Weekly Problems 2022-A

Problems 580-604

580

Norris Easter
Bristol Times and Mirror 1926, 1st Prize

Mate in 2

Solution

The thematic key 1.Sxe6! self-pins the knight and concedes a flight on e4, adding to the one on c4. Two diagonal batteries are now aimed at the black king, with one utilised in the threat, 2.Sbc5. Black has four defences that place a unit on the e-file and unpin the key-piece, which then fires the other battery in distinct ways. 1…Sfe3 (or 1…Rb1/Rc1) 2.Sec5 – White recovers the e4-flight without unpinning the f5-bishop (2.Sg5+?). 1…Sge3 activates the g1-rook but also allows the h4-rook to guard e4, forcing 2.Sg7 (again not 2.Sg5+?). After 1…Se5 the black knight controls the Q + S battery, but the self-block on e5 permits 2.Sc7 (not 2.Sf4+? cutting off the h4-rook). The fourth main variation sees the e6-knight unpinned by the black king itself, with 1…Ke4 2.Sg5 – an indirect battery mate (2.Sec5+? Kf3). Even the by-play is excellent, making further use of the B + S battery: 1…Kc4 2.Sbxd4 and 1…Sxd2 2.Sxd2. Also, 1…Rxe6+ 2.Qxe6.

Andy Sag: The key deals with the unprovided flight, adds a second flight, self-pins the key piece and allows a check. Lots of variation (horse) play including changed mates, pin mates, battery mates, unpins, a double-check and a switchback.
Jacob Hoover: Because the key is a self-pin, the second battery needs help in order to fire. Five defenses unpin the knight.
Nigel Nettheim: The unprovided flight on c4 leads to a good key.
Thomas Thannheiser: Nice switchback in 1…Sfe3 2.Sec5.
Andrew Buchanan: Lots of line opening and closing to ensure uniqueness for every mate.
Dennis Hale: After staring intensely at the diagram position for about 30 minutes, frankly I am confused, and have to throw in the towel. Doubtless adding to this confusion is that 580 and 579 have Easter coming only one week after Christmas. Heaven knows what Pentecost will bring!

581

Shlomo Seider
Spiros Bikos Memorial Tourney 1991, 5th Prize

Helpmate in 2, Twin (b) Qh7 to b1

Solution

White’s c3-rook and e6-bishop are each half-pinning a pair of black units, but are themselves directly pinned. These black pins hamper attempts by these white pieces to capture and mate along the half-pin lines, even with the white queen’s assistance, e.g. 1.Sh1 Qxf5 2.Kg3 Rxd3??, 1.Rg7 Qxg7 2.Sh1 Bxf5?? Instead, the two sides organise to use the queen for such capture-mates on the two lines. The first part is solved by 1.Sxe4 Qxf5 2.Sxc3 Qxd3. The queen’s initial capture on f5 paves the way for the e6-bishop to pin the black rook. The black knight clears the e4-square for the queen to pass through, and captures the c3-rook as the only hideaway move. The twin shifts the queen to b1, ironically giving the piece immediate access to d3, its mating square in the first solution; but now the queen aims to mate on f5, which was accessible from the piece’s original square on h7. 1.Rxe4 Qxd3 2.Rxe6 Qxf5. Here the queen’s capture on d3 prepares for a pin of the g3-knight by the white rook. The black rook removes the e4-pawn for the white queen and does a hideaway by capturing the e6-bishop. The brilliant solutions incorporate multiple exchanges of functions between the pairs of half-pinned pieces and between the white rook and bishop. The white queen’s reversal of moves across the two parts seems very original.

Andy Sag: A tale of two half-pins; in each case the queen captures to unravel a half-pin then moves again to discover the pin by a piece which is itself pinned. Note that Black moves the same piece twice and ends up capturing the remaining pinned piece, not that it needs to be removed but because it is the only square Black can go to that allows the pin-mate to occur!
Nigel Nettheim: I found solving difficult. The twins match marvellously, with a rank versus diagonal pairing as well as the white queen’s double-hop in opposite directions. Only 5th prize? Sadly, the composer died in the year of this tourney.
George Meldrum: First sight of the diagram I said “Oh my goodness,” trying to solve I said “Oh my goodness,” and finally solving I said “Oh my goodness.” Worth more than 5th Prize.

582

Alexandar Atanasijevic
The Problemist 1972

Solution

Although Black could arrange for the king to be mated on b8 or d8 by the knight (Sa6/Se6), these options would take too long, and the king in fact stays on f8 for a knight mate on e6. This plan requires Black to promote all four pawns and self-block on e8, e7, f7, and g8. The only suitable piece for blocking e7 without disturbing the mate is a knight, and starting with 1.a5 2.a4 3.a3 4.a2 5.a1=S carries the advantage that the piece can remove two obstructing white pawns en route: 6.Sc2 7.Sxd4 8.Sxf5 9.Se7. Now 10.f5 11.f4 12.f3 13.f2 14.f1=Q costs the new queen only one extra move to remove the h-pawn before blocking e8: 15.Qxh3 16.Qh5 17.Qe8. Then 18.h5 19.h4 20.h3 21.h2 22.h1=B, avoiding a queen promotion as such a piece would check White on the next move: 23.Bxd5 24.Bg8. Finally, 25.d5 26.d4 27.d3 28.d2 29.d1=R ensures that the blocking piece on f7 doesn’t guard e6: 30.Rf1 31.Rf7 Se6. A classic Allumwandlung in which all four promoting pawns execute an Excelsior (they commence from their home rank), shown with perfect black economy.

Andy Sag: All four possible promotions are featured. The 31-move limit ensures a unique sequence of moves.
Jacob Hoover: Here all four possible pawn promotions are seen, so it’s an AUW. Also, each black pawn does an Excelsior.
George Meldrum: Very frustrating when you mistakenly put the white pawn on f4 instead of f5 and find you can only solve in 33 moves instead of 31.
Nigel Nettheim: One first had to find out why the pawn is on h3 rather than on h5; then the solution fell out. Adding a white pawn on h4 would have lengthened the solution but made it more obvious. A nice “packing” problem with all four promotion pieces.

583

Ado Kraemer
Deutsche Tageszeitung 1929

Mate in 4

Solution

Since the black king is confined by the g3-rook and knight, it seems straightforward to bring the h3-rook to the f-file and mate on f1, e.g. 1.Rh5 b2+ 2.Kb1 Bc3 3.Rf5 Be1 4.Rf1. However, Black has a self-stalemating defence: 1.Rh5? Ba1! 2.Rf5 b2+ 3.Kb1 stalemate (else 3…b1=Q+). To thwart this idea, White moves the rook across h6, to prepare for the knight to interfere on that square and release stalemate. 1.Rh7! (threat: 2.Rf7 and 3.Rf1) – not 1.Rh8? because of 1…Bc5! 1…Ba1 2.Kb1 (2.Sh6? is too early as Black could change plans: 2…Bc3/Bd4!). 2…b2 3.Sh6 Kxh2 4.Sf5. (If 2…B~ then 3.Rf7 and 4.Rf1.) Hence White exploits Black’s immobilisation by performing the Indian manoeuvre: critical play over h6 followed by a self-interference and a battery mate.

Andy Sag: A Meredith setting which is all about simultaneously avoiding checks and stalemates. A minor dual in 1…Bb2+ 2.Kb1/Kd1 B~ 3.Rf7.
George Meldrum: It’s all about the tricky one-line defence, without which you would have multiple keys by the rook.
Bob Meadley: A beauty and I think even Sam Loyd would like it.
Nigel Nettheim: The theme is “turning stalemate into zugzwang.” A rank-threat becomes a file-mate. It’s surprising that the h2-pawn is eventually captured. Altogether marvellous.

584

Mikola Vasyuchko
Ideal-Mate Review 1996, Commendation

Helpmate in 2, Twin (b) Remove Pc7

Solution

Despite the small white force present, Black proceeds to capture a white unit in each part: (a) 1.c6 Kb4 2.cxb5 axb5, and (b) 1.Qd7 Rb4 2.Qxa4 Rxa4. The white unit that’s sacrificed in one solution delivers mate in the other – the Zilahi theme. The thematic rook and pawn also exchange their positions, by giving mate on each other’s starting square. This light problem is capped off by a pair of ideal mates.

Andy Sag: Part (b) was easy to solve so I did that first. Part (a) was much harder to see. In each case Black moves the same unit twice and captures a white unit on second move, only to be then captured by the mating unit.
Bob Meadley: Pretty little thing and (b) was pretty hard for me.
Nigel Nettheim: The ideal mates with twinning were presumably the main aims. Those aims are impressively achieved here.
Andrew Buchanan: I found these surprisingly hard. I thought there had to be a discovered check from bishop battery in at least one twin. Naughty composer put the bishop on h1, when it would have worked perfectly well on a8!

585

Leonid Makaronez
OzProblems.com 5 Feb. 2022

Mate in 3

Solution

The key 1.Bf8!, by controlling e7, sets up an unobvious threat, 2.Sf6+ Kxe6 3.Bd7, in which the e6-knight is passively sacrificed. The same knight handles three defences in a variety of ways, actively sacrificing itself twice. 1…Rxe4 self-blocks e4 and also clears the square for the white queen, allowing 2.Sg5 with two threats, 3.Qd6 and 3.Qxe4 (White avoids 2.Sc5? Rg4!). The queen mates are separated thus: 2…hxg5/R~/Sg3/Sc5/Bh7 3.Qd6, 2…Kd4/Be6/Bb4 3.Qxe4. 1…Rd4 prepares to stop the threatened mate on d7, but interferes with the c3-bishop and also self-blocks: 2.Sf4+ exf4 3.Qf5, 2…Sxf4 3.Sxc3. And 1…Sb8/Sc5 unguards c7: 2.Sc7+ Kd4 3.Bc5. A minor defence, 1…Bxe6, doesn’t disable the threat-move, but the finish is different: 2.Sf6+ Kd4 3.Qe4/Qd3. There’s a good try, 1.Kf8?, which leads to similar play as the key but it fails to 1…Rxe4! 2.Sg5 Rf4+. A lot of subtleties are packed in the three main variations.

Andy Sag: Placing a guard on e7 enables an effective threat but the trick is to find a way to do it without letting Black off the hook (thematic try 1.Kf8?). When you see the name Makaronez, expect a challenge.
George Meldrum: Even though all the variations were in set play it did not make it any easier to find such a subtle and unlikely looking key-move.
Karel Hursky: Great key, great play, and well-placed white king to prevent duals. Bravo, Mr. Makaronez.

586

K. R. Chandrasekaran
The Problemist 2012, 3rd Prize

Mate in 2

Solution

White’s b5-rook executes three tries and the key, all of which threaten 2.Bb5. 1.Ra5? abandons the potential rook mate on b4 and releases the black bishop, which refutes by cutting off the f5-rook’s control of d5: 1…Be5! (2.Bb5+ Kxd5). 1.Rb6? obstructs a knight mate on b6 and is defeated by 1…Sde5!, a similar interference with the white rook. And 1.Rxb3? prevents the a-pawn from mating on b3, and so unties the black rook: 1…Ra5! The key 1.Rb7! avoids these weaknesses, and it provokes six defences on e5 that yield different mates. 1…Be5 2.Rb4, 1…Sde5/Sb6 2.Sb6, 1…e5 2.Sxd6, 1…Sge5 2.Sxe3, 1…Qe5 2.Sd2, and 1…Re5 2.Qd3. Further, 1…Ra5 2.axb3 and 1…Rxc3 2.Qxc3. An intensive demonstration of multiple defences on the same square, enhanced by the rook tries (though it’s a pity that 1.Rxb3? isn’t refuted by a move to e5).

Andy Sag: Easy to see that any threat must deal with removal of the c2-pawn, there to guard d3 so the key must vacate b5 to allow the bishop to go there. All eight variations are set.

587

Hartmut Laue
StrateGems 2003, 3rd Prize

Solution

White wants to play Sg5+ to induce …Bxg5 mate, but this works only if the black king is prevented from escaping to d6. Three thematic tries by the queen attack d6 for this purpose, and Black defeats them in a uniform manner. After 1.Qd3?, 1…g5! anticipates that 2.Sxg5+ will open a queen line to h7 and hence disrupt the battery mate. Likewise, 1.Qb4? is answered by 1…g3!, when the threat-move gives the queen unwanted access to h4. And 1.Qc6? f2! sees a third black pawn prospectively clearing a line for the queen, this time to h1. White needs to place the queen carefully with 1.Qxb6! (same threat), which triggers a couple of knight defences. 1…Scxb6 2.Re7+ Bxe7 and 1…Sdxb6 2.Rf6+ Bxf6. The main tries display an attractive geometry – the queen lines up with the e4-knight and a black pawn in three directions.

Andy Sag: There are many ways to add a second guard to d6 (to free up the e4-knight) including six tries; however, the unlikely looking queen-sacrifice does in fact work.
George Meldrum: It all looked too easy, but after three failed queen moves, a less obvious queen move came as a surprise.

588

Norman Macleod
feenschach 1989, 1st Prize

Helpmate in 3, 2 solutions

Solution

If Black self-blocks one of the two flight-squares on g3 and g5, the other one can be covered by the white knight when it mates. But 1.Be1? tempo?? 2.Bg3 Sh3 and 1.Rc5? tempo?? Rg5 Se2 fail because in each case White lacks a waiting move that could preserve the mating scheme. In the solutions, Black contrives to give the white king room to move by interfering with the black queen. 1.Bc3+ Kd3 2.Be1+ Kd4 3.Bg3 Sh3 and 1.Rc4+ Kd5 2.Rc5+ Kd4 3.Rg5 Se2. Black’s hesitating first moves with the bishop and rook are well motivated, allowing the white king to perform a tempo-switchback.

Andy Sag: As set, the knight can mate as soon as either g5 or g3 is self-blocked but first Black must lose a tempo to allow the white king a move followed by a switchback.
Thomas Thannheiser: White has no waiting moves, so Black has to help on the way to g5 or g3.
Jacob Hoover: Black needs to give up a tempo with a move that blocks a black guarding line in order to give the white king a legal move.
Andrew Buchanan: Very nice indeed. (1+2) x 2 = 6 tempos with just twelve units.

589

Günther Flad
Deutsche Schachzeitung 1937

Mate in 4

Solution

White has several ways to threaten a rook mate on either the h-file or the eighth rank, and two such attempts are logical tries. 1.Kg6? threatens 2.Rd8, but 1…g1=Q+! refutes. White could aim to force the g-pawn to promote to the “wrong” piece by playing 1.Rd3? first – it threatens 2.Rh3 and hence provokes 1…g1=S. However, 1…Bc8! also guards h3 and defeats that try. White therefore starts with 1.Rd5! to threaten 2.Rh5 and induce 1…f5, a defence that closes the c8-h3 diagonal. Now that 2…Bc8 is ineffective against 2.Rd3 (not 2.Rxf5? f1=Q!), Black must play 2…g1=S instead. And with the queening defence by the g-pawn no longer available, White proceeds with the original plan, 3.Kg6 followed by the unstoppable 4.Rd8. A tricky example of the Holst theme, in which White eliminates an adequate promotion defence by compelling Black to promote to a different type of piece.

Andy Sag: A one-liner with mate threats in the correct sequence that forces Black to first block the bishop from guarding h3 and then promote the g-pawn to a knight, precluding its promotion to a queen to control the g-file.
Andrew Buchanan: Surprisingly simple, and definitely fun.
George Meldrum: Cute.
Bob Meadley: Very good; simple but devious four-mover.

590

Milan Vukcevich
Chess Life 1985, 3rd Prize

Helpmate in 2, 4 solutions

Solution

It takes Black two moves to capture either white pawn and open a line for the b6-rook to deliver the eventual mate. Meanwhile White allocates the first move to the g2-rook for controlling the appropriate flights, and to simultaneously close at least one black line of guard that’s hindering the mate. In the first pair of analogous solutions, 1.Bxg6 Rg5 2.Be4 Rh6 and 1.Rxb4 Rd2 2.Re4 Rb1, Black’s bishop and rook make a Grimshaw interference on e4, while the g2-rook shuts off the black queen in two directions. In the second pair, 1.Sxg6 Rg4 2.Se7 Rh6 and 1.Sxb4 Rc2 2.Sd5 Rb1, it’s the two black knights that interfere with the queen, while the g2-rook effects a sort of Novotny by cutting off both the black bishop and rook at the same time. Wonderful line-opening and -closing effects, shown with near-perfect economy.

Andy Sag: As set, Black has three ways to stop mates on first rank or h-file. In each of the four solutions, Black removes an obstructive pawn, White blocks one or two black lines with the g2-rook, the same black piece then moves a second time to clear a line for the b6-rook and self-interfere with the remaining black line(s).
Jacob Hoover: In the first two solutions, we see the bishop and rook interfere with each other – a black Grimshaw. In the last two solutions, the g2-rook blocks these two pieces’ control of the mating line.
Andrew Buchanan: Very pleasant and quick to solve.

591

Unto Heinonen
Springaren 1996, 2nd Prize

Solution

Black is missing the queen, two bishops, and h-pawn, and since the c8-bishop never escaped from its home square, it’s the queen and f8-bishop that account for White’s doubled pawns on the e- and g-files. White is missing three queen-side pawns, all captured on their original files, including the b-pawn on b6. What piece(s) captured Black’s h-pawn and c8-bishop? It would take too long to use the queen or a knight, but the h1-rook has an efficient route for doing both jobs – h1-h6-c6-c8 (hence the sixth rank cannot be blocked by Black prematurely). For removing White’s a- and c-pawns, the best candidate is the a8-rook. 1.b4 c5 2.b5 Qc7 3.b6 Qg3 4.hxg3 h6 5.Rxh6 axb6 6.Rc6 Rxa2 7.Sa3 Rxc2 8.Bb2 Rc4. Black is soon able to develop on the king-side and allow White to play Rxc8 without a disruptive check(mate). However, once on c8 the rook would take too many moves to return to h1; it’s actually faster for White to castle and shift the other rook to h1, meaning the rook on c8 needs just two more moves to finish on a1. This plan precludes 8…Rc3? where Black tries to return the rook to a8 in a conventional way, e.g. 9.Sc2 g6 10.e4 Bh6 11.Bc4 Be3 12.Se2 e6 13.dxe3 Se7 14.Qd3 Kf8 15.0-0-0 Kg7 16.Rxc8 Sbc6 17.Ra8 Sb8 18.Ra1 Kg8 19.Rh1 Ra3 – one move too slow. Instead, Black echoes White’s idea by castling and transferring the other rook to a8, leaving the original one to occupy h8. 9.Sc2 Rch4 10.e4 g6 11.Bc4 Bh6 12.Se2 Be3 13.dxe3 e6 14.Qd3 Se7 15.0-0-0 0-0 16.Rxc8 Sbc6 17.Ra8 Rh8 18.Ra1 Ra8 19.Rh1 Sb8. The two pairs of rooks are thus interchanged – a masterful accomplishment.

Andy Sag: As I suspected, castling enables each king to get to its destination in one move and the rooks all do a shuffle! Nice one!
Jacob Hoover: It didn’t take me very long to figure out that each pair of rooks would end up swapping positions. It took a little more time to figure out that both sides would castle. From the final position White has mate in one with 20.Rxh8!

592

Valentin Rudenko & Viktor Chepizhny
Problem 1958, 4th Prize

Mate in 2

Solution

The black king has two flights in the initial position, both provided with a mating response: 1…Kxc5 2.Qxb4 and 1…Ke4 2.Qxe3. In each case White exploits a different self-pin of the c2-knight. Shifting the d5-knight will create the threat of 2.Qd5, but any such move will also abandon the knight’s control of b6 and e3, and thus spoil the set variations. The key 1.Sf4! compensates by (1) opening a potential pin-line on the fifth rank and (2) guarding d3. Now 1…Kxc5 immobilises both the e5-rook and c2-knight to allow 2.Qxe3, while 1…Ke4 self-pins the knight for 2.Qxb4. The two queen mates of the set play reappear but are swapped against the two flight-moves, to produce a striking reciprocal change. There’s by-play with 1…Rxc5 2.Qd3. A bonus try 1.Se7?, by controlling f5, generates another changed mate, 1…Ke4 2.Rh4 (1…Kxc5 2.Qxe3), but it’s defeated by 1…Rxc5!

Andy Sag: The two set flights are provided for by pin-mates. The key also threatens a pin-mate and leaves both flights available but the mates are reversed!
Henryk Kalafut: Reciprocal exchange of mates between set and actual play including three changed mates after 1…Ke4.
Andrew Buchanan: It was only after the fact that I noticed the beautiful way that the queen mates Qxb4 and Qxe3 are reversed between set play and actual play. That really is lovely.

593

Emil Plesnivý
Národní politika 1938, 1st Prize

Mate in 3

Solution

White has two significant tries that threaten immediate mates: 1.Sf6? (2.Sxd7) Rd2! (incidentally, if 2.Qd3 to shut off the d2-rook and c2-bishop for 3.Sxd7/Sh7, then 2…Rd1+!), and 1.Bf6? (2.Be7) Bh4! (if 2.Bxh4 then 2…Rg5!). With 1.Qe3!, which threatens 2.Qxh6+ Rg7 3.Qxg7/Bxg7, White sacrifices the queen to the two black pieces that had refuted the tries, drawing them to a weaker position. After 1…Rxe3, 2.Sf6 (3.Sxd7) compels 2…Rd3, which interferes with the c2-bishop and permits 3.Sh7. Similarly, 1…Bxe3 is answered by 2.Bf6 (3.Be7), when 2…Bg5 interferes with the g2-rook and allows 3.Bg7. Each of these matching variations shows the Roman theme: a black piece is decoyed and forced to play a substitute defence (2…Rd3/Bg5) that’s analogous to an original refutation (1…Rd2/Bh4!), but the new defence contains a weakness (self-interference) that White can exploit. (1…Rg5 leads to 2.Qxg5 with numerous threats.) Here the rich three-move strategy is unusually combined with four model mates.

Andy Sag: The key is like a Novotny-style manoeuvre, followed by different interferences depending on which piece takes the queen.
Thomas Thannheiser: The Novotny try 1.Qg3? (2.Bg7/Bf6) could be averted by 1…Bd4! The key 1.Qe3! provokes a double Roman decoy.

594

Hermann Seitz
Rochade Europa 1995, Special Hon. Mention

Solution

White could promote the e- or g-pawn on various squares in six moves, but such a plan fails as the black king would take too long to get in position for any promotion mate. We aim for a pawn mate instead and a likely final square for the king is e5, given that a diagonal path from a1 facilitates a unique move order plus that many white pawns are already controlling its surrounding squares. A king trek to e5 requires the white d2-pawn to unguard c3 and the black d4-pawn to be removed. 1.e4 g3 2.e3 dxe3 3.Kb2 exd4 4.Kc3 d5 5.Kd4 e3+ 6.Ke5 gxf4. A delightful mating picture in which all eight white pawns contribute to the model mate.

Andy Sag: Tried promotions first to no avail. The correct pawn play is quite difficult and involves the d2-pawn moving to d5 to guard e6.
Andrew Buchanan: Really quite hard to get the idea out of one’s head that pawns promote, although there’s no way this would have gotten a Special Mention with that, because there’s no interesting way to promote. Then there’s the clue that the black king can’t move uniquely off the main diagonal. But still hard to see the solution, even with the idea that eight white pawns can provide a model mate in the middle of the board.

595

Andy Sag
OzProblems.com 16 Apr. 2022

Mate in 2

Solution

The flight-giving key 1.Rh5! threatens 2.Rh8. Black has four defences that stop this mate on the top rank but they yield to other mates on the same line. 1…Kg8 2.Qe8, 1…Bh7 2.Rd8, 1…Rh1/Rg1 2.Qc8, and 1…Rb8+ 2.axb8=Q. Including the threat, each of these five mates delivered by various pieces occurs on a different square along the rank. A good try, 1.Rh2? Bh7!, entails a thematic threat, 2.Rh8, though it’s a pity that 1.Qh1? threatening 2.Qh8 cannot be incorporated as a proper try that’s uniquely defeated. All units in the diagram take part in the variation-play directly, with no plugs, in this nicely constructed Meredith.

Composer: Theme is mates on the eighth rank. The key gives a flight, leaves the white queen en prise and set check in place. Thematic tries 1.Rh2? Bh7! and 1.Rxd3? Re1!
Jacob Hoover: Two defenses involve line play of some sort. 1…Bh7 closes the h-file, but it also opens the d-file for 2.Rd8, while 1…Rh1/Rg1 gains control of one mating line and loses control of another, allowing 2.Qc8.
Andrew Buchanan: At last one where I could straightforwardly see the theme! Clean, economic demonstration of multiple back rank mates with flight-giving key.

596

James Malcom
OzProblem.com 23 Apr. 2022

Mate in 8, 3 solutions

Solution

This is an improved version of Weekly Problem No.548 from last year, by the same composer. Two of the solutions are identical to those of the earlier problem, commencing with a queen and a knight promotion. They show different ways by which White deals with Black’s impending stalemate and the threat of …g1=Q when the king is released to the h-file. 1.f8=Q! g5 2.Rg8 Kh2/Kh1 3.Qh6+ Kg1 4.Qxg5 Kh2/Kh1 5.Qh4+ Kg1 6.Qxg3 Kh1 7.Rh8+ Kg1 8.Qh2. 1.f8=S! g5 2.Sh7 Kh2/Kh1 3.Sxg5+ Kg1 4.Sh3+ Kh2/Kh1 5.Sxf2+ Kg1 6.Sh3+ Kh2/Kh1 7.Sg5+ Kg1 8.Sxf3. The addition of two units on the b-file cleverly brings about a third solution that involves promoting to yet another type of piece. 1.f8=B! g5 2.Bh6 Kh2/Kh1 3.Bxg5+ Kg1 4.Bh6 Kh2/Kh1 5.Bxe3+ Kg1 6.Bbd4 b2 7.Bxf2+ gxf2+ 8.Bxf2. A terrific three-fold promotion change.

Composer: The two additional pieces make all the difference. Also, the e3-pawn now serves as more than a minor cook stopper. I dare not dream of an Allumwandlung!
Andy Sag: Each solution starts with a different promotion followed by a one-liner sequence featuring alternate self-interferences (or unguards) and checks to ensure stalemate avoidance. Also in each case the promoted (or identical) piece mates.
Mark Salanowski: Very enjoyable more-mover. I found the bishop promotion solution the trickiest to find the correct sequence. Extraordinary that all three solutions require exactly eight moves.
Bob Meadley: It’s a beauty!

597

Raúl Ocampo
OzProblems.com 30 Apr. 2022

Helpmate in 3, 2 solutions

Solution

Inspired by Norman Macleod’s Weekly Problem No.588 (see above), this helpmate involves similar tempo tries where Black blocks a flight-square without considering White’s lack of waiting moves, e.g. 1.Bc7? tempo?? 2.Bd8 tempo?? 3.Bg5 Se2. The two problems, however, are quite distinct in showing different ways of overcoming this tempo difficulty. The solutions are 1.Kg3 Sh3 2.Qg5 Sg1 3.Kf4 Se2 and 1.Kg5 Se2 2.Bg3 Sg1 3.Kf4 Sh3. In each part, the black king opens a line for another black unit to travel through before returning to its initial square – such a three-move manoeuvre is called the Klasinc theme. The side-stepping king also enables the white knight to make a unique tempo-switchback without disturbing Black’s plan. In addition, the knight’s play shows an AB-BA reversal of the first and last moves across the two phases.

It is possible to combine the play of both the Macleod prize-winner and this problem into a four-part helpmate. Refer to the next diagram.

Composer: Another way of playing with White’s time, using the knight. Apparently I have combined the Klasinc and Zalokotsky themes. [The Zalokotsky theme is enacted by the white knight, which visits squares ABC in one phase and the same squares in reverse order CBA in another.]
Andy Sag: The black king and white knight perform coordinated switchbacks to temporarily clear lines for the required self-blocks on g5 and g3 respectively.
Thomas Thannheiser: Reminded me strongly of No.588. But here the knight is the only white actor.
George Meldrum: Similar structure to No.588 but with a different form of implementation; cleverly done.
Andrew Buchanan: Very nice indeed! It solved very quickly, but the clarity and harmony makes it very satisfying.

597a

Norman Macleod (†) & Raúl Ocampo
Version by Peter Wong
OzProblems.com 7 May 2022

Helpmate in 3
Twin (b) Bc1 to b8, (c) Bc1 to a5, (d) BRc1

This setting incorporating the play of both earlier problems is still a Meredith, although it requires twinning (and two of the mating positions are the same). The attribution of such a version isn’t very clear, as one composer is deceased while another (me) is not responsible for any of the solutions. But Raul has agreed to my proposal as given above the diagram.

(a) 1.Kg3 Sh3 2.Bg5 Sg1 3.Kf4 Se2.
(b) 1.Kg5 Se2 2.Bg3 Sg1 3.Kf4 Sh3.
(c) 1.Bc3+ Kd3 2.Be1+ Kd4 3.Bg3 Sh3.
(d) 1.Rc4+ Kd5 2.Rc5+ Kd4 3.Rg5 Se2.

598

Leonid Makaronez
OzProblems.com 7 May 2022

Mate in 3

Solution

The key 1.d5! threatens 2.Qd4+ which produces two good pin-mates, 2…Be4 3.Sd3 and 2…Re4 Se6. Black answers by occupying e4 immediately (after which 2.Qd4? is too slow), but these defences now cause a Grimshaw interference between the two black pieces. 1…Be4 2.Qe3+ Ke5 3.Bg3 finishes with another pin-mate, while 1…Re4 2.Rxf5+ Kxf5 3.Qf6 involves a rook sacrifice. In the by-play, 1…g3 self-blocks and permits 2.Bxf6, which carries two threats (3.Qxf3 and 3.Rh4) separated by 2…Rh7/Rg7/g2 3.Qxf3 and 2…Re3/Re2/Bd3/Be4 3.Rh4. A curious change of effects for the two defences on e4, from self-pins in the threat play to mutual interference post-key.

Andy Sag: The key opens a queen-diagonal and threatens a check only answerable by self-pins allowing two pin-mates. If e4 is blocked (Grimshaw style) to avoid the threatened check, it allows different checks, in one case forcing a third pin-mate and, in the other, a surprise rook sacrifice leading to a queen mate on f6 which was difficult to spot.
George Meldrum: A good try with 1.Bf2? (threat: 2.Qe3+ Rxe3 3.dxe3), refuted by 1…Bd3!

599

Aleksandr Mochalkin & Paul Valois
The Problemist 1991, 2nd Commendation

Solution

Black has four legal moves to g5, three of which form a battery with the h6-bishop. Set play is prepared for these three moves, where White compels Black to fire the batteries by various means. The unit on g5 gets deflected with a check in 1…Rxg5 2.Qg4+ Rxg4 and 1…fxg5 2.Qh4+ gxh4. On 1…Kxg5, White hides the queen with 2.Qa3 to force 2…Kh5 through zugzwang. However, the strong defence 1…Bxg5+ has no adequate response. The hideaway key 1.Qa3! ensures that 1…Bxg5? would mate immediately, but now White must find new ways to deal with the other three defences. 1…Rxg5 is met by a changed deflection, 2.Bg4+ Rxg4, while 1…fxg5 requires a waiting reply, 2.fxg6 g4. After 1…Kxg5, White only needs to maintain the existing zugzwang, but remarkably all possible moves by the white pieces on the h-file (not to mention the three mobile pawns) would spoil the battery mate. Rather, White must find another unique hideaway move for the queen: 2.Qa1 Kh5.

Andy Sag: Black has only four moves, each capturing on g5. Moving the queen out of play leaves Black the choice of mating straight away or setting up a battery which can then be forced to fire by a check or a zugzwang.
Andrew Buchanan: There is the selfmate equivalent of an unprovided check: …Bxg5+. The blocking pawns obviously are to constrain white queen moves, and it’s no surprise to find there is only 1.Qa3!
Jacob Hoover: The key completely abandons the set play lines for new variations. Amazingly, after the key hides the queen away, White can still find a move that keeps her hidden away!

600

Karl Fabel
Die Welt 1951

White to play and not win

Solution

In this joke problem, how does White avoid mating the black king? Of the six possible moves, five would mate immediately, leaving 1.c4+! Rxc4 as the sole way to begin. White then has two candidate moves, both checks defendable by the same rook. If 2.Sc7+? then 2…Rxc7 3.Se7+ Rxe7 4.e4+ Rxe4 and now 5.fxe4 mate is forced by zugzwang. The correct sequence is similar but reverses the order of the checking moves – 2.e4+ Rxe4 3.Se7+ Rxe7 4.Sc7+ Rxc7, which results in a stalemate. An amusing piece of work that deserves to be better known.

Andy Sag: White must force the black rook to go anti-clockwise (from c4) in order to force a stalemate.
Jacob Hoover: I usually don’t do problems like this, but this one seems pretty easy…
Andrew Buchanan: I know this problem – it’s a good one.
George Meldrum: Great problem, quirky and crowd pleaser.

601

William Whyatt
British Chess Federation Tourney 1969
2nd Prize

Mate in 2

Solution

The black king has two flights to c5 and d3, neither of which is provided with a set mate. Moves by the c4-knight would threaten 2.Qxc3, and the key 1.Sb6! places another guard on d5. Not counting the king moves, Black has three defences that capture the c5-rook and two that capture the d3-pawn, all of which are exploitable as self-blocks. Splendidly, White fires the same B + R battery against these five defences in a variety of ways. 1…Sxc5 is a generic block that allows 2.Rg3. 1…dxc5 activates the b8-bishop but enables a double-check, 2.Rd7. 1…Rxc5 controls the battery directly but opens a line for the a6-bishop, resulting in another double-check, 2.Rg4. The generic block on d3, 1…Qxd3, admits 2.Rc7. 1…Bxd3 activates the h5-rook and requires a shut-off mate, 2.Rg5. The flight-moves add even more coherence to the problem by serving as extra defences to the same squares: 1…Kxc5 2.Qxc3 (a distinct mate from the threat) and 1…Kxd3 2.Rd5.

Andy Sag: The key provides for the two set flight-captures, one answered by the threat, the other by a pin-mate.
Jacob Hoover: Five different battery plays in one problem. Another gem from the late, great Bill Whyatt.
Michael McDowell: The Whyatt brings to mind a famous problem by the Maslar brothers with six self-blocks on two flights given by the key, but the novelty here is that all five thematic mates are delivered by the battery. An impressive task.

602

Michael McDowell
OzProblems.com 4 Jun. 2022

Mate in 2, Empress d6

The empress is a fairy piece that combines the powers of a rook and a knight.

Solution

The black king is totally confined by White’s pieces, including the empress which also forms a battery with the d5-knight. The fine key 1.Eb5! surprisingly concedes two flights and abandons the battery. Now White threatens four empress mates, in which the mating squares make a diamond shape: 2.Ed4/Ec3/Eb4/Ec5. Black has four possible moves, where each thwarts a different triplet of threat-moves and thus compels White to play the remaining viable one. 1…Ke4 2.Ed4, 1…Kc4 2.Ec3 (model mate), 1…exf5 2.Eb4, and 1…exd5 2.Ec5. The separation of the four threats with no extraneous variations generates the total Fleck theme.

Composer: The position can be moved down two squares, with the white king moved to, say, h1. Now the two mates following the pawn captures are models of the inferior sideboard variety, complementing the existing model. It depends on whether or not you think this is worth the useless king.
Andy Sag: The key poses a quadruple threat neatly separated by each of the four black moves. The empress does all the action.
Paz Einat: An amazing problem! A precise four-fold Fleck using the empress in its full capacity. And all this with a great thematic key providing two flights.
George Meldrum: This problem blows my mind.

603

Vilmos Schneider
Magyar Sakkélet 1962, 2nd Prize

Helpmate in 2
Twin (b) Pc6 to d6, (c) Pc6 to e6

Solution

It’s tempting to queen one or both white pawns on the seventh rank, but all such strong promotions fail, e.g. in (a) 1.Qe8 dxe8=Q 2.Bd7 exf8=Q hampered by the g5-knight, and 1.Qc7 d8=Q 2.Be6/Rg6 exf8=Q double-check but one flight remains. White actually promotes both pawns to knights in all three parts: (a) 1.Ke6 dxc8=S 2.Rf6 exd8=S, (b) 1.Qc7 d8=S 2.Rg6 e8=S, and (c) 1.Qe8 dxe8=S+ 2.Kg6 exf8=S. In every solution the pair of pawns move in parallel, leading to a complete change of promotion squares for each pawn. The routine black moves are more than justified by this formidable knight-promotion task.

Andy Sag: Interesting triplet. In each case both seventh-rank pawns promote to knights on adjacent squares resulting in the same mating configuration following the sixth-rank pawn from left to right.

604

György Bakcsi & Laszlo Zoltan
Great Britain–Hungary Composing Match 1995
3rd Place

Solution

The black king has just two flights in position (a) and none at all in (b) when a white queen stands on c2, but Black cannot set up any mate-in-1 for White in four moves with the king remaining on c4. Instead, the king aims for d5 in both phases, even though that square is controlled by three white line-pieces – the d2-rook, g2-bishop, and g5-rook. In (a), Black employs three units to interpose on the white lines converging on the target square: 1.Qe4 2.Sf5 3.Bd4 4.Kd5 to enable 4…Se3, a mate that requires the pinning powers of all three white pieces. By contrast, in (b) Black removes the three line-pieces altogether: 1.Qxg5 2.Qxd2 3.Qxg2 4.Kd5 to allow 4…Qd3. This witty problem exemplifies the humorous style of the late Hungarian duo.

Andy Sag: In each case the king moves to d5 after that square is unguarded by (a) three line-blocks that turn out to be self-pins, and (b) capturing all three guarding pieces in the correct order.
George Meldrum: Hilarious triple pin-mate in the first part, and elimination strategy in the second.