Then the expansion permutation process expands the bit rpt bits
For the second part of this lab, using your original plaintext and key values, you should calculate the value of round two by hand, (ie computing all steps in a DES round) using the value of X and the sub-keys as given by the DES Calculator, and verify that you obtain the same value of X as the trace shows at the start of round three. You will find this relatively simple if you use a scientific calculator with the ability to enter and display numbers in various bases and to perform logical operations (eg. the Windows Calculator in Scientific Mode, or kcalc on Linux).
Plaintext: be480047cce655db
Step-2: Expansion Permutation –
Recall that after initial permutation, we had two 32-bit plain text
areas called as Left Plain Text(LPT) and Right Plain Text(RPT). During
the expansion permutation, the RPT is expanded from 32 bits to 48 bits.
Bits are permuted as well hence called as expansion permutation. This
happens as the 32 bit RPT is divided into 8 blocks, with each block
consisting of 4 bits. Then, each 4 bit block of the previous step is
then expanded to a corresponding 6 bit block, i.e., per 4 bit block, 2
more bits are added.
This process results into expansion as well as permutation of the input bit while creating output. Key transformation process compresses the 56-bit key to 48 bits. Then the expansion permutation process expands the 32-bit RPT to 48-bits. Now the 48-bit key is XOR with 48-bit RPT and resulting output is given to the next step, which is the S-Box substitution.
'1' : "0001",
'2' : "0010",
'7' : "0111",
'8' : "1000",
'D' : "1101",
'E' : "1110",
return bin
# Binary to hexadecimal conversion
"0011" : '3',
"0100" : '4',
"1001" : '9',
"1010" : 'A',
"1111" : 'F' }
hex = ""
ch = ch + s[i + 2]
ch = ch + s[i + 3]
binary1 = binary
decimal, i, n = 0, 0, 0
i += 1
return decimal
div = len(res) / 4
div = int(div)
# Permute function to rearrange the bits
def permute(k, arr, n):
# shifting the bits towards left by nth shifts
def shift_left(k, nth_shifts):
s = s + k[0]
k = s
ans = ""
for i in range(len(a)):
return ans
# Table of Position of 64 bits at initail level: Initial Permutation Table
57, 49, 41, 33, 25, 17, 9, 1,
59, 51, 43, 35, 27, 19, 11, 3,
6 , 7 , 8 , 9 , 8 , 9 , 10, 11,
12, 13, 12, 13, 14, 15, 16, 17,
per = [ 16, 7, 20, 21,
29, 12, 28, 17,
19, 13, 30, 6,
22, 11, 4, 25 ]
[15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13 ]],
[[15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10],
[13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1],
[13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7],
[3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14] ],
[ [2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9],
[10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8],
[9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6],
[6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12] ],
[ [13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7],
final_perm = [ 40, 8, 48, 16, 56, 24, 64, 32,
39, 7, 47, 15, 55, 23, 63, 31,
34, 2, 42, 10, 50, 18, 58, 26,
33, 1, 41, 9, 49, 17, 57, 25 ]
print("After inital permutation", bin2hex(pt))
# Splitting
right_expanded = permute(right, exp_d, 48)
# XOR RoundKey[i] and right_expanded
row = bin2dec(int(xor_x[j * 6] + xor_x[j * 6 + 5]))
col = bin2dec(int(xor_x[j * 6 + 1] + xor_x[j * 6 + 2] + xor_x[j * 6 + 3] + xor_x[j * 6 + 4]))
# XOR left and sbox_str
result = xor(left, sbox_str)
print("Round ", i + 1, " ", bin2hex(left), " ", bin2hex(right), " ", rk[i])
# Combination
pt = "123456ABCD132536"
key = "9f4f190e284c81cf"
keyp = [57, 49, 41, 33, 25, 17, 9,
1, 58, 50, 42, 34, 26, 18,
14, 6, 61, 53, 45, 37, 29,
21, 13, 5, 28, 20, 12, 4 ]
2, 2, 2, 2,
1, 2, 2, 2,
23, 19, 12, 4, 26, 8,
16, 7, 27, 20, 13, 2,
# Splitting
left = key[0:28] # rkb for RoundKeys in binary
# Shifting the bits by nth shifts by checking from shift table
left = shift_left(left, shift_table[i])
round_key = permute(combine_str, key_comp, 48)
rkb.append(round_key)
print("Decryption")
rkb_rev = rkb[::-1]
