Probability of SHA1 collisions

HashSha1Probability

Hash Problem Overview


Given a set of 100 different strings of equal length, how can you quantify the probability that a SHA1 digest collision for the strings is unlikely... ?

Hash Solutions


Solution 1 - Hash

alt text

> Are the 160 bit hash values generated > by SHA-1 large enough to ensure the > fingerprint of every block is unique? > Assuming random hash values with a > uniform distribution, a collection of > n different data blocks and a hash > function that generates b bits, the > probability p that there will be one > or more collisions is bounded by the > number of pairs of blocks multiplied > by the probability that a given pair > will collide.

(source : http://bitcache.org/faq/hash-collision-probabilities)

Solution 2 - Hash

Well, the probability of a collision would be:

1 - ((2^160 - 1) / 2^160) * ((2^160 - 2) / 2^160) * ... * ((2^160 - 99) / 2^160)

Think of the probability of a collision of 2 items in a space of 10. The first item is unique with probability 100%. The second is unique with probability 9/10. So the probability of both being unique is 100% * 90%, and the probability of a collision is:

1 - (100% * 90%), or 1 - ((10 - 0) / 10) * ((10 - 1) / 10), or 1 - ((10 - 1) / 10)

It's pretty unlikely. You'd have to have many more strings for it to be a remote possibility.

Take a look at the table on this page on Wikipedia; just interpolate between the rows for 128 bits and 256 bits.

Solution 3 - Hash

That's Birthday Problem - the article provides nice approximations that make it quite easy to estimate the probability. Actual probability will be very very very low - see this question for an example.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestioneastafriView Question on Stackoverflow
Solution 1 - HashPeterView Answer on Stackoverflow
Solution 2 - HashAnthony MillsView Answer on Stackoverflow
Solution 3 - HashsharptoothView Answer on Stackoverflow