How many trials does it take to break HMAC-MD5? – crypto.stackexchange.com

How many trials does it take to break HMAC-MD5? – crypto.stackexchange.com

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I know that you can find collision in MD5 with 2^64 trials using Birthday paradox. Now everyone is saying that HMAC-MD5 is significantly more secure. How can I quantify this security? My question is ...

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