Discrete Mathematical Structures

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DISCRETE MATHEMATICAL STRUCTURES [As per Choice Based Credit System (CB CS) scheme] (Effective from the academic year 2015 -2016)

Subject Code

SEMESTER – III IA Marks 15CS36

20

Number of Lecture Hours/Week

04

Exam Marks

80

Total Number of Lecture Hours

50

Exam Hours

03

CREDITS – 04

Course objectives: This objectives: This course will enable students to •

Prepare for a background in abstraction, notation, and critical thinking for the mathematics most directly related to computer science.

•

Understand and apply logic, relations, functions, basic set theory, countability and counting arguments, proof techniques,

•

Understand and apply mathematical induction, combinatorics, discrete probability, recursion, sequence and recurrence, elementary number theory

•

Understand and apply graph theory and mathematical proof techniques.

Module -1

Teaching Hours

Fundamentals of Logic: Basic Connectives and Truth Tables, Logic Equivalence – The Laws of Logic, Logical Implication – Rules of Inference. The Use of Quantifiers, Quantifiers, Definitions and the Proofs of Theorems, Textbook 1: Ch 2

10Hours

Module -2

Properties of the Integers: Mathematical Induction, The Well Ordering Principle – 10 Hours Mathematical Induction, Recursive Definitions. Fundamental Principles of Counting: The Rules of Sum and Product, Permutations, Combinations – The Binomial Theorem, Combinations with Repetition, Textbook 1: Ch 4: 4.1, 4.2 Ch 1. Module – 3

Relations and Functions: Cartesian Products and Relations, Functions – Plain and One-toOne, Onto Functions. The Pigeon-hole Principle, Function Composition and Inverse Functions. Properties of Relations, Computer Recognition – Zero-One Matrices and Directed Graphs, Partial Orders – Hasse Diagrams, Diagrams , Equivalence Relations and Partitions. Textbook 1: Ch 5:5.1 to 5.3, 5.5, 5.6, Ch 7:7.1 to 7.4

10 Hours

Module-4

12 | P a g e

DISCRETE MATHEMATICAL STRUCTURES [As per Choice Based Credit System (CB CS) scheme] (Effective from the academic year 2015 -2016)

Subject Code

SEMESTER – III IA Marks 15CS36

20

Number of Lecture Hours/Week

04

Exam Marks

80

Total Number of Lecture Hours

50

Exam Hours

03

CREDITS – 04

Course objectives: This objectives: This course will enable students to •

Prepare for a background in abstraction, notation, and critical thinking for the mathematics most directly related to computer science.

•

Understand and apply logic, relations, functions, basic set theory, countability and counting arguments, proof techniques,

•

Understand and apply mathematical induction, combinatorics, discrete probability, recursion, sequence and recurrence, elementary number theory

•

Understand and apply graph theory and mathematical proof techniques.

Module -1

Teaching Hours

Fundamentals of Logic: Basic Connectives and Truth Tables, Logic Equivalence – The Laws of Logic, Logical Implication – Rules of Inference. The Use of Quantifiers, Quantifiers, Definitions and the Proofs of Theorems, Textbook 1: Ch 2

10Hours

Module -2

Properties of the Integers: Mathematical Induction, The Well Ordering Principle – 10 Hours Mathematical Induction, Recursive Definitions. Fundamental Principles of Counting: The Rules of Sum and Product, Permutations, Combinations – The Binomial Theorem, Combinations with Repetition, Textbook 1: Ch 4: 4.1, 4.2 Ch 1. Module – 3

Relations and Functions: Cartesian Products and Relations, Functions – Plain and One-toOne, Onto Functions. The Pigeon-hole Principle, Function Composition and Inverse Functions. Properties of Relations, Computer Recognition – Zero-One Matrices and Directed Graphs, Partial Orders – Hasse Diagrams, Diagrams , Equivalence Relations and Partitions. Textbook 1: Ch 5:5.1 to 5.3, 5.5, 5.6, Ch 7:7.1 to 7.4

10 Hours

Module-4

12 | P a g e

The Principle of Inclusion and Exclusion: The Principle of Inclusion and Exclusion, Generalizations of the Principle, Derangements – Nothing is in its Right Place, Rook Polynomials. Recurrence Relations: First Order Linear Recurrence Relation, The Second Order Linear Homogeneous Recurrence Relation with Constant Coefficients.

10 Hours

Textbook 1: Ch 8: 8.1 to 8.4, Ch 10:10.1 to 10.2 Module-5

Introduction to Graph Theory: Definitions and Examples, Sub graphs, Complements, and Graph Isomorphism, Vertex Degree, Euler Trails and Circuits , Trees: Definitions, Properties, and Examples, Routed Trees, Trees and Sorting, Weighted Trees and Prefix Codes

10 Hours

Textbook 1: Ch 11: 11.1 to 11.3, Ch 12: 12.1 to 12.4 Course outcomes: After studying this course, students will be able to: a n argument using propositional and predicate logic and truth tables. 1. Verify the correctness of an

2. Demonstrate the ability to solve problems using counting techniques and combinatorics in the context of discrete probability.

3. Solve problems involving recurrence relations and generating functions. 4. Construct proofs using direct proof, proof by contraposition, proof by contradiction, proof by cases, and mathematical induction.

5. Explain and differentiate graphs and trees Graduate Attributes (as per NBA)

1. Engineering Knowledge 2. Problem Analysis 3. Conduct Investigations of Complex Problems Question paper pattern: The question paper will have ten questions. There will be 2 questions from each module. Each question will have questions covering all the topics under a module. The students will have to answer 5 full questions, selecting one full question from each module. Text Books:

1. Ralph P. Grimaldi: Discrete and Combinatorial Mathematics, , 5th Edition, Pearson Education. 2004. Reference Books:

1. Basavaraj S Anami and Venakanna S Madalli: Discrete Mathematics – A Concept based approach, Universities Press, 2016

2. Kenneth H. Rosen: Discrete Mathematics and its Applications, 6th Edition, McGraw Hill, 2007. 3. Jayant Ganguly: A Treatise on Discrete Mathematical Structures, Sanguine-Pearson, 2010. 4. D.S. Malik and M.K. Sen: Discrete Mathematical Structures: Theory and Applications, Thomson, 2004.

5. Thomas Koshy: Discrete Mathematics with Applications, Elsevier, 2005, Reprint 2008. 13 | P a g e

DISCRETE MATHEMATICAL STRUCTURES

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QUESTION PAPER S SOLUTION MODULE-1 Set Theory 1. Def ine power set of a set. Determine power sets of the f ollowing sets. (4m)

Jun 2015 / Jan 2016

Sol: Power Set: The collection of all su bsets of a set set A is called the power set of A, and is re pr esented P(A).

For instance, if A = {1, 2, 3,4}, then P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},{1,2,3},{2,3,4},

{1, 3,4},{1,2, 4} . A}. 2. Using laws of set theory show that A – (A – B) = A – (A ∩ B) - (5m) Jun 2015/Jan 2016 Sol:A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1]

= A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1] = A ∩ (A1 U B) [Demor gan’s Law] = (A ∩ A1) U (A ∩ B) [Distr i butive Law] =ø U (A ∩ B) [i.e. A ∩ A1 = ø] = A ∩B 3. In a survey of 260 college students, the following data were obtained: 64 had tak en a mathematics course, 94 had tak en a computer science course, 58 had tak en a business course, 28 had tak en both a mathematics and a business course, 26 had tak en both a mathematics and a computer science course urse,, 22 had had taken both a computer science and a business course, and 14 had taken all three types i. ii.

How many of these students had tak en none of the three courses ses? How many had tak en only a computer science courses ? (11m) Jun 2015 / Jan 2016

Solution: Given: U=260 = 14

|A|=64 |B|=94 |C|=58 |A ∩ C| = 28 |A ∩ B| = 26 |B ∩ C| = 22 |A ∩ B ∩ C|

|A U B U C| =? |B1| =? |A U B U C| = A + B + C – |A ∩ B| – |B |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C| = 64 + 94 + 58 – 26 – 22 – 28 + 14 = 154 i. |A U B U C| = U – |A U B U C| = 260 – 154 = 106 ii.|B1| = |B| - |B ∩ C| - |B ∩ A| + |A ∩ B ∩ C| = 94 – 22 – 26 + 14 = 60 Dept . of CSE, SJBIT


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DISCRETE MATHEMATICAL STRUCTURES 4. For any two sets A and B, prove the following

15CS35 (4m) June 2016

Sol:

A – (A – B) = A ∩ B Solution: A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1] = A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1] = A ∩ (A1 U B) [Demor gan’s Law] = (A ∩ A1) U (A ∩ B) [Distr i butive Law] =øU (A ∩ B) [i.e. A ∩ A1 = ø] = A ∩B 5. State and prove De – Morgan’s law of set theory. Solution:

(6m)

June 2016

The com plement of union of 2 sets is equal to the inter section of comple ment of

•

the sets. AUB=A∩B Consider RHS, A B = { x / x ε A and x ε B } = { x / x ¢ A and x ¢ B } ={x /x ¢ (A U B) } =AUB The union of com pliment of 2 sets is equal to the com pliment of inter section of t he

•

2 sets.

AUB=A ∩ B Consider LHS, A U B = { x / { x / x ε A or x ε B } = { x / x ¢ A or x ¢ B } ={x /x ¢ (A ∩ B) } 6. In a survey of 260 college students, the following data were obtained: 64 had tak en a mathematics course, 94 had tak en a computer science course, 58 had

tak en a business course, 28 had tak en both a mathematics and a business course, 26 had tak en both a mathematics and a computer science course, 22 had taken both a computer science and a business course, and 14 had ta ken all three types of courses.

i.

How many of these students had tak en none of the three courses?

ii. How many had tak en only a computer science courses Solution: Given: U=260 = 14

(8m) june 2016

|A|=64 |B|=94 |C|=58 |A ∩ C| = 28 |A ∩ B| = 26 |B ∩ C| = 22 |A ∩ B ∩ C|

Dept . of CSE, SJBIT


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|A U B U C| =? |B1| =? |A U B U C| = A + B + C – |A ∩ B| – |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C| = 64 + 94 + 58 – 26 – 22 – 28 + 14 = 154 iii. |A U B U C| = U – |A U B U C| = 260 – 154 = 106 iv. |B1| = |B| - |B ∩ C| - |B ∩ A| + |A ∩ B ∩ C| = 94 – 22 – 26 + 14 = 60 7. For any two sets A and B, prove the following

(2m)

june 2016

A – (A – B) = A ∩ B Solution: A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1] = A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1] = A ∩ (A1 U B) [Demor gan’s Law] = (A ∩ A1) U (A ∩ B) [Distr i butive Law] =øU (A ∩ B) [i.e. A ∩ A1 = ø] = A ∩B 8.

Determine the sets A and B given that A – B = {1, 2, 4}, B – A = {7, 8} and AUB = {1, 2, 4, 5, 7, 8, 9} (4m) jan 2015

Solution: set A A = (A U B) – (B – A) = [{1,2,4,5,7,8,9}] –[{7,8}] = {1, 2, 4, 5, 9} Set B B = (A U B) – (A – B) = [{1,2,4,5,7,8,9}] –[{1,2,4}] = {5, 7, 8, 9} 9. Let M, P and C be the sets of students taking Mathematics courses, Physics courses and Computer Science courses respectively in a university. Assume |M| = 300, |P| = 350, |C| = 450, |M \ P| = 100, |M \ C| = 150, |P \ C| = 75, |M \ P \ C| = 10.

How many students are tak ing exactly one of those courses? Solution:



(7m)

Jan 2015

Page 3

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We see that |(M\P)-(M\P\C)| = 100-10 = 90, |(M\C)-(M\ P \C)| = 150-10 = 140 and |(P \C)-(M \P \C)| = 75-10 = 65. Then the region corr es ponding to students tak ing Mathematics courses only has car dinality 300-(90+10+140) = 60. Analogously we com pute the num ber o f students tak ing Ph ysics cour ses only (185) and tak ing Com puter Science cour ses only (235). The sum 60 + 185 + 235 = 480 is the num ber of students tak ing exactly one of those cour ses. 10. For any three sets A,B and C prove that (A-B)-C =A – (BUC) = (A-C) – (B-C) -6M jan 2015 Solution: (A-B) – C = (A∩B) ∩C = A∩ (B∩C)

= A∩ (B∩C) = A – (B∩C) (A – C) – (B – C) = (A –C) ∩ (B – C) = (A ∩ C) ∩ (B – C1) = (A ∩ C) ∩ (BU C) = [(A ∩ C) ∩ B)] U [(A ∩ C) ∩C] = [(A ∩ (C∩B)] U [A ∩ (C∩C)] = [A ∩ (CUB) ] U (A∩Ø) = [A ∩ (BUC) ] UØ = [A ∩ (BUC) ] = [A - (BUC)] 11. Explain the laws of set theory:



(8m)

jun 2014

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12. Determine the sets A and B given that A – B = {1, 3, 7, 11}, B – A = {2, 6, 8} and A∩B = {4, 9} (5m) june 2014 Solution: set A

A = (A – B) + (A ∩B) = [{1, 3, 7, 11}] + [{4, 9}] = {1, 3, 4, 7, 9, 11} Set B B = (B-A) + (A∩B) = [{2, 6, 8}] – [{4, 9}] = {2, 4, 6, 8, 9} 1

1

13. Prove that: A▲B= (B∩A ) U ( A∩B ) = ( B-A) U (A-B). 1

(4m)

june 2014

1

Solution: A▲B= (B∩A ) U ( A∩B ) = (B-A) U (A-B). Let A=P1 U P2 B=P2 U P3

A▲B = (AUB)-(A∩B) = (P1 U P2 UP3) – (P2) = P1 U P3 ………… (1)



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B – A = P3, A – B = P1 Ther ef ore (B – A) U (A – B) = P3 U P1……………………… (2) A=U - A = P3 U P4, AND B=U – B = P1U P4 B ∩ A= P3, A ∩ B = P1 Ther ef or e (B ∩ A1) U (A ∩ B1) = P3 U P1……………………

(3) Fr om (1), (2) and (3), A▲B= (B∩A) U (A∩B) = (B-A) U (A-B). 14. Using Venn diagram, prove the following property of the symmetric difference: (4m) jan 2015 A ▲ (B▲C) = (A▲B) ▲ C

Solution: By using Venn diagr am,

Wher e, A = P1 U P2 U P3 U P4 B = P2 U P3 U P5 U P6 C = P3 U P4 U P6 U P7 L.H.S (B▲C) = (B U C) – (B∩C) = P2 U P4 U P5 U P7 A U (B▲C) = P1 U P3 U P2 U P4 U P5 U P7 A ∩ (B▲C) = P2 U P4 A ▲ (B▲C) = A U (B▲C) - A ∩ (B▲C) = P1 U P3 U P5 U P7. R.H.S (A▲B) = (A U B) – (B ∩ A) = P1 U P4 U P5 U P7 (A ▲ B) U C) = P1 U P4 U P5 U P6 U P3 U P7 (A ▲ B) ∩ C) = P4 U P6 (A ▲ B)▲C) = (A ▲ B) U C) - (A ▲ B) ∩ C) = P1 U P5 U P3 U P7 So that L.H.S = R .H.S Hence pr oved

15. Thirty cars are assemb led in a factory. The options available are a transistor, an air conditioner and power windows. It is known that 15 of the cars have transistor, 8 of them have conditioners and 6 of them have power windows.

Moreover, 3 of them have all three options. Determine at least how many cars do not have any options at all. (5m) Jan 2016 Solution: Given data:

|U| = 30 Total car s |A| = 15 Tr ansistor s |B| =8 Air conditioner s |C|= 6 Power windows |A ∩ B ∩ C| = 3 All o ptions i) |A U B U C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C| Her e A ∩ B, B ∩ C, C ∩ A is a su bset of A ∩ B ∩ C So |A U B U C| ≤ |A| + |B| + |C| - 2|A ∩ B ∩ C| ≤ 15 + 8 + 6 -2(3) ≤ 29 – 6 = 23 Dept . of CSE, SJBIT


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1

(A U B U C) is the set of car s that do not have any o ption. 1

|A U B U C| ≥ |U|- |A U B U C| ≥ 30 – 23 = 7 Ther ef ore a minimum of 7 car s has none of the options. 16. A survey on a sample of 25 new cars showed that the cars had the following 15 cars had air condit ioners 12 cars had radios 11 cars had power windows 5 cars had air conditioners and power windows 9 cars had air conditioners and radios 4 cars had radios and power windows 3 cars had all the three options Find the number of cars that had i) only power windows ii) at least one option Solution: Given data: |U| = 25 Total car s

(7m)

Jan 2016

|A| = 15 Air conditioner s |R| = 12 Radios |W|= 11 Power windows |A ∩ R| = 9 Air conditioner s and Radios |A ∩ W| = 5 Air conditioner s and Power windows |R ∩ W| = 4 R adios and Power windows |A ∩ R ∩ W| = 3 All options i) only power windows: |W| = |W – A – R | = |W| - |A ∩ W| - |R ∩ W| + |A ∩ R ∩ W| = 11 – 5 – 4 + 3 =5 5 CAR S HAVE ONLY POWER WI NDOWS ii) At least one option: |W U A U R| = |W| + |A| + |R| - |A ∩ R| - |A ∩ W| - |R ∩ W| + |A ∩ R ∩ W| = 11 + 15 + 12 – 5 – 9 – 4 + 3 = 23 17. A survey of 500 television viewers of sports channel produced the following information: 285 watch crick et, 195 watch hock ey, 115 watch foot ball, 45 watch Dept . of CSE, SJBIT


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crick et and foot ball, 70 watch cricket and hocke y, 50 watch hockey a nd foot ball and 50 do not watch any of the thre e kinds of games i) How many viewers in survey watch all three kinds of games? ii) How many viewers watch exactly one sport? Solution: At least one:

(8m)

Jan 2016

|CU H U F| = |U| - |C U F U H|1 =500 – 50 = 450 do watch any of the games |C U H U F| = |C| + |F| +|F| -|C∩F|-|C∩H|-|H∩F|+|C∩F∩H| But |C∩F∩H| means those viewer s who watch all the 3 games. So, |C∩F∩H|= |C U H U F| -|C| - |H| -|F| +|C∩F|+|C∩H|+|H∩F| = 450 – 285 – 195 – 115 +45 +70 +50 = 20 Ther ef ore 20 viewer s watch all the 3 games. 2) Only cr ick et viewer s: |C1| = |C – H - F| = |C|-|C∩F|-|C∩H|+|C∩F∩H| = 285 -70 -45 +20 =190 3) only hock ey viewer s: |H1| = | H - C - F| = |H|-|H∩F|-|C∩H|+|C∩F∩H| = 195 -50 -70 +20 = 95 4) only football viewer s: |F1| = | F- C - H| = |F|-|H∩F|-|C∩F|+|C∩F∩H| =115 – 45 -50 +20 = 40 Ther ef or e num ber of viewer s those who watch exactly 1 game is |C1| + |H1| + |F1| = 190 + 95 + 40 = 325. 18. The freshman class of a private engineering college has 300 students. It is known that 180 can program in PASCAL, 120 in FORTR AN, 30 in c++, 12 in PASCAL and c++, 18 in FOR TR AN and c++, 12 in PASCAL and FORTRAN, and 6 in all three languages If two students are selected at random, what is the probability

that they can i) Both program in PASCAL? ii) Both program only in PASCAL?



(6m)

Jan 2016

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300

Solution: the num ber of ways of selecting 2 students fr om 300 is

c

2.

180

c.

The num ber of ways of selecting 2 students fr om those who can progr am P ascal is

2

162

The num ber of ways of selecting 2 students who can pr ogr am only in Pascal is

c. 2

The pr o ba bility of selecting two students both of whom can pr ogr am in Pascal is 180

300

c /

c

2

2

= 0.359 OR 35.9% The pr o ba bility of selecting two students both of whom can pr ogr am in Pascal only is 162

300

c / 2

c

2

=0.291 OR 29.1%. 19. In a survey of 120 passengers, an airline found that 48 en joyed wine with their meals, 78 en joyed mixed drink s, 66 en joyed iced tea. In addition, 36 en joyed any given pair of these beverages and 24 en joyed them al l. If two passengers are selected at random from thee survey sample of 120, what is the probability that they both want only iced tea with their meals? (7m) Jan 2016 Solution: fr om the inf ormation pr ovided, we constr uct the Venn diagram .The sample s pace constants of the pairs of passenger s we can s elect fr om the sample of 1 20. 120

So |S| =

c =7140 2

A) Only iced tea = T = | T – W – C | 1

= |T| -|T∩W| - |T∩C| - |T∩W∩C| = 66 – 36 – 36 + 24 = 18



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The Venn diagr am indicates that ther e ar e 18 passenger s who drink only ice tea. 18

So |A| =

c = 153 2

Ther ef or e p (A) = |A| / |S| = 153 / 7140 = 51 / 2380. B) Exactly 2 of the 3 type Only T and C = |T∩C| - | T∩W∩C| ………………….i 36 – 24 = 12 Only C and W = |W∩C| - |T ∩W∩C| ………………..ii 36 – 24 = 12 Only W and T =|W∩T| - |T∩W∩C| ………………iii 36 – 24 = 12 Adding i, ii, iii we get = 12 + 12 + 12 = 36 = |B| 120

36

Ther ef or e p (B) =

c / 2

c

2

=3/4. 20. Find the probability of getting a sum dif ferent from 10 or 12 after rolling two dice. (5m) Jan 2015 Solution : We can get 10 in 3 dif f erent ways: 4 + 6, 5 + 5, 6 + 4, so P (10) — 3/36. Similar ly we get that P (12) — 1/36. Since they are mutually exclusive events, the pr o ba bility of getting 10 or 12 is P (10) + P (12) — 3/36 + 1/36 — 4/36 — 1/9. So the pr o ba bility of not getting 10 or 12 is 1 — 1/9 — 8/9. 21. Explain set operations: Solution: 1. Inter section : The common elements of two sets:

(6m)

Jan 2015

A ∩ B = {x | (x e A) ∧ (x e B)} . If A ∩ B = ∅ , the sets ar e said to be disjoint. 2. Union : The set of elements that belong to either of two sets: A ∪ B = {x | (x e A) ∨ (x e B)} . 3. Com plement : The set of elements (in the univer sal set) that do not b elong to a given set:

A = {x e U | x /e A} .



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4. Dif f er ence or Relative Com plement : The set of elements that belong to a set but not to another:

A - B = {x | (x e A) ∧ (x /e B)} = A ∩ B . 5. Symmetr ic Dif f er ence : Given two sets, their symmetric dif f er - ence is the set of elements that b elong to either one or the other set but not both. A ⊕ B = {x | (x e A) ⊕ (x e B)} . 22. A compuer services company has 300 programmers. It is known that 180 of these can program pascal, 120 in FOR TR AN, 30 in c++, 12 in pascal and c++, 18 in FOR TR AN and c++, 12 in pascal and FOR TR AN and 6 i n all the three.

a) If a programmer is selected at random what is the probability that she can program in exactly two languages? b) If two programmers are selected at random what is the probability that they can both program in pascal? (10m) Jan 2016 Solution:

a) We note that the num ber of pr ogrammer s who can pr ogr am in 1) pascal and FOR TR AN only is 6 2) pascal and C++ only is 6 3) FOR TR A N and c++ onl y is 12 Thus 6+6+12=24 can pr ogr am in exactly two languages. Pr (A)=24/300=8% b) the num ber of ways of selecting 2 progr ammers fr om 300 is 300 c 2. Same way selecting 2 in pascal is 180 c 2. Same way selecting 2 p rogr ammer s who can pr ogr am in pascal is 162 c 2. The pr o ba bility of selecting two who can pr ogr am in pascal is Pr (B)=180 c 2/300 c 2= 36% The pr o ba bility of selecting 2 who can pr ogr am in both is Pr ©=162 c 2/300 c 2=29% 23. Define power set of a set. Obtain all the power sets of A={1,2,3,4}. Sol:

(3m)

Jan 2016 Power Set: The collection of all su bsets of a set A is called the power set of A, and is re pr esented P(A). For instance, if A = {1, 2, 3,4}, then P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},{1,2,3},{2,3,4},

{1, 3,4},{1,2,4} . A} . 24. Simplif y the following expression: Dept . of CSE, SJBIT


(4m)

Jan 2016 Page 11

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Sol:

25. Let p, q be primitives statements f or which implication p → q is f alse. Determine the Jun 2015 / Jan 2016 truth values of the f ollowing.

Sol:

5 marks q

p v q

T

T

T

T

T

T

F

T

T

T

F

T

T

T

T

F

F

F

F

T

p

qv p

( p v q) ↔ (q v p)

26. By constructing the truth table. Show that the compound propositions pAND(~ q V r) Jun 2015 / Jan 2016

Sol:

5 mark s

27. Prove the following logical statement is a tautology: Jun 2015 / Jan 2016 .

sol:



5 mark s

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28. Let p, q be primitives statements f or which implication .June 2015 / Jan 2016

Sol:

5 mark s

29. Discuss the basic connectives that are used in logic. (6m) Jun 2016 Solution: new pr o position ar e of ten for med by star ting with given pr o positions with the

aid of wor ds or phr ases lik e “not”, “and”, “if …then”, and “if and only if ” such wor ds or phr ases w=ar e called connectives a) Negation: a pr o position o btained by inser ting the word “ not” at an a p pr o pr iate places in a given pr o positions called negation of given pr o position and is denoted by ~p ( p is any pr o position). b) Con junction : a compound pr o position o btained by com bining two given pr o positions by inser ting the wor d “and” in between them is called con junction (denoted by p ^ q) c) Dis junction: a com pound pr o position o btained by inser ting the wor d “or ” in between them is called dis junction (denoted by p v q) d) Conditional: a compound pr o position o btained by com bining two given pr o position by using the wor d “if and then” at a p pr o pr iate place is called a conditional pr o position and is denoted by “ p →q” 30. Given p and q statements, explain the following terms

(7m) Jun 2016

a) Con junction b) dis junction c) logically Equivalence d) tautology Solution:

a) con junction: a com pound pr o position o btained by com bining two given pr o positions by inserting the wor d “and” in between them is called con junction (denoted by p ^ q) b) Dis junction: a com pound pr o position o btained by inser ting the word “or ” in between them is called dis junction (denoted by p v q) c) Logical equivalence : Two pr o positions p and q ar e said to be logical equivalent wher e p and q have the same tr uth value or equivalently the biconditional p ↔ q is tautology. Then we denote p ≡ q.



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d) Tautology: a com pound pr o position which is always tr ue regar dless of the tr uth values of its com ponents is called a Tautology. 31. show that (p v q) ↔ (q v p) is a t autology.

(3m)

Jun 2016

Solution: we have to show that (p v q) ↔ (q v p) is Tautology we can write the tr uth ta ble as f ollows. p

q

p v q

qv p

( p v q) ↔ (q v p)

T

T

T

T

T

T

F

T

T

T

F

T

T

T

T

F

F

F

F

T

32. Define converse, inverse and contra positive of a statement:

(4m) Jun 2016

Solution:

Consider a conditional (p →q) , Then : 1) q→ p is called the conver se of p→q 2)

� p→� q

3)

� q→� p

is called the inverse of p→q

is called the contra positive of p →q 4) The conver se of a conditional pr op osition p → q is the prop osition q → p . As we have seen, the biconditional pr op osition is equivalent to the con junction of a conditional pr op osition an its conver se. p ↔ q ≡ (p → q) ∧ (q → p)

So, for instance, saying that “John is mar r ied if and only if he has a spouse” is the same as saying “if J ohn is mar r ied then he has a sp ouse” and “if he has a spouse then he is mar ried” . Note that the conver se is not equivalent to the given conditional pr op osition, for instance “if John is fr om Chicago then J ohn is fr om Illinois” is tr ue, but the conver se “if John is fr om Illinois then John is fr om Chicago” may be false. 33. Find the truth value of p,q,r for the foll owing using truth tables: (5m) Sol:



Jun 2016

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34. Prove the following tautologies: Sol:

35. Prove the following:

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(5m)

(6m)

Jan 2015

Jan 2015

Sol:

36. Find the truth values for the following logical expressions: Dept . of CSE, SJBIT


(4m)

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Sol:

37. Write the truth table for the following: Sol: Dept . of CSE, SJBIT


(8m)

jun 2014

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38. Simplify the following compound statements: Sol:

(6m) jun 2014

39. Verify whether the following logical expressions are tautology or contradiction using truth tables: Sol:



(6m)

jun 2014

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40. Prove the following logical statement is a tautology: Sol:





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(5m)

(5m)

(5m)

jun 2014

jun 2014

Jan 2016

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DISCRETE MATHEMATICAL STRUCTURES 43. Prove the following logical statement is a tautology. Sol:

15CS35 (5m)

Jan 2016

MODULE-2 Fundamentals of Logic contd Find inverse, converse and contra positive of the following (6m) Jan 2016

1.

Sol:

2. Find inverse, converse and contra positive of the following:

Sol:

(6m)

3. Simplify the following with reasons: Sol:

(4m)



Jan 2016

Jan 2016

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4. Prove below open statements: sol:



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(4m)

Jun 2016

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5. Prove below quantifiers:

(6m)

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Jun 2016

Sol:

6. Write inverse, converse and contra-positive:

(5m)

Jun 2016

Sol:



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7. Prove the below open statements:

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(7m) jan 2015

Sol:

8. Check the validity of the following arguments:

(3m) jan 2015

Sol:



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9. Prove the following quantifiers:

10. Prove the following rules of inf erences:

11. Prove the following rules of inferences Sol:

12. Prove the following rules of inferences Sol:

(3m) jan 2015

(4m)

jan 2015

(6m)

jun 2014

(4m) jun 2014

13. Verify the rules of inference from the following truth tables: (6m) jun 2015 Dept . of CSE, SJBIT


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Sol:

14. Prove the following open statements: Sol:

(4m)

jun 2015

15. Find converse inverse and contra positive of the logical expressions given below: (4m) jun 2015 Dept . of CSE, SJBIT


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Sol:

Properties of the Integers (6m)

16.

Jan 2016

Sol: basic ste p n=6 24<(36-7) = 29 is tr ue Induction step su p pose s(n) is true

4n<(n2-7) Consider 4(n+1)= 4n+4 <(n2-7)+4 <(n2-3) tr ue (8m)

17

Jan 2016

Sol:



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(6m)

18.

Jan 2015

Sol:

19. For all positive integers n, prove that if n>=24, then n can be written as a sum of 5s and 7s. (6m) Jun 2016

Sol: i ) Basic Step: 24= (5+5) + (7+7)

This shows that s (24) is tr ue ii) Induction step: su p pose s(n) is tr ue m=(5+5+……….)+(7+7+……….) if su p pose m has r num ber s of 5 and s num ber s of 7.then we r e pr esent this as f ollows. M+1= [(5+5+……….) + (7+7+……….) + 1] r s = [(5+5+……….) + (7+7+………) + (7+7) + 1] Dept . of CSE, SJBIT


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r s-2 = [(5+5+……….) + (7+7+………)]

r +3 s-2 This shows that Q (m+1) is a sum of 7 and5.Thus s (m+1) is true.

20.

Prove by induction:1 2 + 32

+………..+ (2n-1)2=n (2n-1) (2n+1) ⁄ 3 (6m)

Jun 2016

Sol: Let s (n) =given statement, i) Basic Step: Putting s (1) in the statement 2 1 =1(2(1)-1) (2(1) +1) ⁄ 3 1 =3 ⁄3 1 =1→Tr ue ii) Induction step: S (n) is tr ue for n=k wher e k≥1.su bstitute n=k in the statement. 2 2 2 +3 +………..+ (2k -1) = k (2k -1) (2k +1) ⁄ 3 1 2 Add (2k +1) on both sides 2 2 2 2 2 1 +3 +………..+ (2k -1) + (2k +1) = k(2k -1)(2k +1)(2k +1) ⁄ 3 = (2k+1)[k (2k -1)+3(2k +1)]/3 2 = (2k +1) (2k +5k +1)/3 = (2k +1) (k +1) (2k +3)/3 This statement of s (k +1) is tr ue whenever the statement s (k ) is tr ue. n-1 for all integers n≥1. 21. By induction prove that !n ≥ 2 n-1 Solution: Let p (n) be the pr edicate !n≥ 2 wher e n0 =1.

(7m)

Jan 2015

i) Basic Step: We have p (1) is the statement !1≥ 21-1 is clear ly tr ue. ii) Induction step: Wher e k≥1, if p(k ) is tr ue, then p(k +1), must also be true. Assume that for some f or ced k ≥1, !k ≥ 2k-1

Then p (1+1) is, by the r ecur sive def inition, the lef t side of p(k +1) is

!k +1 = (K +1)! k ≥ (k +1)2k-1 Using p (k )

≥ 2*2k-1 k+1≥ 2 since k ≥ 1 ≥2k right hand side of p (k +1)



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Ther ef or e p (n+1) and p (n) ar e tr ue for all n, n≥1. 22. Prove by mathematical induction. Sol:

(5m)

Jan 2015

23. Prove by induction:

(6m)

Jun 2014

1

2

2

2

+2 +3

+………..+n 2

=n (2n+1) (n+1) ⁄ 2

Solution: Let s (n) denote the given statement i) Basic Step: S (1) is the statement

12 = (1*2*3)/6 1 which is tr ue? ii) Induction step: If n=k Then s (k ) f or any k≥1 = 12 +22 +32 ……...+k 2 = (k ) (2k +1) (k +1)/6

Using this we shall f ind

12 +22 +32 +………+k 2 + (k +1)2 =k (2k +1) (k +1)/6 + (k +1)2 = (k +1) [k (2k +1)/6 + (k +1)] = (k +1) (k +2) (2k +3)/6 → S (k +1) Hence S (k +1) is tr ue whenever s (k ) is tr ue f or k ≥1 Hence it is pr oved. 24. Prove by mathematical induction 1.3+2.4+3.5+…n(n+2)= n (n +1) (2n + 7) / 6 (6m) Jun 2014 Solution:

Basis Step: S (1): 1 (1 + 2) = 1 (2) (9) / 6 3 = 18 / 6 3 = 3 S (1) = Tr ue. Induction Step: Assume S (k ) is tr ue 1.3 + 2.4 + 3.5 +….k (k +2) = k (k+1) (2k +7) / 6 Now, R e place k by (k + 1) (k + 1)(k + 1 + 2) = (k + 1) (k + 3)  Add this on both sides. 1.3 + 2.4 + 3.5+…..k (k + 2) + (k + 1) (k + 3) = k (k + 1) (2k + 3) / 6 + (k + 1) (k + 3)

= (k + 1) [k (2k + 7) / 6 + k + 3] = (k + 1) [k (2k + 7) / 6 + k + 3] = (k + 1) [(2k ^2 + 7k +6k +18) / 6] = (k + 1) / 6 * (4(k + 2) + 9(k + 2))



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= (k + 1) / 6 * (2k + 9) (K + 2) 25. A sequence an is def ined by a1=3, an=an-1+an+1, for n>=2, find an ex plicit for m: Sol: (4m) Jun 2015

26. For n>=0 let fn denote the nth Fibonacci number. Prove that F0+f 1+f2+……….+f n= Summation Fi= fn+2 -1

(4m) Jun 2015

Sol:



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MODULE-3 Relations and Functions 1. Let A= {1,2,3,4} and let R be the relation defined by R = {(x,y)| x,y belongs to A, X<=y}. Determine whether R is ref lexive, symmetric, Anti symmetric or transitive. (5m) Jan 2016

Sol: Ref lexive if for all x e A, x R x. For instance on Z the relation “equal to” (=) is ref lexive. 2. Tr ansitive if for all x, y, z e A, x R y and y R z im plies x R z . For instance

equality (=) and inequality (<) on Z ar e tr ansitive rela tions . 3. Symmetr ic if f or all x, y e A, x R y = y R x. For instance on Z, equality (=) is s ymmetric, but str ict inequality (<) is not. 4. Antisymmetr ic if f or all x, y e A, x R y and y R x im plies x = y. For instance, non-str ict inequality (≤) on Z is antisymmetric. 2. What is the partition of a set? If R = {(1,1),(1,2),(2,1),(2,2),(3,4)(4,3),(3,3),(4,4)} def ined on the set A = {1,2,3,4}. Determine the partition induced.

(5m)

Jan 2016

Sol: Def n: A par tition of a set X is a set of nonempty su bsets of X such that every

element x in X is in exactly one of these subsets (i.e., X is a dis joint union of the su bsets). Equivalently, a family of s ets P is a p art ition of X if and on ly if all of the following conditions hold: 1. P does not contain the empty set. 2. The union of the sets in P is equal to X . (The sets in P ar e said to cover X .) 3. The inter section of any two distinct sets in P is empty. (We say the elements of P ar e pair wCSE dis joint.) 3. Define partial order. If R is a relation on A ={1,2,3,4} defined by X R Y if x|y.

prove that (A,R) is a POSET. Draw its Hasse diagram. (6m) Sol:



Jan 2016

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4. Let A={2,3,4,6,8,12,24} and let<= denotes the partial order of divisibility that

is x<=y means x|y. Let B = {4,6,12}. Determine: a) All upper bounds of B

(4m) Jan 2016

b) All lower bounds of B c) Least upper bound of B d) Greatest lower bound of B Sol:

5. Let A= {1,2,3,4,5} . Def ine a relation R on AXA by (x1,y1)R(x2,y2) if and only if x1+y1=x2+y2 (7m) Jun 2016 Solution: 1) Ver if y that R is an equivalence relation on A

For all (x,y) belongs to AXA . We have x+y=x+y that is (x,y)R (x,y) Ther e R is symmetr ic. Next tak e any (x1,y1),(x2,y2) belongs to AXA. So x1+y1=x2+y2 Ther ef or e R is symmetr ic. Next tak e any (x1,y1),(x2,y2),(x3,y3) belongs to AXA. So x1+y1=x2+y2 and x2+y2=x3+y3. Ther ef ore R is tr ansitive. 2) We note that [(1,3}]= {(x,y) belongs to AXA | (x,y) R (1,3)} = {(x,y) belongs to AXA | x+y=1+3} ={(1,3),(2,2),(3,1)}, Similar ly [(2,4)]={(1,5),(2,4),(3,3),(4,2),(5,1)} [(1,1)]={(1,1)} 6. Let A= {1,2,3,4,6} and r be the relation on A defined by(a,b) belongs to R if

and only if a is a multiple of b. write down R as a set of ordered pairs. (4m) Jun 2016 Solution: Fr om the way R has been def ined

R= {(a b , ) | a, b belongs to A and a is a multiple of b } = {(1,1), (2,1), (2,2), (3,1), Dept . of CSE, SJBIT


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(3,3),(4,1),(4,2),(4,4),(6,1),(6,2),(6,3),(6,6)} 7. Def ine 1) ref lexive 2) symmetric 3) Irref lexive 4) Anti symmetric 5) transitive relations: (5m) Jun 2016

Sol: 1. Ref lexive if for all x e A, x R x. For instance on Z the relation

“equal to” (=) is ref lexive. 2. Tr ansitive if for all x, y, z e A, x R y and y R z im plies x R z. For instance equality (=) and inequality (<) on Z ar e tr ansitive relati on s . 3. Symmetr ic if for all x, y e A, x R y = y R x. For instance on Z, equality (=) is symmetric, but str ict inequality (<) is not. 4. Antisymmetr ic if f or all x, y e A, x R y and y R x im plies x = y. For instance, non-strict inequality (≤) on Z is anti s ymmetri c. 8. A set of 3 members is (A, B, C). Brotherhood is the relation among them.

Discuss whether the relation is equivalence. Sol: Reflexive: a R a V a Є ¸ A

(4m)

Jun 2016

=> a is br other of himself, which is not true Hence, not ref lexive Symmetric: Let a R b => a is br other of b => b is br other of a

=> b R a Hence, R is symmetric Transitive: Let a R b and b R c =>a is br other of b and b is br other of c =>a is br other of c =>a R c. Hence R is transitive. Conclusion: Since R is not ref lexive it is not an equivalence relat ion.

9. Define a relation R on B as (a, b) R (c, d) if a + b = c + d. show that R is an

equivalence relations.

(6m)

Jan 2015

Sol: a. Let (a, b) R (a, b) =>a + b = a + b, which is true Ther ef or e, R is ref lexive. b. Let (a, b) R (c, d) =>a + b = c + d

=>c + d = a + b (since, addition is commu tative) => (c, d) R (a, b) Dept . of CSE, SJBIT


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Ther ef or e, is symmetr ic c. Let, (a, b) R (c, d) and (c, d) R (e, f) => a + b = c + d and c + d = e + f . =>a + b = e + f => (a, b) R (e, f) Ther ef or e, R is tr ansitive and Ther ef or e, R is an equivalence R elation. 10. A = {1, 2, 3} find a. R 1 = {(1, 1) (2, 2) (3, 3)} b. R 2 = {(1, 2) (2, 1) (1, 3) (3, 1) (2, 3), (3, 2)} c. R 3 = A x A Sol: We have, A= {1, 2, 3} a. R 1 = {1, 1) (2, 2), (3, 3)} Reflexive: a R a, V a Є A. (1, 1) (2, 2) (3, 3) Є R Hence, R is r ef lexive.

(7m)

Jan 2015

Symmetric: Let a R b => Є 1R2

2 is not Related to 1 and also b is not Related to a Hence, R is not symmetr ic Transitive: Let a R b and b R c => 1 R 2 and 2 R 3 but, 1 is not Related t o 3 and also a is not Related to c Hence, R is not tr ansitive. Ther ef ore, R is not an equivalence Relation. b. R = {(1, 2), (2, 1) (1, 3) (3, 1) (2, 3) (3, 2)} Reflexive: a R a V a Є A

=> 1 R1, 2 R 2, 3 R 3 not true, Hence, R is not ref lexive Symmetric: Let a R b => 1 R 3 => 3 R 1 => b R a

Hence, R is symmetr ic. Transitive: Let a R b and b R c => 1 R 2 and 2 R 3 => 1 R 3 => a R c Hence, R is tr ansitive Dept . of CSE, SJBIT


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Ther ef or e, R is not an equivalence R elation. c. A = {1, 2, 3} R = A x A = {(1, 1)(1, 2)(1, 3)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(3, 3)} It is ref lexive, symmetric and tr ansitive and hence R is an equivalence Relation. 11. If R is a relation on I, set of integers such that, x R y holds true if (x - y) is divisible by 5, show that R is an equivalence relation. (6m) Jan 2015 Sol: a. Let x R x => x – x = 0 which is d ivisible by 5 Hence, R is r ef lexive. b. Let x R y

=> x - y is divisi ble by 5 => y - x is divisi ble by 5 => y R x Hence, R is symmetr ic. c. Let x R y and y R z. => x - y is divisi ble by 5 and => y - z is divisi ble by 5 => x - z also divisi ble by 5 => x R z So, R is tr ansitive. Ther ef ore, R is an equivalence relation. 12. If R 1 and R 2 are equivalence relations defined on the same set A. prove that R 1 n R 2 is an equivalence relation. (7m) Jun 2015 Sol: Pr ove: R 1 n R 2 is also equivalence relations. (a). Since R l and R 2 ar e r ef lexive, [since R 1 and R 2 ar e equivalence relations] => (a, a) R 1 and (a, a) Є R 2, V a Є A => (a, a) Є R 1 n R 2, V a Є A => R l n R 2 is ref lexive. ( b). Let (a, b) Є R 1 n R 2 => (a, b) Є R 1 and (a, b) Є R 2 => ( b, a) Є R 1 and ( b, a) Є R 2, (since R 1 and R 2 ar e s ymmetric) => ( b, a) Є R 1 n R 2 => R l n R 2 is symmetr ic. (c). Let (a, b) Є R l n R 2 and ( b, c) Є R l n R 2. => (a, b) Є R 1 and (a, b) Є R 2 and ( b, c) Є R 1 and ( b, c) Є R 2 => [(a, b) Є R 1 and ( b, c) Є R 1] and [(a, b) Є R 2 and ( b, c) Є R 2] => (a, c) Є R 1 and (a, c) Є R 2 (since R l and R 2 ar e tr ansitive) => (a, c) Є R l n R 2 Dept . of CSE, SJBIT


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Hence, R 1 n R 2 is tr ansitive. Ther ef or e, R 1 n R 2 is an equivalence R elation. 13. Let A = {I, 2, 3, 4} and B = (A x A). Def ine a Relation R on B as (a, b) R (c, d) if a + b = c +d. S.T. R is an equivalence relation and compute B/R. Sol: (4m) Jun 2015 R {(1, 1)} = {(1, 1)} and B = {(1, 1) (1, 2) (1, 3) (1, 4) R {(1, 2)} = {(1, 2), (2, 1)} R {(1, 3)} = {(1, 3), (3, 1), (2, 2)} R {(1, 4)} = {(1, 4), (4, 1), (2, 3), (3, 2)} R {(2, 4)} = {(2, 4), (4, 2), (3, 3)} R {(3, 4)} = {(3, 4), (4, 3)} R {(4, 4)} = {(4, 4)} P = B/R = {R {(1, 1)}, R {(1, 2)}, R {(1, 3)}, R {(1, 4)}, R {(2, 4)}, R {(3, 4)}, R {4, 4)} 14. Let, A = {a, b, c}, B = {1, 2, 3}, R = {(a, 1) (b, 1) (c, 2) (c, 3)} S = {(a, 1), (a, 2) (b, 1) (b, 2)} Compute R ~ , S~ , R U S, R n S, R -1, S-1 where(R ~ is R compliment) sol: (6m) Jun 2015

A x B = {(a, l) (a, 2) (a, 3) ( b, 1) ( b, 2) ( b, 3) (c, l) (c, 2) (c, 3)} R = {(a, 2) (a, 3) ( b, 2) ( b, 3) (c, 1)} S = {(a, 3) ( b, 3) (c, 1) (c, 2) (c, 3)} R U S = {(a, 1) ( b, 1) (c, 2) (c, 3) (a, 2) ( b, 2)} R n S = {a, 1) ( b, 1)} R -l = {(1, a) (1, b) (2, c) (3, c)} S-1 = {(1, a) (2, a) (l, b) (2, b)} 15. Let A = {1, 2, 3} Rand S be relations on A whose matrices are, (8m) Jun 2014 Sol: � 101 �

MR =

� 011� � � � 0 0 0 � � �

�

and Ms =

011�

� 110� �

� � 0 0 0 � � �

Deter mine relations R, R U S, R n Sand S-l and their matr ix re pr esentation. R = {(1, 1) (1, 3) (2, 2) (2, 3)} S = {(1, 2) (1, 3) (2, 1) (2, 2) (3, 2)} R = {(1, 2) (2, 1) (3, 1) (3, 2) (3, 3)} �

MR~ =

S~ = {(1, 1) (2, 3) (3, 1) (3, 3)}

010�

� 100� �

� � 1 1 1 � � �

(s com pliment)



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� 1 0 0�

Ms~ =

� 001�

� � 1 0 1 � � � �

R U S = {(1, 1)(1,2)(1,3)(2, 1)(2,2)(2,3)(3, 2)} � 111 �

MRUS =

� 111� �

� � 0 1 0 � � �

R n S = {(1, 3) (2, 2)} �

MRnS =

001�

� 010� � � � 000� � �

S-1 = {(2, 1) (3, 1) (1, 2) (2, 2) (2, 3)}

MS-1 =

� 010� � 111� � � � 1 0 0 � � �

16. Let A = {a, b, c} and Rand S be relations on A whose matrices are given below. Find the composite relation S o R, R o R, R o S, S o S and their matrices. (8m) Jun 2014 Sol: �

MR =

010�

� 111� �

� � 0 1 0 � � �

� 1 00�

MS =

� 011� �

� � 101� � �

R = {(a, a) (a, c) ( b, a) (b , b) ( b, c) (c, b)} S = {(a, a) ( b, b) ( b, c) (c, a) (c, c)} S o R = {(a, a), (a, c) ( b, a) ( b, b) ( b, c) (c, b) (c, c)} Dept . of CSE, SJBIT


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R o S = {(a, a) (a, c) ( b, a) ( b, b) ( b, c) (c, a) (c, c) (c, b)} R 2 =R o R = {(a, a), (a, c) (a, b) ( b, a) ( b, c) ( b, b) (c, a) (c, b) (c, c)} S2 = S o S = {(a, a) ( b, b) ( b, c) ( b, a) (c, a) (c, c)} 17. Let R = {(1, 2) (3, 4) (2, 2)} and S = {(4, 2) (2, 5) (3, 1) (1, 3)} be relations on the set A {2, 3, 4, 5} find S o R, R o S, Ro(S o R), So(R o S), Ro(R o R), So(R o R ), S(S o S) (5m) Jun 2014 Sol: R = {(1, 2) (3, 4) (2, 2)} S = {(4, 2) (2, 5) (3, 1) (1, 3)} a. S o R = {(1, 5) (3, 2) (2, 5)} b. R o S= {(4, 2) (3, 2) (1, 4)} c. Ro(S o R) = {(3, 2)} d. So(R o S) = {(4, 5), (3, 5), (1, 2)} e. R o R = {(1, 2), (2, 2)} f. S o S= {(4, 5) (3, 3) (1, 1)} g. Ro(R o R) = {(1, 2) {2, 2)} i. So(S o S) = {(3, 1) (1, 3)} 18. If A= {1,2,3,4} B={2,5} C= {3,4,7} Determine: 1) AXB 2) BXA 3) AU (BXC) 4) (AUB)XC 5) (AXC)U(BXC)

(5m) Jun 2014

Sol:

19. Def ine ref lexive transitive and symmetric relations with respect to quantif iers. (5m) Jan 2014 Sol:



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20. Draw the hasse diagram for the poset (p(u)) where u={1,2,3,4} (5m) Jan 2016 Sol:

21. Let A={1,2,3,6,9,18} and define R on A by xRy if x|y. Draw hasse diagram of the poset. (5m) Jan 2016 Sol:

22. Prove any R is a partial order. Sol:



(5m)

Jan 2016

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MODULE-4 Relations contd 1. Def ine the Cartesian product of two sets. For any non-empty sets A,B,C prove that A X (B U C) = (A X B) U (A X C) (5m) Jan 2016 Sol:

Def n: In mathematics, a Cartesian

product is a mathematical operati on which

retur ns aset (or product set) fr om multi ple sets. That is, for sets A and B, the

Car tesian pr oduct A × B is the set of all or der ed pair s (a, b) wher e a � A and b � B. Assume (x,y) is an element of A X (B U C). This means x is an element of A and y is an element of B or y is an element of C. Since (x,y) can be x as an element of A and y as an element of B, (x,y) is an element of A X B. Since (x,y) can also have x as an element of A and y as an element of C, (x,y) is an element of A X C. A X (B inter sect C)= (A X B) inter sect (A X C) 2. Define the following with one example for each i) Function ii) one-to one function iii) onto function. (6m) Jan 2016 Sol: function def : a function is a relation between a set of in puts and a se t of

per missi ble out puts with the pr o per ty that each in put is related to exactly one out put. Or Definition (f unction): A function, denote it by f , from a set A to a set B is a relation f r om A to B that satisf ies 1. for each element a in A, ther e is an element b in B such that is in the

relation, and 2. if and ar e in the r elation, then b = c . The set A in the a bove def inition is called the domain of the function and B its codomain. Thus, f is a function if it cover s the domain (ma ps every element of the domain) and it is si n g l e valued . One-one def n: one-to-one function is (i) afunction that pr eser ves distinctness: it never ma ps distinct elements of its domain to the same element of its codomain.



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DISCRETE MATHEMATICAL STRUCTURES (ii)

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Onto function def n: onto or a sur jection, if every element y in Y has a

cor r es ponding element x in X such that f ( x) = y. The f unction f may map mor e than one element of X to the same element of Y .

3. State the pigeonhole principle. An of f ice employs 13 clerk s. Show that at least 2 of them will have birthdays during the same month of the year.

(4m)

Jan 2016 Sol: I f m pi g eons ar e put int o m pi g eonhol e s, ther e is an empty hole i f f ther e's a

hol e wit h mor e than one pi g eon Pr oof . The num ber of fr iends of a per son x is an integer k with 0 · k · n ¡1. If ther e is a per son y whose num ber of fr iends is n ¡ 1, then ever yone is a fr iend of y, that is, no one has 0 friend. This means that 0 and n ¡ 1 cannot be simultaneously the num ber s of fr iends of some peo ple in the gr oup. The pigeonhole pr inci ple tells us that ther e ar e at least two peo ple having the same num ber of fr iends. R be def ined by f(x) = X2 and g(x) = x+5. Determine fog and gof show that he composition of two function is not commutative. (5m) Jan 2016

4. Let f: R

R g: R

Sol: gof (-2)= g[f( -2)] = g[- 2+1] = g[-1] = (-1) 2 + 2 = 1 + 2 =3 fog(-2) = f[g( -2)] = f[4+2] = f (6) = 7 gof (x) = g[f ( x)] = g(x+1) = =(x + 1) 2 + 2 = x2 + 2x + 3

gog(x) = g[g(x)] = g(x2 + 2) = (x2 + 2) 2 + 2 = x4 + 4x2 + 6 5. Find the nature of each of the function.

(5m)

Jun 2016

Sol:



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B are given. 6. In each of th e following cases sets A and B and a function f: A Determine (in each case) whether f is one—to-one or onto (or both) (or neither)

(5m)

Jun 2016

Sol:1. A =B= {1, 2, 3, 4}

f = {(1, 1), (2, 3), (3, 4), (4, 2)} f is onto and one-one

1

1

2

2

3

3



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2. A = {a, b, c} B= {1, 2, 3, 4} f = {(a, 1), ( b, 1), (c, 3)} neither onto nor one-one

3. A = {I, 2, 3, 4} B = {a, b, c, d} f = {(1, a), (2, a), (3, d), (4, c)}

4. let A = {1,2,3,4} B = {a, b, c, d} f: A ( B is given as { (1 , a), (2 , b)}

f -1: B ( A = {( a , 1) , (b , 2 )} f -1 is a function and hen ce f is inver ti ble 7. let A,B,C be any three non-empty sets and A=B=C={set of real numbers} f: A ( B , g: f: B ( C be f unction defined by f(a) = a+1 and g(b) = b2 + 2, find a. gof (-2), b. fog (-2), c. gof(x) , d. gog(x)

(8m)

Jun 2016

sol:

a.

gof (-2)= g[f(-2)] = g[-2+1] = g[-1] = (-1) 2 + 2 = 1 + 2 =3

b. fog(-2) = f[g(-2)] = f[4+2] = f (6) = 7 c. gof (x) = g[f (x)] = g(x+1) = =(x + 1) 2 + 2 = x2 + 2x + 3 d. gog(x) = g[g(x)] = g(x2 + 2) = (x2 + 2) 2 + 2 = x4 + 4x2 + 6 8. Let X = {1, 2, 3} and f , g, h and s be funct ion fr om X to x given by f = {(1, 2), (2, 3), (3, 1)},

g = {(1, 2), (2, 1), (3, 3)}, h = {(1, 1), (2, 2), (3, 1)} Dept . of CSE, SJBIT


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s = {(1, 1), (2, 2), (3, 3)} Solution:

(6m)

Jun 2016

a. fog = f[g] = {(1 , 3), (2 , 2), (3 , 1)} b. gof = g[f ] = {(1 , 1), (2 , 3), (3 ,2)} c. fohog d. hog = {(1, 2), (2, 1), (3, 1)} e. fohog = {(1, 3), (2, 2), (3, 2)} f . sog = {(1 , 2), (2 , 1), (3 , 3)} g. gos = {(1 , 2), (2 , 1), (3 , 3)} h. sos = {(1 , 1), (2 , 2), (3 , 3)} fos = {(1, 2), (2 , 3), (3 , 1)} 9. Let R denote the set of all real numbers.

(5m)

Jan 2015

Sol:

Let f: R ( R be a function def ined by f (x) = x 2 Is f an inver ti ble function? Solution: Given f(x) = =x 2

� x � R

f = {(1 , 1), (2 , 4), (3 , 9),………(-1 , 1), (-2 , 4), (-3 , 9),……..} Her e f(2) = 4 and f(-2) = 4 �

� Both 2 and -2 ar e r elated to 4 under f .

i.e. (2, 4) f and (-2, 4) f Similar ly, (4, 2) f -1 and (4, -2) f -1 (i.e.) 4 is related to two d istinct elements 2 and -2 under f -1 � The relation f

-1

is not a function Hence f is not inver ti ble. 10. Find the inverse function f -1, of f: A→B given by Sol: A = B = {1, 2, 3, 4, 5}

(6m)

Jan 2015

f = {(1, 3), (2, 2), (3, 4), (4, 5), (5, 1)} Solution: Given f = {( 1 , 3), (2 , 2), (4 , 3) ,(5 , 4) , (1 , 5)}

f -1 = {(3 , 1), (2 , 2), (4 , 3) ,(5 , 4),(1 , 5 )} Her e, f -1 s relation fr om ‘B’ to ‘A’ and a f unction as well. �

11. Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Determine whether the following functions from A to B are invertible or not a.

f = {(1 , a), (2 , a), (3 , c), (4 , d)}



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ii. g = {(1 , a), (2 , c), (3 , d), ( 4 , d)} Solution: i. Given f = {(1, a), (2 , a), (3 , c), (4 , d)}

(8m) Jun 2014

f -1 = {(a , 1), (a , 2), (c , 3), (d , 4)} Her e, f -1 is a relation fr om B to A but it is not a function because the element a has two images 1 and 2 under f -1 (i.e.) f -1 is not a f unction ii. Given g = {(1, a), (2, c), (3, d), (4, d)} �

�

Her e, g

-1

g -1 = {(a , 1), (c , 2), (d , 3), ( d , 4)} is a relation fr om B to A, but not a function because the element d has two

images 3 and 4 under g-1 . Hence g-1 is not a function. � g is not inver tible.

12. Prove that the symmetric dif f erence is associative on sets. Solution:

(5m)

Jun 2014

To pr ove, (A � B) � C = A � (B � C) we have to pr ove, f A�(B� C) (x) = f (A�B)� C (x) Now, f

(A�B)� C (x)

= fD� C wher e D = A � B

= f D + f C - 2 f D fC ( by pr o per ty 3 of char acter istic function) = f C + f D (1- 2 f C ) = f C ) (b y pr o per ty 3) C + (f A + f B - 2 f A fB ) (1 - 2 f = f C+ f A+ f B- 2 f Af B - 2 f Af C -2 f Bf C+ 4 f Af Bf C ) -2 f A(f B+ f – ) = f A + (f B+ f – C 2 f Bf C C 2 f Bf C = f A + (f B+ f – f C) (1-2f A) C 2 f B =f A + f B� C (1-2fA ) = f A + f B�C -2f A f

B�C

= f A�(B� C)

R.H.S � = f A�(B� C) (x) = f (A�B)� C (x) � (A � B) � C = A � (B � C)

13. Suppose the addresses of customers of a bank are recorded in 101 files on the basis of hashing function .With the account number. As keys, determine the file in which the address of the customer with the acc ount no. 2473876 is recorded Solution: Given No of f iles = 101 n

(6m)

Jun 2014

The keys whose file is being sear ched is a = 2473876 r =? We know that, h n (a ) = r



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�

r = a o n = 2473876 / 101 =83

�

r = 83 (a % n)

�

The address of the customer with account no. 2473876 is recor ded

in file 83. 14. Determine whether f(a,b)=[a+b) is commutative or assosiative. (5m) Sol:

Jun 2015

15. Prove that 151 integers are selected from {1,2,3,….3000} then the selection must include two integers x,y where x|y or y|x. (4m) Jun 2015



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Sol:

16. Let f,g:Z+->Z+ where for all x belongs to Z+ f(x)=x+1 and g(x)=max{1,x-1} the maximum of 1 and x-1. State few properties. (8m) Jun 2015 Sol:

17. Let f: Z->N be def ined by f(x)= 2x-1 if x>0 and -2x for x<=0 Prove that f is one one and onto determine f-1 (6m) Jun 2015 Sol:



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18. Let f,g,h:Z->Z be defined by f(x)=x-1 and g(x)=3x h(x)= 0 even and 1 odd Determine the f ollowing: Sol:



(6m)

.

Jan 2015

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MODULE-5 Groups 1. Define abelian group. For any group G pT G is abelian if and only if (8m) Jan 2016 (ab)2=a2b2 Sol:

2. Define homomorphism and isomorphism in group. Let f be homomorphism from a group G1 to group G2. Prove that i) If e1 is the identity in G1 and e2 is

the identity in G2, then f(e1) =e2 ii) f(a-1)= [f(a)]-1 for all a belongs to G1. (4m) Jan 2016 Sol:

3. State and prove lagranges theorem. Solution: Lagr ange's theor em

(8m)

Jan 2016

Lagrange's theorem, in the mathematics of gr oup theor y, states that for any finite gr oup G, the or der (num ber of elements) of every subgr oup H of G divides the or der of G. The theor em is named af ter Jose ph Lagr ange.

Pr oof of Lagr ange's Theorem: Dept . of CSE, SJBIT


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This can be shown using the concept of lef t cose ts of H in G. The lef t cosets ar e the

equivalence classes of a cer tain equivalence relation on G and ther ef or e for m a par tition of G. S pecif ically, x and y in G ar e related if and only if there exists h in H such that x = yh. If we can show that all cosets of H have the same num ber of elements, then each coset of H has pr ecCSEly |H| elements. We ar e then done since the or der of H times the num ber of cosets is equal to the num ber of elements in G, ther eby pr oving that the or der H divides the or der of G. No w, if aH and bH ar e two lef t cosets of H, we can def ine a map f : aH → bH by setting f(x) = ba-1x. This map is bi jective because its inver se is given by f -1 (y) = a b-1y. This pr oof also shows that the quotient of the ord er s |G| / |H| is equal to the index [G : H] (the num ber of lef t cosets of H in G). If we wr ite this statement as |G| = [G : H] · |H|, then, seen as a statement a bout car dinal num ber s, it is equivalent to the Axiom of choice. 4. Define abelian group. P T a group is abelian if and only if for all a,b belongs (5m) Jun 2016 to G (a,b)-1=a-1b-1

Solution: A gr oup G is said to be commutative or a belian if ab= ba for all a, be G. (a b)(b-1a-1)=a( b b-1)a-1 We have

=a(e)a-1=(ae)a-1=aa-1=e Hence (a b , )-1=a-1 b-1 5. Define a cyclic group. P T every cyclic group is abelian but converse is not true. (6m) Jun 2016

Sol: A cyclic group is a gr oup that can be gener ated by a single element, in the sense that the gr oup has an element g (called a "generator " of the gr ou p) such that, when wr itten multi plicatively, every element of the gr oup is a power of g (a multi ple of g when the notation is additive). Let G be a c yclic gr oup and g be a gener ator of G. for some integer s m and m A b=gm=gn=gm+n=gn+m=gngm=ba This shows G is a belian. Hence pr oved.

6. Let f:G->H be a homomorphism from G to H. If G is abelian P T H is also Dept . of CSE, SJBIT


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abelian.

(4m)

7. P T every subgroup is a cyclic group of itself .

(5m)

Jun 2016

Jun 2016

Solution:

8. Def ine homomorphism and isomorphism.

(5m)

Jan 2015

Sol: Gr ou p homomor phism:

Image of a Gr oup homomor phism(h) fr om G(le ft) to H(r ight). The smaller oval inside H is the image of h. N is the k er nel of h and aN is a coset of h. In mathematics, given two gr ou ps (G, *) and (H, ·), a group homomorphism fr om (G, *) to (H, ·) is a function h : G → H such that for all u and v in G it holds that wher e the gr oup o per ation on the lef t hand side of the equation is that of G and on the right hand side that of H. Fr om this pr o per ty, one can deduce that h ma ps the identity element eG of G to the identity element eH of H, and it al so ma ps inver ses to inver ses in the sense that

h(u - 1) = h(u) - 1. Hence one ca n say that h "is com pati ble with the gr ou p structur e". Dept . of CSE, SJBIT


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Isomor phism: If the homomor phism h is a bi jection, then one can show that its inver se is also a gr oup

homomor phism, and h is called a gr oup isomor phism 9. Define the binary operation on o on Z by x o y=x+y+1. Verif y that (Z,o) is an abelian group. (4m) Jan 2015 Solution:

10. For any group G pT G is abelian if and only if (ab)2=a2b2 (5m) Sol:

11. If a group G P T (a-1)-1=a and (ab)-1=b-1a-1 Sol:

12. P T for all groups (ab)-1=a-1b-1 Sol:



(4m)

Jun 2015

Jun 2015

(4m)

Jun 2014

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DISCRETE MATHEMATICAL STRUCTURES 13. Find every subgroup of S5 for all 2<=n<=5. Sol:

15CS35 (4m)

Jun 2014

14. Let g= S4 for a=[ 12 3 4][2 3 4 1 ] Find a subgroup and all the left cosets. (8m) Jun 2014 Sol:



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Groups Codes (6m) Jan 2016 15. Define a ring and integral domain. Solution: A ring is an A belian gr oup (R , +), together with a second binary oper ation *

such that for all a, b and c in R, a * (b * c) = (a * b) * c a * (b + c) = (a * b) + (a * c) (a + b) * c = (a * c) + (b * c) A ring is said to be an integr al domain if R is a commutative ring with unity. 16. Let r be a commutative ring with unity. Prove that if and only if for all a,b,c belongs to r where a=/ 0 1b=ac=b=c. (8m) Jan 2016

Sol: R is an integr al do main A b=ac=(a b-ac)=0

=a( b-c)=0 = b-c=0 = b=c Conver sely Xy=0=xy=x.0 =y=0 This pr oves that R is an integral domain. 17. Prove that every field is an integral domain. (6m) Jan 2016 Sol: Let f be a field. Then f is a commutative ring with unity. Tak e any a,b belongs to F

such that a b=0. If a=/0 a-1 exists in f and we get B=1 b=(a-1a) b=a-1(a b)=a-10=0 Similar ly if b=/=0 then a=0 Thus a b=0 in f im plies a=0 or b=0. Hence F is an integr al domain. (5m) Jun 2016 18. Show that Z5 is an integral domain. Solution: We note that Z5 is a commutative ring with unity. The multi plicative table for Z5 is given belo w. .

0

1

2

3

4

0

0

0

0

0

0

1

0

1

2

3

4

2

0

4

1

4

3

3

0

1

4

1

2

4

0

3

2

3

1

Z5 has no pro per zer o divisor s. Ther ef or e Z5 is an integr al domain. 19. Prove that Zn is a field if and only if n is prime. (3m)

Jun 2016

Solution: Su p pose n is pr ime then for any integer a such that 0

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Accor dingly [a] is a unit of Zn. This means that every non zer o element of Zn has a multi plicative inver se. Ther ef or e Zn is a f ield. 20. Def ine ring with an example:

(4m) Jun 2016 Sol: A ring is an A belian gr oup (R , +), together with a second binary op er ation * such that for all a, b and c in R, a * (b * c) = (a * b) * c

a * (b + c) = (a * b) + (a * c) (a + b) * c = (a * c) + (b * c) (8m) Jan 2015 21. Def ine hamming metric with example: Sol: Hamming codes can be com puted in linear alge br a ter ms thr ough matr ices because Hamming codes ar e linear codes. For the pur poses of Hamming codes, two Hamming matrices can be def ined: the code generator matrix

and the parity-check matrix

:

and

22. Explain decoding with coset leaders. Sol:

(8m)

Jan 2015

In the field of coding theory, a coset leader is def ined as a word of minimum weight in any par ticular coset - that is, a word with the lowest amount of non-zer o entr ies. Sometimes ther e ar e sever al wor ds of equal minimum weight in a coset, and in that case,

any one of those word s may be chosen to be the coset leader . Coset leader s ar e used in the constr uction of a standar d ar r ay for a lin ear code, which can then be used to decode received vector s. For a received vector y, the decoded message is y - e, where e is the coset leader of y. Coset leader s can also be use d to constr uct a fast decoding str ategy. For each coset leader u we calculate the syndr ome uH′. When we receive v we evaluate vH′ and find the matching syndr ome. The cor r es ponding coset

leader is the most lik ely er r or p atter n and we assume that v+u was the codewor d sent. Exam ple:A standar d ar r ay for an [n,k ]-code is a q n - k by q k array wh er e: 1. The fir st row lists all codewor ds (with the 0 codewor d on the extreme lef t) 2. Each row is a coset with the coset leader in the first column 3. The entry in the i-th row and j-th column is the sum of the i-th coset leader and th e Dept . of CSE, SJBIT


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j-th codewor d. For exam ple, the [n,k ]-code C3 = {0, 01101, 10110, 11011} has a standar d array as follows: 0

01101 10110 11011

10000 11101 00110 01011 01000 00101 11110 10011 00100 01001 10010 11111 00010 01111 10100 11001 00001 01100 10111 11010 11000 10101 01110 00011 10001 11100 00111 01010 23. Define ring with unity and ring with zero divisor: Sol:

(6m)

Jan 2015

For mally, a ring is an A belian gr oup (R , +), together with a second binary oper ation * such that for all a, b and c in R,

a * (b * c) = (a * b) * c a * (b + c) = (a * b) + (a * c) (a + b) * c = (a * c) + (b * c) also, if ther e exists a multi plicative identity in the ring, that is, an element e such that for all a in R, a *e = e *a = a then it is said to be a r ing with unity. The num ber 1 is a common example of a unity. 24. S is a subring of R if and only if for all a,b belongs to S we have a+b belongs to S and ab belongs to S Solution:

(6m)

Jun 2015

25. If R is a ring wiyh unity and a,b are units of R, P T ab is a unit of R and (ab)-1=b-1a-1 (6m) Jun 2015 Sol:



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26. Prove that a unit in a ring R cannot be a proper divisor of zero. (5m) Jun 2014 Sol:

27. If a is a unit of ring R P T –a is also a unit of ring R. Sol:

(6m)

Jun 2014

A set S is an associative ring if ther e are two operator s + and · such that: 1. S is an a belian gr oup under +. 2. S is closed under ·. 3. For any a, b, c � S, (a · b) · c = a · ( b · c). In other words, multi plication in S is associative. 4. For any a, b, c � S, a · ( b + c) = a · b + a · c a nd (b + c) · a = b · a + c · a . In other wor ds, multi plication distr i butes over addition in S.

(-1) · a = -a if (-1) · a + a = 0. Because 1 is the unit element, a = 1 · a and so (-1) · a + a = (-1) · a + 1 · a = ( -1 + 1) · a = 0 · a = 0 . Theor em: If R is a ri ng with unit element 1, th en (-1) · (-1) = 1. Pr oof . This follows fr om the pr evious theor em for a = -1. 28. Let S and T be subrings of a ring R. P T S intersection T is a subring of R . (3m) Jun 2014 Sol:



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ASSIGNMENT QUESTIONS MODULE 1: Set Theory

1) For any thr ee sets A,B and C pr ove that (A-B)-C =A – (BUC) = (A-C) – (B-C) 2) Deter mine the sets A and B given that A – B = {1, 2, 4}, B – A = {7, 8} and AUB = {1, 2, 4, 5, 7, 8, 9} 3) Deter mine the sets A and B given that A – B = {1, 3, 7, 11}, B – A = {2, 6, 8} and A∩B = {4, 9} 4) Pr ove that: A▲B= (B∩A1) U (A∩B1) = (B-A) U (A-B). 5) Deter mine the sets A and B given that A – B = {1, 2, 4}, B – A = {7, 8} and AUB = {1, 2, 4, 5, 7, 8, 9} 6) Using Venn diagr am, pr ove the following pr o per ty of the symmetr ic dif f er ence: 7) Thir ty car s ar e assem bled in a factory. The o ptions availa ble ar e a tr ansistor , an air conditioner and power windows. It is known that 15 of the car s have tr ansistor , 8 of them have conditioner s and 6 of them have power windows. Mor eover , 3 of them have all thr ee o ptions. Deter mine at least how many car s do not have any o ptions at all. 8) Ex plain what you mean by a pr oba bility of an event E and then solve the following pr o blem. If one tosses a fair coin four times, what is the pr o ba bility of getting two heads and two tails? 9) The fr eshman class of a pr ivate engineer ing college has 300 students. It is known that 180 can pr ogr am in PASCAL, 120 in FOR TR A N, 30 in c++, 12 in PASCAL and c++, 18 in FOR TR AN and c++, 12 in PASCAL and FORTRA N, and 6 in all three languages If two students ar e selected at random, what is the pr o ba bility that they can i) Both pr ogr am in PASCAL? ii) Both pr ogr am only in PASCAL? 10) In a sur vey of 1 20 passenger s, an air line found that 48 enjoyed wine with their meals, 78 enjoyed mixed dr ink s, 66 enjoyed iced tea. In addition, 36 enjoyed any given pair of these bever ages and 24 en joyed them all. If two passenger s ar e selected at random fr om the sur vey sam ple of 120, what is the pr o ba bility that they both want only iced tea with their meals? 11) The Ack er mann’s num ber s A m, n ar e def ined r ecur sively f or m, n€ N as f ollows. A0,n = n+1

for n≥ 0, A m,0 = Am-1,1

Am,n =Am-1,p

wher e p = Am ,n-1,

for m≥1 for m,n≥1

Pr ove that A1, n = n + 2 f or all n ≥ 0. 12) Show that ( p v q) ↔ (q v p) is a tautology.



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13) Out of 30 students in a dor mitor y, 15 take an ar t cour se, 8 take a biology cour se, and 6 tak e a chemistry cour se. It is known that 3 students tak e all thr ee cour ses. Show that 7 or mor e students tak e none of the cour se. 14) If N is the set of positive integer s and R is the set of real numbers, ex amine which of the following sets is empty : {x | x� N, 2x + 7 = 3} {x | x� R, x2 + 4 = 6} {x | x � R, x2 + 3x + 3 = 0}. 15) Using Venn diagr ams, investigate the tr uth or falsity of : i) A ∆(B ∩C) = (A ∆ B) ∩ (A ∆ C) ii) A – (B ∪ C) = (A – B) ∩ (A – C) for any three sets A, B, C. 16) Sim plify the following : i) A ∩(B–A)

-

ii) (A ∩ B) ∪ (A ∩ B) ∪ (A ∩ B ∩ C ∩ D). 17) A fair coin is tossed five times. What is the pro ba bility that the num ber of heads always exceeds the num ber of tails as each outcome is obser ved. 18) If statement q has the truth value 1, deter mine all tr uth value assignments for the pr imitive statements p, r, and s for which the tr uth value of the statement e 19) Esta blish the validity or pr ovide a counter exam ple to show the invalidity of th following ar guments : i) p�q � p�r �r �q ii) p p�r p�(q��r ) �q��s �s 20) Ver if y that [p�(q�r )]�[( p�q)�( p�r )] is a tautology. 21) Wr ite dual, negation, conver se, inver se and contr a positive of the statement given below : 22) If Ka bir wear s br own pant, then he will wear white shir t. Test whether the following ar gument is valid: If Ram’s com puter pr ogr am is cor r ect then he will be a ble to complete his com puter science assignment in at most two hour s. It tak es Ram over two hour s to com plete his com puter science assignment. Ther ef ore Ram’s com puter pr ogram is incor r ect. 23) Test whether the following ar gument is valid: If Ram’s com puter pr ogr am is cor rect then he will be a ble to com plete his com puter Dept . of CSE, SJBIT


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science assignment in at most two hour s. It takes Ram over two hour s to com plete his com puter science assignment. Ther ef or e Ram’s com puter pr ogr am is incor r ect. 24) Wr ite down the following pr o position in sym bolic for m, and find its negation, (i) For all integer s n, if n is not divisible by 2, then n is odd. (ii) All integer s ar e rational num ber s and some rational num ber s ar e not integers. (iii) For the univer se of integer s, if r(x): 2x +1 = 5; s(x): x2 =9 ar e o pen sentences. O btain the negation of a quantified statement x  [ r(x) �s(x)]. 25)Wr ite the following in sym bolic for m and esta blish if the argument is valid : If A gets the su per visor ’s position and wor k s har d, then he wi ll get a raCSE. If he gets a raCSE, then he will buy a new car . He has not bought a new car . Ther ef or e A did not get the su per visor ’s position or he did not wor k har d.

26) Ver ify the following without, using tr uth ta bles : [( p → q) � (

r � s) � (p � r)]

�

q → s.

27) Def ine tautology. Show that [(p � q) � (p → r) � (q → r)] → r i s a tautology, by constr ucting a tr uth ta ble. 28) Show that the following argument is invalid by giving a counter example : 29) [( p �

q) � {p → (q → r)}] →

r.

30) Def ine (p�q) � �(p�q). Re pr esent p�q and p�q using only �.

MODULE 2: Fundamentals of Logic contd

1) For the univer se of all peo ple, find whether the following is a valid statement: All mathematics pr of essor s have studied calculus Ramanu jan is a mathematics pr of essor Ther ef or e Ramanu jan has studied calculus. 2) Def ine the Rule of Univer sal s pecif ication and the Rule of Univer sal Gener alization. 3) Ver ify if the following is valid : � x[ p(x) � q(x)] ; � x p(x) �x[

g(x) � r(x)]

� x[s(x) →

r(x)]

� � x s(x).

4) Pr ove that for all real num ber s x and y, if x + y > 100, then x > 50 or y > 50. Dept . of CSE, SJBIT


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5) Deter mine if the argument is valid or not. All peo ple concer ned a bout the envir onment, recycle their plastic container s. B is not concer ned a bout the envir onment. Ther ef or e, B does not recycle his plastic container s. 6) Negate and sim plif y : i) � x[ p(x) �

q(x)]

ii) � x [( p(x) � q(x)) → r (x)]. 7) Pr ove that for every integer n, n2 is even if and only if n is even. 8) Let p(x, y) denote the open statement x divides wher e the universe consists of a ll integer s. Deter mine the truth values of the following statements. Justif y your answer s. 9) x y [ p ( x, y) � p(y , x) � (x = y )] 10) x y [ p ( x , y ) � p( y , x ) ] 11) State the following: (i) Well-Or der ing Pr inci pal (ii) Pr inci pal of mathematical induction. 12) Pr ove by mathematical induction For every positive integ er n, 3 divides (n3 – n). For all positive integer s n>=6, 4n < ( n2-7) 13) If n� N, pr ove that 5Fn+2 =Ln+4 - Ln. 14) Def ine the following : i) Well or der ing princi ple .

15) ii) Pr inci ple of mathematical induction. 16) Esta blish the following by mathematical induction : n

∑i

(2i)

=

2 + (n

-1 ) 2 n + 1

.

t=1

17) Find a unique solution for the recur r ence relation : 4an – 5an–1 = 0, n ≥ 1, a0 = 1. 18) Let Fn denote the nth Fi bonacci num ber : n

Pr ove that

∑ i=1

F(i-1) 2

F(n+ 2) =1 - n . 2

19) 20)

21)



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MODULE 3: Relations and Functions

1) Let R and S ar e symmetr ic relations on the set A. Pr ove that inter section of these two is also symmetr ic. 2) Def ine stir ling num ber of the second kind . Let a={1,2,3,4,5,6,7} and B={w,x,y,z} find the num ber of onto f uncti ons fr om A to B 3) Let T be the set of all tr iangles. Def ine a relation R on T by t1 R t2 if t1 and t2 have an angle of same measur e. Ver ify whether R is an equivalence r elation. 4) Def ine Car tesian pr oduct of two sets. For non–empty sets A, B, C pr ove that, A ×(B ∩ C) = (A × B) ∩ (A × C). 5) For each of the following functions, deter mine whether it is 1 – 1 : 6) i) f : Z → Z, f(x) = 2x + 1

ii) f : Z → Z, f(x) = x3 – x.

7) Let A = B = C = R, f : A → B, f(a) = 2a + 1 ; g : B → C, g(b) = b/2. Com pute gof and show that it is inver tible. 8) Let ∆ABC be an equilater al tr iangle of side 1. Show that if we select 10 points in the inter ior , ther e must be at least two points whose distance apar t is less than 1/3. 9) Give an ex am ple of a relation fr om A to B � B which is no t a function. 10) How many onto functions ar e there fr om (i) A to B, (ii) B to A? 11) Wr ite a function f : A�C and a f unction g : C�A. Find gof : A�A. 12) Wr ite an inver ti ble function f : A�C and find its inver se.

MODULE 4: Relations contd

1) Pr ove that a function f : A�B is inver ti ble if f it is one-one and onto. 2) Def ine the dif f erent pr oper ties of relation with exam ples. 3) Ver ify that (A, R) is a poset and draw its Hasse diagram wher e u= {1,2,3}, A = p(u) and R is a su bset relation on A. Let A = {1,2,3,4,5}. Def ine a relation r on A x A as (x1, y1)) R (x2, y2) if x1 + y1 = x2 + y2 4) 5) Ver ify that R is an equivalence relation 6) Deter mine the equivalence classes {(1, 3)],[(2, 4)], and [(1,1)] 7) Deter mine the par tition on A induced by R. 8) For each of the following relations, deter mine if the relation R is ref lexive, symmetric, antisymmetr ic or transitive : 9) On the set of all lines in the plane, l1R l2 if line l1 is per pendicular to live l2 10) On Z, xR y if x – y is even. 11) For A = {1, 2, 3, 4}, let R = {(1, 1), (1, 2), (2, 3), (3, 3), (3, 4)} be a relation on A. Dr aw the digr a ph of R 2 and find the matrix M(R 2). Dept . of CSE, SJBIT


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12) Dr aw the Hasse diagr am for all the positive integer division of 72. 13) Let A = {1, 2, 3, 4, 5, 6} × {1, 2, 3, 4, 5, 6}. Def ine R on A by (x1, y1) R(x2, y2) if x1y1 = x 2y2. Ver ify that R is an equivalence relation on A. 14) Let A = {1, 2, 3, 4, 5} � {1, 2, 3, 4, 5} and def ine R as i) Ver i fy that R is an equivalence r elation on A. ii) Deter mine the equivalence class [(1, 3)]. iii) Deter mine the partition induced by R. 15) Let A = {1, 2, 3, 6, 9, 12, 18} and def ine a r elation R on A as xRy if f x |y. D r aw the Hasse diagram for the poset (A, R) 16) Let A = {1, 2, 3, 4}, B = {w, x, y, z} and C = { p, q, r, s}. Consider R 1= {(1, x), (2, w), (3, z)}a relation fr om A to B, R2 = {(w, p), (z, q), (y, s), (x, p)} a relati on from B to C. What is the com posite relation R1� R2 for m A to C?

MODULE 5: Groups

1) Def ine acyclic gr oup with an exam ple. Pr ove that every cyclic gr oup is abelian.6 mar k s 2) If (G, *) and (H,*) ar e gr ou ps with res pective identities eG and eH , and f : G � H is an isomor phism, pr ove that f (eG) = eH. 3) Pr ove that in a gr oup code the minimum distance between distinct code wor ds is the minimum of the weights of the nonzer o elements of the code 4) For a gr ou p(G1′), pr ove that it is a belian if (a, b)2 = a2 b , 2 for all a, b�G. �

5) Let A = ��

�

0 1� -1

�

0�

. Ver ify that (A, A2, A3, A4) for m an a belian gr oup under matrix

multi plication. 6) Def ine a cyclic gr ou p. Ver ify that

( Z , �) �

5

is cyclic. Find a gener ator of this gr oup.

Examine if it has any subgr ou ps. 7) D ef ine a binary o p er atio n  o n Z a s x  y = x + y – 1. Ve r if y tha t ( Z, ) is a n a b e lia n gr o up . 8) Let f : G �H be a gr oup homomor phism onto H. If G is an a belian gr ou p, prove that H is also a belian. 9) The encoding f unction is given by the gener ator matr ix 10) Deter mine all the code wor ds. 11) Find the associated par ity-check matr ix H of the tr i ple 12) Def ine a gr oup code. Consider the encoding function E: Z2 > Z 2 2 re petition code wher e E(00)=000000, E(10) = 101010, E(01) = 010101, E(11)=111111. Pr ove that C = E(Z2 2) is a gr oup code. 13) Def ine a r ing. If R is a ri ng with unity and a,b ar e units in R Prove that ab is a unit in R and that (a b)-1= b-1a-1 Dept . of CSE, SJBIT


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14) Deter mine whether (Z, �,�) i s a ring with the binar y oper ations x�y = x + y –7, x�y = x + y – 3xy for all x,y � Z.

15) The (5m, m) five times re petition code has encoding function E : Z2m → Z25m. Decoding with D : Z25m→ Z2m is done by ma jor ity rule. With p = 0.05, what is the

pr o ba bility for the tr ansmission and cor r ect decoding of the si gnal 110. 16) What is the minimum distance of a code consisting of the following code wor ds : 17) 001010, 011100, 010111, 011110, 101001? What kind of er r or s can be detected? 18) The encoding function E : Z22 → Z25 is given by the gener ator matr ix �

G = ��

1

0 1 1 0�

0

1 0 1 1�

� �

. What is the er r or detection ca pacity of the code?

19) Let R be a ring with unity and a, b be units in R. Pr ove that ab is a unit of R and that 20) Find multi plicative inver se of each (non-zer o) element of Z7

21)


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CBCS STUDENTS ARE ADVISED TO REFER THEIR SYLLABUS COMPARE & STUDY THE FOLLOWING QUESTIONS PAPERS ARE FROM 2010 SCHEME

10CS34

USN

Third Semester B.E. Degree Examination, June 2012 Discrete Mathematical Structures Time: 3 hrs. . e c i t c a r p l a m s a d e t a e r t . s e e b g l a l p i k w , n 0 a 5 l b = g 8 n i + n 2 i a 4 , m g e r e e n e h t t t n i r o w s s e n n o i l i s t s a u o r q c e l r a / o n o d g n a a i d r o w t a a u r l d a y v e l i r t o o s l l u a p e p m p o a , c n , o s i r t e a c w s i i n f t a n r e u d o i y f g o n g i n t i e l l p a e m v o e c r n y n O A . . 1 2 : e t o N t n a t r o p m I

Max. Marks:100

Note: Answer FIVE full questions, selecting at least TWO questions from each part. PART – A 1

a. Let S = {21, 22, 23, ….., 39, 40}. Determine the number of subsets A of S such that : i) |A| = 5 ii) |A| = 5 and the largest element in A is 30 iii) |A| = 5 and the largest element is at least 30 iv) |A| = 5 and the largest element is at most 30 v) |A| = 5 and A consists only of odd integers. (10 Marks) b. Prove or disprove: For non-empty sets A and B, P(AB) = P(A)P(B) where P denotes (05 Marks) power set. c. In a group of 30 people, it was found that 15 people like Rasagulla, 17 like Mysorepak, 15 like Champakali, 8 like Rasagulla and Mysorepak, 11 like Mysorepak and Champakali, 8 like Champakali and Rasagulla and 5 like all three. If a person is chosen from this group, what is the probability that the person will like exactly 2 sweets? (05 Marks)

2

a. Verify that [p(q r)][(pq)(pr)] is a tautology. (05 Marks) b. Write dual, negation, converse, inverse and contrapositive of the statement given below : If Kabir wears brown pant, then he will wear white shirt. (05 Marks) c. Define (pq)  (pq). Represent pq and pq using only . (05 Marks) d. Establish the validity or provide a counter example to show the invalidity of the following arguments : (05 Marks) ii) p i) pq pr  pr p(q r) r q s q s

3

a. For the universe of all polygons with three or four sides, define the following open statements: i(x): all the interior angles of x are equal h(x): all sides of x are equal s(x): x is a square t(x): x is a triangle Translate each of the following statements into an English sentence and determine its truth value: x [s(x)  (i(x)  h(x))] i) x [t(x)  (i(x)  h(x))] ii) Write the following statements symbolically and de termine their truth values. iii) Any polygon with three or four sides is either a triangle or a square iv) For any triangle if all the interior angles are not equal, then all its sides are not equal. (08 Marks)

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Discrete Mathematical Structures

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