[Retros] Example 3 with e.p. and repetition from Yefim Treger 08/22/2004
TregerYefim at aol.com
TregerYefim at aol.com
Sun Aug 22 09:26:50 EDT 2004
Once again and some new simple positions for this topic (3-times
repetition):
1. The first example was the game: 1.e4 e5 2.Qe2 Qe7 3.Qd1 Qd8 4.Ke2 Ke7
5.Kd1 Ke8 - all agreed there is no 3-repetition for position after 1.e4 e5
2. The second example was: White: Ke1;Bb2, Rh1 (castle is saved) Black:
Kd6 pawn e3. Black's last move was 1...e2 after which a game goes as follows:
2.Bc3 Kc7 3.Bb2 Kd6 4.Rh2 Kc7 5.Rh1 Kd6. The problem is in 3-times repetition
for
position(s) after 1...e2; after 3....Kd6 and after 5...Kd6. When Rook had
moved nobody argued that right of castle is lost, but positions after 1...e2 and
3...Kd6 can
be understood for chess rules ambiguously. One can agree that if King (Rook)
had not been moved then castle is always possible (but it is not math
correct?) or use logic, math and retro knowledge and (the second opinion):
Count that the given above positions are the same because mathematically
speaking a tree after them is the same and the whole set of possible games is
the same! That is a good definition for equality of positions (what others
think?) but
I will give new example about en passant.
3. Position (picture is below if any) White:Ka5 p. d5 Black: Kf8, Qg5;p.c7;
Black's move which is 1...c7-c5. Assuming that this position pretends to be
repeated White played next as game follows: 2. Ka4 Qg4 3.Ka5 Qg5 (second time?)
4.Ka4 Qg4 5.Ka5 Qg5 -third time! Can he claim a draw?
That is math simple logic: all positions are the same, but I feel that
somebody
thinks in other way, counting that after 1...c7-c5 position has en passant.
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