4 ms·
>By taking advantage of the birthday paradox, and a collision search algorithm that let me search in n(log n) time instead of the naive n^2 n(log n)? I thought
by aaaaaaaaaaab 5y ago
>By taking advantage of the birthday paradox, and a collision search algorithm that let me search in n(log n) time instead of the naive n^2
n(log n)? I thought the birtday paradox lets you expect collisions after sqrt(n) tries…
- yeldarb 5y agoDifferent things; birthday paradox was testing them against each other, turning it into a sorting problem was the other.