Chess

No. 181. C. MANSFIELD (First Prize, 'Good Companions,' 1917)

BLACK (5 men) WHITE (I I men)

wurrE to play and mate in two moves; solution next week. Solution to No. 180 (Loschinski): B—K 4,

threat R—Q 3. 1 Kt—K 4; 2 B—QB 5 (set Kt— K 2). I . . . Q—K 4; 2 R—R 4 (set Q x Kt). 1 . . . R—Q 4; 2 R X R. 1 . . . B x B (K 5); 2 Kt—K 2.

1 R x B (K 5); 2 Q x Kt. If we add to these varia-

tions the two 'set' (pre-key) lines I B x B'(Q 4);

2 B—B 5 and 1. R x B (Q 4); 2 R—R 4 we see the whole complex themes with the four set mates reappearing against different defences after the key.

This week's game shows modern masterplay at its best. Both players go all out to win, both take great risks and until the final combination it is far from clear who is goingto get home. A game worthy of the great event in which it was played, the seven- player double-round event to determine three Soviet players from the Interzonal tournament now taking place in Amsterdam. The winner of this game won the event with seven out of twelve, the loser was last with five; if this game had gone the other way they would have tied in the middle of the table!

White, GELLER; Black, SPASSKY; Opening, 1st kJ)/

LOPEZ.

P—K 4 P—K 4 2 Kt—KB 3 Kt—QB 3

3 8—Kt 5 P-,--OR 3

4 B—R 4 P—Q 3 5 0-0 B—Kt 5 6 P—KR 3 11—R 4 7 P-13 3 Kt—B3 8 P—Q 4 P—QKt 4 9 3 B—K 2 10 B—K 3 0-0 It QKt—Q 2 P—Q 4!? Objectively, this is probably wrong: practically, even against a great master like Geller. its surprise value makes it a good bet. 12 P—Kt 4 B—Kt 3 13 QPxP . . . White now embarks on a highly ori-

ginal plan which just fails to come off. 13 Kt x P, Kt x Kt; i4 PXKt, KtxP; 15 KtxKt, PxKt: 16 B—Q 5!, R—Kt t, with some positional advantage was probably better—but this quieter method is not in Geller's, style.

13 . . . KKt x P 14 Kt—Kt I . . . the point! How can Spassky save the QP?

14 . Q—B I!! An extraordinary resource. 1( now 15 111xP, R—Q 1 White is unable to escape from the PO on the Q file (16 Q—Kt 3?, Kt—R 4) and if 15 Q x P Melt

15 . . . Kt—R 4: 16 Q—Q 1, KtxB; 17 PxKt (17 Qxfflt Kt—B 4), R—Q I; 18 Q—K 2, Kt—B 4: 19 B x Kt, B x B and with threats of B—Q 6 and P—KR 4 White has more than enough for the pawn. 15 Kt—Q 4! KtxKP 16 P—KB 4 . . . now again Black seems to be in MO trouble. '

16 . . . P—QB 4! ... and again he has a counter..

17 PxKt . . . 17 Kt—B 2. P—B 5 and 17 Kt Kt x Kt eh; IS Q x Kt, P—B 5 and 17 Kt—K 2?, Kt x KIP! are all very satisfactory for Black.

17

18 I'Xi) Q—Qt 2

19 Kt—Q 2 P—B 3!

20 R—B l K—R I 21 QB—B 4? . . . Too ambitious: he should first simplifl by KtxKt and then play B—B 4 when Black's advantage (due to weakness of White's king side) would be less marked.

21 . P X P 22 BxKP B—Kt 4! White probably expected 22 . • ' QR—B 1: 23 Kt—s 3.

23 R—B 7 QxR!

24 B X0 B—K 6 ch 25 K—Kt 2 Kt XKt 26 RxR ch RxR 27 B xP R—B 7 ch 28 K—Kt 3 .. otherwise a piece is lost.

28 Kt—B 8 ch

29 K—R 4 P—R 3 threat 30 ... B—Kt 4 male.

30 B—Q S R—B I threat 31 B—B 7 mate.

31 Resigns . . . 31 P—Kt 5. R—B 5 ch: 32 Q—Kt B—B 7 mate.