CSE 311 Lecture 11: Modular Arithmetic

Topics

Sets and set operations

A quick wrap-up of Lecture 10.

Modular arithmetic basics

Arithmetic over a finite domain (a.k.a computer arithmetic).

Modular arithmetic properties

Congruence, addition, multiplication, proofs.

Modular arithmetic and integer representations

Unsigned, sign-magnitude, and two’s complement representation.

Applications of modular arithmetic

Hashing, pseudo-random numbers, ciphers.

Sets and set operations

A set is a collection of elements.

Write $a\in B$ (or $a \not\in B$) to say that $a$ is (or isn’t) an element in the set $B$.

The order of elements doesn’t matter, and duplicates don’t matter.

Sets $A$ and $B$ are equal if they have the same elements.

$A=B \equiv \forall x. x\in A \leftrightarrow x\in B$

$A$ is a subset of $B$ if every element of $A$ is also in $B$.

$A \subseteq B \equiv \forall x. x\in A \rightarrow x\in B$

Sets can be built from predicates or using set operations.

$A\cup B = \{x : (x\in A)\vee(x\in B)\}$

$A\cap B = \{x : (x\in A)\wedge(x\in B)\}$

$A\setminus B = \{x : (x\in A)\wedge(x \not\in B)\}$

$A\oplus B = \{x : (x\in A)\oplus(x\in B)\}$

$\overline{A} = \{x : x\not\in A\} = \{x:\neg(x\in A)\}$

This is Boolean algebra again

Union $\cup$ is defined using $\vee$.

$A\cup B = \{x : (x\in A)\vee(x\in B)\}$

Intersection $\cap$ is defined using $\wedge$.

$A\cap B = \{x : (x\in A)\wedge(x\in B)\}$

Complement works like $\neg$.

$\overline{A} = \{x : x\not\in A\} = \{x:\neg(x\in A)\}$

This means that all equivalences from Boolean algebra translate directly into set theory, and you can use them in your proofs!

Power set

Power set of a set $A$ is the set of all subsets of $A$.

$\mathcal{P}(A) = \{B : B\subseteq A\}$

Examples

Let $\text{Days}=\{M, W, F\}$.

$\mathcal{P}(\text{Days}) = $ $\{\varnothing,\{M\},\{W\},\{F\},\{M,W\},\{M,F\},\{W,F\},\{M,W,F\}\}$

$\mathcal{P}(\varnothing) = $ $\{\varnothing\} \neq \varnothing$

Cartesian product

The Cartesian product of two sets is the set of all of their ordered pairs.

$A\times B = \{(a,b): a\in A \wedge b \in B\}$

Examples

$\mathbb{R}\times\mathbb{R}$ is the real plane.

$\mathbb{Z}\times\mathbb{Z}$ is the set of all pairs of integers.

If $A=\{1, 2\}, B = \{a, b, c\}$,

then $A\times B = $ $\{(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)\}$.

$A\times\varnothing = $ $\{(a,b): a\in A \wedge b \in \varnothing\} = \{(a,b): a\in A \wedge \F\}=\varnothing$.

Russell’s paradox

Let $S$ be the set of all sets that don’t contain themselves.

$S = \{x: x\not\in x\}$

The definition of $S$ is contradictory, hence the paradox.

Suppose that $S \in S$. Then, by definition of $S$, $S\not\in S$, which is a contradiction.

Suppose that $S \not\in S$. Then, by definition of $S$, $S\in S$, which is a contradiction too.

Representing sets as bitvectors

Suppose that universe $U$ is $\{1,2,\ldots,n\}$.

We can represent every set $B\subseteq U$ as a vector of bits:

This is called the characteristic vector of set $B$.

Given characteristic vectors for $A$ and $B$, what is the vector for

$A \cup B = $ $(a_1 + b_1) \ldots (a_n + b_n)$

$A \cap B = $ $(a_1\cdot b_1) \ldots (a_n\cdot b_n)$

Unix/Linux file permissions

$ ls -l
drwxr-xr-x ... Documents/
-rw-r--r-- ... file1

Permissions maintained as bitvectors.

Letter means the bit is 1.

”-“ means the bit is zero.

Bitwise operations

Private key cryptography

Alice wants to communicate a message $m$ secretly to Bob, so that eavesdropper Eve who sees their conversation can’t understand $m$.

Alice and Bob can get together ahead of time and privately share a secret key $K$.

How can they communicate securely in this setting?

One-time pad

Alice and Bob privately share a random $n$-bitvector $K$.

Eve doesn’t know $K$.

Later, Alice has $n$-bit message $m$ to send to Bob.

Alice computes $C = m \oplus K$.

Alice sends $C$ to Bob.

Bob computes $m = C\oplus K$, which is $ (m \oplus K)\oplus K = m$.

Eve can’t figure out $m$ from $C$ unless she can guess $K$.

And that’s very unlikely for large $n$ …

Modular arithmetic in action

What does this Java program output?

public class Seconds {
   final static int SEC_IN_YEAR = 364*24*60*60;
   public static void main(String args[]) {
       System.out.println("Number of seconds in 10100 years: " +
                          (SEC_IN_YEAR*10100));
   }
}

$javac Test.java
$java -Xmx128M -Xms16M Test
Number of seconds in 10100 years: -186619904

It all starts with divisibility …

Divisibility is the core concept behind modular arithmetic.

Definition: $a$ divides $b$, written as $a \vert b$.

For $a\in\Z, b\in\Z$, $a \vert b \leftrightarrow \exists k\in\Z. b = ka$.

Division theorem

Division theorem

For $a\in\Z, d\in\Z$ with $d > 0$,

there exist unique integers $q, r$ with $0 \leq r < d$

such that $a = dq+r$.

If we divide $a$ by $d$, we get a unique quotient $q = \div{a}{d}$ and non-negative remainder $r = \mod{a}{d}$.

public class NotMod {
   public static void main(String args[]) {
       System.out.println("-5 mod 2 = 1.");
       System.out.println("-5 % 2 = " + (-5 % 2));
   }
}

$javac NotMod.java
$java -Xmx128M -Xms16M NotMod
-5 mod 2 = 1.
-5 % 2 = -1

Example: arithmetic mod 7

Congruence modulo a positive integer

Definition: $a$ is congruent to $b$ modulo $m$, written as $\congruent{a}{b}{m}$

For $a,b,m\in\Z$ with $m>0$, $\congruent{a}{b}{m} \leftrightarrow m \vert (a-b)$

Congruence and equality

Congruence property

Let $a, b, m \in \Z$ with $m>0$.

Then, $\congruent{a}{b}{m}$ if and only if $\mod{a}{m} = \mod{b}{m}$.

The $\mod{}{m}\ $ function vs the $\congruent{}{\!}{m}$ predicate

The $\mod{}{m}$ function takes any $a\in\Z$ and maps it to a remainder $\mod{a}{m}\in\{0, 1, \ldots, m-1\}$.

In other words, $\mod{}{m}$ places all integers that have the same remainder modulo $m$ into the same “group” (a.k.a. “congruence class”).

The $\congruent{}{\!}{m}$ predicate compares $a,b\in\Z$ and returns true if and only if $a$ and $b$ are in the same group according to the $\mod{}{m}$ function.

Modular addition property

Modular addition property

Let $m$ be a positive integer ($m \in \Z$ with $m>0$).

If $\congruent{a}{b}{m}$ and $\congruent{c}{d}{m}$, then $\congruent{a+c}{b+d}{m}$.

Modular multiplication property

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 $\congruent{ac}{bd}{m}$.

Example: a proof using modular arithmetic

Let $n\in\Z$, and prove that $\congruent{n^2}{0}{4}$ or $\congruent{n^2}{1}{4}$.

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 $b_{n-1}\ldots b_2 b_1 b_0$.

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.

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 $2^n - |x|$.

Computing the two’s complement representation

For $-2^{n-1} \leq x < 0$, $x$ is represented using the $n$-bit unsigned representation of $2^n - |x|$.

Here is an easy way to compute this value:

• Compute the $n$-bit unsigned representation of $|x|$.
• Flip the bits of $|x|$ to get the representation of $2^n-1-|x|$.
  • This works because the string of all 1’s represents $2^n-1$.
• Add 1 to get $2^n - |x|$.

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 random $x_0,a,c,m$ and produce a long sequences of $x_n$’s.

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:

$f(p) = \mod{(ap + b)}{26}$

$f^{-1}(p) = \mod{(a^{-1}(p - b))}{26}$