Supervisor: Mrs. Pomares

Mathematic HL

Portfolio

Analysis of the stream cipher

symmetrical cryptography

encryption technic

Saturnin Pugnet Session: May 2014

Candidate Number: 001386-0055 IBS of Provence

2 * : see bibliography

Index 1. Introduction------------------------------------------------------3

2. Rational-----------------------------------------------------------3

3. Definition----------------------------------------------------------4

3.1 Basic definitions--------------------------------------------------4

3.2 Introduction to matrix-------------------------------------------4

4. Foreword---------------------------------------------------------7

5. Stream cipher---------------------------------------------------9

5.1 Presentation------------------------------------------------------9

5.2 PNRG---------------------------------------------------------------9

5.3 LSFR-------------------------------------------------------------10

5.3.1 Explanation-------------------------------------------------------10

5.3.2 Example of breaking an LSFR-------------------------------13

5.4 Algebraic Normal Form---------------------------------------16

5.5 A5/1 encryptions-----------------------------------------------18

5.6 RC4 encryptions-----------------------------------------------19

5.6.1 KSA-----------------------------------------------------------------19

5.6.2 PRGA----------------------------------------------------------20

6. Conclusion------------------------------------------------------22

7. Bibliography----------------------------------------------------23

3 * : see bibliography

1. Introduction

Ever since the Roman Empire, man has been using encrypted messages to

communicate, becoming more and more complicated over time. Nowadays with the

development of computer technologies, encrypting data in an efficient way has

become more and more important. Thus the mathematical technic used to encrypt

the message has also become more and more complicated.

Today, there are two types of encryptions: symmetrical encryption and asymmetrical

encryption. We are only going to study symmetrical encryption during this

exploration. In symmetrical encryption, there are two ciphers types:

- The block cipher which consists in encoding a message by blocks of characters

- The stream cipher that consists of encrypting the message character by character.

The goal of this exploration is to understand how symmetrical encryption works, but it

is too vast to be entirely study therefore, we are going to analyze only some stream

cipher encryption types :A5/1, RC4, LSFR.

2. Rationale For many years I have been interested by the computer sciences and I saw this math

portfolio as a good opportunity to develop my knowledge in this subject. Although the

mathematics used in cryptography is not part of the IB program I did my own

research to be able to study it. This study was a really interesting and enriching

project which allows me to develop my understanding of security networks in

computers and the way it is link to mathematics.

4 * : see bibliography

3. Definitions 3.1 Basic definitions

Binary: It is the computer’s language; it is in base 2 using 0 and 1.

Bit: It is binary (0 or 1), it represents one figure. To convert from bit to number I have

used an online converter1.

Octet: It is a sequence of 8 bits.

Key: It is the series that is used to produce an encrypted message by making

operation between the message and the key; it can be numbers, letters, signs etc.

Hexadecimal: It is numbers in base 16 were the 10, 11, 12, 13, 14 and 15 are

represented by A, B, C, D, E and F.

Example: From 1 to 16 it is 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F.

⊕ (XOR): The XOR is a simple binary operation (addition or subtraction) without

taking into account the “borrow” or “carry” addition or subtraction

give the same result in this case, that is why it is considered as a

single operation in cryptography (XOR).

Ex: 1 + 1 = 10 => without the carry = 0 = 1 XOR 1,

1-1=0 it is the same as 1 XOR 1

⊗ (AND): The AND is a binary multiplication of one figure (1 or 0) with another.

Ex: 10101111

⊗ 11

= 010101111

⊕ 101011110

= 111110001

Modulo (mod): Considering A and B two integers, A mod B gives the remainder of an

Euclidian division of A with B. It is often used to set an upper limit for the results in

cryptography. For example when dealing with letters the upper limit must be 26

because there are only 26 letters in the alphabet.

Ex: 5 mod 2 = 5-2×2 = 1; 51 mod 26 = 51-26×1 = 25

: This represents the multiplication from i=1 to i=N of Ci.

Ex: =21 22 23 24 25 26 27 28 29 210

: Galois field9 (2) which is a finite field composed of two elements (0 and 1) where

the addition and subtraction are XOR and the multiplication are AND; most of the

calculi are done in this field as it is the one used in cryptography.

A B A XOR B

0 0 0

1 0 1

0 1 1

1 1 0

A B A AND B

0 0 0

1 0 0

0 1 0

1 1 1

5 * : see bibliography

3.2 Introduction to matrix2

Matrix: It is an array of number

this is an example of matrix with 2 columns and 2 rows.

A matrix can be multiplied by a scalar. Every number in the matrix is multiplied to the

scalar.

X ×

=

where X

A matrix can be added or subtracted to another matrix but it must have exactly the

same number of rows and columns. When you add them together the number in the

first row, first column of the first matrix will be added to the number first row and first

column of the second matrix.

+

=

A matrix can be multiplied to another matrix but the number of column of the first

matrix must be equal to the number of rows of the second one. You multiplied each

rows of the first matrix by the columns of the second matrix. The first number of the

first row of the first matrix is multiplied with the first number of the first column of the

second matrix, the second number with the second number…etc. and they are finally

added together to give the number at the position of the row of the first one and the

column of the second one.

×

=

Ex:

× =

=

Finally, the identity matrix is a matrix I such as A × I = I × A = A and it is characterize

by the diagonal from the top left to the bottom right of the number 1 and 0 in all other

positions. An identity matrix is always a square matrix (same numbers of rows than

columns). We usually write In where n is the number of rows and column.

Example: I2 =

, I3 =

and I4 =

Inverse matrix: The inverse Matrix of a matrix A is A-1 such as A× A-1 = I = A-1 × A

to find the inverse matrix of a 2×2 matrix you must use this formula

A-1 =

×

where A =

Example: A =

A-1 =

×

=

×

=

6 * : see bibliography

For bigger matrix we can find the inverse by using an equation system as we know I

and A, during my analysis I won’t show my calculations to find inverse matrix but I

calculated it myself using equations.

7 * : see bibliography

4. Foreword 4.1 Caesar

3 cipher

The use of cryptography is really old and has become more and more complex over

time. At the beginning it was simple mathematic technic such as the Caesar cipher,

used by Caesar during war time. It was not really secure but most of Caesar’s

enemies were illiterate and would assume that the message was written in an

unknown language. But as it is a simple mathematical model it had quickly become

obsolete.

Caesar cipher works on the principle that each letter correspond to a number (A=0

B=1 C=2 D=3 E=4 F=5 G=6 H=7 I=8 J=9 K=10 L=11 M=12 N=13 O=14 P=15 Q=16

R=17 S=18…) and that you change the value of each letter according to the key with

a maximum value of 26 as there is 26 letters in the alphabet.

Assuming Sn is the series of letters corresponding to the message, Cn is the series

corresponding to the encoded message and K is the key: Cn = (Sn+K) mod 26

Ex: message=HELLO with key=4:

(H+4) mod 26 = (7+4) mod 26 = 11 = L

(E+4) mod 26 = (4+4) mod 26 = 8 = I

(L+4) mod 26 = (11+4) mod 26 = 15 = P

(L+4) mod 26 = (11+4) mod 26 = 15 = P

(O+4) mod 26 = (14+4) mod 26 = 18 = S

LIPPS is the encrypted message of HELLO using Caesar cipher with a key of 4 in the

Caesar cipher.

4.2 Vinegère3 cipher

The Vinegère cipher has been the first one to introduce keys of more than one

character into the encryption. It was created at the 16th century and was known as

the strongest encryption technic. Nevertheless, it was broken by skilled cryptanalyst

during the 17th century but declared officially broken at the 19th century only and out-

of-date.

The key is not as long as the message but is repeated to get the same length (key

abc for a message of 7 letters the key becomes abc abc a). To find the cipher

number you associate the letters to number (A=0, B=1, C=2…) like for the Caesar

cipher and then you add them to the key:

Assuming Sn is the series of letters corresponding to the message, Cn is the series

corresponding to the encoded message, Kn is a series representing the key and L is

the length of the key: Cn = (Sn + (Kn mod L )) mod 26

8 * : see bibliography

Ex: message=HELLO key=ABC

(H+A) mod 26 = (7+0) mod 26 = 7 = H

(E+B) mod 26 = (4+1) mod 26 = 5 = F

(L+C) mod 26 = (11+2) mod 26 = 13 = N

(L+A) mod 26 = (11+0) mod 26 = 11 = L

(O+B) mod 26 = (14+1) mod 26 = 15 = P

therefore the encrypted message would be HFNLP for HELLO using Vinegère cipher

with the key ABC.

4.3 Vernam3 cipher

A more recent encryption technic is the Vernam cipher. The main characteristic of the

Vernam cipher is that the key has the same length than the message. It was created

in 1917, and it was used during the war. The new thing about this kind of cipher is

that it is unbreakable if the key is generated randomly and if it is used only once. It is

the ancestor of stream cipher that we are going to study.

The principle is that each letter of the message corresponds to one character of the

key as in the Caesar cipher the message is modify by adding the value of the

message and the key.

Assuming Sn is the series of figure corresponding to the message, Cn is the series

corresponding to the coded message, Kn is the series corresponding to the key: Cn =

(Sn + Kn) mod 26

Ex: message=HELLO key=18654

(H+1) mod 26 = (7+1) mod 26 = 8 = I

(E+8) mod 26 = (4+8) mod 26 = 12 = M

(L+6) mod 26 = (11+6) mod 26 = 17 = R

(L+5) mod 26 = (11+5) mod 26 = 16 = Q

(O+4) mod 26 = (14+4) mod 26 = 18 = S

At the end you get IMRQS with the message HELLO using the Vernam cipher and

the key 18654.

We are often going to use binary during this study as the computer work with it. The

binary numbers after it has been encrypted by the computer sending the message

and decrypted by the computer receiving the message, are converted into letters,

informations or instruction. Moreover the binary model is just one way of studying it,

that is why sometimes number in base 10, 15, 64 or 264 are used.

9 * : see bibliography

5. Stream Cipher 5.1 Presentation

A scheme of how the streams cipher4,6 work

(⊕=XOR)

Fig 1: Scheme of a stream cipher

We define K as the series representing the key that is known by both the one that

sends the message and the one that receives it.

The Initialization Vector is a series that is changed between each encryption to avoid

that the PRNG (Pseudo Random Number Generator) is found as it is not completely

cryptographically sure (we are going to demonstrate this after). The initialization

vector can be added in several ways. For example it could use the function XOR with

the key as a parameter.

5.2 PNRG

The PRNG is a function that creates random numbers (0 or 1), it works according to

the Golomb’s criteria5 of randomness which are:

For the series , ….

1) The number of 1 and 0 must be almost equal:

0

2) A series is a suite of identical bit next to each other between two opposite bits

(example of series 1 10 01; ). There are as much series of 0

as 1.

There is S/2 series of length 1

Initialization

Vector Pseudo Random

Number

Generator

Encrypted

message

0 or 1

0 or 1

K

Message

10 * : see bibliography

There is S/4 series of length 2

There is S/8 series of length 3

…

There is S/ series of length k

3) And finally there must be no correlation between two bits.

5.3 LSFR5

The PNRG can be an LSFR (Linear Feedback Shift Register) which is a function that

follows the Golomb’s criteria. 5.3.1 Explanation

LSFR (Linear Feedback Shift Register) are used in many stream ciphers such as

A5/1(part 5.5 of the explorations) or GSM(Global System Of Mobile Communication).

It can be represented by a series where the first bit is the output and the n last bit of

the series are used to define the next term of the series, by using the function XOR to

them. Each time an output is defined a new number is added to the series. The

following diagram is a scheme of how a LFSR work :

We defined S the series that represent the LSFR where the initial state is =0, =1,

=0, =1, =1, =1, =0 and the coefficients that represents which number are

going to be used to create the new number of the series S (1 means that they are

used, 0 means they are not used) are =1, =0, =1, =0, =1, =1, =0.

Output: GREEN

Bit used to define the new bits of the series S: RED

Result of the XOR operation of the RED bits: BLUE

EXAMPLE 1

Time

1

0

1

1

1

0

1

0/0

Output

11 * : see bibliography

Time

2

Time 3

The relation between the bits can be represented by that equation in the Galois Field

(2), the matrix is called the multiplication matrix6 where t is the number of round done

(number of output), S the series created by the LSFR, I a positive integer and C is the

series representing the bits used to create the next bits in S (coefficients):

=

.

1

0

0

1

1

1

0/0

Output 10

0

0

1

1

1

0

1/1

Output 0

12 * : see bibliography

For the EXAMPLE 1:

=

.

=

.

-Period of an LSFR

For a series of n terms we know that there are different possibilities as there are

only 2 figures (0 and 1); therefore the series S must be periodic. We also know

according to the formulae above that if everything is initialized at 0 it is going to stay

at zero: = 0. Therefore the maximum length of a period of an LSFR is -1.

-Feedback Polynomial

We define the feedback polynomial of an LFSR of length L and with the coefficient C

= , … as: feedback polynomial = +

, It is what represents an

LSFR.

For example the feedback polynomial of the EXAMPLE 1 ( =1, =0, =1, =0,

=1, =1, =0) L=6:

+

= + 1× + 0× + 1× + 0× + 1× + 1× + 0×

= + + + + 1

13 * : see bibliography

5.3.2 Example of breaking an LSFR6

A LFSR (which is a PRNG) is not cryptographically sure. Assuming L is the number

of bits when it initialized, we can find the complete series knowing only 2L bits

consecutive. I propose a demonstration:

(All the multiplications and additions are made in F2)

The series created by the LSFR initialized with 4 bits:

???? ???? ???? ???? 0111 1011

The goal is to find the inverse matrix of the multiplication matrix A, the matrix to

create the next bit of the serie, to be able to go back in the series

1. First we want to find the coefficients.

We define M a matrix representing the 8 bits M =

If we multiply M by the coefficients we get the bit following each rows of M:

M .

=

= M-1 .

M . M-1 = I4

.

=

M-1 =

Therefore the coefficients are

=

.

=

Now, we know that C1=1, C2=1, C3=1, C4=1

And thus the Feedback Polynomial of that LSFR is X4+ X3+ X2 +X+1 (explanation

feedback polynomial part 5.3.1)

14 * : see bibliography

2. Now we want to find the multiplication matrix A.

We consider A the multiplication matrix of the LSFR and Sn the series created by it:

. A = .

And we know that S5 is the XOR operation of the term of the series which are

selected by the coefficient

S5 = .

= .

= ×S1 + ×S2 + ×S3 + ×S4

= S1 + S2 + S3 + S4

We consider A =

and we know that . A =

For the first column of the matrix ×S1+ ×S2+ ×S3+ ×S4 = S2 therefore =0, =1,

=0, =0

For the second column of the matrix ×S1+ ×S2+ ×S3+ ×S4 = S3 therefore =0,

=0, =1, =0

On the third column of the matrix ×S1+ ×S2+ ×S3+ ×S4 = S4 therefore =0,

=0, =0, =1

On the last column of the matrix ×S1+ ×S2+ ×S3+ ×S4 = S5 = S4 + S3 + S2 + S1

therefore =1, =1, =1, =1

Therefore A=

3. Thirdly we find the inverse matrix A-1 from A.

A . A-1 = I4

.

=

15 * : see bibliography

A-1 =

4. And finally we use the inverse matrix A-1 to find the preceding bits.

We consider x a positive integer

. A =

. A-1 =

We multiply by the inverse matrix to go back in the LSFR

. A-1 =

. A-1 =

. A-1 =

. A-1 =

???? ???? ???? 1111 0111 1011

To break the code further you simply have to continue to multiply the four last digits

by the inverse matrix A-1 that we have found.

To conclude, we have proved that it is possible to break an LSFR knowing 2L bits

where L is the number of bits when initialized, which is why it is never used alone

because it would be too easy to break.

We are now going to analyze how it is used to overcome that problem.

16 * : see bibliography

5.4 Algebraic Normal Form

A way to use LSFR is to combine them by using the following technic

Fig 2: Scheme of an algebraic normal form function

To be able to explain F(x) we have to first of all define f(x1,x2,…xn) and Mb1b2…bn(x1,

x2… xn), two functions.

Note: do not confound F(x) and f(x)

We consider x1, x2… xn a series representing the output of the LSFR L1, L2… Ln (on

Fig 2) and we define the function f(x1, x2…xn) as a function defined by a table that

can vary depending on the encrypting method.

For this example we consider three LSFRs L1, L2, L3 and the three outputs x1, x2, x3s

The function f(x1, x2, x3) can be represented by the following table:

x1x2x3 000 001 010 100 011 101 110 111

f(x1, x2, x3) 1 0 1 1 0 0 0 1

Remember that the calculus are done in F2 (+XOR and ×AND)

For the second function which composes the Algebraic Normal Form we define the

function Mb1b2…bn (x1, x2… xn) that can be defined by the following equation:

Mb1b2…bn(x1, x2… xn) =

LSFR L1

LSFR L2

LSFR Ln

F(x) Output

17 * : see bibliography

Where we define b1, b2… bn as a series of bits and x1, x2… xn as the bits given by the

LSFRs

For the example we consider again three LSFRs L1, L2, L3 and the three outputs x1,

x2, x3

The function M010(x1, x2, x3) can be represented by the following table:

x1x2x3 000 001 010 100 011 101 110 111

M010(x1x2x3) 0 0 1 0 0 0 0 0

The algebraic normal form F(x) corresponds to the sum of the function Mb1b2…bn for

which f(b0, b1…bn) = 1.

We assume b is a series of bits, n is a positive integer (ex: b=001) and x is the series

representing the LSFRs outputs.

F(x) =

=

Example:

We consider the following truth table for three LSFRs:

x1x2x3 000 001 010 100 011 101 110 111

f(x1x2x3) 1 0 1 1 0 0 0 1

F(x) = 1 × M000(x1x2x3) + 0 × M001(x1x2x3) + 1 × M010(x1x2x3) + 1 × M100(x1x2x3) + 0 ×

M011(x1x2x3) + 0 × M110(x1x2x3) + 0 × M101(x1x2x3) + 1 × M111(x1x2x3)

F(x) = M000(x1x2x3) + M010(x1x2x3) + M100(x1x2x3) + M111(x1x2x3)

Therefore,

F(x) = (x1+1)(x2+1)(x3+1) + (x1+1)x2(x3+1) + x1(x2+1)(x3+1) + x1x2x3

That we can simplify by

F(x) = (x1x2x3 + x1x2 + x2x3 + x1x3 + x1+x2 + x3 + 1) + (x1x2x3 + x1x2 + x2x3 + x2) + (x1x2x3

+ x1x2 + x1x3 + x1) + x1x2x3

= x1x2 + x3 + 1

= x1 × x2 + x3 + 1

If three LSFRs where added using the Algebraic Normal Form with the table of truth

above the results would be x1 × x2 + x3 + 1

For example for the output of LSFR1 = 10111, LSFR2 = 11010, LSFR3 = 10100 the

total output (after the use of the algebraic normal form) would be:

x1 × x2 + x3 + 1

1×1+1+1=1

0×1+0+1=1

1×0+1+1=0

18 * : see bibliography

1×1+0+1=0

1×0+0+1=1

If we had the following outputs values LSFR1 = 10111, LSFR2 = 11010, LSFR3 =

10100, the total output of those three LSFRs would be 11001 with the table of truth

above and the Algebraic Normal Form x1x2+x3+1.

19 * : see bibliography

5.5 A5/17

A5/1 is a recent method of encryption which is another example of how to use the

LSFR in a secure way. It is used for phone (GSM).

This is a scheme of how does A5/1 works

This scheme represent three LSFR (each line of cells) with feedback polynomials

(explain in part 5.3) X18+X17+X16+X13+1, X21+X20+1, X22+ X21+X20+X7+1. Those are

represented by the numbers which are the cells of the coefficients (defined part 5.3) .

Those LSFRs are arranged in this way, adding the outputs together to make one

single output. That increases the difficulty to break the LSFRs.

A series of steps are done before using the output to encrypt the message in order to

include the key in the LSFRs and to randomize the three LSFRs. Those steps are too

long to be explained in this study. However, this technique is another way to increase

the complexity of the result making it more difficult to solve.

1 19 18 17 14

1 22 21

8 1 23 22 21

Output

20 * : see bibliography

5.6 RC48

RC4 (rivest cipher 4) is a recent encryption technic which is composed of two

algorithms: the KSA and the PRGA. This type of encryption is one example of stream

cipher that doesn’t use the LSFR. It is used for the WEP (Wi-Fi Encryption Technic)

or SSL (Secure Sockets Layer).

5.6.1 KSA

KSA works as following:

We assume that:

- L is the length of the key

- K is a series representing the key such as K1=1st term of the key

- n is a positive integer representing the number of round done (it starts at 2)

- S is a series of 256 octets where S1=0, S2=1, S3=2… S256=255 (each number is one

octet which is 8 bits)

- j is a series used with the first number of the series j1 = 1.

jn= ((jn-1 + + Kn-1 mod L-1) mod 256)+ 1, the nth term of the series j is defined by

the precedent term, the n-1th value of the key series and the jn-1th value of the S

series

Sn-1 , we exchange the two values of the numbers Sn-1 and the jn number of S

These operations are made 256 times for n=2, n=3 … n=257

To simplify in this example we put only 8 octets therefore it is mod 8 and not mod 256

and the key K values are [K1=5, K2=5, K3=6, K4=2, K5=7, K6=0, K7=6, K8=3] (the

binary values are converted into decimal number to simplify) S would be S1=0, S2=1,

S3=2, S4=3, S5=4, S6=5, S7=6, S8=7

First round:

1. j2 = ((1+0+5-1) mod 8) +1 = 6, We define j2 the second number of the series j by

adding the value of j1 which is one, the value of the last term of the series S which

was exchanged (as it is the first step this value is 0) and the first value of the key

series . Then we do “-1) mod 8)+1” at the end in order to have a result which is

always between 1 and 8, it is just a mathematical trick.

21 * : see bibliography

2. S1 S6, The value of the S series at the first place is then replaced by the value of

the number we found, 6th place.

3. We then repeat these steps 8 times.

n values jn= (jn-1 + + Kn-1 mod L) mod 8 Sn-1 S(0 1 2 3 4 5 6 7)

2 j2 = ((j2-1+ + K2-1 mod 8-1) mod 8)+1

j2 = ((1+0+5-1) mod 8) +1 = 6

S2-1

S1 S6

S6 S2 S3 S4 S5 S1 S7 S8 5 1 2 3 4 0 6 7

3 j3 = ((6+0+5-1) mod 8) +1 = 3 S2 S3 5 2 1 3 4 0 6 7

4 j4 = ((3+1+6-1) mod 8) +1 = 2 S3 S2 5 1 2 3 4 0 6 7

5 j5 = ((2+1+2-1) mod 8) +1 = 5 S4 S5 5 1 2 4 3 0 6 7

6 j6 = ((5+3+7-1) mod 8) +1 = 7 S5 S7 5 1 2 4 6 0 3 7

7 j7 = ((7+3+0-1) mod 8) +1 = 2 S6 S2 5 0 2 4 6 1 3 7

8 j8 = ((2+0+6-1) mod 8) +1 = 8 S7 S8 5 0 2 4 6 1 7 3

9 j9 = ((8+3+3-1) mod 8) +1 = 6 S8 S6 5 0 2 4 6 3 7 1

The final value of S after 8 round will be S1=5, S2=0, S3=2, S4=4, S5=6, S6=3, S7=7,

S8=1 with the key values [K1=5, K2=5, K3=6, K4=2, K5=7, K6=0, K7=6, K8=3]

5.6.2 PRGA (pseudo random generation algorithm)

We assume that:

- n is a positive integer representing the number of round done (starting at 2)

- S is the series from the KSA

- A is a series with the first number of the series A1 equal to 1

- O is a series which represent the output

- M is the series that represent the initial message

- E is a series representing the encrypted message.

An = ((An-1 + Sn -1) mod 256) + 1, the nth term of the series A is defined by the

precedent term and the nth term of the series from the KSA

Sn , the nth is exchanged with the Anth term of the S series

On = +1 , the output is defined by the value of the “Sn + mod

256”th term of the S series

On-1⊕Mn-1 = En-1 , The encrypted series is define by the function XOR of the “n-1”th

term of the output and the message series

To simplify the example we put only 8 octet therefore it is also mod 8 and we use the

result of the KSA example as S (S1=5, S2=0, S3=2, S4=4, S5=6, S6=3, S7=7, S8=1)

22 * : see bibliography

and the message M=101, 111, 110, 011, 000 (as it is one octet it should be 8 bits but

101 = 00000101)

First round:

1. A2 = ((1+0-1) mod 8)+1 = 1, We define the value of A2 the second number of the

series A by adding the first number of the series A, the value of the last term of the

series S that was exchanged. Then we do “-1) mod 8 )+1” at the end in order to have

a result which is always between 1 and 8, it is the same mathematical trick as before.

2. S1 S1, The value of the S series at the first place is then replaced by the value of

the number we found (which is also the 1th place this time).

3. O2-1 = S((0+5-1) mod 8)+1 = S5 = 6 = 110, We define the value of the first term of the O by

adding the two values of the number which were exchanged and then doing the “-1)

mod 8)+1” in order to have a result between 0 and 8. Finally we take the number in

the S series at this value.

4. The last step is to use the XOR function with the first octet of the message and add

it to the value of the first term of the series O.

5. Those steps are then repeated as long as there is a message to encrypt.

n

An=(An-1+Sn-1)mod 8+1 Sn S(5 0 2 4 6 3 7 1) On-1= +1 On–1⊕Mn-1

2 A2 = (A2-1+S2-1)mod 8+1 A2 = ((1+0-1)mod8)+1 = 1

S2

S2 S1

S2S1S3S4S5S6S7S8 0 5 2 4 6 3 7 1

O2-1= +1

S((0+5-1) mod 8)+1 = S5 = 6 = 110

O2–1⊕M2-1

110⊕101=011

3 A3 =((1+2-1) mod 8)+1= 3 S3 S3 0 5 2 4 6 3 7 1 S((2+2-1) mod 8)+1 = S4 = 4 = 100 100⊕111=011

4 A4 =((3+4-1) mod 8)+1= 7 S4 S7 0 5 2 7 6 3 4 1 S((7+4-1) mod 8)+1 = S3 = 2 = 010 010⊕110=100

5 A5 =((7+6-1) mod 8)+1= 5 S5 S5 0 5 2 7 6 3 4 1 S((6+6-1) mod 8)+1 = S4 = 7 = 111 111⊕011=100

6 A6 =((5+3-1) mod 8)+1= 8 S6 S8 0 5 2 7 6 1 4 3 S((1+3-1) mod 8)+1 = S4 = 7 = 111 111⊕000=111

To conclude, if your input in a RC4 is 101 111 110 011 000 and your key 5 5 6 2 7 0

6 3 you would obtain 011 011 100 100 111 if it uses 8 octets and not 256.

23 * : see bibliography

6. Conclusion Nowadays cryptography is used everywhere and the techniques that we have seen

during this study are used in many different domains: WEP, WPA, Bluetooth and

many other transmission systems. Therefore cryptography techniques are changed

regularly in order to avoid way to break the algorithm to be found. As we are

increasingly using transmission of information since the beginning of the internet and

as the encryption techniques are known by everyone the technique must be more

complicated and re-actualize really often in order to be up-to-date. The fact that the

technic of encryption can be known by anyone makes the algorithm even harder to

make, which is why mathematicians are employed to create such complex

algorithms. During this portfolio we have seen a fraction of the encryption technique

which itself is a small part of the computer security system. This domain has many

possibilities and there are still plenty of things to discover and analyze.

To conclude, we have seen that for symmetrical encryption by stream cipher

mathematic is really present. It allows creating technic to encrypt message as well as

technic to break the encrypted message such as for LSFR. During our exploration,

we used matrix, however, it is only one way to represent the encryption making it

easier to solve as equations with unknown numbers however it is not the only

mathematical way to do it. We have also seen that the main difference with other

areas of mathematic is that the operations are done in F2 (galois field (2)) which

make the calculus different. Overall, this exploration allowed me to discover a vast

new field of mathematic.

24 * : see bibliography

7. Bibliography

1Convertissor :

http://sebastienguillon.com/test/javascript/convertisseur.html

2Matrix :

Book: Mathematics for the IB diploma Higher Level 2 Cambridge Hugh Neill and Douglas

Quadling page from page 385 to 404

http://www.unilim.fr/pages_perso/jean.debord/math/matrices/matrices.htm

3Caesar, Vinegère and Vernam:

http://fr.openclassrooms.com/informatique/cours/les-premiers-algorithmes-de-

chiffrement/exemples-d-algorithmes-de-chiffrement

4Basis on stream cipher :

http://www.di.ens.fr/~bresson/P12-M1/P12-M1-Crypto_8.pdf

5LSFR :

http://fr.wikipedia.org/wiki/Registre_%C3%A0_d%C3%A9calage_%C3%A0_r%C3%A9troa

ction_lin%C3%A9aire

http://iml.univ-mrs.fr/~rodier/Cours/LFSR.pdf

http://www.academia.edu/1613421/An_analysis_of_linear_feedback_shift_register_in_stream

_ciphers

http://comsec.uwaterloo.ca/~ece493t/A1.pdf (Golomb randomness criteria)

6General stream cipher + breaking LSFR :

http://perso.univ-perp.fr/christophe.negre/Enseignements/Cryptographie/Master1/slide-stream-

cipher1.pdf

7A5/1 :

http://en.wikipedia.org/wiki/A5/1

http://www.youtube.com/watch?v=LgZAI3DdUA4

8RC4 :

http://www.math.washington.edu/~nichifor/310_2008_Spring/Pres_RC4%20Encryption.pdf

9Galois field :

http://en.wikipedia.org/wiki/Finite_field#F2