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MatPlus.Net Forum Twomovers Miodrag Mladenovic - 2nd Pr. The Problemist 2005/II [#2]
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(1) Posted by Miodrag Mladenović [Saturday, Jan 27, 2007 11:48]; edited by Miodrag Mladenović [07-01-27]

Miodrag Mladenovic - 2nd Pr. The Problemist 2005/II [#2]

I just got January issue of "The Problemist". I won the 2nd Pr. with the following problem:
(= 12+8 )

1...Qa8 2.Qc3#
1...Rg8 2.Be3#
1...Bxc4 2.Qe3#

1.Ke7? [2.S~#] but 1...f4!

1.Sf4? [2.Ke7#]
1...Qa8 2.c3#
but 1...Rg8! [2.Be3?]
1.Sc3? [2.Ke7#]
1...Qa8 2.Sb5#
1...Bxc4! [2.Qe3?]

1.Se3! [2.Ke7#]
1...Qa8 2.Qb2#
1...Rg8 2.Sxf5#
1...Bxc4 2.Qxc4#
1...f4 2.Re4#

Judge Philippe Robert did not like wBh6 because it does not play in solution (yes, I know this is not good but I could not give hime a role in solution). So he recommended new version:

Version by Philippe Robert:
(= 10+10 )

1...Qb8 2.Qd3#

1.Sg6? Bxd4/Sg5 2.Qf3/Rxf4# but 1...Qb8!

1.Sg4? Qb8/Bxd4 2.d3/Qf3# but 1...Sg5!

1.Sd3? Qb8/Sg5 2.Sc5/Rxf4# but 1...Bxd4!

1.Sf3! [2.Kd8#] Qb8/Sg5/Bxd4 2.Qc2/Sxg5/Qxd4#

In my opinion I like my version much more even if wBh6 does not have role in the solution. My main idea was to have set mates by wB and wQ on e3 and then two tries that are cutting one of the pieces. Then in solution both set mates are prevened but there are new mates. Actually I do have two white knight second degree corrections. I am wondering what is the opinion of other probemists? Which version do you prefer?
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(2) Posted by Zalmen Kornin [Sunday, Jan 28, 2007 16:21]

Mr. Mladenovic - I would like to express my believing that, as a general rule, when a composer is developing something whith originality in form and idea, the simplified versions of this same presentation are to be considered already seen and discarded, in most of the instances, by the composer. I'm seeing that the triple set-play and it's link to the tries is something that strongly caracterizate this presentation. If Mr. Robert have another opinion, it would be interesting if he joined this discussion too and explained his reasons for that.
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(3) Posted by Hauke Reddmann [Monday, Jan 29, 2007 12:03]

I'm a defender of economy myself, but NEVER at the cost
of any thematic action. Even if it's only ornamental like
different things happening on a single field, or such.

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(4) Posted by Jacques Rotenberg [Friday, May 25, 2007 14:58]

However, it seems, that the theme you intended is known as "French-Letton" theme (or paradox) :
One try is refuted because white key provoked a weakness
Another try is refuted because white key provoked another weakness
The key of the solution provokes the two weaknesses.
This idea have been worked out by Dombrovskis, and also by Savournin in the 60's

It seems that the judge missed it, it happens !

Another surprise : the version proposed by the judge has less richness in the changed mates.

Or perhaps you may say that the judge wanted to demonstrate that the unusefull Bishop is of no matter, because, when he tried to build the problem whithout it, he only managed to reach the weaker version he indicates?

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(5) Posted by Jacques Rotenberg [Friday, May 25, 2007 16:16]

Now, about thematic content, you speak of 2nd degree white correction.

I think it is a 3rd degree !

1.S~? (underlined by 1.Sf6?) Qa8!
1.Sc3!? Qa8 2.Sb5# but 1...Bxc4! second degree
1.Sf4!? Qa8 2.c3# but 1...Rg8! second degree
1.Se3!! Qa8 2.Qb2# 1...Bxc4 2.Qxc4# 1...Rg8 2.Sxf5# Third degree (twice)
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(6) Posted by Miodrag Mladenović [Sunday, May 27, 2007 08:46]


I think you are right. With your explanation it looks like third degree correction. When it comes to definition of corrections it's always little bit confusing and I am never absolutely sure about degree of correction. Anyway I always enjoy looking into the problems showing black/white correction moves.
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(7) Posted by Milan Velimirović (+) [Sunday, May 27, 2007 12:09]; edited by Milan Velimirović (+) [07-05-27]

As far as I know this is only the second problem of this kind. The first one took part in a tourney I judged:
Marjan Kova?ević Miroslav Stošić MT 1995
(= 12+9 )
#2* v...

1... Sf6,Bc4:,gf5 2.Qf6:,Re3,Qe3#
1.Sd~? Sf6!
1.Sc3!? Sf6 2.Bg4#, 1... Bc4:!
1.Sf4!? Sf6 2.Bg6:#, 1... gf5!
1.Se3!! ~ 2.Rd5#
1... Sf6,Bc4:,gf5 2.Qg3:,Sc4:,Qf5:#
(1... de6,fe6 2.Ra5,Bg6:#)

I will quote my comments from the award:

"This is one of the most complex twomovers I have ever seen, and I've seen a lot! It shows so-called 'Franco-Russe' theme, or change of mates with respective tries which prevent one of thematic set mates (note: this is not quite accurate definition of theme - MV, 2007). And what kind of tries we can see here! In the initial position mates are set for 1... Sf6 2.Qf6:#, 1... Bc4: 2.Re3# and 1... gf5 2.Qe3#. Random removal 1.Sd~? (2.fg6#) is defeated by the utilization of general white error, unguard of f4, 1... Sf6! 2.Qf6? - which prevents one of set mates. Corrective moves compensate that error but close one of white lines and prevent, one by one, remaining set mates: 1.Sc3!? Sf6 2.Bg4:# but 1... Bc4:! 2.Re3? because a3-e3 is closed, or 1.Sf4!? Sf6 2.Bg6# but 1... fg5! 2.Qe3? because g5-e3 is closed. Key move 1.Se3!! closes both lines and changes all thematic mates: 1... Sf6 2.Qg3:#, 1... Bc4: 2.Sc4:#, 1... gf5 2.Sf5:#. So, double White tertiary correction which probably have never been presented in such way! The content is beautifully supplemented by side variations 1... de6 2.Ra5# which gives WRa3 the role after the key and 1... fe6 2.Bg6:# with transferred mate from try 1.Sf4?. With such contents any discussion about technical part is almost superfluous, but it is worth to mention that in spite of unavoidable plugs on c2, h3 and e7 the position is natural and economical. A true masterpiece!"

One digression:
Recently I saw another compound-word-name for "Franco-Russe" theme, a kind of "theme emigration". Do we really need a "dynamic" terminology, sufficiently flexible to adjust itself to any geo-political change in the history?! What should we do about all the indexes in older books? Burn the books and reprint them?
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(8) Posted by Jacques Rotenberg [Sunday, May 27, 2007 13:49]; edited by Jacques Rotenberg [07-05-27]

In the early 70's there was a complaint from Dombrovskis about the name "Franco-Russe" he suggested "Franco-Letton". Russia and Lettonia were both states of the late USSR. In the western countries, USSR used to be called improperly Russia.

Did the problem of Marjan inspire the one of Miodrag ? It seems possible.

Marjan made some great problems with Franco-Letton/Franco-Russe with or without white correction
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(9) Posted by Miodrag Mladenović [Sunday, May 27, 2007 16:01]

Yes, I knew Marjan's problem at the time when I composed mine. Since matrix is completely different I did publish my problem. But I agree with Milan. I do not know any other but these two problems showing this content. If somebody knows more examples I'd like to see them.

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(10) Posted by Miodrag Mladenović [Sunday, Sep 16, 2007 20:32]

Jacque proposed few new versions of my #2. Here is the version in more classical way:

(= 9+11 )


It does contain same number of pieces but there are two more battery mates by wK. However now after thematic try 1.Se7? there is no defense by bRg5 because knight closed line g7-e7. Although it's nice to have three more checkmates it's good to have try that repeats set-play. Actually I know that in Russia they do not accept set-play without try that is justifying it (at least it used to be like that). I like the fact that bunch of white powns are replaced with black powns but on the other hand there is no try that fully repeats set-play so I am still wondering if this is improvement or not.

I am wondering what is an opinion of other members of this forum?
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(11) Posted by Marcos Roland [Wednesday, Sep 19, 2007 18:41]

Worthy to be noted, in the test of today (Pastime), there are two problems of Norman MacLeod which won first prize: problem 5, British Chess Federation, 1973; and problem 6, British Chess Magazine, 1961. In both of them, there's a rook which doesn't have a role in the solution.
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(12) Posted by Jacques Rotenberg [Wednesday, Sep 19, 2007 20:49]; edited by Jacques Rotenberg [07-09-19]

Hello Marcos,

Only on the 6 you are right, on the 5, there is a use. However, because the key closes the line of the Rook a4 and because this is essential to the mechanism (otherwise you have a dual) you can consider it as a very very light defect, in any case this Rook contributes to make the key difficult.

Because you underline this pecular point that was the criticism of the judge to the problem of Miodrag (useless pieces in the actual play), here is another of the versions I shew to Miodrag

(= 11+8 )

This gives a use for the Bishop h6 : 1...Rg7+ 2.Bxg7#
the multiple threat is, for me, a very small defect, but on the other hand, the useless Bishop is also a small defect, so...
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(13) Posted by Jean-Marc Loustau [Thursday, Sep 20, 2007 12:49]

Let me say my opinion… And sorry for the “talkative post” :-)
My favourite versions are either the published original version or the last one Jacques proposed (which is very close to the 1st setting). The reason is the following one:

I think here it is very important to get a try as 1 Ke7; not because it underlines the set play (I don’t care), and surely not because Russians think like this or like that. It is important because this try can be seen as a “thematic try”, showing a “random attempt” to play the half-battery; this random attempt is defeated by 1… f4, and thus all the following tries by the white S, which have the same purpose (playing the half-battery), clearly providing this defence, are corrections (I know! It’s already difficult when corrections are played by a single unit, so when it is by different units it becomes a bit intellectual :-)… For an equivalent with black correction see for example the famous 5th degree by Casa). So to me the problem shows not a 3rd degree correction but a 4th degree correction

Among these 2 settings my favourite is not clear, but perhaps the last one by Jacques, because to me a useless unit is not a so small defect and because the multiple threat is without importance (by the way it can be written as a single one 2 K ~ #); but I am not sure because this version has another defect which is not in the first published setting. In the Jacques version the “random” move by white S (the 2nd degree) can be underlined only by the try 1 Se7, in which there is a dual after the variation 1… Rg8 (In the published problem, the random move is 1 Sf6, as Jacques said, and of course no dual there). This is not so important because there is no change from the 1st degree and thus it is not useful to have a look at this variation, but nevertheless it is not a good point (also because this dual is a thematic mate of one of the 3rd degree corrections); so, particularly with this version, I would write the solution like this:
1… Rg7+, 2 Bxg7#
1 K ~? 1… Qa8 (Rg8, Bxc4) 2 Qc3 (Be3, Qe3)# 1… f4!
1 S ~!? 1… f4, 2 Re4#; 1… Qa8!(2 Dc3?)
1 Sf4 !!? 1… Qa8 2 c3#; 1… Rg8!(2 Be3?)
1 Sc3 !!? 1… Qa8 (f4) 2 Sb5(Re4)#; 1… Bxc4!(2 Qe3?)
1 Se3 !!!! threats 2 K ~# 1…Rg8(Bxc4, Qa8, f4) 2 Sxf5(Qxc4, Qb2, Re4)#

I think it is not useful to write the set play! May be it is also better to write 1 K ~? than 1 Ke7?, showing that this try has the value of a random move, but this is minor.

Some other theoretical remarks (sorry!); if you forget the try by white King some analysts (as Boyer for example) consider you have nevertheless a 4th degree (they call this a “4th degree correction from the set play”… Again a bit intellectual…).
Some other analysts (from Netherlands for example… I think to Visserman, Goldschmeding…) could say that the key 1 Se3 has a 5th degree level, being a correction of 2 different 3rd degree corrections, so the number of effects is the same as for a 5th degree… But of course we have not a “chain” of 5th degree correction.

Well, finally it is a very good problem :-)
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(14) Posted by Jacques Rotenberg [Friday, Sep 21, 2007 01:53]; edited by Jacques Rotenberg [07-09-21]

...and a very nice and deep comment !

by the way, how many levels of correction will you obtain if you combine the different conceptions ?
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(15) Posted by Miodrag Mladenović [Saturday, Sep 22, 2007 11:09]

Yes, very nice comment. I must admit that I have not seen the fourth degree correction myself but it does make sense. However, somehow I do not like threat K~#. I do prefer single threat. Yes, I know that there is multple threat after 1.Kc7? [2.S~#] but I think it's unavoidable.
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MatPlus.Net Forum Twomovers Miodrag Mladenovic - 2nd Pr. The Problemist 2005/II [#2]