World Problem Solving Championship – Solutions

by ChessBase
11/11/2010 – Last week John Nunn showed us four of the problems that had been given to the participants of this competition that was held on October 19th and 20th in Hersonissos, Crete. They were not easy (it was, after all, a world championship!) and John describes how he went about solving them. And he presents another stunning selection of astronomical pictures. All highly instructive.

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World Problem Solving Championship – Solutions

By John Nunn

I hope that readers will have tried to solve at least some of the problems I gave in the first part of my article. Now it’s time to give the solutions. In addition to explaining the solutions, I will also try to give an idea as to how I went about solving those problems, at least in those cases where I was successful. Some of the problems are quite difficult (it was, after all, a world championship!) so don’t be dismayed if you couldn’t work them out.

G. Heathcote, 9th Honourable Mention,
Olympic Tourney, 1948


White to play and mate in 3

Problems like this can be hard to solve, because there’s no obvious starting point. True, the a6-pawn indicates that Black’s king will move to c6 at some point, but this doesn’t help very much. If nothing else suggests itself, it often helps to look for a pretty mate and see how it can be forced. Here I noticed that if the white rook is not on the second rank and Black could be induced to play 1...d6, then White could play 2 Qe2! (threat 3 Bc4#) 2...Bxe2 3 Rf5#. The main problem here is that a rook move along the f-file doesn’t set up any obvious threats, and the clue lies in the fact that, whatever the threat is, we know that 1...d6 will defeat it.

A bit of imagination then leads to the key: 1 Rf6!, with the tricky threat of 2 Rb6! (threat 3 Rb5#) and mate after 2...Kxd4 3 Rd6#. The key lines are 1...Nxf6 2 Qc7 (threat 3 Qc5#) 2...Ne4 (or 2...d6 3 Qa5#) 3 Bc4#, 1...d6 2 Qe2! Bxe2 3 Rf5# and finally 1...Be6 2 Nc2! followed by 3 Nb4#. It is the last line which makes use of the a6-pawn. Note also that 2 Nc2! only works once Black has blocked e6, for otherwise Black can take on f6 and give his king an escape square on e6. Three-movers with difficult quiet threats can be quite tricky! [Click to replay]


Y. Bazlov, 5th Prize, Gusev Jubilee Tourney 1994

White to play and win

I didn’t solve this and it’s interesting to see why. White starts with 1 Ra2+ Kb3 2 Bd1+! Kc3 (2...Kc4 3 Be2+ Kb3 4 Rb2+ Ka3 5 Bd1 Qg8 6 Nd4 gives White decisive threats) and now I didn’t see anything for White. The brilliant idea is 3 Be2! (threatening 4 Rc2+ Kb3 5 Bc4+ Ka4 6 Ra2#) which I actually thought lost to 3...Qb7+ 4 Kc1 Qb3. I didn’t see the remarkable move 5 Bc4!, which turns the tables, although there are still a few twists: 5...Qa4! (this stalemate defence is the best chance; 5...Kxc4 loses to 6 Nxa5+ and 5...Qxc4 6 Rc2+ drops a whole queen after 6...Kd3 7 Ne5+ or 6...Kb3 7 Nxa5+) 6 Ne5! Qe8 (or 6...Qb4 7 Rc2+ Kd4 8 Nc6+) 7 Rc2+ Kd4 (7...Kb4 8 Nd3+ Ka3 9 Ra2#) 8 Nf3+ and the queen falls after 8...Ke3 9 Re2+ or 8...Kc5 9 Bf7+. [Click to replay]


G. Sobrecases, Original for diagrammes, 2010

Helpmate in 5.5 moves:
a) Diagram; b) Add a black rook on b5

In helpmates, spotting the mating position is paramount, and often it is a case of you either see it or you don’t.  Here the idea is that Black plays his king to c3 and his bishop to d2, when White mates by Bxb2. At first it looks easy, as playing Black’s pieces into position takes four moves, leaving him with a spare move, while White’s bishop has plenty of time to reach b2. The problem is that the only reasonable square for White’s bishop to set up Bxb2# is c1, and this is currently occupied by Black’s own bishop. This explains where Black’s spare move is used: the c1-bishop has to travel along the c1-h6 diagonal to clear the way for White’s bishop to reach c1. It turns out that there are currently no suitable squares on the diagonal for this manoeuvre to take place, so White has to spend some time removing the g5-pawn to give Black’s bishop access to h6. This leads to the unique sequence 1 Bh4 Kxe3 2 Bxg5+ Kd4 3 Bh4 Bh6 4 Bg5 Kc3 5 Bc1 Bd2 6 Bxb2#. [Click to replay]

I didn’t solve the second part. I spotted that the addition of the rook on b5 allows a new mating possibility in which Black’s king is on d4, the b5-rook moves to d5 and his bishop moves to e3 (of course, the previous line is not feasible as b2 is defended). White, in the meantime, plays his king to b4 and bishop to c3. The problem this time is the knight on c6, which prevents White’s king occupying b4. I was just working on how to eliminate this knight when I ran out of time. The key idea is to use the e3-pawn to take the knight, and this exactly consumes White’s spare tempi. The solution runs 1 Be1 Rd4 2 exd4 Ke3 3 d5 Kd4 4 dxc6 Rd5 5 Kb4 Be3 6 Bc3#. [Click to replay]


F. Richter, Original for MatPlus 2011

Selfmate in 7 moves

The only real mating possibility is to force Black to move the g3-rook, delivering mate from the e1-bishop. The first two moves are fairly easy. White starts with 1 Rd4, threatening 2 Bg4+ forcing 2...Rxg4#. The only defence is 1...Bxf3. Now any move by the g7-bishop will threaten 3 Rg5+, but it’s impossible to choose the correct square for this bishop unless one sees the main line. This continues 2 Bh8! Bg4 (the only defence) 3 Nd8, which threatens 4 Nd6+ Kxe5 5 Rg5+ Bf5 6 Rxf5+ exf5 7 f7+ (making use of the bishop’s position on h8) 7...Rg7#. Black’s only way to meet this threat is to play 3...bxc4, but then White plays 4 Nf7 and now Black has no defence to the new threat of 5 Nh6+ Kxe5 6 Rg5+ Bf5 7 Ng4+ Rxg4#. [Click to replay]

If you want to see the other problems from the event, they are available on this PDF document.

Leading scores


The 2010 World Chess Problem Solving Champion

Rank

Name

Nat.

Title

Rating

Pts.

360

1

Nunn, John

GBR

GM

2747

71.0

357

2

Murdzia, Piotr

POL

GM

2796

69.5

354

3

Pfannkuche, Michael

GER

GM

2616

64.0

356

4

Evseev, Georgy

RUS

GM

2809

63.5

359

5

Van Beers, Eddy

BEL

GM

2671

61.5

349

6

Selivanov, Andrey

RUS

GM

2544

61.5

353

7

Vuckovic, Bojan

SRB

GM

2617

61.5

354

8

Kopyl, Valery

UKR

IM

2503

61.0

360

9

Piorun, Kacper

POL

FM

2575

60.0

359

10

Viktorov, Evgeny

RUS

2432

59.5

352

11

Zude, Arno

GER

GM

2692

58.0

359

12

Fomichev, Evgeny

RUS

2349

56.5

341

13

Comay, Ofer

ISR

GM

2621

56.5

352

14

Tummes, Boris

GER

GM

2685

56.5

357

15

Paavilainen, Jorma

FIN

GM

2548

55.5

360

16

Kovačević, Marjan

SRB

GM

2564

54.5

358

17

Limontas, Martynas

LTU

2353

54.5

360

17

Mestel, Jonathan

GBR

GM

2581

54.5

360

19

Prentos, Kostas

GRE

IM

2500

52.5

357

20

Kolčák, Marek

SVK

FM

2376

52.5

360

21

Javadov, Ramil

AZE

2353

51.0

360


Once again a few pictures having nothing whatsoever to do with chess problem solving. These were all taken using the remotely-controlled telescopes operated by Global Rent-a-Scope. You can click on the images to see larger versions.

If only our eyes were more sensitive the night sky, instead of being black with a sprinkling of dots, would be full of the colourful clouds of nebulae and the fuzzy blobs of other galaxies, but these hidden objects can only be viewed through a telescope or, even better, using long-exposure astrophotography. The red trail snaking through this image is part of enormous structure called Barnard’s Loop, which extends over a massive 10 degrees of the sky (20 times the diameter of the full moon) and covers much of the constellation of Orion. It’s thought to be the remains of a supernova that exploded about two million years ago, and it’s believed that the reddish glow is the result of the material in the loop being ionised by radiation from the stars embedded in the Orion Nebula. Barnard’s Loop is about 1,600 light-years from Earth. Near the lower left of the image one can see the smaller nebula M78, while at the top, slightly left of centre, is the open cluster NGC 2112. As regards the starfish-shaped object near the top, right of centre, I have no idea what it is.

This image was a combined effort between myself and Dr Christian Sasse. Continuing our theme of supernova remnants, this is probably the most famous remnant of all, the Crab Nebula. The distance isn’t known very accurately, but estimates range around the figure of 6,500 light-years. The Crab Nebula is the result of a supernova explosion which was recorded by Chinese and Arab astronomers in 1054; it’s currently about 11 light-years across and is expanding at about 1,500 kilometres per second. The star that exploded almost 1,000 years ago afterwards collapsed to form a neutron star, that is a star in which intense gravitational pressure has caused its atoms to collapse, leaving only a mass of tightly-packed neutrons. The Crab neutron star rotates rapidly, and emits a beam of radiation (including visible light) which sweeps across our solar system 30 times a second, so that it appears to be flashing very rapidly. Of course, you can’t see this effect on a long-exposure photograph which averages the light received during the exposure. Neutron stars which exhibit this behaviour are called pulsars and if you want to identify it in the image, look for two stars very close together at the centre of the nebula. The pulsar is the fainter of these two (the one at the lower right). The actual pulsations, as recorded by the Kitt Peak Photon Counting Array, can be seen here. And a time-lapse video with the pulsations converted into sound may be found here:

The Crab Pulsar is about 25 kilometres across and is so dense that a teaspoonful weighs about a billion tons. It seems absolutely remarkable to me that an object only 25 kilometres across should be visible from a distance of roughly 60,000,000,000,000,000 kilometres, but this relatively tiny object is putting out 100,000 times as much energy as our Sun and is the power source that illuminates the entire Crab Nebula.

Finally, and moving away from supernovae, here is one of the most magnificent objects in the sky, M31, the Andromeda Galaxy. This is the nearest large galaxy to the Milky Way and is ‘only’ 2.5 million light-years distant. It is a bit larger than the Milky Way, and contains roughly a trillion stars, significantly more than our galaxy. However, astronomers are undecided as to which of the two galaxies is the more massive, since the Milky Way appears to contain more dark matter than Andromeda. It is apparently visible to the naked eye in perfect conditions (although I have never managed to spot it) and can be seen with binoculars even in light-polluted urban conditions. I like this image, which provides a good view of Andromeda’s spiral arms, and its two satellite galaxies, M32 and M110.

Click on the image above to retrieve a 1920 x 1200 version of the Andromeda Galaxy that you can use as a desktop background on the now common high resolution monitors.

Finally, remember the image of Omega Centauri we showed you in a recent article? It is one of the around 150 globular clusters that surround our Milky Way – in fact the the largest, containing 10 million stars, 16,000 light-years away and roughly 85 light-years across.

Well, here's a new picture of the Omega Centauri Globular Cluster, which was caught by NASA's Cassini spacecraft while it was pointed to study Saturn's F-ring. If you click the picture you will see a concatenation of 13 images, each taken about three minutes apart, that show Omega Centauri move by. Some of the cluster's stars can even be seen through the ring's narrow Keeler Gap near the end of the movie. It was produced by the Cassini Imaging Central Laboratory for Operations (CICLOPS).


Feedback

Of the many feedback messages we received – including a number referring to the astronomical pictures, there is one we want to share with you. It is from an old friend and remarkable puzzle expert:

Dear ChessBase,

I enjoyed reading John Nunn's ChessBase notes on the recent solving championship that he again won. The problems are unattributed, but the long selfmate has all the dread hallmarks of Camillo Gamnitzer; do I guess correctly? I think I finally solved it -- took me more than 25 minutes, but if I'm right (about both author and solution) it's the first time I've cracked a Gamnitzer at all.

For the other problems – I completely failed on the mate in 3 and still have no idea what's going on in the study. For the helpmate, I solved the first part reasonably fast: 1 ... Bh4! 2 Kxe3 Bxg5+ 3 Kd4 Bh4! 4 Bh6!! Bg5! (second switchback, this time without capturing) 5 Kc3 Bc1 6 Bd2 Bxb2#. Since the second part's solution has a much more routine motivation (the White pawn must march to remove the Nc6, getting out of the way of ...Rd5 as soon as possible), it must have been included to provide an "echo" of the first solution (the mating picture: Ka3,Bb2 vs. Kc3,Rc4,Bd2, pawns c2,d3, has been exactly shifted up one diagonal space, though it is obtained with rather different play) plus, of course, to add to the solving challenge!

Sincerely, Noam D. Elkies

In a later email he wrote:

I see now that I got the selfmate's solution correctly, but not its composer's name. It's still much in the Gamnitzer style, though one might have preferred the threat made by the initial 1 Rd4 to be longer, ideally finishing on the 7th move as does the threat after 3 Nd8.

Noam D. Elkies, 44, is an American mathematician and chess master. At 14 he received a gold medal with perfect score at the International Mathematical Olympiad, and at 16 he won the Putnam competition. He graduated as valedictorian at age 18, in Mathematics and Music, and earned his Ph.D. at the age of 20 at Harvard University. In 1987 he proved that an elliptic curve over the rational numbers is supersingular at infinitely many primes, and in 1988 he disproved Euler's sum of powers conjecture for fourth powers. His work on these problems won him recognition and a position as an associate professor at Harvard in 1990. In 1993, he was made a full, tenured professor at the age of 26. This made him the youngest full professor in the history of Harvard, surpassing the record previously held by Alan Dershowitz and Lawrence Summers (who were made full professors at age 28).

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