5 replies. Last post: 2005-03-11Reply to this topic Return to forum
michael wrote in his analysis of his game against knox:
"Since winning parity means getting even amount of chains as red (first to move) odd amount of chains as blue (2nd to move) on 6x6."
Now look at the two links:
1.five chains – red to move
2.1.four chains – red to move
red is lost in both cases.
This question came to me by looking at a lost game where I believed to have won the parity with ceteris paribus.
Is the rule uncomplete or wrong? Didn’t I understand the rule?
Looking forward to replies. :)
In the second example there are three chains and one cycle – the cycles don’t affect the parity so there is an odd number of chains in both cases and blue wins.
Is there some change of parity, when red or blue lost one square ?
I’m not sure what you mean by lost one square.
The only time the parity changes is when you make a
“double-cross” move, that is, a single move that
completes two boxes. E.g. if you leave the last
two boxes of a long chain, then responding by
taking those two boxes is a double-cross move.
If a double-cross move happens before you reach
an endgame position, then you need to keep in mind
that the desired parity has been switched (also
any long chain that may have just been taken doesn’t
count as a long chain anymore).