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Lecture 6.1: Misc. Topics:
Number Theory
CS 250, Discrete Structures, Fall 2011
Nitesh Saxena
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Lecture 6.1 -- Number Theory
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Lecture 6.1 -- Number Theory
Course Admin -- Final Exam
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Thursday, December 8, 10:45am1:15pm, lecture room
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Emphasis on post mid-term 2 material
Coverage:
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65% post mid-term 2 (lectures 4.*, 5.*, 6.*), and
35% pre mid-term 2 (lecture 1.*. 2.* and 3.*)
Our last lecture will be on December 6
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We plan to do a final exam review then
Lecture 6.1 -- Number Theory
Outline
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Number Theory
Modular Arithmetic
Application -- cryptography
Lecture 6.1 -- Number Theory
Divisors
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x divides y (written x | y) if the remainder is 0
when y is divided by x
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1|8, 2|8, 4|8, 8|8
The divisors of y are the numbers that divide
y
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divisors of 8: {1,2,4,8}
For every number y
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1|y
y|y
Lecture 6.1 -- Number Theory
Prime numbers
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A number is prime if its only divisors are 1
and itself:
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2,3,5,7,11,13,17,19, …
Fundamental theorem of arithmetic:
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For every number x, there is a unique set of
primes {p1, … ,pn} and a unique set of positive
exponents {e1, … ,en} such that
x  p1
e1
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* ... *
pn
en
How to prove? We studied it earlier!
Lecture 6.1 -- Number Theory
Common divisors
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The common divisors of two numbers x,y are
the numbers z such that z|x and z|y
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common divisors of 8 and 12:
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intersection of {1,2,4,8} and {1,2,3,4,6,12}
= {1,2,4}
greatest common divisor: gcd(x,y) is the
number z such that
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z is a common divisor of x and y
no common divisor of x and y is larger than z
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gcd(8,12) = 4
Lecture 6.1 -- Number Theory
Recall: Recursive Functions: gcd
Euclid’s algorithm makes use of the fact that
gcd(x,y ) = gcd(y, x mod y)
 x , if y  0
gcd( x , y )  
gcd( y , x mod y ), otherwise
(here we assume that x > 0)
10/04/2011
Lecture 3.3 -- Recursion
9
Euclidean Algorithm: gcd(r0,r1)
Main idea: If y = ax + b then gcd(x,y) = gcd(x,b)
r0  q1r1  r2
r1  q2 r2  r3
...
rm  2  qm 1rm 1  rm
rm 1  qm rm  0
gcd( r0 , r1 )  gcd( r1 , r2 )  ...  gcd( rm 1 , rm )  rm
Lecture 6.1 -- Number Theory
Example – gcd(15,37)
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37 = 2 * 15 + 7
15 = 2 * 7 + 1
7=7*1+0
gcd(15,37) = 1
Lecture 6.1 -- Number Theory
Relative primes
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x and y are relatively prime if they have no
common divisors, other than 1
Equivalently, x and y are relatively prime if
gcd(x,y) = 1
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9 and 14 are relatively prime
9 and 15 are not relatively prime
Lecture 6.1 -- Number Theory
Modular Arithmetic
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Definition: x is congruent to y mod m, if m
divides (x-y). Equivalently, x and y have the
same remainder when divided by m.
Notation: x  y (mod m)
Example: 14  5(mod 9)
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We work in Zm = {0, 1, 2, …, m-1}, the group
of integers modulo m
Example: Z9 ={0,1,2,3,4,5,6,7,8}
We abuse notation and often write = instead
of 
Lecture 6.1 -- Number Theory
Addition in Zm :
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Addition is well-defined:
if
x  x' (mod m)
y  y ' (mod m)
then
x  y  x' y ' (mod m)
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3 + 4 = 7 mod 9.
3 + 8 = 2 mod 9.
Lecture 6.1 -- Number Theory
Additive inverses in Zm
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0 is the additive identity in Zm
x  0  x(mod m)  0  x(mod m)
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Additive inverse of a is -a mod m = (m-a)
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Every element has unique additive inverse.
4 + 5= 0 mod 9.
4 is additive inverse of 5.
Lecture 6.1 -- Number Theory
Multiplication in Zm :
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Multiplication is well-defined:
if
x  x' (mod m)
y  y ' (mod m)
then
x  y  x' y ' (mod m)
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3 * 4 = 3 mod 9.
3 * 8 = 6 mod 9.
3 * 3 = 0 mod 9.
Lecture 6.1 -- Number Theory
Multiplicative inverses in Zm
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1 is the multiplicative identity in Zm
x 1  x(mod m)  1 x(mod m)
Multiplicative inverse (x*x-1=1 mod m)
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SOME, but not ALL elements have unique
multiplicative inverse.
In Z9 : 3*0=0, 3*1=3, 3*2=6, 3*3=0, 3*4=3,
3*5=6, …, so 3 does not have a multiplicative
inverse (mod 9)
On the other hand, 4*2=8, 4*3=3, 4*4=7,
4*5=2, 4*6=6, 4*7=1, so 4-1=7 (mod 9)
Lecture 6.1 -- Number Theory
Which numbers have inverses?
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In Zm, x has a multiplicative inverse if and
only if x and m are relatively prime or
gcd(x,m)=1
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E.g., 4 in Z9
Lecture 6.1 -- Number Theory
Extended Euclidian: a-1 mod n
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Main Idea: Looking for inverse of a mod n means
looking for x such that x*a – y*n = 1.
To compute inverse of a mod n, do the following:
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Compute gcd(a, n) using Euclidean algorithm.
Since a is relatively prime to m (else there will be no
inverse) gcd(a, n) = 1.
So you can obtain linear combination of rm and rm-1 that
yields 1.
Work backwards getting linear combination of ri and ri-1 that
yields 1.
When you get to linear combination of r0 and r1 you are
done as r0=n and r1= a.
Lecture 6.1 -- Number Theory
Example – 15-1 mod 37
37 = 2 * 15 + 7
 15 = 2 * 7 + 1
 7 = 7 * 1 + 0
Now,
 15 – 2 * 7 = 1
 15 – 2 (37 – 2 * 15) = 1
 5 * 15 – 2 * 37 = 1
So, 15-1 mod 37 is 5.
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Lecture 6.1 -- Number Theory
Modular Exponentiation:
Square and Multiply method
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Usual approach to computing xc mod n is
inefficient when c is large.
Instead, represent c as bit string bk-1 … b0
and use the following algorithm:
z = 1
For i = k-1 downto 0 do
z = z2 mod n
if bi = 1 then z = z* x mod n
Lecture 6.1 -- Number Theory
Example: 3037 mod 77
z = z2 mod n
if bi = 1 then z = z* x mod n
i
b
z
5
1
30
=1*1*30 mod 77
4
0
53
=30*30 mod 77
3
0
37
=53*53 mod 77
2
1
29
=37*37*30 mod 77
1
0
71
=29*29 mod 77
0
1
2
=71*71*30 mod 77
Lecture 6.1 -- Number Theory
Euler’s totient function
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Given positive integer n, Euler’s totient
function  (n) is the number of positive
numbers less than n that are relatively prime
to n
Fact: If p is prime then  ( p )  p  1
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{1,2,3,…,p-1} are relatively prime to p.
Lecture 6.1 -- Number Theory
Euler’s totient function
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Fact: If p and q are prime and n=pq then
(n)  ( p  1)( q  1)
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Each number that is not divisible by p or by q
is relatively prime to pq.
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E.g. p=5, q=7: {1,2,3,4,-,6,-,8,9,-,11,12,13,-,,16,17,18,19,-,-,22,23,24,-,26,27,-,29,,31,32,33,34,-}
pq-p-(q-1) = (p-1)(q-1)
Lecture 6.1 -- Number Theory
Euler’s Theorem and Fermat’s Theorem
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If a is relatively prime to n then
a
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(n)
 1 mod n
If a is relatively prime to p then
ap-1 = 1 mod p
Proof : follows from a well-known theorem --
Lagrange’s Theorem (we won’t study in this
course)
Lecture 6.1 -- Number Theory
Euler’s Theorem and Fermat’s Theorem
EG: Compute 9100 mod 17:
p =17, so p-1 = 16. 100 = 6·16+4. Therefore,
9100=96·16+4=(916)6(9)4 . So mod 17 we have
9100  (916)6(9)4 (mod 17)  (1)6(9)4 (mod 17)
 (81)2 (mod 17)  16
Lecture 6.1 -- Number Theory
An Application of Number Theory
Cryptography: foundation of secure
communication. EX: Public-Key Cryptography
Lecture 6.1 -- Number Theory
RSA Crypto: Key Generation
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Alice wants people to be able to send her encrypted
messages.
She chooses two (large) prime numbers, p and q and
computes n=pq and  (n) . [“large” =512 bits +]
She chooses a number e such that e is relatively
prime to  (n) and computes d, the inverse of
e in Z  (n )
(i.e., ed =1 mod  (n) )
She publicizes the pair (e,n) as her public key.(e is
called RSA exponent, n is called RSA modulus).
She keeps d secret and destroys p, q, and  (n)
Plaintext and ciphertext messages are elements of Zn
and e is the encryption key.
Lecture 6.1 -- Number Theory
RSA: Encryption
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Bob wants to send a message x (a number
relatively prime to n) to Alice.
He looks up her encryption key, (e,n), in a
directory.
The encrypted message is
y  E ( x)  x e mod n
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Bob sends y to Alice.
Lecture 6.1 -- Number Theory
RSA: Decryption
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To decrypt the message
y  E ( x)  x mod n
e
she’s received from Bob, Alice computes
D( y)  y mod n
d
Claim: D(y) = x
Lecture 6.1 -- Number Theory
Why does it all work?
D ( y )  y mod n
d
 ( x e mod n) d
 ( x ) mod n
e d
 x mod n
ed
x
t ( n ) 1
 (x
mod n
Because ed  1 mod  ( n)
(n) t
) x mod n
 1t x mod n  x mod n
From Euler’s Theorem
Lecture 6.1 -- Number Theory
Tiny RSA example.
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Let p = 7, q = 11. Then n = 77 and
(n)  60
Choose e = 13. Then d = 13-1 mod 60 = 37.
Let message = 2.
E(2) = 213 mod 77 = 30.
D(30) = 3037 mod 77=2
Lecture 6.1 -- Number Theory
Today’s Reading
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Rosen 4
Lecture 6.1 -- Number Theory