[Retros] How many times could a Diagram be repeated during a proofgame?
raosorio at fibertel.com.ar
raosorio at fibertel.com.ar
Sun Jun 10 08:31:03 EDT 2007
Hi,
Werner Keym indicated to me (many thanks Werner) that this point was already discussed by Karl Fabel:
"in the booklet of Karl Fabel, "Am Rande des Schachbretts" (1948), there is an article on "Nicht-identische
Stellungen". Fabel calculates exactly the 22 moves as you do. He uses a position by Hans Kluever
(Schachspiegel, Jan. 1948) as an example:
r1b1k2r/7p/4p1pP/1p1pPpP1/pPpP1P2/P1P5/8/R3KB1R The last move was d7-d5. Kings and rooks have not yet moved. "
I t would be interesting to know the context of Fabel's discussion. Anyway, this is another example of
how indispensable is to build a site concentrating all the existing retro material in order to avoid the wheel
invention again and again.
Best,
Roberto Osorio
Hi,
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The point has been cleverly discussed. Here are my calculations,
a) How many times could the Diagram have been repeated during the proofgame?
In addition to "Who's to move" there are 6 different right status that make the positions
to be different on the same diagram (example),
1) 4 castlings + e.p.
2) 4 castlings
3) 3 castlings
4) 2 castlings
5) 1 castling
6) 0 castling
Status 1) is instantaneous, for instance after b7-b5, first time the diagram appears.
Status 2) to 6) are 5 that can be repeated twice both for WTM and BTM. Finally, status
6) can be repeated a third time reaching a draw position (as Guus set, a retro problem
can start from this situation). Then, the answer is 1 + 5x2x2 +1 = 22
b) How many retractions are needed to show these repetitions?
Let's describe it as forward play. First move is b7-b5, showing status 1) followed by
bishop switchbacks to reach the first time status 2), 1 + 2x2 = 5 single moves.
The maneuver to "squeeze" each status from 2) to 6) is: bishop switchbacks to repeat (4) +
tempo loss to change the turn to move (5) + bishop switchbacks to repeat (4) = 13
On the other hand, each status changing requires a rook switchback (4). Finally, the third
repetition of 6) requires an additional switchback (4).
- single moves to reach status 2) 5
- to squeeze the other 5 status 13x5= 65
- to change the status 4 times 4x4= 16
- to repeat the third time status 6) 4
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TOTAL 90
The 1 single move difference with the 89 calculated by Francoise and Guus is b7-b5. Perhaps
this is tricky but it's not the main point here. Let's say that under retro conventions the
e.p. right has to be shown to be taken into account.
I wold like to remark why I considered interesting to present this matter here. First, the
22 repetitions is a surprissing result. When I asked for an intuitive answer to skilled
problemists they were so far as to answer 8. Second, it's useful to verify that the 50 moves
rule is not in the way.
Best,
Roberto
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PRESENTATION
Let's define "Diagram" just as a board showing pieces on it, i.e., no matter at all anything regarding the right's status, not even who's to move. I present the following Diagram,
r3k2r/8/2b5/Pp6/5BpP/8/8/R3K2R (just an example with the mimimal material)
Questions:
a) How many times could the Diagram have been repeated during the proofgame?
b) How many retractions are needed to show these repetitions?
(all the above following the retro conventions)
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