color = color / blurval;
Is blurval between some known boudries? Are you always going to get one of the same, say, 100 values when you do the division on this line? If so, you could make an array of the possible answers to this division and store them in memory instead of doing the divide.
I tried looking at the for loop but it looks like some code was snipped out (there are more closing brackets at the bottom of your function than there are opens). So either something is missing, or i'm just *really* not awake yet enough to understand your for loop. If there is more code to it, there may be a way to "unroll" that second for() statement some and shave a few other instructions off. Anyway, hope this helps. --Nolan
[This message has been edited by nolan (edited November 08, 1999).]