Code: Select all
a1 += ~(h1->type - exact) & 1;
a2 += ~(h2->type - exact) & 1;
a3 += ~(h3->type - exact) & 1;
a4 += ~(h4->type - exact) & 1;
Moderators: hgm, Rebel, chrisw
Code: Select all
a1 += ~(h1->type - exact) & 1;
a2 += ~(h2->type - exact) & 1;
a3 += ~(h3->type - exact) & 1;
a4 += ~(h4->type - exact) & 1;
But how did you test that? From the graph you can see the equi-distributed draft did 25% better than equidistributed only at an overload of a factor 100. How much was the overload factor in your Elo measurement?cdani wrote:Of the examples shown in this thread, I tried hgm approach but was maybe 2 elo worst.
Code: Select all
a1 += (h1->type & 1);
...
Copying exactly your code. The other things I have done is to initialize to 0 the little hashCount array and to treat the quiescence/static eval entries. Maybe I did not understood something? Thankshgm wrote:But how did you test that? From the graph you can see the equi-distributed draft did 25% better than equidistributed only at an overload of a factor 100. How much was the overload factor in your Elo measurement?cdani wrote:Of the examples shown in this thread, I tried hgm approach but was maybe 2 elo worst.
Yes thanks. I have yet to optimize the whole function.stegemma wrote:Sorry, if exact is bit 0, it must be:
Code: Select all
a1 += (h1->type & 1); ...
Well, you did not understand my question.cdani wrote:Maybe I did not understood something? Thanks
Ah!hgm wrote:Well, you did not understand my question.cdani wrote:Maybe I did not understood something? Thanks
I asked how large your hash table was (number of entries), and how large your typical search tree (time per move and nps).
Mmmm... The worst attemps where clearly outside the error margins, so there was some big influence of the changes. Anyway I will try again the two or three best ideas with maybe 16 or 32 MB. Thanks!hgm wrote:Well, if I do the math it seems that your typical search tree measures at most about 1.7M nodes, and later in the game only 0.23M nodes. (I don't know if you count hash cutoffs in this). The hash table is 4 to 40 times larger.
So basically you are trying to measure the differences between replacement schemes under conditions where virtually no replacement takes place...