Chess no. 429

PHILIDOR

White

Black 10 men C. Mansfield (Hon. Mem., B.C.F. Tourney 116, 1967/68). White to play and mate in two MIMS; solution next week.

Solution to no. 428 (Anderssen and Ekstrom): KI (4) - K 6, threat R Q 4. 1 ... Kt (B3) x R; 2 Q - Kt 1. 1 . . . Kt (Kt 3) x R; 2 B - Kt. 6.

1...R x R; 2 Q - K 1. 1 K x R; 2 B - B 6. To see the full point of the problem, examine why the tries Kt - B 6, Kt - B 2, Kt - K 2 and Kt - B 5 fail.

Tlke following entertaining game was played in the 1968 Skopje tournament, won by the Hun- garian grandmaster Lajos Portisch.

White, Savon. Black, Uhlmann. Opening. French Defence.

1 P-K 4 P-K 3 2 P-Q 3 . . . Uhl- mann has for the last twenty years or more been a leading exponent of the French Defence. having • virtually founded an East German school _ of players. So White plays a move

designed to get away from the standard French Pines.

P–Q 4 3 Kt – Q 2 Kt–KB3 4 P–KKt3 B–K 2 5 K Kt– B 3 P–B4 6 B–Kt 2 Kt–Q133 7 0-0 0-0

8 R–K 1 P–QKt 4 Now we have a game of the same general type as one gets in the slow forms of the Sicilian: White plays for a gradual build-up on the king's side, Black for a queen's side counter.

9 P–K5 Kt–Q2 10 Kt–B1 P–QR 4 11 P–KR 4 P–Kt5 12 B–B4 B–R 3 13 Kt–K3 P–R 5 14 P–B4?

Looks, and is, bad. He must try to continue with his king's side advance e.g. by P – R 5 and – Kt 4, when chances would be about equal.

14 . . . P x Pe.p 15 P x P P–Q 5!

16 P X P P x P 17 Kt–B4 Kt–B4

18 Kt – Kt 5 Kt–Kt5! 19 B x R . . . He may as well accept. 19 B – KB 1 or B – K 4 leaves him a bad position with no compensating material advantage.

19 . . . Q x B 20 Q – R 5 B X Kt 21 B X B Kt (Kt 5) x QP

22 B– B 6 . . . An illusory attack, but he has nothing better.

22 . . . PxB 23 PXP K–R1 24 Kt – Q 6 . . . Threatening 25 Q – Kt 5, R – Kt 1; 26 Kt x P mate.

24 . . . Kt–B 5! A shattering reply; now

25 Q – Kt 5?? is met by 25 Q – Kt 7 mate and 25 P X Kt by 25. . R – Kt I ch mating or winning the queen.

25 Kt x P ch R x Kt 26 P X Kt R x P 27 Q – K 5 Kt – Q 2 28 Q x QP Q – KKt 1 ch!

29 K–R 2 Q–Kt 5! 30 R–K3 . . .30 R – KKI I, Q X RP ch; 31 K – Kt 2,P– K 41; 32 Q X Kt, R – Kt 3 ch; 33 K – B 3, Q X P mate.

30 . . . P – K 4! 31 Resigns . . . 31 R x P. Q X RP ch; 32 K – Kt I, Q – Kt 5 ch:

33 K – R 2, Kt X R: 34 Q X K:,Q x Pch winning easily.