Nc5 3. a6 Nh6 4. ab7 Nf5 5. b8=N Bb7 6. Nc6 Rc8 7. Ra6 Ba8 8. Rb6 a5 9. g4 a4 10. Na5 c6 11. g5 Qc7 12. g6 Kd8 13. gf7 g5 14. Nc3 Bg7 15. Nd5 Bc3 16. Nf3 Bb4 17. c3 g4 18. d4 g3 19. Bh6 g2 20. Qb1 Qg3 21. Ne5 Qg8 22. fg8=N g1=N 23. Ngf6 Nh3 24. Nh5 Ng5 25. Bh3 a3 26. OO a2 27. Rd1 ab1=N 28. Rd3 Na3 29. Re3 Nb5
All the knights are on the 5th rank! A nice extra touch is the promotion squares of the extra knights: on the squares of the original ones!

R065v - S. Hashimoto
1. h4 e6 2. Rh3 Bc5 3. Ra3 Bb6 4. Ra7 c5 5. Rb7 Ra2 6. Ra7 Rb2 7. R1a3 Qc7 8. Rh3 Kd8 9. Rh1 Qh2 10. Ra1 d6
The rooks switch places. This is, as opposed to R065 a shortest proof game.

Probleemblad, March 2000

R079 - Donati, Gianni
1. Nc3 d5 2. Ne4 Bg4 3. Ng5 Be2 4. N1f3 Bc4 5. d3 Ba2 6. Ke2 Bb1 7. Ra6 h6 8. Rh6 Nf6 9. Rh8 Ng4 10. Nh7 Nh2 11. Bh6 Ng4 12. Nh2 Ne3 13. Kf3 Ng2 14. Kg4 g5 15. Kh5 g4 16. Qg4 Ne1 17. Qh4 Kd7 18. Bh3 Kc6
The difference with the other problems with all officers on the a- (or h-) line is that in this case, the position of the officers is turned 90 degrees, instead of mirrored.
?#8240;tienne Dupuis found the following cook: 1. Nc3 Nf6 2. d3 g5 3. Ne4 Rg8 4. Ng5 Nd5 5. Nh7 Nc3 6. Nf3 Rg2 7. Bh2 Rh2 8. Bh3 Ne2 9. Nh2 Nf4 10. Qh5 Ng2 11. Ke2 d6 12. Kf3 Be6 13. Qh4 Ba2 14. Rag1 Ne1 15. Kg4 Kd7 16. Kh5 Kc6 17. Rg8 Bb1 18. Rh8 d5

R080 - Hashimoto, Satoshi
1. Nc3 h5 2. Ne4 h4 3. Ng3 hg3 4. h3 Rh4 5. Rh2 gh2 6. a3 hg1=B 7. a4 Bh2 8. a5 Bd6 9. g3 Ba3 10. Bg2 c5 11. Bd5 Qb6 12. Ba2 Qb3 13. Bb1 b5 14. Ba2 Bb7 15. Bb1 Bd5 16. Ba2 Nc6 17. Bb1 OOO 18. Ba2 Kb7 19. Bb1 Rc8 20. Ba2 Nd8 21. Bb1 Rc6 22. Ba2 Rg6 23. Bb1 Rgg4 24. Ba2 g5 25. Bb1 Bg7 26. Ba2 Bc3 27. dc3 Be6 28. Qd5 Ka6 29. Bd2 f6 30. OOO

R081 - Caillaud, Michel
1. d4 h5 2. d5 h4 3. d6 h3 4. dc7 hg2 5. h4 a5 6. h5 a4 7. h6 a3 8. h7 ab2 9. hg8=R Rh4 10. a4 Rg4 11. Rhh8 Ra5 12. Nh3 Rh5 13. a5 d5 14. a6 d4 15. a7 Qd5 16. a8=R Nd7 17. Rb8 g1=B 18. Raa8 Bh2 19. Na3 b1=R 20. e4 Rb6 21. e5 Rbh6 22. e6 Nb6 23. ef7 Kd7 24. c4 e5 25. c5 Bd6 26. f8=R e4 27. Re8 Bdg3 28. c6 Kd6 29. Bb5 Be6 30. c8=R e3 31. c7 e2 32. Bd7 ed1=B 33. Kf1 d3 3 4. f4 d2 35. f5 dc1=Q 36. f6 Qe3 37. f7 Ba4 38. f8=R Bf7 39. Rcd8 Qee6 40. c8=R Bb5 41. Nc4

R082 - Verduin, Jan Hein
1. Nc3 2. Ka4 Now black wants to play 3. Ka5, and white has only got b2-b4 as last move (the king didn't move into check, because the pawn is paralysed due to the Madrasi condition!). But the position before b2-b4 is illegal, since the black pawn on a2 had to pass a3 or b3, where it would've been paralysed. So black needs to remove the c2-pawn first. 3. Kb5 4. Kc6 5. Kd5 6. Ke4 7. Kf3 8. Ke2 9. Kd1 10. Kc2 11. Kb3 12. Ka4 13. Ka5! Now white's last move must've been b2-b4. 14. cb3 ep 15. Ka4 16. Ka3 17. Na4 Bb4#

R083 - Volet, Thomas
To unlock the position, black needs to bring a bishop to f8 and a rook to h8, or a bishop to c8 and a rook to a8. So white needs to uncapture a rook first.
The retro-play is: Black uncaptures a bishop on b6, unpromotes the bishop currently on h2 on c1, and brings back the pawn to c4. Then retract Nc7-b5 h7-h6 Nd5-c7 f7x[Q/R]g6 Nb4-d5 c5-c4 Nd3-b4 Bc2-d1 Ra1-g1 Bd1-c2.
Now the position can be unlocked. There are two cases now:

So in both cases, a rook was captured on b3!

R084 - Goldsteen, Harry
All (but one) pawns promoted to a bishop. There are three possible squares where one can add a mating bishop, namely a white one on h8 or g7, and a black one on g5.
First some general remarks: At least seven captures are needed to get seven pawns past each other. Pawns that started on the same line promote on the same colour. For different coloured promotions, extra captures are needed.
Add a white bishop on h8: The last moves were g7xh8=B Ng6-h8, or g7xf8 Re8-f8. In the position with the white pawn on g7, 5 white and 5 black officers are missing. Seven black pawns promoted on a black square. An example of the promotion squares is white: b8 (2x), d8 (2x), e8 (2x), g8; black: a1 (2x), c1 (2x), e1, g1 (2x). The pawn captures needed for this are: white: axb, cxd, fxe (captures the black e-pawn), hxg; black: bxa, dxc, fxe, hxg, gxhxg (the g-pawn must get past the white g-pawn), which results in too many captures. Other promotion arrangements result in the same number of captures. So this case is illegal.
Add a white bishop on g7: The last moves were: Bh8xNg7 Ne8xNg7 (this must be an uncapture, or we'd end up in the previous case). This is possible: the white promotion squares are b8 (2x), d8 (2x), e8 (2x), g8, h8, and the black ones are a1 (2x), c1 (2x), e1, g1 (2x). This can be reached with the pawn captures white: axb, cxd, fxe (captures e-pawn), gxh, hxg; black: bxa, dxc, fxe, hxg.
Add a black bishop on g5: The last move is Bxg5. Five white and five black pieces are captured. All black pawns promoted on a black square. Possible promotions are: white: b8 (2x), d8 (2x), e8 (2x), g8; black: a1 (2x), c1 (2x), e1 (2x), g1(2x). The requires captures are: white: axb, cxd, fxe, exfxe, hxg; black: bxa, dxc, fxe, hxg (captures g-pawn). This is one white capture too many. Other capture combinations don't work either.

So the only possibility is adding a bishop on g7!

R085 - Goldsteen, Harry:
(a): There are a few possibilities: First look at the case when a black bishop is added on a5/b4/d2/d4/e5/f6/g7. This requires that the last black move was a capture. There are now five black and five white pieces available for capture. Possible promotion squares are: white: a8 (2x), c8, e8 (2x), g8 (2x); black: a1 (2x), c1 (2x), f1 (2x), h1 (2x). The needed pawn captures are: white: bxa, dxc, fxe, hxg; black: axbxa (to get past the white a-pawn), bxa, dxc (captures the c-pawn) exf, gxh. This is one black capture too many. Other tries result in too many captures too. So these cases are illegal. The next possibility is a black bishop on e1. The last moves were d2xe1=B Bc1xb2 (this must be a capture, otherwise the previous move was b2xa1=B and white was retropat!) Now there are five white and four black pieces left for capture. This is possible with the following captures: white: bxa, dxc, fxe, hxg; black: bxa (captures a-pawn), cxdxc, exf, gxh. The last possibility is a white bishop on d2. The last move was Be1-d2, and the retroplay is easy. So a legal mating bishop can be added on d2 (white) or e1 (black).
(b): Due to the same reasoning as in (a), adding a bishop on b7/c6/d5/e4/e2/g4 is illegal. Adding a bishop on d1 gives a legal mate, the last move was c2xd1=B, e2xd1=B or e2xf1=B (not d2-d1=B due to retropat). Adding a white bishop on g4 gives a legal mate too. So a legal mating bishop can be added on d1 (black) or g4 (white).
(c): In this case, all ten possible additions are legal: Adding a black bishop on a1/b2/c3/d4/e5/e7/g5/h4 doesn't require a capture now, since white has e7-e8=B as retromove. Adding a white bishop on f8 (last move was e7-e8=B, not e7xf8=B!) and e7 (last move was Bd8-e7) are legal too.
(d): Due to the same reasoning as in (a) and (b), a black bishop on b5/d7/e8/d5/e4/f3/g2 results in an illegal position. A black bishop on h1 is legal though, and the last move was g2xh1=B. A white bishop on b5 gives legally mate too.

Shorties II

1 - Martin W. Hoffmann
(a): 1. d3 e5 2. Qd2 Ba3 3. Qb4 f6 4. Qf8 Bf8
(b): 1. d3 e5 2. Bh6 Qg5 3. Qc1 Qc1 4. Bc1 f6
Switchbacks with capturing return-moves easily charm the solver; can two or more such switchbacks be shown in a single solution of no more than 7.0 moves?

2 - Gianni Donati
1. b4 h5 2. b5 h4 3. b6 h3 4. bc7 hg2 5. cbq=Q Rh2 6. Qh2 gh1=R 7. Qh8 Rh8
Two promotions, one with the Pronkin theme, the other with the Ceriani-Frolkin theme, in one short line of play.

3 - Gianni Donati
1. Nc3 d5 2. Nd5 f6 3. Nf6 Kf7 4. Ne4 Qd2 5. Nd2 Ke8 6. Nb1
The knight's circuit describes a parallellogram-shaped hexagon.

4 - Peter Wong
1. Nf3 e5 2. Ne5 Qf6 3. Nd7 Qb2 4. Ne5 Qa2 5. Nf3 Qd5 6. Ng1 Qd8
1. b3 d6 2. Ba3 Be6 3. Bd6 Bb3 4. Be7 Ba2 5. Ba3 Be6 6. Bc1 Bc8
Four switchbacks, counting the queen circuit. The second solution repeats no move of the first solution, nor is there even any unit that is moved in both solutions!

5 - Peter Wong
1. Nc3 d5 2. Nd5 Be6 3. Ne7 Bd5 4. Nd5 Bd6 5. Nc3 Bg3 6. Nb1 Qd3 7. c3
1. Nf3 e5 2. Ne5 Bd6 3. Nd7 Bg3 4. Ne5 Bg4 5. c3 Bf3 6. Nf3 Qd3 7. Ng1
1. c3 e6 2. Qa4 Bd6 3. Qd7 Kf8 4. Qc8 Bg3 5. Qe6 Qd3 6. Qb3 Ke8 7. Qd1
Again, four switchbacks, three white and one black. Part of the fun for me is watching how the move c3 arrives at various moments. Not only are there three solutions, meaning that the first white move is different each time, but the same is also true for the first black move. Can anyone compose another 3-solution position?!

6 - Mark Kirtley
1. d3 c6 2. Be3 Qb6 3. Bc5 Qc5 4. e3 b6 5. Be2 Ba6 6. Bf3 Bc4 7. Bd5 Bd5
1. e3 b6 2. Bc4 Bb7 3. Bd5 Bd5 4. d3 c6 5. Bd2 Qc7 6. Bb4 Qd6 7. Bc5 Qc5
Contrasting orders in captures.

7 - Gianni Donati
1. d4 e5 2. d5 Ke7 3. d6 Ke6 4. dc7 d5 5. cb8=Q/R/B/N Bd7 6. Qc7/Rc8/Bd6/Na6 Bb5 7. Q/R/B/Nc5 Bc5
The differentiation between variations is minimal, but elegant!

8 - Joost de Heer, Gianni Donati & Michel Caillaud
1. a3 e6 2. Ra2! Ba3 3. Ra3 f6! 4. Rh3 f5 5. Rh6 Nh6 6. h3! Rf8 7. h4 Ng8
The white rook sweeps over the third rank before the second white tempo to prevent a switchback dual by a white knight.

9 - Gerd Wilts
1. e3 Nf6 2. Qg4 Nd5 3. Qe6 fe6 4. f4 Kf7 5. f5 Qe8! 6. f6 ef6 7. d3 Qe7
1. f4 e6 2. f5 Ke7 3. f6 Nf6 4. e3 Nd5 5. Qh5 f6 6. Qf7 Kf7 7. d3 Qe7
The black Queen takes a two-move path in the first solution, performing a tempo. In the other solution it is the black king that moves twice, although there is no tempo here, since the white Queen must not pin the black Pawn f7.

10 - Pascal Wassong
1. Nh3 d6 2. Rg1 Bh3 3. gh3 g5 4. Rg3 g4 5. Rg2 g3 6. Rg1 g2 7. Rh1 gf1=B
The phoenix promotion may well lure the solver away from the Rook trek, making it more pointed when found.

11 - Mark Kirtley & Michel Caillaud
1. b4 a6 2. b5 ab5 3. a4 Ra4 4. f4 Rf4 5. e4 Re4 6. Kf2 Rh4 7. Qg4 b4
1. b3 a5 2. b4 ab4 3. a3 Ra3 4. f3 Rf3 5. e3 Re3 6. Kf2 Rh3 7. Qg4 Rh4
In both solutions, the length of White's opening Pawn move is later matched by three of the Pawn's colleagues, while the black Pawn answers with a contrary first step.

12 - Michel Caillaud
1. e4 d5 2. e5 Qd6 3. ed6 Nh6 4. dc7 Nf5 5. cb8=Q Nd4 6. Qf4 Be6 7. Qf7 Bf7
1. e3 Nf6 2. e4 Nd5 3. ed5 Nc6 4. d6 Nd4 5. dc7 d5 6. cd8=N Be6 7. Nf7 Bf7
The Ceriani-Frolkin theme is doubled, while the identity of the surviving black Knight switches!

Probleemblad, May 2000

R086 - Joost de Heer
1. d3 h6 2. Bh6 b6 3. Bc1 Rh2 4. Rh2 Ba6 5. Rh1 Bd3 6. Qd3 b5 7. Qd1
Three switchbacks in a minimal setting.

R087 - Satoshi Hashimoto
1. a4 g6 2. a5 g5 3. a6 ba6 4. Ra4 Bb7 5. Rf4 Bf3 6. e4 g4 7. Bb5 Be2 8. Bc6 dc6 9. f3 Qd2 10. Kf2 Qc1 11. Na3 Bc4 12. Qd3 Be6 13. Nc4 Bc8

R088 - Gianni Donati
1. f4 c6 2. Kf2 Qa5 3. Kf3 Qc3 4. dc3 h5 5. Qd6 Rh6 6. Qa3 Rd6 7. Be3 Rd2 8. Bc5 d5 9. e3 d4 10. Bc4 Re2 11. Nd2 d3 12. Nf1 d2 13. Bb3 d1=Q 14. c4 Qd8 15. Rd1 h4 16. Rd5 Rd2 17. Rf5 Rd4 18. ed4

R089 - ?#8240;tienne Dupuis & Michel Caillaud
1. a4 e6 2. Ra3 Bb4 3. Rg3 Bc3 4. bc3 h5 5. Ba3 h4 6. Bc5 h3 7. Bb6 hg2 8. Nh3 gh1=B 9. Rg1 Be4 10. Rh1 ab6 11. Ng1 Ra5 12. Bh3 Rah5 13. a5 Qg5 14. a6 Qb5 15. a7 c5 16. a8=R Nc6 17. Rc8 Ke7 18. Ra8 Nf6 19. Ra1 Ra8 20. Bg4 Rhh8

R090 - Unto Heinonen
1. e4 h5 2. Qg4 hg4 3. Na3 Rh3 4. Nc4 Rf3 5. h4 Nh6 6. h5 Nf5 7. h6 d5 8. h7 dc4 9. Rh6 Qd4 10. Rf6 ef6 11. e5 Bd6 12. e6 Bg3 13. e7 Qf4 14. d4 Be6 15. d5 Kd7 16. d6 Kc6 17. h8=R Nd7 18. Rh1 Rh8 19. e8=N Nf8 20. d7 Rh3 21. Nd6 Bh4 22. Ne4 Nd6 23. d8=Q Ne8 24. Qd1 Rhg3 25. Nd2 Kd6 26. Nb1

R091 - Unto Heinonen
1. h4 a5 2. h5 a4 3. h6 a3 4. hg7 ab2 5. Rh6 Ra3 6. Rg6 h5 7. d4 h4 8. d5 h3 9. d6 h2 10. de7 d5 11. g3 d4 12. Bg2 d3 13. Kf1 d2 14. Qe1 d1=B 15. Nd2 b1=B 16. c4 Be4 17. e3 Rd3 18. a4 h1=B 19. a5 Rh2 20. a6 Nh6 21. g8=B Bg7 22. a7 Bb2 23. Raa6 f6 24. Bd5 Kd7 25. Re6 b6 26. a8=B Bb7 27. e8=B Kc8 28. Bec6 Bdf3

R092 - Nikita Plaksin & Alexander Zolotarev
Retro-play: -1. .. f7xNe6 -2. Nd8-e6 e5xNf4 -3. d7-d8=N d6xRe5 -4. c6xRd7, and further retro-play involves a3->a8=N->f4; a4xRb3; g3->g8=R->b3; g4->g8=R->e5; h2xBg3; Bf8->g3; g7xNf6 and the whole position unlocks.
So yes, the game could have started with 1. c2-c4.

This problem was first published in Europe Echecs, Oct. 1998

R093 - Gianni Donati
Start from the following position:

[8/pppp1pp1/4p3/8/8/2Pr4/1PpPPPPP/2K1kBnR]

Before this position, at least one bishop move occurred. (Bf8 was captured by Pa2, who promoted on e8) From the starting position, play forward: 1. ed3 Nf3 2. Be2 Ng1 3. Bf3 Kf1 4. Be4 Ke2 5. Bd5 Nf3 6. Rd1 Ne1 7. Bf3 Kf1 8. Be2 Kg1 9. Bf1 Nf3 10. Re1 Nd4 11. Re5 Kh1 12. Be2 Kg1 13. Bf3 Kf1 14. Bc6 Ke1 15. Bb5 Kf1 16. Ba6 ba6 17. Rd5 a5 18. Rd6 cd6 19. cd4 and the diagram position is reached. In total there were at least 1+12=13 bishop moves.

R079c - Gianni Donati
1. Nc3 d5 2. Ne4 Bg4 3. Ng5 Be2 4. N1f3 Bc4 5. d3 Ba2 6. Ke2 Bb1 7. Ra6 h6 8. Rh6 Nf6 9. Rh8 Ng4 10. Nh7 Nh2 11. Bh6 Ng4 12. Nh2 Ne3 13. Kf3 Ng2 14. Kg4 g5 15. Kh5 g4 16. Qg4 Ne1 17. Qh4 Kd7 18. Bh3
The only difference between this problem and the original version is the omission of the last move.

Probleemblad, July 2000

R094 - Satoshi Hashimoto
1. Nf3 e5 2. Ne5 c6 3. Nc6 a5 4. Na5 Bc5 5. Nb7 Ne7 6. Nc5 OO 7. Nd7 f6 8. Nf6 Kh8 9. Nh7 g5 10. Ng5 Bh3 11. Nh3 Rf3 12. Ng1

R095 - Michel Caillaud
1. b4 Nh6 2. b5 Nf5 3. b6 h6 4. ba7 Rh7 5. ab8=B Ra6 6. d4 Re6 7. d5 b6 8. d6 Bb7 9. de7 Bf3 10. ef8=B Bh5 11. Bfa3 c5 12. Bcf4 Ke7 13. Bbd6 Kf6 14. Bf8 Qa8 15. Bb8 Kg5 16. Bc1 Kg4
A cyclical interchange between the three white bishops in a minimal settting, where only the thematical white pieces move. Also notice that after white's 10th move, the bishops are already on the same squares as in the diagram position!

R096 - Gianni Donati
1. a4 f5 2. Ra3 f4 3. Rc3 f3 4. Rc6 fg2 5. f4 d5 6. Nf3 g1=Q 7. Nc3 Qb6 8. Rg1 d4 9. Rg5 d3 10. Ra5 de2 11. f5 ef1=B 12. Qe2 e5 13. Qg2 Bc4 14. d3 e4 15. Bf4 e3 16. Nd1 e2 17. Kd2 e1=R 18. Kc3 Re5 19. dc4 Rb5 20. cb5 Qa6 21. ba6

R097 - Unto Heinonen
The intention was: 1. b4 a5 2. b5 a4 3. b6 Ra5 4. bc7 b5 5. e4 Ba6 6. c8=N b4 7. Nd6 ed6 8. e5 Qg5 9. e6 Kd8 10. e7 Kc8 11. e8=B f6 12. Bg6 hg6 13. c4 Rh3 14. c5 Rc3 15. h4 Nh6 16. h5 Nf7 17. h6 Be7 18. h7 Qh6 19. c6 Rh5 20. c7 Kb7 21. c8=R Bb5 22. Rc5 Na6 23. Rg5 fg5 24. h8=Q Bf6 25. Qg8 Nh8 26. Qgb3 Kc8 27. Qa3 ba3
But Andrei Frolkin gave the following cook: 1. b4 a5 2. b5 a4 3. b6 h6 4. bc7 b5 5. c4 Ba6 6. c8=Q b4 7. Qc5 Bb5 8. Qg5 hg5 9. c5 Rh3 10. c6 Rc3 11. c7 Na6 12. c8=N Rc2 13. Nd6 ed6 14. h4 Qf6 15. h5 Ne7 16. h6 Ng6 17. h7 OOO 18. h8=Q Be7 19. Qh3 Rh8 20. Qa3 Bd8 21. e4 Rh5 22. e5 Nh8 23. e6 Qh6 24. e7 ba3 25. e8=Q Rc3 26. Qe4 Bf6 27. Qg6 fg6. And Olli Heimo gave another cook: 1. b4 a5 2. b5 a4 3. b6 Ra5 4. bc7 b5 5. c4 Ba6 6. c8=N a3 7. Nd6 ed6 8. c5 Qg5 9. c6 Be7 10. c7 Bf6 11. c8=N Ne7 12. Qa4 Nec6 13. Ne7 Kd8 14. Ng6 hg6 15. e4 Rh3 16. e5 Rc3 17. h4 Kc8 18. h5 Nd8 19. h6 ba4 20. h7 Qh6 21. e6 Rh5 22.h8=N g5 23. Ng6 fg6 24. e7 Bb5 25. e8=Q Na6 26. Qe2 Nf7 27. Qd1 Nh8

R098 - Andrei Frolkin & Andrei Kornilov
The last moves were: -1. .. Bg7-h8 -2. Qh8-f8 Ra8-b8 -3. Qg8-h8 Bh8xNg7 -4. Kf8xBe8 Nb6-c8 -5. a5-a6 Rd8-d7 -6. a4-a5 Rb8-d8 -7. a3-a4 Nc8-b6 -8. a2-a3 Ba4-e8 -9. Ke8-f8 Bb3-a4 etc. This shows that white needs all the tempi with his a-pawn, so (since a5-a6 could've been the only last white move) white has the move.
Mate in 2 with 1. f7 Kh7/Bf6 2. Qg8/Qg7.

R099 - Alexander Zolotarev
Try: the last moves were: -1. .. e7xNf6 -2. Nd7-f6 g3-g2 -3. Nb6-d7 g4-g3 -4. Ra8-a7 Ba7-b8 -5. Rh8-a8. Further retractions are: h4-h8=R; h5xNg4; Nf3->g4; Rd3-d2 Nd2-f3; Ra8->e3; Ne3->b6; Be4-c2 Nc2-e3; Bc8->e4; d7xRe6 and the whole position is unlocked.
But the rook can't go back to a8!

Solution: Last moves were: -1. Ra8-a7 e7xNf6 -2. Nd7-f6 g3-g2 -3. Nb6-d7 Ba7-b8 -4. Rh8-a8 Further retractions are: h4-h8=R; h4xNg3; Nf1-g3; Rd3-d2 Nd2-f1; Ra8->e3; Ne3->b6; Be4-c2 Nc2-e3; Bc8->e4; d7xRe6 and the whole position is unlocked.
In this case, the rook can go to a8 via h5 (while the white knight screens the white king on e5)
So black has the move, and mate in 1 with 1. .. Ba7#! and not with 1. Qb4#?

R100 - Frank Christiaans
The last moves were: -1. Kf3xPe3 f4xe3 ep -2. e2-e4 Bc6xXd5.
The black bishop on d5 is a promoted one. If the black knight on h1 is a promoted one, then the black g- and h-pawn need 5 captures to promote on f1 and h1. Together with Bc1 and axb6 this is one capture too many. So the knight on h1 is an original one.
If the bishop on h5 is an original one, then it moved after g2-g3, so white couldn't have castled kingside.
If the bishop on h5 is a promoted one, then it is the a-pawn which promoted on e8. White needs 4 captures for this, so both the black g- and h-pawn promoted. This requires 4 captures. One of these captures was the original white bishop on white squares, and this capture took place after white played g2-g3, so white couldn't have castled kingside in this case either.
So white castled queenside. The fastest way to do this is: 1. a4 Nc6 2. Nc3 Nd4 3. Nd5 Nb3 4. Nb6 Nc1 5. Nc8 Nb3 6. Qb1 Nd4 7. Qa2 Nf5 8. OOO

R101 - Goldsteen, Harry
All pawns promoted to bishops. To get the pawns past each other, at least 8 captures are needed, one from each line. The pawns that come from the same line promote on a same coloured square then. To account for the different coloured bishops, extra captures are needed.
The bishop to be added is white. There are two candidates for adding a mating bishop: c7 and d8.
d8: The last move must've been c7xNd8=B (c7xXb8=B isn't possible since the piece on b8 couldn't have moved last, and b2xc1=B isn't possible either, due to the same reasoning why d8 won't work).
Before c7xd8=B, five white and four black pieces are missing. For the bishop promotions, white needs to capture an odd number of times, and black needs to capture an even number of times, and this isn't possible.
c7: The last moves were: Bd8xNc7 Nd5xXc7 (must be a capture, or we end up in the previous case!). The 8 promotions can be arranged easily now (black captures axb, cxd, fxe, hxg and white captures bxa, dxc, exf and gxh)

R102 - Goldsteen, Harry
The bishop to be added must be black.
(a): There are two candidates: +bBc1 and +bBd2. In the latter case, the last move must've been Bxd2 (and not B-d2 because white has no previous move). There are 6 black bishops on black squares, so there is an extra ninth capture needed by black for promotion. This results in an even number of captures for white and an odd number of captures for black, which isn't possible. Adding a bishop on c1 is legal though: The last moves were d2xc1=B Bc3xXd4. The white captures were bxa, dxc, exf and gxh, while the black captures were axb, cxd, fxe and hxg.
(b): Only one square possible to add a mating bishop, namely g4. This is legal, because black doesn't need to uncapture anything on his last move, since white has g7-g8=B as last move.
(c): Again, only one square possible: e7. White's last move was b7-b8=B.
(d): And here, b5 is the only possibility, with h7-h8=B as last white move.

Probleemblad, Sep. 2000

R063c - Unto Heinonen
1. a4 b5 2. ab5 Na6 3. Ra6 d5 4. Rg6 hg6 5. b4 Rh3 6. Ba3 Rg3 7. hg3 Nh6 8. Rh6 Bh3 9. gh3 gh6 10. Bg2 Bg7 11. Be4 Bc3 12. dc3 a5 13. Kd2 de4 14. Kc1 e3 15. Qe1 Qd1 16. Kb2 OOO 17. ba5 Rd3 18. Bd6 ed6 19. cd3

R103 - Peter van den Heuvel
(a): 1. g4 b5 2. g5 Bb7 3. g6 Bf3 4. ef3 b4 5. Bc4 b3 6. Ne2 bc2 7. Rg1 cb1=B 8. Rg2 Be4 9. Qc2 Bb7 10. Qe4 Bc8 11. Bf7
(b): 1. g4 f5 2. g5 f4 3. g6 f3 4. ef3 b5 5. Bc4 b4 6. Ne2 b3 7. Rg1 bc2 8. Rg2 cb1=Q 9. Qb4 Qf5 10. Qb7 Qf7 11. Bf7

R104 - Satoshi Hashimoto
1. d3 a5 2. Kd2 Ra6 3. Kc3 Rb6 4. Bd2 Rb3 5. ab3 Na6 6. Ra4 Nc5 7. Rg4 a4 8. Rg6 hg6 9. Kb4 Rh3 10. Nc3 Rf3 11. gf3 a3 12. Bh3 a2 13. Be6 a1=R 14. Nh3 Ra8 15. Qa1 f6 16. Qa7 Na4 17. Qb8 Ra5 18. Ra1 Rh5 19. Nb1 Rh8

R105 - Joost de Heer
1. Nc3 Nh6 2. Nd5 Rg8 3. Ne7 d5 4. Ng8 Qf6 5. Rb1 Qb2 6. d4 Qa1 7. Ra1 Ng8

R106 - Michel Caillaud
1. f4 Na6 2. f5 Nc5 3. f6 Ne4 4. fe7 f5 5. d4 Kf7 6. e8=R Be6 7. Re6 Bg3 8. hg3 f4 9. Rg6 hg6 10. d5 Rh5 11. d6 Rg5 12. dc7 d5 13. e3 Bg4 14. c8=Q Qb6 15. Qc4 Rc8 16. Qa6 ba6
1. d4 Na6 2. d5 Nc5 3. d6 Ne4 4. dc7 d5 5. f4 Bg4 6. c8=B Qb6 7. Bf5 Rc8 8. Bg6 hg6 9. f5 Rh5 10. f6 Rg5 11. fe7 f5 12. e3 Kf7 13. e8=N Be6 14. Nc7 Bg3 15. hg3 f4 16. Na6 ba6

R107 - Gianni Donati
1. Nc3 Nc6 2. Nd5 Nd4 3. Ne7 Ke7 4. Nf3 Ke6 5. Ne5 Qf6 6. Nc4 Qf3 7. Nb6 ab6 8. ef3 Ra5 9. Ba6 Rg5 10. Kf1 Bc5 11. Kg1 d6 12. Qf1 Bd7 13. Qd3 Bc6 14. Qg6 hg6 15. Kf1 Rhh5 16. Ke1 Nh6 17. Bf1

R108 - Zvonimir Hernitz
Retract -1. Kh3-g2 h4xg3 ep -2. g2-g4 Bg6xQf5, and h#1 with 1. Ke3 Qf3#.

R109 - Henrik Juel
The last moves were: -1. Kb8-c8 Kd8-e8 -2. Rc7-d7 Ke8-d8. Black now plays his king back and forth between d8 and e8, while white retracts: g2-g3, Bc1-b2-e5-h2-g1, b2-b3, Nh8-f7-g5-f3-d4-b3-a1, Nf7xRh8. Now black can shuffle with his rook, while the black king is on e8: Kc8-b8, Qg3-b8-a8, Ka5-b6-c7-c8, Nd8-f7, Rd6-d7, Rd7-c7-c8-a8-a7. Black can retract c7-c6 now, and the two white rooks can leave the cage. Now the white bishop retracts Bh7-g8-f7-e8-a4-b3-a2-b1, black uncaptures f7xNg6, and the white bishop can finally go to f1. White uncaptures e2xRd3, that rook goes to d8, and black takes back d6-d5. And the whole position unlocks.

R110 - Alexander Zolotarev
This problem is cooked. Start from the following position:

[qQrRb3/pnnN1ppp/1BKpp3/2p5/Pk5P/Rp1r3N/P1PPPPP1/5B2]

and play forward:
(i): 1. ed3 Kb4 2. Be2 h6 3. Bg4 h5 4. f3 hg4 5. Nf2 g3 6. h5 gf2 7. h6 f1=B 8. h7 Be2 9. h8=N Bf1 10. Ng6 Be2 11. Ne7 Bf1 12. Nf5 Be2 13. Ne3 Bf1 14. Nd1 Be2 15. Nc3 Bd1 16. Nb5 Nd5 17. Nc7 Ne3 18. f4 Nc4 19. dc4 Bf3
(ii): 1. ed3 Kb4 2. Be2 h5 3. Bg4 hg4 4. f3 g3 5. Nf2 gf2 6. h5 f1=B 7. h6 Be2 8. h7 Bd1 9. h8=N Be2 10. Ng6 Bf1 11. Ne7 Be2 12. Nf5 Bd1 13. Ne3 Be2 14. Nd1 Bf1 15. Nc3 Be2 16. Nb5 Nd5 17. Nc7 Ne3 18. f4 Nc4 19. dc4 Bf3
So both Be2-f3 and Bd1-f3 could've been the last move.
For (b), a similar cook exists: Last moves were -1. .. Bd1-f3/Be2-f3 -2. f3xNe4 Nf6-e4 -3. Ne5-d7 Nd7-f6, and the knight unpromotes on h8. Black can choose between h7-h6-h5 and h7-h5.

R028cv - Thierry le Gleuher
1. e4 d5 2. e5 Kd7 3. e6 Kc6 4. ef7 e5 5. Ke2 Qf6 6. Kd3 Be7 7. f8=Q Be6 8. Qb8 b5 9. Qa8 Kb6 10. Qe8 c5 11. Qg6 hg6 12. Kc3 e4 13. d4 ed3 14. Kb3 d4 15. c4 dc3 16. Ka3 c4 17. b4 cb3

Probleemblad Nov. 2000

R097c - Unto Heinonen
The intention was: 1. b4 a5 2. b5 a4 3. b6 Ra5 4. bc7 b5 5. e4 Ba6 6. c8=N b4 7. Nd6 ed6 8. e5 Qg5 9. e6 Kd8 10. e7 Kc8 11. e8=B f6 12. Bg6 hg6 13. c4 Rh3 14. c5 Rc3 15. h4 Nh6 16. h5 Nf7 17. h6 Be7 18. h7 Qh6 19. c6 Rh5 20. c7 Kb7 21. c8=R Bb5 22. Rc5 Na6 23. Rg5 fg5 24. h8=Q Bf6 25. Qg8 Nh8 26. Qgb3 Kb8 27. Qa3 ba3
But this correction of R097 is cooked too, as Ton van Namen found out: 1. b4 a5 2. b5 a4 3. b6 Ra5 4. bc7 b5 5. c4 Ba6 6. c8=N b4 7. Nd6 ed6 8. c5 Qg5 9. c6 Be7 10. c7 Bf6 11. c8=N Kd8 12. Ne7 Kc7 13. Ng6 hg6 14. h4 Ne7 15. h5 Rc8 16. h6 Rd5 17. h7 Bb5 18. h8=Q Na6 19. Qh3 Qh6 20. Qa3 ba3 21. e4 Rh5 22. e5 g5 23. e6 Ng6 24. e7 Nh8 25. e8=Q Kb8 26. Qe4 Rc3 27. Qg6 fg6 R111 - Mario Parinello
1. e4 e6 2. Bb5 Ba3 3. d3 Qe7 4. Bh6 Nh6 5. Qf3 OO 6. Qf6 Rd8 7. f3 Kf8 8. Kf2 Ke8 9. Kg3 Qf8 10. Kh4 Qh8 11. g3 Bf8 12. a4 Ng8

R112 - ?#8240;tienne Dupuis
1. f3 h5 2. Kf2 Rh6 3. Kg3 Rf6 4. Kh4 Rf4 5. Kg5 f6 6. Kg6 d6 7. Kh7 Qd7 8. Kg8 Kd8 9. Kf8 Qb5 10. Kf7 Nd7 11. Ke6 Nc5 12. Kd5 Bf5 13. h3 Be4 14. Kd4 Nb3 15. Ke3 Qd3 16. Kf2 b5 17. Ke1 b4

R113 - Satoshi Hashimoto
1. c4 f5 2. Qb3 f4 3. Qb6 cb6 4. e3 Qc7 5. Ke2 f3 6. Kd3 Qg3 7. fg3 f2 8. Be2 f1=Q 9. c5 Qc1 10. c6 Qc5 11. Bf1 Qe5 12. Ne2 Qc7 13. Nc1 Qd8 14. c7 Kf7 15. cd8=Q Kg6 16. Qc8 Kh5 17. Qc2 Nc6 18. Qd1

R114 - Gianni Donati
1. b3 c5 2. Ba3 c4 3. Qc1 c3 4. Qb2 cb2 5. Kd1 ba1=Q 6. Bc1 Qc3 7. a4 Qh3 8. gh3 f6 9. Bg2 Kf7 10. Be4 Ke6 11. Bg6 hg6 12. Nf3 Rh5 13. Rg1 Rc5 14. Rg4 d5 15. Ng1 Ke5 16. a5 Be6 17. Ra4 Bf7 18. Ra1 Kd4 19. Ke1 e5

R115 - Pascal Wassong
1. a4 Nc6 2. a5 Na5 3. b4 Nb3 4. Ra6 c5 5. Rc6 dc6 6. Nf3 Bg4 7. Nh4 Bf3 8. gf3 e5 9. Bh3 e4 10. OO e3 11. Kg2 ed2 12. e4 f6 13. Qe2 d1=R 14. Bg5 Qd2 15. Kg3 Qc1 16. Nd2 Kf7 17. Kf4 fg5 18. Ke5 Nf6 19. Nc4 Rd7 20. f4 Re7 21. Kd6 Rg8 22. Bd7 g4 23. Nf5 Kg6 24. Ne7

R116 - Tom Volet
The white captures are: a6xRb7, and the g- and h-pawn on their own file. The position unlocks when white can retract d2-d3. For this, the white rook needs to go back to the d1-h1 cage.
The last moves were for instance: -1. Qb8-a7 Qa7-a6 -2. Bc4-a2 Qa6-a7 -3. Bb3-c4 Qa7-a6 -4. Ra3-a1 Qa6-a7 -5. Ba2-b3 Qa7-a6 -6. Rc3-a3 Qa6-a7 -7. Rc5-c3 Qc4-a6 -8. Bb3-a2 Qd4-c4 -9. Rc3-c5 Qg4-d4 -10. Bc4-b3 Qd1-g4 -11. Bd2-c1 Qc1-d1 -12. Be1-d2 Qd2-c1 -13. Ra3-c3 Qc3-d2 -14. Ra1-a3 Qd2-c3 -15. Rd1-a1 Qc3-d2 -16. Bd2-e1 Qd4-c3 -17. Bc1-d2 Qh8-d4 -18. d2-d3 and the rook is back behind the bishop. Further retracting involves bringing the black king to a7, and only then the white king can escape.

Twice, the white rook takes over the role of king-protector from another piece.

R117 - Alexander Zolotarev
Last moves were: -1. .. Nd7-b6 -2. e5xBd6 d5-d4 -3. e4-e5 Bh2-d6 -4. h3-h4 Bg1-h2 -5. h2-h3 g2-g1=B -6. d4xNc5 g3-g2 -7. e3-e4 h4xBg3. Further retractions are: Bf8-g3; Bg7xQf8, (Qf1-f8) (f7-f1=Q); (Nf1-c5); (f6-f1=N); f2xRe3; (Rb1-e3); (e7-e2); e3xRd4; (b3-b1=R); (Rb1-d4); Bc1-g7; (b4-b3); (Bh3-g4); (b2-b1=R); Qd1-h5; e2-e3; (b3-b2); b2xNc3, etc.

So the longest queen move is Qf1-f8.

R118 - Andrei Frolkin
The position unlocks when black creates a screen on e1. This can only be a knight. But the knights currently on the board need 5 moves to reach e1, while the white pawn has only 4 retro moves. However, if white captures g2xNf3xe4xd5xb6xc7, this knight can reach e1 in time.

What were the white captures? The extra knight is the promoted black a-pawn, which promoted on a1 without capture. So the white a-pawn, which promoted to a rook, must've made one capture. The 5 captures by the white pawn were all on white squares, and the black h-pawn was captured on its own file. So the only possibility is that white captured the black-squared bishop on b4 or b6, and then promoted on b8. So white didn't uncapture a black pawn on b7, but a black rook. On c6, the other rook must've been captured, and the captures on the d- and e-file are the missing black pawns. The position unlocks without problems once the black knight is on e1.

So the route of the white pawn currently on b7 is: g2xNf3xPe4xPd5xRc6xRb7.

Shorties III

1 - Gianni Donati
(a): 1. f4 Nc6 2. f5 Ne5 3. f6 Ng4 4. fe7 Nh2 5. ef8=N Nf3 6. gf3 Ke7 7. f4 Qf8
(b): 1. Nf3 e5 2. Ne5 Nc6 3. Nc6 Bd6 4. Ne7 Bh2 5. f4 Bf4 6. g3 Ke7 7. gf4 Qf8
The first solution shows was could be called a "Schnoebelen" promotion - the promotee never moves but gets captured.

2 - Gianni Donati
1. d3! e6! 2. d4 Bc5 3. dc5 e5 4. c6 dc6 5. Nd2 Be6 6. Nb3 Bb3
1. Nf3 e5 2. Nd4 Bb4 3. Nc6 Bd2 4. Nd2 dc6 5. Nf3 Be6 6. Ng1 Bb3
Both of the first two moves in the first solution are tempo losses - a nice paradox in what is a shortest game to the diagram. In the second solution no time is wasted and White must use the king's Knight right away, and later have the queen's Knight be an imposter for it. Can a shorty with two such imposter knighs be composed, wheter in one or two equal or unequal length solutions, or two equal-length variations or twins?

3 - Gianni Donati
1. d4 Na6 2. Bd2 Nc5 3. Ba5 Nb3 4. Bb6
1. d3 Nc6 2. Be3 Na5 3. Bb6 Nb3 4. d4
No captures and no repeated moves.

4 - Peter Wong
(a): 1. c3 Nh6 2. Qa4 Nf5 3. Qd7 Qd7 4. Nf3 Kd8 5. Nd4 Qd4 6. e3 Nd7 7. Bd3 Qd3
(b): 1. Nf3 Nh6 2. Ne5 Nf5 3. Nd7 Qd7 4. e3 Kd8 5. Bb5 Qb5 6. c3 Nd7 7. Qb3 Qb3
(c): 1. e3 Nh6 2. Bb5 Nf5 3. Bd7 Qd7 4. c3 Kd8 5. Qa4 Qa4 6. Nf3 Nd7 7. Nh4 Qh4
Despite the repeated black moves, this is a remarkable work. The white Queen is captured on three different squares, and the same is true for the white Bishop, and the white Knight. Also, the order for these captures is changed cyclically in the three solutions (QNB/NBQ/BQN).

5 - Gianni Donati
1. Nf3 e5 2. Ne5 Bd6 3. Nf3 Bh2 4. d3 Bg1 5. Rh7 Qh4 6. Ng1 Qh1 7. Rh1
The white pawn at d3 is a red herring that may lead the solver to try using the white Queen or Bishop!

6 - Noam Elkies
1. e4 a6 2. Ba6 d5 3. Bb7 Ra6 4. Bd5 Bb7 5. Bb7 Qd3 6. Ba6 Qf1 7. Bf1
Under one reasonable way to define switchbacks, there are three here, all with captures on the return moves.

7 - Gianni Donati
1. b3 e5 2. Bb2 Bc5 3. Qc1 Bf2 4. Kd1 Be3 5. Ke1 Ke7 6. Qd1 Kd6 7. Bc1
Three returns to home. Based on the efforts of several composers, I'd say that four returns in one line of play is impossible in an spg within 7.0 moves. Or is it?

8 - Noam Elkies
1. e4 Nf6 2. Qh5 Ne4 3. Qh7 Nc3 4. Qh5 Na2 5. Qd1 Nc1 6. Qc1 Rh2 7. Qd1 Rh8
Again, three returns to home. And the white Queen does two of them! The diagram shows all units home. Can an all-units-home shorty (possible with imposter units apparently home) be composed in which Black is missing the same units as White, but with non-symmetrical play?

9 - Thierry le Gleuher
1. e4 d5 2. ed5 Nc6 3. dc6 Qd5 4. cb7 Qa2 5. ba8=R Qb1 6. R8a7 Qa1 7. Ra1
A Rook Phoenix-Pronkin in a diagram with all units home (apparently). A two-variation shorty by Henrik Juel (Probleemblad 1997-1) shows how a similar Rook Pronkin can be changed to a game showing a Rook circuit.

10 - Mark Kirtley & Michel Caillaud
1. h4 e6 2. h5 Qf6 3. h6 Qf3 4. hg7 Nh6 5. g8=R f5 6. Rg6 Bg7 7. Re6 de6
1. h3! e6 2. h4 Qf6 3. h5 Qf3 4. h6 f5 5. hg7 Nh6 6. g8=B Bg7 7. Be6 de6
Rook and Bishop Ceriani-Frolkin promotions.

11 - Noam Elkies
1. d4 h5 2. Qd2 Rh6 3. Qh6 Na6 4. Qb6 Nb8 5. Qa7 b6 6. Qb8
A tempo-switchback, and the switchbacker is gone! Can some reader compose a tempo switchback by one unit, with a tempo loss of any sort by another?

12 - Michel Caillaud
1. Nc3 Nh6 2. Nd5 Nf5 3. Ne7 Nd4 4. Ng8 Ba3 5. ba3 Nb3 6. Bb1 Nc1 7. Qc1
Black performs a quintuple tempo loss, an equivalence of passing the five knight moves.

13 - Michel Caillaud
1. d3 g6 2. Bh6 Bh6 3. a3 Bc1 4. e3 d6 5. Be2 Bg4 6. Bg4 a6 7. Bc8
Two single tempo losses, one in each side.

14 - Mark Kirtley & Michel Caillaud
1. g3! e5 2. g4 Be7 3. g5 Bf6 4. gf6 h5 5. fg7 Ke7 6. gh8=N Ke6 7. Ng6 fg6
1. g4 h5 2. g5 Rh6 3. gh6 e5 4. hg7 Ke7 5. gf8=Q Kf6! 6. Qg7 Ke6 7. Qg6 fg6
In the second solution Black wastes time while White doesn't, even though in the first solution it was the other way around. Knight and Queen Ceriani-Frolkin promotions.

Probleemblad Jan./Feb. 2001

R110c - Alexander Zolotarev
(a): Last moves were: -1. .. Bd1-h5 -2. f3xNe4 Nf6-e4 -3. Ne5-d7 Nd7-f6 -4. Ng6-e5 Be2-d1 -5. Nh8-g6 Bf1-e2 -6. h7-h8=N Be2-f1 -7. h6-h7 Bf1-e2 -8. h5-h6 f2-f1=B -9. h4-h5 g3xNf2 -10. Nh3-f2 g4-g3 -11. f2-f3 h5xBg4 -12. Be2-g4 h6-h5 -13. Bf1-e2 h7-h6 -14. e2xQd3 and the position unlocks.
(b): Last moves were: -1. .. Be2-h5 -2. f3xNe4 Nf6-e4 -3. a4-a5 Nd5-f6 -4. Nb5-c7 Nc7-d5 -5. Nd4-b5 Bf1-e2 -6. Nf5-d4 Be2-f1 -7. Ne7-f5 Bf1-e2 -8. Ng6-e7 Be2-f1 -9. Nh8-g6 Bf1-e2 -10. h7-h8=N Be2-f1 -11. h6-h7 Bf1-e2 -12. h5-h6 f2-f1=B -13. h4-h5 g3xNf2 -14. Nh3-f2 g4-g3 -15. f2-f3 h5xBg4 -16. Be2-g4 h6-h5 -17. Bf1-e2 h7-h6 -18. e2xQd3 and the position unlocks.

R119 - Alexander Zolotarev
Last moves: -1. .. f7xRe6 -2. Rb6-e6 a7-a6 -3. Rb8-b6 Na6-c7 -4. b7-b8=R e5xRf4 -5. b6-b7 d5xBe4 -6. a5xBb6. Further retractions are: c2-c8=B-e4; c3-c8=R-f4; (c6xQd5); d3-d7; (d6xQe5); g6-g8=Q-e5; (d7-d6); g7-g8=Q-d5 (Bh6-b6); g5-g6; (c7-c6); g6-g7; (Bf8-h6); (g7xPf6).
Six knightmoves are needed: Nb8-a6-c7; Ng8-f6-e8; Nb1-d2; Ng1-f3. The knight on h1 is the promoted h-pawn, the extra whitesquared bishop is the promoted b-pawn.
Ceriani-Frolkin theme with QQRRB.

R120 - Gerald Ettl
Last moves were: -1. .. b5-b4 -2. e6-e7 b4-b3 -3. d5xBe6[+bBc8] Bc8-e6 -4. c4xBd5[+bBc8] Ba8-d5 -5. b3xRc4 b7-b5 -6. a2xPb3[+bPb7] b7xPa6[+wPa2] -7. a5-a6 Rc8-c4 -8. a4-a5 Rf8-c8 -9. c3xBd4 Rh8-f8 -10. d2xRc3[+bRh8] c4xPd3[+wPd2] -11. d2-d3 and the position is released.
But this is cooked: Black can take back -8. .. Rh8-c8 too, and white uncaptures this rook on e3.

R121 - Michel Caillaud
If black may castle, then there are two possible helpmates in 2:
1. Ba7 Ba7 2. OOO Rb8#
1. cb6 d6 2. bc5 Rb8#
Suppose black may castle. The position unlocks when white unpromotes a piece on h8, and retracts this pawn to h4, so black can uncapture h5xg4.
From the position r3k1K1/B1p2p1P/Bb6/2PP1PP1/2Prbrp1/1Prp1p2/2p2P2/N7, play forward:
1. h8=R Ba5 2. Rh6 Bb4 3. Rb6 Ba5 4. Kg7 Bb4 5. Kh6 Ba5 6. Kh5 Bb4 7. Kh4 Ba5 8. Kg3 Bb4 9. Kh2 Ba5 10. Kg1 Bb4 11. Kf1 Ba5 12. Ke1 Bb4 13. Kd2 Ba5 14. Kc1 Bb4 15. Kb2 Ba5 16. Ka3 Bb4+ 17. Ka4 Ba5 18. Kb5 Bb4 19. Kc6 Ba3 20. Bb5 Bc1 21. Ba4 Bd2 22. Rb5 Be1 23. Kb7 Bd2 24. Ka6 Be1 25. Ka5 Bd2 26. Kb4 Be3 27. Ka3 Bc1+ 28. Ka2 Ba3 29. Bb6 Bb4 30. Kb2 Ba5 31. Kc1 Bb4 32. Kd2 Ba3 33. Ke1 Bc1 34. Kf1 Be3 35. Ba5 Bd2 36. Bb4 Be1 37. Ba3 Bd2 38. Bc1 Be1 39. Be3 Bd2 40. Kg1 Bc1 41. Kh2 Ba3 42. Bc1 Bb4 43. Ba3 Ba5 44. Bb4 Bb6 45. Ba5 Ba7 46. Bb6 Bb8 47. Kg3 Ba7 48. Kh4 Bb8 49. Kh5 Ba7 50. Kh6 Bb8 51. Kh7
But then Ba7 would be the 50th move without capture, pawn move, or castling! So the castling solution isn't a legal solution. So this helpmate has only one legal solution!

R122 - Olli Heimo
1. b4 h5 2. Bb2 h4 3. Bf6 h3 4. Nc3 hg2 5. Rb1 gh1=B 6. Bg2 Rh3 7. Bc6 Rd3 8. ed3 Bd5 9. Ke2 Bc4 10. Ke3 Ba6 11. b5 Nh6 12. Rb4 Ng8 13. Rh4 Nh6 14. f4 Ng8 15. Rh8 Nh6 16. Qh5 Ng8 17. Qh7 Nh6 18. h4 Ng8 19. h5 Nh6 20. Bh4 Ng8 21. Rg8

R123 - Satoshi Hashimoto
1. h4 a5 2. Rh3 a4 3. Rb3 ab3 4. a4 Ra6 5. a5 Rf6 6. a6 h5 7. a7 Rhh6 8. Ra6 Rhg6 9. Rc6 dc6 10. a8=R Qd4 11. Ra1 Qa7 12. d4 Bg4 13. d5 Bxe2 14. d6 Bb5 15. Qg4 hg4 16. h5 Rf3 17. h6 Nf6 18. h7 Nd5 19. h8=R f6 20. d7+ Kf7 21. d8=Q Nb6 22. Qd1 Ke8 23. Rh1

R124 - Henrik Juel & Bernd Schwarzkopf
(a): Add Rb8, Rc8, Rd8, Rc7, Rd7, Re7, Rb6, Rc6, Rd6, Rb5.
(b): Add Ra8, Rc8, Rd8, Re8, Rd7, Re7, Rf7, Ra6, Rc6, Rd6.
(c): Add Rd8, Rf8, Rg8, Rh8, Rg7, Rh7, Rd6, Rf6, Rg6, Rh6.
(d): Add Rc8, Rd8, Re8, Rg8, Rb7, Rc7, Rd7, Rc6, Rd6, Rg6.

F233 - Fernand Joseph
(a): Black may not castle, since his last move must've been with his rook or king.
1. .. Ed2 2. Eg1 Eg4 3. Ee5 Kf6# and 1. .. Kf6 2. Kf8 Eh4 3. Kg8 Ea2#
(b): 1. .. Eg4 2. Ea1 Ea6 3. OO Kg6# and 1. .. Kf6 2. Eg7 Ee6 3. OO Ef7#

167.i - Frank Christiaans
Last black move was Kb1-c2 or Kc1-c2, so white may not castle. #2 with 1. Sc3 Kd3/Kb3 2. Be4/Ba4#