CRYPTOGRAPHY. In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of 0's and 1's. The input is the 64 bit block 00000000000000100000000000000010000001000000010001000000000000000000000000000010000000000000001000000100000001000100000000000000 Suppose that the subkey for the current round is this 48 bit number: 000000000000000001000000000000000000000000000000000000000000000001000000000000000000000000000000 What does the 64 bit state look like after the IP transformation is applied to the input? Now find L0L0 and R0R0, the left and right halves of the state. L0=L0= R0=R0= What is the result of applying the expansion box to R0R0? E(R0)=E(R0)= What is the result of XORing the subkey with E(R0)E(R0)? k1⊕E(R0)=k1⊕E(R0)= We now apply the S-box transformation. S(k1⊕E(R0))=S(k1⊕E(R0))= Finally we apply the permutation box to complete the function ff. f(R0)=P(S(k1⊕E(R0)))=f(R0)=P(S(k1⊕E(R0)))= We can now compute the state of DES going into the next round. L1=L1= R1=R1=

CRYPTOGRAPHY. In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of 0's and 1's. The input is the 64 bit block 00000000000000100000000000000010000001000000010001000000000000000000000000000010000000000000001000000100000001000100000000000000 Suppose that the subkey for the current round is this 48 bit number: 000000000000000001000000000000000000000000000000000000000000000001000000000000000000000000000000 What does the 64 bit state look like after the IP transformation is applied to the input? Now find L0L0 and R0R0, the left and right halves of the state. L0=L0= R0=R0= What is the result of applying the expansion box to R0R0? E(R0)=E(R0)= What is the result of XORing the subkey with E(R0)E(R0)? k1⊕E(R0)=k1⊕E(R0)= We now apply the S-box transformation. S(k1⊕E(R0))=S(k1⊕E(R0))= Finally we apply the permutation box to complete the function ff. f(R0)=P(S(k1⊕E(R0)))=f(R0)=P(S(k1⊕E(R0)))= We can now compute the state of DES going into the next round. L1=L1= R1=R1=

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter8: Data And Network Communication Technology

Section: Chapter Questions

Problem 41VE

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In cryptography, a Caesar cipher is a very simple encryption techniques in which each letter in the plain text is replaced by a letter some fixed number of positions down the alphabet. For example, with a shift of 3, A would be replaced by D, B would become E, and so on. The method is named after Julius Caesar, who used it to communicate with his generals. ROT-13 (“rotate by 13 places”) is a widely used example of a Caesar cipher where the shift is 13. In Python, the key for ROT-13 may be represented by means of the following dictionary: key = {‘a’:’n’, ‘b’:’o’, ‘c’:’p’, ‘d’:’q’, ‘e’:’r’, ‘f’:’s’, ‘g’:’t’, ‘h’:’u’,‘i’:’v’, ‘j’:’w’, ‘k’:’x’, ‘l’:’y’, ‘m’:’z’, ‘n’:’a’, ‘o’:’b’, ‘p’:’c’,‘q’:’d’, ‘r’:’e’, ‘s’:’f’, ‘t’:’g’, ‘u’:’h’, ‘v’:’i’, ‘w’:’j’, ‘x’:’k’,‘y’:’l’, ‘z’:’m’, ‘A’:’N’, ‘B’:’O’, ‘C’:’P’, ‘D’:’Q’, ‘E’:’R’, ‘F’:’S’,‘G’:’T’, ‘H’:’U’, ‘I’:’V’, ‘J’:’W’, ‘K’:’X’, ‘L’:’Y’, ‘M’:’Z’, ‘N’:’A’,‘O’:’B’, ‘P’:’C’, ‘Q’:’D’, ‘R’:’E’, ‘S’:’F’, ‘T’:’G’, ‘U’:’H’, ‘V’:’I’,‘W’:’J’, ‘X’:’K’,…