2 ms·
>We don't care that any two images have the same hash, Why not?
by SilasX 5y ago
>We don't care that any two images have the same hash,
Why not?
- snet0 5y agoBased on the pigeonhole principle alone, it will always be the case that collisions exist. The size of the digest is very likely smaller than the size of any given image.
- SilasX 5y agoHow is that relevant to whether “per image” or “Per image pair” is the right metric? Heck, how is it relevant to anything given that hash functions avoid collisions in practice, even given the pigeonhole principle?