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⇤ ← Revision 1 as of 2005-12-04 22:42:03
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| ||Birthday Paradox ||||<(> Let [[latex2($P(n,k)$)]] be the probability that at least one duplicate in ''k'' items exists when each item can take on one of ''n'' different values. So [[latex2($P(365,10)$)]] is the probability that we have a duplicate in 10 peoples birthdays. It is easier to calculate the number of ways we can not have a duplicate [[latex2($N=365 \times 364 \times ... \times (365-k+1) = \frac{365!}{(365-k)!}$)]]. The number of ways with duplicates is [[latex2($N_D = (365)^k$)]]. Then the probability of at least one duplicate is [[latex2($1-\frac{365!}{(365-k)!(365)^k}$)]]|| |
Terms
Authenicator |
Additional information appended to a message to enable the receiver to verify that the message should be accepted as authentic. The authenticator may be functionally independent of the content of the message itself (e.g., a nounce or a source identifier) or it may be a function of the message contents (e.g., a hash value or a cryptographic checksum). |
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Birthday Attack |
When short hash codes are used it is not difficult to find two messages that produce the same hash code, but have slightly different meanings - p332-333. |
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Birthday Paradox |
Let latex2($P(n,k)$) be the probability that at least one duplicate in k items exists when each item can take on one of n different values. So latex2($P(365,10)$) is the probability that we have a duplicate in 10 peoples birthdays. It is easier to calculate the number of ways we can not have a duplicate latex2($N=365 \times 364 \times ... \times (365-k+1) = \frac{365!}{(365-k)!}$). The number of ways with duplicates is latex2($N_D = (365)^k$). Then the probability of at least one duplicate is latex2($1-\frac{365!}{(365-k)!(365)^k}$) |
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