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MatPlus.Net Forum General Good Illegal Positions

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.

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.

I'm too lazy to research :-) - is "good" about proportional to the
minimum length of the required PG?

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?

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.

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.

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.

How easy is it to determine that the following position is illegal?

Arnoldo Ellerman
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)
1…Kxd6 2.Sf7
1…fxg5 2.Re6
1…Sd8 2.Rd5
1…Bxc4 2.Sxc4

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.

Siegfried, what happened to Black's h7 pawn?

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.

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.

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.

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".

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.