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Mathematical Preliminaries. Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques. SETS. A set is a collection of elements. We write. Set Representations C = { a, b, c, d, e, f, g, h, i, j, k } C = { a, b, …, k } S = { 2, 4, 6, … }

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slide2

Mathematical Preliminaries

  • Sets
  • Functions
  • Relations
  • Graphs
  • Proof Techniques

Prof. Busch - LSU

slide3

SETS

A set is a collection of elements

We write

Prof. Busch - LSU

slide4

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 k>0 }
    • S = { j : j is nonnegative and even }

finite set

infinite set

Prof. Busch - LSU

slide5

U

A

6

8

2

3

1

7

4

5

9

10

A = { 1, 2, 3, 4, 5 }

  • Universal Set: all possible elements
      • U = { 1 , … , 10 }

Prof. Busch - LSU

slide6

B

A

  • 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 }

2

4

1

3

5

U

2

3

1

Venn diagrams

Prof. Busch - LSU

slide7

Complement

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

4

A

A

6

3

1

2

5

7

A = A

Prof. Busch - LSU

slide8

{ even integers } = { odd integers }

Integers

1

odd

0

5

even

6

2

4

3

7

Prof. Busch - LSU

slide9

DeMorgan’s Laws

A U B = A B

U

A B = A U B

U

Proof it!

Prof. Busch - LSU

slide10

Empty, Null Set:

= { }

S U = S

S =

S - = S

- S =

U

= Universal Set

Prof. Busch - LSU

slide11

U

A B

U

A B

Subset

A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 }

Proper Subset:

B

A

Prof. Busch - LSU

slide12

A B =

U

Disjoint Sets

A = { 1, 2, 3 } B = { 5, 6}

A

B

Prof. Busch - LSU

slide13

Set Cardinality

  • For finite sets

A = { 2, 5, 7 }

|A| = 3

(set size)

Prof. Busch - LSU

slide14

Powersets

A powerset is a set of sets

S = { a, b, c }

Powerset of S = the set of all the subsets of S

2S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }

Observation: | 2S | = 2|S| ( 8 = 23 )

Prof. Busch - LSU

slide15

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| = |A| |B|

Generalizes to more than two sets

A X B X … X Z

Prof. Busch - LSU

slide16

FUNCTIONS

domain

range

B

4

A

f(1) = a

a

1

2

b

c

3

5

f : A -> B

If A = domain

then f is a total function

otherwise f is a partial function

Prof. Busch - LSU

slide17

RELATIONS

R = {(x1, y1), (x2, y2), (x3, y3), …}

xi R yi

e. g. if R = ‘>’: 2 > 1, 3 > 2, 3 > 1

Prof. Busch - LSU

slide18

Equivalence Relations

  • Reflexive: x R x
  • Symmetric: x R y y R x
  • Transitive: x R y and y R z x R z
  • Example: R = ‘=‘
  • x = x
  • x = y y = x
  • x = y and y = z x = z

Prof. Busch - LSU

slide19

Equivalence Classes

For equivalence relation R

equivalence class of x = {y : x R y}

Example:

R = { (1, 1), (2, 2), (1, 2), (2, 1),

(3, 3), (4, 4), (3, 4), (4, 3) }

Equivalence class of 1 = {1, 2}

Equivalence class of 3 = {3, 4}

Prof. Busch - LSU

slide20

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) }

Prof. Busch - LSU

slide21

Labeled Graph

2

6

e

2

b

1

3

d

a

6

5

c

Prof. Busch - LSU

slide22

e

b

d

a

c

Walk

Walk is a sequence of adjacent edges

(e, d), (d, c), (c, a)

Prof. Busch - LSU

slide23

e

b

d

a

c

Path

Path is a walk where no edge is repeated

Simple path: no node is repeated

Prof. Busch - LSU

slide24

Cycle

e

base

b

3

1

d

a

2

c

Cycle: a walk from a node (base) to itself

Simple cycle: only the base node is repeated

Prof. Busch - LSU

slide25

Euler Tour

8

base

e

7

1

b

4

6

5

d

a

2

3

c

A cycle that contains each edge once

Prof. Busch - LSU

slide26

Hamiltonian Cycle

5

base

e

1

b

4

d

a

2

3

c

A simple cycle that contains all nodes

Prof. Busch - LSU

slide27

Finding All Simple Paths

e

b

d

a

c

origin

Prof. Busch - LSU

slide28

Step 1

e

b

d

a

c

origin

(c, a)

(c, e)

Prof. Busch - LSU

slide29

Step 2

e

b

d

a

(c, a)

(c, a), (a, b)

(c, e)

(c, e), (e, b)

(c, e), (e, d)

c

origin

Prof. Busch - LSU

slide30

Step 3

e

b

d

a

c

(c, a)

(c, a), (a, b)

(c, a), (a, b), (b, e)

(c, e)

(c, e), (e, b)

(c, e), (e, d)

origin

Prof. Busch - LSU

slide31

Step 4

e

b

d

a

(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)

c

origin

Prof. Busch - LSU

slide32

Trees

root

parent

leaf

child

Trees have no cycles

Prof. Busch - LSU

slide33

root

Level 0

Level 1

Height 3

leaf

Level 2

Level 3

Prof. Busch - LSU

slide34

Binary Trees

Prof. Busch - LSU

slide35

PROOF TECHNIQUES

  • Proof by induction
  • Proof by contradiction

Prof. Busch - LSU

slide36

Induction

We have statements P1, P2, P3, …

  • If we know
      • for some b that P1, P2, …, Pb are true
      • for any k >= b that
          • P1, P2, …, Pk imply Pk+1
  • Then
  • Every Pi is true

Prof. Busch - LSU

slide37

Proof by Induction

  • Inductive basis
      • Find P1, P2, …, Pb which are true
  • Inductive hypothesis
      • Let’s assume P1, P2, …, Pk are true,
      • for any k >= b
  • Inductive step
      • Show that Pk+1 is true

Prof. Busch - LSU

slide38

Example

Theorem:A binary tree of height n

has at most 2n leaves.

Proof by induction:

let L(i) be the maximum number of

leaves of any subtree at height i

Prof. Busch - LSU

slide39

We want to show: L(i) <= 2i

  • Inductive basis
        • L(0) = 1 (the root node)
  • Inductive hypothesis
        • Let’s assume L(i) <= 2i for all i = 0, 1, …, k
  • Induction step
        • we need to show that L(k + 1) <= 2k+1

Prof. Busch - LSU

slide40

Induction Step

height

k

k+1

From Inductive hypothesis: L(k) <= 2k

Prof. Busch - LSU

slide41

Induction Step

height

L(k) <= 2k

k

k+1

L(k+1) <= 2 * L(k) <= 2 * 2k = 2k+1

(we add at most two nodes for every leaf of level k)

Prof. Busch - LSU

slide42

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

Prof. Busch - LSU

slide43

Example

  • Theorem: Jika n2 genap maka n genap
  • Proof:
      • Dengan contradiction kita asumsikan
      • jika n2 genap maka n ganjil
      • We will show that this is impossible
slide44

Bukti: Diketahui n2 genap (kita asumsikan benar)

Andaikan  n ganjil (negasinya)

Karena  n ganjil maka nbisa ditulis dengan bentuk

n = 2m + 1 dengan m bilangan bulat

slide45

Diperoleh

n2 = (2m + 1)2

n2 = 4m2 + 4m +1

n2 = 2(2m2 + 2m) + 1

Itu berarti n2 = 2(2m2 + 2m) + 1 ganjil.

Kontradiksi dengan asumsi “n2 genap” karena terjadi kontradiksi maka dapat disimpulkan

jika n2 genap maka n genap

slide46

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

Prof. Busch - LSU

slide47

= n/m 2 m2 = n2

n is even

n = 2 k

Therefore, n2 is even

m is even

m = 2 p

2 m2 = 4k2

m2 = 2k2

Thus, m and n have common factor 2

Contradiction!

Prof. Busch - LSU

latihan1
Latihan
  • Buatlah notasi bagi set Z yang mempunyai anggota {0,1,2,3,4}
    • N mewakili bilangan natural (bilangan asli)
  • Buatlah notasi bagi set Y yang merupakan bilangan ganjil
  • Buatlah notasi bagi set yang terdiri dari 2 pasang (2-tuples) bilangan natural dengan syarat nilai elemen pertama lebih kecil dari elemen kedua pada tuples
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