On May 2, 2007, at 8:17 AM, Peter Drake wrote:
On May 2, 2007, at 8:07 AM, Erik van der Werf wrote:
Don't you determine the sum of visits by adding all values in the
children? I guess it looks like a nice speedup to get the sum
directly
from their parent, but does that really matter in
I'm contemplating making the change you suggest. The following is my
one concern.
Suppose, to keep the example simple, that there are only two choices
at each ply. My tree is originally
ROOT 0
meaning that there is just one node with no playouts.
In the first playout, my first move is A, so
On 5/7/07, Peter Drake [EMAIL PROTECTED] wrote:
When I
look through the children of A and count a total of one playout, it
seems natural that I should update the playout count for A:
I would normally try to save all simulation results, but I can certainly see
the value of forcing invariants
Hi,
see below
On 5/7/07, Peter Drake [EMAIL PROTECTED] wrote:
In the first playout, my first move is A, so then I have:
ROOT 1
A 1
Now I try move B, updating the tree to:
ROOT 2
A 1
B 1
Fine so far. Now UCT likes A better, so the next playout starts with
A, C,
