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please explain the steps to get to the answer no other info was given We define

ID: 3835393 • Letter: P

Question

please explain the steps to get to the answer

no other info was given

We define the simple rotation cipher R by using a shift cipher with shift 0 on the first character in the plaintext, then a shift cipher with shift 1 on the second character in the plaintext, and so on, increasing the shift by 1 each time as we move along the plaintext. Thus, for example, the cipher R encodes plaintext ROTATION as RPVDXNUU. We define a general rotation cipher by repeating the simple rotation, so R^2 = RR, R^3 = RRR, etc., and R^26 = RA0 gives no change in the plaintext. Explain why in a long message encoded by the simple rotation cipher, any given letter in the plaintext will be encoded as a given letter in the about 1/26th of the time. Explain why this means that the resulting will have roughly equal frequencies for each letter. Do the assertions in (a) hold if we replace R by R^k, where k is not divisible by 2 or 1 3? What happens if we use R^k with k divisible by 2 but not 13? What happens if we use R^k with k divisible by 1 3 but not 2? The following has been produced by first applying a simple substitution cipher to the plaintext and then applying either R^3 or R^23 to the result. Decide which of the two general rotation ciphers was used, and then recover the plaintext. The work is made easier by the fact that spaces between words before rotation are replaced by "R". Identify any letters that you think might be transmission errors, and dive the needed corrections to the.

Explanation / Answer

(a) In a given long message, the probability of occurence of R at each position is 1/26. Since in case of simple rotation only R has shift 0, R is not changed to any other letter. Hence, it remains as plaintext even after encoding the message. All other letters are shifted by at least 1 and hence they are changed. Thus,any letter will have the same encoding 1/26th time. Hence the frequency of each letter will be same.

Each letter has equal probability of occurence. Since each letter is shifted by the same letter every time, the encoding of each letter remains same. hence the frquency

(b) For each letter, the number of shifts remain same. Fore example in R^3, R is shifted 0 times thrice, S is shifted 1,2,4 times(S becomes T, then T is shifted twice and becomes V shifts 4 times to become Z), every time. This holds true for any value of k and hence the assertion in (a) is true for R^k