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| (41) Posted by Frank Richter [Tuesday, Oct 9, 2007 08:59] |
Very clear construction. What means the "Barthelemy theme"?
I don't like such duals in Zugzwang problems as 1.Rc8? Ba8,b7,g2 2.Rc1,Bg2#. But here, in try and looking at the only seven pieces, it is possible to tolerate this.
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| (42) Posted by Jacques Rotenberg [Tuesday, Oct 9, 2007 09:38] |
The try is not at all thematic, just nice to notice it because of the quiet and precise refutation.
Barthelemy theme : a single black piece controls a single white battery along two different lines.
two variations cancel each one of these controls, and the battery works with a mate closing the remaining line
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| (43) Posted by Administrator [Tuesday, Oct 9, 2007 12:22] |
QUOTE
What means the "Barthelemy theme"?
Jacques gave the definition, but this is a chance to draw your attention again to the outstanding rendering by Fuchs:
http://www.matplus.net/pub/start.php?app=forum&act=posts&fid=gen&tid=89
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| (44) Posted by Jacques Rotenberg [Tuesday, Oct 9, 2007 13:51]; edited by Jacques Rotenberg [07-10-12] |
Yes indeed, a great classic with twice the theme. One of its unusual features is the anti-critical moves by the thematic black Rook (in two different directions) used as anti-dual.
G. M. Fuchs is not very wellknown, but was a great problemist
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| (45) Posted by Branislav Djurašević [Tuesday, Oct 9, 2007 14:54]; edited by Branislav Djurašević [07-10-09] |
Maybe the phrase "Exceptions prove the Rule" nicely describes the discussion about multiple solutions in orthodox problems with good problems as examples in favor of more solutions.
I support the second answer in the poll. I remember that our outstanding composer Milivoj S. Nesic (1940-2006) was against even twins in orthodox twomover.
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| (46) Posted by Vladimir Tyapkin [Tuesday, Oct 9, 2007 16:14] |
I'd like to discuss different aspect of this topic - competitions. Can multi-solution problems participate in regular tourneys?
My answer is 'no' and that's my primary objection to this kind of problems.
Also, legitimizing it on economy(construction) ground will open a floodgate of other conditions. How about NWK(no white king) problems? If white king is not part of idea, it(and usually some black forces preventing checks) could be easily removed.
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| (47) Posted by Jacques Rotenberg [Tuesday, Oct 9, 2007 20:26] |
OK you say 'no'.
What could be nice, is to explain us why.
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| (48) Posted by Vladimir Tyapkin [Tuesday, Oct 9, 2007 21:16] |
1) I mention one argument - if we allow problems with multiple solutions to compete in regular tourneys, how can we refuse any other condition, NWK is one example.
2) Any new condition in regular competition put all other participants in unequal state.
I have no objection for multiple solution problems per se. I have no objection if such problems compete against each other.
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| (49) Posted by Jacques Rotenberg [Tuesday, Oct 9, 2007 23:53] |
Problems with multiple solutions are much alike try-play problems, nothing else.
Not fairy, not illegal positions, no twins, nothing.
Why this will to mix this with that ???
Moreover, you say you have nothing against !! You are wellcome.
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| (50) Posted by Sarah Hornecker [Wednesday, Oct 10, 2007 00:51] |
QUOTE
in pastime october 3, pb6, there was the following nice diagram :
I solved it yesterday evening (i.e., around one hour ago) so I took a look and reply only now. I did not even really consider 1.Rb4 there. Instead, I considered 1.Bf3 (1...Qh3!) and 1.Bf5 (but much works against it) which led me to the conclusion that only 1.Rb7 can work since the pinning diagonal must stay closed. One of those problems where the real try is easy to overlook. Ok, I was thinking about 1.Rb4 for a short moment but wanted to check 1.Rb7 first, also (without further elaboration on 1.Rb4) I missed the pinned mates then.
Something similar happened in #1 when I decided for 1.Qc4, missing 1.Qd3 Ka4! (refutation, I thought) 2.Qxa3#!
As to FIDE album printing (regarding MV's posting of September 23): It is done by bernd ellinghoven these days who I met last weekend in Forchheim. So this gives no chance to wait ten years for a FIDE album anymore. ;)
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| (51) Posted by Jacques Rotenberg [Wednesday, Oct 10, 2007 01:15] |
This is exactly my purpose.
If you want to show two plays, why should you feel obliged to hide one ?
Perhaps the solver will really miss it, and all you'll win is one more misunderstanding.
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| (52) Posted by Michael McDowell [Wednesday, Oct 10, 2007 17:21] |
Norman Macleod had no objections to multiple solutions, and published an article in The Problemist for May 1980 entitled "Why not multi-solutions?".
Norman wrote: “In the direct-mate I doubt if multi-solutions will ever be very common, because the use of tries is generally more attractive and more interesting for the solver. Multi-solution 2-movers can indeed be boring compared to their try-play counterparts. But sometimes problems with more than one phase do not lend themselves easily to try refutations. If a try is defeated by a defence so obvious that the solver is likely to miss the try play altogether, then surely it is better to have a second solution.”
He also quoted Edgar Holladay, from BCM, March 1980: “The best form for showing a given changed play idea should be determined by unfettered judgment, not by arbitrary rules.”
I think there are some ideas that are naturally better in multi-solution form. The following simple example was composed to show the various types of two-mover which can follow from a block setting. I cannot see that there would be any advantage in presenting this problem in another form.
M.McDowell
Problem Observer, September 1989
 (= 4+3 )
Mate in 2 (4 solutions)
1.Bg8 Complete block
1.Kb6 Added-mate block
1.Be8 Mutate
1.Bh5 Block-threat
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| (53) Posted by Jacques Rotenberg [Friday, Oct 12, 2007 04:12]; edited by Jacques Rotenberg [07-10-12] |
Yossi Retter
Problem 1964
 (= 12+4 )
2#
1.Qd8? blocus
1…S~ a 2.R×é5‡ A
1…Kd3 b 2.Sç5‡ B
1…Kf5 c 2.Bç2‡ C
mais 1…Sf6!
1.Qç8! blocus
1…S~ a 2.Sç5‡ B
1…Kd3 b 2.Bç2‡ C
1…Kf5 c 2.R×é5‡ A
Lacny
Well done, no defect, good refutation.
The two keys are very nice.
it could as well be shown as follow :
 (= 11+4 )
2# 2 Solutions
with slight economy, but, much more important, the fact that the solver cannot miss the theme!
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| (54) Posted by Jacques Rotenberg [Monday, Oct 15, 2007 00:14]; edited by Jacques Rotenberg [07-10-15] |
Michel Caillaud
Kingston T.T. 1999
(1st?) Prize
 (= 7+7 )
2#
1…Se~ 2.Q×f5‡
1…Sd~ 2.Q×f4‡
1.Qg1? blocus
1…Se~ 2.Qd4‡ A
1…Sd~ 2.Q×e3‡
mais 1…Sç2!
1.Qd8! blocus
1…Se~ 2.Q×d5‡
1…Sd~ 2.Qd4‡ A
1…K×e5 2.Re6‡
It could be shown as well as follow :
 (= 7+7 )
2# 3 Solutions
1.Rh4!
1.Qf1!
1.Qc8!
I think this is better.
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| (55) Posted by Jacques Rotenberg [Friday, Oct 26, 2007 09:12] |
Another example is a very nice problem shown today in 'Pastime'.
Yuri Antonov and Sergei Burmistrov
V. Chepizhny Jubilee-50 1985 6th Prize
 (= 9+6 )
2#
(37 p.23, Album FIDE 1983-85)
Two nice tries and a solution, the refutations are of average level but with no defect, all is OK.
However, I think the following way of building would have been better :
 (= 9+5 )
2# 3 Solutions
with richer play (one more changed/transfered mate on 1...Qg7) and better evidence for the content.
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| (56) Posted by Sarah Hornecker [Friday, Oct 26, 2007 11:23]; edited by Sarah Hornecker [07-10-26] |
1.g4 (or 1.f4) is clear to me. With the bottom one I have to search two additional mates now... I don't think I like more solutions if I have to solve. :-(
Uhh... maybe 1.c6 is one... or maybe not, it's still refuted by 1...Qxf6
Give me a hint, please!
uh, as a solver I prefer one solution, as a composer more solutions :-)
It's not a good problem at all. The Pg2 gives out the solution since there's no other reason why it should be on g2 than to walk. Leaving the bK to f4 wouldn't allow mating.
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| (57) Posted by Jacques Rotenberg [Friday, Oct 26, 2007 15:42] |
Cause you give the key, I can give the whole solution :
1.f4! [2.Qd4‡]
1…d×c5 a 2.Qe5‡
1…Q×e6 b 2.Qd1‡
1…Qg7 c 2.Qh1‡
1…Q×f4 2.S×f4‡ C
1…Qd1 2.Q×d1‡
1.Qf6! [2.Bc6‡ A]
1…d×c5 a 2.Sc7‡ B
1…Q×e6 b 2.B×e6‡
1…Qg7 c 2.Sf4‡ C
1…Qa4 2.Qf5‡
1.Qc3! [2.Sc7‡ B]
1…d×c5 a 2.Q×c5‡
1…Q×e6 b 2.Bc6‡ A
1…Qc4 2.b×c4‡
1…Qd4/Q×g3 2.Q(×)d4‡
1.Rc6? [2.Sc7‡, 2.R×d6‡] but 1…Q×e6!
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| (58) Posted by Sarah Hornecker [Friday, Oct 26, 2007 18:26]; edited by Sarah Hornecker [07-10-26] |
QUOTE
Cause you give the key
No, no, no! The author gives the key as I told above! Don't blame me now! :-)
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| (59) Posted by Paz Einat [Friday, Oct 26, 2007 19:23] |
Jacques, your version is very nice! You added 2 changes and a transfer while removing a piece. Adding refutations to any of 2 phases looks cumbersome, and actually, not needed. This is a very good example of a case in which the 3 solution version is superior. The need to have a single solution is based on our pre-conditioning but should not be regarded as inferior.
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| (60) Posted by Administrator [Saturday, Oct 27, 2007 11:56] |
QUOTE
Jacques:
Another example is a very nice problem shown today in 'Pastime'.
Now it can be solved in 1 second, thank you very much Jacques :(
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