Mathematical Preliminaries 1 Mathematical Preliminaries Sets Functions Relations

Mathematical Preliminaries 1

Mathematical Preliminaries • Sets • Functions • Relations • Graphs • Proof Techniques 2

SETS A set is a collection of elements We write 3

Sets – Cardinality of a set is the number of elements in the set |{a, b, c}| = 3 |{{a, b}, c}| = 2 ({a, b} and c) 4

Sets – Subsets & Supersets A is a subset of B, A ⊆ B if: ∀x, x ∈ A ⇒ x ∈ B ∀(x ∈ A), x ∈ B A is a proper subset of B, A ⊂ B if: A ⊆ B ∧ (∃x, x ∈ B ∧ x A) {} is a subset of any set (including itself) {} is the only set that does not have a proper subset 5

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

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

Set Operations A = { 1, 2, 3 } B = { 2, 3, 4, 5} B A • Union A U B = { 1, 2, 3, 4, 5 } 2 3 1 4 5 • Intersection U A B = { 2, 3 } 2 3 • Difference A-B={1} B - A = { 4, 5 } 1 Venn diagrams 8

• Complement Universal set = {1, …, 7} A = { 1, 2, 3 } 4 A = { 4, 5, 6, 7} A 1 5 A 2 6 3 7 A=A 9

{ even integers } = { odd integers } Integers 1 odd 2 3 even 0 4 5 6 7 10

De. Morgan’s Laws U AUB=A B B=AUB 11

Empty, Null Set: ={} SU =S S = U S- = Universal Set =S -S= 12

Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } B U Proper Subset: A U A B B A 13

Disjoint Sets A = { 1, 2, 3 } A U A B = { 5, 6} B= B 14

Set Cardinality • For finite sets A = { 2, 5, 7 } |A| = 3 (set size) 15

Powersets A powerset is a set of sets S = { a, b, c } Powerset of S = the set of all the subsets of S 2 S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Observation: | 2 S | = 2|S| ( 8 = 23 ) 16

Relations A relation R is a set of ordered pairs For example the relation < over the Natural Numbers is the set: { (0, 1), (0, 2), (0, 3), . . . (1, 2), (1, 3), (1, 4), . . . (2, 3), (2, 4), (2, 5), . . . } 17

Relations Often, relations are over the same set that is, a subset of A × A for some set A Not all relations are over the same set, however Relation describing prices of computer components {(Hard dive, $55), (WAP, $49), (256 M DDR, $44), . . . } 18

Relations Q and R are two relations The composition of Q and R, Q ◦ R is: {(a, b) : (a, c) ∈ Q, (c, b) ∈ R for some c} Q = {(a, c), (b, d), (c, a)} R = {(a, c), (b, c), (c, a)} Q ◦ R = {(a, a), (c, c)} Q ◦ Q ? (Q ◦ R) ◦ Q? 19

Relation Graph Each element is a node in the graph if (a, b) ∈ R, then there is an edge from a to b in the graph R = {(a, b), (a, c), (c, a), (b, b), (b, d)} 20

Functions A function is a special kind of relation (all functions are relations, but not all relations are functions) A relation R ⊆ A × B is a function if: For each a ∈ A, there is exactly one ordered pair in R with the first component a 21

Functions A function f that is a subset of A × B is written: f: A→B (a, b) ∈ f is written f(a) = b A is the domain of the function if A′ ⊆ A, f(A′) = {b : a ∈ A′ ∧ f(a) = b} is the image of A′ The range of a function is the image of its domain 22

Functions A function f : A → B is: one-to-one if no two elements in A match to the same element in B onto Each element in B is mapped to by at least one element in A a bijection if it is both one-to-one and onto The inverse of a binary relation R ⊂ A × B is denoted R− 1, and defined to be {(b, a) : (a, b) ∈ R} A function only has an inverse if it is a bijection 23

FUNCTIONS domain 4 5 A 1 2 3 If A = domain range B f(1) = a a b c f : A -> B then f is a total function otherwise f is a partial function 24

GRAPHS A directed graph e b node d a edge c • Nodes (Vertices) V = { a, b, c, d, e } • Edges E = { (a, b), (b, c), (b, e), (c, a), (c, e), (d, c), (e, b), (e, d) } 25

Labeled Graph 2 6 a b 1 5 3 e 6 2 d c 26

Walk e b d a c Walk is a sequence of adjacent edges (e, d), (d, c), (c, a) 27

Path e b d a c Path is a walk where no edge is repeated Simple path: no node is repeated 28

Cycle base a 3 2 e b 1 d c Cycle: a walk from a node (base) to itself Simple cycle: only the base node is repeated 29

Euler Tour 8 b 4 a 7 3 6 5 base e 1 2 d c A cycle that contains each edge once 30

Hamiltonian Cycle 5 b 4 a 3 base e 1 2 d c A simple cycle that contains all nodes 31

Finding All Simple Paths e b d a c origin 32

Step 1 e b d a c (c, a) origin (c, e) 33

Step 2 e b d a (c, a), (a, b) c origin (c, e), (e, b) (c, e), (e, d) 34

Step 3 e b d a (c, a), (a, b) c origin (c, a), (a, b), (b, e) (c, e), (e, b) (c, e), (e, d) 35

Step 4 e b (c, a) d a (c, a), (a, b), (b, e) c origin (c, a), (a, b), (b, e), (e, d) (c, e), (e, b) (c, e), (e, d) 36

root Trees parent leaf child Trees have no cycles 37

root Level 0 Level 1 Height 3 leaf Level 2 Level 3 38

Binary Trees 39

PROOF TECHNIQUES • Proof by induction • Proof by contradiction 40

Induction We have statements P 1, P 2, P 3, … If we know • for some b that P 1, P 2, …, Pb are true • for any k >= b that P 1, P 2, …, Pk imply Pk+1 Then Every Pi is true 41

Proof by Induction • Inductive basis Find P 1, P 2, …, Pb which are true • Inductive hypothesis Let’s assume P 1, P 2, …, Pk are true, for any k >= b • Inductive step Show that Pk+1 is true 42

Example Theorem: A binary tree of height n has at most 2 n leaves. Proof by induction: let L(i) be the maximum number of leaves of any subtree at height i 43

We want to show: L(i) <= 2 i • Inductive basis L(0) = 1 (the root node) • 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 44

Induction Step height k k+1 From Inductive hypothesis: L(k) <= 2 k 45

Induction Step height k L(k) <= 2 k k+1 L(k+1) <= 2 * L(k) <= 2 * 2 k = 2 k+1 (we add at most two nodes for every leaf of level k) 46

Remark Recursion is another thing Example of recursive function: f(n) = f(n-1) + f(n-2) f(0) = 1, f(1) = 1 47

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 48

Example Theorem: is not rational Proof: Assume by contradiction that it is rational = n/m n and m have no common factors We will show that this is impossible 49

= n/m Therefore, 2 m 2 = 4 k 2 n 2 2 m 2 = n 2 is even m 2 = 2 k 2 n is even n=2 k m is even m=2 p Thus, m and n have common factor 2 Contradiction! 50