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MatPlus.Net Forum Retro/Math Frank Christiaans, 6637 feenschach 111 - 1994
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(1) Posted by Juraj Lörinc [Friday, Dec 21, 2012 23:34]

Frank Christiaans, 6637 feenschach 111 - 1994

Browsing through the older feenschach issues I have uncovered the following Illegal cluster problem that was found to be incorrect by solvers (surprise, surprise).

Frank Christiaans, 6637 feenschach 111 - 1994
wKb5 - bKb3
Add wS, bS, 6 bP for IC.
2 solutions

In the solutions issue there are given 2 intended solutions and 2 cooks.

See if you want to see the solutions, however only one of cooks is given there.

Anybody dares to find the other one?
Or the new one?
Can the problem be saved by stipulating pawn in the intersection of intended solutions?
Which IC positions among intended solutions and cooks are the most interesting in your view?
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(2) Posted by Kevin Begley [Saturday, Dec 22, 2012 07:28]

(= 2+8 )

Last move must be -1.a7-a6+ and white is left with no retraction.
The position must be illegal, and removal of any one unit seems to legalize everything.

Unless I'm mistake, this is a second cook.
Was it the one you intended?
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(3) Posted by Mario Richter [Saturday, Dec 22, 2012 09:44]

After the removal of wNa2 the position is still illegal. (Btw., the question "Why does the removal of the wN lead to a legal position?" is a good starting point for finding both the intended solutions as well as the cooks.)
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(4) Posted by Sarah Hornecker [Saturday, Dec 22, 2012 19:30]; edited by Sarah Hornecker [12-12-22]

This is given as a cook. How does it get legal after removing wSa4? What is the retraction? 1...a7-a6+, and then...?
(= 2+8 )

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(5) Posted by Kevin Begley [Saturday, Dec 22, 2012 19:44]; edited by Kevin Begley [12-12-22]

Exactly -- this is the recipe I was following.
I never considered that this cook might not be valid to begin with.

After the wNa4 is removed, black can not retract an uncapture (-1.a7-a6+ is the only option), and since white is left no retraction, the position must remain illegal.
Thus, this is not an illegal cluster -- no cook!

Now, I'm intrigued.
Is there only one cook (the one we are searching for)?
Or, is there no cook at all?

Juraj, without providing us your second cook, can you assure us that it does constitute a valid cook?
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(6) Posted by Nikola Predrag [Sunday, Dec 23, 2012 01:46]

Could bPa6 be a typo?
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(7) Posted by Kevin Begley [Sunday, Dec 23, 2012 03:49]

For this scheme to be effective, it seems you must satisfy three constraints:
1) white must be prevented from any legal retraction,
2) black's retraction must not provide white with a legal retraction (especially via uncapture!), and
3) removal of white's Knight must provide the white King with a legal retraction.

The point of the black Pawn a6 was to economically force black into a non-capturing retraction.
I see no way to force the uncapture of a specific white unit (which has no retraction itself).
So, unless you can force a non-capturing retraction, the entire scheme looks dubious.
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(8) Posted by Joost de Heer [Sunday, Dec 23, 2012 21:33]; edited by Joost de Heer [12-12-23]

A slight variation (probably cooked too, as there are only ICs that are cooked, and ICs that haven't been cooked yet):
W: Kb5
B: Kb3
Add wB, wS, bS, 5bP for an IC
(Read Only)pid=9342
(9) Posted by Joost de Heer [Sunday, Dec 23, 2012 22:20]; edited by Joost de Heer [12-12-23]

(intented solution: 3s4/pS1p4/Bppp4/1K6/8/1k6/8/8)
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(10) Posted by Henry Tanner [Monday, Dec 24, 2012 01:09]

@ Joost

There is a variant of this idea:
(Kb5) Ba6, Sb7 - (Kb3) Sd8, Pa5, c5, c6, d6, d7
- last move c7-c6+

I thought *this* was the author's solution (as it seems to be dualfree):

(Kb5) Bb8, Sa6 - (Kb3) Sb7, Pa7, b4, b6, c6, d7
- last move c7-c6+

Your "variation" gets rid of the cook in the original IC
(to other readers of this thread: yes, there is a real cook, which is not very difficult to see)
as there's one bP less, but there remains the original solution with wSc5 (checking) with the 7 "flights"
of the wS taken with wB, bS, bPPPPP, but now (unlike in the original) there are various ways to arrange the 7 men.

"Can the problem be saved by stipulating pawn in the intersection of intended solutions?"

It seems to me that having Kb5 - Kb3, Pa4 initially would be enough to save the problem.

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(11) Posted by Juraj Lörinc [Wednesday, Dec 26, 2012 23:21]

Well, I admit I was slightly careless with the cook provided in the Retro Corner, but seemingly I was not the only one. Based on the available issue 120 with solutions, the non-cook was claimed by Juha Saukkola and the solutions editor nor the retro editor or whoever involved did not spot the issue with removing wSa4.

To my surprise (right in this moment, I did not notice it a week ago during the first browsing of magazine), based on the solutions text the other cook was seemingly found by me. (I really do not remember exactly everything, that I had solved more than 16 years ago...) It is based on different retro idea. I wonder why it did not pass into Retro Corner. It is not unique.

I was expecting some new cooks could be found. If not, then the Henry's suggestion for saving the problem is identical to mine.
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(12) Posted by Kevin Begley [Thursday, Dec 27, 2012 00:50]; edited by Kevin Begley [12-12-27]

>"I wonder why it did not pass into Retro Corner. It is not unique."

I expect it may have been thought, back then, that it was only necessary to report a single cook.

As for the suggested correction, it should be noted that addition of a unit into the diagram of an Illegal Cluster problem is (generally) the very last method for salvation that should be considered (after all else is exhausted).
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MatPlus.Net Forum Retro/Math Frank Christiaans, 6637 feenschach 111 - 1994