Prof. Busch - LSU1 Mathematical Preliminaries. Prof. Busch - LSU2 Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques.

Slides:



Advertisements
Presentasi serupa
Pengujian Hipotesis untuk Satu dan Dua Varians Populasi
Advertisements

Mata Kuliah : ALGORITMA dan STRUKTUR DATA 1.
Peta Kontrol (Untuk Data Variabel)
Kinematics in Two Dimension - Kinematika dalam Dua Dimensi -
Slide 3-1 Elmasri and Navathe, Fundamentals of Database Systems, Fourth Edition Revised by IB & SAM, Fasilkom UI, 2005 Exercises Apa saja komponen utama.
Introduction to The Design & Analysis of Algorithms
Penerapan Fungsi Non-Linier
PROSES PADA WINDOWS Pratikum SO. Introduksi Proses 1.Program yang sedang dalam keadaan dieksekusi. 2.Unit kerja terkecil yang secara individu memiliki.
1-Sep-14 Analisis dan Perancangan Algoritma Kuliah 3 : Proof by induction E. Haodudin Nurkifli Teknik Informatika Universitas Ahmad Dahlan.
Bilqis1 Pertemuan bilqis2 Sequences and Summations Deret (urutan) dan Penjumlahan.
Implementing an REA Model in a Relational Database
MEMORY Bhakti Yudho Suprapto,MT. berfungsi untuk memuat program dan juga sebagai tempat untuk menampung hasil proses bersifat volatile yang berarti bahwa.
LOGO Manajemen Data Berdasarkan Komputer dengan Sistem Database.
Soal No 17 halaman 66 Find a) the coordinates of the foci and vertices for hyperbola whose equations given, b) equation of the asymptotes. Sketch the curve.
Linked List dan Double Linked List
Red -BlackTrees Evaliata Br Sembiring.
Lecture 2 Introduction to C# - Object Oriented Sandy Ardianto & Erick Pranata © Sekolah Tinggi Teknik Surabaya 1.
Relation
STRUCTURAL CONTROL STATEMENT  If  If…..else….  If ….elseif…else.
Lecture 5 Minimax dengan αβ Pruning Erick Pranata
ADT Tree 2007/2008 – Ganjil – Minggu 8.
K-Map Using different rules and properties in Boolean algebra can simplify Boolean equations May involve many of rules / properties during simplification.
TEKNIK PENGINTEGRALAN
Rumus-rumus ini masihkah anda ingat?
BLACK BOX TESTING.
Presented By : Group 2. A solution of an equation in two variables of the form. Ax + By = C and Ax + By + C = 0 A and B are not both zero, is an ordered.
Agenda Deskripsi perkuliahan Matematika Diskrit Topik Minggu 1:
1 Diselesaikan Oleh KOMPUTER Langkah-langkah harus tersusun secara LOGIS dan Efisien agar dapat menyelesaikan tugas dengan benar dan efisien. ALGORITMA.
1 Pertemuan Tree Matakuliah: T0026/Struktur Data Tahun: 2005 Versi: 1/1.
Pertemuan 13 Graph + Tree jual [Valdo] Lunatik Chubby Stylus.
Teori Matematika terhadap materi teori bahasa dan automata
Ruang Contoh dan Peluang Pertemuan 05
Pewarnaan graph Pertemuan 20: (Off Class)
1 Pertemuan 7 Name and Address Conversion Matakuliah: H0483 / Network Programming Tahun: 2005 Versi: 1.0.
KELOMPOK 7 PEMBAHASAN DAN. Pertanyaan Kelompok 1 Hlm An architect is calculating the dimensions for a regular hexagon shaped window. If the height.
1 Pertemuan 10 Fungsi Kepekatan Khusus Matakuliah: I0134 – Metode Statistika Tahun: 2007.
1 Pertemuan 17 Heaps Matakuliah: T0026/Struktur Data Tahun: 2005 Versi: 1/1.
1 Pertemuan 15 Game Playing Matakuliah: T0264/Intelijensia Semu Tahun: Juli 2006 Versi: 2/1.
Mata kuliah :K0362/ Matematika Diskrit Tahun :2008
1 Pertemuan 12 B-Tree Matakuliah: T0534/Struktur Data Tahun: 2005 Versi: September 2005.
Pertemuan 9 : Pewarnaan graph
9.3 Geometric Sequences and Series. Objective To find specified terms and the common ratio in a geometric sequence. To find the partial sum of a geometric.
Binary Search Tree. Sebuah node di Binary Search Tree memiliki path yang unik dari root menurut aturan ordering – Sebuah Node, mempunyai subtree kiri.
Binary Tree.
STATISTIKA CHATPER 4 (Perhitungan Dispersi (Sebaran))
KOMUNIKASI DATA Materi Pertemuan 3.
Induksi Matematika.
07/11/2017 BARISAN DAN DERET KONSEP BARISAN DAN DERET 1.
Menyelesaikan Masalah Program Linear
Statistika Chapter 4 Probability.
Kode Hamming.
Pengujian Hipotesis (I) Pertemuan 11
Parabola Parabola.
Significantly Significant
Materi 11 Teori Graf.
BILANGAN REAL BILANGAN BERPANGKAT.
REAL NUMBERS EKSPONENT NUMBERS.
Fungsi Kepekatan Peluang Khusus Pertemuan 10
Disusun oleh : KARLINA SARI ( ) ALIFA MUHANDIS S A ( )
Self-Organizing Network Model (SOM) Pertemuan 10
THE INFORMATION ABOUT HEALTH INSURANCE IN AUSTRALIA.
Algoritma & Pemrograman 1 Achmad Fitro The Power of PowerPoint – thepopp.com Chapter 3.
Lesson 2-1 Conditional Statements 1 Lesson 2-1 Conditional Statements.
Al Muizzuddin F Matematika Ekonomi Lanjutan 2013
Splay Trees.
"More Than Words" Saying I love you, Is not the words, I want to hear from you, It's not that I want you, Not to say but if you only knew, How easy, it.
CSCE 355 Foundations of Computation
Warm UP Write the definition of a triangle as a conditional.
Lecture 8 Normal model.
Wednesday/ September,  There are lots of problems with trade ◦ There may be some ways that some governments can make things better by intervening.
Transcript presentasi:

Prof. Busch - LSU1 Mathematical Preliminaries

Prof. Busch - LSU2 Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques

Prof. Busch - LSU3 A set is a collection of elements SETS We write

Prof. Busch - LSU4 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 - LSU5 A = { 1, 2, 3, 4, 5 } Universal Set: all possible elements U = { 1, …, 10 } A U

Prof. Busch - LSU6 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 A B Venn diagrams

Prof. Busch - LSU7 A Complement Universal set = {1, …, 7} A = { 1, 2, 3 } A = { 4, 5, 6, 7} A A = A

Prof. Busch - LSU even { even integers } = { odd integers } odd Integers

Prof. Busch - LSU9 DeMorgan’s Laws A U B = A B U A B = A U B U Proof it!

Prof. Busch - LSU10 Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set

Prof. Busch - LSU11 Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } A B U Proper Subset:A B U A B

Prof. Busch - LSU12 Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B = U AB

Prof. Busch - LSU13 Set Cardinality For finite sets A = { 2, 5, 7 } |A| = 3 (set size)

Prof. Busch - LSU14 Powersets A powerset is a set of sets 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 )

Prof. Busch - LSU15 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 - LSU16 FUNCTIONS domain 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

Prof. Busch - LSU17 RELATIONS R = {(x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ), …} x i R y i e. g. if R = ‘>’: 2 > 1, 3 > 2, 3 > 1

Prof. Busch - LSU18 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 - LSU19 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 - LSU20 GRAPHS A directed graph 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) } node edge a b c d e

Prof. Busch - LSU21 Labeled Graph a b c d e

Prof. Busch - LSU22 Walk a b c d e Walk is a sequence of adjacent edges (e, d), (d, c), (c, a)

Prof. Busch - LSU23 Path a b c d e Path is a walk where no edge is repeated Simple path: no node is repeated

Prof. Busch - LSU24 Cycle a b c d e Cycle: a walk from a node (base) to itself Simple cycle: only the base node is repeated base

Prof. Busch - LSU25 Euler Tour a b c d e base A cycle that contains each edge once

Prof. Busch - LSU26 Hamiltonian Cycle a b c d e base A simple cycle that contains all nodes

Prof. Busch - LSU27 Finding All Simple Paths a b c d e origin

Prof. Busch - LSU28 (c, a) (c, e) Step 1 a b c d e origin

Prof. Busch - LSU29 (c, a) (c, a), (a, b) (c, e) (c, e), (e, b) (c, e), (e, d) Step 2 a b c d e origin

Prof. Busch - LSU30 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)

Prof. Busch - LSU31 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)

Prof. Busch - LSU32 Trees root leaf parent child Trees have no cycles

Prof. Busch - LSU33 root leaf Level 0 Level 1 Level 2 Level 3 Height 3

Prof. Busch - LSU34 Binary Trees

Prof. Busch - LSU35 PROOF TECHNIQUES Proof by induction Proof by contradiction

Prof. Busch - LSU36 Induction We have statements P 1, P 2, P 3, … If we know for some b that P 1, P 2, …, P b are true for any k >= b that P 1, P 2, …, P k imply P k+1 Then Every P i is true

Prof. Busch - LSU37 Proof by Induction Inductive basis Find P 1, P 2, …, P b which are true Inductive hypothesis Let’s assume P 1, P 2, …, P k are true, for any k >= b Inductive step Show that P k+1 is true

Prof. Busch - LSU38 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

Prof. Busch - LSU39 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

Prof. Busch - LSU40 Induction Step From Inductive hypothesis: L(k) <= 2 k height k k+1

Prof. Busch - LSU41 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)

Prof. Busch - LSU42 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

43 Example Theorem: Jika n 2 genap maka n genap Proof: Dengan contradiction kita asumsikan jika n 2 genap maka n ganjil We will show that this is impossible

44 Bukti: Diketahui n 2 genap (kita asumsikan benar) Andaikan n ganjil (negasinya) Karena n ganjil maka n bisa ditulis dengan bentuk n = 2m + 1 dengan m bilangan bulat

45 Diperoleh n 2 = (2m + 1)2 n 2 = 4m 2 + 4m +1 n 2 = 2(2m 2 + 2m) + 1 Itu berarti n 2 = 2(2m 2 + 2m) + 1 ganjil. Kontradiksi dengan asumsi “n 2 genap” karena terjadi kontradiksi maka dapat disimpulkan jika n 2 genap maka n genap

Prof. Busch - LSU46 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 - LSU47 = 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 common factor 2 Contradiction!

Latihan 48

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 49