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| | | (1) Posted by Kevin Begley [Monday, May 10, 2010 13:39] | rules question
Maybe a rules expert can answer this question...
Does an equivalent rotation of a (pawnless) position count as a repetition?
| | | (2) Posted by Hauke Reddmann [Monday, May 10, 2010 17:32] |
No. Nothing about "equivalent" in the FIDE rules,
positions must be identical, including ep and OO.
(Identical, of course, includes two identical
pieces switching place. man Bose-Einstein :-)
Hauke
| | | (3) Posted by Ian Shanahan [Tuesday, May 11, 2010 15:01] |
Moreover, the position is regarded as "repeated" even if the move-potential is different - as in 000 being possible in the first position, but not after the relevant King has moved then switched back to his home square (all else unchanged)!
| | | (4) Posted by Bojan Basic [Tuesday, May 11, 2010 15:39] |
QUOTE
Moreover, the position is regarded as "repeated" even if the move-potential is different - as in 000 being possible in the first position, but not after the relevant King has moved then switched back to his home square (all else unchanged)!
Hauke has already said that this is not correct, and here is the relevant paragraph from the official rules (bold is mine):
QUOTE
The game is drawn upon a correct claim by the player having the move, when the same position, for at least the third time (not necessarily by a repetition of moves):
a. is about to appear, if he first writes his move on his scoresheet and declares to the arbiter his intention to make this move, or
b. has just appeared, and the player claiming the draw has the move.
Positions as in (a) and (b) are considered the same, if the same player has the move, pieces of the same kind and colour occupy the same squares, and the possible moves of all the pieces of both players are the same. Positions are not the same if a pawn that could have been captured en passant can no longer be captured in this manner. When a king or a rook is forced to move, it will lose its castling rights, if any, only after it is moved.
| | | (5) Posted by Sarah Hornecker [Tuesday, May 11, 2010 16:49] |
This was the case when Nenad Petrović made his famous problem, though.
 (= 7+12 )
Nenad Petrović
The Problemist 1959
1st prize
Mate in 8
So while the problem today is correct, it was incorrect at the time when it appeared.
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