Issue Reviews 2010 The Problemist, March, 2010
 

 

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The Problemist, March, 2010
Written by Michael McDowell   

In the March issue Steve Giddins reported on the final of the 2009-10 British Chess Solving Championship, which resulted in a narrow victory for Jonathan Mestel, a point ahead of David Friedgood with World Champion Piotr Murdzia third. Two fairy awards were included, the Cedric Lytton 70JT and the 2005 informal award, by Hubert Gockel. Browsing in the library featured the 1924 Christmas book Simple Two-Move Themes. Chris Reeves presented an appreciation of the late Theodor Steudel, while in the Supplement David Shire examined the two-movers of Andrey Lobusov, John Rice showed a selection of serieshelpmates and Geoff Foster discussed how to develop a helpselfmate idea.

William A. Whyatt

The Problemist, 1959

1R6/6p1/3S2B1/B1p2s1R/P1P1s2P/p1PkP3/p2P4/rbKQ2S1

Selfmate in 2

1.Rd8 (>2.Qc2+ Bxc2)
1...Sfxd6 2.Rxc5 Bc2
1...Sd4 2.Rd5 Bc2
1...Sxe3 2.Qf3 Bc2
1...Sexd6 2.Qg4 Bc2
1...Sxd2 2.Re8 Bc2

A beautiful selfmate (by a composer who is much better known for his directmate three-movers) which entertained the solvers at the British Chess Solving Final.

Alberto Mari

Alfiere di Re, 1922

sB1R4/r1q4p/4pp1Q/1RS1k3/6p1/7S/5Ps1/1BK5

Mate in 2

Set 1...Qxb8 2.Sb7
1...Qd6 2.Sb3
1.Sg5 (>2.Sf7)
1...Qxb8 2.Sd3
1...Qd6 2.Scxe6
1...Kf4 2.Qxf6

Changed mates following withdrawal unpins of the c5 knight.

Michel Caillaud

2nd HM., Lytton-70 JT, 2009-2010

5Q2/3R1sp1/2SS4/4P1P1/5k1K/B2P2Rp/7r/8

Mate in 2: Circe

1.Qg8 (>2.Rxf7)
1...S random 2.Qc4
1...Sxd6 [Sg1] 2.Rf3
1...Sxg5 [Pg2] 2.Bc1
1...Sxe5 [Pe2] 2.e3

The thematic key occupies g8 to threaten 2.Rxf7 by preventing the knight’s rebirth. A random move of the unpinned S opens the line g8-c4. 1...Sxd6 corrects by guarding the fourth rank, but places a white guard on f3 for 2.Rf3. 1...Sxg5 again guards the fourth rank and places a white guard on f3 but corrects by directly guarding f3. However, it cuts the rook guard of d2, allowing 2.Bc1. 1...Sxe5 repeats all of the preceding elements but corrects against 2.Bc1? because of 2…Sxd3[Pd2]! This time the decisive error is to return the wP to e2 for 2.e3. A wonderfully clear example of quaternary correction.

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