Weekly Problems 2021-B

Problems 554-579

554

Toma Garai
feenschach 1984

Helpmate in 2, 2 solutions

Solution

The black king on its original square has three flights on the c-file, too many to cover here. Instead the king aims to be mated on two e-file squares, which are initially controlled by the white rooks. Thus Black begins by capturing one of these rooks, and the resulting discovered check forces the other rook to sacrifice itself on the king’s target square, as the sole legal way to unguard it. 1.Rxe7+ Re6 2.Kxe6 Sf4 and 1.Bxe3+ Re4 2.Kxe4 Sc3. Black’s first move in each solution also acts as a self-block. Splendid demonstration of capture of white force in a helpmate, incorporating both passive and active sacrifices.

Andy Sag: I got nowhere until my suspicions were aroused by noticing the two batteries. Perfectly matched pair of solutions each starting with a battery check, dispensing with both white rooks and finishing with a model mate delivered by the knight.
Andrew Buchanan: The clues here are the separation of the kings, which invites the black one to step forward to two squares, and some curious pieces (e.g. white pawn on c6 and black pawn on d3) which only make sense if the black king moves. Still it took some time to see Black’s first moves.
Ian Shanahan: What an astonishingly paradoxical blood-bath (i.e. capture of white force) exhibiting a perfect harmony of effects and reciprocity between the two solutions! Only the white pawns on the c-file remain unused in the second solution here – a lamentable flaw…

555

Miroslav Havel
Besedy lidu 1920

Mate in 3

Solution

Of the two flights available to the black king, one is provided: 1…Ke1 2.Se3 d1=Q 3.Sg2. The white king can attempt various waiting moves, but if it stays on the first rank, then equivalent play on the queen-side fails: 1.Kh1/Kf1? Kc1 2.Sc3 d1=Q+! White avoids the queen check with 1.Kf2?, but that runs into 1…Kc1 2.Sc3 d1=S+! The final try 1.Kg2? evades both types of promotion checks, but it’s refuted by 1…Ke1! 2.Se3 d1=Q when the king on g2 obstructs the knight’s mating square. Only the furthest withdrawal move 1.Kh2! works, after which 1…Ke1 2.Se3 d1=Q 3.Sg2 and 1…Kc1 2.Sc3 d1=Q 3.Sa2 produce a pair of reflected echo model mates.

Andy Sag: The clue is in the symmetry of the set position. After the black king moves, White can force the pawn to promote and self-block, but first White must move the king so it cannot be checked by the promoted piece and also keep g2 vacant.
Ian Shanahan: Although the two variations after the key are perfectly symmetrical (ending in reflected model mates), only that which follows 1…Ke1 precisely determines the white king's destination-square in the key – a subtle form of asymmetry. Very pretty, and typical of its famous Bohemian composer.

556

Andrew Buchanan
Die Schwalbe 2013, Commendation

Solution

All units in the diagram are on their original squares, hence not many clues remain as to what transpired in the game. The missing h-pawn, g-pawn and f8-bishop suggest that the white pawn could have promoted by capturing the black units on their initial squares, before getting eliminated itself. The black queen seems the best candidate for removing White’s queen-side pawns efficiently. 1.h4 e6 2.h5 Qf6 3.h6 Qxb2 4.hxg7 Qxc2 5.gxf8=S! White avoids promoting to a queen/rook as that would check and disrupt Black’s plan of freeing the a1-rook next move. A bishop promotion also fails but for a reason that’s not apparent until two moves later. 5…Qxa2 6.Rxa2 Sa6 7.Rxa6. Now only the black pawn on e6 and the white pieces on a6 and f8 remain to be captured. The a6-rook is poised to take that pawn followed by a sacrifice on e8 to the black king, a scheme consistent with the latter capturing the f8-knight. However, 7…Kxf8? 8.Rxe6 would leave Black stuck without a waiting move, so Black wastes a tempo with 7…Ke7! first – which would have been impossible had White promoted to a bishop earlier. 8.Rxe6+ Kxf8 9.Re8+ Kxe8 sees the black king completing a round-trip. This terrific homebase proof game incorporates the Schnoebelen theme, in which a promoted piece is captured without making any moves. Such an occurrence always raises the question of why a specific promotion choice is required; that the motivation here is to provide Black with a tempo move gives the problem an extra spark.

Composer: This one is one of my personal favourites, but to my mind overshadowed by P1274528, which was technically much harder to compose.
Andy Sag: I needed a computer hint to solve this one. Each side can only have four non-capture moves and it is clear that White’s a-, b-, and c-pawns must be captured by the black queen. But who in their wildest imagination would see the h-pawn promoting to a knight on f8 to allow the black king to have a tempo move!

557

Leonid Makaronez
OzProblems.com 24 Jul. 2021

Mate in 3

Solution

The key 1.e6! threatens 2.e5+ Kxd5 3.Bxc4. Since that mate requires the rook on c6 to block that square, any move by that piece would defend. 1…Rd6 2.Qe3+ Ke5 3.Bc3 shows White exploiting the distant self-block on d6. 1…Rxe6 is answered by the threat-move but followed by a different mate: 2.e5+ Kxd5 3.Bf3, where the rook blocks another flight. The a4-bishop makes a third distant self-block with 1…Bc2 2.Be3+ Kc3 3.Qe5. If the same bishop guards c4 with 1…Bb5/Bb3, that prospectively closes the b-file and after 2.Kb2, Black can no longer give a discovered check; now the d2-bishop has two threats that are separated by 2…c3+ 3.Bxc3 and 2…Ba5 3.Be3.

Andy Sag: Leonid’s three-movers are usually hard to solve but this one is straightforward. The dual threat after 1…Bb5/Bb3 2.Kb2 is a minor blemish. The f6-pawn is purely there to stop 3.Qf6 in the 1…Bc2 variation. The g3-pawn appears to have no purpose.
George Meldrum: The true task is cleverly buried with the placement of the white pawns and for a long time I was looking for mates along the black squares. The involvement of the white king in the solution is brilliant. A superb problem.

558

Hilding Fröberg
Vart Hem 1939, 1st Prize

Mate in 2

Solution

A fine sacrificial key 1.Qc3! guards e5 to threaten 2.Scxd8. Taking the offered key-piece with the bishop or the queen leads to different promotion-captures on d8: 1…Bxc3 2.cxd8=S and 1…Qxc3 (or 1…Qd4/Qf7/Qxg7) 2.exd8=S. Capturing with the black rook opens a line for the white bishop to d6 and activates the b7-knight: 1…Rxc3 (or 1…Rxc6) 2.Sbxd8. If 1…Sxc6, that self-pins the defending knight and allows another promotion mate on the same square, 2.d8=S. Thus five knight mates occur on d8 (including the threat), the maximum possible. This impressive task problem is capped off by 1…Sf7/Bf7 2.exf8=S, yet another knight-promotion mate.

Andy Sag: The sacrificial key unpins the f5-rook, threatens a double-check mate and unleashes a variety of knight mates from d8 and f8 including promotions.
Karel Hursky: Wonderful geometrical key on the intersection square of pathways of a5-bishop, c5-rook, f6-queen and h3-queen. However, I didn’t find it easily because of a scrummage of pawns and pieces around the kings.
Nigel Nettheim: A magnificent key and four different knight-promotion mates. Somehow there is a good mixture of simplicity and complexity. The self-pinning 1…Sxc6 2.d8=S is excellent.
Andrew Buchanan: Five knight mates on d8 – and a bonus one on f8! All carefully delineated. Good key, with a couple of tries 1.Qh2/Qg3? Rf4! Very entertaining!
George Meldrum: A gem, and funny too.
Ian Shanahan: After a spectacular sacrificial key, one savours an extravaganza of knight-mates – two by pre-existing knights (including the threat), and four via distinct pawn-promotions. An astonishing task, and an unforgettable problem!

559

Vladimír Kočí
Pat a Mat 2012, 1st Prize

Solution

If the black king remains on its original square, each of White’s minor pieces has a potential mating move that would cover the g5-flight. In two solutions, Black prepares for these final moves by self-blocking on the other flight-square g6, using a promoted piece that won’t disrupt the mate. 1.a5 2.axb4 3.bxc3 4.c2 5.c1=R 6.Rg1 7.Rg6 Sf7 and 1.exd6 2.d5 3.d4 4.dxc3 5.cxb2 6.b1=B 7.Bg6 Bd2. Since the white knight and bishop exchange their roles of giving mate and getting captured, the problem illustrates the Zilahi theme (an idea usually seen in helpmates). A contrasting third solution involves a trip by the black king and here both thematic white pieces take part in the mating net: 1.Kg5 2.Kf4 3.Kxe5 4.Kd5 5.Kc6 6.Kc7 7.Kd8 Ba5. All three phases end nicely with a model mate.

Andy Sag: In each case, one black unit makes all seven moves and check avoidance ensures all sequences are unique.
Jacob Hoover: We have a double Excelsior and the Zilahi theme.
Andrew Buchanan: Well the third solution was a lot harder to find than the first two! It's a nice problem to solve to test lateral thinking.
George Meldrum: The two solutions involving pawn moves provided a perfect pair. The king tour was so left field from the other solutions that it made it difficult to see. Whilst I like this deception, I wonder, was this the purposeful intent of the composer?
Nigel Nettheim: The promotion solutions make a fine pair. The other solution is non-thematic, and in my opinion detracts from the effect.
Karel Hursky: The black king’s march to d8 is a bit of a surprise. Why is there a knight on g7 and not a black pawn I couldn’t see. Entertaining problem; it was a pleasure solving it.
Ian Shanahan: It's astonishing that there are three accurate sequences of exactly the same length, the last of which – a king-trek – took me ages to discover, being so very different to the first two solutions that are related by the Excelsior motif (i.e., a pawn begins from its home-square and marches onwards to promote).

560

Slobodan Mladenović
The Problemist 1987, 3rd Prize

Solution

Black has only three legal moves, all with white replies provided. In 1…Qxh5 2.Qg6+ Qxg6 and 1…Qxg5 2.Qe7+ Qxe7 (not 2.Qg6+? Qf6+), the black queen is deflected to give a mate that covers the d6-flight. If 1…Qxf4, the discovered check 2.Kd6+ compels 2…Qxe5. White has no waiting move that could preserve the set play, however, e.g. 1.Qg7? Qxf4! and 2.Kd6+ doesn’t work because the white queen is guarding e5. The key 1.d6! (waiting) blocks the initial flight but creates another one on d5. Now White must induce the black queen to mate on a diagonal line that covers the new flight. 1…Qxh5 2.Qf7+ Qxf7 and 1…Qxg5 Qg8+ Qxg8. After 1…Qxf4, the altered discovered check 2.Kd5+ Qxe5 means the white king gets mated on a different square. Three neat changes are brought about in this selfmate mutate.

Andy Sag: White must find a key that replaces the set play with changed play.
Jacob Hoover: A clever mutate.
Nigel Nettheim: Marvellous change-play.
Andrew Buchanan: A beautiful problem and very satisfying to solve.
Michael McDowell: Not difficult to solve, as having seen the set play the key is the only way to involve the pawns on the b-file, but changes after all three moves is quite an achievement. Given the rich content the ugly knight at g1 can be tolerated.
Ian Shanahan: A complex selfmate mutate (after a quiet key, all continuations are changed) in a tight and rather heavy setting. Very satisfying.
Thomas Thannheiser: Very nice selfmate this week. Truly worthy of a prize!

561

James Malcom
OzProblems.com 21 Aug. 2021

Mate in 8

Solution

To pre-empt the strong defence 1…Bf2+, White starts with 1.0-0-0!, pinning the bishop. Now Black’s pawns have four successive moves before they all become immobilised, and White must use the available time to prepare against the impending stalemate. The plan is to execute an Indian manoeuvre by bringing the rook back across the critical square c1, which the bishop can then occupy to cause an interference that would release the pinned black piece. This scheme requires the white king to make a Bristol clearance along the rank to give the rook access to another critical square, b1. 1…a5 2.Kb1 f6 3.Ka1 g4 4.Rb1 g3 5.Bc1 Bf2 6.Bd2+ (6.Bxe3+? Bg1 7.Bc1 Bf2!) Bg1 (6…Be1 7.Rxe1) 7.Be1 Bf2 8.Bxf2. The thematic clearances are followed by a four-move zigzag of the white bishop.

Composer: A demented Indian + Bristol combo!
Andy Sag: Queen-side castling is clearly necessary but what next? I spent a long time messing with the idea of a battery on the long white diagonal but to no avail. A battery is required but on the first rank after a Bristol style manoeuvre by king and rook.
George Meldrum: Love the setting, love the position, love the layout! There are enough alternate ways to try and solve this that makes it tricky with a fine solution.
Bob Meadley: A classic I reckon. Who would believe the white king ends on a1?

562

Sergey Rumyantsev
Mat 1989, 3rd Prize

Helpmate in 5, Set play

Solution

The interlocked units on the queen-side have to be untangled to allow at least one to aid in the mate. Since a freed white king would still be too far away to be useful, the aim is to release the black rook which can quickly obstruct a flight-square. The set play is 1…Rh2 2.Sg7 Rxb2 3.Kh8 Rxb3 4.Rh2 Rf3 5.Rh7 Rf8. When Black begins, the set play cannot be retained, e.g. 1.Kg7? Rh2 and the king prevents 2.Sg7. Instead, the black rook is unblocked on the a-file and that entails a different mating arrangement. 1.Sg7 Ra5 2.Kh7! Rxa3 3.Kh8 Rxb3 4.Ra8 Rb6 5.Rg8 Rh6. The black king makes an unexpected tempo move, and the two parts yield a pair of reflected echo mates. The black bishop is needed in the diagram to stop an alternative line, 1.Sg7 Rb5 2.Kh8 Rxb3 3.Ra1+ Kc2 4.Rh1 Rf3 5.Rh7 Rf8.

Andy Sag: In the main play, the black king must do a tempo move to allow White to unobstruct the black rook. In the set play, the tempo move is unnecessary and the mate is on the eighth rank instead of the h-file but conceptually the same configuration.
Jacob Hoover: Both mates are models. The presence of the black bishop totally threw me off. When I realized that it doesn't actually participate in either mate, the problem got a great deal easier to solve.
George Meldrum: Nice symmetry between solution and set play.
Nigel Nettheim: A very nice matching pair. The focussing on opposite corners is attractive. Not hard to solve because the black rook had to be freed as early as possible.

563

Comins Mansfield
Schach-Echo 1970

Mate in 2

Solution

Two prominent black checks in the diagram initiate the set play, 1…Qxc6+ 2.Rxc6 and 1…Bxg6+ 2.Rxg6, in which White employs the B + R battery. The thematic try 1.Se7? threatens 2.Bd3 by controlling d5, but since f4 is now solely guarded by the f6-rook, the set mates don’t work against the checking defences and are replaced by 1…Qxc6+ 2.Sxc6 and 1…Bxg6+ 2.Sxg6. The try is foiled by 1…c3! After the key 1.Sc3!, threatening 2.Bd3/Bd5, the set mates fail for the same reason. However, because the key-piece has cut off the b3-rook, White can utilise the R + B battery to deal with the black checks: 1…Qxc6+ 2.Bxc6 and 1…Bxg6+ 2.Bxg6. The three pairs of changed mates bring about the Zagoruiko theme, with the bonus feature that in each triplet of mates, different white pieces play to the same square. There’s by-play with 1…Rxc3 2.b8=Q and 1…Sf2/Sg5 2.f4.

Andy Sag: The key generates a double threat. The set checks are provided for by the diagonal battery but post key the vertical battery must be played instead.
Andrew Buchanan: The basic idea of two change mates between the two batteries is excellent, and the implementation is clean.
George Meldrum: Changed mates for Black’s checks are the main thing with the transition smooth and without a lot of distractions.
Nigel Nettheim: It’s surprising that the set discovered checks from h8 never materialise; they are replaced by those from e3 (thus file instead of diagonal play). The dual threat can hardly be called a weakness.

564

Enguelberto Berlingozzo &
Felix Sonnenfeld
Boletim da UBP 1981
J. Figueiredo Memorial Tourney, 1st Prize

Helpmate in 2, 3 solutions

Solution

A mate with the B + R battery is hindered by Black’s knight on d3 and three line-pieces converging on the same square. The knight needs two moves to unguard the diagonal and it uses them to simultaneously interfere with one of the line-pieces. White can shut off a second line-piece with the queen, and that leaves the third one to be handled by the rook when it fires the battery. 1.Se1 Qf5 (Qd4?) 2.Sc2 Rd4. 1.Sdxc5 Qd4 (Qc2?) 2.Se4 Rc2. 1.Sf4 Qc2 (Qf5?) 2.Sd5 Re4. An elaborate but clear demonstration of cyclic play, in which the three main actors (one black piece and two white ones) rotate their functions in closing three black lines. A white queen dual on the first move is avoided for disparate reasons.

Jacob Hoover: A curious cyclic shift is seen among the three solutions.
George Meldrum: Difficult to solve.
Andy Sag: White’s guard on the b4-pawn is maintained. The white king on f8 prevents a cook, 1.Sdxc5 Qf3 2.Se6 Qxc6. The black rook on a2 prevents a similar one, 1.Sxb4 Qc2 2.Sa6 Qxa4.
Nigel Nettheim: A great three-some. The d3-knight is the hero. Refuting the many temptations, including double-checks, was time-consuming.
Andrew Buchanan: Enjoyable with three thematic solutions and natural dual elimination.

565

Leonid Makaronez
OzProblems.com 18 Sep. 2021

Mate in 3

Solution

The key 1.e3! entails a sacrificial threat, 2.Rf5+ exf5 3.Qe7. The black bishop has three defences that are answered by different white queen checks. The self-block 1…Bxe4 enables 2.Qc3+ Kd5/Kd6 3.Rd7, 2…Sd4 3.Qxd4. A second self-block, 1…Bd5, gives 2.Qb8+ Kxe4 3.Qf4. And 1…Bc8 unguards the fifth rank to allow 2.Qc5+ Kxe4 3.Rf4. A good duel between the black bishop and white queen gives rise to a variety of mates, in a Meredith setting.

Andy Sag: The threat involves a second-move rook sacrifice to force Black to open the e-file. The set self-block on d5 is a clever touch.
Jacob Hoover: White response to each of three bishop defenses is different.
Nigel Nettheim: The key further surrounds the enemy king, although it does forfeit the two pawn-captures. But the variations are the highlight, and they resisted finding for some time.
George Meldrum: Beautiful.
Bob Meadley: A lovely light problem worthy of LM. The white king is close enough to tease.

566

Joost de Heer & Michel Caillaud
Probleemblad 2001

Solution

White is missing five units and since only six moves are available, all black moves except the first must have been captures. Given the five visible black pawn moves in the diagram, including the initial non-capturing one, four of these captures were made by the pawns. The main questions are thus how to quickly place four of the missing units in the paths of the black pawns, and how to eliminate the fifth white unit with Black’s remaining move. Only the missing black queen could have executed this capture, and time is saved if it removed the d2-pawn on its original square, on which the black piece could also be sacrificed to the b1-knight. 1.e4 d5 2.Bc4 dxe4 3.Be6 Qxd2+ 4.Sxd2 fxe6 5.Sdf3 exf3 6.Qd6 exd6. The try 1.e4 f5? 2.Bc4 fxe4 3.Be6 dxe6 4.?? opens the d-file too slowly for the black queen. The deceptive diagram position conceals the fact that Black’s original d-, e-, and f-pawns have cyclically switched their files.

Andy Sag: A unique solution involving capture of five white units, a black queen check and three black pawns ending up on different files.
Karel Hursky: I enjoyed this sparkling gem. A key to a quick solution is to realize how the pawns from d7 and e7 moved.
Jacob Hoover: We see three line-opening moves: (1) Black capturing on e4 to open the d-file; (2) the same move opens the c4-e6 line; and (3) the knight opening the d-file yet again later on.
Andrew Buchanan: Very tricky for a dozy Saturday morning!
George Meldrum: With five black pawn moves already on the table the last thing to choose is where the black queen moves to. At first d6 looked good, then d4, and at last d2, where the position finally played itself. Really weird and wonderful.

567

Udo Degener
Springaren 1988, 5th Prize

Mate in 2

Solution

The white king has four possible moves, any of which would threaten 2.Sg6. Since the threat-move interferes with the h7-bishop’s control of f5, Black can defend by cutting off (or removing) the a5-rook, anticipating its required guard on the same square when the fifth rank is opened. This type of defence is employed to defeat three of the white king’s moves. 1.Kf7? Rb5! (2.Rf8?), 1.Kh5? Qb5! (2.Sxd3??), and 1.Kh6? c5! (2.Qh6??). The key 1.Kg7! avoids the self-obstructions and the black queen's pin seen in the tries. 1…Rb5 2.Rf8, 1…Qb5 2.Sxd3, and 1…c5 2.Qh6. Also, 1…Bxa5 2.Rxd4. The subtle line-play involved – (1) White’s threat closes a white line of guard and (2) each of Black’s defences closes another white line to the same flight-square – is termed Theme A, one of the eight (A to H) lettered line themes.

Andy Sag: The vacation of g6 allows a knight to threaten mate relying on the a5-rook to guard f5. Four defences remove this guard but allow different mates. Incorrect king moves are tries.
George Meldrum: A strange setting with many pieces out of play. The variation involving the white queen is a gem.
Nigel Nettheim: Good tries and play – clear, if not specially deep.
Jacob Hoover: I’m kind of disappointed that there wasn’t a try that was defeated by 1…Bxa5, but I guess you can’t have everything, can you?
Andrew Buchanan: The bishop on h7 is very suspicious, and the equally suggestive empty space round the white king gives the necessary pointer. It’s a pity there is no try defended by 1…Bxa5. A good one!

568

Zdravko Maslar
feenschach 1986, 2nd Hon. Mention

Helpmate in 3, 2 solutions

Solution

In the first solution, the two sides arrange for the black king to be mated on c6 by the bishop along the long diagonal. 1.Bh3 Bh1 2.Bg2 Ra5+ 3.Kc6 Bxg2. Black’s initial move opens the diagonal for the white bishop, which traverses the line to cross over the critical square, g2. The black bishop then returns to g2, interfering with the white piece so that the black king could access the diagonal. After a rook check to guard the fifth rank, the king goes to c6 and the white bishop mates by capturing the piece on the interference square. The second solution shows analogous strategy aimed at mating the king on a6 with the white rook along the a-file. 1.Sb4 Ra1 2.Sa2 Bc6+ 3.Ka6 Rxa2. Now the black knight opens an orthogonal line on which the rook travels over the critical square, a2. A switchback by the knight to that square closes the same line and allows the black king to reach a6, after a bishop check that controls b5 besides b7. And the white rook mates by capturing the piece that caused the interference on a2. Terrific interplay between the white and black forces in this fine example of the Maslar theme, named after this composer.

Andy Sag: Well matched twin! In each case a black piece opens a line to allow the mating piece to do a critical move and then switches back to allow the black king to go into the mating line. The white king is cleverly placed to ensure only one square is available for the initial black move.

569

Brian Tomson
British Chess Magazine 1983, Version
6th Commendation

Solution

Given the placements of Black’s units and the availability of just three white pieces to self-block and give a final deflecting check, a selfmate can be set up only with the white king placed on h8 and the queen sacrificing itself on h5, to force the g5-rook to mate. The king thus makes a trip to the opposite corner, and it has to be chaperoned by the other white pieces to prevent checks by Black’s major artillery. During the series, White must also be careful to avoid checking the black king. 1.Se3 2.Bg8 3.Qb3. A Q + S battery is formed on the third rank. 4.Kb2 5.Kc3. The king masks the battery and stays put until the queen is off the line. 6.Sf5 7.Qe6. Another Q + S battery is created, now on a diagonal. 8.Kd4 9.Ke5 10.Kf6 11.Qf7. White dismantles the battery, making sure not to fire it with 11.Sh4+? 12.Sh4 13.Sg6 14.Kg7 15.Kh8 16.Se5. The only hideaway move for the knight. 17.Qh5+ Rxh5. Fantastic sequence in which the need to shield both kings from checks intricately determines the move order.

Andy Sag: Although it could get to h5 in one move, the queen (not the bishop) must first do some “heavy lifting.” Check avoidance ensures a unique order of moves.
Jacob Hoover: Zepler doubling [a line-play manoeuvre demonstrated here with Bg8 followed by Qb3 and Qe6], line closures, self-pins, unpins.
Karel Hursky: Amazing march of white king to h8. The white knight is a hero shielding the white king on his march. The 16th move Se5 was hard to see.
George Meldrum: A pity the white queen has its eye on the h5-square at the start as it helps give away the target. Having said that the sequence of moves to get to that point is amazing.
Andrew Buchanan: Three shielding units are perfectly orchestrated. Excellent problem, with a number of subtleties. Worthy of a much higher rating surely?

570

Eeltje Visserman
Sahs 1965, 2nd Prize

Helpmate in 2, 2 solutions

Solution

White has two indirect R + S batteries which, if opened, would control the flight-squares on the second and fourth ranks. Both knights are pinned, however, and neither could be unpinned effectively by Black on the first move. So in each solution, Black assists in opening one of the white rook lines by removing its blocking knight. 1.Bxb2+ Kf7 2.Bf6 Sd5 and 1.Rxf4+ Kg7 2.Rf6 Sd1. Black’s initial capture with a pinning piece checks the white king, which is forced to move along the pin-line of the remaining knight. Black then moves the capturing piece again, not only to open the white line but to unpin that knight, allowing it to execute an indirect battery mate. This great problem features the Zilahi theme and an unpinning Grimshaw on f6, the white king’s original square.

Jacob Hoover: A nice display of the Zilahi theme (the white knights swap roles between getting captured and delivering mate) as well as ODT (orthogonal-diagonal transformation) and unpins.
Andy Sag: Hard to solve because retreating the white king along the pin-line is counter-intuitive, but once you find one solution, the second one becomes obvious.
Nigel Nettheim: Without the c7-pawn, a third solution 1.Rc8 Ke5 2.Rc3 Sd1 would unpin a knight in a different way; but that would criminally distract attention from the wonderful and very closely-matching file/diagonal pair.
Andrew Buchanan: Very nice: clear, harmonious and economical, with natural dual elimination for the knight mates. I might prefer BP on b5 to WP on a4, since then all cook-blockers are black, with White focused on the core mating material.
Paz Einat: What a beauty! A classic.

571

Walther Jørgensen
Arbejder-Skak 1950, 1st Prize

Mate in 2

Solution

The black king has four diagonal moves in the diagram, each of which has been provided with a different mating response. 1…Kc4 2.Qxd4, 1…Kxc6 2.Qd7, 1…Ke4 2.Qe5, and 1…Kxe6 2.Sf4. A set mate is also prepared for the black bishop – the only other mobile unit – 1…B~ 2.Qe5, so the position is a complete block. White has no way of maintaining the zugzwang, however, e.g. 1.c7? Kc6! The key 1.Qf8! sets up another zugzwang and, splendidly, changes the reply to every king move. 1…Kc4 2.Qc5, 1…Kxc6 2.Qa8, 1…Ke4 2.Qf5, and 1…Kxe6 2.Sc7. White’s response to the bishop’s moves is replaced as well, with 1…B~ 2.Qf5, hence this mutate shows a complete change of mates.

Andy Sag: Complete block with changed mates for all five possible black moves including four flights. First prize well deserved for this mutate.
Jacob Hoover: A lovely mutate that also features star-flights.
Paz Einat: A perfect mutate with star-flight changes. The pinned knight is a crucial part of what makes this work.
Nigel Nettheim: “Twinkle, twinkle, little star”: the set “star-flight” variations go out, and then new ones come on. A real “First Prize” problem.

572

Edgar Holladay
The Problemist 1986

Solution

In this shape problem, Black has enough self-blocking material to arrange for either potential mating move, e4 or exf4. The latter option, however, turns out to require 10 moves, so both solutions involve placing the king on d5 for e4 mate. In such a mate, the black pawn can be utilised only on d4, while the c3-bishop is the only suitable piece for blocking e5, as anything else would guard e4 or check the white king. 1.Rd6 2.Rcc6 3.Sc5 4.Se6 5.Qc5 6.Be5 7.d4 8.Kd5 9.Bc4 e4, and 1.Sd6 2.Bf5 3.Qf6 4.Be5 5.d4 6.Kd5 7.Sc4 8.Qd6 9.Be6 e4. The remaining black pieces all change their blocking squares across the two phases (the knight and d3-bishop exchange positions while the three major pieces shift their places cyclically). The striking triangular shape of the diagram is transformed into a square at the end of each part.

Andy Sag: Two ways of self-blocking the king on d5 to allow e4 mate. Check avoidance makes each sequence unique. An exf4 scenario with the king on e5 can only be done in 10 moves, so I guess you would call that a try (1.Re6 2.Sd6 3.Bf5 4.Qf6 5.Be5 6.Bf4 7.d4 8.Kd5 9.Ke5 10.Rd5).
Jacob Hoover: Both mates are ideal mates in which every square around the black king is occupied. It didn’t take me very long to find the first solution, which solves almost like one of those sliding-panel puzzles. The second solution, however, almost had me stumped until I realized the mating square was the same as in the first solution.
George Meldrum: Attractive layout and the rook play solution was challenging to find.
Nigel Nettheim: Extraordinary. A triangle becomes a square. Two 10-move tries with the black king elsewhere, thus providing greater contrast, were very distracting.

573

Leonid Makaronez
OzProblems.com 13 Nov. 2021

Mate in 3

Solution

The black king has two flights initially, of which one is provided (1…Ke3 2.Qh3). The key 1.Se4! removes the c2-flight but concedes two more on e4 and c4. The threat is 2.Qh3+ Kxe4 3.Re2, 2…Kxc4 3.Qb3. If Black takes the offered knight, White proceeds to sacrifice the queen as well: 1…Kxe4 2.Qe5+ Kxe5 3.Re2, 2…Kd3 3.Qe2, with two mates on the same square. After 1…Kxc4 2.Sd6+, the black king again has a choice of flights which leads to different mates, 2…Kb4 3.Qb5, 2...Kd3 3.Qh3. Lastly, 1…Rb7 answers the threat by controlling b3, but now 2.Sd6 and 3.Qh3 is unstoppable due to the lack of 2…Bxc6. Including the threat play, the black king gets mated on five different squares.

Andy Sag: The sacrificial give-and-take key removes an unprovided flight to c2 and is hard to see because of the surprise queen sacrifice on second move. Also noticed how the rook blocks …Bxc6 in the non-checking variation.
Nigel Nettheim: The sacrifice of the knight and queen in the main variation is good and surprising.
George Meldrum: Nice double sacrifice, and the black king gets to transverse over many different squares. The black rook moving to b7 interfering the bishop line is a great subtle defensive line neatly handled.

574

Zivko Janevski
Mat 1978, Prize

Solution

Black’s diagonal half-battery arrangement aimed at the white king suggests that when either black knight moves, White will force the other one to give a battery mate. Two symmetrical queen tries target squares controlled by the knights for this purpose: 1.Qg7/Qh8? (waiting) Sb~ 2.Qe5+ Sxe5, 1…Sd~ 2.Qd4+ Sxd4, but 1…Se5! refutes, and 1.Qf8? (waiting) Sd~ 2.Qd6+ Sxd6, 1… Sb~ 2.Qc5+ Sxc5, but 1…Sd6! refutes. A third try, 1.Qg6? (waiting), takes a slightly different approach: while 1…Sd~ 2.Qd6+ Sxd6 is unchanged from the 1.Qf8? try, 1…Sb~ 2.Qxd3+ Bxd3 sees a new, non-battery mate induced. This try is defeated by 1…Sd4! since 2.Qxd3 can be answered by 2…Bb5/Bc4. The key 1.Qe8! (waiting) works in a similar way: 1…Sb~ 2.Qe5+ Sxe5 repeats a line from the 1.Qg7/Qh8? try, and with 1…Sd~ 2.Qxb5+ Bxb5 the bishop is compelled to mate directly. It’s amusing that 1…Sc5 – analogous to the 1…Sd4! refutation of 1.Qg6? – fails to 2.Qxb5 Bxb5, when the same bishop mate occurs due to zugzwang.

Andy Sag: Black is immobile except for the two knights. The associated half-battery makes the solution rather obvious; just need to find the right moves for the queen.
Jacob Hoover: After the correction 1…Sc5, White still plays 2.Qxb5, only this time it’s a pin rather than a check. However, since the queen is right in the bishop’s face, so to speak, Black’s only legal move is to play 2…Bxb5.
Nigel Nettheim: Quite a lot has been extracted from a fairly simple situation.
Andrew Buchanan: Fun to solve, attractive geometry, quite easy. The black king is in the middle of empty space (I don’t like the usage of the term “mirror”.)

575

Arnoldo Ellerman
Arbejder-Skak 1952, 1st Prize

Mate in 2

Solution

The set play comprises four variations, three of which involve Black blocking a flight-square: 1…Sb3 2.Qf1, 1…Bb5 2.Bf7, and 1…Sd4 2.Sd6. In the fourth, Black commits a simple unguard error, 1…Bb8 2.Rxc5. The excellent key 1.Sd7!, by cutting off the d8-rook and e8-bishop, creates three flights for the king, and threatens 2.Se5. When the king escapes to these new flights, White brings back the mating moves used in the set play against other defences. 1…Kd3 2.Qf1, 1…Kd5 2.Bf7, and 1…Kb5 2.Sd6. Hence three mates are harmoniously transferred from the self-blocking moves to the flight-taking ones. The fourth set variation produces the more standard type of changed play, 1…Bb8 2.Sb6, a new mate rather than a new defence.

Andy Sag: The key gives three flights. The key piece moves again to carry out the threat but stays put in all flight variations. The non-flight variation has a changed mate.
Nigel Nettheim: The h8-bishop had to take part, and the white f-pawns suggested that the black king might approach them; but still the key, interfering with white’s own pieces to give three flights, was not at all obvious. The tries with the d8-rook were very simply refuted.
Jacob Hoover: Each of the black king’s three flights lines the black king up with a battery, but none of these batteries fires.
Brian Stephenson: The h8-bishop must be activated, so the f6-knight must be the key-piece. 1.Sd7 gives three flights, so let’s try it first. A welcome traditional two-mover.
Andrew Buchanan: Finding the key was fairly easy but spotting the pattern much harder.

576

Jonathan Penrose
The Problemist Supplement 2002

Solution

Using the four moves to promote the white pawn seems likely, though the obvious plan of mating with a queen promotion fails, e.g. 1.Kf7 c5+ 2.Kf8 c6 3.Rh7 cxd7 4.Rg7 and the f6-knight prevents 4…d8=Q mate. Instead, White aims for a knight promotion mate on e8 with the black king placed on f6. 1.Se8 c5+. Now if 2.Kf6? c6 3.Rg6 cxd7, Black lacks a waiting move that would permit 4…dxe8=S mate. So Black wastes a tempo with 2.d5! first, banking on the en passant rule to still allow the c-pawn to reach the d-file. 2…cxd6 e.p.+ 3.Kf6 d7 4.Rg6 dxe8=S. A crisp helpmate featuring two of the special chess moves and an ideal-mate, from the famous GM-player who passed away on 30 Nov. 2021

Andy Sag: Sub-promotion to a knight on e8 leaving only two black pieces to self-block adds to the challenge. Strangely, if 2.Kf6 the pawn can still promote on e8 but Black reaches the final position too early, so 2.d5 followed by the en passant capture provides the necessary tempo. This problem has a bit of everything!
Mark Salanowski: Difficult but enjoyable. Hard to see the final position with four possible squares for the c-pawn to promote to.
George Meldrum: A terrific e.p. idea to get the tempo right and a timely inclusion of this problem to celebrate this man’s life.
Karel Hursky: Very witty and more memorable than his win against Tal. RIP Mr. Penrose.

577

Carel Mann
De Amsterdammer 1893

Mate in 4

Solution

Black has only two legal moves in the diagram, both of which would enable an immediate mate: 1…Kf5 2.Qd5 and 1…f5 2.Qh4. So White aims to establish the same position but with Black to play, by means of a triangulating manoeuvre. 1.Qa8! Kh6/Kh5 2.Qh8+ Kg5 3.Qh1 Kf5 4.Qd5, or 3…f5 4.Qh4. There’s by-play with 1…f5 2.Qd8+ Kh6 (2…Kh5 3.Qh4) 3.Qh8+ Kg5 4.Qh4. A classic lose-the-tempo problem, making full use of the board.

Andy Sag: The queen does a wide range triangle to return to the set position with Black to move. Nice miniature!
Jacob Hoover: A pseudo-two-mover. White completes a rundlauf with 3.Qh1.
Karel Hursky: Always enjoy solving problems where a queen can access all the corners of a chessboard.
Nigel Nettheim: Triangulation, not by the white king as in pawn endings but by the queen. Very nice.
Andrew Buchanan: Triangulation doesnt come much bigger than that! :-)

578

Andrey Frolkin & Kostas Prentos
feenschach 2004
Champagne Theme Tourney, 4th Prize

Solution

The two promoted black queens require a minimum of 12 moves to reach their positions: five to promote on g1 plus one diagonal move for each piece. Single moves by the d7-knight and h3-bishop, plus two by the g1-rook account for the remaining four moves available. Given the white pawn on g2, both promotion moves were captures on g1 (from h2 to avoid disturbing the white king), and the checking black rook implies another capture on the same square. White is missing only two knights and two pawns, hence one of the latter must have promoted, to enable a third capture on the first rank. 1.f4 g5 2.f5 g4 3.f6 g3 4.fxe7 gxh2 5.exf8=S. Since Black has no spare tempo to shift the d7-pawn, a promoted knight is the fastest option for removing that unit. 5…hxg1=Q 6.Rh5 Qb6 7.Rc5 h5 8.Sxd7 h4 9.Se5 h3 10.Sf3 h2 11.Sg1 hxg1=Q 12.Sc3 Qd4 13.e3 Rh1 14.Se2 Bh3 15.Sg1 Sd7 16.Ba6 Rxg1+. At move 11, the promoted knight lands on g1, replacing a captured knight on its home square, and that constitutes the Pronkin theme. This knight is then captured by Black, and that effects the Frolkin theme – the capture of a promoted piece. The two proof-game themes are thus rendered with the same piece, and this unusual blend is enhanced by a triplet of knight-captures occurring on the same square.

Andy Sag: Three captures on g1 are needed so one of the captured pieces has to have been promoted. The need to capture the d7-pawn dictates promotion to a knight as using the c5-rook (with a bishop promotion) takes too long.
Jacob Hoover: The hardest part of the problem was figuring out the identity of the promoted white piece. One white underpromotion and two black promotions; pin and unpin.
George Meldrum: Black’s line of play is tight and so it is only the need to find White’s line of play which is by no means easy.

579

Thomas R. Dawson
Falkirk Herald 1914

Mate in 3

Solution

The bishop is preventing the knight from accessing g7 and c7, in each case followed by an unstoppable mate. If Black is to move, the bishop cannot maintain its focus on the two squares, so White wants to preserve the zugzwang with a waiting move. Despite the seemingly symmetrical position and the availability of four mobile pawns, only 1.h5! solves. 1…Bd6/Bc7/Bb8 2.Sg7 B~ 3.Sf5, and 1…Bf6/Bg7/Bh8 2.Sc7 B~ 3.Sd5. Not 1.b5? Bc7! 2.Sg7 Ba5+ (2.Sxc7 stalemate), or 1.g6? Bf6! 2.Sc7 Bxh4+, or 1.c6? Bd6! 2.Sg7 Bxb4+. Good illustration of asymmetry and a Christmas tree problem to boot!

Andy Sag: A Christmas tree asymmetric requiring a tempo key. The correct pawn move prevents the bishop from threatening disrupting checks.
Jacob Hoover: A fine example of focal play.
George Meldrum: Only one solution – remarkable.
Bob Meadley: Haven't seen this cute one by TRD.
Nigel Nettheim: Happy Christmas to all, as the pictorial diagram indicates! The defence 1…Bg7 (2…Bi5+??) is ruled out by the edge of the board.
Andrew Buchanan: A charming problem, very decorative and with three thematic tries.
Dennis Hale: Effortlessly exquisite.