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Diferente intre continut:
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 $p^(i[sqrt(n)] + 1)^ = qp^-j^ modulo n$.
Now the application of meet in the middle becomes obvious. We can brute force through the numbers on each side of the equality and find a match.
Using meet in the middle becomes obvious. We can brute force through the numbers on each side of the equality and find a colision.
The algorithm takes O(sqrt(n)) space and O(sqrt(n)) time.
h2. Bidirectional search(interview question)
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