Tagged Questions
6
votes
1answer
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How does a birthday attack on a hashing algorithm work?
A "normal", brute-force attack on a cryptographic hashing algorithm $H$ should have a complexity of about $2^{n}$ for a hash algorithm with an output length of $n$ bits.
That means it takes about ...
8
votes
2answers
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Why would you expect to find a collision in a hash function after approximately $\sqrt{n}$ hashes?
I can't get an intuitive understanding of why it's $2^{(\frac{n}{2})}$ and not $2^n$, where $n$ is the number of bits of which the key consists.
