All \"experts\" keep copying and pasting the same solution for #2 for this probl

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All "experts" keep copying and pasting the same solution for #2 for this problem, which is not what I'm asking for. I want the solution to #1. Don't be a lazy idiot and paste that solution, because it will obviously be for the wrong problem. I am actually trying to understand how to do #1.

3 Shift Cipher Let N> 1 be any integer. Shift cipher mod N is an encryption scheme which works as follows. The key space of this encryption scheme is K-ZN , so key k is a random element in ZN. The message space are any bitstrings, encoded base-N, i.e. the message m is interpreted as rn-(mn-1 .. . mimo) N , where mn-1 , , mo are all elements of ZN, 1.e. they are digits base N.1 The encryption E and decryption D algorithms are as follows: If m_(mn-1 mimo)N then E(k, m)-(Cn-1 C1C0)N, where ci mit k mod N If c(cn-1...c1CO)N then D(k, c)- (mn-1...mimo)N where m c- k mod N An encryption is perfectly secret on message space M and ciphertert space C if it is the case that for every ciphertext cEC and every message mEM there exists exactly one key k ? K st. c is a sufficient one.2) E(k, m). (This is not a necessary condition for perfect secrecy to hold, but it Answer the following questions: 1. Does D(k, E(k, m))- m for all k and m? Argue why or why not.

Explanation / Answer

YES!

D(k, E(k,m)) = m for all k and m.

This is because of the fact that our keyspace is also modulo N.

Let us do this systematically,

D(k, E(k,m)) = D(k, ((mn-1+k)(mn-2+k) ... (m0+k)) modulo 2N )} (note that here we are taking this operation in modulo 2N. We are doing so because both mi's and k is modulo N and hence each of the mi+k is less than 2N and can be identified uniquely.)

In the second step perform the decryption on the same line.

D(k, E(k,m)) = ((mn-1+k-k)(mn-2+k-k) ... (m0+k-k)) modulo 2N

= (mn-1mn-2 ... m0) modulo 2N

= ( (mn-1mn-2 ... m0) modulo N) (this is obvious because all the mi's are less than N.

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