This document provides an overview of mathematical preliminaries including sets, functions, relations, graphs, and proof techniques. It defines sets, set operations, and set representations. It also covers functions, relations, Cartesian products, graphs including walks, paths, cycles, and trees. Finally, it discusses proof techniques like induction and contradiction and provides examples of proofs using these techniques.

1. Mathematical Preliminaries Hw (p.13) 1, 4, 7, 8, 9, 13, 23, 26, 30, 32

2. Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques

3. A set is a collection of elements SETS We write 1 is a member (or element) of set A ship is not a member (or element) of set B Membership of a given set

4. Set Representations C = { a, b, c, d, e, f, g, h, i, j, k } C = { a, b, …, k } S = { 2, 4, 6, … } S = { j : j > 0, and j = 2k for some integer k > 0 } S = { j : j is nonnegative and even } finite set infinite set

5. A = { 1, 2, 3, 4, 5 } Universal Set : all possible elements U = { 1 , … , 10 } 1 2 3 4 5 A U 6 7 8 9 10

6. Set Operations A = { 1, 2, 3 } B = { 2, 3, 4, 5} Union A U B = { 1, 2, 3, 4, 5 } Intersection A B = { 2, 3 } Difference A - B = { 1 } B - A = { 4, 5 } U 2 3 1 4 5 2 3 1 Venn diagrams A B

7. A Complement Universal set = {1, …, 7} A = { 1, 2, 3 } A = { 4, 5, 6, 7} 1 2 3 4 5 6 7 A A = A

8. 0 2 4 6 1 3 5 7 even { even integers } = { odd integers } odd Integers

9. DeMorgan’s Laws A U B = A B U A B = A U B U

10. Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set

11. Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } Proper Subset: A B A B U A B U

12. Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B A B = U

13. Set Cardinality For finite sets A = { 2, 5, 7 } |A| = 3 (set size)

14. Powersets A powerset is a set of subsets Powerset of S = the set of all the subsets of S S = { a, b, c } 2 S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Observation: | 2 S | = 2 |S| ( 8 = 2 3 )

15. Cartesian Product A = { 2, 4 } B = { 2, 3, 5 } A X B = { (2, 2), (2, 3), (2, 5), ( 4, 2), (4, 3), (4, 5) } A X B is an ordered set, i.e. A X B ≠ B X A |A X B| = |A|·|B| Generalizes to more than two sets A X B X … X Z

16. Relation from sets A to B Ex., A = { 1, 2, …, 7 }, B={ 1, 2, …, 50} {(x, y): x  A, y  B, and y= x 2 } {(x, y): x  A, y  B, and x < y} Are relations from A to B. A relation on set A Equivalence relation a partition on set A

17. FUNCTIONS domain 1 2 3 a b c range f : A -> B A B If A = domain then f is a total function otherwise f is a partial function f(1) = a 4 5 In general, we mean this.

18. GRAPHS A directed graph Nodes (Vertices) V = { a, b, c, d, e } Edges (Ordered pairs) E = { (a,b), (b,c), (b,e),(c,a), (c,e), (d,c), (e,b), (e,d) } node edge a b c d e

19. Labeled Graph a b c d e 1 3 5 6 2 6 2

20. Walk Walk is a sequence of adjacent edges (e, d), (d, c), (c, a) is a walk from e to a of length 3 (or denoted as e-d-c-a ) Length = # of edges a b c d e

21. Path a b c d e Path is a walk where no edge is repeated Simple path : no node is repeated

22. Path a b c d e Path is a walk where no edge is repeated Simple path : no node is repeated (e, b), (b, e), (e, d), (d, c), (c, a) is a path from e to a but it is not a simple path.

23. Cycle a b c d e 1 2 3 Cycle : a walk from a node (base) to itself without repeated edges Simple cycle : only the base node is repeated Loop: an edge from a node to itself base

24. Find All Simple Paths starting from c a b c d e origin The longest simple path has at most length 4. Since every vertex can only be visited at most once, and there are 4 other vertices.

25. (c, a) (c, e) Step 1 a b c d e origin Starting from vertex c, list all outgoing edges as long as they do not lead to any vertex already used in the path. At this point, we have all paths of length one starting at c . For all vertices a , e reached by c , we list all outgoing edges originating at a or e according the same way we did before.

26. (c, a) (c, a), (a, b) (c, e) (c, e), (e, b) (c, e), (e, d) Step 2 a b c d e origin

27. Step 3 a b c d e origin (c, a) (c, a), (a, b) (c, a), (a, b), (b, e) (c, e) (c, e), (e, b) (c, e), (e, d)

28. Step 4 a b c d e origin (c, a) (c, a), (a, b) (c, a), (a, b), (b, e) (c, a), (a, b), (b, e), (e,d) (c, e) (c, e), (e, b) (c, e), (e, d)

29. Trees are connected directed graphs without cycles such that there is a special vertex called “root” having exactly one path to every other vertices. root leaf parent child

30. root leaf Level 0 Level 1 Level 2 Level 3 Height 3 The level associated with each vertex is the number of edges in the path form the root to the vertex. The height of the tree is the largest level number of any vertex.

31. Binary Trees A binary tree is a tree in which no parent can have more than two children. A binary tree is a tree in which no parent can have more than two children. (p.10) Example 1.5. Prove that a binary tree of height n has at most 2 n leaves.

32. PROOF TECHNIQUES Proof by induction Proof by contradiction

33. Induction We have statements P 1 , P 2 , P 3 , … If we know for some m that P 1 , P 2 , …, P m are true for any k >= m that P 1 , P 2 , …, P k imply P k+1 Then Every P i is true

34. Proof by Induction Inductive basis Find P 1 , P 2 , …, P m which are true Inductive hypothesis Let’s assume P 1 , P 2 , …, P k are true, for any k >= m Inductive step Show that P k+1 is true

35. Example Theorem: A binary tree of height n has at most 2 n leaves. (p.10) We want to show: L(n) ≦ 2 n for n = 0, 1, 2,…. Proof by induction: let L(i) be the maximum number of leaves of any subtree at height i

36. We want to show: L(n) ≦ 2 n for n = 0, 1, 2,…. Inductive basis L(0) =1 ≦ 2 0 (the root node : height=0) Inductive hypothesis Let’s assume L(i) ≦ 2 i for all i = 0, 1, …, k Induction step we need to show that L(k + 1) ≦ 2 k+1 , k ≧0

37. Induction Step From Inductive hypothesis: height k k+1 Let’s assume L(i) ≦ 2 i for all i = 0, 1, …, k need to show that L(k + 1) ≦ 2 k+1 0 … L(k) ≦ 2 k

38. L(k) ≦ 2 k L(k+1) ≦ 2 · L(k) ≦ 2· 2 k = 2 k+1 Induction Step height k k+1 (we add at most two nodes for every leaf of level k) … need to show that L(k + 1) ≦ 2 k+1 To get a binary tree of height k+1 from one of height k , we can create at most 2 leaves in place of each previous one

39. Remark Recursion is another thing Example of recursive function: f(n) = f(n-1) + f(n-2) f(0) = 1, f(1) = 1

40. Proof by Contradiction We want to prove that a statement P is true we assume that P is false then we arrive at an incorrect conclusion therefore, statement P must be true

41. Example Theorem: is not rational Proof: Assume by contradiction that it is rational = n/m n, m are nonzero integers and without common factors We will show that this is impossible

42. = n/m 2 m 2 = n 2 Therefore, n 2 is even n is even n = 2 k 2 m 2 = 4k 2 m 2 = 2k 2 m is even m = 2 p Thus, m and n have a common factor 2 Contradiction!

43. Languages

44. A language is a set of strings String: A sequence of letters Examples: “cat”, “dog”, “house”, … Defined over an alphabet: Non-empty and finite

45. Alphabets and Strings We will use small alphabets: Strings

46. Empty String A string with no letters

47. String Operations Concatenation w  = ?  w = ?

48. Empty String Observations:

49. Reverse

50. String Length Length: Examples: |  | = ?

51. Length of Concatenation Example:

52. Example 1.8 (p.17) | uv | = | u | + | v | A recursive definition of the length of a string: | a | =1, | wa | = | w | + 1 For all a  , w is any string from  Fix the string u and consider all possible strings v ( the length of v can be 1, 2, …. ( 0 is trivial) ) The proof is done by induction on the length of v ( for any given u )

53. Example 1.8 (p.17) | uv | = | u | + | v | For a string u , consider the length of uv , concatenation of u with a string v Basis Inductive Step Induction Assumption

54. Substring Substring of string: a subsequence of consecutive letters String Substring

55. Prefix and Suffix Prefixes Suffixes prefix suffix

56. Another Operation Example:

57. Another Operation Definitions:

58. The * Operation : the set of all possible strings from alphabet Example: = { a, b } then = ?

59. The + Operation : the set of all possible strings from alphabet except

60. Languages Fall 2008 Automata

61. Note that: Sets Set size Set size String length

62. Another Example An infinite language

63. Operations on Languages The usual set operations Complement: ??

64. When we talk about a language , we must know what ground does this language stands on ….. Languages We should know the ALPHABETS that constitute the language

65. Complement Example The complement of Universal Set?

66. Reverse Definition: Examples:

67. Reverse Hw # 10 (a) Prove or Disprove: i.e., w R  L  w  L R

68. Concatenation Definition: Example:

69. Concatenation Hw 8. Prove

70. Another Operation Definition: Special case:

71. More Examples

72. Star-Closure (Kleene *) Definition: Example:

73. Positive Closure Definition: If  L then L +  L* - {  } It is not necessary that

74. True or False

75. True or False How to prove your answer?

76. Try Hw# 9 & 10(b) on p.28 What does w  L 2 mean? What does w  L * mean?

77. More Examples Consider a language on  = { a, b } What is L 2 ? L * ? e.g.

78. Grammars

79. Another Example Grammar: Derivation of sentence :

80. Grammar: Derivation of sentence :

81. Other derivations for Grammar:

82. Language of the grammar

83. More Notation Grammar Set of variables Set of terminal symbols Start variable Set of Production rules p.21

84. Example Grammar :

85. More Notation Sentential Form: A sentence that contains variables Example: sentence Sentential Forms

86. We write: Instead of:

87. In general we write: If:

88. By default:

89. Example Grammar Derivations

90. Another Grammar Example Grammar : Derivations: From A  aAb and A   , we know A   , ab , aabb , aaabbb , … * * * *

91. Language defined by a Grammar For a grammar with start variable : String of terminals

92. Example For grammar : Since Pf: show L(G)  {a n b n+1 } & L(G)  {a n b n+1 } from A  aAb we get A  a n Ab n when it is applied n times. Together with A   , we get A  a n b n for n = 0, 1, 2, ….  : w  L(G), i.e. S  * w

93. A Convenient Notation In general, we need to give a proof that a given language indeed generated by a certain grammar. Back to last Example

94. Example For grammar : To show Pf: show L(G)  {a n b n+1 } & L(G)  {a n b n+1 } from A  aAb we get A  a n Ab n when it is applied n times. Together with A   , we get A  a n b n for n = 0, 1, 2, ….  : w  L(G), i.e. S  * w L ( G )  L & L ( G )  L Need to show

95. More Examples on Grammars Find grammars for L on { a, b } and give brief arguments to explain why they work. L contains all strings with exactly one a L contains all strings with at least one a L 3 At Least: S  BaB; B  aB | bB | 

96. 1. Problems on p.27 You should be able to do 2 ~ 17, 21 Hand in: 9, 10, 11c, 14ef, 15c, 17 2. Read P. 37~ 41 , and try to describe L ( M ) in Fig. 2.6. Homework for next week.