Solutions to our 2009 Christmas Puzzles – Part 2 By John Nunn
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December 30, 2010 – Another directmate

Dieter Kutzborski, 1st Prize, Schach-Aktiv 2007

White to play and mate in 15 moves

Solving a long problem such as this often depends on a combination of logical thought and chess analysis. White is heavily outgunned in material, so his attack must proceed with direct threats, either checks or threats of mate in one.

The first step is to look for possible threats of mate in one. One idea is Bb2 followed by Bc1 mate and another is e6 followed by Be5 mate. It is soon apparent that neither of these works at the moment, since 1 Bb2? is met by 1...Qa3! and 1 e6? is met by 1...Qa1! (but not 1...Qb8?, when 2 Bb2 does indeed lead to mate next move).

What other possibilities are there in the position? There are two checks, 1 Rf3+? (bad because the d4-bishop is attacked after 1...Ke4) and 1 Be3+, which forces 1...Kg3. In the latter case, White has a discovered check, but he must cover f2 or the king escapes, so, apart from repeating the position with 2 Bd4+, the only move that makes sense is 2 Bg1+, which forces 2...Kf4. But why is the bishop better on g1 than d4? This is where the various threads in the position start to come together. When the bishop is on g1, it is not attacked by the black king after 3 Rf3+ Ke4, so now White has the opportunity to reposition his rook with gain of tempo by giving another discovered check. Where is the rook best posted? To decide this, think back to the comments at the start and recall that 1 e6? was bad because of 1...Qa1!. This defence is eliminated if the rook stands on c3, so White plays 4 Rc3+ forcing 4...Kf4. The bishop must return to d4 to threaten the mate on e5, so White continues 5 Be3+ Kg3 6 Bd4+ Kf4. The key move is 7 e6, which now forces 7...Qb8.

Now there’s just one problem left to solve. With the rook on c3, White is unable to play Bb2, which would now lead to mate on c1. Thus the whole manoeuvre must be reversed in order to transfer the rook back to b3 with gain of tempo, after which Bb2 forces mate. The complete solution is 1 Be3+ Kg3 2 Bg1+ Kf4 3 Rf3+ Ke4 4 Rc3+ Kf4 5 Be3+ Kg3 6 Bd4+ Kf4 7 e6 Qb8 8 Be3+ Kg3 9 Bg1+ Kf4 10 Rf3+ Ke4 11 Rb3+ Kf4 12 Be3+ Kg3 13 Bd4+ Kf4 14 Bb2 c1Q (or 14...Qe5 15 Bxe5#) 15 Bxc1#. [Click to replay]

Comments

“This was the hardest puzzle for me.” (Paul Monsky, Waltham US)

“Wonderful idea of blocking or opening a diagonal in different positions according to convenience.” (Gervasio Calderón Fernández, Buenos Aires, Argentina)

“The real key is discovering the threats e6 and Bb2 and the relationship between them.” (Bryan Lamb, Toronto, Canada)

“Logical problem, pendulum manoeuvres.” (Zvi Mendlowitz, Petah Tikwah, Israel)

“I’m very sad. I tried to do this exercise for 9 days, and I couldn’t. I’m expecting for the solution... but sad... snif, snif...” (Diego Sumic, Buenos Aires, Argentina).

December 31, 2010– Serieshelpstalemate

Michel Caillaud, 1st Prize,
23rd T.T. Problemkiste 2002-3 (version)

Serieshelpstalemate in 27

In a serieshelpstalemate none of the black pieces can be captured, except possibly on White’s final move. This means that all except one black piece have to be immobilised, either by incarceration or by pinning.

One key step is to determine where the black king will be in the final position. Certainly b8 looks the most promising square, as many of the king’s moves are already prevented and the white rook is in an ideal position to pin a black piece on b6 or b7. One’s first thought is to have a bishop on a8 blocked in by a piece on b7, but what can this piece be? It must be a bishop or knight for the position to be stalemate, but a knight would check the white king while a bishop is impossible as only one black pawn can promote to a light-squared bishop.

Therefore try a black piece on b6, which is likely to be a knight (a bishop would check the white king, assuming that the c7-pawn promotes during the play). Then it’s natural to have an immobilised bishop on a7, which could then block in a rook on a8. That removes almost all Black’s moves, but there are two problems: firstly, Black can play ...Kb7, and secondly there is one black man unaccounted for. Fortunately, both problems can be solved together, by having White take Black’s final piece with the single move dxc8B.

With many problems the pieces would now fall into place very easily, but here there are still some difficulties to overcome in determining the move-order, and it getting the black king across the fifth rank to reach b8. It takes Black six moves to get his king to b8, two to play the bishop to a7 and 11 moves to promote all the pawns, for a total of 19. With two moves to get a promoted rook to a8 and three moves to get a knight from d1 (the quickest square) to b6, that leaves three moves for the last piece to get to c8. Seems easy enough, since really only one move is required (...Qc1-c8+ or ...Rc1-c8+, but the ambiguity here should already raise a doubt in the solver’s mind as to whether we are on the correct path) but when you try to do it you encounter a nasty problem: when the rook arrives on a8 it mustn’t check the white king, so the black king must already be on b8. This implies that the knight must already be on b6, but then the bishop must already be on a7 or the knight would prevent it moving there. Unfortunately, this firmly seals the a-file so that the rook can’t get to a8 after all. Several solvers observed that it’s easy to reach the target position in 28 moves, but saving a move isn’t straightforward.

The solution involves a leap of imagination: a black piece must be used as a temporary shield on b6; then the king moves to b8 and rook to a8. The black piece must then slide along the pin-line (so it must be a queen or rook), allowing the bishop to move to a7, after which a knight moves to b6, leaving just the shield (now on b7) to be disposed of. Suddenly everything fits together; the temporary shield must be provided by a queen which from b7 can self-destruct at c8 on Black’s last move. It is then clear that the queen must appear at g1 and then move to b6 (followed later by b7 and c8). This costs Black two moves, exactly accounting for the two apparently spare moves.

The rest isn’t too difficult. First, Black moves his king to free the g-pawn to promote to a queen which can move to b6. Black must provide a shield for his king to cross the fifth rank without wasting any moves, and this can only be arranged with the help of the promoted d1-knight, which occupies d5 on its way to b6. One of the nice points about the problem is the way the composer has arranged for one of the apparent dual possibilities ...Nd1-c3-d5 and ...Nd1-e3-e5 to be prevented. So the knight moves to d5 and the king to b8. Black can’t push the c-pawn at once as this would be discovered check, so now he has to play ...Qb7. Then the c-pawn promotes to a rook, which moves to a8, and then the other pieces slot into place. 1 Kf4 2 g4 3 g3 4 g2 5 g1Q 6 Qb6 7 Be3 8 d2 9 d1N 10 Nc3 11 Nd5 12 Ke5 13 Kd6 14 Kc6 15 Kb7 16 Kb8 17 Qb7 18 c5 19 c4 20 c3 21 c2 22 c1R 23 Ra1 24 Ra8 25 Ba7 26 Nb6 27 Qc8+ 28 dxc8B. [Click to replay]

Comments

“Before I found the solution I thought I had found a shorter alternative and had to control the impulse to send an email to ChessBase asking if the position was correct, but then I thought: ‘No, this is Dr. John Nunn! I must be wrong!’.”(Robert Stelling, Brazil)

“This one was the toughest” (Twan Burg, Netherlands)

“I solved this one while I was in bed trying to sleep.” (Joan Fluvià Poyatos, Tordera (Barcelona), Spain)

“After some hours in the night and an uneasy sleep I found the solution in the morning of New Years Eve.” (Thomas Thannheiser, Lübeck, Germany)

“It took some hours before I realized that White can in fact use his move to promote a pawn.” (Rune Djurhuus, Oslo, Norway)

“All four pawns are promoted, each of them to different kind of pieces.” (Lars Hafskjær, Fetsund, Norway)

“Very frustrating puzzle, the hardest in my opinion, but I loved solving it!  The sheer amount of brain power required for me to get this made me proud I eventually found the right answer.  Good job on this one chessbase!” (Chris Felix, Halifax, Canada)

“First I had to find the only plausible pattern of the stalemate, which I solved while cleaning my teeth. Managing to put the moves in the right order was another problem which took some time. The final idea came during a trip in the train.” (Walter Schmidt, Oberursel, Germany)

“Now, this is an incredible solution for quite some reasons. I remember discussing this puzzle quite early in the week with Teemu, and we both agreed that the solution couldn't be unique, but boy were we wrong.” (Jarle Kvåle, Trondheim, Norway)

January 1, 2011 – Chess and astronomy

Kostas Prentos, 1st Prize,
Oakham Proof Game Tny 2009-10

Proof game in 10 moves

Here we can make the initial observation that Black’s last move must have been ...Qc1-d2+, so we are really solving a proof game in 9.5 moves. When faced with a proof game, it is often best to start by move-counting, which involves working out how many moves each side takes to position his pieces as in the diagram. How many moves does it take to arrange the black pieces, bearing in mind that the b- and e-pawns must be taken en route? The most obvious sequence is ...c6, ...Qb6, ...Qxb2, ...b6, ...Ba6, ...Bxe2, ...Bd1, ...Bc2 and finally ...Qc1-d2+, which exactly consumes Black’s ten moves.

The second important weapon is the knowledge that the game is totally unique. Thus, while we might speculate that White’s first two moves are a4 and Ra3, we can immediately see that any plausible next move, such as c3 or d3, runs into the problem that White could have played his first three moves in a different order, which cannot happen. If White’s third move were Rb3 then the sequence would be unique, but this prevents Black’s queen taking on b2. The fact that the sequence of black moves given above has uniqueness imposed more or less automatically (except near the end) is a strong hint that we are working on the right lines.

Now we have made progress, but problems suddenly begin to appear. Black’s ...Bxe2 only occurs on move 6, and White cannot play d3 before that. Thus White’s 7th-10th moves must include the moves d3, Bf4 and Ke2, which leaves only one spare. However, the queen must leave d1 to allow ...Bd1, yet return to be in the correct place for the diagram. This presents an apparently insuperable difficulty, but this can be overcome by the application of either logic or imagination. The logical method is to add up White’s essential moves; these are a4, Ra3, Rb3, c3, d3, Ke2 and Bf4, which is seven moves, leaving three spare. This means that White’s queen doesn’t have to make an out- and return, but can take three moves to travel from d1 back to d1. The key point is that this allows the queen to move in a triangle.

Now we are close to the solution. White can only free his queen early on by playing c3, which gives the queen a triangular route via a4 and g4. This enforces a fair amount of uniqueness, as White cannot play a4 until the queen has moved to g4. Checking the line shows that everything now slots together perfectly: 1 c3 c6 2 Qa4 Qb6 3 Qg4 Qxb2 4 a4 b6 5 Ra3 Ba6 6 Rb3 Bxe2 7 d3 Bd1 8 Bf4 Bc2 9 Qd1 Qc1 10 Ke2 Qd2+. [Click to replay]

Comments

“I found the key idea fairly quickly, so this only took about 10 minutes.” (Bobby Cheng, Melbourne, Australia)

“Then, at half past two in the night, the solution almost accidentally dropped into my head.” (Jarle Kvåle, Trondheim, Norway)

“The last puzzle took me a lot of time and I considered quitting. But then I finally spotted the queen manoeuvre which I had in fact seen at the very beginning but never gave it a real thought.” (Willem van Briemen, Leiden, Netherlands)

“…reaching the target position thanks to a well-hidden triangular ‘Rundlauf’ of the wQ.” (Noam D. Elkies, Cambridge USA).

Astronomical tie-breaker

This proved quite popular, in fact a number of entrants were apparently not interested in the chess because they only sent in solutions for the tie-breaker. Readers were creative in their methods of finding the solutions if they didn’t know them already. Many used Google, some used Microsoft’s World Wide Telescope software, while others simply asked a friend. It probably helped that all the objects were from the standard Messier catalogue of 109 (or possibly 110, depending on your point of view) astronomical objects, originally published in 1771.

Object 1 – click to enlarge

This is the Pleiades (or M45), an open star cluster in the Milky Way about 440 light-years away which is sometimes called the ‘Seven Sisters’. It’s visible to the naked eye, but the blue cloud surrounding the stars can only be seen in long-exposure photographs. At one time it was believed that the stars in the Pleiades were formed from this cloud, but more recent observations indicate that the stars are just passing through a random dust cloud.

Object 2 – click to enlarge

The Whirlpool Galaxy (or M51) is a galaxy about 23 million light-years away. Together with its companion (NGC 5195) it can be seen with binoculars and it’s a favourite target for both amateur and professional astronomers.

Object 3 – click to enlarge

This was the toughest image. It’s the starburst galaxy M82 (also called the Cigar Galaxy), about 12 million light-year away. Tidal forces caused by the gravity of the nearby massive galaxy M81 have deformed this galaxy, a process that started about 100 million years ago. This interaction has caused star formation to increase tenfold compared to ‘normal’ galaxies.

Comments

“At last the hardest job, identifying galaxies.” (Thomas Thannheiser, Lübeck, Germany)

“I had to do a lot of work with Google and Wikipedia (I hope this is not cheating) to solve the tie-break pictures.” (Joan Fluvià Poyatos, Tordera, Barcelona, Spain)”

“I had no idea about the astronomical objects (beautiful pictures btw) but I do have a background in physics and it helped me in the sense that I have a friend from University who was able to identify them within minutes.” (Frans Andersson, Reading, UK)

“Marvellous pictures indeed. My favorite is "The Pleiades". I cannot stop to dream inside .. in blue!” (Gerard Gris, Lausanne, Switzerland).

“As a former astronomy student it was very funny to see the questions about the astronomical objects/galaxies. I recognized the pictures, but wasn’t always sure of the name, so it also was a nice reason to open up old studybooks again to look through some galaxy-pictures. Thanks for adding that to the puzzle contest :)” (Arjon Severijnen, Lepelstraat, Netherlands).

“I’m not a big expert in astronomy objects but with a little help of WorldWide Telescope software I came to the next answers.” (Andriy Syvak, Lviv, Ukraine)

“Last but not least, the tiebreakers are especially hard. Especially object 1 and 3 which took me days to google and browse the images.” (Ilyas Ramdzan, Malaysia)

“I don’t know any astronomers, so it took a lot of astronomical googling and searching library books to come up with these answers, which I hope are correct.” (Richard Saunders, London, UK” (They were - JN)

“Some great puzzles!!!  Chess is my hobby and I teach astronomy at a local college, so I’m always glad to learn of other chess players interested in astronomy (such as John Nunn) as well see other articles at Chessbase related to astronomy!!!” (Dean Arvidson, Los Angeles, USA).

These pictures were all taken on remote telescopes provided by Global Rent a Scope, which I have been using with great pleasure for some time now. A description can be seen in my article Chess and Astronomy from June last year.

Prize winners

The four prize winners are Matthew Anderton of Guildford, England; Eloy Martinelli of Fairlawn, USA; Ryuji Nakamura of Tokyo, Japan and Miki Granski, Haifa, Israel. These readers, like a number of others, got every puzzle right and in addition solved the astronomical tiebreaker. They were chosen by a random process (based on the digits of pi) from all entires that were similarly accurate.