DISCRETE COMPUTATIONAL STRUCTURES CS 23022 Fall 2005

DISCRETE COMPUTATIONAL STRUCTURES

CS 23022

Fall 2005

CS 23022 OUTLINE9. Matrices & Closures

10. Counting Principles

11. Discrete Probability

12. Congruences

13. Recurrence Relations

14. Algorithm Complexity

15. Graph Theory

16. Trees & Networks

17. Grammars & Languages

1. Sets

2. Logic

3. Proof Techniques

4. Algorithms

5. Integers & Induction

6. Relations & Posets

7. Functions

8. Boolean Algebra & Combinatorial Circuits

Discrete Mathematical Structures: Theory and Applications 3

Learning Objectives

Learn about Boolean expressions

Become aware of the basic properties of Boolean algebra

Explore the application of Boolean algebra in the design of electronic circuits

Learn the application of Boolean algebra in switching circuits


Two-Element Boolean AlgebraLet B = {0, 1}.


Two-Element Boolean Algebra










Minterm



Maxterm





Boolean Algebra


Boolean Algebra


Logical Gates and Combinatorial Circuits

























The Karnaugh map, or K-map for short, can be used to minimize a sum-of-product Boolean expression.




CS 23022 OUTLINE9. Matrices & Closures10. Congruences

11. Counting Principles

12. Discrete Probability

13. Recurrence Relations

14. Algorithm Complexity

15. Graph Theory

16. Trees & Networks

17. Grammars & Languages

1. Sets

2. Logic

3. Proof Techniques

4. Algorithms

5. Integers & Induction

6. Relations & Posets

7. Functions

8. Boolean Algebra & Combinatorial Circuits


Learning Objectives

Learn about matrices and their relationship with relations

Become familiar with Boolean matrices

Learn the relationship between Boolean matrices and different closures of a relation

Explore how to find the transitive closure using Warshall’s algorithm


Matrices


Matrices


Matrices – terms : equal , square


Matrices- terms: zero matrix, diagonal elements


Matrices- terms: diagonal matrix, identity matrix


Matrices – Matrix Sum

Two matrices are added only if they have the same number of rows and the same number of columns

To determine the sum of two matrices, their corresponding elements are added


Matrices – Matrix Addition Example


Matrices- Multiply a Constant x Matrix


Matrices – Matrix Difference


Matrices - Properties

Commutative and Associative properties of Matrix addition

Distributive property of multiplication over addition ( subtraction )-only holds for a constant time a matrix sum (difference)


Matrices

The multiplication AB of matrices A and B is defined only if the number of columns of A is the same as the number of rows of B


Matrices

Let A = [aij]m×n be an m × n matrix and B = [bjk ]n×p be an n × p matrix. Then AB is defined

To determine the (i, k)th element of AB, take the ith

row of A and the kth column of B, multiply the corresponding elements, and add the result

Multiply corresponding elements as in Figure 4.1

Figure 4.1



Matrices

Note that the dimensions of AB are m x p.Then (AB) x C is defined and has dimensions m x qConvince yourself that A x (BC) is defined and also has dimensions m x q



Matrices – Matrix transpose

The rows of A are the columns of AT and the columns of A are the rows of AT


Matrices - Symmetric


Matrices

Boolean (Zero-One) Matrices

Matrices whose entries are 0 or 1

Allows for representation of matrices in a convenient way in computer memory and for the design and implementation of algorithms to determine the transitive closure of a relation


Matrices

Boolean (Zero-One) Matrices

The set {0, 1} is a lattice under the usual “less than or equal to” relation, where for all a, b {0, 1}, ∈ a ∨ b = max{a, b} and a ∧ b = min{a, b}


Matrices – Logical Operations

Note: join is the OR operation; meet is the AND operation


Matrices


Matrices – Boolean Product



The Matrix of a Relation and Closure









ALGORITHM 4.3: Compute the transitive closure Input: M —Boolean matrices of the relation R

n—positive integers such that n × n specifies the size of M

Output: T —an n × n Boolean matrix such that T is the transitive closure of M 1. procedure transitiveClosure(M,T,n) 2. begin 3. A := M; 4. T := M; 5. for i := 2 to n do 6. begin 7. A := //A = Mi

8. T := T ∨ A; //T= M ∨ M2 ∨ · · · ∨ Mi

9. end 10. end


Warshall’s Algorithm for Determining the Transitive Closure

Previously, the transitive closure of a relation R was found by computing the matrices and then taking the Boolean join

This method is expensive in terms of computer time

Warshall’s algorithm: an efficient algorithm to determine the transitive closure

nRM



Let A = {a1, a2, . . . , an} be a finite set, n ≥ 1, and let R be a relation on A.

Warshall’s algorithm determines the transitive closure by constructing a sequence of n Boolean matrices









ALGORITHM 4.4: Warshall’s Algorithm Input: M —Boolean matrices of the relation R

n—positive integers such that n × n specifies the size of M

Output: W —an n × n Boolean matrix such that W is the transitive closure of M

1. procedure WarshallAlgorithm(M,W,n)

2. begin

3. W := M;

4. for k := 1 to n do

5. for i := 1 to n do

6. for j := 1 to n do

7. if W[i,j] = 1 then

8. if W[i,k] = 1 and W[k,j] = 1 then

9. W[i,j] := 1;

10. end




Learning Objectives

Learn the basic counting principles—multiplication and addition

Explore the pigeonhole principle

Learn about permutations

Learn about combinations


Basic Counting Principles





There are three boxes containing books. The first box contains 15 mathematics books by different authors, the second box contains 12 chemistry books by different authors, and the third box contains 10 computer science books by different authors.

A student wants to take a book from one of the three boxes. In how many ways can the student do this?



Suppose tasks T1, T2, and T3 are as follows:

T1 : Choose a mathematics book.

T2 : Choose a chemistry book.

T3 : Choose a computer science book.

Then tasks T1, T2, and T3 can be done in 15, 12, and 10 ways, respectively.

All of these tasks are independent of each other. Hence, the number of ways to do one of these tasks is 15 + 12 + 10 = 37.




Basic Counting Principles Morgan is a lead actor in a new movie. She needs to shoot a scene in

the morning in studio A and an afternoon scene in studio C. She looks at the map and finds that there is no direct route from studio A to studio C. Studio B is located between studios A and C. Morgan’s friends Brad and Jennifer are shooting a movie in studio B. There are three roads, say A1, A2, and A3, from studio A to studio B and four roads, say B1, B2, B3, and B4, from studio B to studio C. In how many ways can Morgan go from studio A to studio C and have lunch with Brad and Jennifer at Studio B?



There are 3 ways to go from studio A to studio B and 4 ways to go from studio B to studio C.

The number of ways to go from studio A to studio C via studio B is 3 * 4 = 12.





Consider two finite sets, X1 and X2. Then

This is called the inclusion-exclusion principle for two finite sets.

Consider three finite sets, A, B, and C. Then

This is called the inclusion-exclusion principle for three finite sets.


Pigeonhole Principle

The pigeonhole principle is also known as the Dirichlet drawer principle, or the shoebox principle.







Permutations


Permutations


Combinations


Combinations


Generalized Permutations and Combinations



Consider 10 chips of 3 types ( R, W, B) with 5 R, 3 W and 2 B

Then, n = 10, k = 3, and n1=5, n2=3 and n3 =2.

The number of different arrangements of these 10 chips is:

C(10,5) * C(10-5,3) * C(10-5-3,2) = C(10,5) * C(5,3) * C(2,2)

= 10! / 5!(5!) * 5!/ 3!(2!) * 2! / 2!(1!) = 10! / 5!3!2! = n!/ n1!n2!n3!

= 10 * 9 * 8 * 7 * 6 / (3 * 2 * 1 * 2 * 1) = 5 * 9 * 8 * 7 = 2520



Consider an 8-bit string. How many 8-bit strings contain exactly three 1s ?

Using the formula above, with n = 8 and k = 3, the answer is C(8,3).

C(8,3) = 8! / 3!(8-3)! = 8!/ 3!5! = 8 * 7 * 6 / 3 * 2 * 1 = 56

Examples: 11100000 00000111 00011100 01010100 etc



Suppose we have x + y = 3, with x≥ 0, y ≥0.

Then n = 3, k=2 and the number of integer solutions is:

C(3+2-1,2-1) = C(4,1) = 4

(0,3) , (1,2), (2,1) , (3,0)

y = 3 - x

0, 3

1, 2

2, 1

3, 000.5

11.5

22.5

33.5

0 1 2 3 4

x

y



Let objects = {1,2,3,4,5}, n = 5, r=3. Then the number of 3-combinations of these objects ( with repetitions allowed) is C(5-1+3,3) = C(7,3) = 7! / 3!(4!) = 35

111, 222, 333, 444, 555

112, 113,114, 115221, 223, 224 , 225331, 332, 334, 335441, 442, 443, 445551, 552, 553, 554

123, 124, 125 , 134, 135,145234, 235, 245345

Repeat all 3 = 5 combinations

Repeat two of three = 20 combinations

No repeats = 10 combinations


Permutations and Combinations

Permutations and Combinations - Rosen

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DISCRETE COMPUTATIONAL STRUCTURES CS 23022 Fall 2005