Question: Cyber Security Using the extremely simplified ECB example of a substitution and transposition cipher given in the lecture on ciphers week 8, complete the following

Cyber Security

Using the extremely simplified ECB example of a substitution and transposition cipher given in the lecture on ciphers week 8, complete the following cipher rounds (be sure to use XOR, not OR). Supposing we begin with the plaintext byte: lower case s = 01110011 in ASCII binary as follows: Cyber Security Using the extremely simplified ECB example of a substitution and

1. Show the next full substitution and transposition round using the information above if the next subkey is 1110:

transposition cipher given in the lecture on ciphers week 8, complete the

3. Now, ignoring the previous round, say we are on the 16th round and we have the following plaintext byte in binary:

following cipher rounds (be sure to use XOR, not OR). Supposing we

And if you have the following subkey 1010, complete the round:

begin with the plaintext byte: lower case s = 01110011 in ASCII

4. What ASCII character does the above round finally represent: ______ 5. Briefly describe in a few paragraphs what DES is, and how it works (include S-boxes):

We will add a mangler (sort of an S-box) function to our simple cipher. The mangler (a sort of S-box) will take two bits starting at the left and rotate them horizontally, then mash the left and right halves together.

As follows You have a plaintext character lowercase m: 01101101 You have a key character Uppercase Y: 01011001 Split the key into two subkeys: 0101 1001, and complete two rounds with the subkeys including the mangler functions as illustrated below. Finally, determine what ascii character is represented by the binary code after the second mangler function.

Now, You will perform 2 rounds with the subkeys, and denote the character represented by the ascii binary code.

binary as follows: 1. Show the next full substitution and transposition round

L1 L2 L3 L4 R1 R2 R3 R4 PlainText 0 1 1 1 0 0 1 1 L1 L2 L3 L4 R1 R2 R3 R4 PlainText 0 1 1 1 0 0 1 1

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