Weekly Problems 2023-A
Problems 633-643
633
Leonid Makaronez
OzProblems.com 7 Jan. 2023
Mate in 3
Solution
The black king has a provided flight, 1…Kg4 2.Qe6+ Kf3 followed by multiple mates, e.g. 3.Rxh3. The key 1.Be2! removes that flight but grants a new one on e4, creating the threat 2.Qxh3+ Kxe4 3.Qd3. If 1…Bg4, then 2.Bxg4+ gives the black king a choice of two diagonal captures, both answered by queen mates, 2…Kxg4 3.Qh3 and 2…Kxe4 3.Qc4. Similarly, if 1…Qb5 – unguarding the g-file – then 2.Rh5+ leads to two king captures and respective queen mates, 2…Kxg6 3.Qg8 and 2…Kxe4 3.Qe3. Note the three changed mates against 2…Kxe4 including the threat line. The distant self-block 1…Qd7 permits 2.Qf3+ Ke6 3.Sg5. Since the black king makes all of its four possible diagonal moves, with different white responses, the star-flights pattern is produced. One more full-length variation involves an orthogonal flight: 1…Qxe4 2.Qf7+ Kg5 3.Rh5. A short mate results after other queen defences, e.g. 1…Qa6 2.Qd5, and after the immediate king capture, 1…Kxe4 2.Qd3. The two queens and black king are especially active in this three-mover that brings about a wealth of mating nets in precise fashion.
Andy Sag: A sacrificial give-and-take key with threat and six variations involving six different flights and a good variety of play, notwithstanding two short mates.
Rauf Aliovsadzade: A give-and-take key. Nice sub-variations after the defenses 1…Qb5 and 1…Bg4.
George Meldrum: Lots of twists and turns in this one as the black king gets mated on six different squares. A quality problem which is difficult to solve with a great key-move.
Solutions
All black units need to be immobilised, and while the two columns of pawns are well-suited to trap a group of pieces on the d-file, the two knights pose more difficulties. The answer is to arrange pins of both knights on c6 and e6 against the king placed on d5, with White playing Rd7 as the stalemating move, to pin a third piece – the dark-squared bishop – on d6. This plan implies shifting all of the pawns down one square, which is compatible with the idea of confining the light-squared bishop on d1. 1.Scb4 2.c2 3.c3 4.c4 5.Sc5 6.Rd3 7.R1d2 8.Bd1 9.e2 10.e3 11.e4 12.e5 13.Se6 14.Bc5 15.Qd4 16.Kd5 17.Bd6 18.c5 19.Sc6 for Rd7. In the final position, the black force reoccupies the exact same squares as those used in the diagram, but with all sixteen pieces reallocated – a maximum task. There are three separate “chains” of units that cyclically exchange their places: (1) the five on the c-file, (2) the three on d4, d5 and d6, and (3) the remaining eight.
Andy Sag: It is clear that the black-squared bishop and both black knights end up pinned. Check avoidance ensures a unique sequence. The initial position is legal if White’s a-, d- and h-pawns were promoted before being captured by black pawns.
Jacob Hoover: This stalemate works because of the triple pin. I like this kind of problem because it’s like solving a sliding-panel puzzle, which can be fun sometimes.
Andrew Buchanan: Cyclic platzwechsels with eight, five, three black pieces. Cycling n pieces will take a minimum of n+1 moves (because the first move must be a piece moving out of the cycle to create a space, then the others move along the cycle and finally this first piece will slot back in to its new space). This minimum is elegantly achieved thrice with unique ordering of moves. Splendid!
Rauf Aliovsadzade: Wow! The position resembles the Empire State Building in New York city. Even some clouds hovering over! Not easy to solve.
635
Vladimir Erokhin
Thèmes-64 1986
Mate in 2
Solution
By placing an extra guard on e4 or c2, White will free one of the rooks to mate on d1. Two tries and the key create such a threat, and they all unpin the d5-bishop, which then defends accordingly. The first try 1.Qh4? (2.Red1) releases the bishop directly by withdrawing the queen – 1…Bxb3 2.Qe4 and 1…Bf3 2.Qc4, but Black refutes with 1…Rg4! The second try 1.Sd4? (2.Rcd1) unpins the bishop by interference and leads to new mates against the same defences – 1…Bxb3 2.Sxb3 and 1…Bf3 2.Sxf3; also 1…cxd4 2.Sxb4 (and 1…Kxd4 2.Rcd1), but 1…axb3! is spoiling. The key 1.Sd6! (2.Red1) also interposes on the pin-line and generates yet another pair of mates – 1…Bxb3 2.Sc4 and 1…Bf3 2.Se4, with by-play 1…Kd4 2.Rcd1. The black bishop’s defences provoke three pairs of changed mates – including different battery openings by the key-piece – in this crisp example of the Zagoruiko theme.
Andy Sag: The key gives a flight and unpins the black bishop enabling a threat and three defensive variations. Many tries, the most interesting with variation content being 1.Sd4 and 1.Qh4.
Jacob Hoover: An unpinning key followed by battery plays. Very nice.
Solution
It’s not obvious in such an open position where to mate the black king, but the placement of the white bishop pair aiming at the two knights provides a hint. If the king is transferred to either of the knights’ squares, that will nearly set up a mating net with the white force already in place. However, Black has insufficient time to play a square-vacating move with a knight, so in each solution, White proceeds to capture one to clear the square for the king. 1…Bxd4 2.Se5 (distant self-block) Bc3+ 3.Ke3 Be1 4.Kd4 Bf2 and 1…Bxc4 2.Sb3 (another distant self-block) Be6 3.Kd3 Bh3 4.Kc4 Bf1. Besides the attractive round-trips made by the bishops, the problem demonstrates the Kniest theme, in which White captures a unit on a square where the black king will be mated.
Andy Sag: A tale of two bishops! Each bishop captures a knight and then circles back to its original position to deliver mate after the black king moves to the square of the captured knight.
Solution
Black needs all fifteen moves to position the units now on the king-side. With only two moves available to the pawns, Black must play …f5 and …gxf6 – not …exf6 as that would cost the bishop an extra move to reach h4. White’s missing f-pawn cannot be directly sacrificed on f6 since it would obstruct …f5, so it has to promote first (after capturing the e-pawn). 1.f4 Sf6 2.f5 Se4 3.f6 Rg8 4.fxe7 f5. The black bishop will soon play to h4, a potential disruptive check that must be blocked by the g1-knight. 5.Sh3 Kf7 6.e8=S Be7 (6…Qg5/Qh4+? would cause a tangle later in which the queen cannot unpin the knight) 7.Sf2 Bh4 8.Sf6 gxf6. Now to avoid a permanently stuck knight on f2, it has to be unpinned by the black rook on g3. But once released the white piece cannot simply return to g1, because the eventual …Rf3 would cause another disruptive check. So the knight heads to g3 in anticipation. 9.Rg1 Rg3 10.Sh1 – White’s pieces from g1 and h1 have swapped places – Rf3+ 11.Sg3 Rf2. This unpin finally frees the knight to return home. 12.Sh5 Qg8 13.Sf4 Qg3 14.Sh3 Kg6 15.Rh1 Kh5 16.Sg1. White’s knight and rook execute not one but two platzwechsels (exchange of places). The excellent manoeuvres don’t involve captures but are subtly motivated by the need to forestall potential checks.
Andy Sag: Retro-analysis indicates that the white f-pawn must promote after capturing the black e-pawn in situ and then be captured on f6, but the clever gymnastics by the white king’s knight to avoid illegal checks on both sides is the key to solving this one.
Jacob Hoover: Themes seen: underpromotion, switchback, rundlauf.
Andrew Buchanan: The Lois theme [double interchange of two pieces] is more commonly applied to king and queen but here two different units enjoy the dance. It’s incredible that such a PG is actually sound. So many subtle interactions contribute to this. Bravo!
638
Otto Nerong
Die Schwalbe 1937
Mate in 6
Solution
Black’s force is mostly incarcerated but it controls the white knight’s four potential mating squares. Two are guarded by the e4-pawn only, so it’s tempting to target this defender; however, 1.Sd6? f5 2.S~ (random tempo move) f4 3.Sd6 f3 4.Sxe4 results in stalemate. Another mating square c2 is guarded by the black bishop, which is itself well protected. White concocts a plan not to capture that piece but to release it, inducing Black to make self-weakening moves through zugzwang. 1.Sa5! f5 2.Sb3 f4 3.Sxa1 f3 4.Sb3 a1=Q 5.Sxa1 Ba2 6.Sxc2. Cute play and curious diagram position in which White has minimal force while Black has almost the maximum (missing only a knight).
George Meldrum: The elimination of the e4-pawn followed by mate on d3 or f3 seemed the order of the day. This distraction caused confusion and delay. Kudos to Otto for this tricky solution.
Thomas Thannheiser: The black pawn moves are just enough to allow Sxa1, after which zugzwang leads to mate on c2.
Andrew Buchanan: Very comical.
Bob Meadley: Nicely done.
639
Vladimir Zheltonozhko & Valery Kirillov
Shakhmatnaya Kompozitsiya 1993, Special Prize
Helpmate in 2, 4 solutions
Solution
This four-phase problem is solved by 1.Qa3 b3+ 2.Kb4 c3, 1.Qxa4 b4 2.Kb5 c4, 1.Qc3 bxc3 2.Sd3 cxd3, and 1.Ra3 bxa3 2.Bd4 cxb3. Both the white b- and c-pawns execute the Albino theme by playing the maximum four different moves. Furthermore, in every solution the two pawns move in parallel, with the b-pawn leading the way and the imitative c-pawn giving mate. The white rooks in supporting roles are also well utilised in all four parts of this harmonious helpmate.
George Meldrum: A nice set with the b- and c-pawns combining in all mates.
Andrew Buchanan: Great stuff. One of those problems where the number of solutions is a big hint.
Jacob Hoover: Not only do both the white b- and c-pawns make all four possible moves, they move in sync in each solution! So we have a synchronized double Albino! Perfetta!
Karel Hursky: Loved this helpmate.
Paz Einat: Fabulous double WP4!! Looks like a clear 1st prize, not a special one…
640
Rauf Aliovsadzade
OzProblems.com 25 Feb. 2023
Mate in 3
Solution
Black has only three legal moves in the diagram, including a king’s flight to d3 (no set response). The key 1.Sf7! contains a threat, 2.Se5 [A], which removes the flight and leaves Black in zugzwang: 2…exf5 3.Sxf5 [B]. Now curiously all three black moves work as defences. 1…exf5 2.Sxf5+ [B] Kd3 3.Se5 [A] – White plays the same two moves as in the threat line but in reverse order. After 1…e5, 2.Sd6 [C] is a waiting move that covers c4 for 2…Kd3 3.Rf3 [D], a pin-mate. And if 1…Kd3 then 2.Rf3+ [D] forces 2…Kc4 3.Sd6 [C]. In these two variations, White likewise makes the same pair of moves but their order is swapped. A neat doubling of a formal idea: reversal of White’s second and mating moves, with no extraneous play at all.
Composer: White AB-BA and CD-DC involving a ‘twist’ – a pseudo-threat.
Andy Sag: Not quite a Meredith but the variations making use of the forced self-pins are nice.
George Meldrum: Initially observed 1.Rf3+? exf3 2.Bb1 e5 3.Sf5; however, 2…f2! stops the mate. Several other moves come close to solving rather than the more obscure key.
Solution
Many of the black queen’s moves are provided with set responses, such as 1…Qf6 2.Qb2+ Qxb2, and 1…Qh1+ 2.Qd1+ Qxd1. Some defences are not, however, like 1…Qh7 when 2.Qxd3+? Qxd3 fails because of the g3-rook, and 1…Qf4 (prospectively guarding c1) when 2.Qc2? mates instead of forcing 2…dxc2, because the black pawn is pinned. White seems to deal with these drawbacks by making any rook move off the rank, but only 1.Rg8! (waiting) succeeds. 1…Qh8/Qf6/Qd4/Qf2/Qh2 2.Qb2+ Qxb2, 1…Qh7/Qe4/Qg3/Qh3 2.Qxd3+ Qxd3, 1…Qh6/Qg5/Qf4 2.Qc2+ dxc2, and 1…Qh5/Qg4/Qe1+/Qh1+ 2.Qd1+ Qxd1. White must avoid 1.Rg7? Qh8!, 1.Rg6? Qh7!, 1.Rg5? Qh6!, and 1.Rg4? Qh5!, in each case allowing the black queen to hide behind the rook. (1.Rg2? and 1.Rg1? are technically not tries since they are refuted in multiple ways.) A fine duel between the two queens, where White’s play is exact throughout (notably 1…Qd4 2.Qb2+, not 2.Qxd3+? Qc3!).
Andy Sag: In set position the black queen has 14 possible moves and all other black units are immobile. With the g3-rook right out of the way, the d-pawn is unpinned, the black queen is given two more moves but all 16 allow white queen checks (on d3, d1, c2 or b2) which force Black to deliver mate. Prize well deserved!
Rauf Aliovsadzade: Excellent key – waiting move not expected!
George Meldrum: Terrific key move and surprising mate given with the black pawn.
Andrew Buchanan: Luckily I quickly guessed the idea, but it's still very cool.
642
Uri Avner
Springaren 1993, 1st Prize
Mate in 2
Solution
Two black captures on e6 yield important set play: 1…Kxe6 2.Qd7 and 1…Qxe6 2.Se3. The thematic try 1.Qa2? prepares an indirect Q + S battery aimed at the e6-flight, and threatens 2.Sd6 (not 2.Se3?). The flight-move 1…Kxe6 leads to a changed mate, 2.Se3, a direct battery opening – not 2.Sd6? as it would interfere with the d3-rook. The self-block 1…Qxe6 allows another changed mate, 2.Qf2. But the try is defeated by 1…Qxg7! The unexpected key 1.Sf4! removes the flight on e6 but grants another one on f4; furthermore, it opens a black queen-line to d6 while closing another one to e3. The threat 2.Se3 utilises the indirect Q + S battery, now aimed at the f4-flight (not 2.Sd6?). The flight-move 1…Kxf4 permits a new direct-battery mate, 2.Sd6 – not 2.Se3? because of another interference with the d3-rook. Compared with the try play, the reversal of the threat-move and a variation mate (2.Se3/Sd6) after different black defences brings about the pseudo le Grand theme. The self-block 1…Qxf4 enables 2.Qd7, which represents a transferred mate, used in the set play against another defence. Indeed, the other set mate, 2.Se3, is also transferred to 1…Kxe6 in the try play. The numerous analogous effects between the three phases are remarkable, including how the new black defences replicate the motifs of flight-move and queen self-block.
Andy Sag: The give-and-take key shifts the set flight-capture to a different square and creates a threat which can only be defended by f4-captures. These defences are answered by a battery mate or exploiting a self-block. The set flight-capture and self-block are provided for by 2.Qd7 and 2.Se3.
Solution
Both duplex solutions of part (a) seem straightforward; the first player self-blocks while the second arranges extra guards for the mate, with little interaction between the two sides. Black begins: 1.hxg5 Rf3 2.h5 h3; White begins: 1.Rxe3 Rb8 2.Ra3 Rb4. When the kings are exchanged in the twin, a similar description applies to the new duplex solutions. Amazingly, however, each side makes the exact same moves as in part (a), only now their order is changed and their motives of self-blocks and guards/mates are reversed. Black begins: 1.Rb8 Rxe3 2.Rb4 Ra3; White begins: 1.Rf3 hxg5 2.h3 h5. A highly original and humorous idea. The lack of interplay between Black and White is a necessary cost for the scheme to work.
Andy Sag: Easy enough to solve. Interesting to see both sides play the same moves in reverse order, when comparing the original with the twin duplex and vice versa.
Thomas Thannheiser: Very, very nice problem. Same moves in (a) and (b) but different kings to check mate with this move order!
Jacob Hoover: That is super crazy!
Michael McDowell: I greatly enjoyed it. Very imaginative, and just my sort of helpmate!