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Phénix

No. 120, Jul-Aug 2003 - SOLUTIONS

4301 - J-M. Chorein
Phenix 120, Jul-Aug 2003

Black left the right to castle. Indeed, the Ra8 is a promoted one! The promotion and the Pe6 involve 6 captures on black squares; so, the BNg8 has been captured by the Pawn via f7 (check). For example, e4xf5, f5xg6 e.p., g6xf7+, f7xg8=R and ç4xBd5 and d5xRe6.

Solution: 1.a4!

A) 1..Rf8 2.axb5 Rf4 3. Ra2!! (3. Ra4? Rb4!!) Rd4 4.Kxd4 h5 5.Rc2 hxg4 6.Rc8 mat.

B) 1..Rf8 2.axb5 Rf2 3. Ra4!! (3. Ra2? Rb2!!) Rd2 4.Kxd2 h5 5.Rc4 hxg4 6.Rc8 mat.

C) 1..h5 2.axb5 hxg4 3.Ra2!! (3.Ra4? Rh2 4.Rc4 Rd2!) Rh2 4.Rg2!! Rh4 5.Rç2 Rh2 6.Rc8 mat.


4302 - A. Kisljak
Phenix 120, Jul-Aug 2003

Retro:

1.Na6-c5+ h2-h1=N 2.Bg2-f1 h3-h2 3.Be4-g2 h4-h3 4.Bh7-e4 h5-h4 5.Bg8-h7 h6-h5 6.g7-g8=B h7-h6 7.h6xNg7 Ne6-g7 8.h5-h6 Nc5-e6 9.h4-h5 Nd3-c5 10.h3-h4 Nc1-d3 11.Ka1-b1 Na2-c1+ 12.Kb1-a1 g6-g5 13.Ka1-b1 Nc1xBa2 14.Bb1-a2 Na2-c1+ 15.h2-h3 Rc1-c2 16.c2-c3 Rg1-c1 17.Rf3-b3

Phoenix theme (wB+bN)


4303 - Andrew Buchanan
Phenix 120, Jul-Aug 2003

a) Last move can’t be a black one, because the position would be remiss with a bP on h3 and a white Piece on g2. The only possible move would be h3xXg2= (null)!
The last move is White one.

Rh1­-g1 isn’t possible (despite of the last move h3xXg2), because the wK is in check and he must play on g1 or g2 giving pat!

The wK, coming from h1, just captured a black piece on g1. In this position, he could choose to capture on g1 or g2 (blacks can move the g1 piece)!

It’s impossible to give back a bR or bQ because of the imaginary double checks.
Giving back a bB on g1 impose too many Pawn’s captures to resolve the position!
We have to give back a bN!

After: f3xe2, e3xd2, d3xc2, c3xb2, b2xa3 et a2xb3.

The last move is: n.Kh1xNg1 !

b) Here, the count of Pawn’s captures prevents to give back a black piece on g1!
The last move must be black’s move. Pe4xXf3 is the only retro-move giving a position where blacks had choice. But the Pawn’s structure doesn’t correspond to the possible captures.

The position is illegal.


4304 - Andrew Buchanan
Phenix 120, Jul-Aug 2003

a) 1.d3 Nh6 2.Bxh6 g6 3.Qc1 Bxh6 4.Kd1 Bd2 5.Qxd2 00 (C+)

b) Before 5… 00, proof game isn’t unique. Example: 1.d3 Nh6 2.Bxh6 g6 3.Bxf8 Kxf8 4.Qd2 Ke8 5.Kd1!!


4305 (4044bis) - Didier Innocenti
Phenix 120, Jul-Aug 2003

Author solution: Add a wB on f8:

1.g3 g6 2.Bg2=N g5 3.Nh4=P gxh4=N 4.Nf3=P Nxf3=B 5.Kf1 Bxh1=R+ 6.Kg2 Rxd1=Q 7.a3 Qxc1 8.Ra2=B Qxb1 9.Be6=N Qa1=R 10.Nxf8=B Ra2=B 11.Kg1 Be6=N 12.Kh1 Ng7=P

Cooked by Vlaicu Crisan: Add a wN on g8!

1.g3 f5 2.Bg2=N f4 3.Kf1 f3 4.a3 fxg2=N 5.e3 Nxe3=B 6.Ra2=B Bxd2=R 7.Kg2 Rxd1=Q 8.Bg5=N Nf6=P 9.Nd2=P fxg5=N 10.Ne2=P Qa1=R 11.Bg8=N Rxh1=Q+ 12.Kxh1 Nf7=P


4306 - Andrew Buchanan
Phenix 120, Jul-Aug 2003

Typo error. The diagram is a copy of n°4214 Phénix 117.
The problem is updated in Phénix 128.


4307 - Joost de Heer
Phenix 120, Jul-Aug 2003

1.d4 Na6 2.Qd3 Nc5 3.Qg6 hxg6 4.d5 Rh3 5.d6 Rb3 6.axb3 Ne4 7.Ra6 Ng3 8.Rc6 dxc6 9.h4 Bf5 10.h5 e6 11.d7+ Ke7 12.h6 Kf6 13.h7 Bc5 14.h8=R Qe7 15.R8h4 Nh6 16.Ra4 Rh8 17.d8=Q Ng8 18.Qd1 Qd8 19.Ra1 (C+)

2 Pronkin pieces (wR, wQ), 2 switchbacks (bQ, bNg8), 1 sibling Rook (bRa8àh8)