Weekly Problems 2023-A

Problems 633-658

633

Leonid Makaronez
OzProblems.com 7 Jan. 2023

Mate in 3

Solution

The black king has a provided flight, 1…Kg4 2.Qe6+ Kf3 followed by multiple mates, e.g. 3.Rxh3. The key 1.Be2! removes that flight but grants a new one on e4, creating the threat 2.Qxh3+ Kxe4 3.Qd3. If 1…Bg4, then 2.Bxg4+ gives the black king a choice of two diagonal captures, both answered by queen mates, 2…Kxg4 3.Qh3 and 2…Kxe4 3.Qc4. Similarly, if 1…Qb5 – unguarding the g-file – then 2.Rh5+ leads to two king captures and respective queen mates, 2…Kxg6 3.Qg8 and 2…Kxe4 3.Qe3. Note the three changed mates against 2…Kxe4 including the threat line. The distant self-block 1…Qd7 permits 2.Qf3+ Ke6 3.Sg5. Since the black king makes all of its four possible diagonal moves, with different white responses, the star-flights pattern is produced. One more full-length variation involves an orthogonal flight: 1…Qxe4 2.Qf7+ Kg5 3.Rh5. A short mate results after other queen defences, e.g. 1…Qa6 2.Qd5, and after the immediate king capture, 1…Kxe4 2.Qd3. The two queens and black king are especially active in this three-mover that brings about a wealth of mating nets in precise fashion.

Andy Sag: A sacrificial give-and-take key with threat and six variations involving six different flights and a good variety of play, notwithstanding two short mates.
Rauf Aliovsadzade: A give-and-take key. Nice sub-variations after the defenses 1…Qb5 and 1…Bg4.
George Meldrum: Lots of twists and turns in this one as the black king gets mated on six different squares. A quality problem which is difficult to solve with a great key-move.

634

Peter Kahl
Die Schwalbe 1955

Solutions

All black units need to be immobilised, and while the two columns of pawns are well-suited to trap a group of pieces on the d-file, the two knights pose more difficulties. The answer is to arrange pins of both knights on c6 and e6 against the king placed on d5, with White playing Rd7 as the stalemating move, to pin a third piece – the dark-squared bishop – on d6. This plan implies shifting all of the pawns down one square, which is compatible with the idea of confining the light-squared bishop on d1. 1.Scb4 2.c2 3.c3 4.c4 5.Sc5 6.Rd3 7.R1d2 8.Bd1 9.e2 10.e3 11.e4 12.e5 13.Se6 14.Bc5 15.Qd4 16.Kd5 17.Bd6 18.c5 19.Sc6 for Rd7. In the final position, the black force reoccupies the exact same squares as those used in the diagram, but with all sixteen pieces reallocated – a maximum task. There are three separate “chains” of units that cyclically exchange their places: (1) the five on the c-file, (2) the three on d4, d5 and d6, and (3) the remaining eight.

Andy Sag: It is clear that the black-squared bishop and both black knights end up pinned. Check avoidance ensures a unique sequence. The initial position is legal if White’s a-, d- and h-pawns were promoted before being captured by black pawns.
Jacob Hoover: This stalemate works because of the triple pin. I like this kind of problem because it’s like solving a sliding-panel puzzle, which can be fun sometimes.
Andrew Buchanan: Cyclic platzwechsels with eight, five, three black pieces. Cycling n pieces will take a minimum of n+1 moves (because the first move must be a piece moving out of the cycle to create a space, then the others move along the cycle and finally this first piece will slot back in to its new space). This minimum is elegantly achieved thrice with unique ordering of moves. Splendid!
Rauf Aliovsadzade: Wow! The position resembles the Empire State Building in New York city. Even some clouds hovering over! Not easy to solve.

635

Vladimir Erokhin
Thèmes-64 1986

Mate in 2

Solution

By placing an extra guard on e4 or c2, White will free one of the rooks to mate on d1. Two tries and the key create such a threat, and they all unpin the d5-bishop, which then defends accordingly. The first try 1.Qh4? (2.Red1) releases the bishop directly by withdrawing the queen – 1…Bxb3 2.Qe4 and 1…Bf3 2.Qc4, but Black refutes with 1…Rg4! The second try 1.Sd4? (2.Rcd1) unpins the bishop by interference and leads to new mates against the same defences – 1…Bxb3 2.Sxb3 and 1…Bf3 2.Sxf3; also 1…cxd4 2.Sxb4 (and 1…Kxd4 2.Rcd1), but 1…axb3! is spoiling. The key 1.Sd6! (2.Red1) also interposes on the pin-line and generates yet another pair of mates – 1…Bxb3 2.Sc4 and 1…Bf3 2.Se4, with by-play 1…Kd4 2.Rcd1. The black bishop’s defences provoke three pairs of changed mates – including different battery openings by the key-piece – in this crisp example of the Zagoruiko theme.

Andy Sag: The key gives a flight and unpins the black bishop enabling a threat and three defensive variations. Many tries, the most interesting with variation content being 1.Sd4 and 1.Qh4.
Jacob Hoover: An unpinning key followed by battery plays. Very nice.

636

Christopher Jones
Shakhmatnaya Poeziya 2004, 1st Hon. Mention

Helpmate in 3½, 2 solutions

Solution

It’s not obvious in such an open position where to mate the black king, but the placement of the white bishop pair aiming at the two knights provides a hint. If the king is transferred to either of the knights’ squares, that will nearly set up a mating net with the white force already in place. However, Black has insufficient time to play a square-vacating move with a knight, so in each solution, White proceeds to capture one to clear the square for the king. 1…Bxd4 2.Se5 (distant self-block) Bc3+ 3.Ke3 Be1 4.Kd4 Bf2 and 1…Bxc4 2.Sb3 (another distant self-block) Be6 3.Kd3 Bh3 4.Kc4 Bf1. Besides the attractive round-trips made by the bishops, the problem demonstrates the Kniest theme, in which White captures a unit on a square where the black king will be mated.

Andy Sag: A tale of two bishops! Each bishop captures a knight and then circles back to its original position to deliver mate after the black king moves to the square of the captured knight.

637

Jorge Lois & Roberto Osorio
Problemesis 2006

Solution

Black needs all fifteen moves to position the units now on the king-side. With only two moves available to the pawns, Black must play …f5 and …gxf6 – not …exf6 as that would cost the bishop an extra move to reach h4. White’s missing f-pawn cannot be directly sacrificed on f6 since it would obstruct …f5, so it has to promote first (after capturing the e-pawn). 1.f4 Sf6 2.f5 Se4 3.f6 Rg8 4.fxe7 f5. The black bishop will soon play to h4, a potential disruptive check that must be blocked by the g1-knight. 5.Sh3 Kf7 6.e8=S Be7 (6…Qg5/Qh4+? would cause a tangle later in which the queen cannot unpin the knight) 7.Sf2 Bh4 8.Sf6 gxf6. Now to avoid a permanently stuck knight on f2, it has to be unpinned by the black rook on g3. But once released the white piece cannot simply return to g1, because the eventual …Rf3 would cause another disruptive check. So the knight heads to g3 in anticipation. 9.Rg1 Rg3 10.Sh1 – White’s pieces from g1 and h1 have swapped places – Rf3+ 11.Sg3 Rf2. This unpin finally frees the knight to return home. 12.Sh5 Qg8 13.Sf4 Qg3 14.Sh3 Kg6 15.Rh1 Kh5 16.Sg1. White’s knight and rook execute not one but two platzwechsels (exchange of places). The excellent manoeuvres don’t involve captures but are subtly motivated by the need to forestall potential checks.

Andy Sag: Retro-analysis indicates that the white f-pawn must promote after capturing the black e-pawn in situ and then be captured on f6, but the clever gymnastics by the white king’s knight to avoid illegal checks on both sides is the key to solving this one.
Jacob Hoover: Themes seen: underpromotion, switchback, rundlauf.
Andrew Buchanan: The Lois theme [double interchange of two pieces] is more commonly applied to king and queen but here two different units enjoy the dance. It’s incredible that such a PG is actually sound. So many subtle interactions contribute to this. Bravo!

638

Otto Nerong
Die Schwalbe 1937

Mate in 6

Solution

Black’s force is mostly incarcerated but it controls the white knight’s four potential mating squares. Two are guarded by the e4-pawn only, so it’s tempting to target this defender; however, 1.Sd6? f5 2.S~ (random tempo move) f4 3.Sd6 f3 4.Sxe4 results in stalemate. Another mating square c2 is guarded by the black bishop, which is itself well protected. White concocts a plan not to capture that piece but to release it, inducing Black to make self-weakening moves through zugzwang. 1.Sa5! f5 2.Sb3 f4 3.Sxa1 f3 4.Sb3 a1=Q 5.Sxa1 Ba2 6.Sxc2. Cute play and curious diagram position in which White has minimal force while Black has almost the maximum (missing only a knight).

George Meldrum: The elimination of the e4-pawn followed by mate on d3 or f3 seemed the order of the day. This distraction caused confusion and delay. Kudos to Otto for this tricky solution.
Thomas Thannheiser: The black pawn moves are just enough to allow Sxa1, after which zugzwang leads to mate on c2.
Andrew Buchanan: Very comical.
Bob Meadley: Nicely done.

639

Vladimir Zheltonozhko & Valery Kirillov
Shakhmatnaya Kompozitsiya 1993, Special Prize

Helpmate in 2, 4 solutions

Solution

This four-phase problem is solved by 1.Qa3 b3+ 2.Kb4 c3, 1.Qxa4 b4 2.Kb5 c4, 1.Qc3 bxc3 2.Sd3 cxd3, and 1.Ra3 bxa3 2.Bd4 cxb3. Both the white b- and c-pawns execute the Albino theme by playing the maximum four different moves. Furthermore, in every solution the two pawns move in parallel, with the b-pawn leading the way and the imitative c-pawn giving mate. The white rooks in supporting roles are also well utilised in all four parts of this harmonious helpmate.

George Meldrum: A nice set with the b- and c-pawns combining in all mates.
Andrew Buchanan: Great stuff. One of those problems where the number of solutions is a big hint.
Jacob Hoover: Not only do both the white b- and c-pawns make all four possible moves, they move in sync in each solution! So we have a synchronized double Albino! Perfetta!
Karel Hursky: Loved this helpmate.
Paz Einat: Fabulous double WP4!! Looks like a clear 1st prize, not a special one…

640

Rauf Aliovsadzade
OzProblems.com 25 Feb. 2023

Mate in 3

Solution

Black has only three legal moves in the diagram, including a king’s flight to d3 (no set response). The key 1.Sf7! contains a threat, 2.Se5 [A], which removes the flight and leaves Black in zugzwang: 2…exf5 3.Sxf5 [B]. Now curiously all three black moves work as defences. 1…exf5 2.Sxf5+ [B] Kd3 3.Se5 [A] – White plays the same two moves as in the threat line but in reverse order. After 1…e5, 2.Sd6 [C] is a waiting move that covers c4 for 2…Kd3 3.Rf3 [D], a pin-mate. And if 1…Kd3 then 2.Rf3+ [D] forces 2…Kc4 3.Sd6 [C]. In these two variations, White likewise makes the same pair of moves but their order is swapped. A neat doubling of a formal idea: reversal of White’s second and mating moves, with no extraneous play at all.

Composer: White AB-BA and CD-DC involving a ‘twist’ – a pseudo-threat.
Andy Sag: Not quite a Meredith but the variations making use of the forced self-pins are nice.
George Meldrum: Initially observed 1.Rf3+? exf3 2.Bb1 e5 3.Sf5; however, 2…f2! stops the mate. Several other moves come close to solving rather than the more obscure key.

641

Ljubomir Ugren
Drago Makuc & Janez Moder Memorial Tourney 1970,  Prize

Solution

Many of the black queen’s moves are provided with set responses, such as 1…Qf6 2.Qb2+ Qxb2, and 1…Qh1+ 2.Qd1+ Qxd1. Some defences are not, however, like 1…Qh7 when 2.Qxd3+? Qxd3 fails because of the g3-rook, and 1…Qf4 (prospectively guarding c1) when 2.Qc2? mates instead of forcing 2…dxc2, because the black pawn is pinned. White seems to deal with these hindrances by making any rook move off the rank, but only 1.Rg8! (waiting) succeeds. 1…Qh8/Qf6/Qd4/Qf2/Qh2 2.Qb2+ Qxb2, 1…Qh7/Qe4/Qg3/Qh3 2.Qxd3+ Qxd3, 1…Qh6/Qg5/Qf4 2.Qc2+ dxc2, and 1…Qh5/Qg4/Qe1+/Qh1+ 2.Qd1+ Qxd1. White must avoid 1.Rg7? Qh8!, 1.Rg6? Qh7!, 1.Rg5? Qh6!, and 1.Rg4? Qh5!, in each case allowing the black queen to hide behind the rook. (1.Rg2? and 1.Rg1? are technically not tries since they are refuted in multiple ways.) A fine duel between the two queens, where White’s play is exact throughout (notably 1…Qd4 2.Qb2+, not 2.Qxd3+? Qc3!).

Andy Sag: In set position the black queen has 14 possible moves and all other black units are immobile. With the g3-rook right out of the way, the d-pawn is unpinned, the black queen is given  two more moves but all 16 allow white queen checks (on d3, d1, c2 or b2) which force Black to deliver mate. Prize well deserved!
Rauf Aliovsadzade: Excellent key – waiting move not expected!
George Meldrum: Terrific key move and surprising mate given with the black pawn.
Andrew Buchanan: Luckily I quickly guessed the idea, but it's still very cool.

642

Uri Avner
Springaren 1993, 1st Prize

Mate in 2

Solution

Two black captures on e6 yield important set play: 1…Kxe6 2.Qd7 and 1…Qxe6 2.Se3. The thematic try 1.Qa2? prepares an indirect Q + S battery aimed at the e6-flight, and threatens 2.Sd6 (not 2.Se3?). The flight-move 1…Kxe6 leads to a changed mate, 2.Se3, a direct battery opening – not 2.Sd6? as it would interfere with the d3-rook. The self-block 1…Qxe6 allows another changed mate, 2.Qf2. But the try is defeated by 1…Qxg7! The unexpected key 1.Sf4! removes the flight on e6 but grants another one on f4; furthermore, it opens a black queen-line to d6 while closing another one to e3. The threat 2.Se3 utilises the indirect Q +  S battery, now aimed at the f4-flight (not 2.Sd6?). The flight-move 1…Kxf4 permits a new direct-battery mate, 2.Sd6 – not 2.Se3? because of another interference with the d3-rook. Compared with the try play, the reversal of the threat-move and a variation mate (2.Se3/Sd6) after different black defences brings about the pseudo le Grand theme. The self-block 1…Qxf4 enables 2.Qd7, which represents a transferred mate, used in the set play against another defence. Indeed, the other set mate, 2.Se3, is also transferred to 1…Kxe6 in the try play. The numerous analogous effects between the three phases are remarkable, including how the new black defences replicate the motifs of flight-move and queen self-block.

Andy Sag: The give-and-take key shifts the set flight-capture to a different square and creates a threat which can only be defended by f4-captures. These defences are answered by a battery mate or exploiting a self-block. The set flight-capture and self-block are provided for by 2.Qd7 and 2.Se3.

643

Laszlo Zoltan
Sakkélet 1986

Helpmate in 2, Duplex
Twin (b) Swap Ka4 and Kg4

Solution

Both duplex solutions of part (a) seem straightforward; the first player self-blocks while the second arranges extra guards for the mate, with little interaction between the two sides. Black begins: 1.hxg5 Rf3 2.h5 h3; White begins: 1.Rxe3 Rb8 2.Ra3 Rb4. When the kings are exchanged in the twin, a similar description applies to the new duplex solutions. Amazingly, however, each side makes the exact same moves as in part (a), only now their order is changed and their motives of self-blocks and guards/mates are reversed. Black begins: 1.Rb8 Rxe3 2.Rb4 Ra3; White begins: 1.Rf3 hxg5 2.h3 h5. A highly original and humorous idea. The lack of interplay between Black and White is a necessary cost for the scheme to work.

Andy Sag: Easy enough to solve. Interesting to see both sides play the same moves in reverse order, when comparing the original with the twin duplex and vice versa.
Thomas Thannheiser: Very, very nice problem. Same moves in (a) and (b) but different kings to check mate with this move order!
Jacob Hoover: That is super crazy!
Michael McDowell: I greatly enjoyed it. Very imaginative, and just my sort of helpmate!

644

William Whyatt
British Chess Federation Tourney 1965, 3rd Prize

Mate in 3

Solution

If White covers the flight on e5 with the king, that will threaten 2.Qc5. Two such king moves are tries that obstruct a potential queen mate: 1.Kf5? Rc3! and the rook’s self-block cannot be exploited by 2.Qg4??; likewise 1.Kf6? Bc4! and the bishop’s self-block cannot be met by 2.Qh8? The key 1.Bg1! entails the threat, 2.Qg4+ Ke5 3.Bh2, or 2…Kc3 3.Qb4. Black defends by shifting either piece on the a-file to b3 (defusing the Qb4 mate), but such moves cause a mutual interference between the rook and bishop pair. If 1…Rb3, 2.Kf6 threatens 3.Qc5, and since 2…Bc4?? is no longer available, Black must answer with 2…Rc3, allowing 3.Qg4. If 1…Bb3, 2.Kf5 threatens 3.Qc5, and now that 2…Rc3?? is ruled out, Black can only respond with 2…Bc4, admitting 3.Qh8. There’s a subsidiary line, 1…Ke5 2.Bh2+ Kd4 3.Qc5. An elegant work that involves a black Grimshaw, subtle differentiation between two white king moves, and long-range play by the busy white queen.

Andy Sag: The unprovided set flight gives a clue. The Grimshaw on b3 is a subtle touch.
George Meldrum: Super cool key-move.
Jacob Hoover: Quite the interesting way to implement a Grimshaw.
Rauf Aliovsadzade: Excellent! The logical tries 1.Kf5? and 1.Kf6? return as second moves after Grimshaw on b3!

645

Alexandar Atanasijevic
Mat 1973, 2nd Prize

Solution

Since the mating pawn will need to be protected, the only viable option for the final move is e4. With c4 also guarded by the d3-pawn, six remaining flights of the king have to be blocked, so Black must promote all six pawns for this purpose. 1.g1=B 2.Bxh2. The plausible 2.Bxb6? to free the b7-pawn first will cost Black extra moves to capture the h2-pawn later. 3.Bd6 4.h2 5.h1=B 6.Bf3 7.Bxg4 8.Be6. This is the most efficient way to block e6, hence not 5.h1=S? 6.Sf2 7.Sxg4 8.Se5. 9.g4 10.g3 11.g2 12.g1=B 13.Bxb6 14.Bbc5 15.b5 16.b4 17.bxa3 18.a2. The alternative 18.axb2? 19.b1=S 20.Sd2 21.Sc4 22.Se5 is one move too slow. 19.a1=B 20.Bxb2 21.Bbe5 22.b2 23.b1=B 24.Bc2 25.Bxa4 26.Bc6 27.a4 28.a3 29.a2 30.a1=B 31.Bad4 for e4. The total of six bishop promotions is wonderfully achieved in a pawns-only setting. Black’s economy is perfect and only the white king doesn’t take part in the model mate.

Andy Sag: Finally cracked it! All six black pawns must promote taking up 18 moves and at least three white pawns must be captured by promoted pieces (up to three could be taken by pawns). This leaves only four “spare” moves, namely 2.Bxh2, 6.Bf3, 20.Bxb2 and 24.Bc2.
Jacob Hoover: The correct sequence ends with a model mate. I was stumped for quite a while, until I realized I was overthinking it.
Andrew Buchanan: Very nice, solves itself really.

646

Michael McDowell
OzProblems.com 8 Apr. 2023

Helpmate in 2, 2 solutions
Antikings

Solution

Antikings is a fairy chess condition in which a king is in check only when not attacked by the opponent. Checkmate occurs when a player is unable to place their king under attack. In the diagram, both kings are attacked, so neither side is in check. For instance, 1.Kc8?? is illegal since c8 is not covered by White. 1.Bb6+ checks by un-attacking the white king, and the only legal reply is 1…Kf2.

The black king initially has four legal moves to b7, d7, c6, and b6 where it would remain under attack by White’s a7-rook or either bishop. White may check by withdrawing the rook or by interfering with it, e.g. 1.Bb4 Bb7+ 2.Kxb7 Ra1+, but this simply forces the king to the a-file or b6 and isn’t mate. To arrange for mate, the king must move “out of range” of all white pieces, while the latter assist by cutting off one another’s line of attack. 1.Kd7 Bf2 2.Ke7 Bb7 (not 2…Rxa5?? which illegally self-checks by removing the white king’s attacker), and 1.Kc6 Rf2 2.Kd5 Rb7. The two pairs of Grimshaw interferences on f2 and b7 are neatly brought about in a miniature setting with perfect white economy.

Composer: The white antiking may look superfluous, but the black bishop is a useful cook-stopper, e.g. with a white pawn on a5 there would be 1.Kb6 Rf2 2.Kxa5 Rh7, etc.
Jacob Hoover: Here we see a double white Grimshaw (or should we call it an anti-Grimshaw? LOL).
Andrew Buchanan: Easier than I thought it would be, once I got my head around the conditions. Congrats to the composer.

647

Paz Einat
Israel–Italy Composing Match 2015, 2nd Place

Mate in 2

Solution

The white queen has three candidate moves that form batteries with different pieces. The first is the try 1.Qb5?, which creates a Q + S battery to threaten 2.Se5 [A]. If 1…Bxc4 then 2.Qxc4, but Black refutes by playing 1…Bxd2! [a], to give the king an escape to e4. The second attempt 1.Qd5? sets up a Q + B battery and threatens 2.Be5 [B]. Two virtual variations follow with 1…Bxc4 2.Qxc4 and 1…Be3 2.Se5, but now 1…dxe6! [b] defeats the try by attacking the queen. The third queen move 1.Qf5! is the key and its threat 2.e5 employs the new Q + P battery. The principal defences are the black moves [a/b] that worked as try refutations, and the respective mates they now provoke are the very threat-moves [A/B] they previously neutralised: 1…Bxd2 [a]  2.Se5 [A] and 1…dxe6 [b] 2.Be5 [B]. Such a paradoxical pattern makes up the Dombrovskis theme. There is some by-play: 1…Be5+/Be3 2.Se5 and 1…Bf-else 2.Qf3. The main Dombrovskis scheme is attractively complemented by the queen’s varied battery play and the three threat-moves occurring on the same e5-square.

Composer: One of my favorites… The tourney’s theme asked for the threats to be on the same square, and I was able to find this nice scheme that was not so difficult to convert.
Andy Sag: The unprovided capture 1…Bxd2 gives a strong clue.
Jacob Hoover: The key also threatens a battery play; the black moves that defeated the tries defend against this threat, but each of these defenses allows the try-threat it defeated, i.e. the Dombrovskis theme.

648

Michel Caillaud
Problemesis 2000, 2nd Hon. Mention

Solution

The white units take up all 14 available moves to reach their diagram positions, including the last move, Qxe7+ (which cannot be a diagonal non-capture as that would cost the queen an extra move). If this capture was that of the e7-pawn, the only black units capable of moving previously were the knights and rooks, but they couldn’t have lost a tempo and returned to their home squares in exactly 13 moves, an odd number. To get around the tempo issue, Black must promote the pawn and then either return it to e7 or use it to replace another piece that gets captured there. Black also has to capture the e2-pawn which never moved, but the obvious choice of the b8-knight as the capturer fails: 1.f3 Sc6 2.Kf2 Sd4 3.Kg3 Sxe2+ 4.Kh4 Sd4 5.Ba6; now 5…e5+ would be a disruptive check, while 5…Sc6 6.d3 e5+ 7.Bg5 would delay the promotion by one move and hinder White’s development later. Black captures with the queen instead: 1.f3 e5 2.Kf2 Qf6 3.Kg3 Qa6 4.Kh4 Qxe2 5.g3 Qf2 6.Ba6 e4 7.d3 e3 8.Sd2 e2 9.Sb3 e1=Q 10.Bg5. With four black moves left, the original queen can’t head for home with 10…Qe3 11.Qe2 Qe7 when the e1-queen is stuck and prevents 12.Rf1. So Black vacates the first rank immediately with 10…Qe7, which leads to the promoted queen replacing the original one on d8 – the Pronkin theme. 11.Qe2 Qfe3 12.Rf1 Qd8 13.Sc1 Qee7 14.Qxe7+. The original queen lands on the pawn’s starting square (where it’s captured), hence the two thematic units have swapped places.

Andy Sag: Hard work this one! The black e-pawn must promote and either be captured on the final move or replace an identical piece captured. A pair of knights would take too many black moves so it has to be a pair of queens. The need to avoid illegal checks leads to a unique sequence.

649

Chris Feather
The Problemist 2001, 1st Prize

Helpmate in 3, 2 solutions

Solution

The white force surrounding the black king is considerable, but surprisingly these units cannot coordinate to give mate themselves. Rather, White aims to promote the distant e7-pawn on a dark square and mate with the new queen. This plan requires Black to fire one of the R + B batteries, so that a rook can be sacrificed on f8 or d8. The black king also needs to shift to d4 or c4 to facilitate the queen mate; thus both players conspire to remove the white piece(s) controlling each square. 1.Bxe6+ Rf8 2.Rxf8+ exf8=Q 3.Kd4 Qc5 and 1.Bxf4+ Sd8 2.Rxd8+ exd8=Q 3.Kc4 Qd3. Spectacular captures of White’s only officers (in the diagram) in both solutions, with model-mate finishes.

Andy Sag: The two batteries look like they are there for a reason! In each case, Black starts with a battery check and removes both white pieces but White promotes the e-pawn and the resultant queen gets the job done.
Karel Hursky: Simple and delightful.
Andrew Buchanan: Very good but quite easy – telegraphed by the white pawns. Still the double sacrifice by White is very impressive.

650

Milan Vukcevich
Sah 1950, 2nd Prize

Mate in 4

Solution

It’s a plausible idea to bring the queen to the b-file, e.g. 1.Qxh2? threatens 2.Qb8 ~ 3.Rb1+ cxb1=Q+ 4.Qxb1, but White can’t afford to unpin the black bishop immediately: 1…B~! 2.Qb8 d1=Q creates a flight on d2. The queen can aim for b7 with 1.h4? e3 2.Bh3 e2 3.Qb7, but now the 4.Rb1+ threat is too slow. Such a try would actually work if the b2-rook were absent, since then 3.Qb7 carries the threat of 4.Qb2. Hence the point of the key 1.Rb8! lies in clearing the b-file for the eventual queen mate. 1…e3 2.Ba8 (again not unpinning the black bishop with 2.Bf1?) e2 3.Qb7, and with the bishop trapped by the e2-pawn, there’s no defence against 4.Qb2. An excellent combination of two clearance themes: the Turton doubling executed by the rook and queen, plus the Bristol carried out by the bishop and queen.

Andy Sag: One-liner with double Bristol clearance allowing the queen to mate on b2.

651

Leonid Makaronez & Rauf Aliovsadzade
OzProblems.com 13 May 2023

Mate in 3

Solution

The strong defence 1…Bxd4 gives the king a flight on e5 and isn’t provided with a white response, though 1…Rxd4 has a short set mate, 2.Qh5. The key 1.gxf4! adds a guard on e5 and threatens 2.Qh5+ Kxd4 3.Qd1 – a nice switchback. Furthermore, White has two knight “try-threats”: 2.Sd7? for 3.Sf6 but 2…Sxd7!, and 2.Sc6? for 3.Se7/Qh5 but 2…Sxc6! These black knight “refutations” serve as first-move defences against the real threat, and they provoke the white knight “try-moves” that failed before: 1…Sd7 2.Sxd7 and 3.Sf6, and 1…Sc6 2.Sxc6 and 3.Se7/Qh5. Two pairs of reciprocal knight captures are thereby produced when compared with the “virtual” threat-play – an unusual scheme. The 1…Bxd4 defence can now be exploited as a self-block: 2.f3 (threats: 3.Bxe4/fxe4) exf3 3.Qxf3. If 1…Sxb3 then 2.Qxb3+ Kxd4 3.Qd1 – another switchback – or 2…Rc4 Qxc4. Lastly, 1…h1=Q 2.Qxh1 (threat: 3.Qxe4) Rxd4 3.Qh5. It’s a slight pity that in this variation, 2…Kxd4 doesn’t stop the 3.Qxe4 threat and so 3.Qd1 isn’t forced.

Composers: Logical tries in threat, with the idea of mutual captures between tries’ refutations and the two main variations.
Jacob Hoover: The try 1.f3? threatens a couple of short mates, 2.Bxe4 and 2.fxe4, with the idea being that 1…exf3 allows 2.Qxf3+ Kxd4 3.Qe4, but 1…Rxd4! defeats this try.
Andy Sag: A few tries, e.g. 1.Sg6? Bxd4!, 1.f3? Rxd4! Black can take the d4-pawn creating a flight to f6 via e5, so finding the correct way to put a second guard on e5 solves this one. Nice switchback in the threat play!
George Meldrum: I found this problem difficult to solve with so many tempting tries being presented.

652

Vitaly Medintsev
5th FIDE World Cup 2017, 1st Hon. Mention

Helpmate in 2, 3 solutions

Solution

Three pins by White’s rooks and f1-bishop are arranged and these pieces could potentially mate by capturing the pinned unit. The black king has three flights, two of which the g1-bishop can cover if Black removes the f2-pawn, while the third on c6 can be guarded by a white piece not needed on a pin-line. (1) 1.Sxf2 unguards the mating square b2 and forces 1…Bg2, since 1…Rh6? would unpin the e5-bishop and let it defend b2 again; now the knight on f2 has only one safe landing square while opening the bishop-line: 2.Sh1 Rxb2. (2) 1.Qxf2 unpins the h5-rook and forces 1…Rc1, since 1…Bg2? would unpin the e2-rook which then controls the mating square e5; here the queen has fourteen legal moves but only one hideaway square: 2.Qe1 Rxe5. (3) 1.Rxf2 prepares the rook to unguard e2 (not possible when another piece captures on f2 and opens the rank) and forces 1…Rh6, since 1…Rc1? would unpin the b2-knight and allow it to access the mating diagonal; then the rook on f2 must choose its spot carefully with 2.Rf5 Bxe2. Top-notch cyclic change of functions between three white pieces, which take turns to guard, pin, and mate, while different black pieces perform hideaway strategies.

Andy Sag: In each case a different black piece captures the white pawn, White moves a different piece to guard the c6-knight, then the same black piece moves again to clear a diagonal for the g1-bishop to guard c5/b6 and allow White to capture a different black piece to do a pin-mate.
Jacob Hoover: In each solution there is only one possible square for the capturing black unit to go to after removing the f2-pawn that doesn't screw the mate up. There is also a cyclic relationship between the unpins and white captures.
George Meldrum: A very nice complementary set of solutions each with initially covering the flight squares of the black king then providing mate by the capture of a pinned piece.

653

Garen Yacoubian
La Tribune de Genève 1979

Mate in 2

Solution

Black’s queen is pinned on the first rank but it’s also half-pinning the two rooks. The loose black king has strong defences like 1…Kg2 that cannot be provided for, so White must resort to a flight-taking key. 1.Rb2? threatens 2.Qxb1 and leads to 1…Q~ 2.QxQ, 1…Qg1 2.Rh6, but 1…Qxa1! is an obvious refutation. Much more difficult is choosing between the two rook moves to the g-file. 1.Rg5? self-pins the b6-rook and waits for the queen to unpin it: 1…Q~ 2.Rh6 (with some duals, 1…Qc1 2.Rh6/Qh8 and 1…Qg1 2.Rh6/Qxg1). But Black can preserve the rook’s pin with 1…Kh2!, after which 2.Qh8+ would be handled by 2…Qh7. The key 1.Rg6!, cuts off the black queen’s diagonal in anticipation of its unpinning, and threatens 2.Qh8. Now that 1…Kh2 doesn’t answer the threat, Black must defend with the queen, either by capturing the white one or by sliding on the pin-line, to be ready to intercept on the h-file when it’s unpinned. All such defences, however, unpin the b5-rook: 1…Q~ 2.Rh5 (with a dual, 1…Qg1 2.Rh5/Qxg1). A fine miniature with plenty of pinning and unpinning effects. The two tries and key also produce dual-free changed mates for 1…Qd1/Qe1/Qf1.

Andy Sag: The key confines the black king to the h-file but leaves the remaining rook pinned, so White must take care to use the 6th rank rook to block the queen if it is unpinned.
Jacob Hoover: A wonderful display of dual avoidance in a miniature setting.

654

Zivko Janevski
Liga Problemista 1982, 2nd Place

Solution

If White could pin or capture the c1-bishop without checking, Black will be forced by zugzwang to play …Bxd7 mate. Set play of this type is prepared for all three bishop moves: 1…Bb2 2.Rxb2 Bxd7, 1…Bxd2 2.Bxd2 Bxd7, and 1…Bxa3 2.Qxa3 Bxd7. In three thematic tries (all waiting moves), White places a line-piece on b4 which interferes with the two remaining ones on the fifth rank. In each case the try-piece manages to deal with one of these weaknesses with a new continuation, but not the other. 1.Bb4? Bxa3 2.Bxa3 Bxd7, but 1…Bb2!, 1.Rb4? Bxd2 2.Rd4 Bxd7, but 1…Bxa3!, and 1.Qb4? Bb2 2.Qxb2 Bxd7, but 1…Bxd2! The key 1.Bc3! (waiting) cuts off the queen’s control of the potential flight on c1 and brings about yet another change: 1…Bb2 2.Bxb2 Bxd7. The other two variations are as set: 1…Bxd2 2.Bxd2 Bxd7 and 1…Bxa3 2.Qxa3 Bxd7. Great example of cyclic refutation, where three black moves rotate their functions as defences and refutations after three try-moves, with changed play involved in four phases including post-key.

Andy Sag: Additional tries: 1.d3? Be3!, 1.Qc3? Bxd2!, 1.Qd4? Bxa3! Black can only move the bishops so to force mate White must pin or capture the c1-bishop, but there is only one way to do it without messing up the mate.
Andrew Buchanan: Simple to solve and then harder to see the clean try play!
Rauf Aliovsadzade: The thematic tries are nice. The impression is that this selfmate is a reworking from a direct stalemate.
Jacob Hoover: White has three thematic tries, each of which involves the square b4 and has one changed continuation. Note the cyclic pattern seen with the labelled defences and refutations. A very nice block-mutate.

655

Anatoly Styopochkin
Šahmatnaja Poezia 2002
Rashid Usmanov-40 Jubilee Tourney, 1st-2nd Prize =

Solution

The black king has two flights on d6/d5, and one could be covered by the potential mating moves Bxe7/Rxf5. But Black has insufficient time to unguard the mating line and block the remaining flight; e.g. the white king is carefully placed to prevent 1.Rg7 2.Rxg5 3.Q~ 4.Be4 5.Bd5 (check!) Bxe7?? and 1.Sc8 2.Sd6 3.Q~ 4.Rb7 (check!) 5.Rb6 Rxf5?? (There are additional tries where the black king goes to d6/d5, aiming for the same mating moves.) The actual solutions, though, surprisingly involve moving the king to each of the white knights’ starting squares. To legalise these king captures, Black begins by capturing the other knight with a promoted piece. 1.h1=B (not 1.h1=Q as the piece would guard the mating square) 2.Bxc6 3.Kxd4 4.Bd3 5.Re4 Bc5 and 1.d1=R 2.Rxd4 3.Kxc6 4.Rb7 5.Bd7 Rc5. In both parts, the king’s move simultaneously unpins the e7-rook and f5-bishop, which then self-block two new flights; even the pair’s move orders are forced for analogous reasons, as they need to avoid interfering with each other. The two solutions end with model mates delivered by different pieces on c5, the black king’s original square.

Andy Sag: The two white knights look essential but who would have thought you need to get rid of both of them (then use the unpinned pieces as self-blockers)!
Andrew Buchanan: One of the most difficult to solve of the recent problems.

656

Jean Haymann
Al Hamishmar 1977
Israel Ring Tourney, 2nd Prize

Helpmate in 2
Twin (b) Qf5 to f6

Solution

The black king has access to e2 and d2 which prove too difficult to deal with, and this combined with the prospective half-pin arrangement of the a4-rook and two black knights suggest that the king will be mated on the fourth rank. White’s queen and d7-rook are respectively the sole pieces guarding f4/d4, and in each phase Black begins with a knight move that cuts off one of these units. In part (a), the king goes to f4 to enable a knight mate on d3 – firing an indirect B + S battery – and this requires the black queen to be pinned as well. 1.Se5 Rxf7 2.Kf4 Sxd3. Black’s knights and queen are all necessarily pinned in the mating position, with all three pins forming during the course of play. In (b) with the f5-queen starting on f6, the king goes to d4 to prepare for a direct B + S battery mate, and the queen must be pinned again. 1.Sd5 Qh8 2.Kd4 Sxg4. We see another triple pin-mate involving Black’s knights and queen, plus the same white pinning units as before, yet strikingly each thematic black piece gets pinned on a different line compared with the first part. In fact, there’s a cyclic pattern of the three white pieces altering which of the three black pieces they pin.

Andy Sag: The two white pawns give a clue that the black king moves to an adjacent file but it is hard to see that both black knights and the black queen must all be simultaneously self-pinned by that move.

657

Jean Morice
Journal de Genève 1977

Mate in 3

Solution

The four white-pawn moves are notable tries; all aim to promote but Black defeats them with a lateral rook move followed by a check or a pin on the seventh rank, e.g. 1.b7? Rb1! 2.e7 Rxb7. Also tempting is 1.Rxg6? (threat: 2.Rg8), which fails to 1…Rxf5! and if 2.Rg8+ Kxh7, the g4-pawn disables 3.Bg5. A fairly obvious key 1.Rh2! activates the white rook and threatens 2.Rxa2 and 3.Ra8. Now almost all legal moves by Black’s rook and bishop counter the threat, and we can divide these defences into three groups. (1) If the rook moves along the file, it is decoyed to interfere with the a1-bishop: 1…Rf2 2.b7 (3.b8=Q) Rb2 3.Bxf6, 1…Rf3 2.c7 (3.c8=Q) Rc3 3.Bxf6, 1…Rf4 2.d7 (3.d8=Q) Rd4 3.Bxf6, and 1…Rxf5 2.e7 (3.e8=Q) Re5 3.Bxf6. These variations demonstrate the Roman decoy theme. (2) Each bishop move prospectively interferes with the rook by closing a file that needs to be accessed: 1…Be5 2.e7 (3.e8=Q) Bxd6 3.Bxf6, 1…Bd4 2.d7 (3.d8=Q) Bc5/Bxb6 3.Bxf6, 1…Bc3 2.c7 ~ 3.c8=Q, and 1…Bb2 2.b7 ~ 3.b8=Q. (3) The rook’s moves on the rank unguard the f5-bishop: 1…Rb1/Rc1/Rd1/Re1 2.Rxg6 (3.Rg8) Kxh7 3.Bg5/Bxf6. In the first two groups, a sort of delayed Grimshaw interference between the two black pieces occurs no less than four times on different squares.

Andy Sag: Defences using bishop moves or vertical rook moves feature black line interferences. Unfortunately, defences using horizontal rook moves each contain a minor dual.
Jacob Hoover: Four rook defenses illustrate the Roman decoy. Two bishop defenses illustrate the Dresden theme [i.e. after 1…Be5 and 1…Bd4, the prior strong defences (respectively) …Re1 and …Rd1 are replaced by weak ones, 2…Bxd6 and 2…Bc5/Bxb6]. A masterful display!
Rauf Aliovsadzade: Very strange that this great three-mover wasn't awarded.
Bob Meadley: My kind of problem!

658

Michael Lipton
British Chess Problem Society-40 Jubilee Tourney 1959-60, 3rd Prize

Mate in 2

Solution

White’s B + S battery aimed at the black king is controlled by the three black major pieces. The same black pieces are also attacking the e6-knight, the front piece of the Q + S battery. If White shifts the d4-bishop, that will threaten 2.Sd4, and in two tries and the key, this bishop cuts off a different black piece to remove its access to the B + S battery. When Black defends by capturing the e6-knight with a second piece, it becomes pinned, and White delivers a battery mate by shutting off (or capturing) the third black piece. 1.Bb2? gives 1…Rcxe6 [x] 2.Sxe1 [A], 1…Rexe6 [y] 2.Sc5 [B], but 1…Qxe6! [z] refutes. 1.Bc5? (or 1.Bc3?) leads to 1…Rexe6 [y] 2.Sb2 [C], 1…Qxe6 [z] 2.Sxe1 [A], but 1…Rcxe6! [x] refutes. Correct is 1.Be3!, which yields 1…Qxe6 [z] 2.Sc5 [B], 1…Rcxe6 [x] 2.Sb2 [C]; here 1…Rexe6?? [y] is prohibited. Also, 1…fxe6 2.Qf8. A lucid display of a cyclic mating pattern, where the black defences [x/y/z] take turns in provoking the white mates [A/B/C] – effectively a mix of changed and transferred mates. The eminent British problemist and IM of composition Michael Lipton passed away in April this year.

Andy Sag: It is clear that either d4 or g7 can be vacated to threaten a double-check, but Black has four ways to capture the e6-knight and then three ways to defend the remaining battery mate so the solution must deal with all of this. Also the key adds a queen check (1…Qa1/Qb2+) but fortunately that does not defend the threat.
Jacob Hoover: In each of the thematic tries the black unit that gets cut off is the one which refutes the try. A shining example of line play which also features battery plays.
Paz Einat: A neat early “carousel” also known as cyclic mating permutations.