Lecture 3: Block Ciphers and The Data Encryption Standard Lecture Notes On "Computer and Network Security"

Lecture 3: Block Ciphers and the Data Encryption
Standard
Lecture Notes on “Computer and Network Security”
by Avi Kak (kak@purdue.edu)
January 15, 2019

11:29am
c
2019 Avinash Kak, Purdue University
Goals:
• To introduce the notion of a block cipher in the modern context.

• To talk about the infeasibility of ideal block ciphers
• To introduce the notion of the Feistel Cipher Structure
• To go over DES, the Data Encryption Standard
• To illustrate important DES steps with Python and Perl code

CONTENTS
Section Title Page
3.1 Ideal Block Cipher 3
3.1.1 Size of the Encryption Key for the Ideal Block Cipher 6
3.2 The Feistel Structure for Block Ciphers 7
3.2.1 Mathematical Description of Each Round in the 10

Feistel Structure
3.2.2 Decryption in Ciphers Based on the Feistel Structure 12
3.3 DES: The Data Encryption Standard 16
3.3.1 One Round of Processing in DES 18
3.3.2 The S-Box for the Substitution Step in Each Round 22
3.3.3 The Substitution Tables 26
3.3.4 The P-Box Permutation in the Feistel Function 33
3.3.5 The DES Key Schedule: Generating the Round Keys 35
3.3.6 Initial Permutation of the Encryption Key 38
3.3.7 Contraction-Permutation that Generates the 48-Bit 42

Round Key from the 56-Bit Key
3.4 What Makes DES a Strong Cipher (to the 46

Extent It is a Strong Cipher)
3.5 Homework Problems 48
2
Computer and Network Security by Avi Kak Lecture 3
3.1: IDEAL BLOCK CIPHER
• In a modern block cipher (but still using a classical encryption

method), we replace a block of N bits from the plaintext with a
block of N bits from the ciphertext. This general idea is illustrated
in Figure 1 for the case of N = 4. (In general, though, N is set
to 64 or multiples thereof.)
• To understand Figure 1, note that there are 16 different possible

4-bit patterns. We can represent each pattern by an integer be-
tween 0 and 15. So the bit pattern 0000 could be represented by
the integer 0, the bit pattern 0001 by integer 1, and so on. The
bit pattern 1111 would be represented by the integer 15.
• In an ideal block cipher, the relationship between the input blocks

and the output block is completely random. But it must be
invertible for decryption to work. Therefore, it has to be one-to-
one, meaning that each input block is mapped to a unique output
block.
• The mapping from the input bit blocks to the output bit blocks
can also be construed as a mapping from the integers correspond-
3
ing to the input bit blocks to the integers corresponding to the

output bit blocks.
• The encryption key for the ideal block cipher is the codebook
itself, meaning the table that shows the relationship between the
input blocks and the output blocks.
• Figure 1 depicts an ideal block cipher that uses blocks of size 4.

Each block of 4 bits in the plaintext is transformed into a block
of 4 ciphertext bits.
4
b b1 b2 b
Plaintext bit block: 0 3
Convert 4 incoming bits to one of 16 integers
Random 1−1 mapping of 16 input integers

to 16 output integers
Convert integer to a 4−bit pattern
Ciphertext bit block: c c c c

0 1 2 3
Figure 1: The ideal block cipher when the block size equals
4 bits. (This figure is from Lecture 3 of “Lecture Notes on Computer and Network Security” by
Avi Kak)
5
3.1.1: The Size of the Encryption Key for the Ideal

Block Cipher
• Consider the case of 64-bit block encryption.
• With a 64-bit block, we can think of each possible input block

as one of 264 integers and for each such integer we can spec-
ify an output 64-bit block. We can construct the codebook by
displaying just the output blocks in the order of the integers cor-
responding to the input blocks. Such a code book will be of size
64 × 264 ≈ 1021.
• That implies that the encryption key for the ideal block cipher
using 64-bit blocks will be of size 1021.
• The size of the encryption key would make the ideal block cipher
an impractical idea. Think of the logistical issues related to the
transmission, distribution, and storage of such large keys.
6
3.2: The Feistel Structure for Block Ciphers
The DES (Data Encryption Standard) algorithm for encryption and

decryption, which is the main theme of this lecture, is based on what
is known as the Feistel Structure. This section and the next two
subsections introduce this structure:
• Named after the IBM cryptographer Horst Feistel and first im-
plemented in the Lucifer cipher by Horst Feistel and Don Cop-
persmith.
• A cryptographic system based on Feistel structure uses the same

basic algorithm for both encryption and decryption.
• As shown in Figure 2, the Feistel structure consists of multiple

rounds of processing of the plaintext, with each round consisting
of a substitution step followed by a permutation step.
• The input block to each round is divided into two halves that I
have denoted L and R for the left half and the right half.
7
Plaintext block
(Divide into two halves, L and R)
Round Keys
L R
K
Round1 1
F(K,R)
L R
K
2
Round
2 F(K,R)
L R
K
n
Round n
F(K,R)
Ciphertext block
Figure 2: The Feistel Structure for symmetric key cryptog-

raphy (This figure is from Lecture 3 of “Lecture Notes on Computer and Network Security” by Avi
Kak)
8
• In each round, the right half of the block, R, goes through un-
changed. But the left half, L, goes through an operation that
depends on R and the encryption key. The operation carried out
on the left half L is referred to as the Feistel Function.
• The permutation step at the end of each round consists of swap-

ping the modified L and R. Therefore, the L for the next round
would be R of the current round. And R for the next round
be the output L of the current round.
• The next two subsection present important properties of the Feis-

tel structure. As you will see, these properties are invariant to
our choice for the Feistel Function.
• Besides DES, there exist several block ciphers today — the most
popular of these being Blowfish, CAST-128, and KASUMI —
that are also based on the Feistel structure.
9
3.2.1: Mathematical Description of Each Round in the

Feistel Structure
• Let LEi and REi denote the output half-blocks at the end of the
ith round of processing. The letter ’E’ denotes encryption.
• In the Feistel structure, the relationship between the output of

the ith round and the output of the previous round, that is, the
(i − 1)th round, is given by
LEi = REi−1
REi = LEi−1 ⊕ F (REi−1, Ki)
where ⊕ denotes the bitwise EXCLUSIVE-OR operation. The symbol

F denotes the operation that “scrambles” REi−1 of the previous
round with what is shown as the round key Ki in Figure 2. The
round key Ki is derived from the main encryption key as will
be explained later.
• F is referred to as the Feistel function, after Horst Feistel natu-

rally.
• Assuming 16 rounds of processing (which is typical), the output

of the last round of processing is given by
10
LE16 = RE15
RE16 = LE15 ⊕ F (RE15, K16)
11
3.2.2: Decryption in Ciphers Based on the Feistel

Structure
• As shown in Figure 3, the decryption algorithm is exactly the

same as the encryption algorithm with the only difference that
the round keys are used in the reverse order.
• The output of each round during decryption is the

input to the corresponding round during encryption
— except for the left-right switch between the two
halves. This property holds true regardless of the
choice of the Feistel function F .
• To prove the above claim, let LDi and RDi denote the left half
and the right half of the output of the ith round.
• That means that the output of the first decryption round con-
sists of LD1 and RD1. So we can denote the input to the first
decryption round by LD0 and RD0. The relationship between
the two halves that are input to the first decryption round and
what is output by the encryption algorithm is:
LD0 = RE16
RD0 = LE16
12
Encryption Decryption
Plaintext block Round Keys Plaintext block

(Divide into two halves, L and R) (Divide into two halves, L and R)
K
1
LE L R RE LD = RE RD = LE
0 0 16 0 16 0
F(K,R)
Round1
F(K,R)
LE RE
1 1 K
2
LD = RE RD = LE
1 15 1
15
F(K,R)
Round
2 F(K,R)
LE RE K
15 15 16
LD = RE RD = LE
1 15 1 15
F(K,R)
Round 16
F(K,R)
LE RE
16 LD = RE RD = LE
16 0 16
0 16
Ciphertext block Ciphertext block
Figure 3: When a Feistel structure is used, decryption

works the same as encryption. (This figure is from Lecture 3 of “Lecture
Notes on Computer and Network Security” by Avi Kak)
13
• We can write the following equations for the output of the first
decryption round
LD1 = RD0
= LE16
= RE15
RD1 = LD0 ⊕ F (RD0, K16)

= RE16 ⊕ F (LE16, K16)
= [LE15 ⊕ F (RE15, K16)] ⊕ F (RE15, K16)
= LE15
This shows that, except for the left-right switch, the output of
the first round of decryption is the same as the input to the last
stage of the encryption round since we have LD1 = RE15 and
RD1 = LE15
• The following equalities are used in the above derivation. Assume

that A, B, and C are bit arrays.
[A ⊕ B] ⊕ C = A ⊕ [B ⊕ C ]
A ⊕ A = 0
A ⊕ 0 = A
14
• The above result is independent of the precise nature

of the Feistel function F . That is, the output of each round
during decryption is the input to the corresponding round during
encryption for every choice of the Feistel function F .
15
3.3: DES: THE DATA ENCRYPTION

STANDARD
• Adopted by NIST in 1977.
• Based on a cipher (Lucifer) developed earlier by IBM for Lloyd’s

of London for cash transfer.
• DES uses the Feistel cipher structure with 16 rounds of process-

ing.
• DES uses a 56-bit encryption key. (The key size was apparently
dictated by the memory and processing constraints imposed by
a single-chip implementation of the algorithm for DES.) The key
itself is specified with 8 bytes, but one bit of each byte is used as
a parity check.
• DES encryption was broken in 1999 by Electronics

Frontiers Foundation (EFF, www.eff.org). This resulted
in NIST issuing a new directive that year that required organiza-
tions to use Triple DES, that is, three consecutive applications
16
of DES. (That DES was found to be not as strong as originally

believed also prompted NIST to initiate the development of new
standards for data encryption. The result is AES that we will
discuss later.)
• Triple DES continues to enjoy wide usage in commercial ap-

plications even today. To understand Triple DES, you must first
understand the basic DES encryption.
• As mentioned, DES uses the Feistel structure with 16 rounds.
• What is specific to DES is the implementation of the F function

in the algorithm and how the round keys are derived from the
main encryption key.
• As will be explained in Section 3.3.5, the round keys are generated

from the main key by a sequence of permutations. Each round
key is 48 bits in length.
17
3.3.1: One Round of Processing in DEA
• The algorithmic implementation of DES is known as DEA for

Data Encryption Algorithm.
• Figure 4 shows a single round of processing in DEA. The dotted

rectangle constitutes the F function.
• The 32-bit right half of the 64-bit input data block is expanded
by into a 48-bit block. This is referred to as the expansion
permutation step, or the E-step.
• The above-mentioned E-step entails the following:

– first divide the 32-bit block into eight 4-bit words
– attach an additional bit on the left to each 4-bit word that is
the last bit of the previous 4-bit word
– attach an additional bit to the right of each 4-bit word that is
the beginning bit of the next 4-bit word.
Note that what gets prefixed to the first 4-bit block is the last bit
of the last 4-bit block. By the same token, what gets appended
to the last 4-bit block is the first bit of the first 4-bit block. The
18
reason for why we expand each 4-bit block into a 6-bit block in
the manner explained will become clear shortly.
• The 56-bit key is divided into two halves, each half shifted sep-
arately, and the combined 56-bit key permuted/contracted
to yield a 48-bit round key. How this is done will be explained
later.
• The 48 bits of the expanded output produced by the E-step are

XORed with the round key. This is referred to as key mixing.
• The output produced by the previous step is broken into eight

six-bit words. Each six-bit word goes through a substitution step;
its replacement is a 4-bit word. The substitution is carried out
with an S-box, as explained in greater detail in Section 3.3.2.
[The name “S-Box” stands for “Substitution Box”.]
• So after all the substitutions, we again end up with a 32-bit word.
• The 32-bits of the previous step then go through a P-box based

permutation, as shown in Figure 4.
• What comes out of the P-box is then XORed with the left half
of the 64-bit block that we started out with. The output of this
19
XORing operation gives us the right half block for the next round.
• Note that the goal of the substitution step implemented by the

S-box is to introduce diffusion in the generation of the output
from the input. Diffusion means that a change in any plaintext
bit must propagate out to as many ciphertext bits as possible.
• The strategy used for creating the different round keys from the
main key is meant to introduce confusion into the encryption
process. Confusion in this context means that the relation-
ship between the encryption key and the ciphertext must be
as complex as possible. Another way of describing confusion
would be that each bit of the key must affect as many bits as
possible of the output ciphertext block.
• Diffusion and confusion are the two cornerstones of block cipher

design.
20
LE i−1 RE i−1
32 bits 32 bits
The Feistel Function
F( RE i−1 , K i )
Expansion Permutation
48 bits
Round Key K
i
48 bits
Substitution with 8 S−boxes
32 bits
Permutation with P−Box
LE RE
i i
Figure 4: One round of processing in DES. (This figure is from

Lecture 3 of “Lecture Notes on Computer and Network Security” by Avi Kak)
21
3.3.2: The S-Box for the Substitution Step in Each

Round
• As shown in Figure 5, the 48-bit input word is divided into eight

6-bit words and each 6-bit word fed into a separate S-box. Each
S-box produces a 4-bit output. Therefore, the 8 S-boxes together
generate a 32-bit output. As you can see, the overall substitution
step takes the 48-bit input back to a 32-bit output.
• Each of the eight S-boxes consists of a 4 × 16 table lookup for an

output 4-bit word. The first and the last bit of the 6-bit input
word are decoded into one of 4 rows and the middle 4 bits decoded
into one of 16 columns for the table lookup.
• The goal of the substitution carried out by an S-box is to enhance

diffusion, as mentioned in the previous subsection. As you will
recall from the E-step described in Section 3.3.1, the expansion-
permutation step (the E-step) expands a 32-bit block into a 48-bit
block by attaching a bit at the beginning and a bit at the end of
each 4-bit sub-block, the two bits needed for these attachments
belonging to the adjacent blocks.
• Thus, the row lookup for each of the eight S-boxes becomes a
function of the input bits for the previous S-box and the next
22
48 bits produced by XORing the output of the Expansion

Permutation and the Round Key
48 bits
S1 S2 S3 S4 S5 S6 S7 S8
32 bits
Figure 5: The 48 bits coming out of the expansion permu-

tation are first XORed with the round key and then, as
shown, fed into the 8 S-boxes of DES. (This figure is from Lecture 3 of
“Lecture Notes on Computer and Network Security” by Avi Kak)
23
S-box.
• In the design of the DES, the S-boxes were tuned to enhance the
resistance of DES to what is known as the differential cryptanal-
ysis attack, or, sometimes more briefly as differential attack. [As
will be explained in much greater detail (and also demonstrated) in Section 8.9 of Lecture 8, differential
cryptanalysis of block ciphers consists of presenting to the encryption algorithm pairs of plaintext bit
patterns with known differences between them and examining the differences between the correspond-
ing cyphertext outputs. The outputs are usually recorded at the input to the last round of the cipher.
Let’s represent one plaintext bit block by X = [X1 , X2 , ...., Xn ] where Xi denotes the ith bit in the
block, and let’s represent the corresponding output bit block by Y = [Y1 , Y2 , ..., Yn ]. By the difference
between two plaintext bit blocks X ′ and X ′′ we mean ∆X = X ′ ⊕ X ′′ . The difference between the
corresponding outputs Y ′ and Y ′′ is given by ∆Y = Y ′ ⊕ Y ′′ . The pair (∆X, ∆Y ) is known as a dif-
ferential. In an ideally randomizing block cipher, the probability of ∆Y being a particular value for a
given ∆X is 1/2n for an n-bit block cipher. What is interesting is that the probabilities of ∆Y taking
on different values for a given ∆X can be shown to be independent of the encryption key because of the
properties of the XOR operator, but these probabilities are strongly dependent on the S-box tables. By
feeding into a cipher several pairs of plaintext blocks with known ∆X and observing the corresponding
∆Y , it is possible to establish constraints on the round key bits encountered along the different paths
in the encryption processing chain. (By constraints I mean the following: Speaking hypothetically for
the purpose of illustrating a point and focusing on just one round of DES, suppose we can show that
the following condition can be expected to be obeyed with high probability: ∆Xi ⊕ ∆Yi ⊕ Ki = 0
for some bit Ki of the encryption key, then it must be the case that Ki = ∆X ⊕ ∆Y .) Note that
differential cryptanalysis is a chosen plaintext attack, meaning that the attacker will feed known
plaintext bit patterns into the cipher and analyze the corresponding outputs in order to figure out the
encryption key. In a theoretical analysis of an attack based on differential cryptanalysis, it was shown
by Eli Biham and Adi Shamir in 1990 that the DES’s encryption key could be figured out provided one
24
could feed known 247 plaintext blocks into the cipher. For a tutorial by Howard Heys on differential
cryptanalysis, see http://www.engr.mun.ca/~howard/PAPERS/ldc_tutorial.pdf. The title of the
tutorial is “A Tutorial on Linear and Differential Cryptanalysis.” ]
25
3.3.3: The Substitution Tables
• Shown on the next page are the eight S-boxes, S1 through S8,
each S-box being a 4×16 substitution table that is used to convert
6 incoming bits into 4 outgoing bits.
• As mentioned earlier, each row of a substitution table is indexed

by the two outermost bits of a six-bit block and each column by
the remaining inner 4 bit.
26
The 4 × 16 substitution table for S-box S1

14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 7
0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 8
4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 0
15 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13
S-box S2
15 1 8 14 6 11 3 4 9 7 2 13 12 0 5 10
3 13 4 7 15 2 8 14 12 0 1 10 6 9 11 5
0 14 7 11 10 4 13 1 5 8 12 6 9 3 2 15
13 8 10 1 3 15 4 2 11 6 7 12 0 5 14 9
S-box S3
10 0 9 14 6 3 15 5 1 13 12 7 11 4 2 8
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
1 10 13 0 6 9 8 7 4 15 14 3 11 5 2 12
S-box S4
7 13 14 3 0 6 9 10 1 2 8 5 11 12 4 15
13 8 11 5 6 15 0 3 4 7 2 12 1 10 14 9
10 6 9 0 12 11 7 13 15 1 3 14 5 2 8 4
3 15 0 6 10 1 13 8 9 4 5 11 12 7 2 14
S-box S5
2 12 4 1 7 10 11 6 8 5 3 15 13 0 14 9
14 11 2 12 4 7 13 1 5 0 15 10 3 9 8 6
4 2 1 11 10 13 7 8 15 9 12 5 6 3 0 14
11 8 12 7 1 14 2 13 6 15 0 9 10 4 5 3
S-box S6
12 1 10 15 9 2 6 8 0 13 3 4 14 7 5 11
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
4 3 2 12 9 5 15 10 11 14 1 7 6 0 8 13
S-box S7
4 11 2 14 15 0 8 13 3 12 9 7 5 10 6 1
13 0 11 7 4 9 1 10 14 3 5 12 2 15 8 6
1 4 11 13 12 3 7 14 10 15 6 8 0 5 9 2
6 11 13 8 1 4 10 7 9 5 0 15 14 2 3 12
S-box S8
13 2 8 4 6 15 11 1 10 9 3 14 5 0 12 7
1 15 13 8 10 3 7 4 12 5 6 11 0 14 9 2
7 11 4 1 9 12 14 2 0 6 10 13 15 3 5 8
2 1 14 7 4 10 8 13 15 12 9 0 3 5 6 11
27
• The Python code shown below illustrates how you can use the
eight S-boxes for the substitutions you need for the right half of
the input in each round:
#!/usr/bin/env python
## illustrate_des_substitution.py
## Avi Kak
## January 21, 2017 (made Python3 compliant January 12, 2018)
## This is a demonstration of how you can carry out S-boxes based substitution
## in DES. The code shown implements the "Substitution with 8 S-boxes" step
## that you see inside the dotted Feistel function in Figure 4 of Lecture 3 notes.
## IMPORTANT: This demonstration code does NOT include XORing with the round
## key that must be carried out on the expanded right-half block
## before it is subject to the S-boxes based substitution step
## shown here.
from BitVector import *
expansion_permutation = [31, 0, 1, 2, 3, 4,
3, 4, 5, 6, 7, 8,
7, 8, 9, 10, 11, 12,
11, 12, 13, 14, 15, 16,
15, 16, 17, 18, 19, 20,
19, 20, 21, 22, 23, 24,
23, 24, 25, 26, 27, 28,
27, 28, 29, 30, 31, 0]
s_boxes = {i:None for i in range(8)}
s_boxes[0] = [ [14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7],
[0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8],
[4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0],
[15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13] ]
s_boxes[1] = [ [15,1,8,14,6,11,3,4,9,7,2,13,12,0,5,10],
[3,13,4,7,15,2,8,14,12,0,1,10,6,9,11,5],
[0,14,7,11,10,4,13,1,5,8,12,6,9,3,2,15],
[13,8,10,1,3,15,4,2,11,6,7,12,0,5,14,9] ]
s_boxes[2] = [ [10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8],
[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],
[1,10,13,0,6,9,8,7,4,15,14,3,11,5,2,12] ]
s_boxes[3] = [ [7,13,14,3,0,6,9,10,1,2,8,5,11,12,4,15],
[13,8,11,5,6,15,0,3,4,7,2,12,1,10,14,9],
[10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4],
28
[3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14] ]
s_boxes[4] = [ [2,12,4,1,7,10,11,6,8,5,3,15,13,0,14,9],
[14,11,2,12,4,7,13,1,5,0,15,10,3,9,8,6],
[4,2,1,11,10,13,7,8,15,9,12,5,6,3,0,14],
[11,8,12,7,1,14,2,13,6,15,0,9,10,4,5,3] ]
s_boxes[5] = [ [12,1,10,15,9,2,6,8,0,13,3,4,14,7,5,11],
[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],
[4,3,2,12,9,5,15,10,11,14,1,7,6,0,8,13] ]
s_boxes[6] = [ [4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1],
[13,0,11,7,4,9,1,10,14,3,5,12,2,15,8,6],
[1,4,11,13,12,3,7,14,10,15,6,8,0,5,9,2],
[6,11,13,8,1,4,10,7,9,5,0,15,14,2,3,12] ]
s_boxes[7] = [ [13,2,8,4,6,15,11,1,10,9,3,14,5,0,12,7],
[1,15,13,8,10,3,7,4,12,5,6,11,0,14,9,2],
[7,11,4,1,9,12,14,2,0,6,10,13,15,3,5,8],
[2,1,14,7,4,10,8,13,15,12,9,0,3,5,6,11] ]
def substitute( expanded_half_block ):

’’’
This method implements the step "Substitution with 8 S-boxes" step you see inside
Feistel Function dotted box in Figure 4 of Lecture 3 notes.
’’’
output = BitVector (size = 32)
segments = [expanded_half_block[x*6:x*6+6] for x in range(8)]
for sindex in range(len(segments)):
row = 2*segments[sindex][0] + segments[sindex][-1]
column = int(segments[sindex][1:-1])
output[sindex*4:sindex*4+4] = BitVector(intVal = s_boxes[sindex][row][column], size = 4)
return output
# For the purpose of this illustration, let’s just make up the right-half of a
# 64-bit DES bit block:
right_half_32bits = BitVector( intVal = 800000700, size = 32 )
# Now we need to expand the 32-bit block into 48 bits:

right_half_with_expansion_permutation = right_half_32bits.permute( expansion_permutation )
print("expanded right_half_32bits: %s" % str(right_half_with_expansion_permutation))
# The following statement takes the 48 bits back down to 32 bits after carrying
# out S-box based substitutions:
output = substitute(right_half_with_expansion_permutation)
print(output)
• By visually examining each line of the expansion permutation array

shown at the beginning of the script, you can tell how each 4-bit
29
segment of the input 32-bit block is going to be given at its left

the last bit of the previous 4-bit segment and, at its right, the
first bit of the next 4-bit segment. For example, the first 4-bit
segment of the input 32-bit block will be at index positions 0, 1,
2, and 3. At its left, we attach the bit position at index 31, which
is the last bit of the previous (circularly speaking) 4-bit block.
And, at its right, we attach the bit position at index 4, which the
first bit of the next 4-bit block.
• Shown below is a Perl implementation of the same script. As

with the Python code, the script that follows illustrates how you
can use the eight S-boxes for the substitutions you need for the
right half of the input in each round. [Note that the Perl script shown
below will only work if the Algorithm::BitVector installed in your computer is of
version 1.26 or higher.]
#!/usr/bin/perl -w
## illustrate_des_substitution.pl
## Avi Kak
## January 15, 2018
## This is a demonstration of how you can carry out S-boxes based substitution
## in DES. The code shown implements the "Substitution with 8 S-boxes" step
## that you see inside the dotted Feistel function in Figure 4 of Lecture 3 notes.
## IMPORTANT: This demonstration code does NOT include XORing with the round
## key that must be carried out on the expanded right-half block
## before it is subject to the S-boxes based substitution step
## shown here.
use strict;
use Algorithm::BitVector 1.26;
my $expansion_permutation = [31, 0, 1, 2, 3, 4,
3, 4, 5, 6, 7, 8,
30
7, 8, 9, 10, 11, 12,

11, 12, 13, 14, 15, 16,
15, 16, 17, 18, 19, 20,
19, 20, 21, 22, 23, 24,
23, 24, 25, 26, 27, 28,
27, 28, 29, 30, 31, 0];
my $s_boxes = { map {$_ => undef} 0..7 };
$s_boxes->{0} = [ [14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7],
[0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8],
[4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0],
[15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13] ];
$s_boxes->{1} = [ [15,1,8,14,6,11,3,4,9,7,2,13,12,0,5,10],
[3,13,4,7,15,2,8,14,12,0,1,10,6,9,11,5],
[0,14,7,11,10,4,13,1,5,8,12,6,9,3,2,15],
[13,8,10,1,3,15,4,2,11,6,7,12,0,5,14,9] ];
$s_boxes->{2} = [ [10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8],
[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],
[1,10,13,0,6,9,8,7,4,15,14,3,11,5,2,12] ];
$s_boxes->{3} = [ [7,13,14,3,0,6,9,10,1,2,8,5,11,12,4,15],
[13,8,11,5,6,15,0,3,4,7,2,12,1,10,14,9],
[10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4],
[3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14] ];
$s_boxes->{4} = [ [2,12,4,1,7,10,11,6,8,5,3,15,13,0,14,9],
[14,11,2,12,4,7,13,1,5,0,15,10,3,9,8,6],
[4,2,1,11,10,13,7,8,15,9,12,5,6,3,0,14],
[11,8,12,7,1,14,2,13,6,15,0,9,10,4,5,3] ];
$s_boxes->{5} = [ [12,1,10,15,9,2,6,8,0,13,3,4,14,7,5,11],
[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],
[4,3,2,12,9,5,15,10,11,14,1,7,6,0,8,13] ];
$s_boxes->{6} = [ [4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1],
[13,0,11,7,4,9,1,10,14,3,5,12,2,15,8,6],
[1,4,11,13,12,3,7,14,10,15,6,8,0,5,9,2],
[6,11,13,8,1,4,10,7,9,5,0,15,14,2,3,12] ];
$s_boxes->{7} = [ [13,2,8,4,6,15,11,1,10,9,3,14,5,0,12,7],
[1,15,13,8,10,3,7,4,12,5,6,11,0,14,9,2],
[7,11,4,1,9,12,14,2,0,6,10,13,15,3,5,8],
[2,1,14,7,4,10,8,13,15,12,9,0,3,5,6,11] ];
# For the purpose of this illustration, let’s just make up the right-half of a
# 64-bit DES bit block:
my $right_half_32bits = Algorithm::BitVector->new( intVal => 800000700, size => 32 );
# Now we need to expand the 32-bit block into 48 bits:

my $right_half_with_expansion_permutation = $right_half_32bits->permute( $expansion_permutation );
31
print "expanded right_half_32bits: $right_half_with_expansion_permutation\n";
# The following statement takes the 48 bits back down to 32 bits after carrying
# out S-box based substitutions:
my $output = substitute($right_half_with_expansion_permutation);
print "$output\n";
## This method implements the step "Substitution with 8 S-boxes" step you see inside
## Feistel Function dotted box in Figure 4 of Lecture 3 notes.
sub substitute {
my $expanded_half_block = shift;
my $output = Algorithm::BitVector->new( size => 32 );
my @segments = map $expanded_half_block->get_slice([$_*6..($_+1)*6]), 0..7;
foreach my $sindex (0..@segments-1) {
my $row = 2*int($segments[$sindex]->get_bit(0)) + int($segments[$sindex]->get_bit(-1));
my $column = int( $segments[$sindex]->get_slice([1..5]) );
$output->set_slice([$sindex*4..$sindex*4+4],
Algorithm::BitVector->new(intVal => $s_boxes->{$sindex}->[$row][$column], size => 4));
}
return $output;
}
• You can download both the Python and the Perl scripts shown
in this section from the lecture notes website.
32
3.3.4: The P-Box Permutation in the Feistel Function
The last step in the Feistel function shown in Figure 4 is labeled

“Permutation with P-Box”. The permutation sequence is shown
below. [It looks like a table, but it is not — as explained below]
P-Box Permutation
15 6 19 20 28 11 27 16
0 14 22 25 4 17 30 9
1 7 23 13 31 26 2 8
18 12 29 5 21 10 3 24
• This permutation ‘table’ says that the 0th output bit will be the
15th bit of the input, the 1st output bit the 6th bit of the input,
and so on, for all of the 32 bits of the output that are obtained
from the 32 bits of the input.
• Do NOT associate any meaning with the row-organization of the

table — except for the following: Each row of the table tells us
how to select the input bits for the output byte corresponding to
the row. For example, for the second output byte, the first entry
in the second row means that the 0th bit of the second output
byte — meaning the 8th bit of the output — will be the 0th bit
33
of the 32-bit input. Note that bit indexing is 0-based — as it

would be in your Perl or Python script
• Keep in mind the fact that, when using the BitVector module
in Python or the Algorithm::BitVector module in Perl, a permu-
tation such as the one shown above can be carried out with a
one-line command. For example, in Python, the code fragment
would look like:
sboxes_output = BitVector representation of the
output of the S-Boxes
right_half = sboxes_output.permute( pbox_permutation )
where permute() is a method defined for the BitVector class.

The argument pbox permutation you see above is the sequence
of all the entries in the ‘table’ on the previous page expressed as
a one-dimensional array.
34
3.3.5: The DES Key Schedule: Generating the Round

Keys
• The 56-bit encryption key is represented by 8 bytes, with the last

bit (the least significant bit) of each byte used as a parity bit.
• The relevant 56 bits are subject to a permutation at the beginning

before any round keys are generated. This is referred to as Key
Permutation 1 that is shown in Section 3.3.6.
• At the beginning of each round, we divide the 56 relevant key bits

into two 28 bit halves and circularly shift to the left each half by
one or two bits, depending on the round, as shown in the table
on the next page.
• For generating the round key, we join together the two halves and
apply a 56 bit to 48 bit contracting permutation (this is referred
to as Permutation Choice 2, as shown in Section 3.3.7) to the
joined bit pattern. The resulting 48 bits constitute our round
key.
• The contraction permutation shown in Permutation Choice 2,

along with the one-bit or two-bit rotation of the two key halves
35
prior to each round, is meant to ensure that each bit of the original
encryption key is used in roughly 14 of the 16 rounds.
• The two halves of the encryption key generated in each round are
fed as the two halves going into the next round.
• The table shown below tells us how many positions to use for
the left circular shift that is applied to the two key halves at the
beginning of each round:
Round Number Number of left shifts

1 1
2 1
3 2
4 2
5 2
6 2
7 2
8 2
9 1
10 2
11 2
12 2
13 2
14 2
15 2
16 1
• When using the BitVector module for programming in Python,

or the Algorithm::BitVector module for programming in Perl,
the steps described above for splitting the 56-bit key, circular-
shifting each half separately, and then rejoining the two halves
36
can be carried simply by a command sequence that in Python

looks like
[left,right] = key_bv.divide_into_two()
left << shifts[i]
right << shifts[i]
rejoined_key_bv = left + right
where key bv is the BitVector representation of the 56-bit key

entering the round and shifts is the array that consists of the
second column entries in the table shown on the previous page.
The method divide into two() is defined for the BitVector
class.
• The Python code shown in Section 3.3.7 is an illustration of how

you can implement the steps described above.
37
3.3.6: Initial Permutation of the Encryption Key
Key Permutation 1
56 48 40 32 24 16 8
0 57 49 41 33 25 17
9 1 58 50 42 34 26
18 10 2 59 51 43 35
62 54 46 38 30 22 14
6 61 53 45 37 29 21
13 5 60 52 44 36 28
20 12 4 27 19 11 3
• The bit indexing is based on using the range 0-63 for addressing
the bit positions in an 8-byte bit pattern in which the last bit of
each byte is used as a parity bit. [Note that each row shown above has only
7 positions — the positions corresponding to the parity bit are NOT included above.
That is, you will NOT see the positions 7, 15, etc., listed in the permutations shown.
Nevertheless, the bit addressing spans the full 0-63 range.]
The permutations
shown above do not constitute a table, in the sense that the
rows and the columns do NOT carry any special and separate
meanings. The permutation order for the bits is given by reading
the entries shown from the upper left corner to the lower right
corner.
• This permutation tells us that the 0th bit of the output will be
38
the 56th bit of the input (in a 64 bit representation of the 56-bit
encryption key), the 1st bit of the output the 48th bit of the input,
and so on, until finally we have for the 55th bit of the output the
3rd bit of the input.
• When programming in Python using the BitVector module, or in

Perl using the Algorithm::BitVector module, the permutations
shown on the previous page can be carried out trivially by call-
ing the permute() method of the modules. Using Python to
illustrate, you could call
user_key_bv = BitVector( textstring = user-supplied_key )
key_bv = user_key_bv.permute( initial_permutation )
where, as mentioned earlier, permute() is a method defined for

the BitVector class and initial permutation is the permu-
tation shown on the previous slide expressed as a 1-dimensional
array of integers.
• The code snippet shown below illustrates how you can create the
56-bit key from the eight characters supplied by the user.
## get_encryption_key.py
## Avi Kak
## January 21, 2017
## This scripts asks the user to supply eight characters (exactly) for
## the encryption key needed for DES based encryption/decryption.
39
import sys
key_permutation_1 = [56,48,40,32,24,16,8,0,57,49,41,33,25,17,
9,1,58,50,42,34,26,18,10,2,59,51,43,35,
62,54,46,38,30,22,14,6,61,53,45,37,29,21,
13,5,60,52,44,36,28,20,12,4,27,19,11,3]
def get_encryption_key():
key = ""
while True:
if sys.version_info[0] == 3:
key = input("Enter a string of 8 characters for the key: ")
else:
key = raw_input("Enter a string of 8 characters for the key: ")
if len(key) != 8:
print("\nKey generation needs 8 characters exactly. Try again.\n")
continue
else:
break
key = BitVector(textstring = key)
key = key.permute(key_permutation_1)
return key
key = get_encryption_key()
print("Here is the 56-bit encryption key generated from your input:\n")
print(key)
• Shown below is a Perl script for doing the same thing — it il-
lustrates how you can create the 56-bit encryption key from the
eight characters supplied by a user.
#!/usr/bin/perl -w
## get_encryption_key.pl
## Avi Kak
## January 14, 2018
use strict;
use Algorithm::BitVector;
my $key_permutation_1 = [56,48,40,32,24,16,8,0,57,49,41,33,25,17,
40
9,1,58,50,42,34,26,18,10,2,59,51,43,35,
62,54,46,38,30,22,14,6,61,53,45,37,29,21,
13,5,60,52,44,36,28,20,12,4,27,19,11,3];
my $key = get_encryption_key();
print "Here is the 56-bit encryption key generated from your input:\n";
print "$key\n";
sub get_encryption_key {
my $key = "";
print "\nEnter a string of 8 characters for the key: ";
while ( $key = <STDIN> ) {
chomp $key;
if (length $key != 8) {
print "\nKey generation needs 8 characters exactly. Try again: ";
next;
} else {
last;
}
}
$key = Algorithm::BitVector->new( textstring => $key );
$key = $key->permute($key_permutation_1);
return $key
}
41
3.3.7: Contraction-Permutation that Generates the

48-Bit Round Key from the 56-Bit Key
Key Permutation 2
13 16 10 23 0 4 2 27
14 5 20 9 22 18 11 3
25 7 15 6 26 19 12 1
40 51 30 36 46 54 29 39
50 44 32 47 43 48 38 55
33 52 45 41 49 35 28 31
• As with the Key Permutation 1 shown in the previous section,

bit addressing shown above for Key Permutation 2 uses the full
0-63 range in an 8-byte pattern. Since the last bit of each byte is
used as a parity bit, you will not see the bit positions 7, 15, 23,
etc., in the permutation shown above.
• As with permutation shown on the previous page, what is shown

above is NOT a table, in the sense that the rows and the columns
do not carry any special and separate meanings. The permutation
order for the bits is given by reading the entries shown from the
upper left corner to the lower right corner.
42
• Since there are only six rows and there are 8 positions in each
row, the output will consist of 48 bits.
• When programming in Python using the BitVector class, the

permutations shown on the previous page can be carried out triv-
ially by calling the permute() method of the class, as mentioned
earlier.
• The Python code shown below illustrates how you can generate
all 16 round keys using the BitVector module:
## generate_round_keys.py
## Avi Kak
## January 21, 2017
import sys
key_permutation_1 = [56,48,40,32,24,16,8,0,57,49,41,33,25,17,
9,1,58,50,42,34,26,18,10,2,59,51,43,35,
62,54,46,38,30,22,14,6,61,53,45,37,29,21,
13,5,60,52,44,36,28,20,12,4,27,19,11,3]
key_permutation_2 = [13,16,10,23,0,4,2,27,14,5,20,9,22,18,11,
3,25,7,15,6,26,19,12,1,40,51,30,36,46,
54,29,39,50,44,32,47,43,48,38,55,33,52,
45,41,49,35,28,31]
shifts_for_round_key_gen = [1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1]
def generate_round_keys(encryption_key):
round_keys = []
key = encryption_key.deep_copy()
for round_count in range(16):
[LKey, RKey] = key.divide_into_two()
shift = shifts_for_round_key_gen[round_count]
LKey << shift
RKey << shift
43
key = LKey + RKey

round_key = key.permute(key_permutation_2)
round_keys.append(round_key)
return round_keys
def get_encryption_key():
key = ""
while True:
if sys.version_info[0] == 3:
key = input("\nEnter a string of 8 characters for the key: ")
else:
key = raw_input("\nEnter a string of 8 characters for the key: ")
if len(key) != 8:
print("\nKey generation needs 8 characters exactly. Try again.\n")
continue
else:
break
key = BitVector(textstring = key)
key = key.permute(key_permutation_1)
return key
encryption_key = get_encryption_key()
round_keys = generate_round_keys(encryption_key)
print("\nHere are the 16 round keys:\n")
for round_key in round_keys:
print(round_key)
• And the Perl code shown below illustrates how you can generate
all 16 round keys using the Algorithm::BitVector module:
#!/usr/bin/perl -w
## get_encryption_key.pl
## Avi Kak
## January 14, 2018
## It subsequently generates the round keys for each of the 16 rounds
## of DES.
use strict;
use Algorithm::BitVector;
my $key_permutation_1 = [56,48,40,32,24,16,8,0,57,49,41,33,25,17,
9,1,58,50,42,34,26,18,10,2,59,51,43,35,
62,54,46,38,30,22,14,6,61,53,45,37,29,21,
44
13,5,60,52,44,36,28,20,12,4,27,19,11,3];
my $key_permutation_2 = [13,16,10,23,0,4,2,27,14,5,20,9,22,18,11,
3,25,7,15,6,26,19,12,1,40,51,30,36,46,
54,29,39,50,44,32,47,43,48,38,55,33,52,
45,41,49,35,28,31];
my $shifts_for_round_key_gen = [1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1];
my $encryption_key = get_encryption_key();
my @round_keys = generate_round_keys($encryption_key);
print "\nHere are the 16 round keys:\n";
foreach my $round_key (@round_keys) {
print "$round_key\n";
}
sub generate_round_keys {
my $encryption_key = shift;
my @round_keys = ();
my $key = $encryption_key->deep_copy();
foreach my $round_count (0..15) {
my ($LKey, $RKey) = $key->divide_into_two();
my $shift = $shifts_for_round_key_gen->[$round_count];
$LKey = $LKey << $shift;
$RKey = $RKey << $shift;
$key = $LKey + $RKey;
my $round_key = $key->permute($key_permutation_2);
push @round_keys, $round_key;
}
return @round_keys;
}
sub get_encryption_key {
my $key = "";
print "\nEnter a string of 8 characters for the key: ";
while ( $key = <STDIN> ) {
chomp $key;
if (length $key != 8) {
print "\nKey generation needs 8 characters exactly. Try again: ";
next;
} else {
last;
}
}
$key = Algorithm::BitVector->new( textstring => $key );
$key = $key->permute($key_permutation_1);
return $key
}
45
3.4: WHAT MAKES DES A STRONG

CIPHER (TO THE EXTENT IT IS A
STRONG CIPHER)
• The substitution step is very effective as far as diffusion is con-

cerned. It has been shown that if you change just one bit of
the 64-bit input data block, on the average it propagates out to
affect 34 bits of the ciphertext block.
• The manner in which the round keys are generated from the
encryption key is also very effective as far as confusion is con-
cerned. It has been shown that if you change just one bit of
the encryption key, on the average that affects 35 bits of the
ciphertext.
• Both effects mentioned above are referred to as the avalanche

effect.
• And, of course, the 56-bit encryption key means a key space of

size 256 ≈ 7.2 × 1016.
46
• Assuming that, on the average, you’d need to try half the keys
in a brute-force attack, a machine able to process 1000 keys per
microsecond would need roughly 13 months to break the code.
However, a parallel-processing machine trying 1 million keys si-
multaneously would need only about 10 hours. (EFF took
three days on a specially architectured machine to
break the code.)
• The official document that presents the DES standard can be

found at:
http://www.itl.nist.gov/fipspubs/fip46-2.htm
47
3.5: HOMEWORK PROBLEMS
1. A text file named myfile.txt that you created with a run-of-

the-mill editor contains just the following word:
hello
If you examine this file with a command like
hexdump -C myfile.txt
you are likely to see the following bytes (in hex) in the file:
68 65 6C 6C 6F 0A
Let’s now try to encrypt the contents of this text file with a 4-bit
block cipher whose codebook contains the following entries:
6, 0, 13, 4, 3, 1, 14, 8, 7, 12, 9, 15, 5, 2, 11, 10
Let’s say that I write the encrypted output into a different file and
then examine this new file with the ‘hexdump -C’ command.
What will I see in the encrypted file?
2. In general, in a block cipher, we replace N bits from the plaintext

with N bits of ciphertext. What defines an ideal block cipher?
48
3. Whereas it is true that the relationship between the input and

the output is completely random for an ideal block cipher, it must
nevertheless be invertible for decryption to work. That implies
that the mapping between the input blocks and the output blocks
must be one-to-one. If we had to express this mapping in the form
of a table lookup, what will be the size of the table?
4. What would be the encryption key for an ideal block cipher?
5. What makes ideal block ciphers impractical?
6. What do we mean by a “Feistel Structure for Block Ciphers”?
7. Are there any constraints on the Feistel function F in a Feistel

structure?
8. Explain the concepts of diffusion and confusion as used in DES.
9. If we have all the freedom in the world for choosing the Feistel
function F, how should we specify it?
10. How does the permutation/expansion step in DES enhance dif-

fusion? This is the step in which we expand by permutation and
repetition the 32-bit half-block into a 48-bit half-block
49
11. DES encryption was broken in 1999. Why do you think that
happened?
12. Since DES was cracked, does that make this an unimportant
cipher?
13. Programming Assignment 1:

Write a Perl or Python script that implements the full DES. Use
the S-boxes that are specified for the DES standard (See Section
3.3.3). Make sure you implement all of the key generation steps
outlined in Section 3.3.5. For the encryption key, your script
should prompt the user for a keyboard entry that consists of at
least 8 printable ASCII characters. (You may choose to either
use the first seven or the last seven bits of each character byte for
the 56-bit key you need for DES.)
What makes this homework not as difficult as you think is that
once you write the code that carries out one round of processing,
you basically use the same code in a loop for the whole encryp-
tion chain and the decryption chain. Obviously, you will have
to reverse the order in which the round keys are used for the
decryption chain.
Although you are free to write your own code from scratch, here
are some recommendations: If using Python, you might want to
start with the my BitVector class. To help you get started with
the Python implementation, please see the hw2_starter.py
50
file. If using Perl, use my Algorithm::BitVector module from

www.cpan.org. It is a popular Perl module for manipulating
bit arrays. It is also well documented. To help you get started
with the Perl implementation, please see the hw2_starter.pl
file. You can download both these starter files through the code
archive for Lecture 3.
14. Programming Assignment 2:

Now modify the implementation you created for the previous
homework by filling the 4 × 16 tables for the S-boxes with ran-
domly generated integers. Obviously, each randomly generated
entry will have to be between 0 and 15, both ends inclusive. Cal-
culate the avalanche effect for this implementation of DES and
compare it with the same effect for your previous implementation.
(See Section 3.3.1 for the avalanche effect.)
51

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Lecture 3: Block Ciphers and the Data Encryption

Lecture Notes on “Computer and Network Security”

by Avi Kak (kak@purdue.edu)

January 15, 2019

• To introduce the notion of a block cipher in the modern context.

• To introduce the notion of the Feistel Cipher Structure

• To go over DES, the Data Encryption Standard

• To illustrate important DES steps with Python and Perl code

Section Title Page

3.1 Ideal Block Cipher 3

3.2 The Feistel Structure for Block Ciphers 7

3.2.1 Mathematical Description of Each Round in the 10

3.2.2 Decryption in Ciphers Based on the Feistel Structure 12

3.3 DES: The Data Encryption Standard 16

3.3.1 One Round of Processing in DES 18

3.3.2 The S-Box for the Substitution Step in Each Round 22

3.3.3 The Substitution Tables 26

3.3.4 The P-Box Permutation in the Feistel Function 33

3.3.5 The DES Key Schedule: Generating the Round Keys 35

3.3.6 Initial Permutation of the Encryption Key 38

3.3.7 Contraction-Permutation that Generates the 48-Bit 42

3.4 What Makes DES a Strong Cipher (to the 46

3.5 Homework Problems 48

3.1: IDEAL BLOCK CIPHER

• In a modern block cipher (but still using a classical encryption

• To understand Figure 1, note that there are 16 different possible

• In an ideal block cipher, the relationship between the input blocks

ing to the input bit blocks to the integers corresponding to the

• Figure 1 depicts an ideal block cipher that uses blocks of size 4.

Convert 4 incoming bits to one of 16 integers

Random 1−1 mapping of 16 input integers

Convert integer to a 4−bit pattern

Ciphertext bit block: c c c c

3.1.1: The Size of the Encryption Key for the Ideal

• Consider the case of 64-bit block encryption.

• With a 64-bit block, we can think of each possible input block

3.2: The Feistel Structure for Block Ciphers

The DES (Data Encryption Standard) algorithm for encryption and

• A cryptographic system based on Feistel structure uses the same

• As shown in Figure 2, the Feistel structure consists of multiple

Figure 2: The Feistel Structure for symmetric key cryptog-

• The permutation step at the end of each round consists of swap-

• The next two subsection present important properties of the Feis-

3.2.1: Mathematical Description of Each Round in the

• In the Feistel structure, the relationship between the output of

where ⊕ denotes the bitwise EXCLUSIVE-OR operation. The symbol

• F is referred to as the Feistel function, after Horst Feistel natu-

• Assuming 16 rounds of processing (which is typical), the output

3.2.2: Decryption in Ciphers Based on the Feistel

• As shown in Figure 3, the decryption algorithm is exactly the

• The output of each round during decryption is the

Plaintext block Round Keys Plaintext block

Ciphertext block Ciphertext block

Figure 3: When a Feistel structure is used, decryption

RD1 = LD0 ⊕ F (RD0, K16)

• The following equalities are used in the above derivation. Assume

• The above result is independent of the precise nature

3.3: DES: THE DATA ENCRYPTION

• Adopted by NIST in 1977.