6 ms·
Those should only differ by a factor of 2 [the collision rate for an image is the number of collisions it has divided by the number of other images, so the aver
by LodeOfCode 5y ago
Those should only differ by a factor of 2 [the collision rate for an image is the number of collisions it has divided by the number of other images, so the average collision rate is the total collisions divided by n(n-1) vs. n(n-1)/2 pairs] which isn't particularly relevant at this scale
- SilasX 5y agoYou care about the total number of collisions, not collisions for a specific image, so they differ by a square -- hence the 1 in a trillion vs million difference.