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Say I have a hash algorithm, and it's nice and smooth (The odds of any one hash value coming up are the same as any other value).
Now say that I know that the odds of picking 2 hashes and there being a collision are (For arguments sake) 50000:1.
Now say I pick 100 hashes. How do I calculate the odds of a collision within that set of 100 values, given the odds of a collision in a set of 2?
What is the general solution to this, so that I can come up with a number of hash attempts after which the odds fall below some acceptable threshold? E.g. I can say things like "A batch of 49999 hash value creations has a high chance of collision".
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The net probability of collision is derived by multiplying them together. The method for calculating probability of collision is appropriate for most encounters that occur in near-circular orbits. 1r + σ2 2r.
Conclusions. So the probability of no collisions is exp(-1/2) or about 60%, which means there's a 40% chance of at least one collision. As a rule of thumb, a hash function with range of size N can hash on the order of √N values before running into collisions.
One method for resolving collisions looks into the hash table and tries to find another open slot to hold the item that caused the collision. A simple way to do this is to start at the original hash value position and then move in a sequential manner through the slots until we encounter the first slot that is empty.
SHA256: The slowest, usually 60% slower than md5, and the longest generated hash (32 bytes). The probability of just two hashes accidentally colliding is approximately: 4.3*10-60. As you can see, the slower and longer the hash is, the more reliable it is.
This is a generalization of the Birthday problem.
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First calculate the probability that there is not a collision:
hashes_picked = 100
single_collision_odds = 50000
# safe_combinations is number of ways to pick hashes that don't overlap
safe_combinations = factorial(single_collision_odds) / factorial(single_collision_odds - hashes_picked)
# all_combinations is total number of ways to pick hashes
all_combinations = single_collision_odds ** hashes_picked
collision_chance = (all_combinations - safe_combinations) / all_combinations
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