## Friday, November 27, 2009

### Multiple-twin problem

The problem we present has a peculiarity. It is a multiple problem, but not exactly a twin, because the produced problems have different number of moves in their solutions.
The American Joseph Wainwright (1851 – 1921) is the composer, known for his tasks with two-mover problems.

Each problem is producing the next one just after the key-move is played, while the number of moves for the solution is increased by 1. To be exact...
...the initial position is Mate in 2 moves,
after the key is Mate in 3 moves,
after the key is Mate in 4 moves,
after the key is Mate in 5 moves.

 (Problem 388) J. C. J. Wainwright, American Chess Bulletin, 1910, Mate in 2 moves. (a) #2 (10 + 9), (b) after the key of (a) #3, (c) after the key of (b) #4, (d) after the key of (c) #5 [8/2p1p1p1/p1PkP1P1/B1p2K2/2P5/pPP4p/P6p/7B]

In the initial position Black is stalemated. The solutions are simple (with possible exception the five-mover) :

(a) 1.b4! (zugzwang) cxb4 2.Bxb4#

(b) 1.b5! (zz) axb5 2.cxb5 (zz) c4 3.Bb4#

(c) 1.b6! (zz) cxb6 2.Bxb6 a5 3.c7 a4 4.c8=S#

(d) 1.Kg5! cxb6 2.Bxb6 a5 3.c7
3...a4 4.c8=Q/B Ke5 5.Bc7#
3...Kxe6 4.c8=Q+ Kd6/Ke5 5.Bc7#
3...Ke5 4.c8=Q a4/Kd6 5.Bc7#

27-11-2009 : The friend reader Alotan has posted a comment :
Nice problem. The mate in 5 had many variations and I had to set it on the chessboard. The reason for comment, however, is that it reminded me a nice helpmate problem by Caillaud, with similar twinning mechanism :

 (Problem 389) Michel Caillaud, First prize, Pitlochry TT 2003 (a) h#2 (5+2), (b) Position of (a) before the mating move and h#2, (c) Position of (b) before the mating move and h#2. [8/4p3/3S4/8/SRBk3K/8/8/8]

It is not exceptional or difficult, but it belongs to those problems that remain carved in the memory of the solver.

## Thursday, November 19, 2009

### Composers cooperating, (No.1)

Today's post is the first of a new series. We will present problems created by cooperating composers. Initially all composers will be Greek, later on only one composer will be Greek.

We believe that when two composers try to cooperate, the final result comes more quickly, because both composers work more intensely. The cooperation is a factor of motivation.

In problem-387 the composers are George Georgopoulos and Efthimios Papakonstantinou.

 (Problem 387)George Georgopoulos & Efthimios Papakonstantinou, First Prize, Die Schwalbe, 1990, Mate in 4 moves.#4 (9 + 12) [q1s5/pb5s/1SP1Sp2/r5b1/r5p1/p1QB2Pk/5P2/5RK1]

If you try to move first the Rook to make room for the Bishop, ( 5 tries : [1.R~? Ba6!] ), the problem will not be solved. With your key-move you must sacrifice the white Queen! Anyway, the sacrifice threatens Mate in 3, and Black can not be indifferent.

Key : 1.Qxf6! ( > 2.Sxg5 Rxg5 3.Qh6+ Rh5 4.Qxh5# )

1...Bxf6 2.Re1 ( > 3.Bf1# )
___2...Ba6 3.Sc4 ( Novotny sacrifice on c4, > 4.Sf4# / Bf1# )
___2...Bxc6 3.Re4 ( Novotny sacrifice on e4, > 4.Sf4# / Bf1# )
1...Sxf6 2.Rb1 ( > 3.Bf1# )
___2...Ba6 3.Rb5 ( Novotny sacrifice on b5, > 4.Sxg5# / Bf1# )
___2...Bxc6 3.Sd5 ( Novotny sacrifice on d5, > 4.Sxg5# / Bf1# )

For the moves 2...Ba6/Bxc6 the White continues with Knight/Rook in one variation, and with Rook/Knight in the other.

(The problem is included in the edition "Selected Chess Compositions by Greek Composers", prepared for the 47_th World Congress of Chess Composition, Halkidiki, Greece, September 4-11, 2004. Editor : Harry Fougiaxis).

## Thursday, November 05, 2009

### How does a machine think?

http://turbulence.org/project/thinking-machine-4/

To be honest, I do not want to know.
I like the way I think (steepest descend method – and all the rest).

But there is a chess playing machine which lets you see, in an artistically interesting manner, the moves as they are calculated.