Weekly Problems 2023-B

Problems 659-668

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Reto List
feenschach 1979, 3rd Prize


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!
Rauf Aliovsadzade: Easy to solve as castling is obvious. Ideal mate.
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.
Bob Meadley: A beauty!


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

Mate in 2


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.


Philippe Schnoebelen
Probleemblad 1996, 1st Hon. Mention


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.


Viktoras Paliulionis
Rozmaitosci szachowe 2012, 2nd Prize

Helpmate in 2
4 set plays, 4 solutions


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.


Rauf Aliovsadzade
OzProblems.com 5 Aug. 2023

Mate in 3


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
Rochade Europa 1995


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.


Andy Sag
OzProblems.com 19 Aug. 2023

Mate in 2


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.


Sergey Sinakevich & Yury Fokin
Schach 1973, 1st Prize

Helpmate in 3
Twin (b) Sc5 to b4


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.


Waldemar Tura
Wola Gułowska 2012, 1st Hon. Mention


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!


Bernardus Postma
Parallèle 50 1948, 1st Hon. Mention

Mate in 6


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 fail: 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.