You might be speaking in general terms, I'm talking about a specific test for full-board repetition due to triple ko. In this case the sequences that you are
comparing would be the same length. So the test would look something like this (forgive me if I screw up the indexes):
if (move[n-1]==move[n-7] && move[n-2]==move[n-8] && move[n-3]==move[n-9] &&
move[n-4]==move[n-10] && move[n-5]==move[n-11] && move[n-6]==move[n-12])
{
// We have seen this board before
// Let's abort the playout and score it as-as.
}
Note that I'm only talking about this in the context of playouts, so it is ok if a position repeated and this test didn't catch it. This is just a possible way
to shorten the playouts. I haven't tried this to see if it matters, I'm just thinking out loud here.
The question is, does the test above ever cause a false-positive? In other
words, is the first comment line ever incorrect?
A van Kessel wrote:
So it's possible to create a triple-ko repetition, take that move sequence and
find
a non-triple-ko situation that uses the exact same repeated move sequence ?
I am afraid I don't follow. Please rephrase.
In my words: you have a sequence of moves (M0) leading,
to a certain position (P0). After P0 , you continue with
a sequence of moves (M1) leading to position (P1)
Now if P0==P1, this means that the move leading to P1 (the last move of M1 is
invalid.
If the sequence M1 would occur anywhere inside of M0, it would cause
no harm, EXCEPT when it would be the final part of M0. But in that case,
P0 would itself would be a repetition of some position length(M1) *before*
P0.
Am I missing something ?
AvK
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