Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Material Type: Notes; Professor: Gaj; Class: Advanced Applied Cryptography; Subject: Electrical & Computer Enginrg; University: George Mason University; Term: Unknown 1989;
Typology: Study notes
1 / 19
This page cannot be seen from the preview
Don't miss anything!
Studied the problem of finding algebraic solutions for the general
equations of the degree 5, e.g.,
f(x) = a
5
x
5
4
x
4
3
x
3
2
x
2
1
x+ a
0
Answered definitely the question which specific equations of
a given degree have algebraic solutions
On the way, he developed group theory ,
one of the most important branches of modern mathematics.
1829 Galois submits his results for the first time to
the French Academy of Sciences
Reviewer 1
Augustin-Luis Cauchy forgot or lost the communication
1830 Galois submits the revised version of his manuscript,
hoping to enter the competition for the Grand Prize
in mathematics
Reviewer 2
Joseph Fourier – died shortly after receiving the manuscript
1831 Third submission to the French Academy of Sciences
Reviewer 3
Simeon-Denis Poisson – did not understand the manuscript
and rejected it.
May 1832 Galois provoked into a duel
The night before the duel he writes a letter to his friend
containing the summary of his discoveries.
The letter ends with a plea:
“ Eventually there will be, I hope, some people who
will find it profitable to decipher this mess. ”
May 30, 1832 Galois is grievously wounded in the duel and dies
in the hospital the following day.
1843 Galois manuscript rediscovered by Joseph Liouville
1846 Galois manuscript published for
the first time in a mathematical journal
Group
i) · is associative e.g., (5 · 7) · 13 = 5 · (7 · 13)
ii) Identity element = 1 a · 1 = 1 · a = a
iii)
No inverse of a for any a 1 or - 1
iv) · is commutative e.g.,
e.g., there is no integer x , such that 5 · x = 1
Group
n
i) + mod n is associative
e.g., (((5+7) mod 16) + 13) mod 16 = (5+((7 + 13) mod 16)) mod 16
ii) Identity element = 0 (0+ a ) mod n = ( a + 0 ) mod n = a
iii) Inverse of a = 0 for a =
n - a otherwise
e.g., 7 + (16-7) =
7 + 9 mod 16 = 0
iv) + mod n is commutative e.g., 5 + 8 mod 16 = 8 + 5 mod 16
Group
n
- {0} = {1, 2, …, n - 1}, · mod n : multiplication modulo n )
i) · mod n is associative
e.g., (((5·7) mod 16) · 4) mod 16 = (5 ·((7 · 4) mod 16)) mod 16
ii) Identity element = 1 ( a ·1) mod n = (1 · a ) mod n = a
iii) There is no inverse of a for any a
that is not relatively prime with n
iv) · mod n is commutative e.g., (5 · 8) mod 16 = (8 · 5) mod 16
e.g., there is no x Z
n
such that
(2 · x) mod 16 = 1
Group
n
__*
i) · mod n is associative
e.g., (((4 ·7) mod 15) · 2) mod 16 = (4 ·((7 · 2) mod 15)) mod 16
ii) Identity element = 1
( a ·1) mod n = (1 · a ) mod n = a
iii)
There is an inverse for every
element of the group
iv) · mod n is commutative e.g., (5 · 8) mod 15 = (8 · 5) mod 15
e.g., (2 · 8) mod 15 = 1
(4 · 4) mod 15 = 1
(7 ·13) mod 15 = 1
(11 ·11) mod 15 = 1
For n = 15, Z
n
*
Group
p
__*
i) · mod n is associative
e.g., (((4 ·7) mod 11) · 2) mod 11 = (4 ·((7 · 2) mod 11)) mod 11
ii) Identity element = 1 ( a ·1) mod p = (1 · a ) mod p = a
iii) There is an inverse for every
element of the group
iv) · mod n is commutative e.g., (5 · 8) mod 11 = (8 · 5) mod 11
e.g., (2 · 6) mod 11 = 1
(3 · 4) mod 11 = 1
(5 · 9) mod 11 = 1
(7 ·8) mod 11 = 1
For p = 11, Z
p
*
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (11)= 11-1=
Cyclic Group
Test for a generator of a cyclic group
11
*
10/
5
10/
2
Result: 2 is a generator of Z
11
__*
10/
5
10/
2
Result: 3 is NOT a generator of Z
11
__*
Subgroups of a Cyclic Group
Performing operations in the cyclic
groups using logarithms and
antilogarithms (1)
i
g
j
g
i
j
i+j
log a + log b
(log a + log b) mod |G|
Performing operations in the cyclic
groups using logarithms and
antilogarithms (2)
k
g
i
j
i+j
log a + log b
(log a + log b) mod |G|
Example:
1
mod 11 = 2
2
mod 11 = 4
3
mod 11 = 8
4
mod 11 = 5
5
mod 11 = 10
6
mod 11 = 9
7
mod 11 = 7
8
mod 11 = 3
9
mod 11 = 6
10
mod 11 = 1
k alog[k]
1 2 3 4 5 6 7 8 9
10
2
4
8
5
10
9
7
3
6
1
k log[k]
1 2 3 4 5 6 7 8 9
10
10
1 8 2 4 9 7 3 6 5
3 · 5 mod 11 = alog[(log[3]+log[5]) mod 10] =
= alog[(8+4) mod 10] = alog[2] = 4
Definition of a Ring
Field
No inverse of a for any a 1 or - 1
e.g., there is no integer x , such that 5 · x = 1
n
No inverse of a if a is not relatively prime with n
e.g., there is no x Z
n
, such that 2 · x = 1 mod 16
Field
p
p
, + mod p , mod p ) is a commutative ring
There is multiplicative inverse for all numbers
from Z
p
i)
ii)
e.g.,
(2 · 6) mod 11 = 1 2
mod 11 = 6
(3 · 4) mod 11 = 1 3
mod 11 = 4
(5 · 9) mod 11 = 1 5
mod 11 = 9
(7 · 8) mod 11 = 1 7
mod 11 = 8
Field
p
Sets of polynomials
Sets of polynomials
3
2
n
n
3
2
2
2
3
2
3
Polynomial rings
n
2
2
i) (Z 2
[x], +) is an abelian group with identity element 0
ii) · is associative
iii) · has an identity element = 1
f(x) · 1 mod n = 1 · f(x) mod n = f(x)
iv) · is distributive over +
e.g., (x
2
+x+1) · ((x+1)+(x
2
(x
2
+x+1) · (x+1)+(x
2
+x+1) ·(x+1)
e.g., ((x
2
+x+1) · (x+1)) · (x
2
+1) = (x
2
+x+1) · ((x+1) · (x
2
+1))
Polynomial rings
2
p
(x
3
2
4
= (x
5
3
2
) + (x
3
4
= x
5
2
4
= x · (x
4
2
4
2
(x
3
2
4
3
2
Finite fields
Finite fields
q
m
2
m
q
m
p
m
e.g., for f(x) = x
3
(x+1) · (x
2
3
mod f(x) = x
2
+x
Finite Fields = Galois Fields
m
Polynomial basis
representation
Normal basis
representation
Fast in hardware
Arithmetic
operations
present
in many libraries
Fast squaring
m
p – prime
p
m
- number of
elements in the field
Most significant
special cases
Primitive Element
Primitive Polynomial
m
Binary representation
(used for storing and processing in computer systems):
Polynomial representation
(used for the definition of basic arithmetic operations):
m- 1
m- 2
2
1
0
i
i
i
m- 1
m- 1
m- 2
m- 2
2
2
1
0
i=
m- 1
Addition and Multiplication
in the Galois Field GF(
m
)
Inputs
m- 1
m- 2
2
1
0
m- 1
m- 2
2
1
0
i
i
Output
m- 1
m- 2
2
1
0
i
m- 1
m- 1
m- 1
m- 2
m- 2
m- 2
2
2
2
1
1
0
0
m- 1
m- 1
m- 2
m- 2
2
2
1
0
Addition in the Galois Field GF(
m
)
i
i
i
i
i
m- 1
m- 1
m- 2
m- 2
2
2
1
0
Multiplication in the Galois Field GF(
m
)
m
m
m- 1
m- 1
2
2
1
0
Extended Euclid’s Algorithm
Example z = 20
mod 117
i
r
i
r
r
r
0
r 1
r
2
r
3
r
4
x
i
x
x
x
0
x
1
x
2
x
3
- 1
mod 117
x
4
q
i
q
q
0
q
1
q
2
q
3
q i
r
i - 1
r
i
r
i +
= r
i - 1
- q
i
r
i
x
i +
= x
i - 1
- q
i
x
i
Check:
20 41 mod 117 = 1
