3 ms·
Not an expert but this provably doesn’t add security. In your example, if sha1(foo)==sha1(bar), then foo, bar have a collision regardless of what you do afterw
by wave_function 4y ago
Not an expert but this provably doesn’t add security.
In your example, if sha1(foo)==sha1(bar), then foo, bar have a collision regardless of what you do afterwards.
Maybe there’s some fancier layering scheme that adds security?