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Assignment#1
------------From Name: Waqas ZafarRoll No: 09060619-045Section: ADepartment: BScs 7
th
--------------ToSubmit To: Mam Ayesha AltafSubject: Network SecurityCourse Code: Cs-312University Of Gujrat (HHC)
How DES Works in Detail
DES is a
block cipher
--meaning it operates on plaintext blocks of a given size (64-bits) and returns cipher text blocks of the same size. Thus DES results in a
permutation
among the 2^64 (read this as: "2 to the 64th power") possible arrangements of 64 bits, each of which may be either 0 or 1. Each block of 64 bits is dividedinto two blocks of 32 bits each, a left half block
L
and a right half
R
. (This division is only used in certainoperations.)
Example:
Let
M
be the plain text message
M
= NEVRQUIT,Plain text in ASCII: 78 69 86 82 81 85 73 84Plain text in Hex=4e 45 56 52 51 55 49 54
M
= 01001110 01000101 01010110 01010010 01010001 01010101 01001001 01010100
L
= 01001110 01000101 01010110 01010010
R
= 01010001 01010101 01001001 01010100DES operates on the 64-bit blocks using
key
sizes of 56- bits. The keys are actually stored as being 64 bits long, but every 8th bit in the key is not used (i.e. bits numbered 8, 16, 24, 32, 40, 48, 56, and 64). However, we willnevertheless number the bits from 1 to 64, going left to right, in the following calculations. But, as you will see,the eight bits just mentioned get eliminated when we create sub keys.
Example:
Let
K
be the key
K
= KASHISAB.Key in ASCII= 75 65 83 72 73 83 65 66Key in Hex=4b 41 53 48 49 53 41 42In binary the key isK=01001011 01000001 01010011 01001000 01001001 01010011 01000001 01000010The DES algorithm uses the following steps:
Create 16 sub keys, each of which is 48-bits long.
The 64-bit key is permuted according to the following table,
PC-1
. Since the first entry in the table is "57", thismeans that the 57th bit of the original key
K
becomes the first bit of the permuted key
K
+. The 49th bit of theoriginal key becomes the second bit of the permuted key. The 4th bit of the original key is the last bit of the permuted key. Note only 56 bits of the original key appear in the permuted key.
PC-1
57 49 41 33 25 17 91 58 50 42 34 26 1810 2 59 51 43 35 2719 11 3 60 52 44 3663 55 47 39 31 23 157 62 54 46 38 30 2214 6 61 53 45 37 2921 13 5 28 20 12 4
Example:
From the original 64-bit keyK=01001011 01000001 01010011 01001000 01001001 01010011 01000001 01000010K + =0000000 0111111 1100000 0000010 1010010 1000000 0000011 0010100 Next, split this key into left and right halves,
C
0
and
D
0
, where each half has 28 bits.
Example:
From the permuted key
K
+, we getC0=0000000 0111111 1100000 0000010D0=1010010 1000000 0000011 0010100With
C
0
and
D
0
defined, we now create sixteen blocks
C
n
and
D
n
, 1<=
n
<=16. Each pair of blocks
C
n
and
D
n
isformed from the previous pair
C
n-1
and
D
n-1
, respectively, for
n
= 1, 2... 16, using the following schedule of "leftshifts" of the previous block. To do a left shift, move each bit one place to the left, except for the first bit, whichis cycled to the end of the block.
Iteration Number ofNumber Left Shifts1 12 13 24 25 26 27 28 29 110 211 212 213 214 215 216 1
This means, for example,
C
3
and
D
3
are obtained from
C
2
and
D
2
, respectively, by two left shifts, and
C
16
and
D
16
are obtained from
C
15
and
D
15
, respectively, by one left shift. In all cases, by a single left shift is meant arotation of the bits one place to the left, so that after one left shift the bits in the 28 positions are the bits thatwere previously in positions 2, 3,..., 28, 1.
Example:
From original pair
C
0
and
D
0
we obtain:
C
1
= 0000000 1111111 1000000 0000100
D
1
= 0100101 0000000 0000110 0101001
C
2
= 0000001 1111111 0000000 0001000
D
2
= 1001010 0000000 0001100 1010010
C
3
= 0000111 1111100 0000000 0100000
D
3
= 0101000 0000000 0110010 1001010
C
4
= 0011111 1110000 0000001 0000000
D
4
= 0100000 0000001 1001010 0101001
C
5
= 1111111 1000000 0000100 0000000
D
5
= 0000000 0000110 0101001 0100101
C
6
= 1111110 0000000 0010000 0000011
D
6
= 0000000 0011001 0100101 0010100
C
7
= 1111000 0000000 1000000 0001111
D
7
= 0000000 1100101 0010100 1010000
C
8
= 1100000 0000010 0000000 0111111
D
8
= 0000011 0010100 1010010 1000000
C
9
= 1000000 0000100 0000000 1111111
D
9
= 0000110 0101001 0100101 0000000
C
10
= 0000000 0010000 0000011 1111110
D
10
=0011001 0100101 0010100 0000000
C
11
= 0000000 1000000 0001111 1111000
D
11
=1100101 0010100 1010000 0000000
C
12
= 0000010 0000000 0111111 1100000
D
12
=0010100 1010010 1000000 0000011
C
13
= 0001000 0000001 1111111 0000000
D
13
= 1010010 1001010 0000000 0001100
C
14
= 0100000 0000111 1111100 0000000
D
14
= 1001010 0101000 0000000 0110010
C
15
= 0000000 0011111 1110000 0000001
D
15
=0101001 0100000 0000001 1001010
C
16
= 0000000 0111111 1100000 0000010
D
16
=1010010 1000000 0000011 0010100
We now form the keys
K
n
, for 1<=
n
<=16, by applying the following permutation table to each of theconcatenated pairs
C
n
D
n
. Each pair has 56 bits, but
PC-2
only uses 48 of these.
PC-2
14 17 11 24 1 53 28 15 6 21 1023 19 12 4 26 816 7 27 20 13 241 52 31 37 47 5530 40 51 45 33 4844 49 39 56 34 5346 42 50 36 29 32
Therefore, the first bit of
K
n
is the 14th bit of
C
n
D
n
, the second bit the 17th, and so on, ending with the 48th bitof
K
n
being the 32th bit of
C
n
D
n
.
Example:
For the first key we have
C
1
= 0000000 1111111 1000000 0000100
D
1
= 0100101 0000000 0000110 0101001Which after we apply the permutation
PC-2
, becomes
K
1
= 101000 001001 001011 000010 000010 101011 000101 000000For the other keys we have
K
2
= 101000 000001 001001 010010 010011 000000 000010 101011
K
3
= 001001 000101 101001 010000 000001 100101 100001 001001
K
4
= 000001 100111 000101 010000 000000 101001 000101 110000
K
5
= 000011 100100 010101 010001 100000 011000 110100 100000
K
6
= 010011 110100 000100 001001 010010 000000 111000 010000
K
7
= 000010 111000 000110 001001 010110 010100 000000 011100
K
8
= 000110 010000 100010 001011 000000 010101 000010 001000
K
9
= 000110 010000 101010 001000 000110 000010 111010 010000
K
10
=000100 000011 100010 001100 001110 010100 000000 010001
K
11
=000100 000010 110001 000100 000000 110110 000000 000010
K
12
=010000 000110 110000 100100 101001 000010 000100 000100
K
13
=110000 001010 010100 100100 101000 000000 001011 000110
K
14
=110000 001000 011000 100011 010101 001000 001010 000011
K
15
=111000 011001 001000 100010 000101 100000 010001 001001
And at last
K
16
=101000 001001 001000 101010 011000 000001 010010 000110
Encryption
There is an
initial permutation
IP
of the 64 bits of the message data
M
. This rearranges the bits according to thefollowing table, where the entries in the table show the new arrangement of the bits from their initial order. The58th bit of
M
becomes the first bit of
IP
. The 50th bit of
M
becomes the second bit of
IP
. The 7th bit of
M
isthe last bit of
IP
.
IP
58 50 42 34 26 18 10 260 52 44 36 28 20 12 462 54 46 38 30 22 14 664 56 48 40 32 24 16 857 49 41 33 25 17 9 159 51 43 35 27 19 11 361 53 45 37 29 21 13 563 55 47 39 31 23 15 7
Example:
Applying the initial permutation to the block of text
M
, given previously, we get
M
= 01001110 01000101 01010110 01010010 01010001 01010101 01001001 01010100
IP
= 11111111 10111100 10100111 01110010 00000000 00000000 01000001 00001101 Next divide the permuted block
IP
into a left half
L
0
of 32 bits, and a right half
R
0
of 32 bits.
Example:
From
IP
, we get
L
0
and
R
0
L
0
= 11111111 10111100 10100111 01110010
R
0
= 00000000 00000000 01000001 00001101We now proceed through 16 iterations, for 1<=
n
<=16, using a function
f
which operates on two blocks--adata block of 32 bits and a key
K
n
of 48 bits--to produce a block of 32 bits.
Let + denote XOR addition, (bit-by-bit addition modulo 2)
. Then for
n
going from 1 to 16 we calculate
L
n
=
R
n-1
R
n
=
L
n-1
+
f
(
R
n-1
,
Kn
)This results in a final block, for
n
= 16, of
L
16
R
16
. That is, in each iteration, we take the right 32 bits of the previous result and make them the left 32 bits of the current step. For the right 32 bits in the current step, weXOR the left 32 bits of the previous step with the calculation
f
.
Example:
For
n
= 1, we haveK1=101000 001001 0010110 00010 000010 101011 000101 000000
L
1
=
R
0
= 00000000 00000000 01000001 00001101
R
1
=
L
0
+
f
(
R
0
,
K
1
)It remains to explain how the function
f
works. To calculate
f
, we first expand each block
R
n-1
from 32 bits to48 bits. This is done by using a selection table that repeats some of the bits in
R
n-1
. We'll call the use of thisselection table the function
E
. Thus
E
(
R
n-1
) has a 32 bit input block, and a 48 bit output block.Let
E
be such that the 48 bits of its output, written as 8 blocks of 6 bits each, are obtained by selecting the bitsin its inputs in order according to the following table:
E BIT-SELECTION TABLE
32 1 2 3 4 54 5 6 7 8 98 9 10 11 12 1312 13 14 15 16 1716 17 18 19 20 2120 21 22 23 24 2524 25 26 27 28 2928 29 30 31 32 1
Thus the first three bits of
E
(
R
n-1
) are the bits in positions 32, 1 and 2 of
R
n-1
while the last 2 bits of
E
(
R
n-1
) arethe bits in positions 32 and 1.
Example:
We calculate
E
(
R
0
) from
R
0
as follows:
R
0
= 00000000 00000000 01000001 00001101
E
(
R
0
) = 100000 000000 000000 000000 001000 000010 100001 011010(Note that each block of 4 original bits has been expanded to a block of 6 output bits.) Next in the
f
calculation, we XOR the output
E
(
R
n-1
) with the key
K
n
:
K
n
+
E
(
R
n-1
).
Example:
For
K
1
,
E
(
R
0
), we have
K
1
= 101000 001001 001011 000010 000010 101011 000101 000000
E
(
R
0
) = 100000 000000 000000 000000 001000 000010 100001 011010
K
1
+
E
(
R
0
) = 001000 001001 001011 000010 001010 101001 100100 011010We have not yet finished calculating the function
f
. To this point we have expanded
R
n-1
from 32 bits to 48 bits,using the selection table, and XORed the result with the key
K
n
. We now have 48 bits, or eight groups of six bits. We now do something strange with each group of six bits: we use them as addresses in tables called "
Sboxes
". Each group of six bits will give us an address in a different
S
box. Located at that address will be a 4 bitnumber. This 4 bit number will replace the original 6 bits. The net result is that the eight groups of 6 bits aretransformed into eight groups of 4 bits (the 4-bit outputs from the
S
boxes) for 32 bits total.
Write the previous result, which is 48 bits, in the form:
K
n
+
E
(
R
n-1
) =
B
1
B
2
B
3
B
4
B
5
B
6
B
7
B
8
,where each
B
i
is a group of six bits. We now calculate
S
1
(B
1
)S
2
(B
2
)S
3
(B
3
)S
4
(B
4
)S
5
(B
5
)S
6
(B
6
)S
7
(B
7
)S
8
(B
8
)
where
S
i
(B
i
)
referres to the output of the
i
-th
S
box.To repeat, each of the functions
S1, S2,..., S8
, takes a 6-bit block as input and yields a 4-bit block as output. Thetable to determine
S
1
is shown and explained below:
S1Column NumberRow No. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150 14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 71 0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 82 4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 03 15 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13
If
S
1
is the function defined in this table and
B
is a block of 6 bits, then
S
1
(B)
is determined as follows: The firstand last bits of
B
represent in base 2 a number in the decimal range 0 to 3 (or binary 00 to 11). Let that number be
i
. The middle 4 bits of
B
represent in base 2 a number in the decimal range 0 to 15 (binary 0000 to 1111). Letthat number be
j
. Look up in the table the number in the
i
-th row and
j
-th column. It is a number in the range 0to 15 and is uniquely represented by a 4 bit block. That block is the output
S
1
(B)
of
S
1
for the input
B
. Forexample, for input block
B
= 011011 the first bit is "0" and the last bit "1" giving 01 as the row. This is row 1.The middle four bits are "1101". This is the binary equivalent of decimal 13, so the column is column number13. In row 1, column 13 appears 5. This determines the output; 5 is binary 0101, so that the output is 0101.Hence
S
1
(011011) = 0101.The tables defining the functions
S
1
,...,S
8
are the following:
S1
14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 70 15 7 4 14 2 13 1 10 6 12 11 9 5 3 84 1 14 8 13 6 2 11 15 12 9 7 3 10 5 015 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13
S2
15 1 8 14 6 11 3 4 9 7 2 13 12 0 5 103 13 4 7 15 2 8 14 12 0 1 10 6 9 11 50 14 7 11 10 4 13 1 5 8 12 6 9 3 2 1513 8 10 1 3 15 4 2 11 6 7 12 0 5 14 9
S3
10 0 9 14 6 3 15 5 1 13 12 7 11 4 2 813 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 71 10 13 0 6 9 8 7 4 15 14 3 11 5 2 12
S4
7 13 14 3 0 6 9 10 1 2 8 5 11 12 4 1513 8 11 5 6 15 0 3 4 7 2 12 1 10 14 910 6 9 0 12 11 7 13 15 1 3 14 5 2 8 43 15 0 6 10 1 13 8 9 4 5 11 12 7 2 14
S5
2 12 4 1 7 10 11 6 8 5 3 15 13 0 14 914 11 2 12 4 7 13 1 5 0 15 10 3 9 8 64 2 1 11 10 13 7 8 15 9 12 5 6 3 0 1411 8 12 7 1 14 2 13 6 15 0 9 10 4 5 3
S6
12 1 10 15 9 2 6 8 0 13 3 4 14 7 5 1110 15 4 2 7 12 9 5 6 1 13 14 0 11 3 89 14 15 5 2 8 12 3 7 0 4 10 1 13 11 64 3 2 12 9 5 15 10 11 14 1 7 6 0 8 13
S7
4 11 2 14 15 0 8 13 3 12 9 7 5 10 6 113 0 11 7 4 9 1 10 14 3 5 12 2 15 8 61 4 11 13 12 3 7 14 10 15 6 8 0 5 9 26 11 13 8 1 4 10 7 9 5 0 15 14 2 3 12
S8
13 2 8 4 6 15 11 1 10 9 3 14 5 0 12 71 15 13 8 10 3 7 4 12 5 6 11 0 14 9 27 11 4 1 9 12 14 2 0 6 10 13 15 3 5 82 1 14 7 4 10 8 13 15 12 9 0 3 5 6 11
Example:
For the first round, we obtain as the output of the eight
S
boxes:
K
1
+
E
(
R
0
) = 001000 001001 001011 000010 001010 101001 100100 011010.
S
1
(B
1
)S
2
(B
2
)S
3
(B
3
)S
4
(B
4
)S
5
(B
5
)S
6
(B
6
)S
7
(B
7
)S
8
(B
8
)
=0010 1111 0100 1101 1010 1001 1011 0000The final stage in the calculation of
f
is to do a permutation
P
of the
S
-box output to obtain the final value of
f
:
f
=
P
(
S
1
(B
1
)S
2
(B
2
)...S
8
(B
8
)
)The permutation
P
is defined in the following table.
P
yields a 32-bit output from a 32-bit input by permutingthe bits of the input block.
P
16 7 20 2129 12 28 17
1 15 23 265 18 31 102 8 24 1432 27 3 919 13 30 622 11 4 25
Example:
From the output of the eight
S
boxes:
S
1
(B
1
)S
2
(B
2
)S
3
(B
3
)S
4
(B
4
)S
5
(B
5
)S
6
(B
6
)S
7
(B
7
)S
8
(B
8
)
=0010 1111 0110 1101 1010 1001 1011 0000
f
=
1101 0011 0000 1001 0111 0110 1101 0001R
1
=
L
0
+
f
(
R
0
,
K
1
)= 1111 1111 1011 1100 1010 0111 0111 0010+ 1101 0011 0000 1001 0111 0110 1101 0001
So now
L1
=
0000 0000 0000 0000 0100 0001 0000 1101R1= 0010 1100 1011 0101 1101 0001 1010 0011
In the next round, we will have
L
2
=
R
1
, which is the block we just calculated, and then we must calculate
R
2
=
L
1
+ f(R
1
, K
2
)
, and so on for 16 rounds. At the end of the sixteenth round we have the blocks
L
16
and
R
16
. Wethen
reverse
the order of the two blocks into the 64-bit block
Step2……………….
L2=0010 1100 1011 0101 1101 0001 1010 0011R2=L1 +f (R1, K2)f (R1,K2)=
0010 1100 1010 1000 1011 0000 0101 0011R2=0000 0000 0000 0000 0100 0001 0000 1101+0010 1100 1010 1000 1011 0000 0101 0011=
0010 1100 1010 1000 1111 0001 0101 1110L2=0010 1100 1011 0101 1101 0001 1010 0011R2=0010 1100 1010 1000 1111 0001 0101 1110
Step 3…………………
L3=0010 1100 1010 1000 1111 0001 0101 1110R3=L2+f(R2, K3)f(R2, K3)=
10001000101001010101101111010110R3= 0010 1100 1011 0101 1101 0001 1010 0011+1000 1000 1010 0101 0101 1011 1101 0110=
1010 0100 0001 0000 1000 1010 0111 0101L3=0010 1100 1010 1000 1111 0001 0101 1110R3=1010 0100 0001 0000 1000 1010 0111 0101
Step4……..
L4=1010 0100 0001 000
0 1000 1010 0111 0101R4=L3+f(R3,K4)So R4=1101 0010 1010 1110 1111 0010 0011 1101
Step5…………
L5=1101 0010 1010 1110 1111 0010 0011 1101R5=L4+f(R4,K5)So R5= 0111 1111 1010 0110 1110 1001 1001 0001
Step6………….
L6=0111 1111 1010 0110 1110 1001 1001 0001R6=L5+f(R5,K6)
So R6=1011 0110 1110 1100 0100 0011 0101 1011
Step7…………..
L7=1011 0110 1110 1100 0100 0011 0101 1011R7=L6+f(R6,K7)So R7=0100 0010 0011 0100 1001 1000 1001 1111
Step8…………..
L8=0100 0010 0011 0100 1001 1000 1001 1111R8=L7+f(R7,K8)So R8=1110 1110 0001 1101 0111 1101 0111 0000
Step9……………
L9=1110 1110 0001 1101 0111 1101 0111 0000R9=L8+f(R8,K9)So R9=
1010 1000 0011 0111 0111 0000 1011 1000
Step10……………
L10=1010 1000 0011 0111 0111 0000 1011 1000R10=L9+f(R9,K10)So R10=1110 1110 1100 1010 0100 0111 0101 0001
Step11……………..
L11=1110 1110 1100 1010 0100 0111 0101 0001R11=L10+f(R10,K11)So R11=0000 1010 1011 0001 0001 0111 0010 1000
Step12…………….
L12=0000 1010 1011 0001 0001 0111 0010 1000
R12=L11+f(R11,K12)So R12=0101 1010 0001 1100 0100 1000 1101 1001
Step13………………
L13=0101 1010 0001 1100 0100 1000 1101 1001R13=L12+f(R12,K13)So R13=0110 0001 1001 1010 1100 0010 1011 0110
Step14……..
L14=0110 0001 1001 1010 1100 0010 1011 0110R14=L13+f(R13,K14)So R14=1010 0111 0000 1000 1111 0101 1111 1011
Step15………………..
L15=1010 0111 0000 10
00 1111 0101 1111 1011R15=L14+f(R14,K15)So R15=0001 1000 0000 0011 1101 0100 1011 1001
Step16……………
L16=0001 1000 0000 0011 1101 0100 1011 1001R16=L15+f(R15,K16)S0 R16=0010 1101 0010 1011 1110 0011 0011 1110
Example:
If we process all 16 blocks using the method defined previously, we get, on the 16th round,
L
16
= 0001 1000 0000 0011 1101 0100 1011 1001
R
16
= 0010 1101 0010 1011 1110 0011 0011 1110
We reverse the order of these two blocks and apply the final permutation to
R
16
L
16
= 0010 1101 0010 1011 1110 0011 0011 1110 0001 1000 0000 0011 1101 0100 1011 1001and apply a final permutation
IP
-1
as defined by the following table:
IP
-1
40 8 48 16 56 24 64 3239 7 47 15 55 23 63 3138 6 46 14 54 22 62 3037 5 45 13 53 21 61 2936 4 44 12 52 20 60 2835 3 43 11 51 19 59 2734 2 42 10 50 18 58 2633 1 41 9 49 17 57 25
IP
-1
= 0111011000110101010010011101001110001011010101110000110000001110
Which in hexadecimal format is76 35 49 d3 8b 57 0c 0e
This is the encrypted form ofPlain Text = NEVRQUIT:Key=KASHISABCipher Text = 76 35 49 d3 8b 57 0c 0e.
Decryption
Decryption is simply the inverse of encryption, following the same steps as above, but reversing the order inwhich the sub keys are applied.So now we haveCipher text=76 35 49 d3 8b 57 0c 0eAnd our key is=
KASHISABCipher text in binary is=0111011000110101010010011101001110001011010101110000110000001110After Ip table the cipher text is=0010 1101 0010 1011 1110 0011 0011 1110 0001 1000 0000 0011 1101 0100 1011 1001So now from above textL16=0010 1101 0010 1011 1110 0011 0011 1110R16=0001 1000 0000 0011 1101 0100 1011 1001From now on the process is same but we only use our keys in reverse order.
So the simple formula we use for decryption is
L
n-1
=
R
n
R
n-1
=
L
n
+
f
(
R
n
,
Kn
)And here n goes from 16 to 1
Step 1…………….
From the above function we can write thisL15=0001 1000 0000 0011 1101 0100 1011 1001R15=0010 1101 0010 1011 1110 0011 0011 1110 + f(0001 1000 0000 0011
1101 0100 1011 1001, 101000001001 001000 101010 011000 000001 010010 000110) Now we calculate t
he R15……
By first we expand R16 with ‘E’ table that will expand our 32 bits to 48
bits.
Next we XOR K16 with the output of ‘E’ table.
Then after XOR our output goes to the 8 S BOXES that will convert 48 bits to 32 bits.
Then our output goes to the ‘P’ table.
Then we had the answer of our function.The function answer is=10001010001000110001011011000101Then at last when we XOR with L16 we finally getR15=
1010 0111 0000 1000 1111 0101 1111 1011//All the steps given below undergo with same procedure.
Step 2……………..
L14=1010 0111 0000 1000 1111 0101 1111 1011R14=0001 1000 0000 0011 1101 0100 1011 1001+f(1010 0111 0000 1000 1111 0101 1111 1011, 111000011001 001000 100010 000101 100000 010001 001001)The function answer is=01111001100110010001011000001111So after XOR with L15R14=0110 0001 1001 1010 1100 0010 1011 0110
Step 3……………………
L13=0110 0001 1001 1010 1100 0010 1011 0110R13=1010 0111 0000 1000 1111 0101 1111 1011 + f(0110 0001 1001 1010 1100 0010 1011 0110, 110000 001000 011000 100011 010101 001000 001010 000011)So R13=0101 1010 0001 1100 0100 1000 1101 1001
Step 4………………………..
L12=0101 1010 0001 1100 0100 1000 1101 1001R12=0110 0001 1001 1010 1100 0010 1011 0110 + f(0101
1010 0001 1100 0100 1000 1101 1001, 110000001010 010100 100100 101000 000000 001011 000110)So R12=0000 1010 1011 0001 0001 0111 0010 1000
Step 5………………………….
L11=0000 1010 1011 0001 0001 0111 0010 1000R11=0101 1010 0001 1100 0100 1000 1101 1001 + f(0000 1010 1011 0001 0001 0111 0010 1000, 010000000110 110000 100100 101001 000010 000100 000100)So R11=
1110 1110 1100 1010 0100 0111 0101 0001
Step
6………………………………..
L10=1110 1110 1100 1010 0100 0111 0101 0001R10=0000 1010 1011 0001 0001 0111 0010 1000 + f(1110 1110 1100 1010 0100 0111 0101 0001, 000100000010 110001 000100 000000 110110 000000 000010)So R10=1010 1000 0011 0111 0111 0000 1011 1000
Step 7………………………………….
L9=1010 1000 0011 0111 0111 0000 1011 1000
R9=1110 1110 1100 1010 0100 0111 0101 0001 + f(1010 1000 0011 0111 0111 0000 1011 1000, 000100000011 100010 001100 001110 010100 000000 010001)So R9=
1110 1110 0001 1101 0111 1101 0111 0000
Step 8………………………………
L8=1110 1110 0001 1101 0111 1101 0111 0000R8=1010 1000 0011 0111 0111 0000 1011 1000 + f(1110 1110 0001 1101 0111 1101 0111 0000, 000110010000 101010 001000 000110 000010 111010 010000)So R8=0100 0010 0011 0100 1001 1000 1001 1111
Step 9……………………………………..
L7=0100 0010 0011 0100 1001 1000 1001 1111R7=1110 1110 0001 1101 0111 1101 0111 0000 + f(
0100 0010 0011 0100 1001 1000 1001 1111, 000110010000 100010 001011 000000 010101 000010 001000)So R7=
1011 0110 1110 1100 0100 0011 0101 1011
Step 10………………………….
L6=1011 0110 1110 1100 0100 0011 0101 1011R6=0100 0010 0011 0100 1001 1000 1001 1111 + f(1011 0110 1110 1100 0100 0011 0101 1011, 000010111000 000110 001001 010110 010100 000000 011100)So R6=0111 1111 1010 0110 1110 1001 1001 0001
Step 11………………………………..
L5=0111 1111 1010 0110 1110 1001 1001 0001R5=1011 0110 1110 1100 0100 0011 0101 1011 + f(0111 1111 1010 0110 1110 1001 1001 0001, 010011110100 000100 001001 010010 000000 111000 010000)So R5=1101 0010 1010 1110 1111 0010 0011 1101
Step 12…………………………
L4=1101 0010 1010 1110 1111 0010 0011 1101R4=0111 1111 1010 0110 1110 1001 1001 0001 + f(1101 0010 1010 1110 1111 0010 0011 1101, 000011100100 010101 010001 100000 011000 110100 100000)So R4=
1010 0100 0001 0000 1000 1010 0111 0101
Step 13……………………………………..
L3=1010 0100 0001 0000 1000 1010 0111 0101R3=1101 0010 1010 1110 1111 0010 0011 1101 + f(1010 0100 0001
0000 1000 1010 0111 0101, 000001100111 000101 010000 000000 101001 000101 110000)So R3=0010 1100 1010 1000 1111 0001 0101 1110
Step 14………………………..
L2=0010 1100 1010 1000 1111 0001 0101 1110R2=1010 0100 0001 0000 1000 1010 0111 0101 + f(0010 1100 1010 1000 1111 0001 0101 1110, 001001000101 101001 010000 000001 100101 100001 001001)So R2=0010 1100 1011 0101 1101 0001 1010 0011
Step 15…………………………………
L1=0010 1100 1011 0101 1101 0001 1010 0011R1=0010 1100 1010 1000 1111 0001 0101 1110 + f(0010 1100 1011 0101 1101 0001 1010 0011, 101000000001 001001 010010 010011 000000 000010 101011)So R1=0000 0000 0000 0000 0100 0001 0000 1101
Step 16……………………………….
L0=0000 0000 0000 0000 0100 0001 0000 1101R0=0010 1100 1011 0101 1101 0001 1010 0011 + f(0000 0000 0000 0000 0100 0001 0000 1101, 101000001001 001011 000010 000010 101011 000101 000000)So R0=1111 1111 1011 1100 1010 0111 0111 0010
SoL0=0000 0000 0000 0000 0100 0001 0000 1101R0=1111 1111 1011 1100 1010 0111 0111 0010After 32 bit swapR0L0=1111 1111 1011 1100 1010 0111 0111 0010 0000 0000 0000 0000 0100 0001 0000 1101So after 1p inverseIp Inv=0100 1110 0100 0101 0101 0110 0101 0010 0101 0001 0101 0101 0100 1001 0101 0100
In BinaryL=0100 1110 0100 0101 0101 0110 0101 0010R=0101 0001 0101 0101 0100 1001 0101 0100In HexL=4e 45 56 52R=51 55 49 54And in Text formNEVRQUIT.
That’s our plain text.
Done
