Math Problem Statement

Consider shift cipher So here, we have K = {0,..., 25) with Pr[k +K]= 1/26 for each key k. Now, we are given the following distribution over M: Pr[M=a] = 0.7 and Pr[M= z) = 0.3. What is the probability that the ciphertext is B?

Solution

To solve this problem, we will need to account for both the key distribution and the message distribution. This is essentially a case of calculating the probability that a specific ciphertext (B) is generated, given the key distribution and the distribution over the plaintext messages MM.

Step-by-step process:

1. Ciphertext generation formula:

In a shift cipher, the ciphertext CC is generated from the plaintext MM by shifting it by some key kk modulo 26. That is:

C=(M+k)mod26C = (M + k) \mod 26

Given that we are asked for the probability that the ciphertext is BB, let's translate BB into its corresponding numeric value:

  • B=1B = 1 (since A=0,B=1,C=2A = 0, B = 1, C = 2, etc.)

Now, we need to account for both possible plaintexts and the corresponding shift keys that could lead to a ciphertext of B=1B = 1.

2. Case 1: M=aM = a (where a=0a = 0):

  • The probability that M=aM = a is given as Pr[M=a]=0.7Pr[M = a] = 0.7.

  • We want C=B=1C = B = 1, which means: 1=(0+k)mod261 = (0 + k) \mod 26 This simplifies to k=1k = 1. So, the only key that satisfies this is k=1k = 1.

  • The probability of selecting this key k=1k = 1 is Pr[k=1]=126Pr[k = 1] = \frac{1}{26}.

Thus, the probability that M=aM = a and k=1k = 1 results in C=BC = B is: [ P(C = B | M = a) = Pr[M = a] \times Pr[k = 1] = 0.7 \times \frac{1}{26} ]

3. Case 2: M=zM = z (where z=25z = 25):

  • The probability that M=zM = z is given as Pr[M=z]=0.3Pr[M = z] = 0.3.

  • Again, we want C=B=1C = B = 1, so: 1=(25+k)mod261 = (25 + k) \mod 26 This simplifies to k=2k = 2. So, the only key that satisfies this is k=2k = 2.

  • The probability of selecting this key k=2k = 2 is Pr[k=2]=126Pr[k = 2] = \frac{1}{26}.

Thus, the probability that M=zM = z and k=2k = 2 results in C=BC = B is: [ P(C = B | M = z) = Pr[M = z] \times Pr[k = 2] = 0.3 \times \frac{1}{26} ]

4. Total probability:

Finally, we sum the probabilities from both cases to find the total probability that the ciphertext is BB:

P(C=B)=P(C=BM=a)+P(C=BM=z)P(C = B) = P(C = B | M = a) + P(C = B | M = z) Substituting the values: P(C=B)=0.7×126+0.3×126P(C = B) = 0.7 \times \frac{1}{26} + 0.3 \times \frac{1}{26} P(C=B)=0.7+0.326=126P(C = B) = \frac{0.7 + 0.3}{26} = \frac{1}{26}

Thus, the probability that the ciphertext is BB is 126\frac{1}{26}.


Let me know if you'd like further clarification or have additional questions!

Here are 5 related questions:

  1. How does the key distribution affect the overall probability in shift ciphers?
  2. How would the probability change if the key space were larger?
  3. What is the probability of obtaining any other letter as the ciphertext?
  4. How would the solution change if the message distribution were uniform?
  5. Can this approach be generalized to other ciphers, and how?

Tip: When working with modular arithmetic, always double-check edge cases where values might "wrap around" (e.g., going from 25 back to 0 in the alphabet).

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Math Problem Analysis

Mathematical Concepts

Probability
Modular Arithmetic
Cryptography

Formulas

Ciphertext C = (M + k) mod 26
P(C = B) = P(C = B | M = a) + P(C = B | M = z)
P(C = B | M = a) = Pr[M = a] × Pr[k = 1]
P(C = B | M = z) = Pr[M = z] × Pr[k = 2]

Theorems

Bayes' Theorem
Modulo Operation

Suitable Grade Level

Grades 11-12, Undergraduate Level