6. [10 points] One way to exchange secret messages between two parties is to use so-called permutation ciphers. We let some non-identity permutation ? of the index set In- 11, 2, ..., n} be our key. Given a message, called a plaintext, we break it down into n-symbol blocks, and then permute n symbols in each block using our key , which produces an encrypted message, called a ciphertext. To decrypt such a ciphertext, we break it down into n-symbol blocks, and then permute n symbols in each block using the inverse ?. Prior to encryption, however, we need to ensure that the number of symbols in our original message is a multiple of n, which can be accomplished by simply appending as many extra (dummy) symbols as necessary For example, suppose our plaintext is "MEET.2PM". We shall use I3 {1, 2, 3 and the key ? such that ?(1)-3, ?(2) = 1, and ?(3)-2. Consequently, the inverse of our key is as follows: ?-1(1) = 2, ?-1(2) = 3, ?-1(3) = 1, we break our plaintext into 3-symbol blocks: "MEE', , "T. 2", and "PM-", where "-" is an extra symbol to make the message length a multiple of 3. During encryption, we treat each block a 3-symbol plaintext tuple (x1, X2, X3) and apply t to its indices, which yields the corresponding 3-symbol ciphertext tuple (X3, XI, x2). As a result, we obtain "EME", "27," and "-PM", i.?., our ciphertext is "EME2T.-PM". During decryption, we break "EME2T.-PM', into 3 symbol blocks, treat each block a 3-symbol ciphertext tuple (Yu y2, ys), and apply -1 to its indices, which yields the corresponding 3-symbol plaintext tuple (??, ??, y). Thus, we get "MEE", "T.2", and "PM-", which is our original plaintext "MEET. 2PM-". Now, given I-1, 2, 3, 4, 5} and the plaintext "ECE242.A1.DUE.0523", obtain the ciphertext using ? such that ?(1) 3, ?(2) 4, ?(3)-5, ?(4) 2, and ?(5) = 1. Verify that applying ?-1 to the ciphertext produces the original plaintext. 2