Slide 1

The Integers and Division

CSE 504 Discrete Mathematics & Foundations of Computer ScienceDr. Djamel BouchaffraCh. 2 (Part 1) Section 2.412Example: Let n and d be positive integers. How many positive integers not exceeding n are divisible by d?

Solution: They are of the form {dk}, where k is a positive integer.0 < dk n 0 < k n/d There are n/d positive integers not exceeding n that are divisible by d.3The Division Algorithm

Let a be an integer and d a positive integer. Then !(q, r) Z2; 0 r < d: a = dq +r.4Definition :

d is called the divisor, a the dividend, q the quotient and r the remainder.q = a div d, r = a mod d.

Example: 101 = 11.9 + 2Quotient = 101 div 11Remainder= 2 = 101 mod 115Greatest Common Divisors & Least Common Multiples

Definition:

Let a and b be integers, not both zero. The largest integer d such that d|a and d|b is called the greatest common divisor of a and b. It is denoted gcd (a, b).

Example: gcd (24, 36)Div (24) = {1,2,3,4,6,8,12,24}Div (36) = {1,2,3,4,6,8,9,12,18,36}Com(24,36) = = {1,2,3,4,6,12}gcd(24,36) = 126Definition:

The integers a and b are relatively prime (rp) if gcd(a, b) =1.

Example: 17 and 22 are rp since gcd(17,22) = 1.7Definition:

The least common multiple (lcm) of the positive integers a and b is the smallest positive integer that is divisible by both a and b.

where max(x,y) denotes the maximum of x and y.

Example : What is the least common multiple of: 233572 and 2433?Solution: lcm(233572 ,2433) = 2 max(3,4). 3max(5,3). 7max(2,0)= 243572

8Theorem:

Let a and b be positive integers. Then ab = gcd(a,b).lcm(a.b).

9Modular Arithmetic

Definition:

Let (a, b) Z2,, m Z+ then a is a congruent to b modulo m if m divides a b.Notation: a b (mod m).

Theorem:

Let a and b be integers, and let m be a positive integer. Then a b (mod m) if and only if a mod m = b mod m.

10Example: 17 5 (mod 6)24 14 (mod 6)?Since: 6|(17 5) = 12 17 5 (mod 6) 6 does not divide 10 24 is not congruent to 14 (mod 6) Theorem:

Let m be a positive integer. The integers a and b are congruent modulo m if and only ifk Z; a = b + km11Applications of Congruences

Hashing Functions

Pseudorandom Numbers

Cryptology (Caesar Cepher)12Hashing Functions

Assignment of memory location to a student recordh(k) = k mod m

Example: h (064212848) = 064212848 mod 111 = 14 when m = 111Key: social security ## of available memory location13Pseudorandom Numbers

Needed for computer simulation

Linear congruential method : xn+1 = (axn + c) mod m

Put them between 0 and 1 as: yn = xn/m14Cryptology (Caesar Cepher)

Encryption:

Making messages secrets by shifting each letter three letters forward in the alphabetB E X A

Mathematical expression:f(p) = (p + 3) mod 26 0 p 25

15Example: What is the secret message produced from the message Meet you in the park

Solution: Replace letters with numbers:meet = 12 4 4 19you = 24 14 20in = 8 13the = 19 7 4park = 15 0 17 10Replace each of these numbers p by f(p) = (p + 3) mod 26meet = 15 7 7 22you = 1 17 23 in = 11 16the = 22 10 7park = 18 3 20 13Translate back into letters: PHHW BRX LQ WKH SDUN16Decryption (Deciphering)

f(p) = (p + k) mod 26 (shift cepher) f -1(p) = (p k) mod 26

Caesars method and shift cipher are very vulnerable and thus have low level of security (reason frequency of occurrence of letters in the message) Replace letters with blocks of letters.MatricesCSE 504 Discrete Mathematics & Foundations of Computer ScienceDr. Djamel BouchaffraCh. 2 (Part 4) Section 2.71718Definition:

A matrix is a rectangular array of numbers. A matrix with m rows and n columns is called an m x n matrix. A matrix with the same number of rows as columns is called square. Two matrices are equal if they have the same number of rows and the same number of columns and the corresponding entries in every position are equal.

Example: The matrix is a 3 X 2 matrix.

19Definition:

Let

The ith row of A is the 1 x n matrix [ai1, ai2, , ain]. The jth column of A is the n x 1 matrix

The (i, j)th element or entry of A is the element aij, that is, the number in the ith row and jth column of A. A convenient shorthand notation for expressing the matrix A is to write A = [aij], which indicates that A is the matrix with its (i, j)th element equal to aij.

20Matrix Arithmetic

Definition

Let A = [aij] and B = [bij] be m x n matrices. The sum of A and B, denoted by A + B, is the m x n matrix that has aij + bij as its (i, j)th element. In other words, A + B = [aij + bij].

Example:

21Definition:

Let A be an m x k matrix and B be a k x n matrix. The product of A and B, denoted by AB, is the m x n matrix with its (i, j)th entry equal to the sum of the products of the corresponding elements from the ith row of A and the jth column of B. In other words, if AB = [cij], thenCij = ai1b1j + ai2b2j + + aikbkj.22Example: Let

Find AB if it is defined.

Solution: Since A is a 4 x 3 matrix and B is a 3 x 2 matrix, the product AB is defined and is a 4 x 2 matrix. To find the elements of AB, the corresponding elements of the rows of A and the columns of B are first multiplied and then these products are added. For instance, the element in the (3, 1)th position of AB is the sum of the products of the corresponding elements of the third row of A and the first column of B; namely 3 * 2 + 1 * 1 + 0 * 3 = 7. When all the elements of AB are computed, we see that

Matrix multiplication is not commutative.23Example: Let

Does AB = BA?

Solution: We find that

Hence, AB BA.

24Matrix chain multiplication

Problem: How should the matrix-chain A1A2An be computed using the fewest multiplication of integers, where A1A2An are m1 x m2, m2 x m3, , mn x m n+1 matrices respectively and each has integers as entries?

25Example: A1 = 30 x 20 (30 rows and 20 columns) A2 = 20 x 40 A3 = 40 x 10

Solution: 2 possibilities to compute A1A2A3A1 (A2A3)(A1A2)A3

1) First A2A3 requires 20 * 40 * 10 = 8000 multiplications A1(A2A3) requires 30 * 20 * 10 = 6000 multiplicationsTotal: 14000 multiplications.2) First A1A2 requires 30 * 20 * 40 = 24000 multiplications (A1A2)A3 requires 30 * 40 * 10 = 12000Total: 36000 multiplications.

(1) is more efficient! 26Transposes and power matrices

Definition

The identity matrix of order n is the n x n matrix In = [ij], where ij = 1 if i = j and ij = 0 if i j. Hence

27Definition

Let A = [aij] be an m x n matrix. The transpose of A, denoted At, is the n x m matrix obtained by interchanging the rows and the columns of A. In other words, if At = [bij], then bij = aij for i = 1, 2, , n and j = 1, 2, , m.

Example:

The transpose of the matrix is

28Definition

A square matrix A is called symmetric if A = At. Thus A = [aij] is symmetric if aij = aji for all i and j with 1 i n and 1 j n.

Example: The matrix is symmetric.

29Zero-one matrices

It is a matrix with entries that are 0 or 1. They represent discrete structures using Boolean arithmetic.We define the following Boolean operations:

30Definition

Let A = [aij] and B = [bij] be m x n zero-one matrices. Then the join of A and B is the zero-one matrix with (i, j)th entry aij bij. The join of A and B is denoted A B. The meet of A and B is the zero-one matrix with (i, j)th entry aij bij. The meet of A and B is denoted by A B.31Example: Find the join and meet of the zero-one matrices

Solution: We find that the joint of A and B is:

The meet of A and B is:

32Definition:

Let A = [aij] be an m x k zero-one matrix and B = [bij] be a k x n zero-one matrix. Then the Boolean product of A and B, denoted by A B, is the m x n matrix with (i, j)th entry [cij] wherecij = (ai1 b1j) (ai2 b2j) (aik bkj).33Example: Find the Boolean product of A and B, where

Solution:

34Algorithm The Boolean Product

procedure Boolean product (A,B: zero-one matrices)for i := 1 to mfor j := 1 to nbegincij := 0for q := 1 to kcij := cij (aiq bqj)end{C = [cij] is the Boolean product of A and B} 35Definition

Let A be a square zero-one matrix and let r be a positive integer. The rth Boolean power of A is the Boolean product of r factors of A. The rth Boolean product of A is denoted by A[r]. Hence

(This is well defined since the Boolean product of matrices is associative.) We also define A[0] to be In.

36Example: Let . Find A[n] for all positive integers n.

Solution: We find that

We also find that

Additional computation shows that

The reader can now see that A[n] = A[5] for all positive integers n with n 5.