The Birthday Problem in .NET

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The Birthday Problem
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Suppose that Trudy is in a room containing a total of N people (including herself). What is the probability that at least one of the other N - 1 people has the same birthday as Trudy Assuming that birthdays are uniformly distributed among the 365 days in a year, the answer is not difficult to compute. As with many discrete probability problems, it is easier to compute the probcomplement is that none of the other N - 1 people have the same birt,hday as Trudy. For each person this probability is 364/365, so that for all N - L people, the probability is (364/365)N-1. Consequently, the probability we want is
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1 - (364/365)Ne1.
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ability of the complement and subtract t h e result from one. In this case, the
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(5.1)
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By setting (5.1) equal to 1/2 and solving for N , we can find the number of people that must be in a room before we expect someone to have the same birthday as Trudy. Doing so, we find that if N 2 254, then the probability is greater than 1 / 2 and we therefore expect to find someone with the same birthday as Trudy. Intuitively, the answer should be about the number of days in a year, and since there are 365 days in a year, the answer 254 is reasonable. Note that in this version of the birthday problem we are comparing every birthday to one specific birthday, namely, Trudy s. Also note that, more
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5.2 BIRTHDAYSA N D HASHING
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generally, if there are 111 possible outcomes, we expect to need about 11 1 comparisons before we find a collision . On the other hand, suppose that we want to find the probability that any two (or more) people in a room share the same birthday, where there are N people in the room. It is again easier to compute the probability of the complement and subtract from one. Here, the complement is that all people have different birthdays, so that the desired probability is given by
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1 - 365/365 .364/365 .363/365.. (365 - N
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+ 1)/365,
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(5.2)
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provided that N 5 366. In this case, to find the number of people that must be in the room before we expect two or more to share the same birthday, we set (5.2) equal to 1/2 and solve for N . Doing so, we find that if N 2 23, then the probability in (5.2) is greater than l/2. That is, provided that there are at least 23 people in a room, we expect two or more to share the same birthday. This fact is sometimes referred to as the birthday paradox because, at first glance, it appears paradoxical that only 23 people suffice when there are 365 days in a year. However, this result is not as paradoxical as it might seem. We are comparing every birthday to every other birthday, so with N people in the room, we are making ):( comparisons, and once we have made about 365 comparisons, we expect to find a match. By this logic, the solution to this version of the birthday problem is the smallest value of N for which
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which yields N = 28. This is close to the precise value of N = 23. As an approximation we often use where M is the number of possible outcomes. For actual birthdays, 111 = 365 and we have = 19, which is indeed a good approximation to the precise result N = 23.
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Birthday Attacks on Hash Functions
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Recall that a cryptographic hash function must provide weak collision resistance and strong collision resistance. If we are given a particular hash value, h ( z )and we can find a w such that h ( w ) = h ( z )then we have broken the hash function, since we have violated the weak collision resistance property. The brute force attack is to randomly generate w, compute the hash and compare the result to h ( z ) ,repeating until a collision is found. If the hash function h generates an n-bit output, the first version of the birthday problem discussed above implies that we will need to compute about 2n hashes before we expect to find such a w. Therefore, for h to be considered secure, it is necessary (but not sufficient) that it is infeasible for Trudy to compute 2 hashes. This is comparable to an exhaustive search for a cryptographic key.
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