3. matrix chain multiplication (fajar)
DESCRIPTION
Tugas Analisis & Desain AlgoriitmeTRANSCRIPT
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LOGO
III Perkalian Matriks BerantaiMatrix Chain Multiplication
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Pendahuluan
Matrix Chain MultiplicationPerkalian Matriks Berantai
Biaya Komputasi adalah Banyaknya perkalian skalar dari suatu perkalian matriks
Biaya Komputasi =
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CONTOH 1
Biaya Komputasi = 12
Banyaknya perkalian skalar dari suatu perkalian matriks
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Asosiatif Matriks & Biaya komputasi
=
n = 3
CONTOH 2
TENTUKAN BIAYA KOMPUTASI
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Cara memposisikan penyisipan tanda kurung (parenthesized) dapat mengakibatkan Biaya Komputasi yang berbeda
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Fungsi Parenthesization
Misalkan P(n) adalah fungsi parenthesization yang didefisikan sebagai jumlah cara memposisikan penyisipan tanda kurung dalam suatu perkalian matriks berantai berukuran n.
Dengan
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CONTOH 3
misalkan diberikan 6 buah matriks dengan dimensi seperti pada tabeL berikut :
1. Hitung berapa cara memposisikan penyisipan tanda kurung (perenthesation) !
2. tentukan bagaimana cara perkalian yang memiliki biaya komputasi minimal !
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= 42
Optimal
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)1()( nCnP [Catalan number]
)2()4
(2
1
12/3
nn
nn
n
n
Dynamic Programming
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TAHAP 1 Karakterisasi struktur dari parenthesization yang optimal
; 1 ≤ i ≤ j ≤ n
OPTIMAL
Combine OPTIMALOPTIMAL
Misalkan
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TAHAP 2 Definisikan secara rekursif nilai solusi optimal
m[i, j] adalah nilai minimum yang dibutuhkan untuk perhitungan perkalian skalar matriks
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TAHAP 3 Hitung nilai solusi optimal
Algoritme Matrix–Chain-Order (MCO) p0 p1 p2 p3 p4 p5 p6
p 30 35 15 5 10 20 25n = 7-1
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p0 p1 p2 p3 p4 p5 p6
30 35 15 5 10 20 25
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p0 p1 p2 p3 p4 p5 p6
30 35 15 5 10 20 25
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p0 p1 p2 p3 p4 p5 p6
30 35 15 5 10 20 25
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p0 p1 p2 p3 p4 p5 p6
30 35 15 5 10 20 25
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p0 p1 p2 p3 p4 p5 p6
30 35 15 5 10 20 25
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Bottom
Up
Slide 10
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Optimal
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Bottom
Up
Slide 10
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Tahap 4 Konstruksi solusi optimal
Hasil
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Kesimpulan
Fungsi Parenthesization menghasilkan
Algoritme Matrix–Chain-Order dalam Dynamic Programming menghasilkan kompleksitas sebesar dengan ruang penyimpanan sebesar
)