Suppose h (m) is a collision-resistant hash function that maps a message of arbitrary bit length into an n-bit hash value. is it true that, for all messages x, x' with x ≠ x', we have h (x) ≠ h (x') ? explain your answer.

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Computers & Technology

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24 February, 07:37

Suppose h (m) is a collision-resistant hash function that maps a message of arbitrary bit length into an n-bit hash value. is it true that, for all messages x, x' with x ≠ x', we have h (x) ≠ h (x') ? explain your answer.

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Louis Barron · Answer 1

That can't be true. Collision resistant just means the chance is really low, but not 0. Suppose you enumerate all possible hash values with each their different original message. Since the message length can be larger than n, you can then find a message whose hash is already in the list, ie., a collision!