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Matematika Diskrit (D) Matematika Diskrit (D) KI 1306 KI 1306
Rully SoelaimanRully [email protected]@gmail.com
Jurusan Teknik InformatikaJurusan Teknik InformatikaFakultas Teknologi Informasi Fakultas Teknologi Informasi
Institut Teknologi Sepuluh NopemberInstitut Teknologi Sepuluh Nopember
• Evaluasi Tengah Semester 30%• Evaluasi Akhir Semester 30%• Tugas (Analisis dan Coding) 40%
Kelas atau Take Home
1 The Foundations: Logic and Proofs 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy
2 Basic Structures: Sets, Functions, Sequences and Sums
2.1 Sets 2.2 Set Operations 2.3 Functions 2.4 Sequences and Summations
3 The Fundamentals: Algorithms, the Integers3.4 The Integers and Division 3.5 Primes and Greatest Common Divisors 3.6 Integers and Algorithms 3.7 Applications of Number Theory
4 Induction and Recursion
4.1 Mathematical Induction 4.2 Strong Induction and Well-Ordering 4.3 Recursive Definitions and Structural Induction 4.4 Recursive Algorithms
5 Counting
5.1 The Basics of Counting 5.2 The Pigeonhole Principle 5.3 Permutations and Combinations 5.4 Binomial Coefficients 5.5 Generalized Permutations and Combinations 5.6 Generating Permutations and Combinations
6 Advanced Counting Techniques
7.1 Recurrence Relations 7.2 Solving Linear Recurrence Relations 7.3 Divide-and-Conquer Algorithms and Recurrence Relations 7.4 Generating Functions 7.5 Inclusion-Exclusion 7.6 Applications of Inclusion-Exclusion
7 Relations
8.1 Relations and Their Properties 8.2 n-ary Relations and Their Applications 8.3 Representing Relations 8.4 Closures of Relations 8.5 Equivalence Relations 8.6 Partial Orderings
Contoh Soal 1Contoh Soal 1
Solusi Soal 1Solusi Soal 1
• Diketahui :
P1 : a + 4b + 9c + 16d = 52
P2 : 4a + 9b + 16c + 25d = 150
P3 : 9a + 16b + 25 + 36d = 800• Bentuk persamaan P4 = P2 + P3 – P1
P4 : 4a + 9b + 16c + 25d = 150 9a + 16b + 25c + 36d = 800
-a - 4b - 9c - 16d = -52
+ P4 :12a + 21b + 32c + 45d = 898
Solusi Soal 1Solusi Soal 1
• Diketahui :
P1 : a + 4b + 9c + 16d = 52
P2 : 4a + 9b + 16c + 25d = 150
P3 : 9a + 16b + 25 + 36d = 800• Bentuk persamaan P5 = P1 + P3 – 2 ×P2
P5 : a + 4b + 9c + 16d = 52 9a + 16b + 25c + 36d = 800
-8a - 18b - 32c - 50d = -300
+ P5 : 2a + 2b + 2c + 2d = 552
Solusi Soal 1Solusi Soal 1
• Diketahui :
P4 :12a + 21b + 32c + 45d = 898
P5 : 2a + 2b + 2c + 2d = 552• Bentuk persamaan P6 = P4 + 2 ×P5
P6 : 12a + 21b + 32c + 45d = 898
4a + 4b + 4c + 4d = 1104
+ P6 : 16a + 25b + 36c + 49d = 2002
Soal 2Soal 2
Solusi Soal 2Solusi Soal 2
• Dari sini diperoleh bentuk
• Karena x≠1, maka
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2
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Solusi Soal 2Solusi Soal 2
• Sehingga diperoleh
1997
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SoalSoal
• SPOJ Problem Set (classical) 9034 - Help Tohu
• SPOJ Problem Set (tutorial) 5241. Alchemy
• SPOJ Problem Set (tutorial) 8360. Prime After N
SPOJ 9034SPOJ 9034
SPOJ 9034SPOJ 9034
• Amati beberapa nilai ak
• k = 1 k = 3
• k = 2 k = 4
2
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SPOJ 9034SPOJ 9034
• Buktikan untuk
• Bukti
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SPOJ 9034SPOJ 9034
• Untuk
• Untuk
• Perhatikan bentuk
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SPOJ 9034SPOJ 9034
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SPOJ 9034SPOJ 9034
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SPOJ 9034SPOJ 9034
)2)(1(
1
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SPOJ 9034SPOJ 9034
