# Weekly Problems 2023-B

## 659

Reto List
feenschach 1979, 3rd Prize

## Solution

Although there are near-misses in which the black king almost gets mated on f8 or h8, the most interesting tries and the solution itself target g8. If it were legal for the king to jump to g8 then …Se7 would mate immediately, so it’s surprising there’s a unique sequence of four moves to organise the same finish. Tries include: (1) 1.Rh7 Kf5 2.Kf7 and the pawn prevents the white king from triangulating, 2…Kg5?? 3.Kg8 Kg6 4.Rh8 Se7; (2) 1.f5 Kg5 2.Kf7 Kxf5 3.Kg8+ Kg6 4.tempo?? Se7 – here the white king can lose a tempo but only by capturing the pawn, which deprives Black of the chance to push it (4.f4) as a tempo move! The solution utilises castling: 1.Rf7 Kf5 2.Rfh7 Kxf6 3.0-0+ Kg6 4.Rh8 Se7. The two rooks exchange positions while the white king succeeds in triangulating (i.e. perform a rundlauf that’s tempo-motivated), all of which – plus an ideal-mate finish – is accomplished using only six units.

Andy Sag: No solution unless Black can castle but not until move 3 because Black must allow the white king to capture the pawn as a tempo move while the black king stays on the 8th rank and the rooks change places!
Thomas Thannheiser: A tricky move-order puzzle! The black rook on f8 has to go first to h7 to allow a triangulation of the white king!
Andrew Buchanan: Easy but sublime.

## 660

Antonio Bottachi
L’Italia Scacchistica 1918, 5th Hon. Mention

Mate in 2

## Solution

The black king has a flight-move that’s provided with a set mate, 1…Kc5 2.Qb5. The key 1.Se6! removes that flight but cedes a new one on e6, to threaten 2.Qe4. Black’s thematic pieces are the c6-rook and b7-knight, which are half-pinned by the white bishop. In four variations, when each defender moves, it not only leaves the other piece pinned but also blocks a different flight-square, and both errors are exploited in the mate. 1…Rxe6 2.Qb5, 1…Rc4 2.Qxd7, 1…Sd6 2.S8c7, and 1…Sc5 2.S6c7. The king’s flight-capture doesn’t deal with the threat-move, 1…Kxe6 2.Qe4, while 1…dxe6 permits 2.Qd4. A classic combination of the half-pin theme and self-blocks, shown with an excellent key that sacrifices the knight to three black units.

Andy Sag: Sacrificial key takes the c5-flight (albeit provided by 2.Qb5) but allows a flight-capture which by the way does not defend the threat.
Paz Einat: One of the best early half-pin problems. Highly unified and without any weaknesses; even the set flight is provided.

## 661

Philippe Schnoebelen

## Solution

Black’s pieces have made 15 visible moves to reach their diagram positions, so with no spare moves left, the missing bishops and pawns were captured on their original squares. White is missing the a- and d-pawns, and the most efficient way to enable their removal as well as the captures of Black’s b-pawn, e-pawn, and bishops is to promote both. The two promotions take 10 moves and the remaining 5 are visible ones required by the three white units away from home, hence there’s no time to move the promoted pieces before their captures. 1.a4 g5 2.a5 g4 3.a6 g3 4.axb7 a5 5.d4 Ra6 6.d5 Rh6 7.d6 Rh3 8.dxe7 h5 9.exf8=S! Black needs to develop the queen quickly so that the two queens don’t get in each other’s way on the 4th rank, thus not 9.exf8=R+? forcing 9…Ke7; White also avoids 9.exf8=B? as the bishop would prevent …Ke7 later on. 9…Qh4 10.bxc8=B! Again 10.bxc8=R/Q+? Ke7 disrupts the black queen’s play, while 10.bxc8=S? controls e7. 10…Qc4 11.Qd4 Ke7 12.Qh4+ Kd6 13.Bg5 Se7 14.e3 Sxc8 15.Bd8 Rxf8. Each of White’s promoted pieces never moved from its promotion square where it is captured, and yet the choice of promotion unit is precisely forced. This is the eponymous Schnoebelen theme, and it’s shown twice here with good economy of length.

Andy Sag: Black needs all 15 moves to reach the diagram position so both black bishops must have been captured in situ. The missing pawns on both sides point strongly to White’s a- and d-pawns being the prime suspects (they have just enough moves to promote) but they must sub-promote with care to allow the black king to reach d6 legally.
Mark Salanowski: Black’s final two moves remove the promoted bishop and knight from the board and add the needed difficulty to solving this neat problem.
Jacob Hoover: The only legal way to get the two queens into the positions they have in the diagram is for them to execute a bi-colored Turton maneuver. I really like how the choice of promotions is determined in this problem. It was also cool to see a bi-colored Turton in a proof game.
Andrew Buchanan: The Schnoebelen is one of my favourite themes.

## 662

Viktoras Paliulionis
Rozmaitosci szachowe 2012, 2nd Prize

Helpmate in 2
4 set plays, 4 solutions

## Solution

The set plays comprise two pairs of symmetrical sequences. 1…Sf2+ 2.Kh5 Be8, 1…Sf6+ 2.Kh3 Bf1, 1…Bc6 2.Kh3 Sf2, and 1…Bd3 2.Kh5 Sf6. Comparing the non-equivalent parts, the b5-bishop and e4-knight swap roles in guarding flights and giving mate, with …Sf2/Sf6 repeated but serving these two different aims. Likewise there are two pairs of symmetrical solutions. 1.Kh5 Bg5 2.Rg4 Be8, 1.Kh3 Bg3 2.Rg4 Bf1, 1.Rh3 Be2+ 2.Kh4 Bg5, and 1.Rh5 Bd7+ 2.Kh4 Bg3. Here it’s the two bishops that exchange functions in guarding flights and delivering mate, with …Bg5/Bg3 recurring for these different purposes. Notwithstanding the symmetries, there’s a pleasant amount of variety in this miniature, which technically achieves eight distinct model mates in a single setting without twins.

Andy Sag: In all eight plays, the black king moves to the h-file, the black rook self-blocks and the white king stays out of the way.
Andrew Buchanan: Very easy but quite surprising. Obviously lots of symmetry but the fundamental four pairs are all quite clear, with no black tempo loss possible.

## 663

OzProblems.com 5 Aug. 2023

Mate in 3

## Solution

The black king has a flight-move that’s given a set mate: 1…Ke8 2.Bxg6. Although Black has few possible moves, e7 and f6 are well-protected and White needs to generate a threat on a light diagonal. 1.Qb8! threatens the queen sacrifice 2.Qe8+ [A] Kxe8 3.Bxg6 [B]. If 1…Bg7 to create a potential flight on f8, then 2.Bxg6+ [B] Kxg6 3.Qe8 [A] which sacrifices the bishop instead and swaps the two white moves compared with the threat play. The error committed by the bishop defence is that it unguards g7 and thus allows the sub-variation 2…Ke6 3.Sxg7. A second defence on the same square, 1…Rg7, by obstructing the f8-bishop, enables another sacrificial continuation, 2.Rxe5, which puts Black in zugzwang. Two sub-variations follow with 2…fxe5 3.Sxe5 and 2…Rg6 3.Be6. A formal move-reversal idea is nicely combined with traditional sacrifices.

Composer: At first glance, the flight-taking key may look like a weakness, but there is a nice compensation in the initial position: 1…Ke8 is met by the immediate 2.Bxg6. White AB-BA and black defenses on the same square.
Andy Sag: The flight-taking key leaves Black in a cramped position with only three legal moves. The queen sacrifice in threat play is a nice touch.

Jonathan Levitt

## Solution

No simple sequence can be arranged in which the black king finds a mating square without capturing any of the white force. The only workable scheme involves removing the a8-knight and placing the king somewhere on the top rank for a queen-promotion mate. That knight is protected by a rook, which is guarded by the bishop, and the latter in turn is covered by the other rook. So the king must take a trip to capture all of these pieces. 1.Kf3 2.Kg2 3.Kxh1. Since the e4-knight controls the long diagonal, to save time later Black has to eliminate this unit as well. 4.Kg2 5.Kf3 6.Kxe4 7.Kd3 8.Kc2 9.Kb1 10.Kxa1 11.Kb2 12.Kc3 13.Kd4 14.Ke5 15.Kf6 16.Kg7 17.Kxh8. Now the c6-pawn must be captured to free the b7-square. 18.Kg7 19.Kf6 20.Ke5 21.Kd5 22.Kxc6 23.Kb7 24.Kxa8 25.Kb7 26.Kc8 for a8=Q. The black king visits all four corners of the board while removing all extraneous units, to bring about a model mate. A lovely realisation of the theme in a rex solus setting.

Andy Sag: The only possible mate is promoting the a-pawn with the black king on the 8th rank but first his majesty must visit all four corners of the board, removing six white units in the correct order to keep within the 26-move limit.
Nigel Nettheim: Easy to solve, but nice, with a try 1.Kxe4? (Kxe4 has to be held in reserve to be part of the long leftward journey, thus saving a move).
Andrew Buchanan: This is quick to solve but really fun.

## 665

Andy Sag
OzProblems.com 19 Aug. 2023

Mate in 2

## Solution

A prominent black check is provided with a set mate, 1…Bf3+ 2.Rxf3. Two important tries by the b3-rook generate additional variations while preserving this set line. 1.Rg3? grants two flights to the king and threatens 2.Rg1. Then 1…Kf2 forces the double-checking 2.Rf3, and although 1…Ke1 doesn’t stop the threat-move 2.Rg1, the new king position means a distinct mate. This try is defeated by 1…Bxd1! The second try 1.Re3? threatens 2.Bxe2, and it leads to 1…B~ 2.Re1 and 1…Sxe3 2.Sd2. Here Black refutes with 1…Sd4! The key 1.Sc3! surprisingly interferes with the b3-rook but activates the one on a1, and threatens 2.Bxe2 again (though now it’s a double-check not hindered by 1…Sd4). The checking defence is met by a new mate – 1…Bf3+ 2.Bxf3, and 1…Bxd1 permits 2.Rxd1. The variety of play brought about by the two tries and the key is admirable. The well-utilised b1-knight serves very different roles in the three phases. The piece guards d2 in 1.Rg3? Ke1 2.Rg1, gives a virtual mate in 1.Re3? Sxe3 2.Sd2, and controls e2 after 1.Sc3! for the threat plus 1…Bxd1 2.Rxd1.

Composer: The key changes the mate after the set check in a pawnless Meredith setting. Multiple significant tries: 1.Re3? Sd4!, 1.Rg3? Bxd1!, and 1.Sf3? Bxd1!
Andrew Buchanan: This is a fun aristocrat problem, with some good tries. The key interrupts the provision for the provided check, which is somewhat paradoxical.
George Meldrum: Great tongue-in-cheek problem where you are encouraged to play 1.Rg3!? Nice try Lao Che!
Nigel Nettheim: The demon solver turns composer. The threat is strong, a double-checkmate; but what is appreciated is the change from the set 1…Bf3+ 2.Rxf3: the b3-rook is surprisingly taken out of play while the a1-rook is brought into play.
Bob Meadley: Never too sure with the devious Andy!
Dennis Hale: Andy Sag clearly is in the top echelon of our patrician solvers and now with no.665, he has composed a problem without proletarians. Congratulations to Andy, a truly intrepid supporter of OzProblems.

## 666

Sergey Sinakevich & Yury Fokin
Schach 1973, 1st Prize

Helpmate in 3
Twin (b) Sc5 to b4

## Solution

No rook mate can be arranged in three moves; instead, the white piece’s function is to support the c-pawn when it gives mate. In part (a), the king goes to b4 for the pawn to mate on c3, an aim that requires unpinning the mating unit, interfering with the h8-bishop, and dealing with two flight-squares. 1.Kb4 e3 2.Qb5 d3 3.Rd4 c3. In part (b), the king stays on b5 for a pawn mate on c4, and this necessitates unpinning the c-pawn again, cutting off the h4-rook, and removing three flights. 1.Qf2 e4 2.Qb6 d4 3.Ra6 c4 (Black avoids blocking b6 with the h8-bishop as that would leave the queen guarding the mating square c4). Three single-step pawn moves are neatly changed to three double-steps by the same pawns, and both parts finish with a model mate.

Andy Sag: In each case the white pawns move in the order e, d, c, lining up on a new rank.
Jacob Hoover: In (a), the e-pawn needs to stay out of the h4-rook’s way so that it can block the h8-bishop’s guard of c3; hence why it moves only one square. In (b), the e-pawn needs to move two squares in order to stay out of the queen’s way so that it can self-block; this move also unpins the c-pawn.

## 667

Waldemar Tura
Wola Gułowska 2012, 1st Hon. Mention

## Solution

The thematic try 1.g6? entails a double-threat, 2.Qe3+ Bxe3 and 2.Qd2+ Bxd2. Black can foil one of these queen checks by promoting the a-pawn to a minor piece, but the other threat remains effective. 1…axb1=S [x] 2.Qe3+ [A] Bxe3 (not 2.Qd2+? Sxd2+) and 1…axb1=B [y] 2.Qd2+ [B] Bxd2 (not 2.Qe3+? Bd3+). Here the wrong queen move would force Black to fire the a1-rook’s battery, but it’s not mate because that rook is controlled by the white one on a5. Another defence is 1…b3 which gives a potential flight on b4 but it doesn’t stop 2.Qd2+ Bxd2. This try is defeated by 1…f1=S! The key 1.Rxb5! removes a defender of a4 and threatens 2.Sa4+ Bxa4. Black counters by promoting the a-pawn to a minor piece, to re-guard a4 with the rook. In contrast to the position after 1.g6?, now the h6-bishop is not in play while the key-piece no longer attacks the a1-rook. Consequently, White must induce the rook battery to open, with 1…axb1=S [x] 2.Qd2+ [B] Sxd2 and 1…axb1=B [y] 2.Qe3+ [A] Bd3. Compared with the try play, there’s a reciprocal change of the white queen continuations (A and B) against the same black promotion defences (x and y). These variations also involve a paradoxical element: in the virtual play, 1…axb1=S disables 2.Qd2+ and 1…axb1=B disables 2.Qe3+, whereas in the actual play, 1…axb1=S enables 2.Qd2+ and 1…axb1=B enables 2.Qe3+. A bonus change occurs after 1…b3 when 2.Rxb3+ compels 2…Bxb3.

Andy Sag: White removes Black’s unwanted second guard on a4. Black tries to avoid giving mate by adding an alternative second guard or by blocking the intended guard but White has all the answers.
Jacob Hoover: The promotion defenses on b1 occur again, but the white responses are swapped. It’s a shame that this problem only got an honorable mention.
Andrew Buchanan: Very nice with lots of changed play.
Rauf Aliovsadzade: Excellent problem by one of the world’s renowned composers. The variations for 1…axb1=B and 1…axb1=S are very nice!

## 668

Bernardus Postma
Parallèle 50 1948, 1st Hon. Mention

Mate in 6

## Solution

With strong defences like 1…c1=Q and 1…Bxc4 available to Black, White must threaten a battery mate immediately with 1.Bd5! Now 1…c1=Q+ is inadequate against 2.K~+ Qc6+ 3.Bxc6, while two bishop defences also produce short variations: 1…Bg2 2.Bxg2 c1=Q+ 3.K~+ Qc6+ 4.Bxc6 and 1…Bxa6 2.Kc7+ Bb7 3.Bxb7. The main line begins with 1…Bc4, which cannot be met by 2.Bxc4? because of 2…c1=Q. Here White wants to renew the battery threat by shifting the bishop elsewhere on the long diagonal, but Black seems able to continue to attack that piece, e.g. 2.Be4? Bd3! or 2.Bh1? c1=Q! Correct is 2.Bg2 which decoys the black bishop to the first rank – 2…Bf1. Then 3.Bh1 exploits the self-interference on that line, as 3…c1=Q+ no longer prevents 4.Kd6/Kd7+ Bg2 5.Bxg2+ Qc6+ 6.Bxc6. A good logical-style more-mover, somewhat marred by the king dual on the fourth move.

Andy Sag: A miniature one-liner except for the white king dual. Perhaps a black knight on f8 would avoid the dual but then it wouldn’t be a miniature.

## 669

Valery Gurov & Abdelaziz Onkoud
Green March-40 Jubilee Tourney 2016, 4th Prize

Helpmate in 2, 3 solutions

## Solution

A white king first move seems likely as a way to activate the queen and bishop on the long diagonal, though the royal piece is trapped initially. If we begin with a random 1.Re~? to allow 1…Ke2+, then 2.Kc5 Qc6 fails only because of the black queen. So in one solution Black’s first move aims not only to free the white king but also to cut off that queen. 1.Bd6 Kxf2+ 2.Kc5 Qc6. The cyclic arrangement of Black’s g3-bishop, f2-knight, and g4-rook – they defend one another in a loop – provides a clue for the other two solutions, in which one of these pieces again releases the king and simultaneously closes a black line of guard. 1.Sd3 Kxg4+ 2.Kc4 Qc2 and 1.Rc4 Kxg3+ 2.Ke5 Qd5. Among the many pleasing effects are the two triplets of king moves that complement one another.

Andy Sag: In each case Black simultaneously unguards a piece and (by line interference) the mating square. A well-coordinated triple!
Andrew Buchanan: Great problem. Hint is that Black’s rook, bishop, and knight round the white king protect one another cyclically, suggesting that White’s first and Black’s second moves are with kings.

## 670

Franz Pachl & Mirko Degenkolbe
harmonie 2004, 3rd Prize

## Solution

Although the black king has plenty of flights, it’s already well-placed to be confined by the white line-pieces once the intervening units are removed. The eventual knight mate will open two indirect batteries for the a4-rook and a7-bishop, so Black needs to capture the three pawns and block the remaining flight on h3. 1.Sb3 2.Rxb6 3.Rxc6 4.Rc2 5.Bxe4 6.Bf5 7.Bh3 Se2 and 1.Sc2 2.Bxe4 3.Bxc6 4.Bb5 5.Rxb6 6.Rh6 7.Rh3 Sf5. The two analogous sequences involve shielding the white king from various checks and an exchange of functions between Black’s rook and bishop.

Andy Sag: The trick is to find two ways to remove the white pawns without checking the white king. The white knight mates from different squares but in both cases unleashes the same two indirect batteries.
Jacob Hoover: The f4-knight mates on whichever potential mating square that isn’t guarded by the black bishop. Lots of things going on here: line closures, clearance captures, and line openings.
Andrew Buchanan: Very pleasant and harmonious, and trickier than many series-movers.

## 671

Herbert Hultberg
Suomen Shakki 1987-88, 3rd Prize

Mate in 3

## Solution

The f4-knight seems to be the only white piece not busy restricting the black king, and it makes the key 1.Sg2!, which threatens 2.Sxh4, besides observing another potential mating square, e3. Black has three defences that protect the h4-pawn. 1…Bf2 is answered by 2.dxc3, when Black is left in zugzwang, since the bishop cannot move without allowing one of the two knight mates: 2…Bg3/Be1 2.Sge3 and 2…B~else 3.Sxh4. After 1…Rxh3, 2.axb6 creates a similar zugzwang, one in which the rook must lose its focus on the two mating squares: 2…R~file 3.Sge3 and 2…R~rank 3.Sxh4. Lastly, 1…Rc4 permits White to cut off the two defenders simultaneously with 2.d4, to threaten both knight mates, which are separated by 2…Rxd4 3.Sge3 and 2…Bxd4 3.Sxh4. The first two pairs of variations demonstrate focal play, while the third pair shows the Novotny theme. The same knight mates occur throughout against different black defences; hence a type of mate transference is also produced in this attractive three-mover.

Andy Sag: Meredith setting with short mate threat and three variations, two of them waiters. The second move Novotny is also a nice touch.
George Meldrum: A really nice problem, with the 2.d4 variation being the cream on the pudding.
Andrew Buchanan: Wow, that is amazing. Such flawless harmony and economy is more often associated with helpmates.

## 672

Andy Sag
OzProblems.com 7 Oct. 2023, Version

Mate in 2

## Solution

[This version of the problem omits a black pawn originally placed on f5.] It’s tempting to shift the white bishop in order to threaten a queen mate on d6, but where should it be placed? The piece has numerous options in four directions, as exemplified by the following tries that form a star-pattern: 1.Bb4? Qe6 2.Qxe6 but 1…Qe7!, 1.Bb8? Qe7 2.Qc6 (this mate is an additional threat) but 1…Qe6!, 1.Bf8? Qe7!, and 1.Bf4? Qe6! While these bishop moves grant one flight on either e5 or c5 to the king, the excellent key 1.Be5! concedes two flights and entails a single threat (2.Qd6). The self-block 1…Qxe5 provokes 2.Qc6, which previously functioned as a second threat in some of the virtual play. Taking the key-piece with 1…Kxe5 enables 2.Qg5, whereas the other flight-move 1…Kc5 is answered by 2.Qd6 (the threat-move, though the king’s new position implies a distinct variation). Finally, 1…Sb5/Sc4 2.Rb5 shows the rook recovering the two orthogonal flights.

In the original setting with a black pawn on f5, the post-key mate after 1…Kxe5 is 2.Qd6. Thanks to Michael McDowell who points out that removing the pawn would result in the variation, 1…Kxe5 2.Qg5, which is better by not repeating the threat-move. The composer concurs and has accepted this suggestion. Michael has also arranged an alternative version – diagrammed below – that uses a dark-squared bishop as a defender, instead of the knight. This brings about the extra variation, 1…Bc5 2.Qe6 (self-block play matching 1…Qxe5 2.Qc6). However, this setting eliminates half of the bishop tries, which Andy considers integral to the composition. Hence he prefers to retain his own version, while Michael’s still makes an interesting comparison.

Composer: The sacrificial key grants two flights and vacates d6 in a Meredith setting. The knight prevents an unwanted check on the a-file and also adds a further defensive variation. Many thematic tries by the bishop requiring the black queen to choose the correct square to stop multiple mates depending on which diagonal the bishop uses.
Rauf Aliovsadzade: Several tries by the white bishop – a challenge for solvers.
Nigel Nettheim: Another really good one by Andy.
Bob Meadley: Very tricky and typical Andy. The key is 1.Be5 but what throws one are the number of bishop tries. More from AS.

## 672v

Andy Sag
OzProblems.com 7 Oct. 2023
Version by Michael McDowell

Mate in 2

## Solution

See text above.

Key: 1.Be5! (threat: 2.Qd6). 1…Qxe5 2.Qc6, 1…Kxe5 2.Qg5, 1…Kc5 2.Qd6, 1…Bc5 2.Qe6, and 1…Bb8 2.Ra5.

## 673

Noam Elkies
U.S. Problem Bulletin 1995

## Solution

White’s last move was the double-check(mate) Rc6(x)c8, so the total count of white moves includes a rook stop on c6. Taking into account a queen switchback to let the a-rook out, the white units require all 20 available moves to reach their current places. The shortest routes for the rooks are R(h1)-h4-b4 and R(a1)-e1-e3-f3-f6-c6-c8; that means (1) White can’t play e4 until the h-rook is developed, and (2) Black can’t play …d6 until the a-rook has passed through on the 6th rank, late in the game. 1.Sc3 c6 2.Se4 Qc7 3.Sg5 Qg3 4.hxg3. Now Black seems to have a large number of spare moves, but wasting them with simple switchbacks (like …Sa6-b8) would end up costing an extra move. Instead, Black needs to lose a tempo and does so by finding a spot for the king to triangulate. 4…Kd8 5.Rh4 Kc7 6.Rb4 Kd6 7.e4 Ke5 8.Bb5 Kf6 9.d3 Kg6 10.Be3 Kh6 11.Qd2 – the only choice for the queen to unguard h5 for the king. 11…Kh5 12.Kf1 – not 12.Bd4? preventing the king’s return. 12…Kg6 13.Re1 Kf6 14.Qd1 Ke5 15.Bd4+ Kd6 16.Re3 Kc7 17.Rf3 Kd8 18.Rf6 Ke8 19.Rxc6 d6 20.Rxc8. A fantastic 15-move tempo-losing trek by the black king, with good interplay between the two sides for the theme.

Andy Sag: It became apparent that Black must triangulate the king. Surprisingly the piece must take a long return journey to the h-file in order to do that!
Jacob Hoover: It took me quite a while to figure out that the black king needed to go for a long walk which includes a tempo move. The positioning of the white units cleverly determines the route the king needs to take.
Andrew Buchanan: This is lots of fun. The main implication of checkmate in the final diagram is that it’s more likely to be by Noam who likes that flourish. Since Noam is a mathematician, we may suspect that he has heard of parity and so it proves. It’s cool that Black’s missing pawn and bishop cannot solve the parity problem. It’s also impressive that king must make 15 moves in order to triangulate.

## 674

Jorge Kapros & Jorge Lois
Telescaccho 2000, 1st-2nd Prize =

Helpmate in 2, 2 solutions

## Solution

Potential cooks where White’s queen or bishop mates along the a2-e6 diagonal are prevented by Black’s queen-side pawns and c7-knight. In the actual solutions, the queen does give mate (elsewhere) though it proves inefficient for Black to allocate two moves to unpin the piece; rather, White unpins it by intercepting on the 3rd rank with the f5-rook or d6-bishop. These white moves also serve to vacate the queen’s mating squares, which are guarded from behind by Black’s b1-bishop and d2-rook respectively. Furthermore, in each case the queen’s mating move doesn’t cover a flight (occupied by the remaining white unit) on d6/f5 – that square will need to be blocked by the black queen. Thus Black’s first moves by the b1-bishop and d2-rook have a double motivation: to unguard the eventual mating square and to unpin the black queen. 1.Bxa2 Rf3 2.Qxd6 Qf5 and 1.Rxa2 Bg3 2.Qxf5 Qd6. A host of matching strategic effects between the two solutions, augmented by the capture of white force. White’s rook and bishop swap roles, and so do Black’s on d2 and b1. The latter pair yields dual avoidance: after 1.Bxa2, not 1…Bg3? because the d2-rook disables the queen mate on d6, and likewise after 1.Rxa2, not 1…Rf3? because the b1-bishop stops the queen mate on f5.

Andy Sag: In each part the mating square is unguarded by Black and vacated by White, unpinning both queens in the process. Then the black queen captures to self-block and the white one mates. Curiously the final queen positions are exchanged.
Jacob Hoover: The squares d6 and f5 exchange functions between being the mating square and being a self-block square for the black queen. Various ODT [orthogonal-diagonal transformation] effects for both White and Black.
Nigel Nettheim: Wonderful reciprocation between the two solutions.
Andrew Buchanan: A great problem – very hard to solve, and very harmonious!

## 675

Leonid Makaronez
OzProblems.com 28 Oct. 2023

Mate in 3

## Solution

The black king has a provided flight-move, 1…Kf4 2.Qf6+ Kxe4 3.Sc3. An incidental try 1.Kxg3? threatens 2.Sc5 and 2.Qxh6, but it’s defeated by the subtle 1…f2!, preparing to check. The tough key 1.Bc5! controls d6 to create a sacrificial threat, 2.Rd5+ exd5 3.Qf5, 2…Kf4 3.Qg4. The first two defences remove the vital e4-pawn but are exploitable as self-blocks: 1…Bxe4 2.Qxg3+ Kf6/Kf5 3.Rf8 and 1…Rxe4 2.Qg7+ Kf5 3.Rf8 (2…Kf4 3.Rf8/Qf6), with good differentiation of the two queen checks. If 1…h5 to disable a queen mate in the threat, then 2.Qg5+ Kxe4 3.Sc3. A group of defences protect d5 but commit the error of unguarding other squares. 1…Bb3 2.Qxg3+ Kxe4 3.Bd3, 2…Kf6 3.Rf8. 1…c6 2.Bd6+ Kd4 3.Bf4. 1…Rd4 2.Bxd4+ Kf4 3.Qg4/Rf8. Lastly, the flight variation is unchanged from the set play, 1…Kf4 2.Qf6+ Kxe4 3.Sc3. No fewer than seven variations in this complex 3-mover, which features an active white queen and a surprising battery mate on the d-file after 1…c6.

Andy Sag: Leonid’s 3-movers are always difficult and this one is no exception.
George Meldrum: A barrage of defensive play ends with the black king being mated on six different squares.

## 676

Jean-Marc Loustau
Phénix 1989, 2nd Hon. Mention

Mate in 2

## Solution

Four potential queen mates on d3, e5, c4, and d5 are initially prevented by Black’s rooks and bishops. The thematic try 1.g3? is a Novotny move that cuts off the h3-rook and h2-bishop simultaneously, to threaten 2.Qd3 [A] and 2.Qe5 [B]. Each threat-move is forced individually with 1…Bxg3 2.Qd3 [A] and 1…Rxg3 2.Qe5 [B]. Furthermore, Black can stop both threats with the other rook-and-bishop pair, and these moves provoke the remaining queen mates: 1…Rc4 2.Qxc4 [C] and 1…Bd5 2.Qxd5 [D]. However, this try is refuted by 1…Rc5! The key 1.Rc6! is another Novotny that shuts off the c8-rook and a8-bishop, to threaten 2.Qc4 [C] and 2.Qd5 [D]. The normal separation of these threats occurs with 1…Bxc6 2.Qc4 [C] and 1…Rxc6 2.Qd5 [D]. Now it’s the first rook-and-bishop pair that’s able to deal with both threats, and such defences bring back the other queen mates: 1…Rd3 2.Qxd3 [A] and 1…Be5 2.Qxe5 [B]. Such a reversal of functions between two pairs of mating moves in the virtual and actual play, from threats to variation-mates, is called the Odessa theme, and here it’s nicely combined with Novotny play.

Andy Sag: Novotny solution with partial threat separation and a Novotny try.
Nigel Nettheim: One Novotny fails but another succeeds, with a bonus variation 1…Sxc6.

## 677

Gazeta Czestochowska 1970, 1st Prize, Version

## Solution

The flights on e2 and e4 suggest a rook mate on the e-file, hence Black wants to unguard either e5 or e1. The former would involve switching the bishop with the d2-pawn after the latter promotes to a knight, but that takes too long. The solution arranges the latter by means of two knight promotions. 1.d1=S Rg1 2.f1=S+ Kh1 3.Sd2 Kh2 4.Sf2 Re1. The two black pawns effectively perform a platzwechsel (exchange of places), while the white king does a switchback and the white rook makes a hesitating move. Incidentally, no tempo play is shown here, e.g. 3…Kh2 is a check avoidance.

Nigel Nettheim: An ingenious idea with a clever implementation. The g4-rook, otherwise unneeded, ensures that White can never play Kg1 to guard f2.
Andy Sag: Cute!
Andrew Buchanan: Simple but a lot of fun.

## 678

OzProblems.com 18 Nov. 2023

Mate in 3

## Solution

The key 1.Bb8! sets up a B + R battery, the only way to handle the two strong flight-moves by Black’s king. White has three threats that lead to different mates against …Kxf3: 2.Rd7+ [A] Kxf3 3.Rxd3, 2.Re7+ [B] Kxf3 3.Sh4, and 2.Rh7+ [C] Kxf3 3.Rxh3. Three black defences manage to thwart two of these threats but not the third. 1…Bd7 [x] 2.Rxd7+ [A] Kxf3 3.Rxd3, 1…Kxf3 [y] 2.Re7 [B] ~ 3.Sh4, and 1…Kh2 [z] 2.Rh7+ [C] Kh1 3.Rxh3. When three white threats are separately forced by three black defences, that is known as the Fleck theme. Furthermore, the composer points out a second pattern in which these thematic black defences act as the refutations of various tries. 1.Rh7? ~ 2.Bb8+ Kxf3 3.Rxh3, but 1…Bd7! [x]; 1.Sh5+? Kh2 2.Bg1+ Kh1 3.Sg3, but 1…Kxf3! [y]; and 1.Re7? ~ 2.Bb8+ Kxf3 3.Sh4, but 1…Kh2! [z]. The Fleck theme – not commonly seen in three-movers – is cleanly accomplished, and it’s supplemented by three tries producing function changes of three black moves.

Composer: Free form of Fleck. The tries’ refutations become thematic defenses in the solution.
Andy Sag: A battery must be set up initially so the white rook can choose the correct move depending on how Black plays.
George Meldrum: Black can foil White’s rook moving first. By delaying the rook move White can answer Black’s move with an appropriate rook line of play.

## 679

Mark Erenburg
World Congress of Chess Composition, Dresden 2017

15th Ukrainian Folk Crafts, 2nd Prize

## Solution

Set play is arranged for any black rook move, which unmasks the c8-bishop: 1…R~ 2.Qg4+ Bxg4. White does not, however, have a reply prepared for Black’s only remaining move, 1…d5. Note also how the white queen is guarded by the b5-rook and h7-bishop along two lines. The thematic try 1.Sd5? (waiting) prevents the pawn move and cuts off the white rook, besides opening the 7th rank for the black one. A random rook defence gives 1…R~ 2.Qg4+ Bxg4 as set, while the correction move 1…Rxh7+ enables 2.Bh6+ Bxf5, exploiting the unguarded queen. Black defeats the try with another correction, 1…Rg7! (2.Qg4+ Rxg4). The key 1.Sg6! (waiting) interferes with the white bishop instead. Then 1…d5 permits 2.Qxd5+, forcing the battery mate 2…Rxd5. The set play for 1…R~ no longer works, since 2.Qg4+? would let the king escape to e4; instead, 2.Be5+ – cutting off the white rook – provokes 2…Bxf5. Lastly, the correction 1…Rxh7+ leads to a different cross-check, 2.Sh4+ Rxh4. Fine changed continuations between the try and key, brought about with intricate line strategy.

Nigel Nettheim: Very impressive. It seemed that 1…d5 should be prevented, but 2.Qxd5+! will follow. The key prepares 1…Rxh7+ 2.Sh4+. The d2-pawn allows 2.Be5+ when needed and the f6-pawn disarms 1…Rf7.

## 680

Andy Sag
OzProblems.com 2 Dec. 2023

Mate in 2

## Solution

White’s e7-pawn is utilised in the set line 1…Sd8+ exd8=Q and it also makes a try, 1.e8=Q? (threats: 2.Qe5/Qh8) – one that is subtly defeated by 1…d2! White can provide for 1…d2 by making a clearance move with the f2-rook along the second rank, thereby giving the queen access to d2. Three such rook moves serve as the main tries and the key. 1.Ra2? threatens 2.Ra4, but 1…Bc3! refutes, while 1.Rc2? threatens 2.Rc4 and brings 1…dxc2 2.Qd2, but again there’s no response to 1…Bc3! The key 1.Rb2! surprisingly does not involve a threat, and it leads to a good variety of play. 1…Bxb2 2.Qxb2, 1…d2 2.Qxd2, 1…e3 2.Qd5, and 1…Bg~ Qg7 show four queen mates from different directions. Three more variations are 1…a4 2.Rb4, 1…S~ 2.Rd6, and 1…Sd8+ exd8=Q.

Composer: A waiter with a sacrificial key, seven variations and three significant tries. The theme is queen/rook mates from various angles.
Bob Meadley: Looks like 1.Rb2 cramping the bishop on a1. All the other defensive moves are then catered for. Andy is on a roll.
Nigel Nettheim: I first tried 1.Ra2 (2.Ra4), but 1…Bc3 defends; in fact 1…Bc3 had to be prevented, and the key turned out to be waiting. The h4-pawn prevents 1.Qg3. Quite elaborate, well done!

## 681

Kjell Widlert & Denis Blondel
Mat 1984, Sarajevo Theme Tourney, 2nd Prize

Helpmate in 2
Twin (b) Sb5 to d1

## Solution

White wants to arrange either Be3 or Rxd6 mate (depending on where the impeding b5-knight is placed in the twins), but both pieces are pinned. In part (a), Black can unpin the bishop in two moves (not 1.Sd3? observing f4), e.g. 1.c1=S and 2.Rc2; however, that’s only a try because White lacks a tempo move. The solution does require an unpinning rook move to c2, though that comes about only after a clearance sacrifice by the white rook, which therefore needs to be freed first. 1.Qb8 Rxc2 2.Rxc2 Be3. For part (b), Black can likewise unpin the rook in two moves (not 1.Qb8? observing d6), e.g. 1.a5 and 2.Qa7, but again White has no tempo move. In the analogous solution, the queen does access a7, though only after a clearance sacrifice by the white bishop, which is thus released immediately. 1.Sd3 Bxa7 2.Qxa7 Rxd6. Besides the principal unpinning and tempo strategies, this helpmate shows the Zilahi theme, where two white pieces exchange their functions in getting captured and giving mate.

Andy Sag: In each case an unpinned white piece captures a pawn which was blocking Black’s ability to effectively unpin a second white piece which then delivers mate. Nice twin!
Nigel Nettheim: Very clever, though not hard to solve. In both parts, the unit that was first unpinned is sacrificed to enable a second unpinning.
George Meldrum: Love it!

## 682

Shakhmaty v SSSR 1985, Special Prize

Mate in 4

## Solution

It’s tempting to double the rooks on the first rank, e.g. 1.Rhh1? Kc2 2.Rb1 f3 3.g3 Kd2 4.Rb2, but 1..Ke2! refutes. The key 1.f3! simply activates the half-battery on the second rank and threatens an immediate mate. 1…Kc2 2.g4+ Kb3 3.Ra1 Kb4 4.Rb2, and not 2.g3+? Kb3 3.Ra1 when 3…fxg3! is available. 1…Ke2 2.g3+ Kxf3 3.gxf4 Kxf4 4.Rf2, and not 2.g4+? Kxf3 3.g5 stalemating since the black pawn obstructs 3…Kf4. The two mating configurations – a model and an ideal-mate respectively – show a reflection echo, with the black king placed unusually far apart.

Andy Sag: The key leaves Black with only two legal moves. For the short-mate threat it is not important whether the g-pawn makes a single move or a double move, but it becomes important to choose the correct move depending on Black’s response to the key.
Bob Meadley: The try 1.Rb1? Kc2 2.Rhh1 f3 3.g3 Kd2 4.Rb2 really threw me off the scent. One of the finest miniature four-movers I have seen.

## 683

Jan Hartong
The Observer 1920, 3rd Hon. Mention

Mate in 2

## Solution

The black king has four diagonal moves, each provided with a mating reply: 1…Kd4 2.Sf3, 1…Kf4 2.Sd3, 1…Kd6 2.Bg3, and 1…Kf6 2.Bc3. White has no waiting move capable of preserving all of the set play, however; e.g. 1.Kd7? Kf6!, 1.Bd2? Kd6!, 1.d6? Kxd6! The key 1.Ba5! (waiting) arranges new mates for two of the king defences, 1…Kd6 2.e5 and 1…Kf4 2.Bc7, while 1…Kd4 2.Sf3 and 1…Kf6 2.Bc3 are unchanged. Brilliant star-flights problem where two pairs of symmetrical variations in the set play are converted post-key to four non-symmetrical ones – all rendered in a Christmas tree setting. The star-related theme is also appropriate for the season!

Andy Sag: Symmetrical set play covers all four star-flights. The only move that preserves symmetry is 1.Re3, but this unguards e3 and so is quickly ruled out. The key is therefore strongly suspected to be an asymmetrical move to a-file.
Bob Meadley: 1.Ba5! – the only move that disturbs the Christmas Tree.
Nigel Nettheim: Treemendous! The Bishop leads, fittingly at this time of year. In the changed mate after 1…Kd6 (was 2.Bg3), the e-pawn takes over the role of checking, like an assistant Bishop. The right edge prevents the cooks 1.Bi5?? and 1.Bj6??, and the left edge acts similarly.
George Meldrum: Great Christmas Tree problem.

Kurt Smulders