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Because of the birthday attack, a collision in an n-bit hash function can be found with about how many hash computations?

2374Medium DSSSB TGT Computer Science 0 replies

Because of the birthday attack, a collision in an n-bit hash function can be found with about how many hash computations?

  • A2^(n−1)
  • B2^n
  • C2^(n/2)
  • Dn²
Correct answer: C

Explanation

By the birthday paradox, about 2^(n/2) random messages are enough to find two with the same hash, so a 64-bit hash gives a collision after roughly 2³² trials.

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