Weekly Problems 2020-B

Problems 502-527

502

Karel Hursky
OzProblems.com 4 Jul. 2020

Mate in 2

Solution

The black king has a flight-move that initially produces a double pin-mate: 1…Kxe4 2.Qh4. Another set variation exploits the masked Q + S battery: 1…Bf4/Bg3 2.Sxf5. Both of these lines are abandoned by the key, 1.Qa8!, which threatens 2.Qxd5. Now 1…Kxe4 results in an even better triple pin-mate, 2.Rxc4. Three other captures of the e4-pawn, involving self-blocks and unguards, are met differently: 1…Qxe4 2.Qa1, 1…Sxe4 2.Se6, and 1…dxe4 2.Rxc4. Lastly, the f5-knight prompts two distinct mates on e3: 1…Sxe7 2.e3 and 1…Se3 fxe3. Multiple defences on the same square, and the two queens operate from all four corners of the board. (If the f7-rook is placed on f4 instead, that brings about another thematic variation, 1…Rxe4 2.Sxf5, but the important set mate for the flight would be lost.)

Karel Hursky is an Australian problemist who resides in Sydney. He is an active tournament player and composes only occasionally, but has a number of high-quality selfmates and helpmates under his belt.

Andy Sag: The two exd captures create vertical flights. That and the clear eighth rank and a-file make the key rather obvious. However, the changed mate with the addition of a third pin is quite cute.
Nigel Nettheim: Not hard to solve (…dxe4 is unprovided) but the highlight is the play, especially the four captures on e4 and the white queen’s baseball-like run to second base.
George Meldrum: I love this problem. The queen abandons the set play after the king flight to set up an even more spectacular response to the king move, truly the pinnacle.
Ian Shanahan: The set double pin-mate after the flight-capture is spectacularly changed to a different, triple pin-mate! Then we see three self-blocks on the flight-square and two more pawn mates thrown in for good measure. It’s a pity that 2.Rxc4 is repeated. Anyway, a grandiose two-mover in traditional Good-Companions style.

503

James Malcom
OzProblems.com 11 Jul. 2020

Solution

White’s plan is to add a second attack on f8 so that Rxf8+ would force Black to recapture with the h7-knight and mate by discovery on the h-file. Hence White seeks to double the rooks on the eighth rank, and out of four candidate moves by the c8-rook, only 1.Rd8! works, for a reason not apparent until eight moves later. Black defends by queening various pawns and checking on c1, in an effort to draw one of the rooks away from its threatening position. 1…d1=Q 2.Rcc8 Qc1+ 3.Rxc1 fxg1=Q 4.Rcc8 Qc1+ 5.Rxc1 g1=Q 6.Rcc8 Qc1+ 7.Rxc1 g2 8.Rcc8. Now 8…g1=Q? is too slow against 9.Rxf8+; instead, Black guards the f8-knight directly with the newly mobilised bishop, 8…Bd6. The farsighted key, placing the rook on the d-file, has provided for this defence: 9.Rxd6. Still, the bishop sacrifice has gained a tempo for Black and makes 9…g1=Q viable, to further delay White’s goal. After 10.Rdd8 Qc1+ 11.Rxc1, however, there’s no further stopping 12.Rcc8 and 13.Rxf8+ Sxf8. An impressive selfmate sequence that delivers four black queen promotions, shown with great precision.

Composer: The concept isn't entirely new (see my selfmate in 15 on superproblem.ru) but it has original elements to it, such as the key and the unique order of Black’s play. (The b4-pawn serves two purposes – it prevents 2…Qd2+ 3.e3 Qb4 and gives Black enough moves such that mopping up the black pawns fails in time.)
Andy Sag: White can play Rxf8+ once both rooks are on eighth rank. Black keeps delaying this by checking on c1 or guarding f8 until eventually running out of pieces.
Jacob Hoover: 1.Ra2? is a fourth try. This is the longest selfmate I have attempted to solve by a longshot.
Nigel Nettheim: An imaginative scheme, with c1 visited eight times. White’s first move anticipates black’s eighth.
George Meldrum: With the idea of placing both white rooks on the eighth rank the journey of solving this problem begins. It is much later in the journey that White’s first move is discovered when the bishop move needs to be dealt with in this crazy problem – crazy good.

Update (28 Aug. 2020): Congratulations to James who has further improved the selfmate in 15 linked above. He has achieved seven black queen promotions in a selfmate in 17, an amazing new task record.

504

Anatoly Vasilenko
Olympic Tourney, Thessaloniki 1984-88
2nd Prize

Mate in 2

Solution

Three set variations utilise the white queen to mate on the e-file: 1…fxe6 2.Qxe6, 1…fxe5 2.Qxe5, and 1…Sg4 2.Qxe4. The curious half-battery arrangement on the fifth rank means that shifting the queen will create the threat of 2.exf6. In three thematic tries by the piece, it manages to preserve two of the set mates on the e-file but fails to account for the third defence: 1.Qd4? fxe6!, 1.Qc4? fxe5!, and 1.Qd6? Sg4! Such a scheme, in which three black moves rotate their functions as normal defences and foiling responses against three tries, is called cyclic refutation. The key-move naturally caters for all three defences, but 1.Qd2! does that by setting up new mates for them: 1…fxe6 2.Bg6, 1…fxe5 2.Qg5, and 1…Sg4 2.Qf4.

Andy Sag: The c6-pawn indicates a queen key but all three set mates must be abandoned.
Andrew Buchanan: Various changed mates, etc: good stuff.
Nigel Nettheim: The “she stoops to conquer” theme, though she was sent on her way by the c6-pawn.
Jacob Hoover: An excellent two-mover.
Ian Shanahan: There are three tries where two of the set-mates work, but the third refutes the try (cyclic refutation). Only one destination works, whereby all three defences lead to changed mates. An elegant and clever problem!

505

Robert Burger & Robin Matthews
British Chess Magazine 1988, 1st Prize

Mate in 3

Solution

White cannot open the B + P and R + P batteries aimed at the black king solely because the pawns involved are blocked by another pair of white pawns. If White begins with a king move, that will threaten 2.c8=S and 3.Sb6 or 3.c7, but 1.Kd7? is too weak against 1…Sxf6+ while 1.Kb8? fails to 1…Sxc7! followed by 2…Sa6+. The key 1.Kd8! avoids these spoiling checks, and Black’s only defences are to use the knight to capture the pawns that are obstructing the battery play. After 1…Sxc7, White continues with 2.e8=S to threaten 3.Sxc7, and a random knight defence, 2…S~, allows 3.c7 – White exploits the annihilation of the c7-pawn to fire the B + P battery. The correction move 2…Se6+ activates the R + P battery instead, 3.fxe6. The other thematic defence, 1…Sxf6 (which deters the 2.c8=S threat with 2…Sxe4!), induces a different promotion by the same pawn, 2.e8=Q. Now the threat of 3.Qxf7 is answered by any knight move (as it opens the rank for the black rook), and 2…S~ facilitates 3.f6 – effecting a similar annihilation of the f6-pawn to free the R + P battery. The knight has two corrections: 2…Sd7 triggers the B + P battery, 3.cxd7, and 2…Sxe4 permits 3.Qxe4. Wonderful reciprocal battery play with two pairs of matching mates along diagonal and orthogonal lines, all set off by a single defender.

Andy Sag: Clever play with sub-promotions and battery mates.
Nigel Nettheim: The dual third-move threats are separated in 1…Rh3 2.c8=S Rxd3 3.c7 and 1..Rh1 2.c8=S Rc1 3.Sb6. The different promotions on e8 are attractive too. Not hard to solve because the white bishop was likely to take part.
Jacob Hoover: In the first line the battery on the fifth rank fires in response to the random knight move while a correction forces the firing of the battery on the long diagonal, whereas in the second line the reverse happens.

506

Mieczysław Zalewski
Szachy 1970, 1st-2nd Prize =

Mate in 2

Solution

Two set mates by the e8-knight close white lines of guard but are playable after black self-blocks: 1…Se5 2.Sd6 and 1…Sg5 2.Sg7. In two thematic tries, White threatens each of these knight mates, but Black refutes them by opening lines for the h2-bishop and g1-rook: 1.Qc3? (2.Sd6) but 1…f3! and 1.Qh4? (2.Sg7) Se5 2.Sd6, but 1…Sf2! The Novotny key 1.Qg3! cuts off the same two black line-pieces to threaten 2.Rxf4 and 2.Bxg4. These threats are separated through captures of the key-piece: 1…Rxg3 2.Rxf4 and 1…Bxg3 2.Bxg4. The main variations (where both threats are stopped) bring back the knight mates seen in the virtual play, but in a paradoxical way. The defence 1…f3, which disables the threat of 2.Sd6 after the 1.Qc3? try, now opens a white queen line and enables 2.Sd6. Likewise, 1…Sf2, which disables the threat of 2.Sg7 after the 1.Qh4? try, now opens another queen line and enables 2.Sg7. Such a pattern of recurring moves with ironically changed functions is called the Dombrovskis theme.

Paz Einat: A crisp Dombrovskis. The Novotny is perfectly used to close the thematic lines that feature in the refutations of the two tries.
Nigel Nettheim: Quite intricate. The h6-pawn is justified by turning 1.Qh4? into a try, matching the 1.Qc3? try.
Jacob Hoover: In the set play we have a Grimshaw pair (1…Rg3 2.Rxf4, 1…Bg3 2.Bxg4) and a pair of self-blocks (1…Sg5 2.Sg7, 1…Se5 2.Sd6). The try 1.Ba4? (threat: 2.Bd7) preserves the set self-blocks and knight mates completely but dies to 1…Sf8! The key is a Novotny interference. The two threats are separated and in addition, the key transfers the other two set mates to different defenses (1…Sf2, 1…f3), each one of which opens a guard of one of the squares that were involved in the set self-blocks.
Ian Shanahan: Wow! What a rich melange of thematic ideas! Firstly, we observe mate transference between set and actual play, self-blocks plus white interference in the first half of the set play, and the Grimshaw theme in its second half. The key itself, with its double interference, flaunts the Novotny theme. The mates threatened by the tries follow the very defences which refute the respective tries – a paradox known as the Dombrovskis theme. The only tiny blemish I can see is that in the second try (1.Qc3?), there is no variation leading to 2.Sg7 – as a counterpoise to the first try's 1.Qh4? Se5 2.Sd6. The composer is to be congratulated! This has to be one of the most beautiful two-movers that I've encountered in ages.

507

Michel Caillaud
Orbit 2002, 2nd Prize

Solution

Black has made a total of ten “visible” moves: four by the pawns, four by the g1-knight, and two by the queen to get captured on b3. That leaves four spare moves, at least some of which were used by the king to clear a path for White’s c7-rook, which arrived there after capturing the trapped f8-bishop. This rook trek took six moves (a five-move trek via c8 would fail due to the black bishop’s obstruction by the black king on d7); three other moves by the pawns plus one by the d2-knight total ten white moves. Hence White also has four spare moves, and again the king needed at least some of them, because Black’s g1-knight must have arrived from f3, where it attacked e1. 1.d4 Sc6 2.Kd2 Se5 3.Ke3 Sf3 4.Sd2 Sxg1 5.f3 c6 6.Kf2 Qb6+ 7.Ke1 Qb3 8.axb3 f5 9.Ra5 f4 10.Rf5 d6 11.Rxf8+ Kd7 12.Rd8+ Ke6 13.Rd7 Kf7 14.Rc7 Ke8. The solution thus requires each king to complete a four-move circuit, both of which are remarkably capture-free.

Andy Sag: Ironically, the word “Orbit” is a clue?! If the black king can go round in a circle, so can the white king!
Jacob Hoover: A black king rundlauf is required, and the white king also completes a rundlauf! Amazing!
Nigel Nettheim: Astonishing. Both kings are sent into “Orbit”, which was certainly the right magazine for this one! Those two Rundläufe arise in very different circumstances, and the need for them wasn’t at all obvious. Hard to solve because of the many tempting false lines.
Andrew Buchanan: Amazing problem. It all depends on timing – the lengths of the different strands of moves can only be assembled in a certain way, to force the kings to execute their elegant circuits. Sublime.

508

Aleksandr Korepin
Tchigorin Memorial Tourney 1938, 2nd Prize

Mate in 2

Solution

White’s e5-bishop has five square-vacating moves on the long diagonal that threaten 2.Se5. Black’s d6-knight has four possible moves, all of which stop the threat by activating the c7-bishop, but in each case the knight finds itself closing a black line and enabling a different set mate: 1…Sf7 2.Qxe6, 1…Sf5 2.Rc3, 1…Se4 2.Rd4, and 1…Sb5 2.Sb2. However, four of the white bishop’s moves likewise interfere with a white piece responsible for one of the set mates. Thus Black is able to refute these try moves by choosing the knight defence that cannot be exploited by White: 1.Bf6? Sf7!, 1.Bd4? Se4!, 1.Bc3? Sf5!, and 1.Bb2? Sb5! Only 1.Ba1! solves, avoiding all of the self-interferences. 1…Sf7 2.Qxe6, 1…Sf5 2.Rc3, 1…Se4 2.Rd4, and 1…Sb5 2.Sb2. The by-play repeats some thematic mates: 1…Qxg7 2.Qxe6, 1…Rb2+ 2.Sxb2, and 1…Bxd3+ 2.exd3. A bishop vs knight duel is brought about with intensive strategic play by both sides.

Andy Sag: Clearly the bishop must vacate e5 for the knight but only the long-range move works. Other bishop moves are tries which set defences take turns in refuting. Unpins to activate the d3-rook are also tries.
Jacob Hoover: White must choose the bishop's destination carefully (“safety play”) in order to avoid getting in the way of another white unit (errors committed by the thematic tries).
Nigel Nettheim: An unlikely-looking key following a good set of thematic tries, and a half-wheel with the d6-knight. The plugs on c8, e8, f4 and g7 had to be tolerated.
Andrew Buchanan: Wonderful problem.
Ian Shanahan: A textbook example of white-obstruction tries forcing the key-bishop to move all the way to the corner in order to avoid wrong-footing his fellow white officers. The unpins of the white rook are nicely separated, though there are rather too many plugs (black and white) present for my taste in this congested position.

509

Linden Lyons
The Problemist Supplement 2014, Commendation

Solution

The black king has three diagonal flights in the diagram, and the waiting key 1.Gf5! grants a fourth on c4. Now wherever the king goes, if White places a unit on e5, that will enable the key-grasshopper to attack d5 and prevent the black piece from returning to that square. The right choice of unit to go on e5 depends on the specific diagonal move made and the resulting new flight-square that must be covered. 1…Kxc4 2.Ge5 (guarding c3), 1…Kc6 2.Be5 (c7), 1…Kxe6 2.Sde5 (d7), and 1…Ke4 2.Sce5 (f3). The star-flights theme is hence combined with four distinct white moves to the same square.

Karel Hursky: Delightful “divertimento”! It made my chess day. Love the stalemate after 2.Ge5.
Ian Shanahan: It was immediately clear to me (on thematic grounds!) that the key would yield the fourth of the black king’s diagonal flights, and so that stricture permitted just one possible opening move – 1.Gf5! I then just needed to ascertain White’s stalemating move after each flight was taken. Already we recognize the presence of the venerable star-flight theme, only to be delighted now by the additional thematic idea that White’s second moves are all to the very same square, e5. A most gratifying blend!
Andrew Buchanan: That 1.Gf5 unguards the fourth diagonal square is suspicious. But I spent a long time being distracted by the e7-pawn which turns out to be unthematic. I suggest removing it as seen in the version below. That allows the solvers to focus on the black king flights, and the occupation by four different units of e5.

509v

Linden Lyons
The Problemist Supplement 2014, Commendation
Version by
Andrew Buchanan

Stalemate in 2, Grasshoppers a1, a7, c5, c8, d4

Andrew’s setting (solution unchanged) involves an extra grasshopper but saves two pawns, to attain a striking rex solus position in Meredith.

510

Leonid Makaronez
OzProblems.com 29 Aug. 2020

Mate in 3

Solution

A set Grimshaw occurs on c4, albeit with one short mate: 1…Bc4 2.Qe5+ Se4 3.Qxe4 and 1…Rc4 2.Qxd3. The key 1.Sa3! threatens 2.Sc2+ dxc2 3.Qd2. After 1…Se4, White compels the mutual interference on c4 with a Novotny sacrifice, 2.Sc4+ Bxc4 3.Qxe4 and 2…Rxc4 3.Qxd3. Black has two defences on d4, self-blocking moves that release the white queen from guarding that square: 1…Rd4 2.Qb3 (threat: 3.Sc2) Rc4 3.Qxd3 and 1…Bd4 – also cutting off the a4-rook – 2.Qg5+ Rf4 3.Qxf4. Similar to the set 1…Bc4, 1…Rxa3 permits 2.Qe5+ Se4 3.Qxe4. Immediate mates follow other defences: 1…d2+ 2.Qxd2 and 1…Rc4 2.Qxd3.

Andy Sag: The Novotny on c4 and Grimshaws on c4 and (partially) on d4 make a nice theme.
Jacob Hoover: It seems that the only white unit not participating is the b1-knight, suggesting that this could be the key-piece, but a random move of this knight allows 1…Ra1+! with mate soon to follow. But 1.Sa3! averts this disaster and also prepares the full-length threat.
Andrew Buchanan: The key is easy here; discovering the black defences and their defects is then fun. Very varied play in a light and attractive position.

511

John Ling
British Chess Problem Society Ring Tourney 1962, 4th Prize

Mate in 2

Solution

In this complete block position, set mates are prepared for every possible black move. The black queen is stopping mates on b2 and c3, and the majority of its moves, 1…Q~ (like 1…Qc5/Qxd3), allow the former, 2.R3b2, while 1…Qc3 forces the latter, 2.Sxc3. Three of the queen’s defences prevent both mates; however, 1…Qxb3/Qc1 self-pins the d2-pawn and permits 2.Sc1 – note how the knight opens the B + S indirect battery to give the f5-bishop control of b1 after 1…Qxb3 – and 1…Qxb1 is a self-block that prompts 2.Ra3. Promotion moves by the d2-pawn self-pin the black queen, thereby completing the half-pin on the second rank: 1…d1=Q 2.Sc3 and 1…d1=S 2.Sc1. Numerous attempts by White to preserve the block just fail. 1.e4? or 1.Bg4/Bd7/Bc8?, by discarding the indirect battery, gets refuted by 1…Qxb3! White lets the black queen check with 1.Be4/Bg6/Bh7? Qc8+!, 1.Kg4/Kh4? Qc4+!, and 1.Kg3/Kh2? Qc7+! Other tries are 1.Bc5/Bd6/Be7/Bf8? Qc5! and 1.Qh2? f2! The key 1.Be6! (waiting) does abandon the indirect battery but compensates by forming another masked B + S battery. Now 1…Qxb3 leads to a changed pin-mate, 2.Sc3. The remaining variations follow the set play in this memorable mutate.

Andy Sag: Complete block waiter with one changed mate, good variety including pin-mates and many tries.
Nigel Nettheim: The key-piece takes up a more attacking position so the key, though not quite obvious, is not surprising. The try 1.e4? is good, the other tries more trivial. The mate after 1…Qxb3 is nicely changed and is the main feature. An unimportant dual remains from the set play after 1...Qd1. The f3-pawn neatly prevents some cooks.
Andrew Buchanan: Fortunately, I had already thought 1…Qxb3 should be handled by a pin, so 1.Be6! and 1…Qxb3 now sees 2.Sc3 instead of 2.Sc1 in the set play. Harder to solve if one had already looked at the set play maybe, which fortunately I hadn’t. I love zugzwang positions like this.
Ian Shanahan: What a lovely, piquant half-pinner in mutate form! In this complete block, there is a single mate-change after 1…Qxb3. It’s not very common to see a black pawn on the verge of promotion being employed in a complete half-pin, as here. Great work by Mr Ling!

512

Geoff Foster
The Problemist Supplement 2018

Solution

All units are on their game-array squares, making this helpmate an example of a homebase problem. The white pawn seems likely to promote, but no mating configuration can be arranged in time if it becomes a queen, e.g. 1…e4 2.Kd7 e5 3.Kc7 e6 4.Kb7 e7 5.a6 exd8=Q 6.Ka7 and the c8-bishop hinders 6…Qc7 mate. The solution sees the pawn promoting to a knight instead: 1…e4 2.Qd5 exd5 3.Sc6 dxc6 4.Bd7 c7 5.Rd8 c8=S 6.Se7 Sd6. A particularly eye-catching homebase diagram is complemented by these features: white pawn minimal, Excelsior, underpromotion, and model mate.

Andy Sag: Neat partial game array helpmate where the black pawn ensures a unique solution by avoiding cooks with rook moving via a7 (e.g. 1…e3 2.Ra7 e4 3.Bg7 e5 4.Rf7 e6 5.Qd7 exd7+ 6.Kf8 d8=Q).
Andrew Buchanan: It’s amazing that the solution is unique given the chaos of related positions, and that makes it all the harder to solve.
Mark Salanowski: Thoroughly enjoyable problem. Every black piece apart from the pawn is required to deliver the mate.
Ian Shanahan: One of a series of complete homebase problems Geoff composed at the time. Very pretty, and not too hard to solve!
Karel Hursky: Loved this entertaining problem. Like a chess player’s nightmare coming true! Would like to share a few lines of the poem “Excelsior” by Henry Wadsworth Longfellow.
‘And from the sky, serene and far,
A voice fell, like a falling star,
Excelsior!’

513

Peter Schmidt & Arno Tüngler
Schach in Schleswig-Holstein 2006, 1st Prize

Solution

The black king is already totally confined, with all flight-squares blocked by black pieces or controlled by white ones. The d7-knight seems primed to give a pin-mate on c5 or f6, if either square is unblocked. Black accordingly aims to remove one of the obstructing pawns in each solution. The best candidates for doing this are the d4-queen and the f5-rook, but both are pinned, so Black begins by capturing one of the pinners. 1.Rc3 2.Rxc4 3.Qxc5; now Black re-blocks the two created flights with 4.Qe3 5.Rd4 – a critical move over d4, followed by an interference on that square, to enable 5…Sc5. Similarly, after 1.Bh5 2.Bxg6 3.Rxf6, the self-blocking moves 4.Rf3 5.Bf5 involve critical play across f5 and a self-interference on that square, permitting 5…Sf6. Each of the two analogous solutions also shows a pair of black pieces swapping their positions. (The pawn on h3 is required to stop a cook, 1.Bg2 2.Kf3 3.Kg4 4.Kh3 5.Qc3 Bxf5.)

Andy Sag: In each case a pawn removal vacates a mating square for the d7-knight, two black pieces swap to restore set self-blocks after one of them removes a set pin and the finale uses the remaining set pin for a pin-mate.
Andrew Buchanan: How clever! Seemed to be impossible, then I came back and it was obvious.
Ian Shanahan: During each black sequence, we see the mobile units (all of which serve to blockade the black king) firstly capture a white pinner at the second move, then exchange squares between their initial position and the end of the sequence (platzwechsel), resulting in a pin-mate. Overall, the two solutions exhibit an ODT – ortho-diagonal transformation – with perfectly matching strategy. Flawless harmony!

514

Paz Einat
Variantim 1997, Commendation

Mate in 2

Solution

The black bishop on e5 has two prominent captures that are met with set mates: 1…Bxd4 2.Qxg2 and 1…Bxd6 2.Qf7. In each case the queen needs to regain control of a flight (c6 or c4) that was previously covered by a captured knight. The thematic try 1.Rc8? (threat: 2.Rc5) attacks c6 and c4 but unguards the e-file, and this leads to a reciprocal change of mates after the bishop defences: 1…Bxd4 2.Qf7 and 1…Bxd6 2.Qxg2. Here the queen has to maintain its control of a flight (e6 or e4) that a captured knight no longer observes. This try, however, is defeated by the pinning 1…Rh8! The key 1.Qc2! (2.Qc5) uses the queen instead to cover the c-file and it generates a third pair of mates against the same bishop moves: 1…Bxd4 2.Qc4 and 1…Bxd6 2.Qa2. Now the queen must keep guarding c6 or c4 to compensate for the missing knight. If the black king takes the flight conceded by the key, 1…Kxd6, the threat-move remains effective – 2.Qc5. This clear demonstration of the Zagoruiko theme – three-fold change of mates for each of two black defences – is skilfully combined with reciprocal change.

Andy Sag: The key gives a flight capture which however does not defeat the threat. Both defensive variations feature changed mates.
Nigel Nettheim: A threat was needed, which quickly led to the key. A knight is sacrificed, perhaps not entirely surprisingly. The position is reasonably attractive, despite the many pawns.
Jacob Hoover: The key dispenses with the set mates entirely. The same two defenses that are seen in the set and try-play come up again, but entirely new mates are the responses. So, the theme of this problem is changed mates.
Ian Shanahan: Across all three phases, one sees a pair of thematic defences yielding distinct mates – a 3x2 Zagoruiko pattern. But there’s more: between set- and try-play, the mates are reversed – reciprocal change. Excellent! The key-move is best of all because it grants a flight. Notice the line-openings after the thematic defences serving to differentiate the mates. Overall, a most satisfying problem!

515

Leonid Makaronez
OzProblems.com 3 Oct. 2020

Mate in 3

Solution

The key 1.Qd2! grants a flight on d4 though it removes an unprovided one on f4. The threat is 2.Sf3+ exf3 3.Qe3, and if 1…Be2/Bg2 preparing to answer 2.Sf3+ with 2…Bxf3 then 3.d4. Another defence by the bishop, 1…Bxd3, is a distant self-block that prompts 2.Qg5+ Kd4 3.Qc5. Taking the flight, 1…Kxd4, gives 2.Bc5+ Ke5 3.Qg5, where White occupies g5 and c5 again but in reverse order. 1…d5 stops the threat because of 2.Sf3+? Ke6!, but it allows 2.Sxc6+ Ke6 3.Sc5, another mate on c5. One final variation is 1…Rf8 2.Sxc6+ dxc6 3.d4.

Andy Sag: Whenever I see the name Makaronez I know I am in for a hard time. I could see that 1…Kf4 had to be stopped but 1.Bg5? doesn't work due to 1…Sxg6! (preventing 3.Qf4 after 2.Sf3+ exf3). 1.Se2? doesn't work either due to 1…d5! blocking the guard on e6. Then I thought that no composer worth his salt would use a flight-taking key without giving something back, hence 1.Qd2! taking the f4-flight but giving the flight-capture on d4 in return.

516

Fadil Abdurahmanovic
The Problemist 1993, Version
5th Prize

Mate in 2
Twin (b) Qb7 to d4, (c) Qb7 to b5, (d) Qb7 to g5

Solution

The quartet of positions are solved by (a) 1.f8=Q! Ke5 2.Qe4, (b) 1.f8=R! Ke7 2.Re8, (c) 1.f8=B! Kf6 2.Qf5, and (d) 1.f8=S+! Kd6 2.Qc5. White promotes to four different types of pieces, so the Allumwandlung theme is brought about. The black king moves to each of its four orthogonal flights in turn, effecting plus-flights. This marvellous problem blending two classical ideas requires only six units, and it also features homogeneous twinning (all queen shifts).

Andy Sag: A novelty miniature quadruplet. In each part the f-pawn promotes to a different piece leaving a single flight to a different black square.
Jacob Hoover: This one was rather easy. The clue was that the problem had four parts, leading me to guess a theme with four parts.
Nigel Nettheim: A bit of fun for the solver. A bit of hard work for the composer. A rare case where a checking key is great!
Bob Meadley: Now that is a pretty thing with P=Q/R/B/S as the solutions. Late in the day and FA pulled it off. Nice.
Ian Shanahan: A mere six units. Yet two themes: white AUW across the totality of White's (necessarily aggressive) key-moves; black king plus-flights in the geometry of Black's moves. What an incredible combination! Even so, this impressive miniature is not without its flaws – the white pawn is unused in all but the final phase; and the white king plays no role at all. In the original publication, the king was in place of the pawn on b4 (The Problemist 1979). Alas, it was cooked in part (b). [Editor: Curious that it took the GM composer 14 years to publish the correction!]

517

Eugene Rosner
The Problemist 1997, Version
7th Commendation

Mate in 2

Solution

The white king has two candidate moves that aim to open the Q + K battery on the fourth rank. Approaching the black king with 1.Ke4? – which self-pins the e3-knight – threatens 2.Kxf3. Black has four defences, the first two of which facilitate other openings of the same battery: 1…Qxf5+ 2.Kxf5 (not 2.Sxf5?? because of the pin), 1…Rxe3+ 2.Kxe3, 1…Rxb5 2.Be7, and 1…Qxc5 2.h8=Q. But 1…Rf2! refutes this thematic try. The withdrawal key 1.Kc4! – self-pinning the c5-bishop instead – threatens 2.Kb3. The same four defences still apply but they lead to a new set of mates: 1…Qxf5 2.Sxf5, 1…Rxe3 2.Rh2, 1…Rxb5 2.Kxb5 (not 2.Be7?? because of the pin), and 1…Qxc5+ 2.Kxc5; now the last two defences yield further battery play. Additional tries are 1.d6? (2.Kd5) Rxb5! and 1.e6? (2.Ke5) Qb8! Four brilliant mate changes between virtual and actual play, shown with good strategic effects.

Andy Sag: The key self-pins the c5-bishop giving rise to a clever mate change after 1…Rxb5, which is interesting because that move also refutes the try 1.d6?
Nigel Nettheim: The white king crawls like a worm into a hole. The changed mate after 1…Rxb5 is a highlight.
Ian Shanahan: This rich royal battery problem adds extra phases to a well-known single-phase predecessor by Godfrey Heathcote. All four mates in the main try-play (after 1.Ke4?) are changed, and in each phase there is dual-avoidance due to the white king pinning one of his own pieces. An extremely rich feast indeed! It is a blemish, though, that there are idle white units in some of the phases (e.g. c2-rook in the thematic try).

518

Toma Garai
Probleemblad 1991, 2nd Prize

Solution

In part (a), the solution is 1.Qc4 Bg5 2.Sf3 Be3. Black begins by blocking the flight-square on c4, after which the white bishop unpins the e5-knight. This knight then opens the e-file for the e6-rook and simultaneously closes the 3rd rank controlled by the g3-rook. The white bishop is thus allowed to give a pin-mate on e3. Part (b) has a white bishop on e6 instead, and it solves by 1.Re3 Rc6 2.Sb4 Rc4. Now Black obstructs the flight on e3, while the white rook on d6 unpins the d5-knight. The latter is freed to activate the white-squared bishop on a diagonal and also to interfere with the black queen along the 4th rank. The white rook is then able to deliver a pin-mate on c4. Each of the two harmonious solutions renders an uncommon idea: White’s mating move occurs on a square that’s originally accessible by the black king.

Andy Sag: Black starts with a self-block, then White unpins a black knight which closes a black line and opens a white one allowing a pin-mate. Interesting that the self-block square in (a) is the mating square in (b) and vice versa.
Mark Salanowski: An orthogonal-diagonal transformation, and a delight to solve.
Jacob Hoover: I found it interesting that the final move of each part is to the same square as the first move of the other part and also that a white unit moves twice in each part.
Nigel Nettheim: Excellent thematic twinning. He may first have composed (b) with a black rook on a4; but then (a) has duals which the queen cleverly avoids.
Ian Shanahan: Perfect parity between each phase. Notice the funktionwechsel (exchange of function): the black piece interfered with during one phase executes the self-block during the other phase. A memorable gem by the Hungarian-American grandmaster!

519

Laimons Mangalis
Sahs 1965

Solution

The diagram is a complete block position, where after every possible black move, White can force Black to mate immediately. 1…Sf3 2.Qh4+ Sxh4, 1…Sxf1/Sg4 2.Qxe3+ Sxe3, and 1…Qxf8 2.Qg6+ hxg6 (1…Bxc2? is a short mate). However, White has no way of preserving all of the set play, e.g. 1.Be8? Qxf8+! The key 1.Rh1! (waiting) eliminates the black knight variations by pinning the piece, but frees the f2-pawn to promote. 1…f1=Q/R+ 2.Qf4+ Q/Rxf4, 1…f1=B 2.Qh3+ Bxh3, and 1…f1=S 2.Qxe3+ Sxe3. Only 1…Qxf8 2.Qg6+ hxg6 remains unchanged. The set Q vs S duel is unexpectedly replaced by a Q vs P duel post-key; an unusual demonstration of total change in mutate form.

Andy Sag: A large number of tries, e.g. 1.Be8? Qxf8+!, 1.Ra6? d6!, 1.Rd1? Bxc2+! Changed play is the name of the game. In the set position all knight moves allow queen checks that force mates. The key pins the knight but now the queen checks after all possible f1-promotions force mates.
Nigel Nettheim: Enjoyable. Not hard to solve, but surprising nevertheless.
Jacob Hoover: I liked the idea of the problem, but I feel that the problem is marred slightly by the queen and rook promotions having the same white response (that keeps the problem from being an Allumwandlung).
Ian Shanahan: We see here changed play, so I suppose that this problem is a selfmate mutate. The main theme is, of course, the (near) black AUW.

520

Joseph Warton
The Tablet 1959

Mate in 3

Solution

Two rook tries by White prepare to sacrifice the queen for a mate along the first rank. 1.Rd3? threatens 2.Qxb1+ Kxb1 3.Rd1, but 1…Bb2!, and 1.Rxg2? threatens 2.Qxb1+ Kxb1 3.Rg1/Rh1, 1…Bb2 2.Rxb2 and 3.Rxb1/Qxb1, but 1…Bc3! Another try, 1.Re5?, generates a short threat, 2.Qd4, by cutting off the h8-bishop; however, Black defeats it straightforwardly with 1…Bxe5! The seemingly similar 1.Rg7! entails the same threat of 2.Qd4, but here the analogous defence 1…Bxg7 carries the weaknesses of clearing and unguarding the g7-square. White continues with 2.Rh1, threatening 3.Qxb1, and if Black accepts this second rook sacrifice, 2…gxh1=Q, then 3.Qxg7. There is by-play in which the initial threat gets extended to some full-length variations: 1…Sa4/Sc4/Sd5 2.Qd4+ Sb2/Sc3 3.Qxb2/Qxc3. The brilliant main line of this three-mover exemplifies the Warton brothers’ famously devious style.

Andy Sag: A very difficult to solve one-liner in a Meredith setting and involving sacrifice of both rooks! Out of the 42 possible first moves, Rg7 was the last one I looked at.
Ian Shanahan: Some very cute play here, notably the surprise move 2.Rh1 and ensuing pawn-promotion with line-opening.
Bob Meadley: That's a beauty by the brothers Warton – very clever.

Thanks to Bob who contributes an interesting short article about The Warton Brothers.
Michael McDowell points out that this problem was in fact composed solely by Joseph Warton!

521

Marjan Kovačević
diagrammes 2010, 1st Prize

Mate in 2

Solution

A dual mate is set for 1…Re4 2.Rgxe4/Rbxe4. The white rooks can move along the rank to maintain the double attack on e4, and when such moves also control f5, the queen is freed to threaten mate on d4. A preliminary try 1.Rgf4? preserves the set mates for 1…Re4, but fails to both 1…Bd6! and 1…Bc5! Two thematic tries by the rooks show an anti-Bristol effect in that the pieces interfere with each other on the rank, resulting in the removal of the dual. Thus 1.Rgc4? Re4 2.Rcxe4, and this try also provides for 1…Bc5 with 2.Rxc5, but 1…Bd6! still refutes. Similarly, 1.Rbf4? Re4 2.Rfxe4, and now 1…Bd6 permits 2.Qxd6, but 1…Bc5! instead refutes. The remarkable key 1.Qxf7! (threat: 2.Qe6) grants two flights to the black king and radically changes the idea of the problem. The new flight defences, which give check by discovery on the e-file, require shut-off mates: 1…Kxf5+ 2.Rge4 and 1…Kd6+ 2.Rbe4. The white rook moves seen in the set and virtual play are transferred to the king defences, becoming direct battery mates.

Andy Sag: The key gives two battery-check flights, each answered by a different battery cross-check (mate).
Andrew Buchanan: Hard to find! So many tries!
Ian Shanahan: There is an astonishing density to the numerous themes paraded here. Considering first the dual mates of the set-play, we see them separated across the two tries: this is known as the Makihovi-Ellerman theme. After the generous key offers two flights, the black royal battery yields two cross-checks via two diagonal white batteries. Across all phases, both radical and total change are apparent. Phew! What a gorgeous melange!

522

Wolfgang Pauly
Hamburgischer Correspondent 1932

Helpmate in 4, Set play

Solution

In the set play, White begins and mates on the fourth move: 1…e6 2.Kf5 e7 3.Kg6 e8=Q+ 4.Kh7 Qf7. In spite of the symmetrical position, an equivalent line of play on the left side of the board fails as the king wouldn’t be mated on b7. In the actual solution, Black starts and the set line cannot be replicated because the king is unable to reach h7 in four moves. Instead, the king heads to a8: 1.Kd5 e6 2.Kc6 e7 3.Kb7 e8=B 4.Ka8 Bc6. Once again, there’s no equivalent solution on the other side of the board, because the king can go no further right from h7. The theme of asymmetry is therefore effected twice, along with a promotion change, in a very light setting. The quirky use of two dark-squared white bishops is well justified in this helpmate selected from Anything but Average: Chess Classics and Off-beat Problems.

Andy Sag: In the set play, Black takes the shorter diagonal to h7 and the promoted queen mates on f7 but if Black moves first, losing a tempo to reach h7 in four moves is not legally possible, so the black king must then take the longer diagonal to a8. Now if White promotes to a queen, moving the black king to a8 is illegal but this can be resolved by promoting instead to a bishop.
Nigel Nettheim: The black king can reach h7 in three but not four moves, so the set play has to be abandoned.
Andrew Buchanan: Very pleasant to solve with satisfying asymmetric solutions.
George Meldrum: The set play proved to be a distraction in solving the problem. Concentration was on the right-hand side of the board where underpromotion was quickly discounted. It took a eureka moment with new thinking to solve this one.
Ian Shanahan: A dainty little asymmetric (i.e. a symmetric position leading to distinctly non-symmetric solutions). I found this a tougher nut to crack than expected.
Bob Meadley: Another cutie. The old black king takes a long walk to a8 and White promotes to a bishop for the death blow. Marian Stere's book on this composer is a beauty.

523

Niels Löw
Skakbladet 1925

Mate in 2

Solution

The square-vacating key 1.Bc1! threatens 2.Sd2. Since the threat-move will close the key-bishop’s line to f4, Black can defend by cutting off the white queen’s control of the same square, to create a potential flight. Four such defences on d6 cause a variety of self-interferences by the black pieces, leading to different white mates. 1…d6 2.Sc6 (not 2.Sd5?), 1…Rd6 2.Sd5 (not 2.Sc6?), 1…Sd6 2.Sc5 (not 2.Sc6/Sd5?), and 1…Bd6 2.Qxb7. A fifth defence has the same motivation of shutting off the white queen, though it occurs on another square: 1…Re5 2.Qxe5. Two subsidiary variations increase the utility of the white rooks: 1…Bxb4+ Rxb4 and 1…Sg~ 2.Re3.

Andy Sag: The key-bishop vacates d2 for a knight threat but that relies on the queen to maintain a long-range guard on f4, inviting five line interferences including four Grimshaw-style defences on d6. All seven variations are set.
George Meldrum: Interference play involving the d6-square is pleasing though confusing being arranged as set play.
Nigel Nettheim: Not hard to solve, as a threat was needed. Marvellous set mates, unchanged in the play. The four occupations of d6 (no more seem possible here) are the highlight. The h2-rook was added to let the g2-knight take part in the play. The g5-pawn prevents a dual after 1…Sxf4.
Michael McDowell: Nice problem this week. Four Theme A + interference defences on the same square.
Ian Shanahan: Finding the key was an unexpectedly tough nut to crack. But what a lovely example of four-fold dual avoidance! The complexity of the mechanism, involving a fair amount of line-play, delighted me.

524

Leonid Makaronez
OzProblems.com 5 Dec. 2020

Mate in 3

Solution

The black king has an unprovided capture on d5, and if White precludes it by opening the half-battery on the long diagonal with 1.Rc2? or 1.Rgd2? (threatening 2.Rd6/Rxc5 in either case), Black refutes with 1…e4! The key 1.Rxe5! also removes the flight and activates the half-battery, to threaten 2.Rc2+ Re4 3.Sd4, a double pin-mate. If Black gives a check allowed by the key, 1…Rd4+, White replays the moves seen in the threat but in reverse order, 2.Sxd4+ cxd4 3.Rc2. Two variations see White firing the battery with the g2-rook immediately: 1…Bd1 2.Rge2+ Re4 3.Qxe4 and 1…Sd6 2.Rg7+ Re4/Se4 3.Rxc7. Lastly, 1…Re4 unguards c5 and is exploited by 2.Sa5+ bxa5 3.Rxc5. An intricate battery-play three-mover from the prolific IM of composition.

Andy Sag: The key removes the flight-capture and sets up a battery but allows a check. However, I must admit almost falling victim to the try 1.Ke7? (2.Sd8+ Kxd5 3.Rd2) e4!! then 2.Rd7 (3.Sd8) Sd6!! or 2.Rf5 (3.Sd8) Rd4!! which maintains the flight-capture and almost works but for the innocuous 1…e4.
George Meldrum: The aggressive key was a surprise (taking away the king flight) – a reminder to expect the unexpected. The follow-up play is all new, and who would have anticipated that the d4-square was a weakness for Black with its triple guard.
Nigel Nettheim: Another try is 1.Rg5? (2.Rd6/Rxc5) e4 2.Rxc5+ Rxc5 3.Sd4 or 2…bxc5 3.Sa5, but 1…Re4! [Editor: It’s interesting how the 3rd-move mates Sd4 and Sa5 in this try occur as 2nd-move continuations after the key.]

525

Vladimír Kočí
OzProblems.com 12 Dec. 2020

Mate in 2

Solution

The key 1.Bc4! crosses over d3 to threaten 2.Sd3, which otherwise would cause a self-interference. The white half-battery formed by the f8-bishop and two rooks is controlled by Black’s rooks, either of which can defend by capturing the d6-rook. Such a defence not only opens a line for the remaining white rook to traverse, but acts as an anticipatory self-pin that White exploits by firing the masked battery: 1…Rexd6 2.Re5 and 1…Rdxd6 2.Rc7. The rook passively sacrificed in these thematic lines also delivers a mate with 1…Rxe1/Re3 2.Rc6. There is by-play: 1…Sb4 2.cxb4. The two main variations show good strategy though the matrix is not new.

Andy Sag: The key is rather obvious but the pin-mates after captures of the d6-rook are a nice touch. The h7-knight seems unneeded. [Editor: The composer concurs and the piece is removed.]
Jacob Hoover: Black defends by capturing the d6-rook, but doing so opens a line for the e7-rook to deliver a pin-mate.
George Meldrum: A pair of pleasing pin-mate variations.
Ian Shanahan: Clearly the e1-knight must take part, and this points to the key, which also gives a crucial extra guard to d5. The two thematic variations create a masked battery by capture, which fires to yield pin-mates.

526

Mark Kirtley
The Problemist Supplement 2012

Proof game in 7½, 2 solutions

Solution

Both sides are missing their dark-squared bishops and Black is also missing two pawns. White has made five visible moves with the queen-side units, and two more are necessary for the c1-bishop to capture the g7-pawn, leaving one spare move. In the first solution, White uses this remaining move to capture the f8-bishop on its home square. 1.b4 a5 2.Bb2 a4 3.Bxg7 Sf6 4.Bxf8 Kxf8 5.Sc3 Kg7 6.Sxa4 Rf8 7.c3 Kg8 8.Qb3. The need to remove the white bishop forces Black to “castle” in a roundabout way. In the second solution, White plays a tempo move to use up the extra time. The f8-bishop must then be sacrificed on another square, after capturing the white one placed on g7, and this plan requires Black to lose a tempo as well. 1.b3! a6! 2.Bb2 a5 3.Bxg7 Bxg7 4.b4 Bc3 5.Sxc3 a4 6.Sxa4 Sf6 7.c3 0-0 8.Qb3. The contrasting solutions demonstrate pseudo-castling in one and tempo play twice in the other, plus mutual captures by the bishops.

Andy Sag: In one solution, each side includes a tempo move. In the other, the c1-bishop would take three moves to remove the g7-pawn and f8-bishop in situ and that the king could then capture the white bishop and then the penny dropped that the king could get to g8 without castling.
Jacob Hoover: Black’s king and rook got in their castled positions “the hard way” in one part, but actually castled in the other.
George Meldrum: I found the non-castling solution very hard to find.
Nigel Nettheim: The two tempos lost by slow castling in (a) and by restrained pawn-moves in (b) make a nice contrasting pair.
Andrew Buchanan: Mark Kirtley is one of my favourite composers, and he specializes in these two-solution problems. One half, as is sometimes the case, was much easier than the other half. The castling solution requires tempos for both White and Black: very neat.

527

Ian Shanahan
The Games and Puzzles Journal 1988, Version
“Sword and Shield”

Solution

The only plausible plan for immobilising the whole black force is to lock most of it on the h-file behind the pawns. This requires shifting each pawn one square forward, to let the pieces in, and the resulting g2-pawn needs to be blocked as well. The black king is the sole suitable blocking unit, since any other piece standing on g1 would not be confined by the white king. 1.Rh8 – clearing the file for the king and not 1.Rh1+?? because checking is forbidden – 2.Kh7 3.Kh6 4.Kh5 5.Kh4 6.Kh3 7.Kh2 8.Kg1 9.Rh1 10.Rgh2 11.g2 12.Qh3 13.g3 14.g4 15.Bg5 16.Bh4 17.g5 18.Bh5 19.g6 Ke2. The composer writes in Chess Problems by Dr. Ian Shanahan: “Incarceration with intricate timing, incorporating critical play, in a White Rex Solus setting. Figurative problem: the diagram position resembles a ‘sword’, the stalemate position a ‘shield’!” He also indicates that Black’s e2-pawn in the original version (queried by some solvers) is redundant, as confirmed by computer-testing.

Andy Sag: It is clear that a bishop must go to h5 to achieve a lock-in with black pieces on h-file and that the king must reach g1 before h1 can be occupied to avoid checking.
Nigel Nettheim: Cute! Not hard to solve.
Jacob Hoover: Hilarious!
Andrew Buchanan: Ian’s name on a page of problems always attracts my interest.
Karel Hursky: A wonderful phantasmagoria! Love the way the sword changes to a shield! Very memorable problem!