Weekly Problems 2025-A

Problems 737-742

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737


Abdul Jabbar Karwatkar
The Problemist 1969, 2nd Prize

Solution

White has two apparent tactics or “tries”: 1.Sxf3+? forces 1…Rxf3+ which isn’t mate because of the b3-rook, and 1.Bxe5+? forces 1…Rxe5+ which isn’t mate because of the e1-rook. The key 1.Ba6! unguards d7 and threatens 2.Qxd7+ Bxd7. Since this mate depends on Black’s d2-bishop (or c1-queen) controlling the g5-flight, Black responds by cutting off the piece with a rook on e3. These defences cause a mutual interference between the rooks – this is called the Wurzburg-Plachutta theme – enabling the try-moves to work. 1…Ree3 2.Sxf3+ Rxf3 and 1…R3e3 2.Bxe5+ Rxe5 (Black’s mating moves open an indirect B + R battery that re-guards g5 and f4 respectively). In directmate problems, a Wurzburg-Plachutta requires at least three moves to be implemented (see WP691), but this example demonstrates how in a selfmate the theme can be realised in two moves.

Andy Sag: Obvious key unguards the pinned knight. The variations were a bit harder to spot.
George Meldrum: Very neat.

738


Rudolf Queck
Schach 1951

Solution

The black king is almost totally confined for a diagonal battery mate, 1…Ra2, but the d2-flight cannot be guarded or blocked in time. Alternative mating squares for the king – the most plausible being a4 for a rook mate on a5 (supported by the bishop) – also fail to work within four moves. The surprising solution involves setting up another diagonal battery aimed at the king on c3, but from the opposite direction. 1.Kxd4 Rb5+ 2.Ke3 Bxf6 3.Kd2 Re5 4.Kc3 Re2. The black king carries out a 4-move round-trip (rundlauf) in the shape of a diamond, while White dismantles the B + R battery and re-forms it on another part of the board. The final position is also a model mate with only the white king not taking part.

Andy Sag: The black king does a circular switch-back!
Satanick Mukhuty: Nice round-trip of the black monarch!
George Meldrum: Very tricky problem.

739


Vojko Bartolovic
Problem 1970, 1st Prize

Mate in 2

Solution

The white knight on d5 has eight possible moves, seven of which are thematic tries. These tries generate various threats that are countered by the black knight on d3. 1.Sc3? (2.Se2) bxc3 2.bxc3, but 1…Sxc1! refutes, and similarly 1.Se3? (2.Sc2) fxe3 2.fxe3, but 1…Se1! refutes. A second pair of symmetrical tries are 1.Sc7? (2.Se6) Sc5! and 1.Se7? (2.Sc6) Se5! A third pair contains multiple threats, and these are handled somewhat drastically by capturing the try-piece: 1.Sxb4? (2.Sc2/Sc6/Qxd3) Sxb4! and 1.Sxf4? (2.Se2/Se6/Qxd3/Qe3) Sxf4! The last try is 1.S5xb6? (2.Qd5) Sxf2!, where it’s a slight pity that White brutally captures a black officer. The white knight’s eighth move is the key, 1.S5xf6!, which in contrast carries no threat. 1…Sd~ 2.Be5 and 1…Sb~ 2.Qd5. White executes a knight tour consisting of the key and seven tries, while Black employs one knight to defeat these tries with seven different moves. A fine task problem demonstrating a knight vs knight duel.

Andy Sag: Post key, only the knights can move. The try 1.S5xb6? is brutal but I thought the refutation was subtle. First prize deserved.
Jacob Hoover: Each try is refuted by a different defense. Since all eight possible moves of the d5-knight are seen here, we have a white knight-wheel.

740


Fadil Abdurahmanovic & Boris Shorokhov
Shakhmatnaya Kompozitsiya 2017, 3rd Prize

Helpmate in 2, 3 solutions

Solution

In two of the solutions, the black king remains on e6, meaning the flights on e5, e7, and f7 need to be dealt with. For the first part, a knight promotion mate on d8 occurs after the c8-rook unguards the mating square and simultaneously blocks e7. 1.Re8 Rd5 2.Rxe7 d8=S. White aims for a rook mate on the e-file for the second part, forcing the black queen to find a unique hideaway square. 1.Qa4 e8=B 2.Qxd7 Re1. In the third solution, White arranges a queen promotion mate on e8, which is viable only if the black king goes to d7 (because of the impeding black pieces on g8 and h8). 1.Qxd1 dxc8=R 2.Kd7 e8=Q. Four different types of promotion, or the Allumwandlung theme, are spread over three phases of play. Furthermore, White’s three units rotate their functions to (1) be sacrificed, (2) control flights, and (3) give mate. Hence we see a full 3x3 cycle of function changes, which incorporates a cyclic Zilahi.

Andy Sag: The three solutions are check free and feature all four possible promotions. All three white units take turns to get captured and participate in delivering mate. Difficult to solve.
Jacob Hoover: We have a cyclic change of roles for the three non-royal white units. In addition, since all four possible pawn promotions take place, we have an AUW.
Satanick Mukhuty: Interesting AUW. Excluding the king, White has three units; each of them gets captured in exactly one of the three solutions – perfection!

741


William Whyatt
Weekly Times 1953

Mate in 3

Solution

Black has two legal moves only and set play is prepared for them: 1…Kc5 2.Rc1+ Kd4 3.Rc4 and 1…d5 2.Rh8 dxe4 3.Rd8. Yet White has no pure waiting move capable of preserving both variations, e.g. 1.Rh3? Kc5!, 1.Re1? d5!, 1.Bf8? d5! The fabulous key 1.Rh6! (waiting) changes the reply to 1…Kc5, and it’s aimed at forestalling stalemate – 2.Ke3 gxh6 3.d4. The other set line is retained, 1…d5 2.Rh8 dxe4 3.Rd8. A third variation is added when Black accepts the offered rook, 1…gxh6 2.Bf6+ Kc5 3.d4. The three-move mutate was one of Whyatt’s specialties, and here he demonstrates the form with a first-rate key.

Andy Sag: The 1…Kc5 variation is all about stalemate avoidance. A stroke of genius!
George Meldrum: You’ve got to love White’s first move.
Satanick Mukhuty: Attractive mutate!

742


Francois Labelle
Retro Championnat de France, RIFACE 2017

Proof game in 4½
Twin (b) Proof game in 5½ exactly

Solution

Black’s d-pawn clearly has captured one of White’s missing knights on the c-file. If the b-knight is sacrificed on c4, that leaves three moves for the g-knight to reach d5. All this makes the first solution straightforward: 1.Sa3 d5 2.Sc4 dxc4 3.Sh3 Qd5 4.Sf4 Kd8 5.Sxd5. Part (b), requiring the same position to be reached after White’s 6th move, is trickier. Both sides must “waste” a move somehow, but White cannot lose a tempo with either knight. Suppose the b-knight tries to be captured on c5 instead, to use up three moves; that would be compatible with the d-pawn spending an extra tempo to reach c4 via c5. However, after 1.Sc3? d6 2.Sa4/Se4, Black is stuck since 2…Qd5 is unplayable. The solution utilises the d-pawn route just mentioned, but here the plan is to sacrifice the g-knight instead on the c-file. 1.Sf3 d6 2.Se5 Qd7 3.Sxd7 Kd8 4.Sc5 dxc5 5.Sc3 c4 6.Sd5. The identity of the knight on d5 is thus changed, and the black queen gets captured on a different square. Other than …Kd8, no moves are repeated across the two solutions.

Andy Sag: In (b), to get six white moves, one knight must be captured on a black square.
Jacob Hoover: In part (a) both white knights move toward the edge of the board initially and in part (b) they both move toward the center.
Satanick Mukhuty: The d7-pawn captures the b1-knight in one line and the g1-knight in the other. Enjoyed the unusual twinning.