Weekly Problems 2021-A

Problems 528-541

528

Jean-Marc Loustau
Ideal-Mate Review 1993, Prize

Solution

The diagram is solved by 1.Sxg3 Ke7 2.Ke4 Kd6 3.Bf3 Sxg3. Black’s first move vacates e4 and captures the knight guarding it, allowing the square to be accessed by the black king on the second move. After the white king approaches, the bishop goes to f3 and interferes with the rook, permitting a knight mate on g3, which was cleared by Black’s initial knight move. The solution of part (b), with a white pawn on d4 instead, is perfectly analogous: 1.Sxe2 Kf7 2.Kf4 Kg6 3.Rf3 Sxe2. Here Black first vacates f4 and captures its guarding knight, so that the black king can move to that square. The white king approaches via another path, and the rook interferes with the bishop by landing on f3, enabling a knight mate on e2 – the square cleared by Black’s initial capture. The two parts involve a Grimshaw on f3, a curious capture cycle of the four knights, and ideal mates.

Andy Sag: This tricky twin features delayed knight recapture after self-block preventing the king’s return to f3. The white king looks out of play but don't be fooled. Its placement on e8 ensures its unique path and also prevents a cook, 1.Re3 d3 2.Sg2 Sh5 3.Sf2+ making 3...Sg1 illegal!
Jacob Hoover: Since the white knights swap roles between getting captured and delivering mate, this is a Zilahi. Also, we see the black rook and bishop blocking each other, so there's another instance of role-swapping.
Ian Shanahan: Wonderful, perfectly matching strategy between the two solutions (including paradoxical captures of White’s already meagre force, as well as a black follow-my-leader in each). Note, too, the exchange-of-function between the two white knights, and that for both solutions White's mating square is the same as Black's capture square. Of course – given the name of the publishing journal! – both sequences end in an ideal mate.

529

Charles Promislo
Good Companions 1915, 1st Hon. Mention

Mate in 2

Solution

In this block position, set mates are arranged for all possible black moves: 1…Rh~ 2.Qf4, 1…Rg4 2.hxg4, 1…e5 2.Sg3, 1…exd5 2.Qxd5, 1…Sg~ 2.Rxf6, and 1…Sb~ 2.Sd4. Tries that attempt to maintain the block include 1.Qe3? exd5!, 1.Bg3? e5!, 1.d6? e5!, and 1.h7? Rxh7+! The key 1.Qc3! (waiting) leads to numerous changed mates that result from the queen observing different squares. Black’s h4-rook now generates four distinct variations instead of two: 1…Rh~ 2.Qf3, 1…Rf4 2.Sg3, 1…Rg4 2.hxg4, and 1…Rxh3 2.Qxh3, where the second line also shows a transferred mate (from the set 1…e5 2.Sg3). The e6-pawn provokes a single new mate, down from the two prepared: 1…exd5/e5 2.Qc8. One more change comes from the g8-knight – 1…Sg~ 2.Qxf6, while 1…Sb~ 2.Sd4 is as set. A classic mutate from a great exponent of the form.

Andy Sag: Complete block. The key sacrifices the queen, moves double guard from f4 to e5 and features no less than five changed mates!
George Meldrum: A nice array of changed mates. A breeze in comparison to last week’s problem which was impossible to solve.
Nigel Nettheim: The complete block is handled by changing most mates. The changed play with the h4-rook is a highlight (rather than the sacrificial key).
Andrew Buchanan: Beautiful! Why wouldn't a lovely problem like this get a prize?
Ian Shanahan: The American problemist Charles Promislo was renowned for his sophisticated and elegant mutates (this being one of his very best). As with many mutates – i.e. complete blocks with mate-changes between set- and actual-play – the key is pendular; this time I spotted it instantly! The construction is absolutely perfect. Pure art!

530

Martin Hoffman
Die Schwalbe 2013, 6th Hon. Mention

Mate in 5

Solution

The black rook, while uncapturable due to stalemate, is defending against mating threats by both the c5-rook along the a-file and the d8-knight on c7; thus 1.Se6? Rf8+! or 1.Ra5+? Rxa5 2.Se6 Ra7. No progress is made by 1.Rc4? Rf4! or 1.Rc3? Rf3!, and 1.Re5? simply fails to 1…Rxe5! Correct is 1.Rd5! which puts Black in zugzwang. Short mates follow the weaker defences, 1…Rxd5? 2.Se6 then 3.Sc7, and 1…Re5? 2.Se6 Rxe6 3.Ra5. After 1…Rg5, 2.Rd1 exploits the placing of the g2-pawn, which hinders 2…Rg1, and the only way to prevent 3.Ra1+ is 2…Ra5, permitting 3.Se6 Ra7 4.Ra1 Rxa1 5.Sc7. The thematic defence is actually 1…Rh5, which seems to handle 2.Rd1 with 2…Rh1. But the terrific 3.Rg1! traps the black rook into another zugzwang position, as after 3…Rxg1 4.Se6 the black rook is lodged behind g2-pawn and cannot stop 5.Sc7.

Andy Sag: The black rook obviously can’t be captured so White’s mission is to force it to an obstructed position from where it cannot check after Se6 threatening Sc7 mate. The c2-pawn prevents a dual after 1…Rh5 2.Rd2, etc.
Mark Salanowski: Fantastic interplay between the two rooks. Deceptively hard problem to solve – moving the rook to the g1-square is hard to see as it looks so useless there.
George Meldrum: A true cat and mouse game. Amazing.
Michael McDowell: The Hoffmann is a lovely little problem. Ideal for solving from the diagram, very easy to set up from memory, and just the sort of thing with which to challenge a player.

531

Vladimír Kočí
OzProblem.com 23 Jan. 2021

Helpmate in 3, 2 solutions

Solution

The black king has to approach the centre in order to be confined by White’s king and f/g-pawns. In the first solution, the target square is e4, for a bishop mate from b7, and this plan requires neutralising three black pieces controlling the long diagonal. 1.Sd7 Bc8 2.Kxf3 b5 3.Ke4 Bb7. In the second solution, the black king heads to f4, to be mated by the knight on h5; here the prospective e5-flight can only be guarded by the white king, so two black pieces must cede control of d4. 1.Sd6 Sg7 (not 1…Sxf6?, self-pinning) 2.Kg3 Kd4 3.Kxf4 Sh5. Matching switchbacks by the mating pieces in the two phases, with different knights cutting off the d8-rook on the same line.

Andy Sag: White knight or bishop moves to allow the black king to dig itself into a trap on the 4th rank, and then switches back to deliver mate.
George Meldrum: Like a boomerang the bishop and knight come back to deliver the final blow. Really nice.
Karel Hursky: Two switchbacks. Bravo!
Ian Shanahan: The two solutions’ strategies match, culminating in a switchback mate. However, a small forest of material is required…

532

Michael Lipton
The Problemist 1991, 2nd Hon. Mention

Mate in 2

Solution

The black king has three legal moves, of which two are provided with set mates: 1…Ka8 2.Kc7 and 1…Kxc6 2.Qd5. The principal try 1.Qg3? (waiting) arranges new mates for these two defences: 1…Ka8 2.Qb8 and 1…Kxc6 2.Qc7, but 1…Ka6! remains unaccounted for. The key 1.Qg1! (waiting) changes the responses to the king moves again, to 1…Ka8 2.Qa7 and 1…Kxc6 2.Qb6, while answering 1…Ka6 with 2.Qa7/Qb6. The three pairs of changed mates illustrate the Zagoruiko theme, charmingly realised in miniature. Remarkably, there are three supplementary tries that produce a brand new mate for each king defence: 1.Qf7? Ka8 2.Sb6, but 1…Ka6!; 1.Ke7? Kxc6 2.Qc8, but 1…Kc7!; and 1.Qd5? Ka6 2.Qb5, but 1…Ka8!

Nigel Nettheim: Both set mates are changed and the forces are small, but the dual after 1…Ka6 seems unfortunate.
Andy Sag: I like the try 1.Ke7? Kc7! best.
Ian Shanahan: A 6-man setting of the 3x2 Zagoruiko pattern (i.e., a pair of black defences leads to new mates in each of three phases) embracing nine distinct mates in all across the various phases is a fantastic achievement – although not without flaws. It is a pity that there is a post-key dual; and it is also debatable whether the tries refuted by a move (1…Ka6) for which there is no set mate are genuine tries at all. Nevertheless, a lot of play is extracted from the lone black king.

533

Zivko Janevski
StrateGems 2013, 1st Prize

Helpmate in 2, 2 solutions

Solution

The black king has two horizontal flights, which combined with how the second rank is well-guarded by Black, suggests that White will mate on the first rank with a major piece. The first solution begins with 1.Rxe3+ Kxe3, a passive sacrifice of the e3-bishop designed to clear away two pieces blocking the white queen’s diagonal line to f1. Then 2.Kd1 self-pins the knight along the d-file – which is also opened by the white king’s move – permitting 2…Qf1. The analogous second solution starts with 1.Sxe4 Kxe4, sacrificing the e4-bishop in order to remove two pieces off the d-file for the white rook. Then 2.Kf1 self-pins the e2-rook on the queen’s diagonal – again also opened by the white king – to enable 2…Rd1.

Andy Sag: Well matched solutions where the rook and queen swap pinning and mating roles and the uncaptured bishop controls a potential flight square.
Nigel Nettheim: Brilliant! Either bishop is abandoned, as well as the capturing piece, so that the white king can clear the path for his rook. The d2-knight and the e2-rook are like-minded. The one remaining bishop is enough to help with a pin-mate on the first rank. Marvellous unity of the two solutions.
Ian Shanahan: Perfectly matching strategy and exchanges-of-function between the two solutions, each of which begins with a paradoxical capture of a white piece – a type of clearance-manoeuvre – followed by the white king capturing the B1-capturer (which ends up being pinned in the other solution for the ensuing pin-mate). At B2, the black king moves onto the pin-line, then White mates with the pinning piece from the other solution. Exceedingly rich and beautiful play throughout! Small wonder that the problem, composed by a Grandmaster, won 1st Prize.

534

Alexander Goldstein
Magasinet 1934, 1st Prize

Mate in 3

Solution

If White unpins the c5-bishop in any way to threaten 2.Bb4, the b6-bishop won’t have an adequate defence: 1…Bxc5 2.Qxc5 or 1…Bc7/Bd8 2.Qxc7/Qxd8. But extracting the white king off the pin-line immediately fails, e.g. 1.Ke4? Bg6+! 2.Ke5 Sf3+!, while 1.d4? is defeated by 1…cxd3 e.p.! The key 1.Kd4! threatens 2.Sxc4+ Kb5 3.Qd7. Black has four thematic defences by the g1-knight and h5-bishop on e2 and f3. When either piece plays to one of these squares, that prevents its access by the other piece, and White exploits such an obstruction accordingly. 1…Se2+ 2.Kxc4 and 3.Bb4 (since 2…Be2+?? has become illegal); 2…Bxc5 3.Qxc5, 2…Bc7/Bd8 3.Qxc7/Qxd8. 1…Be2 2.Kc3 and 3.Bb4 (2…Se2+??). 1…Sf3+ 2.Kd5 and 3.Bb4 (2…Bf3+??). 1…Bf3 2.Ke5 and 3.Bb4 (2…Sf3+??). A textbook example of mutual square-obstruction doubled.

Andy Sag: Knight checks or bishop defences on e2 or f3 enable the white king to safely unpin on the second move due to obstruction on those squares precluding further checks.
George Meldrum: The white king moves to five different squares in play.
Nigel Nettheim: The fearless white king leads from the front and dodges all the arrows. Many wonderful features include the interferences on e2 and f3 and the unsocial black queen sidling out to prevent a dual (1.Kd4! Sf3+ 2.Kd5, not 2.Kc3? Qa1+!).

535

Ľudovít Lačný
problem 1969, 1st Prize

Mate in 2

Solution

The key 1.b5! adds an attack on c6 to threaten 2.Qxe5. When Black moves the e5-knight to defend, a random placement – here specifically 1…Sg4 – unguards d7 and allows 2.Qxd8. The knight has six other legal moves, all of which are corrections that disable 2.Qd8, but these defences contain new errors that enable various white responses. 1…Sd7 2.Qd6 and 1…Sf7 2.Qe6 show interferences with the d8-rook and g8-bishop. 1…Sxf3+ 2.Qxf3 and 1…Sxc6 2.Qxc6 see Black clearing squares White wants to occupy. 1…Sxc4 2.Qg5 and 1…Sxd3 2.Qe5 involve self-pins answered by pin-mates (though technically 1…Sxd3 is not a real correction as it doesn’t stop the primary threat on e5). By-play: 1…Re8 2.Qd6. In total, seven knight defences are met by seven different queen mates – a record number for such a duel between the two pieces. The task is enhanced by the three pairs of corresponding defensive motifs, plus mate changes from the set play, 1…Sg4/Sxc6 2.Se3.

Andy Sag: The key frees up the queen for a variety of mates following the (near) e5-knight wheel. A changed mate after 1…Sxc6.
Andrew Buchanan: A clue is to wonder about the b2-pawn which is clearly blocking defence 1…Qb2/Qa1. So White must threaten something on d4 or e5. The first doesn't work, but if 2.Qxe5 is the threat, we had better protect c6.
Nigel Nettheim: The new pin-mate with 2.Qg5 is a highlight, and the change after 1…Sxc6, as well as the near-complete knight-wheel.
George Meldrum: Every now and then you get a problem that twists your brain. This was one of them.
Ian Shanahan: The lack of provision for 1…Sxd3 directly telegraphs the key. Anyway, one then discerns a wonderful duel between the white queen and the black knight, who executes five secondary corrections after his ‘random’ move to g4. There are seven distinct mates by the queen in all (including the threat). Neither of these tallies holds the record [individually], but we do savour a major accomplishment nonetheless. The f3-rook seems expensive.

536

Nigel Nettheim
The Games and Puzzles Journal 1987

Solution

The two kings have to be restricted by a small white force, so the white royal heads to the edge-square a3, to prepare for …cxb4 mate. To induce this mating move, White must either give a deflecting check on b4 with a promoted rook (not a queen since its promotion on e8 would check) or set up a zugzwang position. The former plan requires a second rook promotion aimed at controlling the black king, but it takes just too long, e.g. 1.Kb3 2.Ka3 3.Ba2 4.Bb2 5.Sf1 6.Se2 7.Sb3 8.e5 9.e6 10.e7 11.e8=R 12.Re6 13.Rd6 14.e4 15.e5 16.e6 17.e7 18.e8=R 19.Re4 20.Rb4+ cxb4. To arrange a mate through zugzwang instead, White places a promoted knight on b4 and the e3-knight on c4, where the pieces serve to both divert the black pawn and guard the black king. This scheme, combined with an initial promotion to a rook for a self-block on a2, proves to be the fastest. 1.Kb3 2.Bb2 3.e5 4.e6 5.e7 6.e8=R 7.Ra8 8.Ra2 9.Ka3 10.Bb3 11.Sc4 12.e4 13.e5 14.e6 15.e7 16.e8=S 17.Sf6 18.Sd5 19.Sb4 cxb4. An attractive sequence in which the move order is nicely forced without involving any white captures.

Jacob Hoover: We have two underpromotions (one of which is an Excelsior and both being necessary to avoid checking) and mate by zugzwang.
Andy Sag: After realising that none of the checking finales could be reached in 19 moves, I stopped flogging a dead horse and started looking for a zugzwang.
Karel Hursky: Love how the black king gets stalemated by Ka3, Sb4, Sc4... Not easy to see.
Mark Salanowski: How clever that every white piece must move in a unique series of 19 moves to force the selfmate with no check to the black king at the end. A joy to solve.
Henryk Kalafut: White needs three units to self-block the king’s flight-squares a2, b2, b3 resulting in a clever, unique sequence of moves. All the white pieces shifted their places during the solution.
Andrew Buchanan: Very nice. As usual with these kinds of problems there are plenty of mates in n+1 to taunt the aspirant solver. These are actually logical puzzles in that one can progress systematically and eliminate all possibilities apart from the correct one.
Ian Shanahan: There’s much to admire here: the capture-free white sequence ending with an atypically quiet move; the subtlety of the mechanism forcing total accuracy of the move-order via anticritical play; the economy of means. Bravo!

537

Miroslav Bílý
Problem Observer 1983, 1st Prize

Helpmate in 3½, 2 solutions

Solution

The first solution runs 1…Sh3 2.Kd4 Sf2 3.Qc4 Kb6 4.e5 e3. The final position shows an ideal-mate in that every flight-square around the black king is uniquely guarded by one white unit or blocked by a black one, with all pieces on the board participating in the mate. The second solution is 1…Sf3 2.Qc5+ Kb7 3.Rd6 d3+ 4.Kd5 e4. The same ideal-mate recurs but is shifted one square up, resulting in an exact echo of the chameleon type (the units change their square colours).

Andy Sag: Solutions are matched in the sense that final configurations are identical with all eight pieces shifted one square vertically.
Jacob Hoover: The final positions are ideal-mates and are chameleon echoes of each other.
Andrew Buchanan: Precisely echoed ideal mate – very nice. The longer these things are, the harder they are to make unique.
Ian Shanahan: This helpmate gem parades an absolutely perfect (0,1)-echo of an ideal mate – typical of its composer and his Hungarian compatriots. One can easily see why it gained the top prize.

538

Georges Legentil
Problèmes d'Échecs 1951

Mate in 6

Solution

The black king has no legal moves, so White must take care not to capture the g6-pawn and give stalemate. There are two alternative plans, both using the white king to interfere with a bishop-line to release the black king, followed by a battery mate. In the try 1.Kg8? g5 2.Kf7 g4 3.Kf6, White cuts off the h8-bishop, hoping for 3…Kxd4 4.Ke6, but 3…g3! 4.Bg2 h1=Q 5.Bxh1 g2 6.Bxg2 Kxd4 7.Ke6 is one move too slow. The solution sees the white king shadowing the g6-pawn as it marches down the file: 1.Kh6! g5 2.Kh5 g4 3.Kh4 g3 4.Kh3 g2 – now the white king can capture the pawn without causing stalemate because the h1-bishop’s line is closed – 5.Kxg2 Kxe4 6.Kg3. Thanks to Bob Meadley who sent me this amusing problem with his accompanying poem:

Whilst strolling with a pawn one day,
He turned his head to me to say,
“When I get down to g2
I wonder what you’ll do?”

So I replied, “My friend, you're dead.
And all your talk won't change my head.
By all means go take the knight,
It will be your last delight,
For king to g3 and off you go to bed.”

Andy Sag: A classic royal one liner; all white moves are by the king! To avoid stalemate, the g6-pawn cannot be removed until it reaches g2 forcing the black king to capture the e4-knight whereupon a battery mate is unleashed.
Andrew Buchanan: Very elegant machinery for the effect.
Mark Salanowski: A neat side-by-side march of the king and black pawn down the board to the g2-square where it can be captured without a stalemate. A “pas de deux” with a zugzwang denouement.
Ian Shanahan: I normally don’t solve directmates of such length, but in this case the elegant ratchet mechanism and stalemate-avoidance veritably leapt to my eye! A nice lightweight problem.
Thomas Thannheiser sent this earlier 5-mover by Fritz Giegold for comparison.

539

Kenneth Howard
Die Schwalbe 1929, 1st Prize

Mate in 2

Solution

The square-vacating key 1.Rb8! threatens 2.Qb7, which can be stopped by either of the black knights half-pinned along the diagonal. 1…Sb~ defends by opening a line for the a5-rook to pin the white queen, but the move also self-pins the d3-knight and permits 2.Sb4. 1…Sc5 guards b7 directly and also opens a line for the b1-bishop to pin the white knight, preventing 2.Sb4; but the defence has self-pinned the b5-knight (besides interfering with d4-bishop), enabling 2.Qxd6. There is one subsidiary variation, 1…R~ 2.Qxb5.

Andy Sag: The half-pin of the knights gives a clue.
Nigel Nettheim: A nice scheme of three half-pins. The key is just fair. The d4-bishop prevents 1…Sd4+; it could have been a pawn but, as a bishop, it enhances the variation 1…Sc5 2.Qxd6. Composed before the coming widespread use of try-play.
Andrew Buchanan: Fun to solve, and nicely constructed.
Ian Shanahan: Pins galore! Two thematic variations unfurl from the half-pinned black knights, leading to simultaneous pins of Black and White whereby the free white piece executes a pin-mate that exploits the half-pin action. One savours the very rich strategy (typical of the era), though alas, the key was spotted almost instantly.

540

Vladimír Kočí
OzProblems.com Mar. 27 2021

Helpmate in 5, 2 solutions

Solution

In both solutions, Black arranges various self-blocks while assisting White in promoting the e-pawn. 1.Sd3 e5 2.Qh3 e6 3.Bg3 e7 4.Bf2 e8=B 5.Qf1 Bh5 and 1.Qf5 exf5 2.Sg6+ fxg6 3.Rd3 g7 4.Rdd1 g8=Q 5.d3 Qg2. The first part features a model mate and good interplay between the two sides, as the white king’s position dictates that 4.Bf2 (closing the f-file) must precede 5.Qf1. Nice change of promotion piece and square.

Andy Sag: It is clear that the pawn must promote but the trick is to find what to promote to and which file to promote on to avoid premature checking.
Karel Hursky: Loved the mate with two bishops.
Andrew Buchanan sent an improved version of this helpmate with an additional solution, diagrammed below. The two composers have agreed to a joint authorship.

540v

Vladimír Kočí & Andrew Buchanan
OzProblems.com Apr. 3 2021

Helpmate in 5, 3 solutions

1.Sc3 d5 2.Qg3 d6 3.Bd3 d7 4.Be2 d8=B 5.Qe1 Bg5. 1.Qe5+ dxe5 2.Sf6+ exf6 3.Rc3 f7 4.Rcc1 f8=Q 5.c3 Qf2. 1.Qc5 dxc5 2.Sb6 cxb6 3.Kc3 b7 4.Kb3 b8=Q 5.c3 Qxb5. While dispensing with two black pawns, this setting renders a different queen promotion in the third solution which ends in another model mate.

541

William Whyatt
The Problemist 1967, 5th Prize

Mate in 3

Solution

The key 1.Qg7! grants a flight on d5 and threatens 2.Qe5+ dxe5/Rxe5 3.dxe5. Black has four possible moves to d5, all of which defeat the threat, but these defences also serve to annihilate the d5-pawn. Once that pawn is removed, White proceeds to divert the defender from d5, clearing that square for the d4-pawn to access and unmask the R + P battery. 1…Qxd5 2.Qb7 (threatening 3.Qb1 by exploiting the self-block on d5) Qxb7/Qc6 3.d5. 1…Bxd5 2.Qf7 (guarding f4 to threaten 3.Sf2, but not 2.Qf6? Bxc4!) Bxf7/Be6 3.d5, 2…Rf5 3.Qxf5. 1…Rxd5 2.Qg5 (again threatening 3.Sf2, but not 2.Qf6? Rxc5!), 2…Rxg5/Re5 3.d5, 2…Rf5 3.d5/Qxf5. 1…Kxd5 2.Qxg8+ Ke4 3.d5. A first-class three-mover with four harmonious variations and no by-play.

Andy Sag: Whyatt’s three-movers are consistently excellent with great variety of play; I especially like the switchback after 1…Qxd5. The queen is offered up on the second move in the threat and three of the defensive variations.