CHESS. PROBLEM 1579. PROBLEM 1580. Solutions to Problems. Otago Witness Sixth Solution Tourney. Selected Games.

Otago Witness, Issue 2356, 20 April 1899, Page 48

CHESS.

PROBLEM 1579.

PROBLEM 1580.

Solutions to Problems.

Otago Witness Sixth Solution Tourney.

Selected Games.

.Conducted by Johk Mouat.

Tourney Problem No. 5 (Selected). By J. Buchan, Jun., Sale, Victoria (Pint Special Prize Sydney Morning Herald Fourth Problem Tourney).

4Kt r 2,1 K3^R 2, 6pl,pp 1 P k 1 Xt 1, rq 3 p PI, p383, 2P1Q3.b68. Whito to play and mate in two morts

Tourney Problem No. 6 (Competing). By F. A. L Kuskop, Wellington.

K3Kt 3, 87, 4klKtl. 6p 1, 2 P5,1 b6, 4 B 3,1 Q6. "White to play and mate jn three moves.

Problem* 1575—Tourney Problem No. 1, Selected— (Rowland):— Key move 1 P-K 4. Problem 1576—Tourney Problem No. 2, Competing —(Mackenzie) :— The composer's solution is 1 P-K 6, p-kt 5 ; 2QB 5 eh, etc. Unfortunately this seems to be defeated by 1 b-kt, and the problem has therefore no solution.

In our issue of April 6 the publication of the problems contributed to the sixth tourney began. A solution tourney is held in connection therewith, and will be carried on up to the 30th of June next. In this tho scores of solvers other than the prize winners in the fifth tourney will be brought forward on the continuous principle, and a prize of 10« 6d will be given to ihe solver who, on this principle, obtains the highest number of points. A prize of 10s 6d is^also offered by Mr S. S. Slackburne, of Christchurch, to be competed for in the ordinary way — that is to say, not on the continuous plan. The problems competing in the tourney, together with selected problems, will be submitted for solution. The conditions of the competition are. — Points. Each author's solution and every " cook, ' two-movers 1, three-movers 2; satisfactory criticism of each problem that has a solution and is not cooked, 1. A correct statement that a problem has no solution counts as a solution. As a. part of his criticism, a solver should giva marks, from 0 to 10, according to his estimate of the merits of a problem. Solutions and criticisms must be -completed within one week from the time problems • are received by the solver, and should be- posted or delivered at the first opportunity after they aro ready. Criticisms that aro not sufficiently brief andj to the point will be subject to abridgment. Please condense. .It ha? been arranged that the probTeniß shall be published in the Canterbury Times simultaneously, with their publication in the Otago Witness. The readers of the Canterbury Times may send their solutions and criticisms to the chess editor of that paper.

Game between Mr Thrift (Dublin) and Mr Schott: — Ruy Lopez. White. Blaok. White. Black. (Thrift.) (Schott.) (I'hiift.) (Schott.) 1 P-K 4 P-K 4 15 Kt-K 2 P-B 3 2 X Kt-B 3 Q Kt-B 3 16 Kt-ll 4 Bx B 3 B-Kt 5 P-QR3 17 Q x B P x P 4 B-R 4 BKt 5 18 P x P B x B 5 P-B 3 B-R 4 19 P x B X x R eh 6 0-0 X Kt-K 2 29 Xx R R-B Ich 7 P-Q 4 P x P 21 K-Kt 1 Q-Kt 5 8P x P P-Q 4 22 P-K Xt 3 Xt x P fl P-K 5 OO 23 Q x P Kt-Q 6 10 R-B 2 BB 4 24 K-Kt 1 QxQ Xt 11 Kt-B 3 P-Q Xt 4 25 Qx Xt Q-B 7ch 12 BKt 5 Q-Q 2 26 K-R 1 Kt-Ktß 33 R-B 1 P-R 3 Winß. 148K3 BQKt3 A fine game, in which a young American player defeated Janowski while on his recent tour in the States: — Queen's Gambit Declined. White. Black. White. • Black. (Janowski.) (Marshall.) (Janowski.) (Marshall.) 1 P-Q 4 P-Q 4 22 R x P Kt-B 5 2 P-Q B 4 P-K 4 23 K-R 2 MB 1 3Ql'xP -P-Q5 24R-KI Kt(K6)x 4 Kt-K B 3 Q Kt-B 3 P 5 P-K R 3 B-K B 4 25 R-K Xt 1 Kt-Q 6 6 P-R 3 P-Q R 4 26 Rx Xt Xt x R -7 P-K 3 B-B 4 27 RxP eh R-B 2 8 B-Q 3 B x B 28 R x Rch X x R 9Q x B Px P < 29 Xt x Xt 1" R.Q7 eh 10 QxQ eh 11 x Q 30 K-Kt 3 11 x P 11 B x P B x B 31 Kt-Q 6 eh K-K 2 12 Px B X Kt-K 2 32 Kt(Q6)xP P-R5 13 Kt-B 3 Xt-Xt 3 3J K-Kt 4 R-Kt 6 14 Q Kt-K* 5 R-Q 2 34 P-R 4 R x P 15 P-K 6 P x P 35 K-R 5 R-Kt 6 16 X Xt-Xt 5 K-K 2 36 X x P R-Kt 5 17 0-0 Q Kt-K 4 37 P-R 5 PR 6 18Kt-Q4 P-B4 38 Kt-R5 R-Kt 4 19 Kt-Kts P-R3 39 Xt (B 5)20 Kt-K 4 Xt x P Xt 3 K.-B 3 2J.KR-B1 KtzEF ' Eesiwife