 # 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
Modular arithmetic and integer representations
Unsigned, sign-magnitude, and two’s complement representation.
Applications of modular arithmetic
Hashing, pseudo-random numbers, ciphers.

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## Sets and set operations

A quick wrap-up of Lecture 10.

### 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!

## More sets

Power set, Cartesian product, and Russell’s paradox.

### 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 $A\times B =$ $\{(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)\}$.
$A\times\varnothing =$ $\{(a,b): a\in A \wedge b \in \varnothing\} = \{(a,b): a\in A \wedge \mathsf{F}\}=\varnothing$.

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.
Define $S$ with respect to a universe of discourse.
$S = \{x \in U: x\not\in x\}$
With this definition, $S\not\in S$ and there is no contradiction because $S\not\in U$.

## Working with sets

Representing sets as bitvectors and applications of bitvectors.

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

$java -Xmx128M -Xms16M NotMod -5 mod 2 = 1. -5 % 2 = -1  ### Example: arithmetic mod 7$+$0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5$*$0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 1 3 5 3 0 3 6 2 5 1 4 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 2 6 0 6 5 4 3 2 1 ## Modular arithmetic properties Congruence, addition, multiplication, proofs. ### Congruence modulo a positive integer Definition:$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)$Examples: what do these mean and when are they true?$\congruent{x}{0}{2}$True for every$x$that is divisible by 2, i.e., even.$\congruent{-1}{19}{5}$True because$-1-19=-20$is divisible by 5.$\congruent{y}{2}{7}$True for every$y$of the form$y=2+7k$where$k\in\Z$. ### Congruence and equality Congruence property Let$a, b, m \in \Z$with$m>0$. Then,$\congruent{a}{b}{m}$if and only if$\mod{a}{m} = \mod{b}{m}$. Proof: Suppose that$\congruent{a}{b}{m}$. Then$m \vert a-b$by definition of congruence. So$a-b = km$for some$k\in\Z$by definition of divides. Therefore,$a = b + km$. Taking both sides modulo$m$, we get$\mod{a}{m}=\mod{(b+km)}{m}=\mod{b}{m}.$Suppose that$\mod{a}{m} = \mod{b}{m}$. By the division theorem,$a = mq + (\mod{a}{m})$and$b = ms + (\mod{b}{m})$for some$q,s\in\Z$. Then,$a - b=(mq + (\mod{a}{m})) - (ms + (\mod{b}{m}))=m(q-s) + (\mod{a}{m}-\mod{b}{m})=m(q-s)$, since$\mod{a}{m} = \mod{b}{m}$. Therefore,$m \vert (a-b)$and so$\congruent{a}{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}$. Proof: Suppose that $\congruent{a}{b}{m}$and $\congruent{c}{d}{m}$. By the definition of congruence, there are$k$and$j$such that$a - b = km$and$c - d = jm$. Adding these equations together, we get$(a+c)-(b+d) = m(j+k)$. Reapplying the definition of congruence, we get that$\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}$. Proof: Suppose that $\congruent{a}{b}{m}$and $\congruent{c}{d}{m}$. By the definition of congruence, there are$k$and$j$such that$a - b = km$and$c - d = jm$. So,$a = km + b$and$c = jm + b$. Multiplying these equations together, we get$ac = (km + b)(jm + d) = kjm^2 + kmd + bjm + bd$. Rearranging gives us$ac - bd = m(kjm + kd + bj)$. Reapplying the definition of congruence, we get that$\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}$. Let’s look at a few examples:$0^2 = \congruent{ 0 }{ 0 }{4}1^2 = \congruent{ 1 }{ 1 }{4}2^2 = \congruent{ 4 }{ 0 }{4}3^2 = \congruent{ 9 }{ 1 }{4}4^2 = \congruent{ 16 }{ 0 }{4}$It looks like$\congruent{n}{0}{2} \rightarrow \congruent{n^2}{0}{4}\congruent{n}{1}{2} \rightarrow \congruent{n^2}{1}{4}$Proof by cases: Case 1 ($n$is even): Suppose$\congruent{n}{0}{2}$. Then$n=2k$for some integer$k$. So$n^2 = (2k)^2 = 4k^2$. Therefore, by definition of congruence,$\congruent{n^2}{0}{4}$. Case 2 ($n$is odd): Suppose$\congruent{n}{1}{2}$. Then$n=2k+1$for some integer$k$. So$n^2 = (2k+1)^2 = 4k^2 + 4k + 1=4(k^2 + k) + 1$. Therefore, by definition of congruence,$\congruent{n^2}{1}{4}$. ## Modular arithmetic and integer representations Unsigned, sign-magnitude, and two’s complement representation. ### 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:$99 = 64 + 32 + 2 + 118 = 16 + 2$So for$n = 8$:$99 = 0110\ 001118 = 0001\ 0010$This works for unsigned integers. How do we represented signed integers? ### 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:$99 = 64 + 32 + 2 + 118 = 16 + 2$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:$99 = 64 + 32 + 2 + 118 = 16 + 22^8 - 18 = 256 - 18 = 238 = 128 + 64 + 32 + 8 + 4 + 281 = 64 + 16 + 1$So for$n = 8$:$\ \ \,99 = 0110\ 0011-18 = 1110\ 1110\ \ \,81 = 0101\ 0001$### 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$2^n - |x|$. ## Applications of modular arithmetic Hashing, pseudo-random numbers, 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 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:$f(p) = \mod{(ap + b)}{26}f^{-1}(p) = \mod{(a^{-1}(p - b))}{26}$## Summary Sets can be represented efficiently using bitvectors. This representation is used heavily in the real world. With this representation, set operations reduce to fast bitwise operations. Modular arithmetic is arithmetic over a finite domain. Key notions are divisibility and congruency$\mod{}{m}\$.
Modular arithmetic is the basis of computing.
Used with two’s complement representation to implement computer arithmetic.
Also used in hashing, pseudo-random number generation, and cryptography.