lec03-algorithmanalysis

lec03-algorithmanalysis
September 16, 2018 Algorithm An algorithm is a set of instructions to be followed to solve a problem. There can be more than one solution (more than one algorithm) to solve a given problem. An algorithm can be implemented using different programming languages on different platforms. An algorithm must be correct. It should correctly solve the problem. Once we have a correct algorithm for a problem, we have to determine the efficiency of that algorithm. how much time that algorithm requires. how much space that algorithm requires. We will focus: How to estimate the time required for an algorithm How to reduce the time required 9/16/2018 CS202 - Fundamentals of Computer Science II

Analysis of Algorithms
We can have different algorithms to solve a given problem.  Which one is the most efficient?  Between the given two algorithms, which one is more efficient? Analysis of Algorithms is the area of computer science that provides tools to analyze the efficiency of different methods of solutions. The efficiency of an algorithm means How much time it requires. How much memory space it requires. How much disk space and other resources it requires. We will concentrate on the time requirement. We will try to find the efficiency of the algorithms, not their implementations. An analysis should focus on gross differences in the efficiency of algorithms that are likely to dominate the overall cost of a solution. 9/16/2018 CS202 - Fundamentals of Computer Science II

Analysis of Algorithms (cont.)
How do we compare the time efficiency of two algorithms that solve the same problem? Naïve Approach: To implement these algorithms in a programming language (C++), and run them to compare their time requirements. Comparing the programs (instead of algorithms) has difficulties. How are the algorithms coded? Comparing running times means comparing the implementations. We should not compare implementations, because they are sensitive to programming style that may cloud the issue of which algorithm is inherently more efficient. What computer should we use? We should compare the efficiency of the algorithms independently of a particular computer. What data should the program use? Any analysis must be independent of specific data. 9/16/2018 CS202 - Fundamentals of Computer Science II

Analysis of Algorithms (cont.)
When we analyze algorithms, we should employ mathematical techniques that analyze algorithms independently of specific implementations, computers, or data. To analyze algorithms: First, we start the count the number of significant operations in a particular solution to assess its efficiency. Then, we will express the efficiency of algorithms using growth functions. 9/16/2018 CS202 - Fundamentals of Computer Science II

The Execution Time of Algorithms
Each operation in an algorithm (or a program) has a cost.  Each operation takes a certain of time. count = count + 1;  take a certain amount of time, but it is constant A sequence of operations: count = count + 1; Cost: c1 sum = sum + count; Cost: c2  Total Cost = c1 + c2 9/16/2018 CS202 - Fundamentals of Computer Science II

The Execution Time of Algorithms (cont.)
Example: Simple If-Statement Cost Times if (n < 0) c absval = -n c else absval = n; c Total Cost <= c1 + max(c2,c3) 9/16/2018 CS202 - Fundamentals of Computer Science II

Example: Simple Loop Cost Times i = 1; c sum = 0; c while (i <= n) { c3 n+1 i = i + 1; c4 n sum = sum + i; c5 n } Total Cost = c1 + c2 + (n+1)*c3 + n*c4 + n*c5  The time required for this algorithm is proportional to n 9/16/2018 CS202 - Fundamentals of Computer Science II

Example: Nested Loop Cost Times i=1; c1 1 sum = 0; c2 1 while (i <= n) { c3 n+1 j=1; c4 n while (j <= n) { c5 n*(n+1) sum = sum + i; c6 n*n j = j + 1; c7 n*n } i = i +1; c8 n Total Cost = c1 + c2 + (n+1)*c3 + n*c4 + n*(n+1)*c5+n*n*c6+n*n*c7+n*c8  The time required for this algorithm is proportional to n2 9/16/2018 CS202 - Fundamentals of Computer Science II

General Rules for Estimation
Loops: The running time of a loop is at most the running time of the statements inside of that loop times the number of iterations. Nested Loops: Running time of a nested loop containing a statement in the inner most loop is the running time of statement multiplied by the product of the sized of all loops. Consecutive Statements: Just add the running times of those consecutive statements. If/Else: Never more than the running time of the test plus the larger of running times of S1 and S2. 9/16/2018 CS202 - Fundamentals of Computer Science II

Algorithm Growth Rates
We measure an algorithm’s time requirement as a function of the problem size. Problem size depends on the application: number of elements in a list for a sorting algorithm, the number disks for towers of hanoi. So, we say that (if the problem size is n) Algorithm A requires 5*n2 time units to solve a problem of size n. Algorithm A requires 7*n time units to solve a problem of size n. The most important thing to learn is how quickly the algorithm’s time requirement grows a function of the problem size. Algorithm A requires time proportional to n2. Algorithm A requires time proportional to n. An algorithm’s proportional time requirement is known as growth rate. We can compare the efficiency of two algorithms by comparing their growth rates. 9/16/2018 CS202 - Fundamentals of Computer Science II

Algorithm Growth Rates (cont.)
Time requirements as a function of the problem size n 9/16/2018 CS202 - Fundamentals of Computer Science II

Order-of-Magnitude Analysis and Big 0 Notation
If Algorithm A requires time proportional to f(n), Algorithm A is said to be order f(n), and it is denoted as O(f(n)). The function f(n) is called the algorithm’s growth-rate function. Since the capital O is used in the notation, this notation is called the Big O notation. If Algorithm A requires time proportional to n2, it is O(n2). If Algorithm A requires time proportional to n, it is O(n). 9/16/2018 CS202 - Fundamentals of Computer Science II

Definition of the Order of an Algorithm
Algorithm A is order f(n) – denoted as O(f(n)) – if constants k and n0 exist such that A requires no more than k*f(n) time units to solve a problem of size n  n0. The requirement of in the definition of O(f(n)) formalizes the notion of sufficiently large problems. In general, many values of k and n can satisfy this definition. 9/16/2018 CS202 - Fundamentals of Computer Science II

CS202 - Fundamentals of Computer Science II
Order of an Algorithm If an algorithm requires n2–3*n+10 seconds to solve a problem size n. If constants k and n0 exist such that k*n2 > n2–3*n+10 for all n  n0 . the algorithm is order n2 (In fact, k is 3 and n0 is 2) 3*n2 > n2–3*n+10 for all n  2 . Thus, the algorithm requires no more than k*n2 time units for n  n0 , So it is O(n2) 9/16/2018 CS202 - Fundamentals of Computer Science II

Order of an Algorithm (cont.)
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A Comparison of Growth-Rate Functions

A Comparison of Growth-Rate Functions (cont.)

Growth-Rate Functions
O(1) Time requirement is constant, and it is independent of the problem’s size. O(log2n) Time requirement for a logarithmic algorithm increases increases slowly as the problem size increases. O(n) Time requirement for a linear algorithm increases directly with the size of the problem. O(n*log2n) Time requirement for a n*log2n algorithm increases more rapidly than a linear algorithm. O(n2) Time requirement for a quadratic algorithm increases rapidly with the size of the problem. O(n3) Time requirement for a cubic algorithm increases more rapidly with the size of the problem than the time requirement for a quadratic algorithm. O(2n) As the size of the problem increases, the time requirement for an exponential algorithm increases too rapidly to be practical. 9/16/2018 CS202 - Fundamentals of Computer Science II

Growth-Rate Functions
If an algorithm takes 1 second to run with the problem size 8, what is the time requirement (approximately) for that algorithm with the problem size 16? If its order is: O(1)  T(n) = 1 second O(log2n)  T(n) = (1*log216) / log28 = 4/3 seconds O(n)  T(n) = (1*16) / 8 = 2 seconds O(n*log2n)  T(n) = (1*16*log216) / 8*log28 = 8/3 seconds O(n2)  T(n) = (1*162) / 82 = 4 seconds O(n3)  T(n) = (1*163) / 83 = 8 seconds O(2n)  T(n) = (1*216) / 28 = 28 seconds = 256 seconds 9/16/2018 CS202 - Fundamentals of Computer Science II

Properties of Growth-Rate Functions
We can ignore low-order terms in an algorithm’s growth-rate function. If an algorithm is O(n3+4n2+3n), it is also O(n3). So, even if an algorithm is O(n3+4n2+3n), it is O(n3). We only use the higher-order term as algorithm’s growth-rate function. We can ignore a multiplicative constant in the higher-order term of an algorithm’s growth-rate function. If an algorithm is O(5n3), it is also O(n3). O(f(n)) + O(g(n)) = O(f(n)+g(n)) We can combine growth-rate functions. If an algorithm is O(n3) + O(4n2), it is also O(n3 +4n2)  So, it is O(n3). Similar rules hold for multiplication. 9/16/2018 CS202 - Fundamentals of Computer Science II

Some Mathematical Facts
Some mathematical equalities are: 9/16/2018 CS202 - Fundamentals of Computer Science II

Growth-Rate Functions – Example1
Cost Times i = 1; c sum = 0; c while (i <= n) { c3 n+1 i = i + 1; c4 n sum = sum + i; c5 n } T(n) = c1 + c2 + (n+1)*c3 + n*c4 + n*c5 = (c3+c4+c5)*n + (c1+c2+c3) = a*n + b  So, the growth-rate function for this algorithm is O(n) 9/16/2018 CS202 - Fundamentals of Computer Science II

Cost Times i=1; c1 1 sum = 0; c2 1 while (i <= n) { c3 n+1 j=1; c4 n while (j <= n) { c5 n*(n+1) sum = sum + i; c6 n*n j = j + 1; c7 n*n } i = i +1; c8 n T(n) = c1 + c2 + (n+1)*c3 + n*c4 + n*(n+1)*c5+n*n*c6+n*n*c7+n*c8 = (c5+c6+c7)*n2 + (c3+c4+c5+c8)*n + (c1+c2+c3) = a*n2 + b*n + c  So, the growth-rate function for this algorithm is O(n2) 9/16/2018 CS202 - Fundamentals of Computer Science II

Cost Times for (i=1; i<=n; i++) c1 n+1 for (j=1; j<=i; j++) c2 for (k=1; k<=j; k++) c3 x=x+1; c4 T(n) = c1*(n+1) + c2*( ) + c3* ( ) + c4*( ) = a*n3 + b*n2 + c*n + d  So, the growth-rate function for this algorithm is O(n3) 9/16/2018 CS202 - Fundamentals of Computer Science II

Growth-Rate Functions – Recursive Algorithms
void hanoi(int n, char source, char dest, char spare) { Cost if (n > 0) { c1 hanoi(n-1, source, spare, dest); c2 cout << "Move top disk from pole " << source c3 << " to pole " << dest << endl; hanoi(n-1, spare, dest, source); c4 } } The time-complexity function T(n) of a recursive algorithm is defined in terms of itself, and this is known as recurrence equation for T(n). To find the growth-rate function for that recursive algorithm, we have to solve that recurrence relation. 9/16/2018 CS202 - Fundamentals of Computer Science II

Growth-Rate Functions – Hanoi Towers
What is the cost of hanoi(n,’A’,’B’,’C’)? when n=0 T(0) = c1 when n>0 T(n) = c1 + c2 + T(n-1) + c3 + c4 + T(n-1) = 2*T(n-1) + (c1+c2+c3+c4) = 2*T(n-1) + c  recurrence equation for the growth-rate function of hanoi-towers algorithm Now, we have to solve this recurrence equation to find the growth-rate function of hanoi-towers algorithm 9/16/2018 CS202 - Fundamentals of Computer Science II

Growth-Rate Functions – Hanoi Towers (cont.)
There are many methods to solve recurrence equations, but we will use a simple method known as repeated substitutions. T(n) = 2*T(n-1) + c = 2 * (2*T(n-2)+c) + c = 2 * (2* (2*T(n-3)+c) + c) + c = 23 * T(n-3) + ( )*c (assuming n>2) when substitution repeated i-1th times = 2i * T(n-i) + (2i )*c when i=n = 2n * T(0) + (2n )*c = 2n * c1 + ( )*c = 2n * c1 + ( 2n-1 )*c = 2n*(c1+c) – c  So, the growth rate function is O(2n) 9/16/2018 CS202 - Fundamentals of Computer Science II

What to Analyze An algorithm can require different times to solve different problems of the same size. Ex. Searching an item in a list of n elements using sequential search.  Cost: 1,2,...,n Worst-Case Analysis –The maximum amount of time that an algorithm require to solve a problem of size n. This gives an upper bound for the time complexity of an algorithm. Normally, we try to find worst-case behavior of an algorithm. Best-Case Analysis –The minimum amount of time that an algorithm require to solve a problem of size n. The best case behavior of an algorithm is NOT so useful. Average-Case Analysis –The average amount of time that an algorithm require to solve a problem of size n. Sometimes, it is difficult to find the average-case behavior of an algorithm. We have to look at all possible data organizations of a given size n, and their distribution probabilities of these organizations. Worst-case analysis is more common than average-case analysis. 9/16/2018 CS202 - Fundamentals of Computer Science II

What is Important? An array-based list retrieve operation is O(1), a linked-list-based list retrieve operation is O(n). But insert and delete operations are much easier on a linked-list-based list implementation.  When selecting an ADT’s implementations, we have to consider how frequently particular ADT operations occur in a given application. If the problem size is always small, we can probably ignore an algorithm’s efficiency. In this case, we should choose the simplest algorithm. 9/16/2018 CS202 - Fundamentals of Computer Science II

What is Important? (cont.)
We have to weigh the trade-offs between an algorithm’s time requirement and its memory requirements. We have to compare algorithms for both style and efficiency. The analysis should focus on gross differences in efficiency and not reward coding tricks that save small amount of time. That is, there is no need for coding tricks if the gain is not too much. Easily understandable program is also important. Order-of-magnitude analysis focuses on large problems. 9/16/2018 CS202 - Fundamentals of Computer Science II

Sequential Search int sequentialSearch(const int a[], int item, int n){ for (int i = 0; i < n && a[i]!= item; i++); if (i == n) return –1; return i; } Unsuccessful Search:  O(n) Successful Search: Best-Case: item is in the first location of the array O(1) Worst-Case: item is in the last location of the array O(n) Average-Case: The number of key comparisons 1, 2, ..., n  O(n) 9/16/2018 CS202 - Fundamentals of Computer Science II

Binary Search int binarySearch(int a[], int size, int x) { int low =0; int high = size –1; int mid; // mid will be the index of // target when it’s found. while (low <= high) { mid = (low + high)/2; if (a[mid] < x) low = mid + 1; else if (a[mid] > x) high = mid – 1; else return mid; } return –1; 9/16/2018 CS202 - Fundamentals of Computer Science II

Binary Search – Analysis
For an unsuccessful search: The number of iterations in the loop is log2n + 1  O(log2n) For a successful search: Best-Case: The number of iterations is  O(1) Worst-Case: The number of iterations is log2n +1  O(log2n) Average-Case: The avg. # of iterations < log2n  O(log2n)  an array with size 8  # of iterations The average # of iterations = 21/8 < log28 9/16/2018 CS202 - Fundamentals of Computer Science II

How much better is O(log2n)?
n O(log2n) 1024 (1KB) 16, 131, 262, 524, 1,048,576 (1MB) 1,073,741,824 (1GB) 30 9/16/2018 CS202 - Fundamentals of Computer Science II

Binary Search – Analysis (cont.)
Binary search is more efficient than sequential search. Binary search algorithm is the best searching algorithm which uses key comparisons Binary search requires a sorted list of items 9/16/2018 CS202 - Fundamentals of Computer Science II

Sorting Sorting is a process that organizes a collection of data into either ascending or descending order. An internal sort requires that the collection of data fit entirely in the computer’s main memory. We can use an external sort when the collection of data cannot fit in the computer’s main memory all at once but must reside in secondary storage such as on a disk. We will analyze only internal sorting algorithms. Any significant amount of computer output is generally arranged in some sorted order so that it can be interpreted. Sorting also has indirect uses. An initial sort of the data can significantly enhance the performance of an algorithm. Majority of programming projects use a sort somewhere, and in many cases, the sorting cost determines the running time. A comparison-based sorting algorithm makes ordering decisions only on the basis of comparisons. 9/16/2018 CS202 - Fundamentals of Computer Science II

Sorting Algorithms There are many sorting algorithms, such as: Selection Sort Insertion Sort Bubble Sort Merge Sort Quick Sort First three sorting algorithms are not so efficient, but last two are efficient sorting algorithms. 9/16/2018 CS202 - Fundamentals of Computer Science II

Selection Sort The list is divided into two sublists, sorted and unsorted, which are divided by an imaginary wall. We find the biggest element from the unsorted sublist and swap it with the element at the end of the unsorted data. After each selection and swapping, the imaginary wall between the two sublists move one element back, increasing the number of sorted elements and decreasing the number of unsorted ones. Each time we move one element from the unsorted sublist to the sorted sublist, we say that we have completed a sort pass. A list of n elements requires n-1 passes to completely rearrange the data. 9/16/2018 CS202 - Fundamentals of Computer Science II

Selection Sort (cont.) Unsorted Sorted 9/16/2018 CS202 - Fundamentals of Computer Science II

Selection Sort (cont.) typedef type-of-array-item DataType; void selectionSort(DataType theArray[], int n) { for (int last = n-1; last >= 1; --last) { int largest = indexOfLargest(theArray, last+1); swap(theArray[largest], theArray[last]); } 9/16/2018 CS202 - Fundamentals of Computer Science II

Selection Sort (cont.) int indexOfLargest(const DataType theArray[], int size) { int indexSoFar = 0; for(int currentIndex=1; currentIndex<size;++currentIndex) { if (theArray[currentIndex] > theArray[indexSoFar]) indexSoFar = currentIndex; } return indexSoFar; void swap(DataType &x, DataType &y) { DataType temp = x; x = y; y = temp; 9/16/2018 CS202 - Fundamentals of Computer Science II

Selection Sort -- Analysis
In general, we compare keys and move items (or exchange items) in a sorting algorithm (which uses key comparisons).  So, to analyze a sorting algorithm we should count the number of key comparisons and the number of moves. Ignoring other operations do not affect our final result. In selectionSort function, the for loop executes n-1 times. In selectionSort function, we invoke swap function once at each iteration.  Total Swaps: n-1  Total Moves: 3*(n-1) (Each swap has three moves) 9/16/2018 CS202 - Fundamentals of Computer Science II

Selection Sort – Analysis (cont.)
In indexOfLargest function, the for loop executes the size of the unsorted part minus 1 (from n-1 to 1), and each iteration we make one key comparison.  # of key comparisons = n-1 = n*(n-1)/2  So, Selection sort is O(n2) The best case, the worst case, and the average case of the selection sort algorithm are same.  all of them are O(n2) This means that the behavior of the selection sort algorithm does not depend on the initial organization of data. Since O(n2) grows so rapidly, the selection sort algorithm is appropriate only for small n. Although the selection sort algorithm requires O(n2) key comparisons, it only requires O(n) moves. A selection sort could be a good choice if data moves are costly but key comparisons are not costly (short keys, long records). 9/16/2018 CS202 - Fundamentals of Computer Science II

Insertion Sort Insertion sort is a simple sorting algorithm that is appropriate for small inputs. Most common sorting technique used by card players. The list is divided into two parts: sorted and unsorted. In each pass, the first element of the unsorted part is picked up, transferred to the sorted sublist, and inserted at the appropriate place. A list of n elements will take at most n-1 passes to sort the data. 9/16/2018 CS202 - Fundamentals of Computer Science II

Insertion Sort (cont.) Sorted Unsorted 23 78 45 8 32 56 Original List After pass 1 After pass 2 After pass 3 After pass 4 After pass 5 9/16/2018 CS202 - Fundamentals of Computer Science II

Insertion Sort (cont.) void insertionSort(DataType theArray[], int n) { for (int unsorted = 1; unsorted < n; ++unsorted) { DataType nextItem = theArray[unsorted]; int loc = unsorted; for (;(loc > 0) && (theArray[loc-1] > nextItem); --loc) theArray[loc] = theArray[loc-1]; theArray[loc] = nextItem; } 9/16/2018 CS202 - Fundamentals of Computer Science II

Insertion Sort – Analysis
Running time depends on not only the size of the array but also the contents of the array. Best-case:  O(n) Array is already sorted in ascending order. Inner loop will not be executed. The number of moves: 2*(n-1)  O(n) The number of key comparisons: (n-1)  O(n) Worst-case:  O(n2) Array is in reverse order: Inner loop is executed p-1 times, for p = 2,3, …, n The number of moves: 2*(n-1)+( n-1)= 2*(n-1)+ n*(n-1)/2  O(n2) The number of key comparisons: ( n-1)= n*(n-1)/2  O(n2) Average-case:  O(n2) We have to look at all possible initial data organizations. So, Insertion Sort is O(n2) 9/16/2018 CS202 - Fundamentals of Computer Science II

Insertion Sort – Analysis
Which running time will be used to characterize this algorithm? Best, worst or average? Worst: Longest running time (this is the upper limit for the algorithm) It is guaranteed that the algorithm will not be worst than this. Sometimes we are interested in average case. But there are some problems with the average case. It is difficult to figure out the average case. i.e. what is average input? Are we going to assume all possible inputs are equally likely? In fact, for most algorithms average case is same as the worst case. 9/16/2018 CS202 - Fundamentals of Computer Science II

Bubble Sort The list is divided into two sublists: sorted and unsorted. The largest element is bubbled from the unsorted list and moved to the sorted sublist. After that, the wall moves one element back, increasing the number of sorted elements and decreasing the number of unsorted ones. Each time an element moves from the unsorted part to the sorted part one sort pass is completed. Given a list of n elements, bubble sort requires up to n-1 passes (maximum passes) to sort the data. 9/16/2018 CS202 - Fundamentals of Computer Science II

Bubble Sort (cont.) 9/16/2018 CS202 - Fundamentals of Computer Science II

Bubble Sort (cont.) void bubbleSort(DataType theArray[], int n) { bool sorted = false; for (int pass = 1; (pass < n) && !sorted; ++pass) { sorted = true; for (int index = 0; index < n-pass; ++index) { int nextIndex = index + 1; if (theArray[index] > theArray[nextIndex]) { swap(theArray[index], theArray[nextIndex]); sorted = false; // signal exchange } 9/16/2018 CS202 - Fundamentals of Computer Science II

Bubble Sort – Analysis Best-case:  O(n) Array is already sorted in ascending order. The number of moves: 0  O(1) The number of key comparisons: (n-1)  O(n) Worst-case:  O(n2) Array is in reverse order: Inner loop is executed n-1 times, The number of moves: 3*( n-1) = 3 * n*(n-1)/2  O(n2) The number of key comparisons: ( n-1)= n*(n-1)/2  O(n2) Average-case:  O(n2) We have to look at all possible initial data organizations. So, Bubble Sort is O(n2) 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort Mergesort algorithm is one of two important divide-and-conquer sorting algorithms (the other one is quicksort). It is a recursive algorithm. Divides the list into halves, Sort each halve separately, and Then merge the sorted halves into one sorted array. 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort - Example 9/16/2018 CS202 - Fundamentals of Computer Science II

Merge const int MAX_SIZE = maximum-number-of-items-in-array; void merge(DataType theArray[], int first, int mid, int last) { DataType tempArray[MAX_SIZE]; // temporary array int first1 = first; // beginning of first subarray int last1 = mid; // end of first subarray int first2 = mid + 1; // beginning of second subarray int last2 = last; // end of second subarray int index = first1; // next available location in tempArray for ( ; (first1 <= last1) && (first2 <= last2); ++index) { if (theArray[first1] < theArray[first2]) { tempArray[index] = theArray[first1]; ++first1; } else { tempArray[index] = theArray[first2]; ++first2; } } 9/16/2018 CS202 - Fundamentals of Computer Science II

Merge (cont.) // finish off the first subarray, if necessary for (; first1 <= last1; ++first1, ++index) tempArray[index] = theArray[first1]; // finish off the second subarray, if necessary for (; first2 <= last2; ++first2, ++index) tempArray[index] = theArray[first2]; // copy the result back into the original array for (index = first; index <= last; ++index) theArray[index] = tempArray[index]; } // end merge 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort void mergesort(DataType theArray[], int first, int last) { if (first < last) { int mid = (first + last)/2; // index of midpoint mergesort(theArray, first, mid); mergesort(theArray, mid+1, last); // merge the two halves merge(theArray, first, mid, last); } } // end mergesort 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort - Example 6 3 9 1 5 4 7 2 divide 6 3 9 1 5 4 7 2 divide divide 7 2 6 3 9 1 5 4 divide divide divide divide 6 3 9 1 5 4 7 2 merge merge merge merge 2 7 3 6 1 9 4 5 merge merge 1 3 6 9 2 4 5 7 merge 1 2 3 4 5 6 7 9 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort – Example2 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort – Analysis of Merge
A worst-case instance of the merge step in mergesort 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort – Analysis of Merge (cont.)
k-1 k-1 Merging two sorted arrays of size k Best-case: All the elements in the first array are smaller (or larger) than all the elements in the second array. The number of moves: 2k + 2k The number of key comparisons: k Worst-case: The number of key comparisons: 2k-1 k-1 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort - Analysis Levels of recursive calls to mergesort, given an array of eight items 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort - Analysis 2m level 0 : 1 merge (size 2m-1) 2m-1 2m-1 level 1 : 2 merges (size 2m-2) level 2 : 4 merges (size 2m-3) 2m-2 2m-2 2m-2 2m-2 . . level m-1 : 2m-1 merges (size 20) 20 20 level m 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort - Analysis Worst-case – The number of key comparisons: = 20*(2*2m-1-1) + 21*(2*2m-2-1) m-1*(2*20-1) = (2m - 1) + (2m - 2) (2m – 2m-1) ( m terms ) = m*2m – = m*2m – 2m – 1 = n * log2n – n – 1  O (n * log2n ) 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort – Average Case
There are possibilities when sorting two sorted lists of size k. k=2  = = 6 different cases # of key comparisons = ((2*2)+(4*3)) / 6 = 16/6 = /3 Average # of key comparisons in mergesort is n * log2n – 1.25*n – O(1)  O (n * log2n ) 9/16/2018 CS202 - Fundamentals of Computer Science II

Mergesort – Analysis Mergesort is extremely efficient algorithm with respect to time. Both worst case and average cases are O (n * log2n ) But, mergesort requires an extra array whose size equals to the size of the original array. If we use a linked list, we do not need an extra array But, we need space for the links And, it will be difficult to divide the list into half ( O(n) ) 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort Like mergesort, Quicksort is also based on the divide-and-conquer paradigm. But it uses this technique in a somewhat opposite manner, as all the hard work is done before the recursive calls. It works First, it partitions an array into two parts, Then, it sorts the parts independently, Finally, it combines the sorted subsequences by a simple concatenation. 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort (cont.) The quick-sort algorithm consists of the following three steps: 1. Divide: Partition the list. To partition the list, we first choose some element from the list for which we hope about half the elements will come before and half after. Call this element the pivot. Then we partition the elements so that all those with values less than the pivot come in one sublist and all those with greater values come in another. 2. Recursion: Recursively sort the sublists separately. 3. Conquer: Put the sorted sublists together. 9/16/2018 CS202 - Fundamentals of Computer Science II

Partition Partitioning places the pivot in its correct place position within the array. Arranging the array elements around the pivot p generates two smaller sorting problems. sort the left section of the array, and sort the right section of the array. when these two smaller sorting problems are solved recursively, our bigger sorting problem is solved. 9/16/2018 CS202 - Fundamentals of Computer Science II

Partition – Choosing the pivot
First, we have to select a pivot element among the elements of the given array, and we put this pivot into the first location of the array before partitioning. Which array item should be selected as pivot? Somehow we have to select a pivot, and we hope that we will get a good partitioning. If the items in the array arranged randomly, we choose a pivot randomly. We can choose the first or last element as a pivot (it may not give a good partitioning). We can use different techniques to select the pivot. 9/16/2018 CS202 - Fundamentals of Computer Science II

Partition Function void partition(DataType theArray[], int first, int last, int &pivotIndex) { // Partitions an array for quicksort. // Precondition: theArray[first..last] is an array; first <= last. // Postcondition: Partitions theArray[first..last] such that: // S1 = theArray[first..pivotIndex-1] < pivot // theArray[pivotIndex] == pivot // S2 = theArray[pivotIndex+1..last] >= pivot // Calls: choosePivot and swap. // place pivot in theArray[first] choosePivot(theArray, first, last); DataType pivot = theArray[first]; // copy pivot 9/16/2018 CS202 - Fundamentals of Computer Science II

Partition Function (cont.)
// initially, everything but pivot is in unknown int lastS1 = first; // index of last item in S1 int firstUnknown = first + 1; // index of first item in unknown // move one item at a time until unknown region is empty for (; firstUnknown <= last; ++firstUnknown) { // Invariant: theArray[first+1..lastS1] < pivot // theArray[lastS1+1..firstUnknown-1] >= pivot // move item from unknown to proper region if (theArray[firstUnknown] < pivot) { // belongs to S1 ++lastS1; swap(theArray[firstUnknown], theArray[lastS1]); } // else belongs to S2 } // place pivot in proper position and mark its location swap(theArray[first], theArray[lastS1]); pivotIndex = lastS1; } // end partition 9/16/2018 CS202 - Fundamentals of Computer Science II

Invariant for the partition algorithm 9/16/2018 CS202 - Fundamentals of Computer Science II

Initial state of the array 9/16/2018 CS202 - Fundamentals of Computer Science II

Moving theArray[firstUnknown] into S1 by swapping it with theArray[lastS1+1] and by incrementing both lastS1 and firstUnknown. 9/16/2018 CS202 - Fundamentals of Computer Science II

Moving theArray[firstUnknown] into S2 by incrementing firstUnknown. 9/16/2018 CS202 - Fundamentals of Computer Science II

Developing the first partition of an array when the pivot is the first item 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort Function void quicksort(DataType theArray[], int first, int last) { // Sorts the items in an array into ascending order. // Precondition: theArray[first..last] is an array. // Postcondition: theArray[first..last] is sorted. // Calls: partition. int pivotIndex; if (first < last) { // create the partition: S1, pivot, S2 partition(theArray, first, last, pivotIndex); // sort regions S1 and S2 quicksort(theArray, first, pivotIndex-1); quicksort(theArray, pivotIndex+1, last); } 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort – Analysis Worst Case: (assume that we are selecting the first element as pivot) The pivot divides the list of size n into two sublists of sizes 0 and n-1. The number of key comparisons = n-1 + n = n2/2 – n/2  O(n2) The number of swaps = = n n-1 + n swaps outside of the for loop swaps inside of the for loop = n2/2 + n/  O(n2) So, Quicksort is O(n2) in worst case 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort – Analysis Quicksort is O(n*log2n) in the best case and average case. Quicksort is slow when the array is sorted and we choose the first element as the pivot. Although the worst case behavior is not so good, and its average case behavior is much better than its worst case. So, Quicksort is one of best sorting algorithms using key comparisons. 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort – Analysis A worst-case partitioning with quicksort 9/16/2018 CS202 - Fundamentals of Computer Science II

Quicksort – Analysis An average-case partitioning with quicksort 9/16/2018 CS202 - Fundamentals of Computer Science II

Radix Sort Radix sort algorithm different than other sorting algorithms that we talked. It does not use key comparisons to sort an array. The radix sort : Treats each data item as a character string. First it groups data items according to their rightmost character, and put these groups into order wrt this rightmost character. Then, combine these groups. We, repeat these grouping and combining operations for all other character positions in the data items from the rightmost to the leftmost character position. At the end, the sort operation will be completed. 9/16/2018 CS202 - Fundamentals of Computer Science II

Radix Sort – Example mom, dad, god, fat, bad, cat, mad, pat, bar, him original list (dad,god,bad,mad) (mom,him) (bar) (fat,cat,pat) group strings by rightmost letter dad,god,bad,mad,mom,him,bar,fat,cat,pat combine groups (dad,bad,mad,bar,fat,cat,pat) (him) (god,mom) group strings by middle letter dad,bad,mad,bar,fat,cat,pat,him,god,mom combine groups (bad,bar) (cat) (dad) (fat) (god) (him) (mad,mom) (pat) group strings by middle letter bad,bar,cat,dad,fat,god,him,mad,mom,par combine groups (SORTED) 9/16/2018 CS202 - Fundamentals of Computer Science II

Radix Sort – Example 9/16/2018 CS202 - Fundamentals of Computer Science II

Radix Sort - Algorithm radixSort(inout theArray:ItemArray, in n:integer, in d:integer) // sort n d-digit integers in the array theArray for (j=d down to 1) { Initialize 10 groups to empty Initialize a counter for each group to 0 for (i=0 through n-1) { k = jth digit of theArray[i] Place theArray[i] at the end of group k Increase kth counter by 1 } Replace the items in theArray with all the items in group 0, followed by all the items in group 1, and so on. 9/16/2018 CS202 - Fundamentals of Computer Science II

Radix Sort -- Analysis The radix sort algorithm requires 2*n*d moves to sort n strings of d characters each.  So, Radix Sort is O(n) Although the radix sort is O(n), it is not appropriate as a general-purpose sorting algorithm. Its memory requirement is d * original size of data (because each group should be big enough to hold the original data collection.) For example, to sort string of uppercase letters. we need 27 groups. The radix sort is more appropriate for a linked list than an array. (we will not need the huge memory in this case) 9/16/2018 CS202 - Fundamentals of Computer Science II

Comparison of Sorting Algorithms

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