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For any given pair of people the possibly of them

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Unformatted text preview: ntire year. For any given pair of people, the possibly of them having the same birthday is 1/m = 1/365 ; them Probability of zero collision Probability = Probability that all of the n people have different birthdays = m * (m-1) * (m-2) *…* (m-n+1) / mn (m-1) = 1 – n(n-1)/2m approximately when m>>n n(n-1)/2m Pcollision = 1 – Probability of zero collision = n(n-1)/2m approximately n(n-1)/2m How difficult to find a Hash collision ? How secure is a one-way hash with 64-bit output ? Based on the property of a good hash function, the hash Based output of any input string should be uniformly distributed over the hash output space of size m=264 the This is analogous to the fact that the birthday of any given This person is uniformly distributed over any days within a year (i.e. output space of size m = 365) Thus, according to the Birthday Paradox, if no. of all possible Thus, outcomes = m, we only need to try about n = √ m inputs to we the hash function to have a good chance to f...
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