05 August 2011

Constructional Challenges

Most probably you have heard of the Millennium Prize Problems. These are seven problems in mathematics that were stated by the Clay Mathematics Institute of Cambridge (CMI), Massachusetts, in 2000. The correct solution to any of them is awarded with a prize of 1,000,000 US dollars by the institute. In 2003, Grigori Perelman published a proof of the Poincaré conjecture. He was selected to receive the Fields Medal for this and awarded with the Millenium prize, but he declined both awards — it's quite interesting to read more about him and his work. Anyway, this is the only solved problem so far.

In mathematics, there are lots of such so far unsolved problems — and the same applies to chess compositions. We know several themes which composers all over the world and throughout the years still haven't been able to master. In the March 2006 issue of The Problemist Stephen Rothwell showed seven of those constructional challenges calling them "chess composing millennium problems". Until that time, no correct settings of the themes had yet been composed. Correct means correct and legal position, without promoted pieces. For each of them he stated a conjecture that it is unsolvable. Nevertheless, he offered a book credit of Euro 50 for any correct setting of one of the seven constructional challenges that he mentioned. Closing date was 31 December 2006. Unfortunately, I don't know about the results and I wonder how many entries he received at all ...

In 2010, the German composer Werner Keym published his first book "Eigenartige Schachprobleme" (only available in German) which all members of the Schwalbe received as a present. Interestingly, he also listed several challenges for the readers — without timely limitation and each awarded with Euro 100! Four of the seventeen tasks have already been solved in 2010. You can find these solutions here (the last four diagrams).

Three composing challenges were mentioned by both Rothwell and Keym. Let's have a look at them:

1. Double square vacation with both white castlings in a moremover

  1.1Thomas Rainer Dawson  
Chess Amateur, 1923

  1.2Andreas Thoma  
König und Turm, 2003
    Sam Loyd was the first to show the double square vacation by castling. This was already achieved in 1877 but only with an illegal position. I won't bother you with that, for I prefer to give you a correct problem which you see in diagram 1.1.

The key
1. 0-0!
vacates the squares e1 and h1 for the white queen at once.
1. - b3 2. Qe1 ~ 3. Qa5#
1. - c5 2. Qh1 ~ 3. Qa8#

Diagram 1.2 demonstrates a twofold double castling square vacation. So, we see it is possible! The solution goes like this:
1. Ba4! (threats 2. Nb4 3. Na6#)
1. - d5 2. 0-0 dxc4 / a5 3. Qh1 / Qe1 4. Qxc6# / Qxa5#
1. - aS 2.0-0-0 Nf7 / d5 3. Na1 / Qe1 4. Nb3# / Qxa5#

You may criticize the rather weak key, but consider that the first step is always proving the feasibility. After that is done you can look for improvements. However, there is the slight weakness that the square vacation of e1 is used by the wQ twice. A perfect setting would need the additional variation 3. Ne1 after 2. 0-0-0. This is what we still are trying to construct.

Conjecture: It is impossible to construct a moremover showing double square vacation of a1, e1, h1 by white castling on both sides whereby the square vacation of e1 has to be used by two different white pieces.

2. Oudot task in the orthodox helpmate

  2.1Marcel Tribowski  
StrateGems, 2002
2nd Prize

  2.2Reinhardt Fiebig  
harmonie, September 2005
    A famous, yet unachieved, helpmate task is the so-called Oudot task which is named after the French composer Jean Oudot. It demands the promotion of three black pawns to queen in a single solution in an orthodox helpmate. The minimum length of such a composition is nine moves, as it takes at least three moves for each pawn to substantiate its promotion to a black queen.

Problem 2.1 shows the theme in a helpstalemate:
1. a1=Q! Ke3 2. Qa5 Kf2 3. Qf3+ exf3 4. c1=Q! fxg4 5. Qxc3 g5 6. Qf6+ gxf6 7. b1=Q! fxe7 8. Qd1 e8=Q 9. Qh5 Qxh5.
Bear in mind that the problem could not be fully computer-tested.

Another close try is diagram 2.2 — Do you see why I say close?
1. e1=Q! Ka1 2. Qc3+ Kb1 3. Qb3 axb3 4. f1=Q bxc4 5. Qf4 c5 6. Qd6 cxd6 7. g1=Q d7 8. Qg6 d8=N 9. Qh7 Nf7#
Yes, the composer used nine black pawns!

Conjecture: It is impossible to construct an orthodox helpmate with three promotions to black queen by different pawns in a single solution.

3. 100 Dollar task

  3.1Camil Sénéca  
Thèmes 64, 1956

  3.2Josef Ettner  
Ryszard Nojek  
Werner Keym
Die Schwalbe, 2005
    One of the most spectacular of all challenges is the 100 Dollar task that was defined in 1963. It requires a black and a white excelsior with promotion to knights in an orthodox helpmate in five.

A correct double N-excelsior is shown in diagram 3.1, but it takes seven moves:
1. d5 g4 2. d4 g5 3. d3 g6 4. d2 g7 5. d1=N g8=N 6. Ndc3 Ne7 7. Nb5 Nc6#.

The earliest attempt in five moves dates back to 1965 when Jenö Ban used six promoted pieces. His solution is often used to demonstrate the task. Since then, composers gradually managed to reduce the amount of promoted pieces. The best result so far is diagram 3.2 with only two promoted pieces:
1. b5 e4 2. b4 e5 3. bxc3 e6 4. cxd2 e7 5. dxe1=N e8=N#.

Conjecture: It is impossible to construct an orthodox helpmate in five with a black and a white N-excelsior.

Whereas Keym awards the 100 Euros only for compositions meeting the original requirements of the first two tasks mentioned here, he obviously assumes that the conjecture for the third task is true. He offers 100 Euros for the first construction with only one promoted piece.

Solving these or other tasks might appear to be impossible, the more if one reads how many brilliant composers already made so many incorrect attempts. Still, the very same was true for the famous Babson task. After many decades, someone finally came up with a solution, but this was not one of the big names! Leonid Yarosh was virtually unknown until 1983 when he published the first correct Babson. So, don't give up — at least not before someone proved the impossibility.

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