Guided Chess Problem Composing Competition 2016 Tasks, Answers, and Results by Peter Wong April 2016 This is the second Guided Chess Problem Composing Competition, an open event linked to the 2016 Australian Junior Chess Championships. It is aimed at introducing chess players and problem solvers to the basics of constructing a problem, though the set questions could be of interest to experienced composers as well. Thanks to Nigel Nettheim who initiated the event and invited me to arrange the tasks and adjudicate on the entries. He has written an informative report on its organisational aspects, available from the Events section of the OzProblems site. As in the previous year, the final results of the competition were very close. This time they were especially close because the tasks were generally more straightforward, so fairly fine points separated most of the entries. Here are the prize-winners: 1st Prize: Ralf Krätschmer (Germany) 2nd Prize: Ilija Serafimović (Serbia) 3rd Prize (equal): Dušan Mijatović (Serbia) and Andy Sag (Australia) Ralf and Ilija were also the two top-scoring contestants last year but they have swapped their positions. Third place is shared by Andy and a new participant, Dušan. While Ralf and Andy are established problemists, Ilija and Dušan are both juniors who attend the Chess Academy run by the world-renowned composer, Marjan Kovačević. Congratulations to the winners! Marko Lozajic of Serbia (another student of Marjan Kovačević) and Stefan Felber of Germany also deserve our compliments; both achieved scores very close to the above group. Special mentions go to two young entrants who have not attempted composing before: Danila Pavlov of Russia – a world junior champion in problem solving – who gave consistently good answers, and Erin Dullaway of Australia who is remarkably only seven years of age. The rest of this document provides (1) the four tasks as given in the original questions paper, and (2) a selection of the answers received, along with the maximum points allocated to each part. For Tasks 1 and 4, the positions to be reconstructed are more or less fixed. However, Tasks 2 and 3 – requiring two unsound problems to be corrected – are less restrictive, and they elicited a good variety of responses which are interesting to compare.

Task 1

Mate in 2 (incomplete) The position shows a nearly complete two-move problem. The pieces in the diagram are all placed correctly, but it is missing the white king and a white rook. When these two pieces are added on the appropriate squares, the problem is solved uniquely by the key 1.Nb1!, which threatens 2.Ra3. Black has four bishop moves that disable this threat, and they are answered by various mates delivered by the white pawn: 1…Bb3 2.cxb3, 1…Bd3 2.cxd3, 1…Bb5 2.c3 and 1…Bd5 2.c4. The four variations are thus linked to create two artistic ideas or themes: (1) bishop-star (four black bishop moves that form a star pattern lead to different mates), and (2) Albino theme (a white pawn on its starting square plays all four of its possible moves). Where should the white king and the white rook be placed to enable the above key and variations? Ensure that in the completed position, each of the four bishop moves defeats the threat and so requires White to mate with the respective pawn move.

Task 1 – Answer Add a white king on h1 and a white rook on f3. Camil Seneca Bulletin Ouvrier des Echecs 1949

Mate in 2 The white king and the white rook are placed on the same diagonal so that after 1.Nb1!, 1…Bd5 pins the rook and hence disables the threat of 2.Ra3, leaving White with just one mating move, 2.c4. Total: 5 points

Task 2

Mate in 2 (unsound) If Black were to play in this position, we apparently have these set variations: 1…N3-any 2.Nf5 and 1…N5-any 2.Nf3. When White begins, the intended key is a waiting move, 1.Kd1!, which aims to preserve the set play while avoiding checks by the black knights. However, the problem actually has no solution, because after 1.Kd1 Black has a reply that stops White from mating on the next move. 1. What is the black move that refutes 1.Kd1? 2. Modify the position so that 1.Kd1! does solve the problem as intended. The variations 1…N3any 2.Nf5 and 1…N5-any 2.Nf3 should work both in the initial position and after the key is played. You can add or remove pieces as required, or shift some (or even all) of the existing pieces to other squares – so long as the play remains essentially the same. Various sound settings are possible, based on different ways of confining the black king. Try to find the most economical position.

Task 2 – Answers Problem source: Felix Seidemann, Teplitz-Schönauer Anzeiger 1931. 1. The intended key is defeated by 1…Nxh3. 1 point 2. The most economical corrections of the problem are diagrammed below. Version A is the best entry and it scores 10 points. A white knight and a black pawn are used to confine the black king, and the position is rotated to ensure that the pawn does not prevent either knight mate. The resulting set-up displays an attractive quasi-symmetry. The next two positions differ only in how the h5-flight is covered – blocked by a black pawn or guarded by a white one. Although white economy is generally considered more important, it could be argued that using a white pawn here produces a “cleaner” look with no extraneous black units, so the two settings are ranked equally. Version D is another good correction with a pawnless arrangement. These positions all preserve the miniature (seven-piece) aspect of the original problem and its model mates. Version A Ralf Krätschmer

Version B Dušan Mijatović, Ilija Serafimović

1.Ka5!

1.Kd1!

Version C Andy Sag

Version D Stefan Felber

1.Kd1!

1.Kd1!

Total: 11 points

Task 3

Mate in 2 (unsound) This two-move problem is solved by 1.Qg7! with the threat of 2.Qd4. The thematic variations involve different captures of the white rook: 1…Kxf4 2.Nd5, 1…Bxf4 2.Qa7, and 1…Nxf4 2.Qg1. Additional variations are 1…Nf3 2.Rxf3 and 1…Be5/Bc5 2.Qe5. However, the problem is spoiled by a cook, or unintended solution: 1.Nd5+! Kd3 2.Qxf5. Modify the position so that the problem is solved only by the intended key and not the cook. The corrected position should basically retain the key and variations of the original problem. Minor changes to the play are acceptable but not major ones, such as losing a variation. The flightgiving feature of the key-move should be preserved, i.e. the white queen surprisingly unguards f4, which becomes a flight-square accessible to the black king. For instance, let’s consider the idea of shifting the white knight from b4 to e1. The variation 1…Kxf4 2.Nd5 is now replaced by 1…Kxf4 2.Nxg2, which is an acceptable change. The cook 1.Nd5+ no longer applies, and the similar checks 1.Nxg2+? and 1.Nc2+? fail to 1…Kd3! since the king would escape to c3 after 2.Qxf5+. However, a new cook has arisen: 1.Re4+! Kf2 2.Qxg2. That means further changes to the position would be necessary, if you were to proceed with the idea of starting the knight on e1. (This scheme may or may not lead to the best answer.) Provide two distinct sound settings of this problem, in the order of your preference. Aim for the most economical positions. (Note that two very similar arrangements, such as adding the same piece on different squares to serve the same function, would count as one answer only.)

Task 3 – Answers Problem source: Sámuel Gold, Rochester Herald 1904. Besides eliminating the cook, the best entries actually improve on the economy of the original problem or even enhance its solution a little. Version A is brilliant in accomplishing both and it scores 16 points. After the key, 1…Kxc5 is answered by a fine switchback mate, 2.Qb4, and this means the white knight that mates after the king move in the original can be dispensed with. The two positions following are among three that share second place. Version B and a similar setting by the same composer are the lightest entries received – they require ten units only, compared with twelve in the original. Version C has the same material as the original, but the transfer of the black bishop to c7 is significant. Now 1…Bxf4 opens a line for the white queen (2.Qa7), and this nicely matches the other thematic variation 1…Nxf4 2.Qg1 where the same line effect occurs. For this reason, a similar position submitted by Dušan Mijatović, with the bishop placed on b8 instead, scores one point less. Version A Ralf Krätschmer

Version B Marko Lozajic

1.Qb2!

1.Qg8!

Version C Stefan Felber, Andy Sag, Ilija Serafimović

1.Qg7!

Version D is interesting in its placement of the white queen away from the diagonal where it forms a battery with the white rook. On c7 the queen observes f4, so the key meets the requirement of conceding a flight to the black king. The key is therefore still good, but it’s not as surprising as the original one which abandons the battery. The simplest way to correct the problem is to add a black bishop to protect f5, as Version E demonstrates. Andy Sag sent an equivalent position with the bishop placed on h7. Version F uses a white pawn on f5 to obstruct the queen mate, and this necessitates a white bishop to guard the pawn. While not so economical, this is a laudable effort from a very young entrant! Version D Ilija Serafimović

Version E Stefan Felber, Dušan Mijatović, Danila Pavlov

1.Qg7!

1.Qg7!

Version F Erin Dullaway

1.Qg7! Total: 32 points

Task 4

Mate in 2 (incomplete) This problem position is missing four units (pieces or pawns), but otherwise the diagram is exactly correct. When the missing units are added, the problem is solved by 1.Ba3!, which guards f8 and threatens mate by another white unit. Black’s c7-bishop and the d7-pawn can answer the threat, but in doing so they interfere with each other’s line of defence, resulting in these main variations: 1…d6 2.Qf2 and 1…Bd6 2.Qb3. The key-bishop must go all the way to a3 rather than c5 or b4 because on these squares the piece would disrupt a queen mate and so allow Black to escape: 1.Bc5? d6! (2.Qf2 barred) and 1.Bb4? Bd6! (2.Qb3 barred). Add the four units on the right squares to complete the composition, using the clues listed below. Ensure that only 1.Ba3! solves and that the two main variations work as intended. The tries 1.Bc5? and 1.Bb4? should be uniquely defeated by the thematic defences as given above. Two minor variations will also arise in the completed problem, based on the defences 1…Rxa3 and 1…Be5. The following hints provide the main purposes of the missing units, but these units could have other functions as well. (1) Add a black unit on e8 to prevent the black king from escaping to that square. (2) Add a white unit to guard f6, another flight-square. (3) Add a white unit to deliver the threatened mate when the key is played. (4) Add a black unit to defeat the potential cook, 1.Bd6, a very strong move that not only involves the same threat as the key, but also threatens three other mates, including 2.Qf2 and 2.Qb3.

Task 4 – Answers (1) Add a black bishop on e8. (2) Add a white pawn on g5. (3) Add a white knight on g6. (4) Add a black rook on g1. The key 1.Ba3! threatens 2.Nh8. The two minor variations are 1…Rxa3 2.Qf6 and 1…Be5 2.Nxe5. The potential cook 1.Bd6 threatens 2.Nh8/Ne5/Qf2/Qb3 and it’s defeated by 1...Rg4 or 1…Rxg5. Voldemārs Mačs Padomju Jaunatne 1949

Version A

Mate in 2

1.Ba3!

Most competitors reconstructed the original problem, but two found Version A. The latter is heavier in using a white knight to control f6, and it scores one point less. The deduction is small due to an advantage of the second position, in which 1.Bd6? is technically a try that’s refuted by 1…Rxh3! only. Total: 18 points All tasks: 66 points