26 August 2011

New content in the Chess Informant

I became quite curious when I read the announcement that, beginning with issue 110, there are some new columns in the Chess Informant. Among other new things, it now has a selection of chess problems prepared by the International Solving Grandmaster and Grandmaster of chess compositions Milan Velimirovic. They not only want to offer problems for solving, but also intend "to bring insights from the professional point of view into the secrets of the chess problem world".

The other day, I could use the opportunity to have a look at the selected studies and problems and I looked forward to the promised insights.

Velimirovic summarizes the five World championships in chess composition. These competitions cover three year periods, starting with the years 1989 to 1991 for the first championship. He concentrates on the results in the three sections twomovers, threemovers, and moremovers. Thus, the reader is presented fifteen problems, each composed by the respective winner of a section.

I can only speak for the electronic edition which comes with a PGN file containing the problems. And what I saw there is not what I call professional. Naturally, each composition is stored as a game. But the result of all games is "0-1"!? Of course, that's totally wrong, as in orthodox mate problems it's always White who checkmates. The notation and annotations are not really sufficient to understand clearly what theme is shown and how, though its name and a brief description is given at the end of each virtual game. Moreover, there are obvious errors like an explanation mentioning the white queen, whereas there is none on the board at all!

Here is one of the awarded compositions that Velimirovic picked for the interested reader:

Anatoly Slesarenko
Championship of USSR 1991
1st Place
[2n5/3p1Q2/1r6/3p1p2/2p1k1p1/2R3PN/2Npp1nK/B7]
  #2(7+11)  
    This is the original notation for this problem:

[solution]

I don't think the pattern of the theme is really made clear hereby. However, this table which I found here should help:

  a b c d
  A,B C D    
  C,D     A B

Applying it to the problem gives us a better picture:

  Kxd3 cxd3 gxf3 Kxf3
1. Rd3? threats Nf2, Qxd5 Qxf5 Ng5    
1. Rf3! threats Qxf5, Ng5     Nf2 Qxd5


Another renowned problemist takes care of the endgame studies: Yochanan Afek. Unfortunately, in a quite dry manner he simply shows nine diagrams without further comments.

The studies are also collected in a PGN file, each of them with lots of variations but no further text. Similar to the mate problems, the result tag of each virtual game does not show what you'd expect. I think "1-0" to indicate that White should win or "1/2-1/2" where White is to draw is common usage whereas here all games are marked with "Line". But that's the only thing to criticize. Now, there is one more sample for you:

G. Amann
Schach, 2008
1st Prize
[4kq2/pK3p2/p2Rp3/P3P1N1/8/8/6P1/8]
  Win(6+6)  
    1. Nh7! Qe7+
1. - Qh6 2. Nf6+ Kf8 3. Rd8+ Kg7 4. Rg8#
2. Kc8 f6 3. Ng5!!
3. Nxf6+? Kf8 4. Rd8+ Kg7 5. Rd7 Kh8 6. Rxe7=
3. - fxg5 4. g4! Kf8 5. Rd8+ Kf7 6. Rd7 +-

I hope future installments of these columns will be prepared with more diligence.

19 August 2011

A few more fascinating tasks

Admittedly, not all chess players nor all composers are actually thrilled by task problems. But I am. Therefore, I chose three more of them for you. And I think that you may like at least some of them.

The first three compositions feature the flight task which demands that in the diagram all eight flight squares for the king that is to be mated are accessible for him. No. 1 is sort of an extension of this task as the white king is even the only piece in a 6x7 area. The second problem demonstrates the task in a minimum of two moves. It's notable that the king is mated in the middle of the board and only one flight square is occupied. Finally, No. 3 shows an ideal mate with all four bishops after a series of checks and sacs of the heavy pieces.


  1Manfred Zucker  
Die Schwalbe 06/2002
Commendation
[b1QBR3/pr6/k7/8/4K3/NN6/8/8]
  s#15(6+4)  

  2Alexander Hildebrand  
Eskilstuna Kuriren, 1956
 
[8/Q3R1p1/2K2krp/4p1rp/5b1B/4N2B/8/5R2]
  s#2(7+8)  

  3Gerson Berlinger  
Schach-Echo 07/1988
 
[Q6R/5K2/3B4/b6p/2B3k1/8/8/3b3R]
  s#6(6+4)  

Solutions:
11. Bh4 Kb6 2. Bf2+ Ka6 3. Kd4 Kb6 4. Kd5+ Ka6 5. Re3 Kb6 6. Re4+ Ka6 7. Nc4 Kb5 8. Nd6+ Ka6 9. Re3 Kb6 10. Re6+ Ka6 11. Be1 Kb6 12. Ba5+ Ka6 13. Bc3 Kb6 14. Bd4+ Ka6 15. Qc4+ Rb5#
21. Bc8 e4 2. Bb7 Kxe7#
31. Be6+ Kg5 2. Qd8+ Bxd8 3. Rg8+ Kh6 4. Bf8+ Kh7 5. Rh8+ Kxh8 6. Rxh5+ Bxh5#


The task achieved in the next problems is the maximum effect of eight white promotions to a knight (diagrams 4 and 5) or a rook (diagram 6).


  4W. A. Shinkman  
(Version A. Chéron)  
Deutsche Schachzeitung, 1908
[N7/PPPPPPPP/K1k3rB/b1pnnb1p/8/1r6/8/8]
  #8(11+9)  

  5Jorma Pitkänen  
Problemkiste 05/2010
 
[K7/PPPPPPPP/4N1k1/pqpnn2R/8/Nr6/8/br1B4]
  s#8(13+9)  

  6Bo Lindgren  
feenschach, 1966
2nd Prize
[8/PPPPPPPP/4BB1N/1n4pp/kr6/2r1p3/2b1p3/2K5]
  s#10(12+9)  

Solutions:
41. b8=N+ Rxb8 2. axb8=N+ Kd6 3. c8=N+ Ke6 4. d8=N+ Bxd8 5. exd8=N+ Kf6 6. g8=N+ Rxg8 7. hxg8=N+ Kg6 8. f8=N#
51. h8=N+ Kf6 2. g8=N+ Kxe6 3. f8=N+ Kd6 4. e8=N+ Kc6 5. d8=N+ Kb6 6. c8=N+ Ka6 7. b8=N+ Qxb8+ 8. axb8=N+ Rxb8#
61. a8=R+! Na7 2. Rxa7+ Kb5 3. b8=R+ Kc6 4. c8=R+ Kd6 5. d8=R+ Kxe6 6. Rc6+ Rxc6 7. e8=R+ Kxf6 8. f8=R+ Kg6 9. g8=R+ Kxh6 10. h8=R+ Bh7#


Composition No. 7 pleases us with an alert black king. With each key move he visits another neighbouring square. Similarly, in the last problem the black king is checkmated on each of the eight adjacent squares.


  7Harald Dieffenbach  
Schach, 1962
1st Prize
[4nK2/1k4N1/3p3b/pr6/2p1rn2/3P3P/bR4B1/8]
  h#3(6+10)  
  8.1...

  8Jan Kovalic  
The Problemist 05/2009
 
[7n/2p1p3/3pP1b1/1PPrkqp1/1pn1Nr2/P1b5/1P3PP1/2K1R3]
  h#2(10+13)  
  8.1...

Solutions:
71) 1. Ka6 dxc4 2. Tb7 c5 2. Ta7 Tb6#
2) 1. Ka7 Te2 2. Td4 Txe8 3. Tb6 Ta8#
3) 1. Ka8 Te2 2. Tb7 Txe4 3. Ta7 Txe8#
4) 1. Kb6 dxc4 2. Td4 Ld5 3. Kc5 Txb5#
5) 1. Kb8 Txa2 2. Te7 Txa5 3. Tc7 Ta8#
6) 1. Kc6 Lh1 2. Kd5 Tg2 3. Td4 Tg5#
7) 1. Kc7 Lxe4 2. Tb8 Ke7 3. Tc8 Tb7#
8) 1. Kc8 Te2 2. Te7 Lc6 3. Tc7 Txe8#
81) 1. Kd4 Sd2 2. Dd3 Sb3#
2) 1. Td4 axb4 2. Kd5 Sxc3#
3) 1. dxc5 Sxc5+ 2. Kd6 Sb7#
4) 1. Se3 bxc3 2. Kxe4 Txe3#
5) 1. Kxe6 c6 2. Lf6 Sxg5#
6) 1. Tg4 Sxg5+ 2. Kf4 Sh3#
7) 1. Df6 f3 2. Kf5 Sg3#
8) 1. Sf7 Sxd6+ 2. Kf6 Se8#

12 August 2011

Helpmate records

In the previous post I wrote about the famous 100 Dollar task and the Oudot task. Of course, there are several other challenges for composers of helpmate problems. I want to draw your attention to two of them.

The first one asks for the longest correct orthodox helpmate with no other black piece than the king and no promoted (white) piece. As far as I know, seven moves is still the maximum. There are already lots of h#6.5 which can be extended to a h#7 by forcing the black king to capture a white queen as first move leading to the actual diagram — see No. 1 where we could put the black king on h1 and add a wQh2. Moreover, dozens of "real" h#7 have been published and I picked two for you. Diagram 2 is an extraordinary problem using the complete set of white pieces. The third composition is also something special as it works without promotion.

Lots of attempts have been made but still nobody successfully extended the record to 7.5 or more moves yet. The only thing I know of: In 2004, Noam D. Elkies came up with a h#10 that had seven white bishops.


  1Laszlo Zóltan  
Magyar Sakkélet 04/1958
[8/8/8/1K6/8/8/3P3k/8]
  h#6.5(2+1)  

  2Christopher Jeremy Morse  
British Chess Magazine 06/1971
[1NKRQRBB/k1PPPP2/N1P5/1PP5/1P6/8/8/8]
  h#7(16+1)  

  3Daniel Novomesky  
Pravda, 2007
[5K1k/8/8/8/5L2/8/5B2/8]
  h#7(3+1)  

Solutions:
11. - d4 2. Kg3 d5 3. Kf4 d6 4. Ke5 d7 5. Kd6 Ka5 6. Kc7 Ka6 7. Kb8 d8=Q#
21. Ka8 Bh7 2. Ka7 Rg8 3. Ka8 Qf8 4. Ka7 Re8 5. Ka8 Kd8 6. Ka7 c8=Q 7. Ka8 Qb7#
31. Kh7 Be5 2. Kg6 f4 3. Kf5 Kg7 4. Ke6 Kh6 5. Kf7 f5 6. Kg8 f6 7. Kh8 f7#

Challenge:Compose a correct helpmate in more than 7 moves with black king solus and no promoted piece.


The second record to break is even more fascinating. I don't remember exactly how it happened, but while searching the Internet for interesting helpmates, I found a downloadable article in German that was published in the problem magazine harmonie in June 2006. It dealt with the quest for the longest helpmate. As with other tasks and records, several experts have tried a variety of constructions in vain. There were always duals or other flaws. Apparently, every now and then, some rather obscure (mainly) Russian problem chess media report on a new achievement, but in all cases a closer look reveals cooks. So far, nobody has been able to surpass the record that was established almost 80 years ago:

Bernhard Hegermann
The Problemist FCS 10/1934
[6kR/4p1p1/1p2P1P1/1P2p3/1P2P3/1P2p1p1/4P1P1/5BK1]
  h#28(11+7)  

Solution:
1. Kxh8 Kh1 2. Kg8 Kg1 3. Kf8 Kh1 4. Ke8 Kg1 5. Kd8 Kh1 6. Kc7 Kg1 7. Kd6 Kh1 8. Kxe6 Kg1 9. Kf6 Kh1 10. Kg5 Kg1 11. Kf4 Kh1 12. Kxe4 Kg1 13. Kd4 Kh1 14. Kc3 Kg1 15. Kxb3 Kh1 16. Kxb4 Kg1 17. Kxb5 Kh1 18. Kc4 Kg1 19. b5 Kh1 20. b4 Kg1 21. b3 Kh1 22. b2 Kg1 23. b1=Q Kh1 24. Qf5 Kg1 25. Qf7 gxf7 26. Kc3 f8=Q 27. Kd2 Qc8 28. Ke1 Qc1#

Challenge:Compose a correct helpmate in more than 28 moves with no promoted pieces.


Surely, there aren't so many people being interested in helpmate records. But I hope there'll always be some problemists that still don't give up the search for a new record. Never forget Leonid Yarosh!

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
[3kN3/3Pp3/2p1Pp2/5p2/1p3P1Q/4p3/P3P3/4K2R]
  #3(9+7)  
 

  1.2Andreas Thoma  
König und Turm, 2003
[2N4n/p2p4/2p3p1/1Bk2pPp/2Pp1P1Q/P2Pp3/2N1P3/R3K2R]
  #4(13+10)  
    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
[8/4p2p/7p/8/3K2p1/1PPp4/pppPP2n/7k]
  h=9(5+10)  
 

  2.2Reinhardt Fiebig  
harmonie, September 2005
[6nk/6n1/4p3/5p2/2p4p/p2p4/P3ppp1/1K6]
  h#9(2+12)  
    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
[8/3p4/pp6/k7/n7/8/6P1/7K]
  h#7(2+5)  
 

  3.2Josef Ettner  
Ryszard Nojek  
Werner Keym
  
Die Schwalbe, 2005
[5bnq/1p3nkn/p4pp1/K7/8/1pPp4/1P1PP3/2BbB3]
  h#5(7+13)  
    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.