# CSE 311 Lecture 13: Primes and GCD

## Topics

Modular arithmetic applications
A quick wrap-up of Lecture 12.
Primes
Fundamental theorem of arithmetic, Euclid’s theorem, factoring.
Greatest Common Divisors (GCD)
GCD definition and properties.
Euclidean algorithm
Computing GCDs with the Euclidean algorithm.
Extended Euclidean algorithm
Bézout’s theorem and the extended Euclidean algorithm.

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## Modular arithmetic applications

A quick wrap-up of Lecture 12.

### Modular arithmetic properties

Let $m$ be a positive integer ($m \in \Z$ with $m>0$).
If $\congruent{a}{b}{m}$ and $\congruent{c}{d}{m}$, then $\congruent{a+c}{b+d}{m}$.
Modular multiplication property
Let $m$ be a positive integer ($m \in \Z$ with $m>0$).
If $\congruent{a}{b}{m}$ and $\congruent{c}{d}{m}$, then $\congruent{ac}{bd}{m}$.

### Unsigned integer representation

Represent integer $x$ as a sum of $n$ powers of 2:
If $x = \sum_{i=0}^{n-1} b_i2^i$ where each $b_i\in\{0,1\}$,
then the representation is $b_{n-1}\ldots b_2 b_1 b_0$.
Examples:
$99 = 64 + 32 + 2 + 1$
$18 = 16 + 2$
So for $n = 8$:
$99 = 0110\ 0011$
$18 = 0001\ 0010$

### Sign-magnitude integer representation

If $-2^{n-1} < x < 2^{n-1}$, represent $x$ with $n$ bits as follows:
Use the first bit as the sign (0 for positive and 1 for negative), and
the remaining $n-1$ bits as the (unsigned) value.
Examples:
$99 = 64 + 32 + 2 + 1$
$18 = 16 + 2$
So for $n = 8$:
$\ \ \,99 = 0110\ 0011$
$-18 = 1001\ 0010$
$\ \ \,81 = 0101\ 0001$

The problem with this representation is that our standard arithmetic algorithms no longer work, e.g., adding the representation of -18 and 99 doesn’t give the representation of 81.

### Two’s complement integer representation

Represent $x$ with $n$ bits as follows:
If $0 \leq x < 2^{n-1}$, use the $n$-bit unsigned representation of $x$.
If $-2^{n-1} \leq x < 0$, use the $n$-bit unsigned representation of $2^n + x$.
Key property:
Two’s complement representation of any number $y$ is equivalent to $\mod{y}{2^n}$ so arithmetic works $\mod{}{2^n}$.
Examples:
$99 = 64 + 32 + 2 + 1$
$18 = 16 + 2$
$2^8 - 18 = 256 - 18 = 238 = 128 + 64 + 32 + 8 + 4 + 2$
$81 = 64 + 16 + 1$
So for $n = 8$:
$\ \ \,99 = 0110\ 0011$
$-18 = 1110\ 1110$
$\ \ \,81 = 0101\ 0001$

Arithmetic $\mod{}{2^n}$ is easy in hardware: just throw away bits $n+1$ and higher.

### Computing the two’s complement representation

For $-2^{n-1} \leq x < 0$, $x$ is represented using the $n$-bit unsigned representation of $2^n + x = 2^n - |x|$. To compute this value:

• Compute the $n$-bit unsigned representation of $|x|$.
• Flip the bits of $|x|$ to get the representation of $2^n-1-|x|$.
• Add 1 to get $2^n - |x|$.

This computation works because $|x|+\overline{|x|}$ is all 1s, which represents $2^n-1$. So we have $\overline{|x|} = 2^n - 1 - |x|$ and $\overline{|x|} + 1 = 2^n - |x|$.

Example: -18 in 8-bit two’s complement
18 in 8-bit unsigned: $0001\ 0010$
Flip the bits: $1110\ 1101$
Add 1: $1110\ 1110$

### Applications of modular arithmetic

Modular arithmetic is the basis of modern computing, with many applications.

Examples include

• hashing,
• pseudo-random numbers, and
• simple ciphers.

### Hashing

Problem:
We want to map a small number of data values from a large domain $\{0, 1, \ldots, M-1\}$ into a small set of locations $\{0, 1, \ldots, n-1\}$ to be able to quickly check if a value is present.
Solution:
Compute $\text{hash}(x) = \mod{x}{p}$ for a prime $p$ close to $n$.
Or, compute $\text{hash}(x) = \mod{ax+b}{p}$ for a prime $p$ close to $n$.
This approach depends on all of the bits of data the data.
Helps avoid collisions due to similar values.
But need to manage them if they occur.

### Pseudo-random number generation

Linear Congruential method
$x_{n+1} = \mod{(ax_n + c)}{m}$

Choose $x_0$ randomly and $a, c, m$ carefully to produce a sequence of $x_n$’s.

Example
$a = 1103515245, c = 12345, m = 2^{31}$ from BSD
$x_0 = 311$
$x_1 = 1743353508, x_2 = 1197845517, x_3 = 1069836226, \ldots$

### Simple ciphers

Ceasar or shift cipher
Treat letters as numbers: A = 0, B = 1, …
$f(p) = \mod{(p + k)}{26}$
$f^{-1}(p) = \mod{(p - k)}{26}$
More general version
$f(p) = \mod{(ap + b)}{26}$
$f^{-1}(p) = \mod{(a^{-1}(p - b))}{26}$

$a^{-1}$ is the multiplicative inverse of $a$ modulo 26, and we’ll soon see how to compute these inverses.

## Primes

Fundamental theorem of arithmetic, Euclid’s theorem, factoring.

### Primality

Prime number
An integer $p > 1$ is called prime if its only positive factors are $1$ and $p$.
Composite number
An integer $c > 1$ is called composite if it is not prime.
A prime number is divisible only by itself and 1.
We say that $a$ is a factor of $b$ if $a\vert b$.
Note that 1 is neither prime nor composite.
The above definitions apply only to integers greater than 1.

### A key theorem about all positive integers

Fundamental theorem of arithmetic
Every positive integer greater than 1 has a unique prime factorization.

In other words, every integer $n > 1$ can be written uniquely as a prime, or the product of two or more primes ordered by size.

Examples
$48= 2\cdot 2\cdot 2\cdot 2\cdot 3$
$591 = 3 \cdot 197$
$45,523 = 45,523$
$321,950 = 2 \cdot 5 \cdot 5 \cdot 47 \cdot 137$
$1,234,567,890 = 2 \cdot 3 \cdot 3 \cdot 5 \cdot 3,607 \cdot 3,803$

### A key theorem about primes

Euclid’s theorem
There are infinitely many primes.
Suppose that there are finitely many primes: $p_1, \ldots, p_n$.
Define the number $P = p_1 \cdot \ldots\cdot p_n$, and let $Q = P + 1$.
Case 1: If $Q>1$ is prime, then $Q$ is a prime different from all of $p_1, \ldots, p_n$, since it is bigger than all of them. This contradicts the assumption that the list $p_1,\ldots, p_n$ includes all primes.
Case 2: If $Q>1$ is not prime, then $Q$ has some prime factor $p$, which must be in $p_1, \ldots, p_n$. Therefore $p \vert P$ and $p \vert Q$ so $P = jp$ and $Q = kp$ for some integers $j,k$. We then have $Q- P = (k-j)p = 1$, which means that $p \vert 1$. But no prime divides 1, leading again to a contradiction.
Since both cases are contradictions, the assumption must be false. $\qed$

### Important algorithmic problems

Primality testing
Given an integer $n$, determine if $n$ is prime.
Factoring
Given an integer $n$, determine the prime factorization of $n$.
• We don’t know of an efficient algorithm for factoring large numbers.
• The security of commonly used cryptographic protocols (e.g., RSA) hinges on this fact.
• For example, it took two years and thousands of machine-hours to factor a 232-digit (768-bit) number known as RSA-768.
• But factoring is easy for quantum computers!

## Greatest Common Divisors (GCD)

GCD definition and properties.

### Definition of greatest common divisor (GCD)

Greatest common divisor (GCD)
The greatest common divisor of integers $a$ and $b$, written as $\gcd{a}{b}$, is the largest integer $d$ such that $d\vert a$ and $d\vert b$.

Examples:

 $\gcd{100}{125}$ $=$ $25$ $\gcd{17}{49}$ $=$ $1$ $\gcd{11}{66}$ $=$ $11$ $\gcd{13}{0}$ $=$ $13$ $\gcd{180}{252}$ $=$ $36$

### How can we compute $\gcd{a}{b}$?

A naive approach is to first factor both $a$ and $b$:
$a = 2^3\cdot 3\cdot 5^2\cdot 7\cdot 11 = 46,20$
$b = 2\cdot 3^2\cdot 5^3\cdot 7\cdot 13 = 204,750$
And then compute $\gcd{a}{b}$ as follows:
$\gcd{a}{b} = 2^{\min(3,1)}\cdot 3^{\min(1,2)}\cdot 5^{\min(2,3)}\cdot 7^{\min(1,1)}\cdot 11^{\min(1,0)} \cdot 13^{\min(0,1)}$

But factoring is expensive! Can we compute $\gcd{a}{b}$ without factoring?

## Euclidean algorithm

Computing GCDs with the Euclidean algorithm.

### Euclidean algorithm is based on two useful facts

$\gcd{a}{0}$
If $a$ is a positive integer, then $\gcd{a}{0} = a$.

Proof follows straightforwardly from the definition of GCD and divisibility.

GCD and modulo
If $a$ and $b$ are positive integers, then $\gcd{a}{b} = \gcd{b}{\mod{a}{b}}$.
Proof:
First note that by definition of mod, $a = qb + \mod{a}{b}$ for some integer $q=\div{a}{b}$.
Now, let $d = \gcd{a}{b}$. Then $d\vert a$ and $d\vert b$, so $a=kd$ and $b=jd$ for some $k,j\in\Z$. Therefore, $\mod{a}{b} = a - qb = kd - qjd = d(k-qj)$. So, $d\vert (\mod{a}{b})$ and since $d\vert b$, we have that $d = \gcd{a}{b} \leq \gcd{b}{\mod{a}{b}}$.
Next, let $e = \gcd{b}{\mod{a}{b}}$. Then $e\vert b$ and $e\vert (\mod{a}{b})$, so $b=me$ and $\mod{a}{b}=ne$ for some $m,n\in\Z$. Therefore, $a = qb + \mod{a}{b} = qme + ne$. So, $e\vert a$ and $e\vert b$, we have that $e = \gcd{b}{\mod{a}{b}} \leq \gcd{a}{b}$. The result follows from these cases. $\qed$

### Euclidean algorithm

Apply $\gcd{a}{b} = \gcd{b}{\mod{a}{b}}$ until you get $\gcd{a}{0} = a$.

Example implementation:

// Assumes a >= b >= 0.
public static int gcd(int a, int b) {
if (b == 0)
return a;             // GCD(a, 0) = a
else
return gcd(b, a % b); // GCD(a, b) = GCD(b, a mod b)
}


$\gcd{660}{126}$
$= \gcd{126}{\mod{660}{126}} = \gcd{126}{30}$
$= \gcd{30}{\mod{126}{30}} = \gcd{30}{6}$
$= \gcd{6}{\mod{30}{6}} = \gcd{6}{0}$
$= 6$

In tableau form:

 660 = 5 * 126 + 30 126 = 4 * 30 + 6 30 = 5 * 6 + 0

## Extended Euclidean algorithm

Bézout’s theorem and the extended Euclidean algorithm.

### Bézout’s theorem about GCDs

Bézout’s theorem
If $a$ and $b$ are positive integers, then there exist integers $s$ and $t$ such that $\gcd{a}{b} = sa + tb$.

We can extend Euclidean algorithm to find $s$ and $t$ in addition to computing $\gcd{a}{b}$.

### Extended Euclidean algorithm

1. Compute GCD and keep the tableau.

$\gcd{35}{27} = 35s + 27t$.

 $a$ $=$ $q$ $*$ $b$ $+$ $r$ $35$ $=$ $1$ $*$ $27$ $+$ $8$ $27$ $=$ $3$ $*$ $8$ $+$ $3$ $8$ $=$ $2$ $*$ $3$ $+$ $2$ $3$ $=$ $1$ $*$ $2$ $+$ $1$
$\gcd{a}{b}$   $\gcd{b}{\mod{a}{b}}$   $r=\mod{a}{b}$
$\gcd{35}{27}$ $=$ $\gcd{27}{\mod{35}{27}}$ $=$ $\gcd{27}{8}$
$=$ $\gcd{8}{\mod{27}{8}}$ $=$ $\gcd{8}{3}$
$=$ $\gcd{3}{\mod{8}{3}}$ $=$ $\gcd{3}{2}$
$=$ $\gcd{2}{\mod{3}{2}}$ $=$ $\gcd{2}{1}$
$=$ $\gcd{1}{\mod{2}{1}}$ $=$ $\gcd{1}{0}$

### Extended Euclidean algorithm

1. Compute GCD and keep the tableau.
2. Solve the equations for $r$ in the tableau.

$\gcd{35}{27} = 35s + 27t$.

 $a$ $=$ $q$ $*$ $b$ $+$ $r$ $35$ $=$ $1$ $*$ $27$ $+$ $8$ $27$ $=$ $3$ $*$ $8$ $+$ $3$ $8$ $=$ $2$ $*$ $3$ $+$ $2$ $3$ $=$ $1$ $*$ $2$ $+$ $1$
 $r$ $=$ $a$ $-$ $q$ $*$ $b$ $8$ $=$ $35$ $-$ $1$ $*$ $27$ $3$ $=$ $27$ $-$ $3$ $*$ $8$ $2$ $=$ $8$ $-$ $2$ $*$ $3$ $1$ $=$ $3$ $-$ $1$ $*$ $2$

### Extended Euclidean algorithm

1. Compute GCD and keep the tableau.
2. Solve the equations for $r$ in the tableau.
3. Back substitute the equations for $r$.

$\gcd{35}{27} = 35s + 27t$.

 $r$ $=$ $a$ $-$ $q$ $*$ $b$ $8$ $=$ $35$ $-$ $1$ $*$ $27$ $3$ $=$ $27$ $-$ $3$ $*$ $8$ $2$ $=$ $8$ $-$ $2$ $*$ $3$ $1$ $=$ $3$ $-$ $1$ $*$ $2$
 $1$ $= 3 - 1 * (8 - 2 * 3)$ Plug in the def of 2. $= 3 - 8 + 2*3$ $= (-1) * 8 + 3*3$ Group 8’s and 3’s. $= (-1) * 8 + 3*(27 - 3*8)$ Plug in the def of 3. $= (-1) * 8 + 3*27 + (- 9) * 8$ $= 3 * 27 + (-10) * 8$ Group 8’s and 27’s. $= 3 * 27 + (-10) * (35 - 1*27)$ Plug in the def of 8. $= 3 * 27 + (-10) * 35 + 10 *27$ $= 13 * 27 + (-10) * 35$ Group 27’s and 35’s.

### Multiplicative inverse $\mod{}{m}$

Suppose $\gcd{a}{m} = 1$.

By Bézout’s Theorem, there exist integers $s$ and $t$ such that $sa + tm = 1$.

$\mod{s}{m}$ is the multiplicative inverse of $a$
$1 = \mod{(sa+tm)}{m} = \mod{sa}{m}$

### Using multiplicative inverses to solve modular equations

Solve: $\congruent{7x}{1}{26}$

① Compute GCD and keep the tableau.

%

② Solve the equations for $r$ in the tableau.

 $a$ $=$ $b$ $*$ $q$ $+$ $r$ $26$ $=$ $7$ $*$ $3$ $+$ $5$ $7$ $=$ $5$ $*$ $1$ $+$ $2$ $5$ $=$ $2$ $*$ $2$ $+$ $1$
 $r$ $=$ $a$ $-$ $b$ $*$ $q$ $5$ $=$ $26$ $-$ $7$ $*$ $3$ $2$ $=$ $7$ $-$ $5$ $*$ $1$ $1$ $=$ $5$ $-$ $2$ $*$ $2$

③ Back substitute the equations for $r$.

④ Solve for $x$.

• Multiplicative inverse of 7 mod 26
• $\mod{(-11)}{26}=15$
• So, $x=26k + 15$ for $k\in\Z$.

## Summary

Every positive integer $p>1$ is either prime or composite.
$p$ is prime if its only factors are $p$ and 1.
Otherwise, $p$ is composite.
$\gcd{a}{b}$ is the greatest integer that divides both $a$ and $b$.
It can be computed efficiently using the Euclidean algorithm.
By Bézout’s Theorem, $\gcd{a}{b} = sa + tb\,$ for some integers $s, t$.
$s, t$ can be computed using the extended Euclidean algorithm.
If $\gcd{a}{b} = 1$, $\mod{s}{b}$ is the multiplicative inverse of $a$ modulo $b$.