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MatPlus.Net Forum Retro/Math Castling

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### Castling

(= 3+16 )

ser-00 in how many moves? Anticirce

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

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

It seems possible to remove the Bg8

QUOTE
It seems possible to remove the Bg8

Then f3xBg4[Pg2] could've been the last white move.

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.

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 :)

..."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.

Last move could also have been KxB anywhere, freeing up the king for retromoves.

-"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.

ok! I missed this.

@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 :-)

anyhow, there is perhaps a way to show it lighter :

(= 3+13 )

is it right ?

So the last move was Kb7xRa8, and I just can castle now.

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

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

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 )

No, because now bxc[Pc2] is possible as last mvoe.

ok

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.