Chess 471

PHILIDOR

G. W. A. Eason (Hon. Men., BCE Tourney, 1968/9). White to play and mate in two moves; solution next week.

Solution to No.470 (Loyd): B-Kt5 !,threatening 2 B-K8!, B-B4; 3 B-Kt6!, B x B; 4 Q-Kt4 mate or 2 . . . . K-B4; 3 Q-B3ch, K-K3; 4 Q-B7 mate.

A 1 . . . . P x B; 2 Q-KB1 !, K x P; 3 Q-B6ch, K-B4; 4 Q x BP mate. B 1 . . . . K-B4; 2 Q-B3ch. K-K3; 3 B-K8!, any; 4 Q-87 mate. Very hard. because of the difficulty in seeing the idea.

Sam Loyd: solutions

The first of last week's Loyd problems were the .Charles XII at Bender' set. The initial position was 816R1i7p/5K1k/8/6p1 /5bpp/4N3—mate in three, by I R x P, B x R; 2 Kt-B3, B any; 3 P-Kt4 mate

or 1 B x Kt; 2 R-R3ch, 11-R5; 3 P-104. Now

remove WKt and mate in four by 1 P x P B-K6; 2 R-Kt4, B-Kt4; 3 R-R4ch!, B x R; 4 P-Kt4: and,

finally, remove WKRP as well and mate in five by

1 R-Kt7!, B-K6; 2 R-Ktl, B-Kt4; 3 R-R Ich,

B-125; 4 R-R2% P x R; 5P-Kt4 or 1 .. . 11--Kt8; 2 R-Ktl, B-R7; 3 R-K I !. K-RS; 4 K-Kt6. any; 5 R-K4 mate. Not too difficult, but an attractive set.

Next the Loyd/Steinitz composing-solving con- test; position: 8/6p117pt4p2p,I4KPkpi6p I 6Q1;8— mate in three. Key P x P. A I . . P- Kt3; 2 Q-BI

and now 2 ... P-R6; 3Q-B4, 2 ... P-Kt4; 3 Q-B5,

2 ... P-Kt7; 3 Q x P. ti I . K-Kt4; 2 Q-B3 and

now 2 . . . P-R6; 3 Q x P. 2 . . P-Kt3; 3 Q-134,

2 ... any other; 3 Q-B5. Cl P-Kt4; 2 Q-111 etc.

Again not difficult—I think Loyd's feat in com- posing it in ten minutes much more striking than Steinitz's in solving it in five.

'Stuck Steinitz' was harder. Steinitz found what he thought was the main line viz; 1 P-114. B

r •ves; 2 B B8. any; 3 B x12. any; 4 B x P mate—

rather a dull problem. But he missed the subtle point I P-B4, B -R8! Now 2 B-B8? P- Kin 3 B x P is stalemate—and White must play 2 P- Kt3 (threat Kt-B5). P-Kt3 (2 ... B- K5; 3 Kt x P mate);

3 B-K7, any; 4 B x P (or 0- 116) mate. A diabolical trap into which Steinitz fell headlong.

Lastly, the extraordinary live-mover nIrb4/ 1p3p1prIp6i1R5K /8/p3pIPN/ I PP I R3iN6k where Julien bet Loyd that whatever else gave mate in the main variation it was not White's QKtP: But I P-Q1(t4! (threat 2 R-Q5 and 3 R-Q1),

R H4ch (to cut off the rook); 2 P x R ! (threat R-Ktl mate), P-R7; 3P- B6! (threat 4 R-05 or R-KB5), B-B2 (to meet R-05 with B x P and R-K135 with B-05); 4 P x P!, any; 5 P x

mate. A justly famous and astonishing tour de.

force; set the board up and play through it whether you tried to solve it or not.

Hearty congratulations to any solver who succeeded in making a clean sweep of all the problems --and I hope that those who failed have at least enjoyed the solutions.