The assertion " * No other X’ can have HASH(X’) equal to Y. Its one to one mapping." and the quiz question is unfortunately not correct. Not having collisions in hash calculation is impossible simply because the set of possible inputs is a superset of set of possible hashes. The probability can be reduced arbitrarily with the hash length. Hope it helps.
Hello @GULSEN_BAYIR
Yes, you are right. We have updated the question accordingly. Thank you for pointing it out.
Happy Learning
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