Weekly Problems 2023-A

633

Leonid Makaronez
OzProblems.com 7 Jan. 2023

Mate in 3

Solution

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

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

634

Peter Kahl
Die Schwalbe 1955

Solutions

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

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