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MatPlus.Net Forum Fairies Babson-Zagoruyko with Tertiary Black Correction!
 
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(1) Posted by shankar ram [Friday, Jul 22, 2016 18:15]; edited by shankar ram [16-07-24]

Babson-Zagoruyko with Tertiary Black Correction!


This post is in continuation to: http://www.matplus.net/start.php?px=1469195657&app=forum&act=posts&tid=647&fid=xshow3&page=3&pid=14793#n14793

There I had quoted a fairy 2# by Hubert Gockel showing 3x3 Lacny with Tertiary Black correction.
That problem had 3 solutions with all the keys being made by the WQ.
A somewhat similar problem is given below:

Reto Aschwanden
1st Prize,Springaren 2001
Dedicated to K.Widlert
(= 5+13 )

#2*
3 Solutions
Magic Pieces(4+3): a8,b7,f6,h6;e2,e6,g1

Magic pieces move/capture/check normally but also change the colour of any pieces they observe/guard/attack, both Black and White, other than Ks and other magic pieces.
In the diagram position, after 1.b7-b8=~+?, BK can move to f2, because Magic Ba8 now observes WSh1, which turns into a BS!

Set Play:
1..e1=MS 2.b8=MQ(WSg3,WBb1)#
1..e1=MB(WSg3)! 2.b8=MR(WBb1)# [2.b8=MQ(BSg3,WBb1)+? Se4!]
1..e1=MR(WBb1,WPe3)! 2.b8=MB(WSg3)# [2.b8=MQ(WSg3,BBb1)+? Be4!]
1..e1=MQ(WSg3,WBb1,WPe3)!! 2.b8=MS# [2.b8=MR(BBb1)+? Be4!, 2.b8=MB(BSg3)+? Se4!]

1.MQh7!(WSf7,WBb1) threat 2.Se5#
Black defends by promoting to magic Q/R/B/S on e1. Now 2.Se5+? MSd3/MBc3/MRxe3/MQc3,xe3(BSe5)!
1..e1=MS 2.b8=MB(WSg3)#
1..e1=MB(WSg3)! 2.b8=MS# [2.b8=MB(BSg3)+? Se4!]
1..e1=MR(BBb1,WPe3)! 2.b8=MQ(WSg3,WBb1)# [2.b8=MS+? Be4!]
1..e1=MQ(WSg3,BBb1,WPe3)!! 2.b8=MR(WBb1)# [2.b8=MS+? Be4!, 2.b8=MQ(WBb1,BSg3)+? Se4!]

1.MQg7!(WSf7,WSg3) threat 2.Se5#
1..e1=MS 2.b8=MR(WBb1)#
1..e1=MB(BSg3)! 2.b8=MQ(WBb1,WSg3)# [2.b8=MR(WBb1)+? Se4!]
1..e1=MR(WBb1,WPe3)! 2.b8=MS# [2.b8=MR(BBb1,+? Be4!]
1..e1=MQ(BSg3,WBb1,WPe3)!! 2.b8=MB(WSg3)# [2.b8=MQ(BBb1,WSg3+? Be4!, 2.b8=MS+? Se4!]

1.MQg6!(WSf7,WBB1,WSg3) threat 2.Se5#
1..e1=MS 2.b8=MS#
1..e1=MB(BSg3)! 2.b8=MB(WSg3)# [2.b8=MS+? Se4!]
1..e1=MR(BBb1,WPe3)! 2.b8=MR(WB)# [2.b8=MS+? Be4!]
1..e1=MQ(BSg3,BBb1,WPe3)!! 2.b8=MQ(WBb1,WSg3)# [2.b8=MB(BSg3)+? Se4!, 2.b8=MR(BBb1)+? Be4!]

W has to make the BBb1 and BSg3 white in the final mating position, to prevent them interposing on e4. But before that, they keep changing their colour in bewildering fashion due to W1 and B1 moves. Their penultimate colour determines W's promotion on b8.

4 fold Babson across set and 3 solutions:
set: 1...e1=S/B/R/Q 2.b8=Q/R/B/S
1.Qh7! 1...e1=S/B/R/Q 2.b8=B/S/Q/R
1.Qg7! 1...e1=S/B/R/Q 2.b8=R/Q/S/B
1.Qg6! 1...e1=S/B/R/Q 2.b8=S/B/R/Q

In each phase, two secondary and one tertiary correction by the BPe2:
1...e1=S(Random), e1=B/R(Secondary Corrections), e1=Q(Tertiary correction of both secondary corrections)
Cumulative weaknesses of B moves:
1...e1=S (Line opening a2-f2) - Primary
1...e1=B (Line opening a2-f2/Colour changing of Sg3) - Secondary
1...e1=R (Line opening a2-f2/Colour changing of Bb1) - Secondary
1...e1=B (Line opening a2-f2/Colour changing of Sg3,Bb1/Color changing of Sg3+Bb1)) - Tertiary

12 pairs of reciprocal changes between the 4 phases:
QR--/RQ--, Q-B-/B-Q-, Q--S/S--Q, -RB-/-BR-;
-R-S/-S-R, --BS/--SB, BS--/SB--, B--R/R--B;
-SQ-/-QS-, --QR/--RQ, R-S-/S-R-, -Q-B/-B-Q;

To summarise:
4 fold Babson in #2
random, 2 secondary and 1 tertiary correction by BPe2
12 fold reciprocal change
4x4 Zagoruyko

Needless to say, a remarkable task, achieved with light construction. A Babson, even done with a "helpful" fairy condition is not very easy. To do it in 4 phases is mind boggling.

The similarities with Gockel's problem are: the 3 keys by WQ, which bring about the changes, black tertiary correction, the very intensive use of the fairy element and the light setting. Only instead of a Lacny, there are multiple reciprocal changes.
 
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(2) Posted by Nikola Predrag [Sunday, Jul 24, 2016 11:59]; edited by Nikola Predrag [16-07-24]

Uh, it seems as if I'm expected to "throw" the first comment.
Shankar, since you've provided all the explanations, we can only try to provide some proper "poetic" words.
Well, I'm not a poet so I say "breathtaking", since that's how it affected me.

However, you probably want me to comment the "content-stipulation" complex.
I should say that it's a beautiful example of a fake #2. By "fake" I mean it's a SIMULATION of #2.
And in the complex of "content-stipulation", perhaps it might be called a "hyper-modern stipulation".

A simulation works as a more or less successful/convincing surrogate.
If Gockel's s#2 was half-convincing, the Aschwanden's "simulation" creates the "virtual depth" that matches the "real dimensions".

The apparently bad keys change the color of 2 or even 3 black (relevant) pieces. But now it's Black who can choose which color-change to nullify while defending against the threatening color-change of Sf7.
 
 
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(3) Posted by shankar ram [Sunday, Jul 24, 2016 17:36]

Right, Nikola.. Since the others seem to have taken a weekend off from chess problems!
I had seen this problem some months before the Gockel problem. Normally, I find Reto's task achievements too heavy analyse, so I only look at the end results. But in this case the position was light enough, with only one fairy element, that I looked into the details. The colour toggling mechanism is the heart of the problem. And I thought Reto had done a good job working it into a proper #2 with a threat.

I'm, of course, an unabashed devotee of cyclic themes. But I do feel your opinions provide a good balance!
 
   
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(4) Posted by ichai [Sunday, Jul 24, 2016 20:44]

The mechanism seems more simple than the previous one, but the way it is done is much interesting too, with many nice features.
 
   
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(5) Posted by ichai [Monday, Jul 25, 2016 00:42]

If you add a white pawn e7 you have the 4th solution 1.Qf8!
 
   
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(6) Posted by shankar ram [Monday, Jul 25, 2016 06:28]

Ichai..
Yes.. Looks like a valid 4th solution!
Probably Reto saw it, but didn't add it, since it repeats the set play.
 
   
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(7) Posted by Georgy Evseev [Monday, Jul 25, 2016 08:26]

Some thoughts.

1. Yes, there some issues with Reto's problem - it is much more of static geometry than of chess play.

2. I am not sure it may be called a correction problem - change of color of Sg3 and Bb1 is definitely not an error effect. At least we can have exactly the same foundations to call it a defensive effect. I think that most correct way is to call it anti-dual effect (though I can see objections for this term also). There is also a possibility to call this "a side effect", which may be correct, but somewhat meaningless.

3. Some years ago I had an interesting discussion with Peter Gvozdjak, concerning the question, if such problem _as a whole_ may be called cyclic. The end result was that it depends on the definition of "cycle" for multiphase problem.
 
   
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(8) Posted by ichai [Monday, Jul 25, 2016 12:18]; edited by ichai [16-07-25]

Change of colour is not that big deal.
It looks (here) like (reversible) interceptions.
It recalls somehow what may happen with Lions or Chinese pieces.
 
 
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(9) Posted by Nikola Predrag [Monday, Jul 25, 2016 12:39]

The whole point is about how the solutions refer to the set-play, where White has the exact answer to each of the black promotions.
Apparently, any threat which provokes Black to promote bMPe2, should be a solution with simply repeated set mates. But none of these 4 set mates will answer the same black promotion after any of the 3 keys.
There's obviously some error in each key as referred to the set-play. And obviously there's something that compensates for the error(s) since there are solutions after all.
The question is what may be seen as "Black" and what as "White" correction, and how that all is related between the phases.

(BTW, it seems that one must "guess" a property of the Magic pieces: after bBb1-e4, what's the color of that Bishop in various variations?)
 
   
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(10) Posted by Georgy Evseev [Monday, Jul 25, 2016 15:45]

My main thought was that proposed kind of thinking may convert many cyclic problem into the "correction" one.

Here is the example.

Georgy EVSEEV & Lev GROLMAN
8 WCCT 2007-08
2 Place

(= 13+9 )

‡2 (13+9) C+
Anticirce Calvet
Lions, Bishop-Lions, Rook-Lions

Any random move of Lig8 has a defensive effect (check to white king) and a defensive error (evacuating g8).
Every key evacuates a white piece from one thematic line, so such "random" move potentially allows two mates. But black has no truly random move - two "random" defences have additional defensive effect of
closing another thematic line, so only one mate is possible.

The third defence (marked by exclamation mark) has an additional defensive effect (extra check in tries, removal of guard from b7 in solution)and an secondary defensive error - restoring possibility to move along the third thematic line.

Correction? When composing this problem, it was never thought about.
At the same time this is often somewhat classic approach to 3x3 Lacny cycle. See another similar example below.

1.RLa5? [2.LIa6‡]
1…LIb3+ a 2.LIxg2(LIg8)‡ A
1…LId5+ b 2.LIxg3(LIg8)‡ B
1…LIe6+! c 2.LIxg4(LIg8)‡ C
but 1…BLa4!

1.BLa7? [2.LIa6‡]
1…LIb3+! a 2.LIxg3(LIg8)‡ B
1…LId5+ b 2.LIxg4(LIg8)‡ C
1…LIe6+ c 2.LIxg2(LIg8)‡ A
but 1…Rh2!

1.BLh1! [2.b7‡]
1…LIb3+ a 2.LIxg4(LIg8)‡ C
1…LId5+! b 2.LIxg2(LIg8)‡ A
1…LIe6+ c 2.LIxg3(LIg8)‡ B


Extra example
Michel CAILLAUD
Memorial L. Lehen, Pat a Mat 2014-15
2 Prix
(= 12+13 )

‡2 (12+13) C+
3 solutions
Circe
Fairy Queen c5
Double-Lion g1

1.Qd6! [2.Qd1‡]
1…Qd5 a 2.DLd4‡ S
1…Qxd6(Qd8)! b 2.DLxd6‡ A
1…Qd4 c 2.DLd5‡ B

1.Qd4! [2.Qd1‡]
1…Qd5 a 2.DLd6‡ A
1…Qd6 b 2.DLd5‡ B
1…Qxd4(Qd8)! c 2.DLxd4‡ S

1.Qd5! [2.Qd1‡]
1…Qxd5(Qd8)! a 2.DLxd5‡ B
1…Qd6 b 2.DLd4‡ S
1…Qd4 c 2.DLd6‡ A
 
   
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(11) Posted by shankar ram [Monday, Jul 25, 2016 19:46]; edited by shankar ram [16-07-25]

Georgy, Thank you for your comments and for showing the two new problems!

>> change of color of Sg3 and Bb1 is definitely not an error effect. At least we can have exactly the same foundations to call it a defensive effect.
Georgy, In some fairy types, a defence motif can itself also be an error!

>> Correction? When composing this problem, it was never thought about.
Well.. there are many problems containing some features which were not consciously intended by the composer(s)! Even in the problem by Reto, I don't know whether He intended the correction feature. I've not seen the original source.

>> My main thought was that proposed kind of thinking may convert many cyclic problem into the "correction" one.
I would say "add correction play feature", rather than "convert"!

Problem by Evseev/Grolman:
Quite a fine anti-circe/chinese piece specific Lacny mechanism. Well done!
And you're right. It shows in each phase, a cycle in the defence type between the 3 black moves: 2 moves show dual avoidance and the 3rd is a correction of both.

Problem by Caillaud:
Another fine mechanism. 3 keys/3 defences/3 mates on the same 3 squares! And this too, shows the same dual avoidance/correction cycle as the Evseev/Grolman problem.

>> At the same time this is often somewhat classic approach to 3x3 Lacny cycle.
I think you're referring to the approach of converting a:
1..x/y/- 2.A/B/-
1..-/y/z 2.-/C/A
1..x/-/z 2.C/-/B
pattern (sometimes called a "carousel") to a full fledged:
1..x/y/z 2.A/B/C
1..x/y/z 2.B/C/A
1..x/y/z 2.C/A/B
(sometimes called a.. well, you know!)
 
 
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(12) Posted by shankar ram [Monday, Jul 25, 2016 20:01]

Nikola..

>> (BTW, it seems that one must "guess" a property of the Magic pieces: after bBb1-e4, what's the color of that Bishop in various variations?)

According to the full definition, a non-magic piece observed by an EVEN number of magic pieces, RETAINS its colour, whereas one observed by an ODD number of magic pieces changes its colour.

Therefore, in the set position, 1..Be4 would be illegal since it would be observed by the MSf6, turn white and give a self check. After 1..e1=MS/MB, W has to prevent 2..Be4, because the BB would RETAIN its colour on e4, since it is now observed by MBa8 and MSf6.

After, 1.Qh7 and 1.Qg6, Bb1 turns white, so 1..Be4 would not be playable

After, 1.Qg7, 1..Be4 would again be illegal. But after 1..e1=MS/MB, W again has to prevent 2..Be4, as explained above.

Colour me magic...;-)
 
   
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(13) Posted by Nikola Predrag [Monday, Jul 25, 2016 21:13]; edited by Nikola Predrag [16-07-25]

I guessed so by the result but it's still not clear, after
1.MQh7(WSf7,WBb1) e1=MR(WBb1,WPe3), there's bBb1 again, so why 2.b8=MB(WSg3) doesn't mate.
It seems that the attack of MQh7 on bB counts as permanent as bB "travels" along b1-h7, so the arrival to e4 adds only 2 "new" effects.
Popeye gives e.g.:
Black magic Bc8
Black magic Sd8
Black magic Qa6
White Ba8
Stipulation ~1
1.Ba8-b7[b7=b] !
with MQa6->b6 it's the same,
but with MQa6->c6 it's 1.Ba8-b7!, no color-change

I must be missing something obvious :-(
 
   
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(14) Posted by Joost de Heer [Tuesday, Jul 26, 2016 08:24]; edited by Joost de Heer [16-07-26]

Colour change only happens when the observation is on a new line. With Qa6 the bishop is observed by the queen through the line a6-a8, after Ba8-b7 through the line a6-b7, a new line. With Qc6, the bishop is observed through the line c6-a8, after Ba8-b7 through the same line, so the observation of Qc6 doesn't change, so the queen doesn't start a colour change.

So, with this knowledge, spot the colour change after one of the key moves that isn't written.
 
   
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(15) Posted by Nikola Predrag [Tuesday, Jul 26, 2016 09:04]

Thanks Yoost, I did guess so, but I couldn't find a self-consistent reason for such behavior. It doesn't look as obvious but as ridiculous.
Well, at least I know the rule now.
 
   
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(16) Posted by Joost de Heer [Tuesday, Jul 26, 2016 10:29]

Compare it with pins: Does a Pelle move change the pin? Applied to magic pieces: Does moving a piece along a magic line change the magic influence?
 
 
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(17) Posted by shankar ram [Tuesday, Jul 26, 2016 10:44]; edited by shankar ram [16-07-26]

Nikola..
If I remember correctly, there's another twist in the rules which states that the same magic piece can change the colour of an already observed piece if it observes it from a "different direction" (or square?).

Conversely, this would imply that a piece does not change colour if it continues to be observed by a magic piece "along the same line".


EDIT:
Ok, Joost has already posted an explanation
 
   
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(18) Posted by Nikola Predrag [Tuesday, Jul 26, 2016 14:10]

Thanks Shankar, that makes sense as a property of a Magic piece.
Joost's "explanation" deals with the attacked piece, not with the attacking one. It suggests that the initially attacked piece changes color when the Magic piece attacks it no more, just as an unpin.
 
   
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(19) Posted by Georgy Evseev [Tuesday, Jul 26, 2016 19:11]

>>>> At the same time this is often somewhat classic approach to 3x3 Lacny cycle.
>> I think you're referring to the approach of converting a "carousel" to a full fledged Lacny

Partially. In fact I had more direct approach in mind.
In fact, there are generally two kinds of such approach.

1. "Positive" approach.

There are three obstacles hindering the fulfilment of stipulation.
Each of three keys destroys one of obstacles, each of black defences also destroys one of obstacles and each next white continuation also destroys one obstacle.
Here carousel change is achieved almost automatically, but some ingenuity is required to add "diagonal" variations.

This is the more "popular" approach, but no correction can be found in the solution, however searched for.
I would like to show the following problem as an example)).

Narayan SHANKAR RAM
Probleemblad 1986
2 Prix
(= 10+11 )


r‡2 (11+10) C+
Paralysing pieces

1.Rb7? Z
1…RPh2 a 2.Sf5 A Sxc2‡
1…RPc8 b 2.Sc6 B d4‡
1…BPh7 c 2.Se2 C Sxc2‡
but 1…Rb8!

1.Bh5? Z
1…RPh2 a 2.Sc6 B Sxc2‡
1…RPc8 b 2.Se2 C Sxc2‡
1…BPh7 c 2.Sf5 A d4‡
but 1…Bf7!

1.Qxg3! Z
1…RPh2 a 2.Se2 C d4‡
1…RPc8 b 2.Sf5 A Sxc2‡
1…BPh7 c 2.Sc6 B Sxc2‡
(1…BPf7 2.Sc6 Sxc2‡
1…Sxc2+ 2.Sxc2 d4‡
1…RPb8 2.Sb5 d4‡)

2. "Negative" approach.

This is exactly what was shown in my previous post.
General error of black moves should allow all three white continuations, but one of them is always destroyed by key and another one is destroyed by specific black defence.
Again carousel change is almost automatic, but "diagonal" variations require some ingenuity.

This approach adds "antiduals" and "correction". (I am still not sure if these words may be mentioned without quotes.)
 
   
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(20) Posted by shankar ram [Tuesday, Jul 26, 2016 19:58]; edited by shankar ram [16-07-26]

Thanks for quoting my problem, Georgy!
Yes.. There's no dual avoidance or correction in this kind of mechanism. Each black move has a distinct, different error.
A similar case is the well known "3 BK flights", W1 guards 1, B1 blocks/opens line to 2, W2 guards 3 mechanism pioneered by Milan Velimirovic in #3s and used by Miodrag in R#2s.

The two other problems quoted by you definitely have the dual avoidance/correction feature.

>>... (I am still not sure if these words may be mentioned without quotes.)

I think quotes are used if:
a) One is still not (yet) fully confident about the terminology
b) One feels others may not (yet) agree!

That's OK.. This forum is not the U.N security council.. ;-)
 
   
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MatPlus.Net Forum Fairies Babson-Zagoruyko with Tertiary Black Correction!