Algoritme Struktur Data & Algoritme (Data Structures & Algorithms)
1 Math. a) any systematic method of solving a certain kind of problem b) the repetitive calculations used in finding the greatest common divisor of two numbers (called in full Euclidean algorithm) 2 Comput. a predetermined set of instructions for solving a specific problem in a limited number of steps
Analisa Algoritme
Denny (
[email protected]) Suryana Setiawan (
[email protected])
al•go•rithm n.
Fakultas Ilmu Komputer Universitas Indonesia Semester Genap - 2004/2005
Suatu set instruksi yang harus diikuti oleh computer untuk memecahkan suatu masalah. Program harus berhenti dalam batas waktu yang wajar (reasonable) Tidak terikat pada programming language atau bahkan paradigma pemrograman (mis. Procedural vs Object-Oriented)
Version 2.0 - Internal Use Only
Analisa Algoritme: motivasi
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Expected Outcome
perlu diketahui berapa banyak resource (time & space) yang diperlukan oleh sebuah algoritme Menggunakan teknik-teknik untuk mengurangi waktu yang dibutuhkan oleh sebuah algoritme
mahasiswa dapat memperkirakan waktu yang dibutuhkan sebuah algoritme mahasiswa memahami teknik-teknik yang secara drastis menurunkan running time mahasiswa memahami kerangka matematik yang menggambarkan running time
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Outline
Analisa Algoritme: what?
Apa itu analisa algoritme?- what Bagaimana cara untuk analisa/mengukur? - how Notasi Big-Oh Case Study: The Maximum Contiguous Subsequence Sum Problem
Algorithm 1: A cubic algorithm
Algorithm 2: A quadratic algorithm
Algorithm 3: An N log N algorithm
Algorithm 4: A linear algorithm
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Mengukur jumlah sumber daya (time dan space) yang diperlukan oleh sebuah algoritme
Waktu yang diperlukan (running time) oleh sebuah algoritme cenderung tergantung pada jumlah input yang diproses.
Î Running time dari sebuah algoritme adalah fungsi dari jumlah inputnya
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Selalu tidak terikat pada platform (mesin + OS), bahasa pemrograman, kualitas kompilator atau bahkan paradigma pemrograman (mis. Procedural vs Object-Oriented)
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1
Analisa Algoritme: how?
Analisa Algoritme: how?
Bagaimana jika kita menggunakan jam?
Jumlah waktu yang digunakan bervariasi tergantung pada beberapa faktor lain: kecepatan mesin, sistem operasi (multi-tasking), kualitas kompiler, dan bahasa pemrograman.
Sehingga kurang memberikan gambaran yang tepat tentang algoritme
Notasi O (sering disebut sebagai “notasi big-Oh”) Digunakan sebagai bahasa untuk membahas efisiensi dari sebuah algoritme: log n, linier, n log n, n2, n3, ... Dari hasil run-time, dapat kita buat grafik dari waktu eksekusi dan jumlah data.
Process 1: Wall-Clock time
CPU time
Process 2: Process 3: Idle: Wall-Clock time SDA/ALG/V2.0/7
Analisa Algoritme: how?
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Analisa Algoritme: how?
Contoh: Mencari elemen terkecil dalam sebuah array Algoritme: sequential scan / linear search Orde-nya: O(n) – linear
running time
public static int smallest (int a[]) { // assert (length of array > 0); int elemenTerkecil = a[0]; for (ii = 1; ii < a.length; ii++) { if (a[ii] < elemenTerkecil) { elemenTerkecil = a[ii]; } } return elemenTerkecil ; }
worst case average case
best case
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Big Oh: contoh
n
input size
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Dominant Term
Sebuah fungsi kubic adalah sebuah fungsi yang suku dominannya (dominant term) adalah sebuah konstan dikalikan dengan n3. Contoh: 10 n3 + n2 + 40 n + 80 n3 + 1000 n2 + 40 n + 80 n3 + 10000
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Mengapa hanya suku yang memiliki pangkat tertinggi/dominan saja yang diperhatikan?
Untuk n yang besar, suku dominan lebih mengindikasikan perilaku dari algoritme.
Untuk n yang kecil, suku dominan tidak selalu mengindikasikan perilakunya, tetapi program dengan input kecil umumnya berjalan sangat cepat sehingga kita tidak perlu perhatikan.
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Big Oh: issues
Orde Fungsi Running-Time
Apakah fungsi linier selalu lebih kecil dari fungsi kubic? Untuk n yang kecil, bisa saja fungsi linier > fungsi kubic Tetapi untuk n yang besar, fungsi kubic > fungsi linier Contoh:
Function c log N log2 N N N log N N2 N3 2N N!
• f(n) = 10 n3 + 20 n + 10 • g(n) = 10000 n + 10000 • n = 10, f(n) = 10.210, g(n) = 110.000 Î f(n) < g(n) • n = 100, f(n) = 10.002.010, g(n) = 1.010.000 Î f(n) > g(n)
Mengapa nilai konstan/koefesien pada setiap suku tidak diperhatikan? Nilai konstan/koefesien tidak berarti pada mesin yang berbeda ∴ Big Oh digunakan untuk merepresentasikan laju pertumbuhan (growth rate)
Name Constant Logarithmic Log-squared Linear N log N Quadratic Cubic Exponential Factorial
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Fungsi Running-Time
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Fungsi Running-Time (2)
Untuk input yang sedikit, beberapa fungsi lebih cepat dibandingkan dengan yang lain.
Untuk input yang besar, beberapa fungsi runningtime sangat lambat - tidak berguna.
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Contoh Algoritme
Contoh Algoritme
Mencari dua titik yang memiliki jarak terpendek dalam sebuah bidang (koordinat X-Y) Masalah dalam komputer grafis. Algoritme brute force: • hitung jarak dari semua pasangan titik • cari jarak yang terpendek
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Tentukan apakah ada tiga titik dalam sebuah bidang yang segaris (colinier). Algoritme brute force: • periksa semua pasangan titik yang terdiri dari 3 titik.
Jika jumlah titik adalah n, maka jumlah semua pasangan adalah n * (n - 1) / 2 Orde-nya: O(n2) - kuadratik Ada solusi yang O(n log n) bahkan O(n) dengan cara algoritme lain.
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Jumlah pasangan: n * (n - 1) * (n - 2) / 6 Orde-nya: O(n3) - kubik. Sangat tidak berguna untuk 10.000 titik. Ada algoritme yang kuadratik.
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Studi Kasus
Brute Force Algorithm (1)
Mengamati sebuah masalah dengan beberapa solusi. Masalah Maximum Contiguous Subsequence Sum Diberikan (angka integer negatif dimungkinkan) A1, A2, …, AN, cari nilai maksimum dari (Ai + Ai+1 + …+ Aj ). maximum contiguous subsequence sum adalah nol jika semua integer adalah negatif. Contoh (maximum subsequences digarisbawahi) -2, 11, -4, 13, -4, 2 1, -3, 4, -2, -1, 6
Algoritme: Hitung jumlah dari semua sub-sequence yang mungkin Cari nilai maksimumnya Contoh: jumlah subsequence (start, end) • • • •
(0, 0), (0,1), (0,2), …, (0,5) (1,1), (1,2), …, (1, 5) … (5,5) 0
1
-2
2
11
3
-4
4
13
5
-4
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Brute Force Algorithm (2)
11 ii
-4
13 jj
-4
Loop of size N inside of loop of size N inside of loop of size N means O(N3), or cubic algorithm. Slight over-estimate that results from some loops being of size less than N is not important.
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Actual Running Time
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Analysis
public static int maxSubSum1( int [] A ) { int maxSum = 0; for(int ii = 0; ii < A.length; ii++) { for( int jj = ii; jj < A.length; jj++) { int thisSum= 0; for(int kk = ii; kk <= jj; kk++) thisSum += A[kk]; if( thisSum > maxSum ) maxSum = thisSum; } } return maxSum ; } -2
2
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How To Improve?
For N = 100, actual time is 0.47 seconds on a particular computer. Can use this to estimate time for larger inputs: T(N) = cN 3 T(10N) = c(10N)3 = 1000cN 3 = 1000T(N) Inputs size increases by a factor of 10 means that running time increases by a factor of 1,000. For N = 1000, estimate an actual time of 470 seconds. (Actual was 449 seconds). For N = 10,000, estimate 449000 seconds (6 days).
Remove a loop; not always possible. Here it is: innermost loop is unnecessary because it throws away information. thisSum for next jj is easily obtained from old value of thisSum: Need Aii + A ii+1 + … + A jj-1 + Ajj Just computed Aii +A ii+1 + …+ A jj-1 What we need is what we just computed + Ajj
In other words: • sum (ii, jj) = sum (ii, jj - 1) + Ajj
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4
The Better Algorithm
Analysis
public static int maxSubSum2( int [ ] A ) { int maxSum = 0; for( int ii = 0; ii < A.length; ii++ ) { int thisSum = 0; for( int jj = ii; jj < A.length; jj++ ) { thisSum += A[jj]; if( thisSum > maxSum ) maxSum = thisSum; } } return maxSum ; }
Same logic as before: now the running time is quadratic, or O(N2) As we will see, this algorithm is still usable for inputs in the tens of thousands. Recall that the cubic algorithm was not practical for this amount of input.
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Actual running time
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Recursive Algorithm
For N = 100, actual time is 0.011 seconds on the same particular computer. Can use this to estimate time for larger inputs: T(N) = cN 2 T(10N) = c(10N)2 = 100cN 2 = 100T(N) Inputs size increases by a factor of 10 means that running time increases by a factor of 100. For N = 1000, estimate a running time of 1.11 seconds. (Actual was 1.12 seconds). For N = 10,000, estimate 111 seconds (= actual).
Use a divide-and-conquer approach.
Membagi (divide) permasalahan ke dalam bagian yang lebih kecil
Menyelesaikan (conquer) masalah per bagian secara recursive
Menggabung penyelesaian per bagian menjadi solusi masalah awal
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Recursive Algorithm (2)
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Computing the Third Case
The maximum subsequence either lies entirely in the first half lies entirely in the second half starts somewhere in the first half, goes to the last element in the first half, continues at the first element in the second half, ends somewhere in the second half. Compute all three possibilities, and use the maximum. First two possibilities easily computed recursively.
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Easily done with two loops; see the code For maximum sum that starts in the first half and extends to the last element in the first half, use a right-to-left scan starting at the last element in the first half. For the other maximum sum, do a left-to-right scan, starting at the first element in the first half.
4
-3
5
-2
-1
2
6
-2
4*
0
3
-2
-1
1
7*
5
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5
Recursion version
Recursion Version
static private int maxSumRec (int[] a, int left, int right) { int center = (left + right) / 2; if(left == right) { // Base case return a[left] > 0 ? a[left] : 0; } int maxLeftSum = maxSumRec (a, left, center); int maxRightSum = maxSumRec (a, center+1, right); for(int ii = center; ii >= left; ii--) { leftBorderSum += a[ii]; if(leftBorderSum > maxLeftBorderSum) maxLeftBorderSum = leftBorderSum; } ...
... for(int jj = center + 1; jj <= right; jj++) { rightBorderSum += a[jj]; if(rightBorderSum > maxRightBorderSum) maxRightBorderSum = rightBorderSum; } return max3 (maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBoderSum); } // publicly visible routine (a.k.a driver function) static public int maxSubSum (int [] a) { return maxSumRec (a, 0, a.length-1); }
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Coding Details
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Analysis
The code is more involved Make sure you have a base case that handles zeroelement arrays. Use a public static driver with a private recursive routine. Recursion rules: Have a base case Make progress to the base case Assume the recursive calls work Avoid computing the same solution twice
Let T( N ) = the time for an algorithm to solve a problem of size N. Then T( 1 ) = 1 (1 will be the quantum time unit; remember that constants don't matter). T( N ) = 2 T( N / 2 ) + N Two recursive calls, each of size N / 2. The time to solve each recursive call is T( N / 2 ) by the above definition Case three takes O( N ) time; we use N, because we will throw out the constants eventually.
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Bottom Line
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N log N
T(1) = 1 = 1 * 1 T(2) = 2 * T(1) + 2 = 4 = 2 * 2 T(4) = 2 * T(2) + 4 = 12 = 4 * 3 T(8) = 2 * T(4) + 8 = 32 = 8 * 4 T(16) = 2 * T(8) + 16 = 80 = 16 * 5 T(32) = 2 * T(16) + 32 = 192 = 32 * 6 T(64) = 2 * T(32) + 64 = 448 = 64 * 7
Any recursive algorithm that solves two half-sized problems and does linear non-recursive work to combine/split these solutions will always take O( N log N ) time because the above analysis will always hold. This is a very significant improvement over quadratic. It is still not as good as O( N ), but is not that far away either. There is a linear-time algorithm for this problem. The running time is clear, but the correctness is non-trivial.
T(N) = N(1 + log N) = N + N log N = O(N log N)
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6
Linear Algorithms
Idea
Linear algorithm would be best. Remember: linear means O( N ). Running time is proportional to amount of input. Hard to do better for an algorithm. If input increases by a factor of ten, then so does running time.
Let Ai,j be any sequence with Si,j < 0. If q > j, then Ai,q is not the maximum contiguous subsequence.
Proof:
Si,q = Si,j + Sj+1,q
Si,j < 0 Î Si,q < Sj+1,q
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The Code - version 1
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The Code - version 2
static public int maximumSubSequenceSum3 (int a[]) { int maxSum = 0; int thisSum = 0; for (int jj = 0; jj < a.length; jj++) { thisSum += a[jj]; if (thisSum > maxSum) { maxSum = thisSum; } else if (thisSum < 0) { thisSum = 0; } } return maxSum; }
static public int maximumSubSequenceSum3b (int data[]) { int thisSum = 0, maxSum = 0; for (int ii = 0; ii < data.length; ii++) { thisSum = Max(0, data[ii] + thisSum); maxSum = Max(thisSum, maxSum); } return maxSum; }
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Running Time
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Running Time: on different machines
Cubic Algorithm on Alpha 21164 at 533 Mhz using C compiler Linear Algorithm on Radio Shack TRS-80 Model III (a 1980 personal computer with a Z-80 processor running at 2.03 Mhz) using interpreted Basic
n
10 100 1,000 10,000 100,000 1,000,000 SDA/ALG/V2.0/41
Alpha 21164A, C compiled, Cubic Algorithm 0.6 microsecs 0.6 milisecs 0.6 secs 10 mins 7 days 19 yrs
TRS-80, Basic interpreted, Linear Algorithm 200 milisecs 2.0 secs 20 secs 3.2 mins 32 mins 5.4 hrs SDA/ALG/V2.0/42
7
The Logarithm Algorithm
Running Time: moral story
Even the most clever programming tricks cannot make an inefficient algorithm fast. Before we waste effort attempting to optimize code, we need to optimize the algorithm.
Formal Definition of Logarithm For any B, N > 0, logB N = K, if B K = N. If (the base) B is omitted, it defaults to 2 in computer science. Examples: log 32 = 5 (because 25 = 32) log 1024 = 10 log 1048576 = 20 log 1 billion = about 30 The logarithm grows much more slowly than N, and slower than the square root of N.
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Examples of the Logarithm
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Why log N?
BITS IN A BINARY NUMBER How many bits are required to represent N consecutive integers? REPEATED DOUBLING Starting from X = 1, how many times should X be doubled before it is at least as large as N? REPEATED HALVING Starting from X = N, if N is repeatedly halved, how many iterations must be applied to make N smaller than or equal to 1 (Halving rounds up). Answer to all of the above is log N (rounded up).
B bits represents 2B integers. Thus 2B is at least as big as N, so B is at least log N. Since B must be an integer, round up if needed. Same logic for the other examples.
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Repeated Halving Principle
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Linear Search
An algorithm is O( log N ) if it takes constant time to reduce the problem size by a constant fraction (which is usually 1/2). Reason: there will be log N iterations of constant work.
Given an integer X and an array A, return the position of X in A or an indication that it is not present. If X occurs more than once, return any occurrence. The array A is not altered. If input array is not sorted, solution is to use a linear search. Running times: Unsuccessful search: O( N ); every item is examined Successful search: • Worst case: O( N ); every item is examined • Average case: O( N ); half the items are examined
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Can we do better if we know the array is sorted?
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8
Binary Search
/**
Yes! Use a binary search. Look in the middle Case 1: If X is less than the item in the middle, then look in the sub-array to the left of the middle Case 2: If X is greater than the item in the middle, then look in the sub-array to the right of the middle Case 3: If X is equal to the item in the middle, then we have a match Base Case: If the sub-array is empty, X is not found. This is logarithmic by the repeated halving principle. The first binary search was published in 1946. The first published binary search without bugs did not appear until 1962.
Performs the standard binary search using two comparisons per level. @param a the array @param x the key @exception ItemNotFound if appropriate. @return index where item is found. */ public static int binarySearch (Comparable [ ] a, Comparable x ) throws ItemNotFound { int low = 0; int high = a.length - 1; int mid; while( low <= high ) { mid = (low + high) / 2; if (a[mid].compareTo (x) < 0) { low = mid + 1; } else if (a[mid].compareTo (x) > 0) { high = mid - 1; } else { return mid; } } throw new ItemNotFound( "BinarySearch fails" ); }
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Binary Search
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Big-Oh Rules (1)
Can do one comparison per iteration instead of two by changing the base case. Save the value returned by a[mid].compareTo (x) Average case and worst case in revised algorithm are identical. 1 + log N comparisons (rounded down to the nearest integer) are used. Example: If N = 1,000,000, then 20 element comparisons are used. Sequential search would be 25,000 times more costly on average.
Mathematical expression relative rates of growth
DEFINITION: (Big-Oh) T(N) = O(F(N)) if there are positive constants c and N0 such that T(N) ≤ c F(N) when N ≥ N0. DEFINITION: (Big-Omega) T(N) = Ω(F(N)) if there are constants c and N0 such that T(N) ≥ c F(N) when N ≥ N0. DEFINITION: (Big-Theta) T(N) = Θ(F(N)) if and only if T(N) = O(F(N)) and T(N) = Ω(F(N)). DEFINITION: (Little-Oh) T(N) = o(F(N)) if and only if T(N) = O(F(N)) and T(N) ≠ Θ (F(N)).
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Big-Oh Rules (2)
Big-Oh Rules (3)
Meanings of the various growth functions
T(N) = O(F(N)) T(N) = Ω(F(N)) T(N) = Θ(F(N)) T(N) = o(F(N))
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Growth of T(N) is ≤ growth of F(N) Growth of T(N) is ≥ growth of F(N) Growth of T(N) is = growth of F(N) Growth of T(N) is < growth of F(N)
Jika ada lebih dari satu parameter, maka aturan tersebut berlaku untuk setiap parameter. 4 n log(m) + 50 n2 + 500 m + 1853 Î O(n log(m) + n2 + m)
4 m log(m) + 50 n2 + 500 m + 853 Î O(m log(m) + n2)
Jika ada lebih dari satu parameter, maka aturan tersebut berlaku untuk setiap parameter. 4 n log(m) + 50 n2 + 500 m + 1853 Î O(n log(m) + n2 + m)
4 m log(m) + 50 n2 + 500 m + 853 Î O(m log(m) + n2) SDA/ALG/V2.0/53
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9
Summary
Exercises
more data means the program takes more time Big-Oh tidak dapat digunakan untuk N yang kecil
Urutkan fungsi berikut berdasarkan laju pertumbuhan (growth rate) N, √N, N1.5, N2, N log N, N log log N, N log2 N, N log (N2), 2/N, 2N, 2N/2, 37, N3, N2 log N A5 + B5 + C5 + D5 + E5 = F5
0 < A ≤ B ≤ C ≤ D ≤ E ≤ F ≤ 75
hanya memiliki satu solusi. berapa?
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Further Reading
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What’s Next
Chapter 5 http://telaga.cs.ui.ac.id/WebKuliah/IKI101 00/resources/AlgDesignManual/INDEX.HTM
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Sorting
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10
