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|(1) Posted by Bob Baker [Sunday, Feb 2, 2020 06:08]|
Good Illegal Positions
Can someone point me to a source of good illegal positions? By "good" I mean positions that seem as though they should be legal, and require a detailed analysis to prove their illegality. Thanks.
|(2) Posted by Arno Tungler [Sunday, Feb 2, 2020 10:15]|
If you search in the PDB https://pdb.dieschwalbe.de/index.jsp indicating something like
(g='2#' OR g='3#' OR g='n#' OR g='h#' OR g='s#') AND k='illegal' AND NOT g='retro'
you find more than 500 such positions. Many of those are serious problems where the authors had just overlooked that the position was illegal. Some of the most interesting cases are the P0569668, P1013151, P1016935, P1062393, P1066265, P1079599, P1081412, P1086130, P1099562, P1102599, P1105413.
|(3) Posted by Hauke Reddmann [Sunday, Feb 2, 2020 12:19]|
I'm too lazy to research :-) - is "good" about proportional to the
minimum length of the required PG?
|(4) Posted by Bob Baker [Sunday, Feb 2, 2020 20:37]|
Thanks Arno. Hauke, I'm not familiar with the idea of using a proof game to show that a position is illegal. How does that work?
|(5) Posted by Hauke Reddmann [Sunday, Feb 2, 2020 23:43]|
I was a bit sloppy as usual :-) Of course you can't give a proof
game for an illegal position. The question referred to the *legal*
positions requiring a very tricky retrounknotting, and also I would
just count the backward moves until the position gets "obviously" legal.
Sorry for the confusion.
|(6) Posted by Rosie Fay [Monday, Feb 3, 2020 17:41]|
Here is a retro which needs deep analysis:
(= 15+8 )
(Thierry le Gleuher; Economy Records in Add Unit(s) Problems, 5A.)
The stipulation is "Add the missing unit(s)". A solution may be found in entry P1227636 in PDB. In brief:
To disentangle Black's kingside, White must retract most pieces back home so that White can retract g2xf3. Interleaved with these White moves are 19 Black moves. But, all this time, Black has no retrotempi with units in the diagram, so they must be provided by the added units. The only way is to add black pawns on a3, b3, c3, d3, e3. Each provides 4 retrotempi, except that the b and d pawns can't both retract home, because the bishop on g8 will need to retract to c8 eventually.
Thus the position would be illegal if 5 black pawns were added but any of them were nearer rank 7, because Black would run out of retrotempi, but it would take deep analysis to show that this is the case.
In case you thought that there's a simpler way to add units: A Black piece captured White's c1 bishop on c1. White is not missing anything else, so no Black pawn ever left its file. Thus White's 3 captures g2xf3 and hxgxf were of Black's 3 missing pieces: the f8 bishop, the queen and a knight. So we may add Black pawns but not a piece.
|(7) Posted by Bob Baker [Tuesday, Feb 4, 2020 08:16]|
Thank you Rosie. The position with black pawns added at a4, b3, c3, d3, and e3 would make a very good illegal position, just one move short of enough waiting moves by Black to make the position legal.
|(8) Posted by Michael McDowell [Tuesday, Feb 4, 2020 08:56]|
How easy is it to determine that the following position is illegal?
The Problemist 1969
(= 15+11 )
Mate in 2
Set 1…Rb3 2.Sf7, Re6, Rd5, Sc4
1.Sde4? Rb3 2.Sf7 1…Sxd6!
1.Rxf6? Rb3 2.Re6 1…Kxf6!
1.Qh2? Rb3 2.Rd5 1…Kxf5!
1.Sge4? Rb3 2.Sc4 1…Sxd6!
1.Rc4 (>2.Sf7, Re6, Rd5)
|(9) Posted by Siegfried Hornecker [Wednesday, Feb 5, 2020 11:43]; edited by Siegfried Hornecker [20-02-05]|
Pretty easy. The pieces can mvoe as they want, so we need to take only pawns into account. One possibility: Pg6 comes from the e-file (two captures). b and c pawn are crossed (before Pc2 came from c5: c4xb5 - then before that Pb6-b3, before that b5xc6).
So to untangle the white pawns we need four captures. Then we need one by black to untangle the h-file. So White captured the black a-pawn on the a-file, promoted his a-pawn and Black captured it on the h-file with his g-pawn. Or he captured something else and the a-pawn promoted and replaced it.
So all missing pieces are accounted for and the position is legal.
In short: a2=promoted, b2=c6 (1 capture), c2=b6 (1 capture), e2=g6 (2 captures), a7=captured on a-file (1 capture) - all 5 black captured stones; g7=h5 - the white captured stone
Alternatively e2=f5 and f2=g3 or g6 instead e2=g6, in any case those are two captures.
|(10) Posted by Bob Baker [Wednesday, Feb 5, 2020 13:34]|
Siegfried, what happened to Black's h7 pawn?
|(11) Posted by Bob Baker [Wednesday, Feb 5, 2020 14:39]; edited by Bob Baker [20-02-05]|
Here's my try at an illegality proof.
If White's h-pawn reaches h7 without making a capture, Black's h-pawn must be captured first. White's two remaining king-side captures and two captures to cross pawns on the b and c files leave White unable to capture Black's a-pawn, so White's a-pawn can't promote to replace a piece captured on h6.
If White's h-pawn captures a pawn on g6, White needs four captures on the king side, leaving each player only one capture on the queen side. Then no b-pawn or c-pawn can ever leave its file, since some pawn of the same color would have to capture to return to that file. So the b and c pawns can't cross, and the position is illegal.
|(12) Posted by Siegfried Hornecker [Wednesday, Feb 5, 2020 23:06]; edited by Siegfried Hornecker [20-02-06]|
correction: a captured to b and b captured to c. So the h-pawn was captured on h7.
But what happened to Pd7 then? I guess it IS illegal after all, and I was mistaken.
EDIT: So I retract the "easy" part. It's a normal difficulty IMO.
|(13) Posted by Bob Baker [Thursday, Feb 6, 2020 03:17]|
I haven't yet answered Michael's "how easy" question. It was not easy for me. I had only worked out half the proof before I saw Siegfried's comment.
|(14) Posted by Michael McDowell [Thursday, Feb 6, 2020 08:07]|
The analysis in The Problemist was "Black cannot have made more than one pawn capture. If this was ab, white would need six pawn captures (four on the K side and two on the Q side) but black is only five pieces short. If black's only capture was gh, this must have been of a promoted WP, but retracting four of white's five captures, ef, fg, ab and bc, leaves only the c pawn missing, and this could not have promoted without a further two captures (making six in all) since the black c pawn has not moved off the file."
Only three solvers claimed illegality. Can't say I would have even looked for it. It seems ridiculous to declare such an interesting problem "illegitimate".
|(15) Posted by Bob Baker [Thursday, Feb 6, 2020 16:26]|
I would call the position interesting both as a "mate in 2" problem and a "prove illegal" problem. Which is more interesting depends on one's perspective. At the moment I'm looking for good illegal positions.
|(16) Posted by Guus Rol [Sunday, Nov 29, 2020 12:21]|
The way to measure how good an illegal position is, is to introduce the concept of DTL - distance to legality. Good illegality shows MINIMAL distances to legality in one or more contributing aspects - similar to tries in common compositions which ALMOST work. Minimals refer to quantification but do not provide criteria for deciding on the character of the contributors. For instance, resolving an illegal position might be just "1 grasshopper promotion away" from being legal but we consider grasshoppers as objects outside the language space we want to address. On the other side "1 too many captures required" is in the standard toolkit of every retro-analyst. Probably the most valuable contributors are the ones relating to the dynamic phase of retro-analysis which consists of concrete retractions. Here we get "1 tempo move lacking", "a collision on retraction routes", "an unfixable move order conflict" and such. Many of these can be categorized under the heading of "resolvable by (illegal) pass moves" minimalized as "requiring 1 pass move".
Note that we rate the quantitative aspects of legal positions in similar ways. Instead of "lacking 1 tempo move" for illegal positions, we get "just enough tempo moves" for legal ones. Success and disaster are next door neighbours!
|(17) Posted by Bob Baker [Wednesday, Nov 10, 2021 10:00]|
Maybe someone can help me with a position that is confusing me. It's the position by Frolkin that won 1st prize in Section H--Retros and Proofgames at https://www.wfcc.ch/competitions/composing/fidewcc_2018/ with stipulation "Release the position". The retroplay given as the solution produces the position below. I want to continue the retroplay to make a proof game, but can't find a way to release the position of the five pieces at d1, d2, e1, e2, and f1. Is there a way to do this that I can't find, or is the position resulting from the given retroplay not required to be legal?
(= 15+15 )
|(18) Posted by Hauke Reddmann [Wednesday, Nov 10, 2021 10:49]|
It isn't "trivial" (except for a retro buff, which I am not ;-)
0.Pass with Sh6-g8-h6 if necessary.
1.Unmove away wK.
2.Unmove away wRh4.
3.Unplay wNg1. (Retroshielding!)
6.Unplay wPh7 to h2.
7.Unmove away wRg3.
9.Unmove away wNg2.
10.Unplay sPh3 to h5.
11.Unmove away wQf1.
12.Unmove away wBf6.
13.Unmove away wSg1.
14.Unmove away bKh1.
15.Unmove away sBe2 to h3.
16.Unmove away wBd1 to g2.
17.Unmove away bQe1.
18.Unmove away bRd2.
20.Unmove away bNf3.
(= 16+16 )
The position now look about this. Finally, you can play everything home!
|(19) Posted by Bob Baker [Wednesday, Nov 10, 2021 13:47]|
Thanks, Hauke, that did it. The key point I missed is the requirement to keep the white h-pawn at home until a white rook gets to g3.
|(20) Posted by Andrew Buchanan [Wednesday, Nov 10, 2021 15:35]|
Hi looking at the Arnoldo Ellerman problem from earlier in this thread (post 8 by Michael McDowell), with its superb Pseudo-Karlstrom-Fleck. I completely agree that soundness does not depend on legality, however, it's nice to make a position legal if one can. How about just removing bPh5? As far as a novice like me can see, there is zero impact on set & actual play, but maybe some #2 expert would be kind enough to explain what the problem would lose in terms of quality.
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