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| (1) Posted by Joost de Heer [Wednesday, Oct 27, 2021 15:51] |
Castling
 (= 3+16 )
ser-00 in how many moves? Anticirce
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| (2) Posted by Joost de Heer [Thursday, Oct 28, 2021 10:41] |
White's last move must've been a non-capturing move with either his rook or his king, so 1. 000 is an illegal try.
a) PRA: If the rook moved last: 1. Ra4 2. Rxh4[Ra1] 3. 000. If the king moved last: 1. g3 2. Kxf2[Ke1] 3. 000
b) Since you can't prove whether the king or the rook moved last, you have to reset the castling rights for both: 1. Ra4 2. Rxh4[Ra1] 3. Kxf2[Ke1] 4. 000
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| (3) Posted by Andrew Buchanan [Thursday, Oct 28, 2021 16:56] |
Cool!
I was a bit confused by the stipulation, which said ser-00, and so I thought that the castling goal must be kingside, for which the solution is not unique.
Another thought: if it was consequent seriesmover, then it would be 1. g3,g4 2. 0-0-0.
Thanks,
Andrew
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| (4) Posted by Jacques Rotenberg [Thursday, Oct 28, 2021 19:55] |
It seems possible to remove the Bg8
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| (5) Posted by Joost de Heer [Thursday, Oct 28, 2021 22:44] |
QUOTE
It seems possible to remove the Bg8
Then f3xBg4[Pg2] could've been the last white move.
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| (6) Posted by Joost de Heer [Thursday, Oct 28, 2021 22:47] |
QUOTE
I was a bit confused by the stipulation, which said ser-00, and so I thought that the castling goal must be kingside
Popeye uses '00' for castling as a goal, either king- or queenside castling.
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| (7) Posted by Andrew Buchanan [Friday, Oct 29, 2021 03:02] |
Thank you Joost: I didn't know about the "00" notation.
PRA is defined purely only for castling and e.p. Like you, I think that it could be usefully generalized to other conditional situations (e.g. Fuddled Men). However it's not clear to me that it could be successfully redefined for constituent sub-parts of castling rights. At which level would one now analyze a random PRA problem? Do these different levels of analysis give the same answer? Do the sub-parts decompose into smaller fragments themselves? (E.g. which square did wK come from, if it did just move?) For a moment, I just caught a horrifying glimpse of some Feynmann-type path integral analysis being required to solve any PRA retro, and it frightened me a little. I feel better now :)
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| (8) Posted by Jacques Rotenberg [Friday, Oct 29, 2021 09:46] |
..."Then f3xBg4[Pg2] could've been the last white move."...
two remarks :
a) you can transfer, say g6, to f3
b) you don't need to do so :
before f3xBg4[Pg2] what did black play ? not a capture, because no black piece is on its rebirth square, also not a move of the rook f2 (illegal retro check), so... what was the previous move of white ? necessarily a move of the king or of the rook.
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| (9) Posted by Joost de Heer [Friday, Oct 29, 2021 10:34] |
Last move could also have been KxB anywhere, freeing up the king for retromoves.
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| (10) Posted by Nikola Predrag [Friday, Oct 29, 2021 10:40] |
-"necessarily a move of the king or of the rook"-
Yes, without bBg8, various captures by wK would bring it to e1 with regained castling right, e.g. wKf8xbBg8->wKe1.
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| (11) Posted by Jacques Rotenberg [Friday, Oct 29, 2021 10:45] |
ok! I missed this.
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| (12) Posted by Hauke Reddmann [Friday, Oct 29, 2021 11:21] |
@Andrew: Nothing horrible with Feynman path integrals if you use the knot-theory graphic notation :P
BTW, you were not the first interpreting PRA as "retro twins", and indeed any
underdetermined combination of board and stipulation could be interpreted
as a set of twins. (man Kriegsspiel) For a prominent instance, consider
Fischer 960.
 (= 6+3 )
#2???
This is just a demo, to be valid either for all 960 positions there had to be a solution or a retro setup would prove that at least one of OO or OOO is possible. But I have faith in our retro geniuses :-)
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| (13) Posted by Jacques Rotenberg [Friday, Oct 29, 2021 13:24] |
anyhow, there is perhaps a way to show it lighter :
 (= 3+13 )
is it right ?
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| (14) Posted by Sarah Hornecker [Friday, Oct 29, 2021 14:13] |
So the last move was Kb7xRa8, and I just can castle now.
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| (15) Posted by Jacques Rotenberg [Friday, Oct 29, 2021 14:31] |
All the missing white pieces were taken by the black pawns
All the missing black pieces were taken by the white pawns : 3 captures by white pawns a and b to get to the c file, promote, and be taken by a black pawn
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| (16) Posted by Joost de Heer [Saturday, Oct 30, 2021 16:04] |
Another idea with a bit more variation:
 (= 3+16 )
ser-castling in how many moves? Anticirce
a) with PRA
b) without PRA
White's last move can't have been a capture and black can't uncapture a piece, so either white's rook or white's king moved last
a)
- If the rook moved last, we can assume the king still has castling rights. 1. Ra3 2. Rxh3[Rh1] 3. 00
- If the king moved last, we can assume the rook still has castling rights. 1. c3 2. Kxd2[Ke1] 3. 000
b)
Since we don't know which piece moved last, we have to reset the castling state for both the rook and the king: 1. Rb1 2. Rxb4[Ra1] 3. Kxd2[Ke1] 4. 000
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| (17) Posted by Jacques Rotenberg [Sunday, Oct 31, 2021 03:05] |
A matter of taste. This way of showing the idea looks better
The same trick as above should give a lighter diagram.
For instance :
 (= 3+13 )
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| (18) Posted by Joost de Heer [Sunday, Oct 31, 2021 07:06] |
No, because now bxc[Pc2] is possible as last mvoe.
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| (19) Posted by Jacques Rotenberg [Sunday, Oct 31, 2021 08:55] |
ok
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| (20) Posted by Jacques Rotenberg [Sunday, Oct 31, 2021 09:39] |
a simple way would be to add 2 black pieces with one capture less for black, it still wins a piece.
another way would be something as follows :
 (= 3+13 )
ser-castling in how many moves? Anticirce
a) with PRA
b) without PRA
a)
- If the rook moved last, 1. Ra3 2. Rxh3[Rh1] 3. 00
- If the king moved last, 1. g3 2. Kxf2[Ke1] 3. 000
b)
We don't know which piece moved last, 1. Ra4 2. Rxh4[Ra1] 3. Kxf2[Ke1] 4. 000
it would be slightly better with the two solutions of a) castling on the same side, and the solution of b) on the other side.
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