section 2 - radford · web view(2 and 3 are the divisors or factors of 6) (2 and 5 are factors or...
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Section 2
Section 2.2: Affine Ciphers; More Modular Arithmetic
Practice HW (not to hand in)
From Barr Textbook
p. 80 # 2a, 3e, 3f, 4, 5a, 7, 8
9, 10 (Use Maple)
Extra Problems
Find the following gcd’sAnswers
a. gcd(30, 40)
10
b. gcd(150, 500)
50
c. gcd(187, 455)
1
In shift ciphers, messages are encrypted by using an additive key. To increase security, we can, in addition to an additive parameter, encipher messages using a multiplicative parameter. In affine ciphers, the key used for encipherment involves using both a multiplicative and additive parameter. Before describing affine ciphers, we give some necessary mathematics background.
Mathematics Background for Affine Ciphers
All natural numbers (numbers in the set
}
,
5
,
4
,
3
,
2
,
1
{
K
can be expressed as the product of two or more numbers. For example,
3
2
6
×
=
,
5
2
2
5
4
20
×
×
=
×
=
, and
7
1
7
×
=
.
Two numbers that can be multiplied together to get another number are called the factors or divisors of that number. For example,
3
2
6
×
=
(2 and 3 are the divisors or factors of 6)
5
2
2
5
4
20
×
×
=
×
=
(2 and 5 are factors or divisors of 20).
7
1
7
×
=
(1 and 7 are the divisors or factors of 7).
Note that the only factors of 7 are 1 and itself. A number with this special type of property is said to be prime, which we formally define next.
Definition: A natural number
p
is said to be prime if
1
>
p
and its only divisors are 1 and
p
. A natural number that is not prime is said to be composite.
It can be shown that there are an infinite number of primes. The following set lists the first ten primes:
}
,
29
,
23
,
19
,
17
,
13
,
11
,
7
,
5
,
3
,
2
{
K
The prime numbers provide the building blocks of all numbers. The next theorem illustrates this fundamental fact.
Theorem: The Fundamental Theorem of Arithmetic. Every natural number larger than 1 is a product of primes. This factorization can be done in only one way if is disregarded.
For example, to factor 30, we can compute
4
4
4
4
3
4
4
4
4
2
1
order
ng
disregardi
ion
factorizat
prime
same
3
2
5
5
2
3
5
3
2
5
6
30
×
×
=
×
×
=
×
×
=
×
=
.
An elementary way to obtain prime factorizations with small prime factors involves the use of a calculator and a factor tree. The next two examples illustrate this technique.
Example 1: Factor 90
Solution: Using a calculator and the factor tree (the prime factors are the outer leaves),
we see that
5
3
3
2
90
×
×
×
=
EMBED Equation.3
█
Example 2: Factor 935
Solution: Using a calculator and the factor tree (the prime factors are the outer leaves),
we see that
17
11
5
935
×
×
=
.
█
Greatest Common Divisor
The common prime factors of two numbers can be used to find the greatest common divisor of two numbers, which we define next.
Definition: The greatest common divisor of two natural numbers
a
and
b
, denoted as
)
,
gcd(
b
a
, is the largest natural number that divides
a
and
b
with no remainder.
Elementary Method for Computing the gcd of Two Numbers
Decompose each number into its prime factors. The gcd is obtained by multiplying the prime factors the two numbers have in common. If the two numbers have no common prime factors, then the gcd = 1.
Example 3: Find the gcd(20, 30).
Solution: We first obtain the prime factorizations of 20 and 30 and circle the factors in common.
5
2
2
20
×
×
=
5
3
2
30
×
×
=
The gcd is the product of the common prime factors. Hence,
10
5
2
)
30
,
20
gcd(
=
×
=
. █
Example 4: Find the gcd(1190, 935).
Solution: We again look for the common prime factors.
17
7
5
2
1190
×
×
×
=
17
11
5
935
×
×
=
Hence,
85
17
5
)
935
,
1190
gcd(
=
×
=
.
█
Example 5: Find the gcd(15, 26).
Solution: The prime factorizations of 15 and 26 are
5
3
15
×
=
13
2
26
×
=
15 and 26 have no common prime factors. Thus,
1
)
26
,
15
gcd(
=
. 15 and 26 are said to be relatively prime.
█
Note: Two numbers
a
and
b
where the
1
)
,
gcd(
=
b
a
are said to be relatively prime.
Multiplicative Inverses
In the real number system, every non-zero number has a multiplicative inverse – the number you must multiply to a given number to get 1.
Example 6: Fill in the ( ) for
(
)
1
)
2
(
=
,
(
)
1
)
10
(
=
, and
(
)
1
)
(
=
a
if we are working in the real number system.
Solution: In each case, we are looking for a multiplicative inverse. Hence,
1
2
1
)
2
(
=
÷
ø
ö
ç
è
æ
1(We say
1
2
2
1
-
=
is the multiplicative inverse of 2)
1
10
1
)
10
(
=
÷
ø
ö
ç
è
æ
(We say
1
10
10
1
-
=
is the multiplicative inverse of 10)
1
1
)
(
=
÷
ø
ö
ç
è
æ
a
a
(We say
1
1
-
=
a
a
, where
0
¹
a
, is the multiplicative inverse of
a
)
█
Note: In some number systems, multiplicative inverses in most cases do not exist.
Example 7: Consider the integers
}
,
4
,
3
,
2
,
1
,
0
,
1
,
2
,
3
,
4
,
{
K
K
-
-
-
-
. If we attempt to solve
(
)
1
)
2
(
=
and
(
)
1
)
10
(
=
, no solution exists in either case since
2
1
and
10
1
are not integers. In factor, 1 is the only integer that has a multiplicative inverse.
█
Note: In the modular arithmetic system, a multiplicative inverse may or may not exist, depending on the following fact involving the gcd:
*Fact: If the
1
)
,
gcd(
=
m
b
, then
b
has an inverse with respect to the modulus
m
, that is,
1
-
b
exists.
Example 8: Does 8 have an inverse with respect to the modulus 26?
Solution: No,
1
8
-
MOD 26 does not exist since
1
2
)
26
,
8
gcd(
¹
=
.
█
Example 9: Does 9 have and inverse with respect to the modulus 26?
Solution: Yes, 9 does have an inverse MOD 26 since
1
)
26
,
9
gcd(
=
. What is
1
9
-
MOD 26? To answer this question, we must fill in the blank to solve the problem
)(?)
9
(
MOD 26 = 1.
We claim that
3
9
1
=
-
since
)
3
)(
9
(
MOD 26 = 27 MOD 26 = 1. That is,
1
9
-
MOD 26 = 3.
█
Later in the course, we will see a general mathematical method for computing multiplicative inverses. For now, since we will work with a MOD 26 system, we will display a table showing the numbers in a MOD 26 with their multiplicative inverses:
a
1
3
5
7
9
11
15
17
19
21
23
25
26
1
MOD
a
-
1
9
21
15
3
19
7
23
11
5
17
25
Table 1: Multiplicative Inverses MOD 26
Example 10: Use Table 1 to find
1
7
-
MOD 26.
Solution: Table 1 shows that
1
7
-
MOD 26 = 15. Note that we can prove this by calculating (7)(15) MOD 26 = 105 MOD 26 = 1.
█
Multiplicative inverses expand our ability to solve equations and congruences in modular arithmetic. This is made possible using the multiplicative property of modular arithmetic, which we state next.
Multiplicative Property for Modular Arithmetic
If
m
b
a
mod
º
,
then for any number
k
,
m
kb
ka
mod
º
We illustrate how this property can be used in the following example.
Example 11: Solve
5
1
11
º
-
x
mod 26 for
x
.
Solution: Just like in your regular algebra class, the goal is to isolate
x
on
one side of the equation.
26
mod
5
1
11
º
-
x
26
mod
1
5
1
1
11
+
º
+
-
x
(Add 1 to both sides)
26
mod
6
11
º
x
(Simplify)
26
mod
6
11
11
11
1
1
×
º
×
-
=
x
(Multiply both sides by
1
11
-
)
NOTE!!
26
mod
11
6
11
11
º
x
is NOT correct in Modular
arithmetic
26
mod
6
19
×
º
x
(From the inverse table we see that
1
11
-
MOD 26 = 19.
26
mod
114
º
x
(Simplify))
10
=
x
(Compute the integer remainder)
█
Multiplicative inverses in modular arithmetic can be useful in solving systems of linear equations, which are useful for cryptanalysis. This next example illustrates this fact.
Example 12: Solve the system of equations (congruences)
26
mod
11
17
26
mod
18
8
º
+
º
+
b
a
b
a
Solution: Our goal is to find the values of a and b that simultaneously solve this system of equations. The process is similar to how systems of equations are solved in ordinary algebra. To keep track of our steps, we first number the equations
26
mod
18
8
º
+
b
a
(1)
26
mod
11
17
º
+
b
a
(2)
To eliminate the parameter b, we subtract equation (2) from equation (1):
26
mod
7
9
26
mod
11
17
26
mod
18
8
º
-
º
+
º
+
-
a
b
a
b
a
Since
17
26
mod
9
=
-
, we can write the resulting equation from the subtraction as:
26
mod
7
17
º
a
.
We next solve this result by multiplying both sides by
1
17
-
:
26
mod
7
17
17
17
1
1
×
º
×
-
-
a
.
Noting from the MOD 26 multiplicative inverse table that
23
26
mod
17
1
=
-
, we obtain
5
26
mod
161
26
mod
7
23
º
º
×
º
a
Hence
5
=
a
. We can substitute
5
=
a
into either equation (1) or (2) to find b. Choosing equation (1)
26
mod
18
8
º
+
b
a
, we obtain:
26
mod
18
)
5
(
8
º
+
b
or
26
mod
18
40
º
+
b
.
Subtracting 8 from both sides gives
26
mod
)
40
18
(
-
º
b
or
4
26
mod
22
º
-
º
b
.
Hence
5
=
a
and
4
=
b
solves the above system of equations.
█
Mathematical Description of Affine Ciphers
Given
a
and
b
in
}
25
,
24
,
,
2
,
1
,
0
{
26
K
=
Z
where
1
)
26
,
gcd(
=
a
. We encipher a plaintext letter
x
to obtain a ciphertext letter
y
by computing
)
(
b
ax
y
+
=
MOD 26.
Here, the key is made up of a multiplicative parameter
a
and an additive parameter
b
.
The next example illustrates how a message is enciphered.
Example 13: Encipher “RADFORD” using the affine cipher
)
4
5
(
+
=
x
y
MOD 26.
Solution: Using the MOD 26 alphabet assignment table, we encipher the message by computing
L
MOD
MOD
MOD
y
x
R
Þ
=
=
+
=
+
=
Þ
=
Þ
11
26
89
26
)
4
85
(
26
)
4
)
17
(
5
(
17
E
MOD
MOD
MOD
y
x
A
Þ
=
=
+
=
+
=
Þ
=
Þ
4
26
4
26
)
4
0
(
26
)
4
)
0
(
5
(
0
T
MOD
MOD
MOD
y
x
D
Þ
=
=
+
=
+
=
Þ
=
Þ
19
26
19
26
)
4
15
(
26
)
4
)
3
(
5
(
3
D
MOD
MOD
MOD
y
x
F
Þ
=
=
+
=
+
=
Þ
=
Þ
3
26
29
26
)
4
25
(
26
)
4
)
5
(
5
(
5
W
MOD
MOD
MOD
y
x
O
Þ
=
=
+
=
+
=
Þ
=
Þ
22
26
74
26
)
4
70
(
26
)
4
)
14
(
5
(
14
L
MOD
MOD
MOD
y
x
R
Þ
=
=
+
=
+
=
Þ
=
Þ
11
26
89
26
)
4
85
(
26
)
4
)
17
(
5
(
17
T
MOD
MOD
MOD
y
x
D
Þ
=
=
+
=
+
=
Þ
=
Þ
19
26
19
26
)
4
15
(
26
)
4
)
3
(
5
(
3
Hence, the ciphertext is “LETDWLT”
█
Note: Recall that for an affine cipher
)
(
b
ax
y
+
=
MOD 26 to be defined properly,
1
)
26
,
gcd(
=
a
. Besides allowing a recipient to decipher a message, the next example illustrates another reason why this requirement is essential.
Example 14 Use the affine cipher
)
1
2
(
+
=
x
y
MOD 26 to encipher “AN”.
Solution: Note that
2
=
a
and
1
2
)
26
,
2
gcd(
¹
=
. Enciphering “AN” gives
B
MOD
MOD
MOD
y
x
A
Þ
=
=
+
=
+
=
Þ
=
Þ
1
26
1
26
)
1
0
(
26
)
1
)
0
(
2
(
0
B
MOD
MOD
MOD
y
x
N
Þ
=
=
+
=
+
=
Þ
=
Þ
1
26
27
26
)
1
26
(
26
)
1
)
13
(
2
(
13
The two letters A and N both encipher to the same letter B. If a sender was to transmit this message to a recipient, the recipient will not be able to decipher the message uniquely, which is not desirable.
█
Deciphering an Affine Cipher
For an affine cipher
26
mod
)
(
b
ax
y
+
º
where
1
)
26
,
gcd(
=
a
, decipherment can be done uniquely. Given the numerical representation of the plaintext message
x
and ciphertext message
y
, we take
26
mod
)
(
b
ax
y
+
º
26
mod
y
b
ax
º
+
(Rearrange both sides)
26
mod
)
(
b
y
ax
-
º
(Subtract
b
from both sides)
26
mod
)
(
1
1
b
y
a
x
a
a
-
º
-
-
(Multiply both sides by
1
-
a
)
26
mod
)
(
1
b
y
a
x
-
º
-
(Cancel and simplify)
Hence,
)
(
1
b
y
a
x
-
=
-
MOD 26 is the decipherment formula for affine ciphers. We illustrate this formula with an example.
Example 15: Decipher the message “ARMMVKARER” that was encrypted using the affine cipher
26
)
5
3
(
MOD
x
y
+
=
(*)
Solution: Recall here that x is the numerical representation of the plaintext letter and y is the numerical representation of ciphertext letter. To decipher, we must solve equation (*) for x. We do this using the following steps:
26
5
3
MOD
y
x
=
+
(switch both sides of the equation across the “=” mark)
26
)
5
(
3
MOD
y
x
-
=
(subtract 5 from both sides of the equation)
26
)
21
(
3
MOD
y
x
+
=
(use the fact that
21
26
5
=
-
MOD
)
26
)
21
(
3
3
3
1
1
MOD
y
x
+
=
-
-
(multiply both sides by the multiplicative inverse of 3.
NOTE: In MOD arithmetic the division operation
26
3
21
3
3
MOD
y
x
+
=
is NOT CORRECT!!)
26
)
21
(
9
MOD
y
x
+
=
(use the multiplicative inverse MOD 26 table to see that
9
26
3
1
=
-
MOD
).
Using the decipherment formula
26
)
21
(
9
MOD
y
x
+
=
, we can decipher the message using the following repetitive steps:
H
MOD
MOD
MOD
x
y
A
Þ
=
=
=
+
=
Þ
=
Þ
7
26
189
26
)
21
(
9
26
)
21
0
(
9
0
E
MOD
MOD
MOD
x
y
R
Þ
=
=
=
+
=
Þ
=
Þ
4
26
342
26
)
38
(
9
26
)
21
17
(
9
17
L
MOD
MOD
MOD
x
y
M
Þ
=
=
=
+
=
Þ
=
Þ
11
26
297
26
)
33
(
9
26
)
21
12
(
9
12
L
MOD
MOD
MOD
x
y
M
Þ
=
=
=
+
=
Þ
=
Þ
11
26
297
26
)
33
(
9
26
)
21
12
(
9
12
O
MOD
MOD
MOD
x
y
V
Þ
=
=
=
+
=
Þ
=
Þ
14
26
378
26
)
42
(
9
26
)
21
21
(
9
21
T
MOD
MOD
MOD
x
y
K
Þ
=
=
=
+
=
Þ
=
Þ
19
26
279
26
)
31
(
9
26
)
21
10
(
9
10
H
MOD
MOD
MOD
x
y
A
Þ
=
=
=
+
=
Þ
=
Þ
7
26
189
26
)
21
(
9
26
)
21
0
(
9
0
E
MOD
MOD
MOD
x
y
R
Þ
=
=
=
+
=
Þ
=
Þ
4
26
342
26
)
38
(
9
26
)
21
17
(
9
17
R
MOD
MOD
MOD
x
y
E
Þ
=
=
=
+
=
Þ
=
Þ
17
26
225
26
)
25
(
9
26
)
21
4
(
9
4
E
MOD
MOD
MOD
x
y
R
Þ
=
=
=
+
=
Þ
=
Þ
4
26
342
26
)
38
(
9
26
)
21
17
(
9
17
Hence, the message is “HELLO THERE”.
█
Cryptanalysis of Affine Ciphers
For an affine cipher
)
(
b
ax
y
+
=
MOD 26, an enemy must know the multiplicative parameter
a
and additive parameter
b
in order to decipher and break a message. Once
a
and
b
are known,
)
(
1
b
y
a
x
-
º
-
MOD 26 can be computed and the message broken. Two methods of attack can be used to attempt to break an affine cipher.
Methods for Breaking and Affine Cipher
1. Exhaustion. Note there are 12 possible multiplicative parameters
a
where
1
)
26
,
gcd(
=
a
and 26 possible additive parameters
b
. This gives
312
26
12
=
×
total
)
,
(
b
a
pairs to test.
2.Frequency analysis. Quicker way which involves matching to highly frequently occurring ciphertext letters with two highly frequently occurring plaintext letters. Involves solving a system of equations MOD 26.
The next example illustrates method 2.
Example 16: Suppose we receive a ciphertext that was enciphered using an affine cipher. After running a frequency analysis on the ciphertext, we find out that the most highly frequently occurring letters in the ciphertext are W and H. Assuming that these letters correspond to E and T respectively, find the parameters
a
and
b
that were used in the affine cipher.
Solution: Recall that for an affine cipher
26
mod
)
(
b
ax
y
+
º
,
x
is the numerical representation of the plaintext letter and
y
is the numerical representation of the ciphertext letter. Hence, using the MOD 26 alphabet assignment and the equation
26
mod
)
(
b
ax
y
+
º
, we see that
Plaintext
4
=
Þ
x
E
corresponds to the ciphertext
22
=
Þ
y
W
which gives the equation (1)
26
mod
)
4
(
22
b
a
+
×
º
Plaintext
19
=
Þ
x
T
corresponds to the ciphertext
7
=
Þ
y
H
which gives the equation (2)
26
mod
)
19
(
7
b
a
+
×
º
Rearranging and putting these equations together gives
26
mod
22
4
º
+
b
a
(1)
26
mod
7
19
º
+
b
a
(2)
To find
a
, we must solve this system of equations.
To eliminate the parameter b, we subtract equation (2) from equation (1):
26
mod
15
15
26
mod
7
19
26
mod
22
4
º
-
º
+
º
+
-
a
b
a
b
a
Since
11
26
MOD
15
=
-
, we can write the resulting equation from the subtraction as:
26
mod
15
11
º
a
.
We next solve this result by multiplying both sides by
1
11
-
:
26
mod
15
11
11
11
1
1
×
º
×
-
-
a
.
Noting from the MOD 26 multiplicative inverse table that
19
26
mod
11
1
=
-
, we obtain
25
26
mod
285
26
mod
15
19
º
º
×
º
a
Hence
25
=
a
. We can substitute
25
=
a
into either equation (1) or (2) to find b. Choosing equation (1)
26
mod
22
4
º
+
b
a
, we obtain:
26
mod
22
)
25
(
4
º
+
b
or
26
mod
22
100
º
+
b
.
Subtracting 120 from both sides gives
26
mod
)
100
22
(
-
º
b
or
0
26
mod
78
º
-
º
b
.
Hence
25
=
a
and
0
=
b
solves the above system of equations. Hence, the affine cipher y = (25x + 0) mod 26 = 25x mod 26 was used to encrypt the message.
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