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Design and Analysis of AlgorithmDivide and Conquer Algorithm
Aryo Pinandito, ST, M.MT – PTIIK Universitas Brawijaya
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History Divide and Conquer dulunya adalah strategi
militer yang dikenal dengan nama divide ut imperes.
Sekarang strategi tersebut menjadi strategi fundamental di dalam ilmu komputer dengan nama Divide and Conquer.
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Definisi Divide: membagi masalah menjadi beberapa
upa-masalah yang memiliki kemiripan dengan masalah semula namun berukuran lebih kecil (idealnya berukuran hampir sama),
Conquer: memecahkan (menyelesaikan) masing-masing upa-masalah (secara rekursif), dan
Combine: mengabungkan solusi masing-masing upa-masalah sehingga membentuk solusi masalah semula.
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Definisi (2) Obyek permasalahan yang dibagi : masukan (input) atau instances yang berukuran
n seperti: tabel (larik/array), matriks, eksponen, dll, bergantung pada masalahnya.
Tiap-tiap upa-masalah mempunyai karakteristik yang sama (the same type) dengan karakteristik masalah asal, sehingga metode Divide and Conquer lebih natural jika diungkapkan dalam skema rekursif.
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Skema Umum Algoritma Divide and Conquer
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Jika pembagian selalu menghasilkan dua upa-masalah yang berukuran sama:
,)()2/(2
n ,)()(
0
0
nnnfnTnng
nT
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Divide and Conquer
Minimum dan Maximum
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Contoh-contoh masalah Mencari Nilai Minimum dan Maksimum (Min
Maks)
Persoalan: Misalkan diberikan tabel A yang berukuran n elemen dan sudah berisi nilai integer.
Carilah nilai minimum dan nilai maksimum sekaligus di dalam tabel tersebut.
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Penyelesaian dengan Algoritma Brute Force
T(n) = (n – 1) + (n – 1) = 2n – 2 = O(n)
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Penyelesaian dengan Divide and Conquer
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Penyelesaian dengan Divide and Conquer Ukuran tabel hasil pembagian dapat dibuat
cukup kecil sehingga mencari minimum dan maksimum dapat diselesaikan (SOLVE) secara lebih mudah.
Dalam hal ini, ukuran kecil yang dipilih adalah 1 elemen atau 2 elemen.
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Contoh Algoritma MinMaks(A, n, min, maks) Algoritma: Untuk kasus n = 1 atau n = 2, SOLVE:
Jika n = 1, maka min = maks = A[n]Jika n = 2, maka bandingkan kedua elemen untuk
menentukan min dan maks. Untuk kasus n > 2,
(a) DIVIDE: Bagi dua tabel A menjadi dua bagian yang sama, A1 dan A2(b) CONQUER: MinMaks(A1, n/2, min1, maks1) MInMaks(A2, n/2, min2, maks2)(c) COMBINE:
if min1 <min2 then min <- min1 else min <- min2 if maks1 <maks2 then maks <- maks2 else maks <- maks1
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Review Contoh
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Kompleksitas Waktu Asimptotik
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Brute Force vs Divide and Conquer MinMaks1 secara brute force :
T(n) = 2n – 2 MinMaks2 secara divide and conquer:
T(n) = 3n/2 – 2 Perhatikan: 3n/2 – 2 < 2n – 2 , n 2. Kesimpulan:
Algoritma MinMaks lebih mangkus dengan menggunakan metode Divide and Conquer.
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Divide and Conquer
Merge Sort, Insertion Sort, Quick Sort, Selection Sort
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Sorting dengan Metode Divide and Conquer
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Pendekatan pada Algoritma Sorting
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Pendekatan pada Algoritma Sorting (2)
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(a) Merge Sort Algoritma:1. Untuk kasus n = 1, maka tabel A sudah terurut dengan sendirinya (langkah SOLVE).
2. Untuk kasus n > 1, maka (a) DIVIDE: bagi tabel A menjadi dua bagian, bagian kiri dan bagian kanan, masing-masing bagian berukuran n/2 elemen. (b) CONQUER: Secara rekursif, terapkan algoritma D-and-C pada masing-masing bagian. (c) MERGE: gabung hasil pengurutan kedua bagian sehingga diperoleh tabel A yang terurut.
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Contoh Merge Sort
Contoh Merge: A1 A2 B 1 13 24 2 15 27 1 < 2 1 1 1 13 24 2 15 27 2 <13 2 1 2 1 13 24 2 15 27 13<15 13 1 2 13 1 13 24 2 15 27 15<24 15 1 2 13 15 1 13 24 2 15 27 24<27 24 1 2 13 15 24 1 13 24 2 15 27 27 1 2 13 15 24 27
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Contoh 4.3. Misalkan tabel A berisi elemen-elemen berikut:
4 12 23 9 21 1 5 2 DIVIDE, CONQUER, dan SOLVE: 4 12 23 9 21 1 5 2 4 12 23 9 21 1 5 2 4 12 23 9 21 1 5 2 4 12 23 9 21 1 5 2 MERGE: 4 12 9 23 1 21 2 5 4 9 12 23 1 2 5 21 1 2 4 5 9 12 21 23
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Merge Sort dengan Metode Divide dan Conquer
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Kompleksitas Waktu Merge Sort
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KompleksitasPenyelesaian: T(n) = 2T(n/2) + cn = 2(2T(n/4) + cn/2) + cn = 4T(n/4) + 2cn = 4(2T(n/8) + cn/4) + 2cn = 8T(n/8) + 3cn = ... = 2k T(n/2k) +kcn Berhenti jika ukuran tabel terkecil, n = 1: n/2k = 1 k = 2log n sehingga T(n) = nT(1) + cn 2log n = na + cn 2log n = O(n 2log n)
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(b) Insertion Sort
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Insertion Sort dengan Divide and Conquer
Prosedur Merge dapat diganti dengan prosedur penyisipan sebuah elemen pada tabel yang sudah terurut (lihat algoritma Insertion Sort versi iteratif).
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Divide, Conquer, and Solve
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Merge
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Kompleksitas Insertion Sort
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(c) Quick Sort Termasuk pada pendekatan sulit membagi,
mudah menggabung (hard split/easy join)
Tabel A dibagi (istilahnya: dipartisi) menjadi A1 dan A2 sedemikian sehingga elemen-elemen A1 elemen-elemen A2.
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Quick Sort
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Teknik Partisi Quick Sort
1. pilih x { A[1], A[2], ..., A[n] } sebagai pivot,
2. pindai tabel dari kiri sampai ditemukan A[p] x
3. pindai tabel dari kanan sampai ditemukan A[q] x
4. pertukarkan A[p] A[q]5. ulangi (2), dari posisi p + 1, dan (3), dari
posisi q – 1 , sampai kedua pemindaian bertemu di tengah tabel
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Pseudocode Quick Sort
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Cara Pemilihan Pivot Pivot = elemen pertama/elemen
terakhir/elemen tengah tabel Pivot dipilih secara acak dari salah satu
elemen tabel. Pivot = elemen median tabel
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Kompleksitas Algoritma Quicksort: Kasus terbaik (best case)
Kasus terbaik terjadi bila pivot adalah elemen median sedemikian sehingga kedua upatabel berukuran relatif sama setiap kali pempartisian.
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Recursive Tree Quick Sort: Best Case n n/2 n/2
n/4 n/4 n/4 n/4
n/8 n/8 n/8 n/8 n/8 n/8 n/8 n/8 ... ... ... ... ... ... ... .... 1 1 1 ...................1...1....1......................... 1 1 1
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Kompleksitas Quick Sort: Best Case
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Kompleksitas Algoritma Quick Sort Kasus terburuk (worst case)
Kasus ini terjadi bila pada setiap partisi pivot selalu elemen maksimum (atau elemen minimum) tabel.
Kasus jika tabel sudah terurut menaik/menurun
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Recursive Tree Quick Sort: Worst Case
n 1 n – 1 1 n – 2 1 n – 3 ... 2 1 1
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Kompleksitas Quick Sort: Worst Case
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Kompleksitas Quick Sort: Average Case Kasus rata-rata (average case)
Kasus ini terjadi jika pivot dipilih secara acak dari elemen tabel, dan peluang setiap elemen dipilih menjadi pivot adalah sama.
Tavg(n) = O(n 2log n).
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(d) Selection Sort
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Pseudocode Selection Sort
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Contoh Selection Sort
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Kompleksitas Selection Sort
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Divide and Conquer
Perpangkatan
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Perpangkatan an
Misalkan a R dan n adalah bilangan bulat tidak negatif:
an = a × a × … × a (n kali), jika n > 0 = 1 , jika n = 0
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Perpangkatan dengan Brute Force
Kompleksitas waktu algoritma: T(n) = n = O(n)
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Perpangkatan dengan Divide and Conquer
Algoritma menghitung an:
1. Untuk kasus n = 0, maka an = 1.
2. Untuk kasus n > 0, bedakan menjadi dua kasus lagi:
(i) jika n genap, maka an = an/2 an/2
(ii) jika n ganjil, maka an = an/2 an/2 a
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Menghitung 316 dengan Divide and Conquer
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Pseudocode Perpangkatan Divide and Conquer
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Kompleksitas Algoritma Perpangkatan Divide and Conquer
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Penyelesaian Kompleksitas Algoritma
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Brute Force vs. Divide and Conquer Kompleksitas algoritma perpangkatan dengan
menggunakan metode Brute Force:
TBF(n) = O(n)
Kompleksitas algoritma perpangkatan dengan menggunakan metode Divide and Conquer
TDnC(n) = O(log n)
Metode Divide and Conquer pada algoritma perpangkatan lebih mangkus daripada metode brute force.
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Questions?
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Terima KasihThank
You
Danke Gratias
Merci
ありがとうございます
감사합니다 Kiitos
谢谢ًشكرا
Grazias
धन्यवाद