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| (1) Posted by Hauke Reddmann [Saturday, Oct 10, 2020 17:51] |
Promotions to wrong color
Since this came up in the chat:
1.d4 c5 2.dxc5 b6 3.cxb6 Qc7 4.bxa7 Qxh2 5.axb8=bS
My lame attempt at a SPG, probably not even correct.
Has somebody done this yet? Malcom already did show
some dummy pawn SPG.
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| (2) Posted by edderiofer [Sunday, Oct 11, 2020 05:21] |
I'm not sure what chat you're referring to, unless it's the chatroom for this site, in the sidebar, which appears to be broken for me.
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| (3) Posted by Sarah Hornecker [Sunday, Oct 11, 2020 07:45] |
With all due respect, may I repeat a problem I am very proud of?
 (= 2+2 )
h#3 Köko
FIDE Laws of Chess of 1993
MatPlus, issue 31, no. 1097
The old FIDE Laws of Chess had already fixed the "Pam Krabbé Castling", but not...
1.Bg7 h8R 2.Bf6 Kg6 3.0-0!!! Kh7 mate
Now why do I quote this? Well, there is this young man, America's new talent of chess composition, who was not the first and will not be the last to do something similar. Apologies if I don't repeat the position correctly.
 (= 1+9 )
James Malcom (SuperProblem, I think?)
Last two single moves?
Retraction: 1.0-0-0+ a7-a8bR!
It is in the PDB, so James will later correct my bad memory if the position is wrong. Obviously the idea of castling with a promoted rook, or even to promote to a rook of the other color, is much older, but doing it as a retractor like this is a new idea.
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| (4) Posted by Hauke Reddmann [Sunday, Oct 11, 2020 11:03] |
@eddie: Nope, just a random chat on Zoom, many
of the problemist also being frequent MatPlus posters.
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| (5) Posted by James Malcom [Sunday, Oct 11, 2020 18:42] |
Siegfried, you got the position right, but the source wrong. It was an original for the Die Schwalbe site-https://pdb.dieschwalbe.de/P1375777. It was a correction after Josef Cumpe-see https://pdb.dieschwalbe.de/P1234039. Talk about a Platzwechsel! :-)
On the same note, here is one of my personal favorites of all my problems with promotion to an enemy piece.
James Malcom, SuperProblem, 11/2/2019
s#3 (Version)
b) b7<->f6, c7->f5, a6->g3
 (= 11+6 )
a) dxe8=bB+! Bd7 2. cxb8=bN! f5 3. a8=bR! Ra7#
b) dx=e8=bQ+! Qd7 2. axb8=N g2 3. Nxg2 Qxd8#
To get back to the subject of proof games, great idea Hauke! However, there is a dual in your PG-2... b5 also works. Here's another one based on Siegfried's problem.
PG 5.0
 (= 15+13 )
1. f4 g5 2. fxg5 Nh6 3. gxh6 Bg7 4. hxg7 h6 5. gxh8=bR 0-0
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| (6) Posted by Joost de Heer [Sunday, Oct 11, 2020 21:04] |
Castling with a (formerly) opponent piece: https://pdb.dieschwalbe.de/P1095829
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| (7) Posted by Andrew Buchanan [Monday, Oct 12, 2020 02:56] |
A new version of the Reddmann-Malcom opus, to avoid the dual 3. ... Sf6/Sh6
 (= 15+13 )
PG 5.0
Note that Bb6 (or a5) is required to avoid a different cheeky dual!
Another slant on this:
b) Bb6->h6 Chess960
This has the distinction of being the only Chess960 problem I know with the orthodox starting position! :)
EDIT: Solution:
a) 1. f4 e5 2. fxe5 Sf6 3. exf6 Bc5 4. fxg7 Bb6 5. gxh8=bR 0-0
b) 1. f4 g5 2. fxg5 Sf6 3. gxf6 Bh6 4. fxe7 Rf8 5. exf8=bR 0-0
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| (8) Posted by Hauke Reddmann [Monday, Oct 12, 2020 09:48] |
OK, since even I (almost) could do it - who is the first
to show TWO of this promotions (preferrably different
color, for fairness)?
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| (9) Posted by Rosie Fay [Monday, Oct 12, 2020 15:08] |
Is that legal in Chess960, then, Andrew? So is this a sound Chess960 PG3.5?
 (= 15+16 )
(I didn't think it was.)
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| (10) Posted by James Malcom [Monday, Oct 12, 2020 15:29] |
Here's a simple double for you, Hauke-obvious solution is obvious.
PG 5.0, 2 sol.
 (= 14+14 )
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| (11) Posted by Andrew Buchanan [Monday, Oct 12, 2020 15:48] |
 (= 1+7 )
Last two single moves? (-:
(Promotion to foreign officer or king allowed, but not pawn.)
In Chess960, castling must obey the usual rules about KR not having moved, K not being in check on any square from start to finish, and no square on either piece's route being blocked. There is nothing to say that a promoted foreign R might not be able to castle as long as it hasn't moved. I think this harmonizes quite well with the foreign promotion joke. Not sure what your example shows, James, but it's barking up a different tree.
Here's my attempt at the 2 F.Promotion challenge: please cook it if you can:
 (= 13+13 )
PG in 6.5 (-:
EDIT: Solution 1. b4 g5 2. b5 g4 3. b6 g3 4. bxa7 gxh2 5. axb8=S hxg1=R 6. Rxh7 Rxh7 7. Rh1
Natch says no orthodox solutions
The fopromos (faux promos?) suggest Fopronkin which breaks the symmetry under which the promotions might be a1/c1 & f8/h8.
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| (12) Posted by Hauke Reddmann [Monday, Oct 12, 2020 19:18] |
I think Rewan just misplaced two pawns on the queenside.
As is, 4 solutions.
Of course, you merely can replace the white-field rooks
with e.g. knights but that kinda ruins the joke.
(Can Jacobi implement the condition already?)
Hauke
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| (13) Posted by James Malcom [Wednesday, Oct 14, 2020 17:47] |
Andrew, what are the solutions to your PGs?
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| (14) Posted by Andrew Buchanan [Wednesday, Oct 14, 2020 19:23] |
James: my solutions are added to earlier posts. Natch says no orthodox solutions to these or the ones which follow here, but need to have Jacobi equipped.
Mildly preferred version of one silly idea.
 (= 13+15 )
PG in 5.5 Chess960 (-:
Full homebase, double fopro:
 (= 13+13 )
PG in 6.0 (-:
 (= 12+13 )
PG in 8.0 (-:
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| (15) Posted by Joost de Heer [Wednesday, Oct 14, 2020 22:09] |
Winchloe has the condition 'Promotions adverses' (a pawn promoting changes colour), with it, Andrew's 6.5-mover is correct.
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| (16) Posted by Andrew Buchanan [Monday, Oct 19, 2020 02:55] |
Thanks Joost
That’s great and I’ve now used WinChloe to check another longer PG (not posted here).
However it would be better if the WinChloe condition allowed for promotion into both colours of units. In many joke problems the foreign promotion is only because promotion into one’s own colour doesn’t work. Obviously one can run twice if there is a single promotion at stake, but in these PGs there are multiple promotions so they can’t really be fully tested. To see the limitation, change the PG 6.5 above to make Sg8 White and there is no WinChloe solution, whether promotions adverses is switched on or not.
The other issue is that the condition doesn’t combine with promotion en pions, so the two other problems that I proposed above can’t be tested.
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