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| | | (1) Posted by Joaquim Crusats [Saturday, Jan 10, 2009 23:28] | A problem by Július Sunyer
I need some help with a problem published by Dr. Július Sunyer in 1930.
Jordi Breu, a Catalan composer now aged 81 is preparing a new home-made edition of “Selecció dels Problemes compostos per Josep Paluzie” (in Catalan language, “Selection of Problems composed by Josep Paluzie”) mostly for historical reasons so that some copies can be preserved in the National Library of Catalonia for future chess composers interested in the history of chess composition in Catalonia. The book includes the problems that “El Club d’Escacs Barcelona” (Barcelona Chess club) published in “Els Escacs a Catalunya” (Chess in Catalonia) in 1930 in a homage dedicated to Josep Paluzie. Among the problems there was the one below by Július Sunyer, another well-known Catalan composer of retros (as I have found out in the Retrograde Analysis Corner). The problem, which I did not find in the PDB Server, is the following:
 (= 9+12 )
Add two white pieces (none of them on g3) and then white #1
ELS ESCACS A CATALUNYA, December,1930, pages 670-671.
The solution by Sunyer was published in the same magazine, February 1931, p.718.
Solution: Add a wBd1 and a wRg4 and 1.Rg2#
The main ideas in Sunyer’s solution are the following:
Black played last, the only possible move was b1=B. All captures by black took place in black squares, thus the wB initially in f1 must still be on the board. Since it is not the piece that can deliver the mate it must be on d1 because when black promoted, the bRa1 couldn’t be checking the wK.
Now there is some reasoning to exclude a mate by the wQ. Owing to the fact that all the knights are still on the board, and given the pawn structure, the first capture had to be gxQf6. Since the wB on dark squares can not mate, the only possible solution is the one above.
However, Dr Sunyer was very self critic about the need of the statement (none of them on g3). Paluzie wondered whether or not the problem could be improved.
So I would be most grateful if you could comment on the following points:
a) is the problem cooked? The initial position (including the two additional pieces) seems to be easily legal,
b) has the problem been reprinted somewhere else?
c) can you see any way to improve the problem keeping all the original content in such a way that a mate by a wQ cannot be initially ruled out (and ideally the mate by the wQ should be also unique as the intended one by the wR) without the need of any additional statement added to the stipulation?
J.Breu would like to use the comments in the new edition of the booklet, but of course only if you agreed.
Thanks in advance,
Joaquim Crusats
| | | (2) Posted by Joaquim Crusats [Monday, Jan 12, 2009 11:13] |
I have received a correction by email which seems to be sound.
Since that author might want, perhaps, to publish it I would be most grateful if you didn't anticipate it here. Could you send any possible correction to me via email at:
joaquimchessproblem AT gmail.com
In any case I'll eventually post the new version here.
Thanks.
| | | (3) Posted by Joost de Heer [Monday, Jan 12, 2009 13:47]; edited by Joost de Heer [09-01-12] |
The original position seems to be illegal after the addition of the two white pieces:
White has 11 pieces left (including the two pieces that have to be added). Black's last move was b2-b1=B. This pawn captured 5 times. This includes the b-pawn which was captured on b2. Since the last move was b2-b1=B, the white bishop from c1 could never have left its homesquare.
| | | (4) Posted by Joaquim Crusats [Monday, Jan 12, 2009 14:58] |
The wP initially on b2 need not be captured on b2. In fact, it couldn't or the position would indeed be illegal. It was captured on c3 when captures where possible, that is, after the wQ had been taken on f6.
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