Suppose h(m) is a hash function that maps a message of arbitrary bit length into an n-bit hash value. You are presented with two values xandy, x≠y, for which h(x) =h(y). Does this mean that h is not collision resistant? Explain your answer.
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Answer:Just because a collision has occurred, doesn't mean that the hash function is not collision resistant. It is impossible to have a hash function with 0 collisions, so as long as the collision rate is quite low, the hash function is collision resistant.
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