密码学与网络安全-附录E

First Problem Probability: P = 1 − (1 − 1/N)k Sample size: k ≥ ln 2 × N
Second Problem
The second problem is the same as the first except that the predefined value is one of the samples. This means that we can use the result of the second problem if we replace k with k − 1 because after selecting one sample from the sample set only k − 1 samples are left. Therefore, P = 1 − (1 − 1/N)k−1 and k ≥ ln2 × N + 1.
Q = Q1 × Q2 × Q3 × … × Qk = 1 × (1 − 1/N) × (1 − 2/N) × … × (1 − (k − 1)/N) Q = (e−1/N) × (e−2/N) × … × (e−(k−1)/N) Q = e−k(k −1)/2N Q= e−k /2N
2
Using the approximation 1 − x ≈ e−x with x = i/N Using the relation 1 + 2 + … + (k − 1) = k(k − 1)/2 Using the approximation k(k − 1) ≈ k2
1 − x ≈ e−x 1 + 2 + … + (k − 1) = k (k − 1)/2 k(k − 1) ≈ k2 //Taylor’s series when x is small
E.1 FOUR PROBLEMS
We preseБайду номын сангаасt solutions to four problems discussed in Chapter 11.
P = 1 − (1 − 1/N)k ≥ 1/2 → (1 − 1/N)k ≤ 1/2 (1 − 1/N )k ≤ 1/2
→ (e−k/N) ≤ 1/2 Using the approximation 1 − x ≈ e−x with x = 1/N
(e−k/N ) ≤ 1/2 → ek/N ≥ 2 → k/N ≥ ln 2 → k ≥ ln 2 × N
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APPENDIX E
BIRTHDAY PROBLEMS
4. Finally, if P is the probability that at least one sample is equal to the predetermined value, then P = 1 − Q or P = 1 − (1 − 1/N)k. Sample Size Now we find the minimum size of the sample with P ≥ 1/2 to be k ≥ ln2 × N as shown below:
Sample size: k ≥ (2 × ln2)1/2 × N1/2
Fourth Problem
In the fourth problem, we have two samples of equal size, k. We need to find the minimum value of k, such that, with probability P ≥ 1/2, at least one of the samples in the first set has the same value as a sample in the second set. To solve the problem, we first find the corresponding probability P. We then set the probability to 1/2 to find the minimum size of the sample. Probability We solve this using a strategy similar to the one we used for the first problem: 1. According to the first problem, the probability that all samples in the first set have values different from the value of the first sample in the second set is Q1 = (1 − 1/N)k. 2. The probability that all samples in the first set have values different from the first and second samples in the second set is Q2 = (1 − 1/N)k × (1 − 1/N)k. 3. We can extend the logic to say that the probability that all samples in the first set have values different from any sample in the second set is
First Problem
We have a sample set of k values, in which each sample can take only one of the N equally probable values. What is the minimum size of the sample set, k, such that, with probability P ≥ 1/2, at least one of the samples is equal to a predetermined value? To solve the problem, we first find the probability P that at least one sample is equal to the predetermined value. We then set the probability to 1/2 to find the minimum size of the sample. Probability We follow four steps to find the probability P: 1. If Psel is the probability that a selected sample is equal to the predefined value, then Psel = 1/N because the sample can equally likely be any of the N values. 2. If Qsel is the probability that a selected sample is not equal to the predefined value, then Qsel = 1 − Psel = (1 − 1/N). 3. If each sample is independent (a fair assumption), and Q is the probability that no sample is equal to the predefined value, then Q = Qselk = (1 − 1/N)k.
Second Problem
− Probability: P = 1 − (1 − 1/N )k−1
Sample size: k ≥ ln 2 × N + 1
Third Problem
In the third problem, we need to find the minimum size, k, of the sample set, such that, with probability P ≥ 1/2, at least two of the samples have the same values. To solve the problem, we first find the corresponding probability P. We then set the probability to 1/2 to find the minimum size of the sample. Probability We use a different strategy here: 1. We assign probabilities to samples one at a time. Assume that Pi is the probability that the sample i has a same value as one of the previous samples and Qi is the probability that the sample i has a value different from all previous samples. a. Because there is no sample before the first sample, P1 = 0 and Q1 = 1 − 0 = 1. b. Because there is one sample before the second sample and the first sample can have one of the N values, P2 = 1/N and Q2 = (1 − 1/N). c. Because there are two samples before the third sample and each of the two samples can have one of the N values, P3 = 2/N and Q1 = (1 − 2/N). d. Continuing with the same logic, Pk = (k − 1)/N and Qk = (1 − (k − 1)/N).