Die Schwalbe

66 problem(s) found in 1677 milliseconds (displaying 66 problem(s)). [A='Brobecker' AND NOT COOKED] [download as LaTeX]

1 - P1067883
Alain Brobecker
Messigny 05/2005
P1067883
(8+0) C+
Färbe die Steine!
sKa6, sTc6

R: 1. a5xb6ep+ b7-b5 2. Tb5-d5+,Tb5xd5+
play all play one stop play next play all
The Qa4 and the Bc4 attacks both Kings, so those two pieces have the same color and there was a discovered check. The only possible discovered check is a5xb6+ en passant, so Kb3, Qa4, Bc4 and Pb6 are white while Ka6 is black.
To have the en passant capture, black must have played b7-b5 and we deduce from it that Rc6 is black.
Before black has played b7-b5, the BKa6 was attacked by the WBc4, or it can not have moved, so there was a discovered check of this WBc4, and the only possible discovered check is with Rd5 which is then white.
Last two single moves are: n.a5xb6+ e.p. and n-1... b7-b5
Henrik Juel: The solution text is full of file errors
The retroplay is 1.a5xb6ep+ b7-b5 and e.g. 2.Tb5-d5+ (2021-12-24)
comment
Keywords: Colouring problem, Minimal
Genre: Retro
Computer test: rawbats
FEN: 8/8/KPR5/3R4/Q1B5/1K1R4/8/8
Input: Gerd Wilts, 2005-05-17
Last update: Mario Richter, 2021-12-25 more...
2 - P1068635
Alain Brobecker
R379 The Problemist 11/2006
P1068635
(31+0)
Färbe die Steine. Letzte zwei Einzelzüge?
(von http://abrobecker.free.fr/chess.htm)
The Bd8 attacks both K and since no discovery is possible then it was the last piece to move, and its only legal move was d7-d8=B+. Thus WBd8 and BPd6, otherwise it would attack one of the K and d7-d8=B+ would have been illegal. Since WBd8 is a promoted pawn and there are 3 pawns on the d-file, it has made the only capture of the game (a black S then) and we can conclude that no other pawn has changed of file. Thus WPa5, WPb2, WPd3, WPe2, WPf2, WPg2, WPh2 while BPa7, BPc5, BPd6, BPe6, BPf6, BPg6, BPh6. Hence WBf1, WRh1 and then BBh7. Kc7, Qc8, Rc6, Bb8, Sa8 are of identical color (let's say "red") while Ke7, Qf8, Rf7, Sg8 are "blue", and WSb7. If BKc7, then black would be retro-stalemated since no black piece would have squares to come from. So WKc7 is white as are all other "red" pieces, while the "blue" ones are in fact black. Last BRh8, BBg7 since the white ones are already colored. Before d7-d8=B+, the only possible black move is Qe8-f8.
NBQB1qnr/pNK1krbb/1pRppppp/P1p5/8/3P4/1P2PPPP/5B1R
R: 1. d7-d8=L+ De8-f8
play all play one stop play next play all
hans: R: -1.d7-d8=L+ De8-f8 -2.a4-a5
(White: Kc7,Dc8,Tc6,Th1,Lb8,Ld8,Lf1,Sa8,Sb7,a5,b2,c3,e2,f2,g2,h2) (2012-05-31)
comment
Keywords: Colouring problem, Last Moves? (2)
Genre: Retro
FEN: NBQB1QNR/PNK1KRBB/1PRPPPPP/P1P5/8/3P4/1P2PPPP/5B1R
Input: Gerd Wilts, 2006-11-26
Last update: A.Buchanan, 2023-06-06 more...
3 - P1070107
Alain Brobecker
5335 Phénix 161 04/2007
P1070107
(23+0)
Färbe die Steine!
hans: White: Ke5, Td5, Tf5, b7, c7, c2, d6, e2, f4, f6, g7, (11)
Black: Ke3, Tg3, Th3, Sf1, a2, d2, d4, d7, e7, f2, f7, g2, (12)
captures w(4x) axb, bxc, gxf, hxg. / s(5x) bxcxd, cxd, gxf, hxg. (2012-06-01)
Henrik Juel: Here is an argued solution, verifying Hans' (with Th3 changed to Th2), and showing uniqueness.
The pawn captures were axbxcxd (short for axb, bxc, cxd), bxcxd, hxgxf, hxgxf, so all captures were made by pawns, and Pe2 is white and Pe7 black.
If Ke3 were white and Ke5 black, any coloring of Pd4 and Pf4 would be illegal, so Ke3 is black and Ke5 white; this implies that Pd2, Pf2 are black, Pd6, Pf6, Td5, Tf5 are white, and Tg3, Th2, Sf1 are black.
The f-bishops got out to be captured, so Pg2 is black and Pg7 white.
Pf4 is white and Pf7 is black, because otherwise the initial pawn structure on the king's side could not be reached by retracting hxgxf twice; note that two white pawns above two black ones on a file does not work.
Now consider Pd4, Pd7 on the d-file. If they were white and black, the coloring would be Pd4 white and Pd7 black, and Pb7 white because of [Lc8]; but now the final white pawn would have to sit on a2 or the c-file, with both possibilities contradicting the fact that White pawns made two captures on the queen's side. This fact also rules out the possibility that Pd4, Pd7 were both white, so Pd4, Pd7 are both black. This implies that the three captures by Black pawns on the queen's side were bxcxd and cxd, so Pa2 is black and the remaining Pb7, Pc2, Pc7 white, and [Lc8] could get out.
Nice problem, Alain. (2012-06-01)
hans: Indeed, nice problem, Alain, and thank you, Henrik, for the explanation, although your captures are not right. (cxd for white?) Excuses for the typo (Th2) (2012-06-01)
comment
Keywords: Colouring problem
Genre: Retro
FEN: 8/1PPPPPP1/3P1P2/3RKR2/3P1P2/4K1R1/P1PPPPPR/5N2
Input: Gerd Wilts, 2008-09-15
Last update: Gerd Wilts, 2008-09-15 more...
4 - P1070125
Alain Brobecker
Thierry Le Gleuher

5540 Phénix 168 12/2007
P1070125
(23+0)
Färbe die Steine! Letzte 2 Einzelzüge?
Monochromes Schach

RBh2, RNe8

CASES SOMBRES:
Les 8 pions de cases sombres sont encore sur l'échiquier, donc pas de promotion sur cases sombres, donc CNb8, CBg1, TNh8.
Il manque 3 pièces de cases sombres: DN, TB et un F. Si PBg3 (=1 capture de pièce sombre) alors PNf2 (2 ou 3 captures de pièces sombres car il a pu faire une prise e.p.), mais avec le Pe3 qui a aussi effectué une capture de pièce sombe il y a plus de 3 pièces capturées, ce qui est contradictoire. Donc PNg3+, qui a donc effectué 2 captures sans prise en passant (puisque P?g2), donc l'ordre des captures a été PN?xFBf4 puis PNf4xTBg3 (TBa1 n'aurait pas pu aller en f4). Toutes les captures de pièces B étant utilisées, aucun autre PN n'a dépassé la 5ème, donc PNa7, PNc7, PNe7 et on en déduit PBb2,PBe3, PBf2, PBh4.
On a aussi FNg5 puisque le FBc1 a été pris par PNg5xFBf4.

CASES CLAIRES:
Il y a eu 6 captures sur cases claires, sans prise en passant puisque les 8 pions sombres sont présents. Au plus il y a pu avoir une promotion sur case claire.
Il y a eu au moins 2 captures pour obtenir la configuration de Ps en a4, b3, c2, et au moins un des Ps est N.
Supposons PBb7 qui nécessite 3 captures par PB, donc il n'y a pas eu de promotion, donc PNf et PNh ont été capturés par une autre pièce B (si l'un deux était en g2 cela demande encore davantage de capture) et avec la configuration a4, b3, c2, cela fait 7 captures. Contradiction, donc PNb7 et on en déduit TNa8.
Supposons qu'il y ait deux PN parmi a4, b3, c2, cela nécessite 5 captures par les N. Par ailleurs PNh et FNc8 ayant été capturé il y a au moins 7 captures. Contradiction, donc un seul PN parmi a4, b3, c2 et au moins 2 captures par ce PN.
Supposons TNh1 alors elle est issue de promotion, donc 4 captures + 2 captures pour PNa4,b3,c2 + FNc8 capturé. Contradiction, donc TBh1.
Supposons PNg2 alors 3 captures + 2 captures pour PNa4,b3,c2, donc pas de promotion. Mais seulement 2PB + FB + DB capturés. Contradiction, donc PBg2.
Si PNa4 ou PNc2 alors 3 captures donc pas de promo et les blancs seraient rétropat. Donc PNb3, PBc2, PBa4 et 2 captures.
Si CBb1 et CNg8, les B sont rétropat avant f4xTg3+ car a4 et d2xDe3 ont été joués il y a longtemps pour laisser sortir RN et TBa1.
La seule solution pour lever le retro pat des blancs est CBg8, CBb1, et les deux derniers coups sont:
n.h7xCg8=C f4xTg3+
Mu-Tsun Tsai: Coloring: rn2k1Nr/ppp1p3/8/6b1/P6P/1p2P1p1/1PP2PPK/1N4NR
Last two moves: -1...f4:R -2.N=h7:N (2012-07-06)
comment
Keywords: Colouring problem, Monochromatic Chess, Last Moves? (2)
Genre: Retro, Fairies
FEN: RN2K1NR/PPP1P3/8/6B1/P6P/1P2P1P1/1PP2PPK/1N4NR
Input: Gerd Wilts, 2008-09-15
Last update: A.Buchanan, 2023-06-09 more...
5 - P1088952
Alain Brobecker
R412 The Problemist 07/2009
nach L. Ceriani
P1088952
(11+10)
Position shifted -- what's the a-file?
Vertikaler Zylinder
Paulo Roque: .
Solution: The a-file is n-file.

Demonstration:

1) Change of algebraic notation:
File (i,j,k,l,m,n,o,p)
Rank (1,2,3,4,5,6,7,8)
The squares are yellow or red.
For example: The wK is in square p4, your color is red.

2) Retroplay: 1.i7xXp8=T# Sm6xSn8 (unique possibility, because if 1...o4xXp3?? then pawn not retracts for k7).

3) Analysis of retroplay position:

A) Logical elements in the position (LEP):
(a.1) captured four white pieces(T,2L,D).
(a.2) captured five black pieces.

B) Development:
(b.1) White pieces captured: i7xp6 (one capture) + kxjxixp (three captures) = four captures.
(b.2) Black pieces captured: j2xk3xl4 (two captures) + oxpxi (two captures) = four captures.

C) Reflections:
(c.1) Can not have white bishop captured in original square (by a.1 + b.1).
(c.2) Can not have the two black bishops captured in original square (by a.2 + b.2).

4) Conclusion:
After checking files (for c.1 + c.2) , only answer a-file is n-file. (2009-09-15)
comment
Keywords: Vertical Cylinder (Vertikaler)
Genre: Retro, Fairies
FEN: R4nkR/1p1ppppp/7p/8/N2P3K/7p/P1PPPP1P/8
Input: Gerd Wilts, 2009-07-17
Last update: A.Buchanan, 2023-06-03 more...
6 - P1098614
Alain Brobecker
5632 Phénix 172 04/2008
Pascal Wassong gewidmet
3rd honorable mention
P1098614
(9+12)
Vorgabepartie. Welcher weiße Stein wurde vorgegeben?
Alain Brobecker: The stipulation should read "Welcher weißer Steine..." i think. English stipulation was: "Advantage game. What piece has white given?" (2010-02-01)
Gerd Wilts: Thank you, I changed it. (2010-02-02)
Henrik Juel: White gave [Lf1], I think. Black pawns captured all other missing white officers; White captured [Th8] in its corner, b2xc3, f2xg3, and KxSe2 to allow a retroplay like -1... Th1 -2.Kd1xSe2 Sg1 -3.e2 etc. b2xc3 happened early in the game to free [Ta1,Lc1], so it is not possible to uncapture [Dd8] quickly (with the intention of retracting black KDL to e8,d8,c8 and b7xLc6). (2010-02-05)
comment
Keywords: Odds game
Genre: Retro
FEN: 5b2/2pppppp/2p5/8/4k3/P1P1P1P1/2PPKpPP/1b2r3
Input: Gerd Wilts, 2010-01-31
Last update: Alain Brobecker, 2021-07-19 more...
7 - P1098624
Alain Brobecker
5732 Phénix 175 07-08/2008
vom Computer komponiert
P1098624
(16+16) C+
Wie verlief die Partie mit 4. c5#?
Berolinabauern

1.Nc3 b5 2.Na4 Kd7 3.b4+ Kc6 4.c5#
composed with a homebrew computer program ( http://abrobecker.free.fr/chess/synthetics.htm#berolina )
comment
Keywords: Synthetic problem, Berolina chess
Pieces: bu = Berolina Pawn (BE)
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/*2p*2p*2p*2p*2p*2p*2p*2p/8/8/8/8/*2P*2P*2P*2P*2P*2P*2P*2P/RNBQKBNR
Input: Gerd Wilts, 2010-01-31
Last update: A.Buchanan, 2022-09-13 more...
8 - P1098627
Alain Brobecker
5811 Phénix 179 12/2008
vom Computer komponiert
P1098627
(16+16) C+
Wie verlief die Partie mit 4. ... LxB=T#?
Einsteinschach

1.g4 h5 2.Bg2=N hxg4=N 3.Kf1 Rh4=B 4.Qe1=R Bxf2=R#
vom Computer komponiert ( http://abrobecker.free.fr/chess/synthetics.htm#einstein )
comment
Keywords: Synthetic problem, Einstein Chess, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Gerd Wilts, 2010-01-31
Last update: Alain Brobecker, 2021-12-25 more...
9 - P1098628
Alain Brobecker
5812 Phénix 179 12/2008
Composed with a computer
P1098628
(16+16) C+
Wie verlief die Partie mit 4. ... Txd6# bzw. 4. ... Td6#?
Monochromes Schach
Keywords: Synthetic problem, Monochromatic Chess, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Gerd Wilts, 2010-01-31
Last update: Alain Brobecker, 2021-08-25 more...
10 - P1098637
Alain Brobecker
Thierry Le Gleuher

5900 Phénix 183 04/2009
5th commendation
P1098637
(11+10)
Welche Steine sind auf Brett B?
Alice-Schach

Les Pions b2, c2, d2, b7, c7 et e7 qui n’ont pas bougé sont sur A.
Le PBa6 est sur A et le PBd5 est sur B car ils viennent de e2 et f2 en capturant toutes les pièces noires manquantes.
La position du CN en a1 avec RBb3, PBc2 et PBb2 ne peut se justifier que par la promotion d'un PN par a2-a1=C.
Le RN n'a pu atteindre b1 que si le FBc1 d'origine a été capturé sur place puis a été remplacé par un FBc1 de promotion qui est donc sur A, car il est passé au dessus de PBd2A ou PBb2A.
Donc PNa s'est promu dans l'axe et le PNc4 sur B et PNc6 sur B ont capturé les 4 pièces blanches restantes en venant de f7 et d7.
On voit maintenant que sur les colonnes a, g et h, les PBs ont croisé les PNs sans faire de capture. Cela impose un double pas pour chacun de ces Pions (Noir et Blanc). Par exemple PBg2A-g4B-g5A-g6B puis PNg7A-g5B puis PBg6B-g7A-g8B=C. Donc PBa4B et PNh5B.
Donc le CN promu en a1 est sur B, et il ne doit pas donner un échec impossible au RBb3 qui doit donc être sur A.
Le FNa2 est sur B car il ne doit pas donner d'échec impossible et il n’a pas de dernier coup (RBb3 sur A et PNc4 sur B).
De toute évidence le FB promu l’a été en h8 (case sombre), donc le CB a été promu sur g8 (case claire sur échiquier B). Reprendre un coup de Cavalier (CBa3+ ou CBc3+) ne libère pas un dernier coup noir. Donc les CBa3 et c3 ne doivent pas donner échec au RN. Ils sont donc tous les deux sur le même échiquier. Or les Cavaliers d’origines ne peuvent être tous les deux sur des cases de même couleur du même échiquier. Donc l’un des CBa3 ou c3 est de promotion et se trouve donc sur A (case sombre). Donc CBa3 et CBc3 sont sur A! Et CBe3 est sur B.
Pour ne pas être en échec par les CBs le RNb1 est donc sur B.
Finalement le CNa5 est sur B pour ne pas donner un échec impossible au RBb3.
Le dernier coup est Blanc (d4-d5 ou e4x?d5 avec ? qui n'est pas un F) afin de libérer un denier coup noir!

A: 8/1pp1p3/P7/8/8/NKN5/1PPP4/2B5
B: 8/8/2p5/n2P3p/P1p5/4N3/b7/nk6
Keywords: Alice Chess
Genre: Retro, Fairies
FEN: 8/1pp1p3/P1p5/n2P3p/P1p5/NKN1N3/bPPP4/nkB5
Input: Gerd Wilts, 2010-01-31
Last update: Alain Brobecker, 2021-12-27 more...
11 - P1098638
Alain Brobecker
5901 Phénix 183 04/2009
Composed with a computer
P1098638
(16+16) C+
Wie verlief die Partie mit 4. ... Dxe4#?
Gitterschach
1. e4 d5 2. Dh5 Lg4 3. Dxd5 Dxd5 4. Sh3 Dxe4#
play all play one stop play next play all
Keywords: Synthetic problem, Grid Chess, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Gerd Wilts, 2010-01-31
Last update: Alain Brobecker, 2021-08-25 more...
12 - P1108108
Alain Brobecker
14515 Die Schwalbe 243 06/2010
Composed with a computer
P1108108
(12+4)
Drehe das Brett so, dass die Stellung legal ist. Matt?
hans: Turn board 90° (bKh8)
Mate: 1. Tg7xTg8# is legal.
White pawns made 11 captures (axbxcxdxe, bxcxd cxd, exd, fxe, gxf, hxg,) plus 4 pieces on diagram makes 15, so 1 black piece left over for g8 Not D or L. S is maybe possible, but than I have to count the tempo-moves, but T is the simplest. (2012-06-01)
comment
Keywords: Board Rotation
Genre: Retro
FEN: kB6/R1P3p1/RKP2pp1/1PPP1P2/2PP4/8/8/8
Input: Gerd Wilts, 2010-06-20
Last update: Alain Brobecker, 2011-12-17 more...
13 - P1108594
Alain Brobecker
Michel Caillaud
Axel Gilbert
Thierry Le Gleuher

Messigny 2010
2. ehrende Erwähnung
Märchen-TT
P1108594
(9+9)
Letzte 3 Einzelzüge?
Känguruhs a5, d8, b6 und f6
R: 1. e7xKAd8=KA KAg5-d8 2. a6-a7
play all play one stop play next play all
Keywords: Last Moves?
Pieces: dl = Kangaroo (KÄ)
Genre: Retro, Fairies
FEN: K1k*3QB3/Prrp1p2/1*3qp1p*3q2/*3Q7/8/8/PPP1P3/8
Input: Gerd Wilts, 2010-06-25
Last update: A.Buchanan, 2023-06-02 more...
14 - P1110972
Alain Brobecker
Thierry Le Gleuher

R424F The Problemist 07/2010
P1110972
(8+12)
On which square of the original board is the white king?
Shrink chess

wKc2.
Not a1, c1, d1 (illegal wB).
Not a2, b1, e2 (illegal wP/bP structure).
Not b2 (6 wP captures but only 4 missing black units).
Not d2 (4 wP captures, but Bf8 captured at home).
Not e1. Suppose it were: Black has just captured on f2, and visible bP captures account for 6 of 7 other missing white units. [Pc2,Pd2] have captured all missing black units, so [Pa2,Pb2] have not captured, so one must have itself been captured to let [Pa7,Pb7] pass, accounting for one more. But [Rh1] could not have escaped to be captured, so illegal position.
Alain Brobecker: Shrink Chess (Joseph Boyer, 1954): An *EDGE* file or rank disappears if unoccupied (eg: 1.a4 b5 2.axb5 Nc6 3.Rxa7 Rb8 4.Rb7 a-file disappears).

The printed diagram was only 4*6 and the squares were coloured purely for convenience. (2010-08-01)
comment
Keywords: Shrink Chess, no 8x8 board
Genre: Retro, Fairies
FEN: ppppqqqq/P2nqqqq/pP1pqqqq/rP1rqqqq/PbPPqqqq/KBknqqqq/qqqqqqqq/qqqqqqqq
Input: Gerd Wilts, 2010-07-20
Last update: Alfred Pfeiffer, 2015-02-04 more...
15 - P1110975
Alain Brobecker
F2832 The Problemist 07/2010
dedicated to Florian Bonnet
Composed with a computer
P1110975
(8+8) C+
Find unique Fischer Random Chess initial setup and games ending in
a) 4. Qc1xPc6#
b) 4. Qb1xBf5#
a) STDLKTLS 1. c4 b6 2. c5 Da6 3. c6 dxc6 4. Dxc6#
b) SDLLTKTS 1. c3 d6 2. Lb3 Lf5 3. Lxf7 Kxf7 4. Dxf5#
play all play one stop play next play all
more ...
comment
Keywords: Chess960, Synthetic problem, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: 8/pppppppp/8/8/8/8/PPPPPPPP/8
Reprints: 960/9 feenschach 189, p. 212, 11/2011
Input: Gerd Wilts, 2010-07-20
Last update: Alain Brobecker, 2021-08-25 more...
16 - P1186210
Alain Brobecker
F2863 The Problemist 01/2011
Composed with a computer
P1186210
(16+16) C+
Find the shortest logical progressive chess games leading to
a) 4. ?, Da3-c1#
b) 4. ?, Dc4#
c) 4. ?, Dc5-h5#
Keywords: Synthetic problem, Logical Progressive Chess, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Gerd Wilts, 2011-01-29
Last update: Alain Brobecker, 2021-08-25 more...
17 - P1186654
Alain Brobecker
Thierry Le Gleuher

Retros mailing list 03/2007
after Frank Christiaans
P1186654
(11+3)
6 letzte Einzelzüge?
R: 1. La5xBb6 b7-b6 2. Ld4-a7! Ka7xSa8 3. Sb6-a8+ Ka8-a7 und weiter z.B. 4. Sa4-b6+ Ka7-a8 5. Sc5-a4+
play all play one stop play next play all
Highest number of retro moves with 14 pieces.
See http://abrobecker.free.fr/chess/lengths.pdf

Source is Franck Christiaans, Die Schwalbe, 1995, dedicated to W.Keym:
k7/B1P1p1p1/BQ1B4/1K6/8/4B3/PPPP4/8, 11+3: Last move? (s. P0005981)
Alain Brobecker: Source = P0005981 (2013-03-09)
Henrik Juel: The solution is given in Alain's funny retro notation
Here is the PDB version
R: 1. La5xBb6 b7-b6 2. Ld4-a7! Ka7xSa8 3. Sb6-a8+ Ka8-a7 and e.g. 4. Sa4-b6+ Ka7-a8 -5. Sc5-a4+ etc. (2020-05-02)
comment
Keywords: Last Moves? (6), Type A, Non-standard material (LL), Move Length Record
Genre: Retro
FEN: k1K5/B1P1p1p1/BB1N4/8/8/4B3/PPPP4/8
Reprints: A14 Length Records in Last Single Moves? , p. 4, 11/2009
A14 Phénix 282, p. 11060, 02/2018
Input: Gerd Wilts, 2011-02-05
Last update: Mario Richter, 2020-05-02 more...
18 - P1189477
Alain Brobecker
Phénix 198-199 9-10/2010
Jubilé TLG-50 2010
1. Lob
P1189477
(8+8)
Ergänze einen Stein!
Nikolai Beluhov: According to Thierry Le Gleuher (Phenix 198-199/2010, p. 8312), this is the current record for this type of stipulation with no King in check. For the previous record, see P1000069 (Henrik Juel, Thema Danicum, 1997, 17 pieces). For the records when diagram Kings in check are allowed, see P1189330 and P1189060. (2011-03-07)
Henrik Juel: White pawns captured all missing black men, including [Pc7], which must have captured to promote. Thus seven black pawn captures are accounted for, leaving only [Pa2] to add.
Add wPa4. (2011-03-07)
comment
Keywords: Add pieces
Genre: Retro
FEN: 8/1p5p/6pP/p4pPP/1k4pP/7p/KPP3P1/8
Input: Nikolai Beluhov, 2011-03-07
Last update: Nikolai Beluhov, 2011-03-07 more...
19 - P1191981
Alain Brobecker
Retros mailing list 13/03/2011
Composed with a computer
P1191981
(7+0) C+
Ergänze einen Stein!
Nikolai Beluhov: This problem was composed with the help of a computer (see
http://www.pairlist.net/pipermail/retros/2011-March/003427.html
for the original publication).
As far as I know, this is the economy record for the stipulation "Add one piece" where no King is in check both before and after the addition.
Allowing one King to be in check after the addition, it is possible to replace one of the Queens by a Rook: see Alain Brobecker's compilation of addition economy records at
http://abrobecker.free.fr/chess/addunits.pdf (2011-04-04)
Henrik Juel: In case a solution is needed: Add sKb7. (2011-04-04)
Alain Brobecker: See P1227362 for a more economic version. (2011-12-17)
comment
Keywords: Add pieces, Aristocrat, Miniature
Genre: Retro
FEN: 7Q/8/3Q4/5Q2/2Q5/4Q3/8/Q6K
Input: Nikolai Beluhov, 2011-04-04
Last update: Alain Brobecker, 2022-04-08 more...
20 - P1203837
Alain Brobecker
Messigny 2011
4.ehrende Erwähnung
P1203837
(15+8) C+
Wo wurde der fehlende wL geschlagen?
Die letzten Züge waren
R: 1. Th1-g1 Dg1-f1 2. Tf1-e1 Te1-d1 3. Dd1-c1 Lc1-b2 4. Bb2xBa3 (oder Bb2xYc3)
Um die Stellung zu entwirren, muß im weiteren Rückspiel sBc3xwLd2 zurückgenommen werden.
play all play one stop play next play all
Keywords: Where was piece x captured?, Last Moves? (6), Umnov, RIFACE Retro Composition Tourney (2011)
Genre: Retro
FEN: 8/8/8/8/8/PPPPPPPP/BbKpppkr/NNQrRqR1
Input: Mario Richter, 2011-10-03
Last update: Alain Brobecker, 2012-07-20 more...
21 - P1204451
Alain Brobecker
F2501 The Problemist 09/2006
Composed with a computer
P1204451
(2+1) C+
#5
Monochromes Schach
1. Le6 Zz. Kb5 2. Ld7+ Ka6 3. Kc4 [4. Lc8#] Kb7 4. Kb5 Zz. Ka8 5. Lc6#
play all play one stop play next play all
Keywords: Monochromatic Chess, Rex solus
Genre: Fairies
FEN: 8/5B2/8/8/k7/3K4/8/8
Input: Hans Gruber, 2011-10-09
Last update: Alain Brobecker, 2021-12-27 more...
22 - P1211268
Alain Brobecker
6066 Phénix 11-12/2009
Composed with a computer
P1211268
(2+1) C+
#6
5x6-Brett
Loch a3
wSL=Grenouille
1. Kc1 Ka2 2. Kc2 Ka1 3. GRe6 Ka2 4. GRe2+ Ka1 5. GRe5 Ka2 6. GRa5#
play all play one stop play next play all
Alain Brobecker: The initial problem was given with full 5*6 board and the stipulation was: "White removes a square and mates in 6 moves".
The Forg is a (0;3)+(0;4) leaper. (2021-12-27)
more ...
comment
Keywords: no 8x8 board, Aristocrat, Rex solus (s)
Pieces: sl = Frog ()
Genre: Fairies
FEN: qqqqqqqq/qqqqqqqq/5qqq/5qqq/5qqq/q3*3Nqqq/5qqq/k2K1qqq
Input: Hans Gruber, 2011-10-09
Last update: Alfred Pfeiffer, 2018-12-18 more...
23 - P1211852
Alain Brobecker
Stephen Emmerson

13 The Problemist 01/2010
Neujahrsgruß, dedicated to Eric Angelini
Composed with a computer
P1211852
(2+1)
(#8) siehe Text
Forderung a) Weiß entfernt ein Feld und setzt doppelt so schnell matt. (Entferne c3, dann 1. Tg4 Kb3 2. Kb1 Ka3 3. Kc2 Ka2 4. Ta4#)
Forderung b) Weiß entfernt zwei Felder und setzt achtmal so schnell matt. (Entferne a4/b4, dann 1. Th3#)
Forderung c) Schwarz entfernt zwei Felder und macht remis (Entferne g1/h2!)

(Weiß kann in 8 Zügen mattsetzen: 1. Tb1 2. Ka2 5. Ka5 6. Kb6 Kb8 7. Tc3 Ka8 8. Tc8#)
play all play one stop play next play all
Alain Brobecker: a) White removes a square and mates twice as fast.
b) White removes two squares and mates eight times as fast.
c) Black removes two squares and draws.

Part a) was composed with P1227727 in mind. (2011-12-23)
comment
Keywords: Aristocrat, Minimal, Miniature, Rex solus
Genre: Fairies
FEN: 8/8/8/8/8/k7/8/K6R
Input: Hans Gruber, 2011-10-09
Last update: Alain Brobecker, 2021-12-27 more...
24 - P1227362
Alain Brobecker
N1KAi) Economy Records in Add Unit(s) Problems , p. 4, 13/03/2011
Version by Werner Keym
Composed with a computer
P1227362
(7+0) C+
Add one unit.
+BKh6, other squares attacked twice and no square is unguarded before a promotion.
play all play one stop play next play all
Economy record for an "Add one unit" problem, type A, in which the unit to add is a King.
See http://abrobecker.free.fr/chess/addunits.pdf (2011-12-17)
Alain Brobecker: See P1191981 for the original version. (2011-12-17)
Henrik Juel: Solution: Add the black king on h6 (2011-12-17)
comment
Keywords: Type A, Add pieces, Non-standard material, Economy record, Aristocrat, Miniature
Genre: Retro
FEN: 6Q1/2K5/8/5Q2/3Q4/Q7/4Q3/7R
Input: Alain Brobecker, 2011-12-17
Last update: A.Buchanan, 2023-04-13 more...
25 - P1227370
Alain Brobecker
N2Ai) Economy Records in Add Unit(s) Problems , p. 4, 04/04/2011
Composed with a computer
P1227370
(9+0) C+
Add two units.
All squares are controlled at least twice!
Double check by promotion to queen by n.g7x?h8=Q doesn't work, BK would be on g8 where there's not enough room for the WK to parry attack by WQe6, or on h7 which is too much controlled.
So we must put BK on a square controlled only twice and add WK in order to parry one check.
The squares that are controlled only twice are a5,d1,f3,f4,f7,f8 and h2. But only f4 allows enough room for WK to parry the second check, so +WKd2 +BKf4.
play all play one stop play next play all
Economy record for an "Add 2 units" problem, type A.
See http://abrobecker.free.fr/chess/addunits.pdf (2011-12-17)
Alain Brobecker: Similar problem 4Q3/2Q5/7Q/Q3Q3/4Q3/1Q6/6Q1/5Q2 has the maximum of 11 squares that are controlled twice, but the BK is on the edge. (2011-12-23)
comment
Keywords: Type A, Add pieces, Non-standard material, Economy record, Aristocrat
Genre: Retro
FEN: 7Q/Q3Q3/4Q3/1Q6/7Q/3Q4/6Q1/2Q5
Input: Alain Brobecker, 2011-12-17
Last update: A.Buchanan, 2023-04-13 more...
26 - P1227638
Alain Brobecker
5515 diagrammes 152 01-03/2005
Yves Tallec gewidmet
2nd Prize
P1227638
(8+7) C+
s#13
1. Se2 Kh3 2. Sf4+ Kh2 3. Tg4 Kh1 4. Tg2 d6 5. Tg3 Kh2 6. Tg4 Kh1 7. Tg2 d5 8. Tf2 Kg1 9. Sh3+ Kh1 10. Te2 d4 11. Tf2 d3 12. Sc4 b2+ 13. Sxb2 Sb3#
play all play one stop play next play all
Alain Brobecker: Full solution: 1.Ne2 Kh3 (1...Kh5? 2.Nf4+ Kh4 3.Rg2 d6 4.Rg1 and s#8) 2.Nf4+ Kh2 (2... Kh4? as before) 3.Rg4! (3.Rg2? Kh1 4.Rf2 Kg1 5.Nh3+ Kh1 6.Re2 d5! 7.Rf2 d4 8.Re2 d3 and we must stalemate by 9.Rf2 or let the bK escape) 3...Kh1 4.Rg2 d6 5.Rg3 Kh2 6.Rg4 Kh1 7.Rg2 d5 8.Rf2! (the d pawn can no more choose between single and double step, so all moves are now forced) 8... Kg1 9.Nh3+ Kh1 10.Re2 d4 11.Rf2 d3 12.Nc4 b2+ 13.Nxb2 Nb3# (2011-12-23)
more ...
comment

Genre: s#
Computer test: O. Jenkner: Gustav 3.2 g (15 h)
FEN: 8/3p2p1/6P1/p5P1/P2N3k/1p6/1NpP4/n1K3R1
Input: Alain Brobecker, 2011-12-22
Last update: Frank Müller, 2012-04-16 more...
27 - P1228029
Alain Brobecker
The Problemist Supplement 107 07/2010
Composed with a computer
P1228029
(4+2)
#8 (Giant Frog f4)
1.Kc1 Ka2 2.Kc2 Ka1 3.GFf8 Ka2 4.GFf3 Ka1 5.GFf7 Ka2 6.GFf2+ Ka1 7.GFf6 Ka2 8.GFa6#
play all play one stop play next play all
Pendulum movement. Keymove is the only moment when white can lose a tempo (1.Kc2? and GF can no more access 6th line, so capture of BPb4 and mate with promoted WPb3 takes 9 moves).
comment

Pieces: tu = Giant Frog (GF)
Genre: Fairies
FEN: 8/8/8/8/1p3*2R2/1P6/1P6/k2K4
Input: Alain Brobecker, 2011-12-26
Last update: Alain Brobecker, 2011-12-26 more...
28 - P1230203
Alain Brobecker
14885 Die Schwalbe 249 06/2011
version by Arnold Beine and Peter Heyl
P1230203
(5+3) C+
ser-h#8
1. Lxc3 2. Kb2 3. Ta2 4. Ka1 5. Txc2 6. Ka2 7. La1 8. Tb2 Sc3#
play all play one stop play next play all
R and B exchange places to allow mate, K circuit.
Ravi Shankar made another version without the ideal mate but without capture and using more space from the board.
Alain Brobecker: See P1113513 (2012-01-15)
comment
Keywords: Seriesmover, Pure Round Trip, Interchange
Genre: Fairies
FEN: 8/8/8/8/K7/P1P5/kbP5/rN6
Input: Alain Brobecker, 2012-01-15
Last update: Alain Brobecker, 2021-08-25 more...
29 - P1232420
Alain Brobecker
15002 Die Schwalbe 251 10/2011
Kevin Begley & Noam Elkies gewidmet
P1232420
(9+0)
Drehe das Brett und ergänze einen Stein!
Henrik Juel: Turn the board 90 degrees clockwise (wKg2) and add bKc7, last move b7-b8=D++ (2012-08-25)
comment
Keywords: Add pieces
Genre: Retro
FEN: 8/3Q2K1/5Q2/2Q5/5Q2/8/Q3Q1Q1/1Q6
Input: Nikolai Beluhov, 2012-02-16
Last update: Alain Brobecker, 2012-02-16 more...
30 - P1232427
Alain Brobecker
15009 Die Schwalbe 251 10/2011
P1232427
(16+16) C+
Logisches Progressivschach.
Finde die Partie, in der die vierte Zugserie mit dem vierten Teilzug
a) Dd7-c6#
b) De5xSg3#
c) Lh4-f2#
d) Lb3-d5#
e) Dd7#
endet.
a) 1. d3 2. e6 ... e5 3. Kd2 ... Ke3 ... Ke4 4. Lc5 ... d6 ... Dd7 ... Dc6#
b) 1. Sc3 2. e6 ... Dh4 3. Se4 ... Sg3 ... f3 4. Dxh2 ... Dh5 ... De5 ... Dxg3#
c) 1. f3 2. e6 ... Le7 3. Kf2 ... Kg3 ... Kh3 4. Lh4 ... Dg5 ... Dh5 ... Lf2#
d) 1. d3 2. d6 ... Le6 3. Kd2 ... Ke3 ... Kf3 4. Lb3 ... e6 ... Dg5 ... Ld5#
e) 1. f3 2. d6 ... Lh3 3. Kf2 ... Kg3 ... Kxh3 4. h6 ... h5 ... h4 ... Dd7#
play all play one stop play next play all
Keywords: Synthetic problem, Logical Progressive Chess, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Nikolai Beluhov, 2012-02-16
Last update: Alain Brobecker, 2021-08-25 more...
31 - P1234279
Alain Brobecker
15131 Die Schwalbe 253 02/2012
Composed with a computer
P1234279
(16+16) C+
Find the unique game in kamikaze chess ending with 5.Qd5xPf7#
1. e3 Sf6 2. Le2 Se4 3. Lh5 Sxd2[-sSd2] 4. Dd5 g5 5. Dxf7[-wDf7]#
play all play one stop play next play all
Keywords: Kamikaze, Synthetic problem, Homebase
Genre: Retro, Fairies
Computer test: Created with an homebrew program: http://abrobecker.free.fr/chess/synthetics.htm
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Alain Brobecker, 2012-03-11
Last update: Alain Brobecker, 2021-08-25 more...
32 - P1237298
Alain Brobecker
5912 diagrammes 159 10-12/2006
P1237298
(10+6) C+
s#6
b) -wBe6 = s#5
c) -wBc3 = s#4
a) 1. cxb6ep+ Lc5 2. e8=D fxe6 3. Dcxe6+ Ld6 4. Dc4+ Lc5 5. Dg6+ d6 6. Dge6 cxb6#
b) 1. cxb6ep+ Lc5 2. e8=D f6 3. exf6 Kd6 4. e5+ Kc6 5. f7 cxb6#
c) 1. cxb6ep+ Lc5 2. e8=D f6 3. exd7 f5 4. exf5 cxb6#, 3. ... fxe5 4. De7 cxb6#
play all play one stop play next play all
Anton Baumann: C+ Gustav 4.2a (2022-05-20)
comment
Keywords: En passant as key
Genre: s#
FEN: 8/2ppPp2/P1kbP3/KpP1P3/P1Q1P3/2P5/8/8
Input: Frank Müller, 2012-04-22
Last update: Marcin Banaszek, 2022-05-25 more...
33 - P1244702
Etienne Dupuis
Alain Brobecker

Messigny 2008
3rd place
P1244702
(14+13)
Solve the position
Theme was "a promoted piece captures a promoted piece"
hans: R: -1. Tb7xSb8+ (unpromote Sb8 on h1-h7) h6xSg7 (rearrange black pieces Ta8, Dd8, Ke8) then b7-b8=T c6xLb7 (2014-10-07)
comment
Keywords: RIFACE Retro Composition Tourney (2008), Ceriani-Frolkin Theme, Prentos Theme
Genre: Retro
FEN: 1Rkrqb1n/p1ppprP1/Kp3pp1/8/8/8/PP1PPPP1/1NBQ2NR
Input: Alain Brobecker, 2012-07-16
Last update: A.Buchanan, 2014-06-28 more...
34 - P1247344
Alain Brobecker
15182 Die Schwalbe 254 04/2012
Dedicated to Peter Fayers
P1247344
(15+14)
BP in 5.0 Zügen
Actuated Revolving Center
Keywords: Unique Proof Game, Actuated Revolving Center
Genre: Retro, Fairies
FEN: rn1qkbnr/ppp2ppp/8/8/3pP3/1P6/P1PP1PPP/RN1QKBNR
Input: Mario Richter, 2012-08-20
Last update: Alain Brobecker, 2021-12-27 more...
35 - P1254151
Thierry Le Gleuher
Alain Brobecker

15242v Die Schwalbe 255 06/2012
P1254151
(14+15) C+
Geschichte der Springer?
Taschenspringer
Definition Taschenspringer: Beide Seiten erhalten zu Beginn der Partie einen zusätzlichen Springer der eigenen Farbe, den sie im Laufe der Partie anstelle eines eigenen Zuges auf dem Brett einsetzen können; das ist auch mit Schachgebot (und damit Matt) zulässig.

Im Original waren die Plätze von wDh3 und wTg3 vertauscht, was aber eine alternative Auflösung erlaubt hätte.
comment
Keywords: Pocket Piece (Ss)
Genre: Retro, Fairies
FEN: N7/2pp4/pp6/6P1/3PPPrB/4pqRQ/P1PpBbPR/k1Kbnrnn
Input: Mario Richter, 2012-11-23
Last update: A.Buchanan, 2024-01-18 more...
36 - P1390776
Alain Brobecker
Quartz 44, p. 767, 12/2017
Composed with a computer
6. Lob
P1390776
(31+0) C+
Position after Black's second move, white can mate in 1.
Chess960.
31 unspecified pieces.
BQNBRKNR : 1.Nf3 e5 2.Nxe5 Nce7 then 3.Nxd7#
play all play one stop play next play all
If the piece captured was not a pawn, then it was a white knight captured by a black knight and no other piece moved, but that's not possible. So a pawn was captured. Only possibility
is BPe7-e5 captured by WNf3xe5, and black then played BNc8-d7.
Thus we know the start positions of the Ns.
Only possible mate is WNe5xd7# with BKf8. Since we have BNg8, then we have BRh8.
e8 cannot contain a Queen or Bishop, then BRe8.
d8 cannot contain a Queen, thus BBd8 and the rest is BBa8, BQb8.
Keywords: Chess960, Unique Proof Game, Colouring problem, Carving problem
Genre: Retro, Fairies
Computer test: Created with homebrew program.
FEN: yy1yyyyy/yyyyyyyy/8/4y3/8/8/yyyyyyyy/yyyyyy1y
Input: Alain Brobecker, 2021-06-13
Last update: Alain Brobecker, 2022-04-08 more...
37 - P1390780
Alain Brobecker
Die Schwalbe 286 2017/08
Composed with a computer
P1390780
(30+0) C+
Position after Black's third move, white can mate in 1.
30 unspecified pieces.
1.e3 f5 2.Bd3 f4 3.Bxh7 fxe3 then 4.Bg6#
play all play one stop play next play all
Keywords: Unique Proof Game, Colouring problem, Carving problem
Genre: Retro
Computer test: Created with an homebrew program.
FEN: yyyyyyyy/yyyyy1yy/8/8/8/4y3/yyyy1yyy/yyyyy1yy
Input: Alain Brobecker, 2021-06-13
Last update: Alain Brobecker, 2022-04-08 more...
38 - P1390782
Alain Brobecker
Die Schwalbe 287 10/2017
Composed with a computer
1. Lob
P1390782
(32+0) C+
Position after Black's second move, white can mate in 1.
Chess960.
32 unspecified pieces.
RKNRBNQB : 1.g3 a5 2.Qg2 Ka7 then 3.Qxb7#
play all play one stop play next play all
a2 -> f2 and h2 are WPs. b7 -> h7 are BPs.
g3 cannot be a WN or no mate would be possible. So g3 is a WP.
To have a mate along the diagonal with b8 free, we need the battery WQg2+WBh1 mating by WQg2xb7#.
For this to be mate we need to have BPa5 and not BNa5.
Since we have BQg8+BBh8, then a8 is not BB, c8 is not BB, d8 is not BN, which leads to BBe8 and BRd8.
Black's second move was Xb8-a7, but not BQ or BB already placed.
Thus we have BKa7 and then BRa8.
Last the kNights are in the remaining places and the baseline is RKNRBNQB.
Keywords: Chess960, Unique Proof Game, Capture-free, Colouring problem, Carving problem
Genre: Retro, Fairies
Computer test: Created with homebrew program.
FEN: y1yyyyyy/yyyyyyyy/8/y7/8/6y1/yyyyyyyy/yyyyyy1y
Input: Alain Brobecker, 2021-06-13
Last update: Alain Brobecker, 2022-04-08 more...
39 - P1392476
Christopher Cedric Lytton
Alain Brobecker

R382 The Problemist 01/2007
P1392476
(9+5) C+
Add all remaining units for a legal position with no unit attacking an enemy unit.

The pieces can be found in the following order: WPa2, BPa3, WNb3, BPc7, BPf7.
Then we must note that the whole 4th line is empty because the squares are attacked by both a W piece
and the BRa4, or no remaining piece would fit. After this we have: WPc3, WPb5, BPb6, WPg5, BPg6,
WPe5. Then WPd3 (to block WQ attack on BPg6), BPd7, BQd8 (only square), BBh7 (only square) and
BPh5.
Let’s now suppose we have BPe7, then BKh1 but the position is illegal because after the first BB has
moved after b7-b6 or g7-g6, the other one can’t reach the diagram position. So we have BPe6 and BKe7.
Keywords: Add pieces, Construction task
Genre: Retro
FEN: 1n2r1n1/1B4B1/8/8/r7/p1P1N1RP/2Q1RK2/8
Input: Alain Brobecker, 2021-08-04
Last update: Alain Brobecker, 2022-04-08 more...
40 - P1393033
Alain Brobecker
diagrammes 159 10/2006
Composed with a computer
P1393033
(16+0) C+
Add the 16 black pieces for a legal position with no unit attacking an enemy unit.

Solution is: qrbbN1NQ/1p1p4/3Pp3/pPBnP1pB/P1p2pPp/5P2/1RP3RP/K1nrk3
Keywords: Add pieces, Construction task
Genre: Retro
Computer test: Created with an homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: 4N1NQ/8/3P4/1PB1P2B/P5P1/5P2/1RP3RP/K7
Input: Alain Brobecker, 2021-08-22
Last update: Alain Brobecker, 2021-08-25 more...
41 - P1393034
Alain Brobecker
diagrammes 160 01/2007
Composed with a computer
P1393034
(16+0) C+
Add the 16 black pieces for a legal position with no unit attacking an enemy unit.

Solution is: 1R2Nnqr/QB1ppp1p/7P/Bp1P3R/bPp5/p1P1PPp1/P4nP1/1KN1krb1
Keywords: Add pieces, Construction task
Genre: Retro
Computer test: Created with an homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: 1R2N3/QB6/7P/B2P3R/1P6/2P1PP2/P5P1/1KN5
Input: Alain Brobecker, 2021-08-22
Last update: Alain Brobecker, 2021-08-25 more...
42 - P1393035
Alain Brobecker
Tangente Jeux 23 01/2007
Composed with a computer
P1393035
(16+0) C+
Add the 16 black pieces for a legal position with no unit attacking an enemy unit.

Solution is: q1b1rb1K/rpBpnN2/p1p1ppp1/P6k/1PNPPP1p/2P2R1P/BQ2R1P1/7n
Keywords: Add pieces, Construction task
Genre: Retro
Computer test: Created with an homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: 7K/2B2N2/8/P7/1PNPPP2/2P2R1P/BQ2R1P1/8
Input: Alain Brobecker, 2021-08-22
Last update: Alain Brobecker, 2021-08-25 more...
43 - P1393150
Alain Brobecker
14885v Die Schwalbe 258 12/2012
version by Ravi Shankar
P1393150
(3+5) C+
ser-h#8
1. Df6 2. Kb2 3. Ta2 4. Ka1 5. Tc2 6. Ka2 7. Da1 8. Tb2 Sc3#
play all play one stop play next play all
version of P1230203
comment
Keywords: Seriesmover, Pure Round Trip
Genre: Fairies
FEN: 8/6p1/8/8/8/p3K3/kq1N4/rN6
Input: Alain Brobecker, 2021-08-25
Last update: Alain Brobecker, 2021-08-25 more...
44 - P1394799
Alain Brobecker
JM1 Phénix 223 12/2012
Mémorial Jean-Michel Trillon
2nd honorable mention
P1394799
(24+0)
Colour the pieces.
Bichromatic chess.
(The start and end squares of every move must be of different colours)
b) R: 1. d7-d8=S# c2-c1=L
play all play one stop play next play all
Some notes on Bichromatic chess: Bishops cannot move. Castlings are forbidden. Queens are Rooks with
a different shape (!?). Pawns cannot capture neither make double steps, thus they cannot change column
and there is at most one promotion per file (and both Pawns must be absent).
Nd8 attacks both K and has no square to come from, so thge last move is n.d7-d8=N+, thus WNd8.
The Ps have the natural colour, ie WPa4, WPb2, WPe2, WPf2, WPg2, WPh4 and BPa6, BPb6, BPe6,
BPf7, BPg7, BPh7. We can deduce BBf8 since there was no promotion on this column. But then the
BRh8 was not able to go out and has been captured in its homecage.
Moreover, if there was a promotion on the c-file it’s a promotion to B since there is at most one promotion
per file and the extremities are occupied by Bs. Thus there as been no promotion to R (or Q) during
the game, so there remains 1 BR and 2 WRs, and we can deduce the colours of the Ks and of the pieces
touching them: BRb8, BKb7, BQa7, WKc6, WRd6, WRc5, WNe7. We deduce also WQa8.
If WBc8 then it has been promoted, then WBc1 (one promotion per file) but then black is retro-stalemate
before n.d7-d8=N#. Thus our hypothesis is wrong and we have BBc8.
If WBc1 then black is retro-stalemate before n.d7-d8=N+, so we have BBc1 and the last black move is
n-1... c2-c1=B.
(This was the small idea of this problem, a light trap with a promoted Bishop).
Henrik Juel: The only problem with these two keywords in the PDB, so far
Interesting problem (2021-10-22)
A.Buchanan: Very fun. How about home-and-away-base? I can find a position where the "orthodox" homebase colouring doesn't work, but I can't make the solution quite unique. (2021-10-23)
A.Buchanan: Alain do you have further publication details please? (2021-10-23)
Mario Richter: Alain will surely know better, but according to my archives the wanted details are:
2° Mention d’Honneur : Alain BROBECKER (JM1, Phénix 223 décembre 2012)

The problem is not difficult, but it nicely shows the subtleties of the "Bichromatic chess" condition. Besides the "light trap with a promoted Bishop" there's also a second trap: wKb7, black rooks d6 and c5, which fails only because bRh8 "has been captured in its homecage" (a fact that can be easily overlooked.) (2021-10-23)
A.Buchanan: This problem was featured in yesterday's Monthly International Zoom Call, and was a great success, with both traps being fallen into :) (2021-10-24)
A.Buchanan: In a colouring problem, non-standard material would equate to 3+ queens or bishops on a certain colour square, or 5+ rooks or knights (2021-10-24)
comment
Keywords: Colouring problem, Bichrome Chess, Promotion (Sl), Last Moves? (2)
Genre: Retro, Fairies
FEN: QRBN1B2/QK2NPPP/PPKRP3/2R5/P7/8/1P2PPPP/2B5
Input: Alain Brobecker, 2021-10-22
Last update: Alain Brobecker, 2021-12-24 more...
45 - P1397442
Alain Brobecker
diagrammes 155 2005
P1397442
(7+0)
Colour the pieces. Player to move mates in 2 moves.

The Qb8 attacks both K, so she comes from b7 which is also next to both K. So last move is b7-b8=Q+, thus WQb8. Moreover BPb6 or one of the K would have been attacked before this move. If BKa7 then we would have WKc7, WNa8, WNa8, WRc8, but then B couldn't have played before (retro stalemate). So our hypothesis is wrong and we have WKa7 and then BKc7, BNa8, BNa6, BRc8.
nQr5/K1k5/np6/8/8/8/8/8
The mate is then: 1... Nxb8 2.Kxa8 Nc6#.
Keywords: Miniature
Genre: Retro
FEN: NQR5/K1K5/NP6/8/8/8/8/8
Input: Alain Brobecker, 2021-12-24
Last update: Alain Brobecker, 2021-12-24 more...
46 - P1397443
Alain Brobecker
diagrammes 156 2005
P1397443
(10+0) C+
Colour the pieces.
R: 1. Tb7-b8+,Tb7xb8+ b3-b2,a3xb2,c3xb2
play all play one stop play next play all
Solution: kRN5/r1BN4/RKQ5/8/8/8/1p6/8
w: Tb8 Sc8 Lc7 Sd7 Ta6 Kb6 Dc6
s: Ka8 Ta7 Bb2
Henrik Juel: retroplay 1.Tb7-b8+ or Tb7xDTLSb8+ b3-b2 (2021-12-24)
comment

Genre: Retro
Computer test: rawbats
FEN: KRN5/R1BN4/RKQ5/8/8/8/1P6/8
Input: Alain Brobecker, 2021-12-24
Last update: Mario Richter, 2021-12-25 more...
47 - P1397480
Alain Brobecker
diagrammes 156 2005
d'après Gidéon Husserl
P1397480
(27+0)
Colour the pieces. Last 2 single moves?
R: 1. c7xSd8=T++ La5-b6
play all play one stop play next play all
Autor: Inspired by P0006899

w: La8 Td8 Se8 Tf8 Lg8 Dh8 Ba7 Sf7 Tc6 Kd6 Lf6 Bh6 Be5 Bg5 Bc4 Bf4
s: Tb8 Kc8 Bb7 Bg7 Bh7 Ba6 Lb6 Be6 Bc5 Bd5 Bf5
BrkRNRBQ/Pp3Npp/pbRKpB1P/2ppPpP1/2P2P2/8/8/8

Genre: Retro
FEN: BRKRNRBQ/PP3NPP/PBRKPB1P/2PPPPP1/2P2P2/8/8/8
Input: Alain Brobecker, 2021-12-25
Last update: Mario Richter, 2021-12-26 more...
48 - P1397482
Alain Brobecker
diagrammes 156 2005
P1397482
(8+5) C+
#5

Last move is c7-c5, and then:
1.dxc6! (1.bxc5? Rxa5+ and the rook goes crazy) followed by:
1...dxc6+ 2.Kc4 (to avoid the crazy rook)
2...Rxa5 3.bxa5 c5 4.a6! bxa6 5.Bc6#
2...Rxb6 3.axb6 c5 4.d7 cxb4 5.d8=Q/R#
2...Rxa7? 3.d7 ~ 4.d8=Q/R#
1...Rxb6+? 2.Kxb6 bxc6 3.Bb3 c5 4.Bd5# or 2...dxc6 3.d7 c5 4.d8=Q/R#
1...bxc6+? 2.Kxa6 c5 3.b7#

Genre: n#
FEN: k7/Pp1p4/rP1P4/PKpP4/BP6/8/8/8
Input: Alain Brobecker, 2021-12-25
Last update: Alain Brobecker, 2021-12-25 more...
49 - P1397483
Alain Brobecker
RIFACE 2005
Special Honorable Mention
P1397483
(8+0)
Colour the pieces. Spiders at a6 and b8.

Both spiders are attacking both kings, thus there was a discovery which was a7x?b8=Spider+.
So the b8 Spider is white and we deduce that Kd8, Spider a6, Bc7 and Rd7 are white when Ka8 is black.
Pb7 is black otherwise it would give an illegal check to BKa8.
And Pd6 is black otherwise black would be retro stalemate: BKa8 couldn't come from b8 which is attacked twice, and the black piece captured on b8 had no legal move (a queen, rook or spider would have attacked the WKd8).

Pieces: du = Spider (UD)
Genre: Retro, Fairies
FEN: K*2Q1K4/1PBR4/*2Q2P4/8/8/8/8/8
Input: Alain Brobecker, 2021-12-25
Last update: A.Buchanan, 2023-06-02 more...
50 - P1397484
Alain Brobecker
diagrammes 157 2006
P1397484
(6+0)
Colour the pieces. Last 2 single moves?
Grasshopper at c5.

Composed to have an anti-discovery.
Ra4 and Gc5 attacks both kings, thus this two pieces share the same colour and there was a discovery (only possibility for a double check) or an anti-discovery or both.
Last moves are n-1... b7-b5 n.a5xb6+++ en passant, thus Pb6, Ra4, Ga5 and Ka3 are white when Ka7 is black.
And the Pb4 is white otherwise it would attack WKa3.

Pieces: du = Grasshopper (G)
Genre: Retro, Fairies
FEN: 8/K7/1P6/2*2Q5/RP6/K7/8/8
Input: Alain Brobecker, 2021-12-25
Last update: A.Buchanan, 2023-06-02 more...
51 - P1397485
Alain Brobecker
diagrammes 158 2006
2nd commendation
P1397485
(28+0)
Colour the pieces. Last 4 single moves?
BNQ2b2/1PNppppp/KpkqR3/p1N1BP2/b1pPP3/1R6/2P3PP/8
w: La8 Sb8 Dc8 Bb7 Sc7 Ka6 Te6 Sc5 Le5 Bf5 Bd4 Be4 Tb3 Bc2 Bg2 Bh2
s: Lf8 Bd7 Be7 Bf7 Bg7 Bh7 Bb6 Kc6 Dd6 Ba5 La4 Bc4

R: 1. a7xTb8=S+ Dd5-d6 2. Ld6-e5+ De5-d5
play all play one stop play next play all
Le Cb8 attaque les deux rois et n'a pas de case de provenance, donc le dernier coup est a7x?b8=C+, donc le Cb8 est B. Par ailleurs aucun autre pion ne manque donc il n'y a pas eu d'autre promotion, et comme la pièce prise en b8 ne peut pas être un C qui n'aurait pas eu de case de provenance et aurait déjà attaqué les deux R, on en déduit que le dernier coup est a7xTb8=C+.
Si le Pb7 était noir il aurait attaqué un des deux R avant ce coup, donc le Pb7 est B.
Reprenons le dernier coup a7xTb8=C+, alors il manque une T et deux C et chaque colonne contient deux P, et cela implique un nombre pair de prises par les pions. Comme le Pa7 et le Pb7 sont B, il y a eu deux prises sur les colonnes a et b et donc le Pa5 et le Pb6 sont N, alors que sur toutes les autres colonnes aucun des pions n'ont été échangés, donc sur les colonnes c à h les P du bas sont B alors que les P du haut sont N.
On en déduit que le Ff8 est N, et que la TNh8 a été prise dans la cage nord est. On en déduit que le Fe5 est B et que les Tb3 et Te6 sont B.
Le Rc6 est N car sinon le PNd7 l'attaquerait, donc le Fa4 et la Dd6 sont N alors que les Ra6, Cc7, Cc5 sont B. Et il s'ensuit que la Dc8 et le Fa8 sont B.
Remarquons que les prises ont été: la TNh8 dans le cage nord est, la TNa8 prise en b8 sur promotion, et les deux CN ont été pris sur les colonnes a et b par les pions. Le seul coup noir possible avant a7xTb8=C+ est Dd5-d6, mais alors la TBe6 était déjà sur la 6ème rangée et n'a pas pu prendre de pièce, donc il y a eu une découverte par Fd6-e5 et la encore le coup noir est forcé: 1. ... De5-d5 2.Fd6-e5+ Dd5-d6 3.a7xTb8=C+

Genre: Retro
FEN: BNQ2B2/1PNPPPPP/KPKQR3/P1N1BP2/B1PPP3/1R6/2P3PP/8
Input: Alain Brobecker, 2021-12-25
Last update: Mario Richter, 2021-12-26 more...
52 - P1397506
Alain Brobecker
The Problemist 2007/05
P1397506
(31+0)
?+?=31: Colour the pieces.

All 16 pawns are on (the) board, so no promotion occured.
The only capture of the game was a S, either by BPhxSg or by WPhxSg. Except on the g and h files, the pawns on bottom of the diagram are W while those on top are W. We can deduce that Ra2 and Bc1 are W, that Bc8 is B, then that Bf1 is W and Bb4 is B.
But if the capture was WPhxBSg then the g2 pawn would be W, and the WBf1 couldn't have moved, thus the WK and WQ couldn't have moved either. So the capture was BPh3xWSg2 and Pg3 and Ph2 are W while Pg2 and Pg7 are B.
The WPs are making a gate, so Rg1 is W, and thus Rb3 and Rh7 are B.
Since the BPf7 has not moved, the Kg6 is B and then Kg8 is W. The Qh8 has no square to come from, hence Qh8 has the same color as Kg8, ie W, while Qh3 is B.
Let's suppose the Sf8 is B. If B is to move then W is retro stalemated, since the WKg8 can't retro-play from f8 since this square is heavily controlled: the BBb4 and BSh7 would need three moves retro-played to free the f8 square, no covering is possible and the BPe6 can't be retro played soon (note: Cedric found a cook in previous version where the WS was able to block the WB attack on f8).
If W is to move he is also retro stalemated, since we can't find a black move that gives play to W: The captured WS has been captured long ago, since the BPg2 can't come from h3 right now, and so the captured piece can't have played just before. The Sa1 and Sh1 are caged in a way that would need more than one B move retro played to let them escape. The WK can't come from f8 in this case either. The WQ can't come from h7 where it would have attacked the BK.
So there's a contradiction, and we can conclude our hypothesis was wrong, so the Sf8 is W while Sa1 and Sh1 are B.

Genre: Retro
FEN: 2B3KQ/1P1P1PPN/4P1KR/P7/1B6/PRP3PQ/RPPPPPPP/N1B2BRN
Input: Alain Brobecker, 2021-12-27
Last update: Alain Brobecker, 2021-12-27 more...
53 - P1397507
Alain Brobecker
diagrammes 160 2006
P1397507
(30+0)
?+?=30: Colour the pieces. Last 4 single moves?

Inspiré du problème suivant: Ra6, Pb6, Tc6, Td5, Da4, Fc4, Rb3, Td3 (Alain Brobecker, championnat de France de solution d'analyse rétrograde 2005). La Da4 et le Fc4 attaquent les deux R, donc ces deux pièces sont de la même couleur et il y a eu une découverte par n-1... b7-b5 n.a5xb6+ en passant et on a PBb6, DBa4, FBc4, TBd3, RBb3, CBa1 et RNa6. Avant cela le RNa6 était attaqué par le Fc4 qui n'a pas de case de provenance, il y a donc eu une autre découverte par n-1.Tb5-d5+ et on a TBd5. La découverte ne pouvait pas être bxc6 ou Txd5 car la deuxième pièce manquante est un C qui a été pris par PBfxCNe ou PNfxCBe. Cela implique aussi qu'il n'y a pas eu de promotion et donc DNh8, TNg8, TNb8, FNc8. En dehors des colonnes e et f, les P ont la couleur "naturelle". Mais alors PBe7 car sinon la cage nord est n'aurait pas pu être mise en place, donc PBb2, PBc6, PBd2, PBe3, PBg2, PBh2 et PNa7, PNc7, PNd7, PNe6, PNf7, PNg7, PNh7. On en déduit FBc1 et FNf8. Enfin CNa8 et CBd8 sinon les N seraient rétro pat. Enfin les derniers demi-coups sont: n-2... Cb6-a8 n-1.Tb5-d5+ b7-b5 n.a5xb6+.

Genre: Retro
FEN: NRBN1BRQ/P1PPPPPP/KPP1P3/3R4/Q1B5/1K1RP3/1P1P2PP/N1B5
Input: Alain Brobecker, 2021-12-27
Last update: Alain Brobecker, 2021-12-27 more...
54 - P1397509
Alain Brobecker
diagrammes 160 2007
P1397509
(30+0)
Colour the pieces. Last 6 single moves?

The Qa4 and Rc4 attack both K, so there has been a discovery after n-1... b7-b5 n.c5xb6+ e.p. We deduce from it WPb6, WQa4, WRc4, WKb4 and BKc6, and then WNa2, WPa5, BQa8, BNb8.
Before this two moves the WQa4 was already attacking BKc6, and since she has no square to come from there has been a discovery by n-1.Nb5:a3 (b5xc6 is impossible because Pb5 controls c6) and thus WNa3. Before those moves all 16 pawns were on the board, and they had made no capture, so WPb2, WPd2, WPe4, WPf3, WPg2, WPh2 and BPa6, BPc7, BPd7, BPe7, BPf6, BPg7, BPh7. There has been no promotion, so BNe1. The BRs couldn't go out, so BRa7 and the 2nd BR has been taken in the northeast cage, so WRf2. Last we have BBc8, BBf8 then WBd4, WBh3.
The f7-f6 move has been played a "long" time ago so has to let the BK and BQ out. Similarly e2-e3 has been played "long" ago to let WBh3 out, thus the 6 last single moves are: n-3... Kc6-d5 n-2. e3-e4+ Kd5-c6 n-1.Nb5-a3+ b7-b5 n.c5xb6+.

Genre: Retro
FEN: QNB2B2/R1PPP1PP/PPK2P2/P7/QKRBP3/NP3P1B/N2P1RPP/4N3
Input: Alain Brobecker, 2021-12-27
Last update: Alain Brobecker, 2021-12-27 more...
55 - P1397510
Alain Brobecker
RIFACE 2007
Commendation
P1397510
(7+0)
Colour the pieces.
Doped pawns.

Qb6 and Rc7 are attacking both kings, so there was a discovered check. The only possibility is n... b7-b6=Q+, so we have BQb7, BRc7, BKc6, BNb8 and WKa7, WRa8. Last we have WPa6 otherwise white would have been retro-stalemate.

Pieces: bu = Doped pawn ()
Genre: Retro, Fairies
FEN: RN6/K1R5/*2PQK5/8/8/8/8/8
Input: Alain Brobecker, 2021-12-27
Last update: A.Buchanan, 2023-06-02 more...
56 - P1397511
Alain Brobecker
Variant Chess 56 2008/02
P1397511
(15+12)
PG in 3.5 double moves.
Marseillais Chess.
1. a4 ... 2. Ta3 e6 3. ... Dh4 4. Tf3 ... 5. b3 La3 6. ... Dxf2+ 7. Txf2 ... 8. Txf7 Sf6 9. ... Tf8 10. Lxa3 ... 11. Txf8#
play all play one stop play next play all
1.a4,Ta3 e6,Dh4
2.Tf3,b3 Fa3,Dxf2+
3.Txf2,Txf7 Cf6,Tf8
4.Fxa3,Txf8#
Keywords: Double Move Chess, Unique Proof Game
Genre: Retro, Fairies
FEN: rnb1kR2/pppp2pp/4pn2/8/P7/BP6/2PPP1PP/1N1QKBNR
Input: Alain Brobecker, 2021-12-27
Last update: A.Buchanan, 2023-06-03 more...
57 - P1397526
Alain Brobecker
Shakhmatnaya Kompozitsiya 2008
1st prize
P1397526
(16+16) C+
Find the unique monochromatic chess game terminating with 5... Qf8xPf2#
1. h4 e5 2. Th3 Lc5 3. Tf3 De7 4. Txf7 Df8 5. Txd7 Dxf2#
play all play one stop play next play all
Keywords: Monochromatic Chess, Synthetic problem
Genre: Retro, Fairies
Computer test: created with a homebrew program ( http://abrobecker.free.fr/chess/synthetics.htm#monochromatic )
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Alain Brobecker, 2021-12-28
Last update: A.Buchanan, 2022-09-14 more...
58 - P1397527
Alain Brobecker
The Problemist 2008/03
P1397527
(0+0)
Different letters represent units different in type and/or colour. Recover the position.
a) solution
play all play one stop play next play all
Diagram
Henrik Juel: A is black bishop, B is white pawn, C is white king, D is black king (2021-12-30)
Henrik Juel: The seven black bishops cannot be black pawns, because this would make [Lc8] inaccessible, so White cannot make his eight pawn captures
The five black promotions to bishop are easily explained, e.g. two promotions on b1 and h1 and one on g1 (2021-12-31)
comment
Keywords: Rebus
Genre: Retro
FEN: 8/8/8/8/8/8/8/8
Input: Alain Brobecker, 2021-12-28
Last update: A.Buchanan, 2023-05-18 more...
59 - P1397661
Alain Brobecker
Tangente 122 2008/05
P1397661
(2+3)
This game of Tic-Tac-Toe was played between experts (never playing a losing move).
How did the game go?

Black just played and his last move can't have been c1 (or a2), since b2 (respectively b1) would have won. So black's last move is b3. Black's first move can't have been c1 since white would have played b2 and won. So black started with a2 and white must have answered with a3. So we can conclude that game was a2, a3, c1, c3, b3.
Alain Brobecker: First problem of this kind was made by Les Marvin: P0005297 (2022-01-01)
Alain Brobecker: Full analysis of such Tic-Tac-Toe games with expert players can be found here: http://abrobecker.free.fr/text/tictactoe.pdf (2022-01-01)
comment
Keywords: Tic-Tac-Toe, no 8x8 board
Genre: Retro, Fairies
FEN: qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/PpPqqqqq/p2qqqqq/2pqqqqq
Input: Alain Brobecker, 2022-01-01
Last update: A.Buchanan, 2024-01-18 more...
60 - P1398142
Alain Brobecker
SS009 The Hopper Magazine I01 24/12/2021
P1398142
(31+0) C+
Position after Black's third move, white can mate in 1.
31 unspecified pieces.
Keywords: Unique Proof Game, Colouring problem, Carving problem
Genre: Retro
Computer test: Created with a home-brew program
FEN: yyyy1yyy/yyyyyyyy/8/2y5/8/2y5/yy1yyyyy/yyy1yyyy
Input: Alain Brobecker, 2022-01-09
Last update: Miguel Ambrona, 2023-07-26 more...
61 - P1398145
Alain Brobecker
R581 The Problemist 11/2021
P1398145
(31+0) C+
Position after Black's third move, white can mate in 1.
31 unspecified pieces.
1. c3 d5 2. Db3 Kd7 3. Dxb7 De8 4. Dxd5#
play all play one stop play next play all
Keywords: Unique Proof Game, Colouring problem, Carving problem
Genre: Retro
Computer test: Created with a homebrew program
FEN: yyy1yyyy/yyyyyyyy/8/3y4/8/2y5/yy1yyyyy/yyy1yyyy
Input: Alain Brobecker, 2022-01-09
Last update: A.Buchanan, 2024-01-31 more...
62 - P1400086
Alain Brobecker
R587 The Problemist 2022/03
Dedicated to Mario Richter
P1400086
(2+4) C+
Add all remaining units for a legal position with no unit attacking an enemy unit.
We must have one WP and one BP per column, the BPs being above the WPs since no capture occurred.
On the c column the WP cannot be on c2, where it would attack the square d3 where a black shield would be needed
to protect from BQe4, and BP cannot be on c7 to c5, thus we have WPc3 and BPc4.
This implies that on column b we have BPb3 and WPb2, then due to BPb3 the only possibility on
column a is WPa3 and BPa4.
Now let us look at the shield needed to protect WRb7 from the attack of BQ e4: a black piece on c6 is not
possible, so we must have a black piece on d5. It cannot be a BN which would attack WQb6 or a BB which
still would attack WRb7, and the BK and BRs already are elsewhere, so we have BPd5. We then deduce that
we have WPd2, and this forces WBc1 which has not moved from the game.
We need another shield to protect BPb3 from the attack of WQb6. Now that WBb4 is no more possible since this
bishop is still home, the only possibility is WNb5.
We cannot have a black piece on e3, hence no white piece on e2 and from this we deduce that we have
WPd6 and BPd7. The only possible shield to protect WPe6 from the attack of BQe4 is BNe5 (BBe5 would attack WPc3).
Now we also need a shield on c7 to protect BPd7 from the attack of WRb7, and the only possibility is WKc7.
On the f column we have WPf2 which is the only possibility.
More difficult to see, the only possibility for light squared WB is WBh5. To see this we must acknowledge
that g2 will contain a black piece as a shield and that other squares are attacked and no shield will help:
WBb1 is not possible because no shield can be put in-between it and BQe4, and WBe8 is not possible
since no B piece can be placed in b8-d8 as a shield.
Let's suppose that we have WPh4, then two black shields are needed on on g2 and h3 to
protects WPf2 and WPh4 from BRh2, but one of this shield would be the light squared BB
and the other one would be the remaining BN which would attack on of the WPs.
Thus our hypothesis is wrong, and we have WPh6 then BPh7.
We deduce then that we have BPg5 then WPg3.
What we said about the black shields on g2 and h3 (to protect WBh5 this time)
still holds, but this time we have a possibility which is BNg2 and BBh3.
BBf6 is the only possible square for dark squared BB. Thus we have BPf5.
WNd1 is the only possible square for the remaining WN.
Last we have two possibilities for the remaining WR, namely a1 and f1.
But since WPa3, WPb2, WPd2, WBc1 forms a well know retro-analytical cage
from which the WR cannot have escaped, we know that we have WRa1.
r6k/1RK1p2p/1Q2Pb1P/1N1pnppB/p1p1q3/PpP3Pb/1P1P1Pnr/R1BN4
play all play one stop play next play all
Alain Brobecker: The dedication was forgotten in the magazine! (2022-03-28)
comment
Keywords: Construction task, Add pieces, Aristocrat, Miniature
Genre: Retro
Computer test: Verified with homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: r6k/1R6/1Q6/8/4q3/8/7r/8
Input: Alain Brobecker, 2022-03-28
Last update: Alain Brobecker, 2023-06-16 more...
63 - P1400167
Alain Brobecker
RIFACE 2021
P1400167
(31+0) C+
Position after Black's third move, white can mate in 1.
31 unspecified pieces.
1. c4 g5 2. Da4 Lg7 3. Dxd7+ Kf8 4. Dxd8#
play all play one stop play next play all
Keywords: Unique Proof Game, Colouring problem, Carving problem, RIFACE Retro Composition Tourney (2021)
Genre: Retro
Computer test: Created with homebrew program.
FEN: yyyy1yyy/yyyyyyyy/8/6y1/2y5/8/yy1yyyyy/yyy1yyyy
Input: Alain Brobecker, 2022-04-01
Last update: A.Buchanan, 2023-04-25 more...
64 - P1402532
Alain Brobecker
Michel Caillaud
Eric Huber

RIFACE 2022
P1402532
(16+16)
Find the unique Vogtländer Chess game finishing on white's 3rd move with a mate by double check.
1. e4 d5 2. Ke2 dxe4 3. Kd3#
play all play one stop play next play all
First version was cooked and corrected by Eric Huber and Michel Caillaud
Keywords: Vogtländer Chess, Synthetic problem, RIFACE Retro Composition Tourney (2022)
Genre: Retro, Fairies
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Alain Brobecker, 2022-07-07
Last update: A.Buchanan, 2023-04-25 more...
65 - P1403290
Alain Brobecker
Die Schwalbe 315, p. 570, 2022/06
P1403290
(8+5) C+
Add all the missing units for a legal position with no unit attacking an enemy unit.
Chess960, what was the start position?
(written with help of Mario Richter)
We must have one WP and one BP per column, the BPs being above the WPs since no capture occurred.
From this fact we immediately deduce WPh2, BPf7, WPf3, BPe7, WPe2, BPb7, BPa3 and WPa2.
Now we need a white piece to protect BPa3 from the attack of WQa5, only WNa4 fits.
Let's suppose that we have BPc7, then no white piece could go on b6 to protect it from WQa5, hence this hypothesis is wrong and we have BPc6. From this we deduce that we have WPb6 to protect BPb7 from the attack of WRb4 (WBb6 is not possible since it would attack BNf2).
Now let's consider the initial column of light squared WB and dark squared BB. It cannot be column b or column d otherwise light squared WB couldn't have reached f1, and it cannot be column f since BB would still be trapped there but f8 is attacked by WNh7. So the initial column of light squared WB and dark squared BB is column h and from this we deduce that we have BBh8 and also WPg5 (because WPg2 would trap light squared WB on h1).
We need a white shield to protect BBh8 (or any other black piece on 8th rank) from WRa8, and the only possible shield is WBb8 (on other squares it would attack BPs). This implies that we have BPd7.
The remaining black pieces (BQ, BRs and BB) must be on 8th rank, anywhere else they would attack or be attacked. So they are on c8,d8,e8 and g8. We must have BBc8 since the other pieces would attack WBb8 (this also says that c is the initial column for dark squared WB and for light squared BB), and also we have BQe8
since on the remaining squares BQ would attack a white piece. Then we have BRd8 and BRg8.
According to Chess960 rules the BK must have been between the two BRs. The only possible explanation
is that the Chess960 start position was NNBRQKRB and black made a "right side" castling
in which BKf8 and BRg8 exchanged place to land on the usual BRf8+BKg8 castling positions.

RBbrq1rb/1p1ppppN/1Pp1k3/Q5P1/NR3K1p/p4P2/P1PPPn1P/3n1B2

Note: The problem is still valid without WQa5 as my homebrew verification program shows, but seems too hard to solve for humans (at least for me).
play all play one stop play next play all
Keywords: Chess960
Genre: Retro, Fairies
Computer test: Verified with homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: R7/6pN/4k3/Q7/1R3K1p/8/2PP1n2/3n1B2
Input: Alain Brobecker, 2022-08-02
Last update: A.Buchanan, 2024-01-18 more...
66 - P1403291
Alain Brobecker
The Problemist 2022/07
P1403291
(3+4)
Add all remaining units for a legal position with no unit attacking an enemy unit.
Shrink Chess (An edge file or rank disappears if unoccupied, Joseph Boyer, 1954).
Which ranks were discarded?
(written with help of Mario Richter)}
Since we don't know what ranks have been removed with the Shrink Chess condition, let us start by numbering the ranks 11 to 16 for clarity.
We must keep in mind that some pawns could be on rank 11 or 16, and we have one WP and one BP per column, the BPs being above the WPs since no capture occurred.
We need a white shield on c13 and a black shield on b14 to protect WBd12 and BBa15 from mutual attack. Due to possible attacks on WBb13 and BBb11, only WRc13 and BNb14 can do.
On the d-file the only possible position for BP is BPd16.
On the c-file the BP is on c15 or c16, but in both case we need a white shield to protect it from WRc13. The only possible shield is WPc14, and this implies BPb16.
From all this we deduce that the 7th and 8th ranks were removed, because with only the 8th rank removed the BPs on file b and d would have prevented BBc8 to get out.
Knowing that the remaining ranks are 1st to 6th, we immediately have WPa4 and BPa6, and from this we deduce we have WPb2. We have WPd4 and this implies BPc6.
We also have WPg3 because this pawn has moved to let the WBf1 out, and from this we deduce BPf5.
On the h-file we know we'll have a black shield between WP (on h4 or h5) and BRh1.
Only BNh2 is suitable (because BNh3 would attack the WP from f-file which must stand on f2 or f4). We don't have any more shield available for BRh1, hence we know that no white pieces can stand on squares c1 to g1.
Let's look at the possibilities for WQ. Only two squares are possible: e3 and f2.
But on e3 we would need a white shield to protect BPe from its attack, but no shield can go on e4 or e5. So we have WQf2 which implies WPf4, and also WNg2 as a shield to protect BNh2 from WQ's attack.
Then only e3 is possible for WK and this implies BPe6.
We also have very small space for the remaining pieces. Only g6 and h6 are possible for BQ and BR, and not to attack WPf4 we have BQg6, BRh6, BPh5 and WPh4.
Last, only a1 and a3 are available for the two remaining pieces, and so we finish by WNa1 and WRa3.

8/8/pppppkqr/b4p1p/PnPP1PpP/RBR1K1P1/1P1BPQNn/Nb5r

Note: BPg4 was not needed by my homebrew verification program to solve the problem, but adding it allowed more human-like reasonings.
play all play one stop play next play all
Keywords: Shrink Chess, no 8x8 board
Genre: Retro, Fairies
Computer test: Verified with homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: qqqqqqqq/qqqqqqqq/5k2/b7/8/1B6/3BP3/1b5r
Input: Alain Brobecker, 2022-08-02
Last update: Alain Brobecker, 2023-06-16 more...
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The problems of this query have been registered by the following contributors:

Gerd Wilts (17)
Nikolai Beluhov (4)
Mario Richter (3)
Hans Gruber (3)
Alain Brobecker (38)
Frank Müller (1)