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MatPlus.Net Forum General Third degree/Tertiary play |
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| | (1) Posted by Colin Sydenham [Friday, Jan 16, 2009 16:53] | Third degree/Tertiary play I like to think I understand the concept of tertiary correction, but I'm trying to puzzle out whether there is third degree/tertiary play which is NOT tertiary correction. I remember an exchange with Chris Becker on the subject; he assured me that the distinction exists, and sent me a problem which, he said, demonstrated it. This was about 25 years ago; I have lost the correspondence, and CB is no longer around to help.
Can anyone enlighten me? I'd be very grateful for an illustrative problem.
Colin Sydenham | | (2) Posted by Jean-Marc Loustau [Sunday, Jan 18, 2009 19:21]; edited by Jean-Marc Loustau [09-01-18] | Dear Colin
Of course I cannot be sure about Chris Becker wanted to say, but here is what I suppose. Everything is a matter of definition. Either you decide a tertiary play must be a tertiary correction, either you decide it can be something else. In this 2nd case the level is given by the number of anti-dual effects (+1), assumed that chain(s) of anti-dual with the thematic mates can be built (the level N occurs not as the end of a single level N-1 chain but as the end of several chains). Well, it will be clearer with 2 examples with quaternary play (very famous, I suppose you know them):
E. VISSERMAN
NBvP 1942
2nd Pr.
(= 12+10 )
2#
1 Qb5! (2 Qxd5#)
1… Sd5 plays 2 Se8# A
1… Sc7! 2 Dc5# B (2 Se8? Sxe8!)
Classical 2nd degree chain A->B
1… Se5 plays! 2 Qd7# C (2 Se8? Impossible!)
Again a 2nd degree chain A->C
1… Sc6!!! 2 Sc4# D (2 Se8? Impossible again! 2 Qd7? Impossible! but also : 2 Qc5? bxc5!)
Thus we get here a 3rd degree chain A->C->D; but also another 3rd degree chain A->B->D
1… Sc6 can be seen as a 4th degree defense (matter of definition).
Other example:
R.J. MILLOUR
BCPS 1967 4th Pr.
(= 8+11 )
2#
1 Sh2! (2Bf2#)
1… Sd7 2 Qd1# A
1… Sb3 2 Sf3# B
1… Se4 2 Rxb1# C
1… Sd3!!! 2 Rf1# D (2 Qd1? Impossible; 2 Sf3+? Kd1! 2 Rxb1? Sc1!)
Here 3 2nd degree chains: A->D, B->D, C->D
Again 1… Sd3 can be seen as a 4th degree defense (matter of definition).
Of course these 2 problems don’t show a 4th degree correction; but if a difference can be made between quaternary correction and quaternary play, I think here it is. A 4th degree correction is a particular case of a quaternary play, when the play can be reduced to a single chain...
Of course same things can be also done with white correction. | | (3) Posted by René J. Millour [Monday, Jan 19, 2009 15:00] | Many thanks to JMLoustau for quoting one of my problems.
On this subject, I simply add 2 other examples.
René J. Millour
Stella Polaris, 1967
1st Honorable Mention
(= 12+7 )
12+7
1.Se2! [2.Qd4#]
1...Qf6 2.c4# (2.Sc3?, 2.Sef4?)
1...Re4 2.Sc3# (2.Sef4?, 2.c4?)
1...Sb3 2.Sef4# (2.c4?, 2.Sc3?)
and the 4th degree:
1...Se6 2.Rf5# (2.c4?, 2.Sc3?, 2.Sef4?)
1...Kd6 (triple self-pinning!) 2.d8Q#.
René J. Millour
French Championship, 1983-4
2nd Place
(= 11+11 )
1.Rc7! [2.Qa7#]
1...Sb3 2.f8Q#
1...Se2 2.e8Q#
1...Sf5 2.d8Q#
and the 4th degree:
1...Se6 2.Sxc6# (2.f8Q?, 2.e8Q?, 2.d8Q?)
1...Bxf7+(Rxg7+, Qxh7+) 2.Kxf7(Kxg7, Kxh7)#.
I grab the opportunity to say it is CEDER’s theme, here combined with promotions: “black piece P interferes with black piece X, mate A; then P interferes with black piece Y, mate B; then P interferes simultaneously with X and Y, mate C (A?, B?)”.
In this problem, and also in the “8+11” quoted by JML, the theme is extended to “X,Y,Z and A,B,C,D”. | | (4) Posted by Colin Sydenham [Monday, Jan 19, 2009 15:31] | Dear Jean-Marc,
Thank you for your help. It is an interesting approach, but I'm not quite sure that I follow your maths. In the Millour there are 3 primary interferences on 3 different lines followed by a single secondary interference at the intersection point (ie on all 3 lines). You reach quaternary by adding the 3 primary to the 1 secondary; is that right?
In the Visserman 2 BSs make 2 moves, constituting 2 pairs of primary + secondary. I'm not quite sure where your 4 comes from here. Evidently not 2+2+1. Are you only counting 1 for the second pair (2+1+1), because 1..Se~ does not permit 2.Qc5, and is therefore not secondary to the first pair? Or is there something else I haven't understood?
How would you analyse this, using your counting?:
N.A.Macleod & C.P.Sydenham Probleemblad 1977
(= 7+12 )
1.Qf5
1...Se~ /Sec5/Sed4/Sbc5/Sbd4
2. Bb1 /Re1 /Sc3 / Qf5/Sd6
Best wishes, Colin | | (5) Posted by Jean-Marc Loustau [Monday, Jan 19, 2009 20:07]; edited by Jean-Marc Loustau [09-01-20] | Dear Colin
Just to be accurate first : you write about “my maths”, “my counting” etc. I must say this is not “my analysis”, it is “an analysis”. My personnal opinion about what is or what should be a quaternary play is not the question. So:
1) some analysts think that a quaternary play is mandatorily a 4th degree correction (in the most general meaning of this term); in this approach, the 4th degree is a “theme” (by the way a difficult one) which can be show as a single chain A -> B -> C -> D with at each step the previous mates prevented by anti-dual effect(s). For this group, none of the previous problem is a 4th degree.
2) other analysts think that the degree qualifies a single variation (and not a set of variations). A variation is told to be 4th degree when:
a) it has 3 different anti-dual effects (to mates A, B, C)
b)each of the mates A, B, C occurs in a variation (in the same phasis)
For this group all the previous problems show what we could call a quaternary play (but of course not a 4th degree correction)
3) Among this second group of analysts some are more restrictive, and think that a Nth degree variation must have N different negative effects (and of course N-1 anti-dual effects); in other words each of the mates A, B, C is connected with a specific negative effect. For this group the 1st problem quoted by René-Jean (Stella Polaris 1967) does not have a 4th degree variation (because mates A B C are connected to a same negative effect = guard of e6).
4) I would not be surprized if other analysts exist!
This is just a matter of definition as given above; nothing more (there is no “count”, no “3 primary added to 1 secondary” or things like that). About your beautiful problem with Norman:
1) the 1st group will find 2 3rd degree chains: 1… Se plays/Sec5/Sed4 and 1… Sec5/Sbc5/Sbd4 (so twice the “3rd degree theme”), and moreover an anti-dual couple (1… Sbd4/Sed4)
2) the 2nd group will find 2 4th degree variations : 1… Sed4 2 Sc3# (not 2 Bb1? Re1? Sd6?) and 1… Sbd4 2 Sd6# (not 2 Qf5? Re1? Sc3?)
3) the 3rd group will find 2 3rd degree variations because the mates Sc3# and Sd6# share the same negative effect, so each of the variations 1… Sed4 and 1… Sbd4 has “only” 3 different negative effects
As you see about your problem the 1st and the 3rd group will get the same “label”; in the Visserman 1… Sc6 has 4 different negative effects so 3rd group would get a 4th degree variation (and twice the 3rd degree theme: 1… Sd5 plays/Sc7/Sc6 and 1… Sd5 plays/Se5 plays/Sc6).
So the same words “Nth degree” are applied to very different things: once it is a “theme” (Black correction of Nth degree = a set of N variations with 1 variation of each degree between 1 and N), once it just qualifies technically a variation (so you can use the words “quaternary play” which are different from “quaternary correction”).
I don’t know if I have been clear, I hope so… Thank you for not asking to which group I belong, but if you do, I probably will answer as the late Patrick McGoohan “I am not a number!”. In fact I analyse for my own understanding but I am not at all an analyst :-)
JM | | (6) Posted by Colin Sydenham [Tuesday, Jan 20, 2009 19:18]; edited by Colin Sydenham [09-01-22] | Dear Jean-Marc
You have been very helpful, and I'm sorry if I appeared to attribute to you any view which you were merely describing and do not in fact hold. Up to now I should have located myself in the first analytical group you describe (correction only), and I shall now have to think about the other two. You've opened my eyes, and made yourself very clear. Thank you.
Dear Rene (if you're still attending),
I do admire the elegance of your first problem that Jean-Marc quoted, which was new to me. In much more laboured style I had something on the same theme in The Problemist 1979:
(= 10+14 ) (10+14)
[Set: 1..Se3; 2.Rxb4. Try: Qg4? Se3!]
Qf4!(Sd2/d6)
1..Sg7 /Se3/Sd6/Sd4 (b2/Rxa4
2 cxb4/Qc7/Qf1/cxd4(Sd2/Sd6)
Best wishes to both
Colin | | (7) Posted by Jean-Marc Loustau [Wednesday, Jan 21, 2009 12:26]; edited by Jean-Marc Loustau [09-01-21] | Dear Colin
Thank you to you too for this topic. Moreover all the previous problems, including your last one, are actually very good problems. Let me add another one, close to the Visserman from the point of view of our subject (but from a general point of view, the content is of course different), and even probably more striking because there is only 1 black random move, and only 1 black thematic unit:
Drs C. GOLDSCHMEDING and G.A. CROES
Probleemblad 1946, 3rd Prize
(= 12+10 ) 2#
1 Qe3! (2 Qxe7)#
1… Se7 plays 2 d5#
1… Sg6! 2 Se4# (2 d5? Se5!)
1… Sf5!! 2 Qxg5# (2 d5? Sd4! 2 Se4? Qxe4!)
1… Sc6! 2 Rf7# (2 d5? Se5/d4!)
1… Sd5!!(!) 2 Qe5# ( 2 d5? Sc3! 2 Rf7? Kxf7!... But also, Ra5 being interferred: 2 Qxg5? Ke6!)
So we have the classical task of 2 3rd degree black corrections given by a same black unit; in the last variation there is an extra-anti-dual effect to one of the thematic mates: so 2nd and 3rd groups of analysts would say that this is a 4th degree variation… In fact we get 3 systems of 3rd degree correction (by a same unit): of course (1… Se7 plays/Sg6/Sf5) and (1… Se7 plays/Sc6/Sd5) , but also (1… Se7 plays/Sf5/Sd5). To my knowledge it is the only existing problem with this property…
Best wishes
JM | | (8) Posted by René J. Millour [Thursday, Jan 22, 2009 09:44] | Yes, all these problems are very interesting.
Just a small detail. In the “Millour, BCPS 1967, 4th Prize”, please note that, in my opinion, 1...Rxf8 2.Qd1# is of no interest (same mate than in the variation 1...Sd7). In 1967 I was a young problemist. Today instead of WRf8, I prefer WRf7 (and 1...Rf8 is not any more a defence)!
Best wishes!
René | | (9) Posted by Jean-Marc Loustau [Thursday, Jan 22, 2009 12:45] | Dear René
(should I say René or René-Jean?)
I agree with you that's a little better with Rf7; moreover this gives an extra-use to Bh7. I suppose nevertheless that some people could prefer with Rf8 just for visual reasons: one of the very striking points in this masterpiece, apart of course the thematic content what is the most important, are the very long black and white lines (a beginner or a player, for example, will see that, but not the Ceder theme): so from this point of view, the longer the better. So although I prefer too with Rf7, I think the 2 versions are excellent.
JM | | (10) Posted by Colin Sydenham [Thursday, Jan 22, 2009 16:30]; edited by Colin Sydenham [09-01-22] | Dear Jean-Marc
Yes, this is a problem I met when the Goldschmeding booklet came out in 1976, when I see that I labelled it Triple Tertiary. I'm very glad to have your detailed analysis of it. I don't believe I have ever seen this achievement equalled by a single piece. Of course the great Casa quinary correction problem actually has a second quaternary correction variation, but that's using two BSs.
And Rene
Although I am inclined to agree with the tiny modification of your beautiful problem, I do not think the unnecessary variation in any way detracts from the beauty of the whole.
Best regards to both
Colin | | No more posts |
MatPlus.Net Forum General Third degree/Tertiary play |
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