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6
votes
1answer
966 views

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
193 views

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.
10
votes
1answer
337 views

Security of N bit HMAC

Lets say that I am using 128 bit HMAC. How many operations are needed to find "non secure" message. Is birthday attack possible?