Never mind why I’m doing this — this is mainly theoretical.
If I were MD5 hashing string representations of integers, how high would I have to count before two of the hashes collide?
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Never mind why I’m doing this — this is mainly theoretical.
If I were MD5 hashing string representations of integers, how high would I have to count before two of the hashes collide?
This problem (in generic case) is known as Birthday Paradox
The probability of collision in generic case can be computed easily. However, in your particular case, you have to actually compute (and store!) each MD5.
EDIT @Scott : not really. The Pigeonhole principle (being just a particular case of Birthday problem) would say that having 2^128 possible MD5 values, we surely will have a collision after 1 + 2^128 tries. The birthday paradox says that the probability of collision will be grater than 0.5 for about 2^70 MD5 values.
With these estimates for storage requirements, it’s up to you to decide if the problem worth it. By me it does not.