On Sun, May 30, 2010 at 3:11 AM, bgnj
<address@hidden> wrote:
Øystein Johansen wrote:
>
> Let's first try to get the number of legal positions below 2^64. The
> difference is "only" 8.190836056001331E+16 positions. I could divide this
> differently into contact and non-contact positions.
>
This is an interesting problem. I counted all the non-contact positions in
order to see if that would get the number of positions down to 64 bits but
unfortunately it did not.
Let's keep non-contact positions out and we're still not within 64-bit. However note that the Trice number also counts positions that is game over, and it also counts some obviously illegal positions. We can directly remove all positions where there move than 15 checkers on the bar, since these positions can never be reached. I've not calculated this number, but I guess is that it's not enough to make the grand total suitable for 64 bit.
MaX: I like your idea as well. (And I like your python code). 11 points each should be enough for any normal position, but remember this is theoretical discussion of only academic interest, and you solution is more like a pragmatic solution rather than a theoretical perfect solution. :-)
-Øystein
PS:
Here was my methodology:
1) Assume no pieces are on the bar for a non-contact position.
You can actually also assume there is no checkers on the 24 point, if you want to exclude "game over" positions from your count.
-Øystein