Pagini recente » Atasamentele paginii Profil aurelmva | Atasamentele paginii Profil AlexProbaji | Atasamentele paginii Profil alynmuntean | Istoria paginii utilizator/stefanikrobertz | Diferente pentru blog/meet-in-the-middle intre reviziile 51 si 52
Nu exista diferente intre titluri.
Diferente intre continut:
bq. Given n a prime number and p, q two integer numbers between 0 and n-1 find k such that p^k^ = q modulo n.
This problem can be solved using the baby step, giant step algorithm which uses the meet in the middle trick.
We can write k = $i sqrt(n) + j$
We can write k = $i([sqrt(n)] + 1) + j$
Notice that $i <= sqrt(n)$ and $j <= sqrt(n)$.
Now our equality looks like this $p^(i ([sqrt(n)] + 1) + j)^ = q modulo n$.
We can divide by $p^j^$ and get $p^(i[sqrt(n)] + 1)^ = qp^-j^ modulo n$.
Nu exista diferente intre securitate.
Topicul de forum nu a fost schimbat.