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What's the probability for the clash for the md5 algorithm? I believe it is extremely low.
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MD5: The fastest and shortest generated hash (16 bytes). The probability of just two hashes accidentally colliding is approximately: 1.47*10-29.
As of 2010, the CMU Software Engineering Institute considers MD5 "cryptographically broken and unsuitable for further use", and most U.S. government applications now require the SHA-2 family of hash functions. In 2012, the Flame malware exploited the weaknesses in MD5 to fake a Microsoft digital signature.
Though organizations have nearly two years to move away from SHA-1, the MD5 deprecation date arrives on February 11. Originally published by renowned cryptographer Ron Rivest in 1992, MD5 has long been considered too weak to be used for digital certificates and signatures.
A major concern with MD5 is the potential it has for message collisions when message hash codes are inadvertently duplicated. MD5 hash code strings also are limited to 128 bits. This makes them easier to breach than other hash code algorithms that followed.
You need to hash about 2^64 values to get a single collision among them, on average, if you don't try to deliberately create collisions. Hash collisions are very similar to the Birthday problem.
If you look at two arbitrary values, the collision probability is only 2-128.
The problem with md5 is that it's relatively easy to craft two different texts that hash to the same value. But this requires a deliberate attack, and doesn't happen accidentally. And even with a deliberate attack it's currently not feasible to get a plain text matching a given hash.
In short md5 is safe for non security purposes, but broken in many security applications.
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It generates a 128-bit value. The accidental clash rate should therefore be 2-64 (because of the Birthday Paradox).
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