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| | Page: [Previous] [Next] 1 2 3 4 5 | | (21) Posted by Jacques Rotenberg [Tuesday, Feb 9, 2010 04:47] |
Miodrag,
Your problem looks very nice
| | | (22) Posted by Miodrag Mladenović [Tuesday, Apr 20, 2010 09:26]; edited by Miodrag Mladenović [10-04-20] |
Here is my latest threemover showing this theme:
Miodrag Mladenović
5HM FIDE World Cup 2010
 (= 13+10 )
#3
1.Rd1! [2.Qxd6+ Kxd6 3.Sc4#]
1...Rc~!? 2.Qd4+ Bxd4 3.exd4# (a)
1...Rd5!!? 2.Re8 ~ 2.Rxe6# (ab)
1...Rc4!!!? 2.Sxg3 ~ 3.Sxf3# (abc)
1...Rb5!!!!? 2.Sc4+ Kxe4 3.Sxg3# (abcd)
1...Bd3 2.cxd3 ~ 3.Sxf3#
Black errors:
(a) openning of line b6-d4
(b) closing of breaked line a6-c4-e6
(c) closing of breaked line a6-e2-f3
(d) closing of line a6-c4
I am very happy with a position of this problem. It has very nice threat and variations are OK. Judge mentioned as a dual 1...Rxc7 2.Kxc7 but this defense does not defend against primary threat 2.Qd4+ so I do think that this is a minor dual. I could fix it by changing the threat and adding several more pieces but I did not want to do this. Although it looks like this theme is not very popular I do enjoy compising problems showing it.
| | | (23) Posted by Sathya Narayanan [Thursday, Apr 29, 2010 08:52] |
Excellent problem! Misha trundles out problems of correction play by the dozen!.I have only a few suggestions to make.Never stop composing if it gives you pleasure, dont compose for accolades especially prizes.do not bother about the popularity of the theme which is purely subjective. I would like to see more of Mladenovic.
cgsnarayanan
| | | (24) Posted by Sathya Narayanan [Thursday, Apr 29, 2010 12:59] |
Very elegant and pointed.Hard to beat this in economy.congrats, Mladenovic!
| | | (25) Posted by David Knezevic [Friday, Apr 30, 2010 01:47] |
Another recent excellent example with a complex mechanism:
C. G. S. Narayanan
C10763. The Problemist, March 2010
 (= 11+10 )
#3
1.Qf8! ~ 2.Sf6+
1... S~(=Sxd3) 2.hxg4
1... Sd7! 2.g8=S (2.hxg4?)
1... Sc6!! 2.Bf4 (2.hxg4?,2.g8=S?)
1... Sc4!!! 2.Sxf3 (2.hxg4?,2.g8=S?,2.Bf4?)
There is a (serious?/small?) flaw in a triple/quadruple after 1...Sc4(Sxd3) 2.Sxf3(hxg4) e5 3.Qf7/Qg8/g8=Q/B# which can be easily fixed by addition of the black Pf7. One primary defence (Sf7) is lost, but there is another move of the same weight (Sxd3) so thematically everything remains OK.
I am sure that author knew it, but suppose he didn't want to have too passive WQ in the initial position. However, this can be "amortized" e.g. by shifting everything one file to the left and moving the WQ to h8; or by moving the WQ to f8 and WSe4 to c5 with (unfortunately necessary) addition of BPe4 (key: 1.Scxe4).
Am I missing some important detail which would justify the inaccuracy after 2...e5 ?!
| | | (26) Posted by Miodrag Mladenović [Friday, Apr 30, 2010 07:15] |
Excellent problem. I do agree with Milan’s proposal to add bPf7 to avoid duals in side defense. Also shifting position for one line left and moving wQ to h8 is good idea.
| | | (27) Posted by Sathya Narayanan [Friday, Apr 30, 2010 08:40] |
Both the suggestions of Milan are very good.I thought a good deal about it.From an artistic point of view the random and corrections anti-clockwise f7-d7-c6 and c4 appealed to me.So in this position plugging f7 (taking away a crucial random move which forms part of the bent line as a sequence)just to avoid a minor dual in a non-thematic line is just a matter of taste.
| | | (28) Posted by Jean-Marc Loustau [Friday, Apr 30, 2010 13:17] |
Well… As you know correction is always a matter of discussion. Some analysts could have some thematic criticisms about this nice problem:
1) the reason why 2 hxg4 works after 1… S~ is the fact the S has lost access to d7 (or g4) (in other words 2 hxg4 is not a threat because 2… Sd7/Sxg4 refutes). So it could be fallacious to say that there is again the primary error in move 1… Sd7 (the S doesn’t loose access to d7 when it plays on d7): from this point of view 1… Sd7 does not correct 1… S~;
2) the reason why 2 g8=S works after 1… Sd7 is of course the interference of the Rook on 7th line, but ALSO the fact the S has lost access to c6 (2 g8=S is not a threat because 2… Rxc7 AND 2… Sc6 refutes). So, as previously, it could be fallacious to say that there is the secondary error in move 1… Sc6 (the S doesn’t loose access to c6 when it plays on c6): from this point of view 1… Sc6 does not correct 1… Sd7;
3) there is something similar with 2 Bf4, but less clear because this move is not possible after the key (when the S is on e5): if 2 Bf4 works after 1… Sc6, it’s because 2… Rc6 is impossible, AND the line d6-f4 is opened, AND S has lost access to c4; so this complex tertiary error is not in the variation 1… Sc4 (the S doesn’t loose access to c4 when it plays on c4): from this point of view 1… Sc4 does not correct 1… Sc6;
So, again from this point of view, there is a no 4th degree nor 3rd degree, but several 2nd degree chains: 1...S~->1...Sc6 / 1...S~->1...Sc4, and 1...Sd7->1...Sc4
As we all know, there are many opinions about what is and what is not correction, and it’s just a matter of definition. I don’t want, with this post, to introduce the discussion of the definition of the correction, I give only a point of view among others…
| | | (29) Posted by Jacques Rotenberg [Sunday, May 2, 2010 08:14]; edited by Jacques Rotenberg [10-05-02] |
well well.... Jean-Marc
1) 1...Sd7 can be seen perfectly as a correction move, and your chain of reasoning 'fallacious' (to use your words) How can be played 1...S~ ? The easiest way I know, is to keep the Knight in the hand, and only then -not before- you ask yourself where to put it down. In other words, there is no reason to be interested in why 2 hxg4 is not a threat. (In completely other words, you can also say that 1...Sd7 and 2...Sd7 are different moves.)
2) as you pointed out the fallacious way you discard 1...Sc6 is the same as the previous one, so for the same reason 1...Sc6 is perfect !
3) The same for 1...Sc4 - here you add even a mistake : "....if 2 Bf4 works after 1… Sc6, ...AND S has lost access to c4" you should write "...to f7" even if it disturbs your arguments
All in all, a very nice and clear 4th degree
| | | (30) Posted by Jean-Marc Loustau [Sunday, May 2, 2010 17:25]; edited by Jean-Marc Loustau [10-05-02] |
Jacques...
As I told, I do not want to discuss here about what is or what is not correction. I have just given a point of view, the point of view of the effect analysis. Everybody is free to see what he wants in everything, and the "piece in the hand" analysis is an analysis among other possible analysis (and of course I do not want to discuss if an analysis is better than another one, because it is purely conventional, so not a matter for discussion).
I did not say that "1... Sd7 cannot be seen as a correction move", it depends of the kind of analysis, I just said that from the "effect analysis" it is fallacious (I mean very questionable) to see it as a correction move.
To stay at the 1st step, the reason (the "harmful effect") why 2 hxg4 is possible after 1... S~ must also exist after the wished correction variation, so 1... Sd7. This harmful effect is the fact S has lost access to d7, and a way to prove that, is to test why 2 hxg4 is not a threat (in fact the real harmful effect is the fact the S has lost access to d7 and g4, but the 2 must exist in the correction variation, and the difficulty here is with d7).
(If you wish you can write it otherwise: 1... tempo 2 hxg4? Sd7! Black can exchange the order of their moves: 1... Sd7 2 hxg4? tempo! No difference).
There is a particular case in which it can be admitted that when the S is on a square (here d7) it has lost access to this square: when it will be forced to leave this square after the white move (typically in zugwang case).
About the 3rd point you made the mistake: I did not write anything about f7 because it was not useful; this does not "disturb" my arguments, this changes nothing; I don't know if you are or not familiar with the effect analysis, but the principle is simple: "the reason why a 2nd move is possible must exist in the correction variation"; so let us analyse 1... Sc6 2 Bf4 to see what is this reason: Bf4 works because 1) d6-f4 is opened 2) 2... Rc6 is no more possible 3) Black S has lost access to f7 4) Black S has lost access to c4. I do not see other reasons, but even if there were other reasons this would change nothing. Now let us analyse 1... Sc4: points 1) 2) and 3) are there again, but, as I said, point 4) is highly questionable for the same reasons as above. In other words, in cases of complex harmful effects, that is to say harmful effects which are the sum of several basic harmful effects (as here, with 4 basic effects), the sum of the basic effects must exist in the correction variation: if one is missing, it is not a correction.
All in all, surely a nice problem, but a highly questionable 4th degree from the effects point of view.
Of course, I respect your opinion, there are so many understandings of the correction (again all this is a matter of definition and conventions: we just have not the same conventions and definitions)! The same for the anti-dual: in the "piece in the hand" approach, a classical focal play showes a perfect anti-dual; not in the effect analysis approach...
| | | (31) Posted by Jacques Rotenberg [Monday, May 3, 2010 02:22] |
About the 3rd point, I copy here your full sentence
"if 2 Bf4 works after 1… Sc6, it’s because 2… Rc6 is impossible, AND the line d6-f4 is opened, AND S has lost access to c4"
just to underline that after 2.Bf4 Sc4 you would have 3.Qc5#, so that the loss of access to c4 has no interest here.
For the rest of your (long) post, let's say : you did not convince me, sorry.
| | | (32) Posted by Jean-Marc Loustau [Monday, May 3, 2010 10:46] |
To be short: I don't want to convince anyone, because I do not pretend to hold a truth. I gave just a point of view which can be shared or not.
About the 3rd point: I missed the mate Qc5# (my mistake); so you're right (the access lost to c4 is not a reason why Bf4 works after 1... Sc6). So 1... Sc4 is actually a correction of 1... Sc6. So I correct the analysis:
there is a 3rd degree correction chain: 1... S~ -> 1... Sc6! -> Sc4!! and also a 2nd degree link: 1... Sd7 -> 1... Sc4!; but 1... Sd7 doesn't correct 1... S~ and is not corrected by 1... Sc6 (again from the point of view I gave, other analysis are possible).
| | | (33) Posted by Jacques Rotenberg [Monday, May 3, 2010 17:33] |
This discussion may be illustrated by a small schema (if I understood what you say) :
 (= 7+3 )
2#
1.Bd5 ? block 1...S~ 2.S(x)f3#, 1...Sxh3!? gxh3# but 1..d6!
1.Bd6 ! block 1...S~ 2.g3#, 1...Sxh3!? 2.gxh3#, 1...Sxf3!? 2.Sxf3#
and you say that 1.Bd5? Sxh3!? and 1.Bd6! Sxf3!? are NOT corrections - but 1.Bd6! Sxh3!? yes.
And I say that all these are corrections.
| | | (34) Posted by Jean-Marc Loustau [Monday, May 3, 2010 18:00]; edited by Jean-Marc Loustau [10-05-03] |
I confirm you understand:
from the effect analysis "1.Bd5? Sxh3!? and 1.Bd6! Sxf3!? are "very questionable" corrections" (notice that I prefer to write this way, and not your way).
Also, the case you show is not exactly the same as in the 3#, but quite close, I admit.
Of course we all agree that 1.Bd6! Sxh3! is a correction.
| | | (35) Posted by Jacques Rotenberg [Thursday, May 6, 2010 23:15] |
Now, what do you say of this (very) famous schema ?
 (= 2+2 )
2#
1.Re2! block 1...S~ 2.Re1#; 1...Sg1!? 2.Rh2#
For me : an excellent pedagogic example.
For you ? Not a correction ??
| | | (36) Posted by Sarah Hornecker [Friday, May 7, 2010 00:47] |
Of course it is a correction!
| | | (37) Posted by Jean-Marc Loustau [Friday, May 7, 2010 11:45]; edited by Jean-Marc Loustau [10-05-07] |
Again, I don’t like your way of writing « for you, not a correction ».
About this very famous example (by O. Dehler I think) I answer: “for an effects analyst, a very questionable correction”.
By “questionable” I mean there is a question.
The question is: “does the realization of a possibility contain the (primary) error of the loss of this possibility?"
Here: does the fact to play on g1 contain the error of the loss of access to g1?
The answer is a matter of choice. I can admit that the 2 answers (yes/no) are receivable, so I say it is “very questionable"...
If you don't agree that this question is relevant to think/feel it is or not a correction, that is to say your own definition of the correction is not based on effects (and I don’t know exactly how it is possible to get a definition of a Nth degree without using effects, but let us say it is possible). In this case, this is definitively not debatable because definitions are a matter of conventions: that’s what I said from the beginning.
I have said also, and please respect that: “I don’t want, with this post, to introduce the discussion of the definition of the correction, I give only a point of view among others”. If you wish to debate about the definition of the correction, it should be better to open a new topic, but probably I will not intervene in the discussion, because I am not very interested in definitions.
| | | (38) Posted by Jacques Rotenberg [Friday, May 7, 2010 17:54] |
Dear Jean-Marc,
Here is your wording :
"...So I correct the analysis:
there is a 3rd degree correction chain: ... and also a 2nd degree link: ...."
So you discard the 4th degree (even if in the end you add "...(again from the point of view I gave, other analysis are possible)...".
| | | (39) Posted by Jean-Marc Loustau [Friday, May 7, 2010 18:27]; edited by Jean-Marc Loustau [10-05-07] |
Well, I don't see why it is useful to repeat...
Yes, the analysis given says there is no 4th degree.
Yes, this analysis is from the specific point of view of an effects analyst.
But, may be to make me clearer (sorry if I have not been clear before):
the point of view depends of the definition which is a convention; personaly (if you want my personal view) I am willing to accept any convention of definition, so I have no personal point of view if it is or not a 4th degree: the only thing I can say is that it exists at least one definition/convention (I mean of course a reasonable one) in which there is no 4th degree here; as you expressed you own point of view, it exists also at least one defintion/convention in which there is a 4th degree here. But the fact is: it is not an "universal 4th degree" as most or all the problems seen previously, because at least 2 analysis don't give the same result.
| | | (40) Posted by Miodrag Mladenović [Tuesday, Feb 1, 2011 20:43]; edited by Miodrag Mladenović [11-02-01] |
Here is another good problem showing 4th degree black correction in #3:
C. G. S. Narayanan
Mat Plus 2010
 (= 13+7 )
#3
1.Sa4! (2.Bxg7+ Qxg7 3.Rxd2#)
1...Bc1 2.Sc3 ~ 3.Sb5# (a)
1...Be3 2.Rc3 ~ 3.Rd3# (ab) (2.Sc3? Bxf2!)
1...Bf4 2.Bg2 ~/Be5 3.Qe4/Se6# (abc) (2.Sc3?/Rc3? Be5!)
1...Bg5 2.Qe6 ~ 3.Qd5# (abcd) (2.Sc3?/Rc3?/Bg2? Be7!)
non-thematic:
1...B~ 2.e3+ or 2.Bxg7+ Qxe3/Qxg7 3.Bxg7/e3
Black errors:
(a) unguard of c3
(b) closing of breaked line h6-e3-d3
(c) closing of breaked line h6-f4-e4
(d) closing of breaked line h6-g5-d5
Even with minor dual in non-thematic variation this is an excellent problem. Nice and original matrix.
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