Question

This problem explores the use of a one-time pad version of the Vigenère cipher. In thisscheme, the key is a stream of random numbers between 0 and 26. For example, if the key is 319 5 . . . , then the first letter of plaintext is encrypted with a shift of 3 letters, the second with ashift of 19 letters, the third with a shift of 5 letters, and so on.a) Encrypt the plaintext sendmoremoney with the key stream9 0 1 7 23 15 21 14 11 11 2 8 9b) Using the ciphertext produced in part (a), find a key so that the cipher text decrypts to theplaintext cashnotneeded

Answer 1

To encrypt the plaintext sendmoremoney with the key stream 9 0 1 7 23 15 21 14 11 11 2 8 9, we use the Vigenère cipher. The encrypted ciphertext is ZSVPYQPAIHZWC. To find the key that decrypts the ciphertext cashnotneeded, we apply the reverse process of the Vigenère cipher.

Step-by-step explanation:

To encrypt the plaintext sendmoremoney with the key stream 9 0 1 7 23 15 21 14 11 11 2 8 9, we will use the Vigenère cipher. The Vigenère cipher is a polyalphabetic substitution cipher that uses a keyword to determine the shifting of each letter in the plaintext. Each letter in the keyword corresponds to a number from 0 to 26, which represents the shift. For example, 'A' is shift 0, 'B' is shift 1, and so on. To encrypt, we can follow these steps:

1. Convert the plaintext into its corresponding numerical representation. For example, A=0, B=1, C=2, and so on.
2. Add the corresponding number from the key stream to each numerical representation of the letters in the plaintext.
3. Take the modulo 26 of each sum to ensure it falls within the range of 0-25.
4. Convert the resulting numbers back into their corresponding letters using the reverse mapping.

Using this process, we can encrypt the plaintext sendmoremoney with the key stream 9 0 1 7 23 15 21 14 11 11 2 8 9 to obtain the ciphertext ZSVPYQPAIHZWC.

To find a key that decrypts the given ciphertext ZSVPYQPAIHZWC into the plaintext cashnotneeded, we can use a similar process in reverse:

1. Convert the ciphertext into its corresponding numerical representation.
2. Subtract the corresponding number from the key stream from each numerical representation of the letters in the ciphertext.
3. Take the modulo 26 of each difference to ensure it falls within the range of 0-25.
4. Convert the resulting numbers back into their corresponding letters using the reverse mapping.

By applying this process to the given ciphertext ZSVPYQPAIHZWC, we can find a key that decrypts it to the plaintext cashnotneeded.

Answer 2

Step-by-step explanation:

Take a as the 0th index and z as the 25th index.

send shift with key 9017 will give

s=b

e=e

n=o

d=k.

s shift 9 places lands the letter b.

Similarly for more money, the result is jdmsxzpmh.

the ciphertext is "beokjdmsxzpmh"

b)

take the first 4 ciphertext

"beok".

Calculate the shift from b to s which is 17 shifts. thus beok can be decrypted with key 17 0 24 19 to get "send"

thus the resulting key is 17 0 24 19 3 11 5 12 15 14 2 8 17