The event of this championship is, of course, the "defeat" of Michel Caillaud : he is indeed "only" 2nd, which would be a victory for any normal solutionist, but not for him. He is overtaken by our friend Onkoud, who already has 2 international master titles (for game and composition) and is seeking the 3rd, for solving.
Michel's weak point has always been the 3#, I took advantage of it in 2005, but it is rare that he does not score, moreover, on a 2#, particularly devious it is true. Being the only one to solve the selfmate 5#, which is more usual, was not enough to catch up.
The study of Kraline and Krikheli was fascinating, it was appreciated two and a half years ago by the listeners of the course http://lecoursdumaitre.e-monsite.com/en/pages/lessons/cat-2019/february-5-2019.html but solving it in a competition is another matter: not one competitor scored a single point.
No difficulty for the first 2#. The second one, through which the scandal arrives, reminds us of the famous Mansfield 1956, but reserves us a surprise.
The first 3#, a magnificent "triplet" by the author of "Mostly three-movers", is too well oiled to be difficult, but one can forget one or two secondary variations. The second one is not difficult either, but one can be disturbed by its artificial "set play" (not supported by a try) and... a try, precisely, reversing the continuations of the JA: the "Zagorouiko theme" in 3 moves.
The 11# is fun and very easy: you try moves without knowing where you are going and precisely... you go there!
The 2# helpmate contains 2 solutions in each twin. Unusual statement: sometimes competitors... just forget to do everything. I don't like it, an elegant solution compensated by another one, in which a white Rook is not used for mate. The 3# helpmate is of a much better standard, but also easier.
A 2-moves selfmate is normally easy. There are exceptions, but this is not one of them! And the pretty variant is the icing on the cake. Finally, the highlight of this kind of test, the moremover selfmate (in this case 5), which was supposed to decide between the brilliant candidates until then, was much more accessible than usual: we know from the start that there will be 2, and only 2, variants after 1...hxg1B and 1...hxg1N. In short, find a plausible key and solve two 4# selfmates! Not a big deal. Surprising that only Michel solved it. I had also swallowed it in the past, because it was given at the Ukraïne 2002 championship, then forgotten. I renewed it, then found that I had it in my collection, without any sign indicating the difficulty! Thank you Mr. Petkov.
No comments on the fairy problems, which I consider inappropriate in a solving championship (no matter how interesting the said problems may be). Moreover, even outside a championship, I will not be made to look for a diagram containing an wKh7 and a bPg8-sic!
Conclusion: in 2h30, it was possible to solve most of the problems (selected by V. Crisan) by... keeping a cool head and taking the time to read again!
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