Weekly Problems 2024-A

Problems 685-691

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685


Zvonimir Hernitz & Mato Prikril
British Chess Federation Tourney 1962, 2nd Prize

Solution

In the set play with White to move, …0-0 mates immediately. This pin-mate employs most of the white pieces, including the two knights on guarding duties, though the a1-rook is redundant. In the actual solution, Black cannot avoid capturing the g2-knight on the first move, thereby spoiling the set mate. Black thus organises another mate, one that utilises the a1-rook along the first rank, and this necessitates the surprising removal of two more white pieces. 1.Kxg2 2.Kxh1 3.Qg2 4.Sc3 5.Se2 6.Sxc1 7.Se2 8.Sd4 for 0-0-0. The change from short to long castling is well accomplished, along with a new pin-mate where a different black piece is immobilised by the white bishop.

Andy Sag: The set king-side castling mate cannot be preserved so Black uses the eight moves to unpin the knight, pin the queen and enable queen-side castling mate instead. Very cunning! Note the d3-pawn stops 1.Kxg2 2.Kxh1 3.Kg1 4.Sf2 Se2.
Nigel Nettheim: Brilliant! 3.Qg2 was, for me, the hard move to find, even though the mate was bound to be 0-0-0. Not 1.d2+? for 0-0 because castling out of check is illegal, a seldom-arising rule.
George Meldrum: An intriguing setting where White can provide mate in one by king-side castling.  Maintaining a castling theme a clear task, the scene does not disappoint.

686


Vasil Dyachuk
Schach 1995, Hon. Mention

Mate in 2

Solution

Two checks by the white knight aren’t mates solely because they would cut off the g7-bishop or the c8-rook. White can threaten mate by shifting one of these line-pieces across the critical square, to avoid the interference. Most such anti-critical moves fail, however, because they cause their own obstructions with another white piece. 1.Bc3? (threat: 2.Sf6) is refuted by 1…Bh4! since 2.Qa5?? is prevented, and 1.Bb2? is countered by 1…Qg5! since 2.Qa2?? is disabled. Similarly, 1.Rc3? (2.Sc7) Bg3! and 1.Rc2? Qg3!, while 1.Rcc6? hinders the a6-rook instead of the queen, enabling 1…Se5! (creating a flight on d4) which cannot be exploited by 2.Rad6?? The key 1.Rc1! (2.Sc7) avoids the self-interferences and leaves the set variations intact. 1…Bg3 2.Qa5, 1…Qg3 2.Qa2, and 1…Se5 2.Rd6. The by-play 1…Bf3 2.Qxd3 gives the white queen an extra purpose.

Andy Sag: The tries show no square works for the bishop so it is a matter of finding a suitable square for the c-rook.
Jacob Hoover: Five tries exist, each of which involves anti-critical play but interferes with another white line-piece.
Nigel Nettheim: A number of good tries.
Andrew Buchanan: Quick to solve, but very clean with lots of nice tries.
Paz Einat: The scheme looks like a well-known Dombrovskis, but what we see is white self-interference tries with unified refutations. A good work of art.

687


Pal Benko & Andrew Kalotay
The Problemist 1989, 4th Prize

Helpmate in 3½
Twin (b) Bb8 to a7, (c) Kb3 to c2

Solution

An edge-mate is necessary given this minimal material, so the players coordinate to let the black king reach the first rank in three moves. Part (a) is solved by 1…Re2+ 2.Kd3 Re3+ 3.Kd2 Bf4 4.Kc1 Re1, where White arranges a B + R battery mate. Part (b) steers for a similar ideal-mate, but the roles of the two line-pieces are reversed, to create a R + B battery: 1…Rh2 2.Kd3 Rh1 3.Kd2 Bg1 4.Kc1 Be3. In part (c), the black king heads for the corner while White sets up a R + K battery: 1…Rd1 2.Kf3 Kd2 3.Kg2 Ke1 4.Kh1 Kf2. Hence three different types of battery mates are exemplified in this top-notch Wenigsteiner (a problem with no more than four units).

Andy Sag: In each case a different piece triggers the battery mate. Nice triplet.
Nigel Nettheim: Three different batteries are shown. The Hungarian statistician A. Kalotay (1941-) discusses this helpmate in an article named “Structural Aspects of Chess Problems”.
Jacob Hoover: Three excellent battery-forming maneuvers achieved with only four pieces! Amazing!

688


Alphonse Grunenwald
Süddeutscher Rundfunk 1959, 1st-2nd Prize =

Mate in 3

Solution

If White fires the B + S battery in one of three ways to force the king to d1, the same knight has a potential follow-up mate (currently guarded by a black unit), e.g. 1.Sa4+ Kd1 2.Sc3. Such a device, in which the knight gives a discovered check and then mates from a new direction (avoiding the interference on b2) is called a Siers battery. In the set play, this tactic occurs three times when each black defender – forced by zugzwang – unguards a mating square. 1…Se~ 2.Sa4+ Kd1 3.Sc3, 1…Sf~ 2.Sc4+ Kd1 3.Se3, and 1…g2 2.Sd3+ Kd1 3.Sf2. White aims to preserve these variations by making a pure waiting move with one of the pawns, but three of these moves would block a square needed by the white knight: 1.c3? Se~!, 1.c4? Sf~!, and 1.d3? g2! (although 1.c3? and 1.c4? are refuted by multiple moves, they are tries in spirit!). White therefore must start with 1.d4!, which avoids all the hindrances and keeps the set play unchanged. A very neat combination of Siers battery play and square-obstructing tries, all shown in a Meredith setting.

Andy Sag: There are set mating lines for all black moves with the b2-knight mating on three different squares. In this case White plays a move preserving these lines. An additional try is 1.Ka2? Sc3+!
Nigel Nettheim: We look for a first move that does not harm the set mates. Easy to solve, but nice.
Bob Meadley: Nice little three-mover.

689


Johann van Gool
Tungsram Cup 1978, 5th Prize

Helpmate in 2, 3 solutions

Solution

Each of the white knights takes two moves to arrange a check, but most of the eight potential mating squares are well guarded by Black, while a check on f5 would cut off the white rook. Only e6 seems unprotected, except that when either knight moves, it opens a line of defence for the c4-queen or the e3-rook. To offset this effect, Black employs the e6-knight to close the line again, by placing it on the square just vacated by the white piece. 1.Sc7 Sf4 2.Sd5 Se6 and 1.Sc5 Sg5 2.Se4 Se6. Ironically, the mating square becomes doubly protected by Black after White’s first move. The third solution involves a white promotion, and yet it matches the other two parts precisely, with the black knight closing a line of guard to e6 that White has initially opened. 1.Sd8 f8=S 2.Sf7 Se6. A bi-coloured platzwechsel – exchange of places between a white and a black unit – is presented three times.

Andy Sag: The three solutions are thematically coordinated. Each uses a different knight (two original, one promoted) to mate on e6. Also, in each case, the black knight moves twice to interfere with Black’s guard on e6.
Jacob Hoover: In each solution the black knight switches places with a white unit and in doing so interferes with a black line piece. Also, the black knight move is chosen carefully in order to avoid giving check.
Nigel Nettheim: Marvellous. Black uses a jumping piece (S) to interfere with the action of the line pieces (Q, R, B). The d6-pawn prevents White’s 1…f8=Q followed by 2.Qb4 Qxb4.

690


Satoshi Hashimoto
Problemesis 2003

Solution

White’s seventeen moves are largely visible in the diagram, including a 7-move trek by the rook via the h8-corner. This trip involves the sweeping move Rh8-d8, which means that Black will need to clear that line of all pieces at some point, before returning them to their home squares. Black has made three pawn moves, while the h8-rook needs four (Rh6-f6-h6-h8) to open a path for the white rook. The remaining four pieces from d8 to g8 require a minimum of eight moves for each to make a switchback, leaving two spare moves out of the seventeen total. 1.a4 h5 2.Ra3 Rh6 3.Rg3 Rf6 4.Rg6 e6 5.Rh6. Suppose Black develops the bishop to the queen-side with 5…Bc5, then 6.Rh8 Rh6 7.b3 Qg5 8.Bb2 Ke7 9.Bd4 Sf6 10.Rd8 results in a gridlock, because Black is forced to play …d6 soon to allow Rd7+ Ke8, but that leaves the c5-bishop stranded. So Black develops the bishop to the king-side instead, making use of the two spare moves. 5…Be7 6.Rh8 Rh6 7.b3 Bh4 8.Bb2 Qg5 9.Bd4 Ke7 10.Sc3 Sf6 11.Rd8 d6 12.Rd7+ Ke8 13.Sd5 Sg8 14.c3 Qd8 15.Qc2 Be7 16.h4 Bf8 17.Se7 Rh8. Five switchback manoeuvres arise from an attractive problem position (homebase set-up for the black pieces), with the unusual feature that no captures occur at all.

Andy Sag: The white rook takes seven moves to get from a1 to d7 leaving no spare moves. However Black has two spare moves; the trick is to find out how to use them.
Jacob Hoover: The sequence features a bi-colored Turton between the rooks and a Bristol maneuver with the black bishop and queen.
Nigel Nettheim: Each side’s moves were fairly clear, but the intertwining of them had many nuances. It seems amazing to me that the whole scheme works soundly. Back-rank switchbacks are shown here on the K-side, raising the question whether something similar could be shown on the Q-side.

691


Kenneth S. Howard
To Alain White 1945

Mate in 3

Solution

Black’s queen and a5-bishop are preventing a pair of knight mates on c5 and c3, while the black rooks are stopping another pair on f2 and g3. White induces Black to weaken these guards after the key 1.Rd5!, which threatens 2.Re5+ fxe5 3.Qxe5. This queen mate relies on the pins of both black knights, so Black can defend by unpinning either piece. After 1…Qb4, the queen keeps protecting c5 but by taking over the bishop’s duty in observing c3, it becomes overloaded: 2.Sc3+ Qxc3 3.Sxc5. Similarly, 1…Bb4 maintains control of c3 but interferes with the queen’s guard of c5: 2.Sxc5+ Bxc5 3.Sc3. Such a mutual interference between two defenders that act on the same kind of line (diagonal or orthogonal), and becoming overloaded in turn, is termed a Wurzburg-Plachutta. The two black rooks bring about the same idea on g2 when they try to free the other knight. 1…Rhg2 2.Sg3+ Rxg3 3.Sxf2, and 1…Rgg2 2.Sxf2+ Rxf2 3.Sg3. A doubling of the Wurzburg-Plachutta theme is complemented by matching unpinning play on diagonal/orthogonal lines and two pairs of knight mates, all of which combine to create a harmonious, helpmate-like impression.

Andy Sag: The pinned black knights give a strong clue. Unpinning defences lead to four different knight mates.
George Meldrum: Nice problem with lots of tries.
Nigel Nettheim: The theme is the triple-negative “unsuccessful unpinning defence” x 4.
Jacob Hoover: Four unpinning defenses exist in two Wurzburg-Plachutta pairs. This was an easy solve, but a rewarding one nonetheless.