# Weekly Problems 2022-B

## 606

Viktor Schtscherbina
Shakhmatnaya Kompozitsiya 1992
Hon. Mention

Mate in 5

## Solution

Although it seems likely that the bishop will move along the b8-h2 diagonal several times to threaten mate, so as to decoy the rook to a weak square, such a plan is insufficient. 1.Be5? Rd8! (1…c5! also refutes) or 1.Bg3? Rf8! leaves the rook ready to check, and while 1.Bf4? Re8! looks more promising, no progress is made after 2.Bc7 c5 3.Bb6 Re5. Correct is 1.Bd6! forcing 1…c5, then 2.Be7 sets up a surprising zugzwang. Black has three moves that would keep guarding c5, and White responds to them in different and yet analogous ways, by shifting the bishop along the d8-h4 diagonal. 2…c4 weakens d4 and allows 3.Bf6, a double-attack of d4 and d8, which answers 3…Rd8 with 4.Bxd8 c3 5.Bb6 (3…Re8/Rf8 4.Bd4+ Re3/Rf2 5.Bxe3/Bxf2). If 2…Rc7 then 3.Bg5 similarly controls both e3 and e7, bringing 3…Re7 4.Bxe7 c3 5.Bc5 (3…Rf7 4.Be3+ Rf2 5.Bxf2). And finally 2…Rc6 prompts 3.Bh4, which aims at both f2 and f6 for 3…Rf6 4.Bxf6 c3 5.Bd4. A difficult more-mover that makes interesting use of geometry in the way the white bishop outmanoeuvres the black rook.

Andy Sag: The bishop plays cat and mouse with the c8-rook which cannot escape its fate but White must take care to avoid stalemate, e.g. 1.Bf4? Re8 2.Bd6 Re3! 3.Bc5??
Andrew Buchanan: Looks very simple but really hard to find!
George Meldrum: Great problem!

## 607

Valery Kopyl
Chervoni Girnik 2003
Pidlivailu-75 Jubilee Tourney, 1st Prize

Mate in 2

## Solution

A set mate is arranged for black castling: 2.Qxh7. White has many ways to place a second guard on f7 to threaten 2.Qxf7, which would also deal with 1…Kf8. Three such moves, 1.Rf5?, 1.Be6?, and 1.Sd8? are similarly refuted by 1…Rf8! White castling is tempting, especially since by vacating e1 the move provides for 1…Rf8 with 2.Re1, exploiting the self-block. But 1.0-0? is thwarted by 1…0-0! when the set mate no longer works. The key 1.Se5! counters 1…Rf8 with a new mate, 2.Bd7. Accepting the offered knight, 1…dxe5, opens a line for the d5-rook: 2.Rd8. The set variation 1…0-0 2.Qxh7 remains unchanged. We see thematic castling on both sides and some plausible tries.

Andy Sag: Sacrificial key in Meredith setting.
George Meldrum: A few tries, my favourite being White’s castling.
Bob Meadley: Very clever with the white castling position. Good one.

## 608

Fadil Abdurahmanovic &
Zlatko Mihajloski
Die Schwalbe 2017, 3rd Prize

## Solution

A rook mate on b5 seems probable since the piece would guard the flights on b3 and a5, leaving only the one on c3 to be covered or blocked. But bringing the rook to b5 in four moves is far from an easy task. The king-side pawns prevent the piece from reaching the b-file via the top ranks, so the rook heads for c5 from a lower rank. Many false trails remain, though, when the black units attempt to make way for the white piece, e.g. 1.e2 Kxd2 2.Sf2 Rc1 and the queen cannot clear the c-file, or 1.Ra2 Rg2 2.Qc2 Rxc2 and Black has no suitable way of blocking c3. The solution is 1.Rh2 Rg2 2.Qc1 Rc2 3.Rd2 Rxc5 4.Qc3 Rxb5, which shows matching clearance play on the second rank and the c-file. Both the black rook and queen cross over a critical square, g2 and c2 respectively, on which the white rook lands and then it travels in the opposite direction – the Turton manoeuvre. Each black piece then follows the rook back along the same line – the Bristol theme, with the black rook returning to d2 to stop a prospective queen check, and the queen self-blocking on c3. The bi-coloured versions of two line-clearance themes are thereby fused and doubled.

Andy Sag: Bristol style clearances allow the white rook to reach b5 in four moves and the black rook switchback avoids an unwanted queen check. Brilliant!
George Meldrum: It is plainly obvious that if the queen is placed on c3 then the white rook can mate on the b5-square. No attempt was made to hide this fact yet after many tries, I gave up and looked for many other ways to crack this hard nut. In the end returned to the obvious and found the extremely unobvious beautiful solution.

## 609

Joe Bunting
To Alain White 1945

Mate in 2

## Solution

The black king is constricted in the diagram, but White has difficulty preparing an effective threat. Only 1.Qf6! works, to threaten 2.Qxf5, and surprisingly the key grants three flights to the king. The latter’s moves to f4, d5, and d3 form a Y-pattern and they yield different mates: 1…Kf4 2.Rxg4, 1…Kd5 2.Qe6, and 1…Kd3 2.Qd4. This Y-flights demonstration stands out not only because of the perfect key, but because the three thematic variations all finish with pin-mates. In two cases White’s queen and g5-rook exchange roles in pinning the black bishop and delivering mate. The by-play consists of 1…B~ 2.Bc2 (a transferred mate with respect to the set play, 1…Sf3 2.Bc2) and 1…Sd4 2.Qxd4.

Andy Sag: The key gives three flights answered by pin-mates in a Meredith setting.
Andrew Buchanan: The key offering three flights is impressive. Clean play, and only eleven units.

Milos Tomasevic
Mat 1976

## Solution

White has only one workable idea: place the king on h6 and use a promoted queen to check on either g7 or h7 (covering the g8-flight), thereby forcing the black queen to capture and give mate. To allow the promotion to occur without checking Black, the white king must go to d8, and such a trip requires the pawn to act as a shield as well. 1.Kh6 2.c4 3.c5 4.Kg5 5.Kg4 6.Kh3 7.Kg2 8.Kf1 9.Ke2 10.Kd3 11.Ke4 12.Ke5 13.Kd6 14.c6 15.c7 16.Kc6 17.Kb7 18.Kc8 19.Kd8 20.c8=Q. Now the king returns to h6 via another route, taking advantage of how the white queen can cut off both major black pieces simultaneously on f5. 21.Qd7 22.Kc7 23.Kd6 24.Qb5 25.Ke5 26.Ke4 27.Qf5 28.Kf4 29.Kg5 30.Kh6 31.Qh7+ Qxh7. A long, capture-free rundlauf by the white king which visits three of the board edges, elegantly done in miniature.

Andy Sag: Fairly straightforward. Achieving a unique sequence is remarkable with only six pieces!
Jacob Hoover: Themes are interference and Excelsior.

## 611

Leonid Makaronez
OzProblems.com 6 Aug. 2022

Mate in 3

## Solution

The white queen has four candidate moves, Qg3, Qg5, Qh2, and Qb8, that target g7 or h8 while maintaining the guard on the e5-flight. The threat in each case is 2.Se6+ Bxe6 followed by a queen mate on the long diagonal, exploiting the interference of the e7-pawn. Only 1.Qg3! solves, by handling both of Black’s main defences. 1…Rg1 2.Be3+ Kc3 3.Qc7 shows long-range play by the queen, while 1…e5 2.Qe3+ Rxe3 3.fxe3 sacrifices the piece after a self-block. If 1…e3 then 2.Qf4, a short mate. The tries are 1.Qg5? e5 2.Qe3+, but 1…Rg1!, 1.Qh2? Rg1/Rh1 2.Be3+, but 1…e5!, and 1.Qb8? Rg1/Rh1 2.Be3+, but 1…e5!

Andy Sag: The withdrawal key creates an unusual threat immobilising the e7-pawn. It also unpins the e4-pawn adding a short mate by-play.
George Meldrum: I love the threat in this problem, and the white queen can be placed on many different squares to reach g7 or h8, which made it hard to solve. A black rook could have been used instead of the queen on a1 [either piece stops a dual in 1…e5 2.Qe3+/Bxe1].
Karel Hursky: Very amusing. Upwardly and diagonally mobile queen.

## 612

Zoran Gavrilovski
StrateGems 2016, 5th Prize

## Solution

Black’s R + B battery on the fifth rank is primed to give a double pin-mate, i.e. a mate that relies on the pins of the f6-knight and f4-rook. The key 1.Qc2! threatens 2.Qxd3+, which would compel such a battery opening, 2…Bxd3. Black can defend by unpinning either white piece and thus disrupt the prospective mate. A random move by the queen, 1…Q~, permits 2.Sg4+ Bxg4, when White forces the battery to fire in another way. A correction move 1…Qg8 prepares to meet 2.Sg4+ with 2…Qxg4, but now 2.Sd5+ brings 2…Qxd5. If 1…Bg1, which represents a random unpin of the rook, then 2.Re4+ induces yet another battery mate, 2…Bxe4. A subtle correction, 1…Rg3, clears the f-file for the f1-rook so that 2.Re4+ Bxe4 doesn’t mate; instead White exploits the newly formed B + R battery with 2.R4f3+ Rxf3. Two complex pairs of variations that feature black correction play and four different continuations executed by the unpinned pieces. There’s good by-play with 1…dxc2 2.Sxc2+ Bxc2 and 1…dxe2 2.Qe4+ Bxe4, making further use of the main battery.

Jacob Hoover: A great try 1.Qb5? threatens 2.Qxd3+ Bxd3, but the correction move 1…Qg8! refutes. The try play includes 1…dxe2 2.Sc2+ Bxc2, which shows both a changed mate and a transferred mate compared with the solution.
Andy Sag: Surprisingly large number of variations for a selfmate including a number of unpins. Interesting retro-analysis shows three white pawns must promote (and be captured or replace captured pieces) to get the black pawn configuration on the d-file.

## 613

Tony Lewis
The Problemist 2006

Mate in 2

## Solution

The diagram is a complete block position in which every black move is provided with a mating response. 1…Be~ 2.Qe4, 1…Bxd4+ 2.Rxd4, 1…Bb8+ 2.Qxb8, 1…Bc~ 2.Sd3, 1…Bxe2 2.Sxe2, 1…f5/fxg5 2.Qxe5, and 1…f1=Q 2.Rxf1. However, White has no simple waiting move that could preserve all of the set play. The white queen especially is nicely controlled by its need to observe b8 and several squares on the e-file simultaneously, e.g. 1.Qe7? (waiting) Bb8+!, 1.Qc8? (threat: 2.Qf5/Qg4) Bxd4+! Any king move is also defeated by a check, e.g. 1.Kb6 Bc7+!, 1.Kb7? Bd5+! Another try is 1.gxf6? (threat: 2.Qxe5) gxf6! The key by the queen, 1.Qd8! (waiting), does abandon the e-file but  compensates in various ways, which results in these changed mates: 1…Bd6 2.Qxd6, 1…Bc7 2.Qxc7, 1…Bxd4+ 2.Qxd4, 1…f5 2.Rh4, and 1…fxg5 2.Qxg5. The remaining play is unchanged from the set: 1…Bb8+ 2.Qxb8, 1…Bc~ 2.Sd3, 1…Bxe2 2.Sxe2, and 1…f1=Q 2.Rxf1. A well-constructed mutate that contains nine variations, all dual-free in both phases of play.

Andy Sag: A busy waiter with five changed mates and many tries.
Andrew Buchanan: Fun problem, nice key. I like these kinds of block problems.
Bob Meadley: Tricky.

## 614

Viktor Chepizhny
Shakhmatnaya Kompozitsiya 2011, Prize

Helpmate in 2, 3 set plays, 3 solutions

## Solution

White’s b8-rook, e3-knight, and e2-bishop are each uniquely controlling a flight-square (b7, d5 and b5 respectively), and these pieces are able to make a two-move sequence that delivers mate. However, Black has difficulty in not disturbing these mates, since all legal moves would either give check or capture a piece that’s needed to guard a flight. In the three set plays, Black manages to find a waiting move that neither checks nor captures, by occupying the square just vacated by White’s first move – the Umnov effect. 1…Rb7 2.Rb8 Rc7, 1…Sd5 2.Se3 Sb4, and 1…Bh5 2.Qe2 Be8. When Black begins in the actual solutions, there’s no avoiding the immediate capture of one of the three thematic white pieces. To compensate for creating a flight-square, Black employs the capturing piece to block it. Meanwhile, it seems as though White could proceed with either of the mating sequences by the two remaining pieces. But it turns out that only one works, because the other mate would be spoiled by the self-blocking black piece. 1.Qxe2 Rb7 2.Qb5 Rc7 (not 1…Sd5?), 1.Rxb8 Sd5 2.Rb7 Sb4 (not 1…Bh5?), and 1.Sxe3 Bh5 2.Sd5 Be8 (not 1…Rb7?). Therefore this helpmate demonstrates cyclic dual avoidance, in addition to a cyclic Zilahi where three white pieces rotate their sacrificial and mating duties.

Andy Sag: Only three black pieces can move without checking. In each solution a black piece makes a capture then self-blocks the square previously guarded by the captured piece. In set play a black piece moves to the square vacated by the mating piece.
Andrew Buchanan: A sublime problem. Not hard to solve, but very satisfying to see the relationship between the set and main play, together with the cyclic dual elimination.

## 615

Werner Sidler
Schweizerische Schachzeitung 1969

Mate in 4

## Solution

The sacrificial key 1.Rh2! threatens a short mate (2.Sf3) and also releases the black king, meaning the black rook is now liable to be captured without causing stalemate. After 1…Kxh2, however, White makes no progress with 2.dxc8=Q? Kg1, since 3.Qh3 – the only useful way to place an extra guard on g2 – gives stalemate. So instead 2.dxc8=B Kg1 3.Bch3 Kh2 4.Sf3. Similarly after 1…Rf8, if 2.gxf8=Q? Kxh2 then either 3.Qf2 or 3.Qf1 would be stalemate. Hence 2.gxf8=R Kxh2 3.Rf2 Kg1 4.Sf3. Two analogous underpromotion variations; if only not one but both of them had finished with a model mate!

Andy Sag: The key offers the rook to avoid stalemate and threatens mate. Black has two defences but in each case, White can underpromote to enable guarding the bishop without giving stalemate so the mate can be delayed but not prevented.
Andrew Buchanan: Very nice composition, with the double underpromotion.
Karel Hursky: Who would have thought there will be two underpromotions? Experienced two aha! moments. Very entertaining.
George Meldrum: A fun problem with underpromotion commotion.

## 616

Rauf Aliovsadzade & Leonid Makaronez
OzProblems.com 10 Sep. 2022

Mate in 3
Twin (b) Kg3 to g7

## Solution

Black has only two legal moves by the f6-pawn. The key 1.Rc8! (waiting) preserves a set variation for 1…fxe5 while providing for 1…f5. After the self-block 1…fxe5, the d3-knight surprisingly has only one viable option for releasing the d4-pawn: 2.Sb4 d3 3.exd3. If 1…f5, 2.Rf8 handles the check 2…f4+ with 3.Rxf4. In part (b), White must find new ways to deal with the two pawn defences. The key 1.Bb4! (waiting) prepares to meet 1…fxe5 with 2.Bc3, which frees the d4-pawn for 2…dxc3 3.Rc4. And 1…f5 prompts 2.Kf6, leading to the unavoidable self-block 2…f4 3.Sc5. Good changed play across the two phases. Curiously, not just the keys but all four continuations by White are waiting moves.

Andy Sag: I found part (a) easy enough but part (b) has a bishop sacrifice which took a long time to spot.
George Meldrum: A scattered setting but without any giveaways, making it hard to solve. The two totally unique solutions are to be commended.

## 617

Jean-Marc Loustau
Ideal-Mate Review 1997, Prize

## Solution

The series-helpmate is solved by 1.Kf4 2.Bf3 3.e4 4.e5 g3, with an ideal-mate finish. When the task is changed to helpmate, the same mating position cannot be reached despite White having three extra moves (the restricted king cannot triangulate). Instead, the solution is 1.Qf6+ Kg4! 2.Bf4 Kh5 3.Kf5 g3! 4.Be4 g4, which obtains the same mating configuration as the first part, but with all units shifted up one square. A rare form of twinning generates an exact chameleon echo mate, with the white king and pawn making tempo moves to boot.

Andy Sag: Part (b) is tricky as both white units must find a tempo move. Its finale is identical to (a) but all one square north.
Andrew Buchanan: Easy to solve but still a very beautiful chameleon echo. The need for a second white tempo move in (b) forces the black move order.
Thomas Thannheiser: Nice idea to combine a series-mover with a normal problem. Part (a) is easy because the g3 mate is obvious, but in (b) the white king and pawn have to slow down a little on their way to their final squares.

## 618

C. G. S. Narayanan
2nd World Chess Composition Tournament 1980-83, 5th Place

Mate in 2

## Solution

Set mates are prepared for 1…Kd5 (2.Qc4) and 1…Be4/Bf3+/Bg2 (queen captures bishop), but none for Black’s remaining moves, 1…Kb7, 1…a5, and 1…Bd5. If White shifts the b6-bishop, that will open a number of vital lines and provide for all black defences. But White must take care in choosing the bishop’s landing square – 1.Bg1! (waiting) – because the alternatives would cause various self-interferences. 1…Kd5 2.Qc4, 1…Kb7 2.Qxa6, 1…a5 2.Qb5, 1…Bd5 2.Qxa6, 1…Be4 2.Qxe4, 1…Bf3+ 2.Qxf3, and 1…Bg2 2.Qxg2. Three bishop tries prospectively interfere with a queen’s line-of-guard in the would-be mate. 1.Ba7? Kb7! (2.Qxa6??). 1.Bd4? Kd5! (2.Qc4??). Although 1.Bc5? similarly disrupts 1…Kd5 2.Qc4??, the bishop now guards d6 to set up a changed mate, 1…Kd5 2.Se7; this try is instead refuted by 1…a5! (2.Qb5??). The other three bishop tries directly interfere with a white piece: 1.Ba5? Kb7! (2.Qxa6??), 1.Be3? Be4! (2.Qxe4??), and 1.Bf2? Bg2! (2.Qxg2??). The thematic white bishop curiously serves no positive function other than in the try variation, 1.Bc5? Kd5 2.Se7. If it were legal to simply remove that piece as the first move, that would also have solved the problem!

Andy Sag: The white bishop must move (or disappear!!) to deal with unprovided flight 1…Kb7 but where to?? Apart from g1 the other six are all thematic tries.
Jacob Hoover: A nice problem featuring avoidance of self-injury.

## 619

Dennis K. Hale & Ian Shanahan
StrateGems 2012

Illegal cluster Add BQ, BS, 2 BPs
Progressive twin (b) Pf5 to f3
(c) & Ke2 to f7

## Solution

The three parts involve adding the same group of black units – a queen, a knight, and two pawns – but each solution (diagrammed below) is based on a different impossible check. In (a) the checking queen has no possible last move as it’s constricted by the other black units, including the knight which can’t be put on g3 because that would allow a legal retraction, …Qf3-g4+. The knight also cannot be swapped with either pawn on f4/g5, as it would produce an illegal double-check situation that’s not legalised by the removal of any pawn. In (b) the knight is prevented from giving a legal check not only by the f3-pawn but also by the units added on f2, g2, and h2, all of which rule out a knight promotion as the last move. The queen cannot exchange squares with either pawn on the same rank, similarly because of the double-check brought about. In (c) the knight check is made illegal in a more conventional manner, without involving promotions. Once again, the queen must be placed precisely, avoiding the pawn squares e6/h7 where it would check along with the knight. This neat illegal cluster problem also utilises the black king fully in restricting the checking piece in all three solutions.

Andy Sag: Parts (a) and (c) were reasonably straightforward with queen and knight checks respectively, but (b) was diabolically difficult until I realised that it could legally be a promoted knight on g1 if any of the three units on the second rank were removed.

Solution to (a)

Solution to (b)

Solution to (c)

## 620

Rauf Aliovsadzade & Daniil Yakimovich
OzProblems.com 8 Oct. 2022

Mate in 3

## Solution

The key 1.Sf8! entails a short threat, 2.Se6, against which Black has two defences, 1…Bxd5 and 1…Rxe5. These black moves generate two closely related variations. 1…Bxd5 is exploited as a line-opening by White with 2.Bxb5, which controls d3 to threaten 3.c3. Now if 2…Be6+ then the original threat returns as 3.Sxe6, while the switchback 2…Bc4 permits the capture of the defender, 3.Rxc4. Similarly, 1…Rxe5 opens a diagonal line for White, with 2.Qxg6 guarding d3 to threaten 3.c3 again. Black counters with either 2…Rf5, which enables the original threat, 3.Se6, or 2…Re4, a switchback that allows the defender to be captured, 3.Qxe4. The Visserman type is a scheme found in three-movers (or longer problems), in which two white second moves create changed play, by functioning rather like the key and a try in a two-mover. In most Visserman problems, the two thematic white continuations are followed by changed mates against the same black defence, but in this case, 2.Bxb5 and 2.Qxg6 lead to a transferred mate, 3.Se6, against different defences, 2…Be6+ and 2…Rf5. This unusual idea is accompanied by a wealth of matching effects in the two variations.

Composers: Visserman type, mate transference, and return of black units to their initial squares.
Andy Sag: The initial threat has two defensive lines, each opening a different diagonal allowing White to guard d3 thus freeing up the c-pawn to add a decisive second threat, leaving Black ‘up the creek’.
George Meldrum: Not particularly easy to solve with the c2-pawn being the centre of attention in by-play; who would have thought!

## 621

Krasimir Gandev
Schach-Echo 1970, 2nd Prize

Helpmate in 2, 2 solutions

## Solution

Firing the B + K battery on the long diagonal to give mate (covering the g7-flight) seems a plausible goal. However, currently Black’s queen and c4-knight control the battery, and also the white king has no suitable landing square, as Ke4 and Kd5 would interfere with the b1-bishop and c5-rook respectively and create a flight-square. In the two perfectly analogous solutions, Black and White use opposite strategies to remove these hindrances: 1.Qb8 Bh7 2.Sb6 Ke4 and 1.Qe1 Rh5 2.Sd2 Kd5. The black queen’s initial move in each phase crosses over a critical square (b6 or d2) so that its subsequent occupation by the knight causes a self-interference, preventing the queen from disturbing the mate. White’s first move by the bishop or the rook is anti-critical in the sense that the line-piece crosses over a square (e4 or d5) in order to avoid a later self-interference, when that square is occupied by the king. The two parts of the problem are also connected by a dual avoidance effect: 1.Qb8 Rh5? 2.Sb6 – 2…Kd5?? disabled, and 1.Qe1 Bh7? 2.Sd2 – 2…Ke4?? disabled.

Andy Sag: In each case the black queen moves to the edge of the board to be cut off by the c4-knight, and a white piece also moves to the edge of the board to allow the white king to discover mate without cutting off that piece.
Jacob Hoover: An elegant example of both critical play and anti-critical play. Also, we have an ODT [orthogonal-diagonal transformation] effect between the two solutions.
Andrew Buchanan: Not seen this kind of idea before but it works well. Once solved, I found it impossible to improve on – it seems very well constructed.

## 622

Vladimir Pachman
Národní osvobození 1946

Mate in 4

## Solution

Set play is prepared for the two possible black moves in the diagram, but only 1.Kc3! (waiting) manages to preserve all of these lines. 1…Kg5 enables 2.Qg7+ which branches into 2…Kf5 3.Kd3 e4+ 4.fxe4 (ideal mate) and 2…Kh5 3.Qf6 e4 4.g4 (model mate). If 1…e4 instead then 2.f4 leads to 2…e3 3.g4 e2/e5 4.g5 (model mate) and 2…e5 3.f5 e4 4.Qg6 (model mate). This Bohemian problem thus brings about four ideal/model mates (with no by-play), including a pair of echo mates delivered by the g-pawn.

Andy Sag: A clever miniature waiter with a counter-intuitive key. The sub-variation where White invites (no, forces!) a check on the third move is particularly hard to see.
George Meldrum: A lovely key move. Frustrating to solve being such a ‘simple’ setting with an array of precise lines to discover.

## 623

Satoshi Hashimoto
Problemesis 1999, 2nd Prize

## Solution

Out of twelve moves available to Black, ten are “visible” or taken up by units to reach their present positions in the diagram (including a castling move). That leaves one spare move each for the missing queen and f-pawn to assist with their own captures – in the latter case by White’s g-pawn. White is missing only the d-pawn, but Black has no way of capturing it on the d-file in time. To sacrifice it elsewhere, White must promote the pawn, and that requires it to capture something on c7 – an idea consistent with moving the black queen once. 1.d4 Sc6 2.d5 Sa5 3.d6 c6 4.Qd5 Qc7 5.dxc7 d6 6.Qh5 Be6 7.c8=B. White promotes to a bishop as it’s the only piece capable of a manoeuvre that facilitates black castling as well as its own removal soon after. 7…Bb3 8.Bh3 b6 9.g4 0-0-0 10.g5+ f5 11.gxf6+ e.p. Kb7 12.Bc8+ Rxc8. An attractive example of the Valladao theme that incorporates promotion, castling, and en passant capture. That the promoted bishop gets eliminated illustrates the Ceriani-Frolkin theme, with the bonus feature that the capture takes place on the promotion square.

Andy Sag: An unusual unique sequence involving white sub-promotion to a bishop, black queen-side castling, an en passant capture of a black pawn to give a second battery check, and finally a switchback of the promoted bishop to give a third check and then be captured!
Andrew Buchanan: Very fine Valladao. Always tricky to get the en passant working okay. Note the h5-queen necessary to pin the f7 -pawn until the black king has moved. However, the 2nd prize shows how much the field has moved forward technically since then.

## 624

Gerhard Latzel
Netherlands–Germany Composing Match 1954
1st Place

Mate in 2
Twin (b) Ke8 to h8

## Solution

In both twin positions, the black king initially has four available moves, none provided with a set mate. The give-and-take key in each part alters these black flights into a different group of four legal moves. In part (a), 1.Sg6! (waiting) covers all of the diagonal flights and leaves the king with four orthogonal moves, which are answered by various battery mates. 1…Kf5 2.Rg3, 1…Kxg6 2.Rd5, 1…Kg4 2.Re5, and 1…Kh5 2.Rg2; also 1…h5 2.Rg3. The try 1.Sxh6? (waiting) fails to 1…Kf6! when 2.Be7+ allows the king to escape to g7. By contrast in (b), the key 1.Sxh6! (waiting) removes all of the orthogonal flights but gives the king four diagonal moves. 1…Kxf4 2.Qc1, 1…Kxh6 2.g8=Q, 1…Kf6 2.Be7, and 1…Kh4 2.Be7. Here 1.Sg6? (waiting) becomes the main try, defeated by 1…Kxg6! when 2.Rd5 fails to observe the f7-flight. The two variations ending with 2.Be7 are technically distinct (the king is mated on different squares), but the repeated mating move is still a flaw. Regardless, the problem’s two-phase change from plus-flights to star-flights, resulting in the maximum task of eight different king moves, is outstanding.

Michael McDowell refers to another rendition of the same task by J. J. Gill, in which all eight mating moves are different. A pity this is achieved at the cost of a checking key!

Andy Sag: A Meredith twin waiter, featuring cross-flights in one part and star-flights in the other. Also, the key to each part is a try in the other.
Andrew Buchanan: Wow, eight different flights! I guessed the key to (a) immediately, but then had big problems with (b) as I expected the key to be 1.Sh5. An amazing problem.

## 625

Marjan Kovačević
Orbit 2009, 1st Prize

Helpmate in 3, 2 solutions

## Solution

White could form a B + S battery with 1…Sxb3, and aim for a double-check mate when the black king goes to e6. This scheme requires only one extra self-block on e5 by the bishop, but White runs into a tempo difficulty, e.g. 1.e2 (random waiting move) Sxb3 2.Be5, and now even though the e6-knight is unneeded for the mate, it can’t make any tempo move without preventing 3.Ke6. The first solution sees White wasting a move by forming the battery with the other knight instead. 1.Bxd4 Sc5 2.Be5 Sxb3 3.Ke6 Sd4. The e6-knight ends up on the starting square of its sibling, while the king is mated on the first knight’s original square. The second solution does employ the d4-knight to remove the b3-pawn; however, here the motive is not to create a direct battery but to allow the bishop to guard flights on c4 and d5 when the king goes to d4. The goal is a knight mate on e6, and this necessitates another self-block on e5. 1.dxe6 Sxb3 2.e5 Sc5+ 3.Kd4 Se6. Now the d4-knight finishes on its sibling’s starting square, while the king gets mated on the former’s original square. The actual interchange of the two white knights is wonderfully combined with their Zilahi exchange of functions, the latter indicating they get sacrificed and give mate in turn.

Andy Sag: After finding one solution I struggled to identify a common theme. I first thought it might be a double-check but it wasn’t that. In each case a black unit captures a knight and then self-blocks on e5. The uncaptured knight makes all three white moves ending up on the captured knight’s square after clearing the bishop’s line and allowing the black king to occupy the set position of that knight.

## 626

Felix Sonnenfeld & Almiro Zarur
Sredba na solid 1971, 1st Prize

## Solution

A withdrawal move by the e4-knight will threaten 2.Qf2+, forcing 2…Sxf2 mate. Out of five candidate moves by the knight, four are thematic tries that fail to work for analogous reasons. Black’s a5-bishop and d7-rook are guarding d2 next to the white king. If 1.Sc3? – cutting off one line of guard – then 1…Bd3! – interfering with the other – gives a flight to the king and thus renders the threat ineffective. Likewise, 1.Sd6? Rc3! sees the same two lines closed but in reverse order. The second pair of tries involve interfering with white pieces that are confining the black king. 1.Sc5? shuts off the b5-rook and it’s defeated by 1…Bg6! which cuts off the g8-rook, enabling the king to escape to g5 after 2.Qf2+. Similarly, 1.Sg5? Bf5! creates a flight on g4 by shutting off the g8-rook and e6-bishop. Only 1.Sf6! solves, avoiding the closures of both black and white lines, and Black has no defence against the threat. The harmonious tries demonstrate two selfmate variants of Theme F, one of the lettered line themes.

Andy Sag: Clearly the white knight must move to unguard f2 but where to? Only f6 can avoid unwanted flights and other complications. From there it is a “one-liner”.

## 627

Harri Hurme & Hannu Sokka
Scandinavian Championship 1970, 1st Place

## Solution

In the first solution, the black king remains on h7 to be mated by the rook on g7. This requires the black rook to self-block on h8, and it captures the bishop en route. 1.Rf3 2.Rxf6 3.Rf8 4.Rh8 Rg7. The king aims for h6 in the second part, to facilitate a bishop mate. The black queen, in freeing the h6-pawn, captures the knight as the only hideaway move, avoiding errors like checking White and interfering with the black bishop. 1.Qxe8 2.h5 3.Kh6 4.Bh7 Bg7. In the third part, the king goes to h5 for a knight mate. Now the queen captures the rook, both to give the king access to g6 and to self-block on g4. 1.Qxg4 2.Kg6 3.Kh5 4.Bg6 Sg7. The three white pieces rotate their functions in getting sacrificed and giving mate, to bring about a cyclic Zilahi. This theme, a rare accomplishment in series-movers, is remarkably augmented by three different pieces mating on the same square (g7) and three model mates.

Jacob Hoover: The three solutions exhibit a cyclic variation of the Zilahi idea. I enjoyed this one, although I had difficulty finding the rook-mate solution.

## 628

Rauf Aliovsadzade
OzProblems.com 3 Dec. 2022

Mate in 3

## Solution

There is notable set play where Black’s a8-knight makes two self-blocking moves: 1…Sc7 2.Sb8+ Kc5 3.Be7, and 1…Sb6 2.Sxa5+ Kd6 3.Be7. The principal try 1.Bd8?, by guarding c7 and b6, threatens the knight checks seen in the set variations: 2.Sb8+ Kc5 3.Be7 and 2.Sxa5+ Kd6 3.Be7. But Black escapes with 1…Rf2+! 2.Rxf2 Sc7, preventing 3.Rf6. Another try 1.Bxe5? threatens 2.Sb8+ Kc5 3.Bd6, but it’s also refuted by 1…Rf2+! The key 1.Be7! involves a single threat not seen previously: 2.Sxe5+ Kb6 3.Bc5. Here 1…Rf2+ 2.Rxf2, with the threat of 3.Rf6, leads to 2…Sc7 3.Sb8. The black rook produces a similar second variation: 1…Rxg5 2.Rxg5, threatening 3.Rg6, 2…Sc7 3.Sb8. The active dark-squared bishop delivers mates on e7, d6, and c5 in various phases while its move to e7 also acts as the key. The knight moves to b8 and a5 change their functions from set continuations to virtual threats, and Sb8 becomes a third-move mate as well post-key.

Composer: May be a fresh idea for three-movers: set play variations show up as threats in a try.
Jacob Hoover: The key 1.Be7! abandons the set play entirely. This one was rather easy as three-movers go.
George Meldrum: It is all about the black squares around the black king. The white bishop on f6 can bolster support on these black squares by the tries 1.Bd8? or 1.Bxe5?, both failing to Black’s 1…Rf2+! The key allows a check to the white king and introduces new mates using White’s rooks.

## 629

Comins Mansfield
British Chess Federation Tourney 1968
1st Hon. Mention

Mate in 2

## Solution

The black king has a provided flight on e4: 1…Ke4 2.Rg4. More set play takes the form of five self-blocks on the same square: 1…Re4 2.Rxd6, 1…Be4 2.bxc3, 1…S2e4 2.Sf3, 1…S6e4 2.Sf5, and 1…e4 2.Bg7. The key 1.Qc6!, with the threat of 2.Qd5, surprisingly removes that flight but grants a new one on c4. Now 1…Kxc4 enables 2.Qxc5. Black can also defend by capturing the c4-pawn with other pieces, and three such moves provoke mates that were originally set against other defences – 1…Rxc4 2.Rxd6, 1…Bxc4 2.bxc3, and 1…S2xc4 2.Sf3 – i.e. three transferred mates are shown. When the d6-knight makes the capture, a new mate comes about: 1…S6xc4 2.Bxc5. Lastly, one defence from the set play counters the threat and it generates a changed mate: 1…Be4 2.Qxc5. A lucid demonstration of mate transference, in which the three set defences occur on one square while the post-key ones occur on another, all unified by the self-blocking errors.

Andy Sag: A give-and-take key with a threat defended by five different captures of the c4-pawn.
Jacob Hoover: After the key, three mates are transferred to new defenses. Every defense – both in the set play and in the actual play – that is not a king move is a self-block.
Andrew Buchanan: The set play reveals the idea, and explains the quasi-symmetry of the black units. A set mate exists if and only if Black moves to e4! Sweet!

## 630

Josif Kricheli
Priokskaya Pravda 1968, 1st Prize

Helpmate in 2, 2 solutions

## Solution

The black king has access to e4 and e6, and if either square is blocked or guarded, the other can be covered by the white knight when it mates on f4 or f6. The first solution, in which the black knight blocks a flight, has a corresponding try that fails due to the lack of a white waiting move. 1.Sd4? white tempo?? 2.Se6 Sf6; in particular, 1…gxh6 would’ve allowed the h4-bishop to control the mating square. Black blocks e4 instead with 1.Sd2 gxh6! 2.Se4 Sf4, where the white pawn capture does work as a tempo move. The second solution employs the white queen to guard a flight (besides those on the c-file), and there’s an analogous try too in which Black lacks a waiting move. 1.Se1 Qc2? 2.black tempo?? Sf4; here 2.hxg5 would’ve prevented the mate. Rather, White guards e6 with 1.Se1 Qc8 2.hxg5! Sf6 – now the black pawn capture acts as a tempo. Brilliant tempo play by both sides that incorporates reciprocal pawn captures, complete with matching “logical” tries.

Andy Sag: Interesting that 1.Se1 Qc2? (2…Sf4) doesn’t work because any of Black’s nine possible 2nd moves will stop the intended mate. Also 1.Sxg5? (2.Se4 Sf4) doesn’t work because any of White’s remaining thirteen legal 1st moves will mess up the intended sequence.
Jacob Hoover: Both solutions involve play between the white g5-pawn and the black h6-pawn. As a nice bonus, both mates are models.
George Meldrum: Both solutions carry with them mirrored tries which can throw you off if attempted first. The whole setting is truly remarkable.
Andrew Buchanan: Easy but beautiful tempo problem. In one solution gxh6 while in the other hxg5. Harmonious design dual elimination, and a very natural feeling.

## 631

Bengt Rudolf Giöbel
Polistidningen 1946

Add a white queen, then mate in 1
Twin (b) Reflect position left-to-right

## Solution

Since all eight white pawns are on their initial squares in the diagram, the white queen couldn’t have escaped from the first rank. The thematic try is to add the queen on f1 for 1.e3 mate, but that’s also impossible because the queen began the game on the left side of the king and neither piece (constrained by the pawns) could have leapt over the other. The solution is to add the queen on a1 for 1.b3 mate. For part (b), the position is reflected, but the symmetrical placing of the queen on h1 for 1.g3 mate is only a try, again because the king and queen couldn’t have jumped over each other. Rather, the solution is to add the queen on c1 for 1.d3 mate. Hence two perfectly symmetrical positions paradoxically yield a pair of non-symmetrical solutions, due to retro-analytical considerations.

Andy Sag: Retro-analysis indicates that the white queen cannot have been promoted and so must be on rank one queen-side. As the Meerkat would say: Simple!
George Meldrum: Wonderful Christmas surprise puzzle.
Andrew Buchanan: A charming problem that I’ve known for many years. However it occurs to me today that if we cycle all the pieces one file to the right, then the stipulation can be lightened to say “white unit”, rather than “white queen”. [See diagram below.] Also the solution is in both cases +WQb1 1.c3 mate (not +WQg1? 1.f3 mate) which is slightly amusing. I can’t think of any other problem where the twin has identical solution.

## 631v

Bengt Rudolf Giöbel
Polistidningen 1946
Version by Andrew Buchanan

Add a white unit, then mate in 1
Twin (b) Reflect position left-to-right

## 632

Matti Myllyniemi
Suomen Tehtäväniekat 1967
Osmo Kaila-50 Jubilee Tourney, 1st Prize

Mate in 2
Twin (b) Helpmate in 2
(c) Selfmate in 2
(d) Helpstalemate in 2

## Solution

In the directmate (a), Black’s king is well protected by the pawns and the only way to generate a mating threat is 1.e8=S!, followed by 1…e5 2.Sxc7. Black begins in the helpmate (b) and if 1.e5 e8=S?, the unavoidable 2.e4+ disrupts the knight mate. So 1.e5 e8=Q 2.e4+ Qxe4. In the selfmate (c), White wants to force the e6-pawn to mate through zugzwang. This requires the bishop to self-block on c2 without giving the black king a spare move to c6. 1.e8=B! e5 2.Bc2 e4. The goal is stalemate in (d), and here White must take care of the e6-pawn without checking or releasing other black pawns. 1.e5 e8=R e4+ Rxe4. An original presentation of the Allumwandlung theme, in which White promotes to four different pieces.

Andy Sag: Quite a novelty with the e-pawn promoting to a different piece in each of the four parts.
Andrew Buchanan: Very nice! It took me a few minutes to solve, because I didn’t spot the e7-pawn. After that was identified, the solutions all came in 30 seconds.
Jacob Hoover: Since we have all four possible pawn promotions, this problem exhibits the AUW theme. I enjoyed this immensely.
Thomas Thannheiser: I like this type of problem. A little too easy, but very good to explain problem chess to chess players. AUW: S, Q, B, R.