Download presentation

Presentation is loading. Please wait.

Published byMargaret Baxter Modified over 2 years ago

1
Chapter 3 Brute Force Brute force is a straightforward approach to solving a problem, usually directly based on the problem’s statement and definitions of the concepts involved. Example 1 Compute. Brute force algorithm: A better algorithm: Example 2 Search x from n elements. Brute force algorithm: linear search. A better algorithm: binary search.

2
Example 3 Sort n elements. (1) Selection Sort Find the smallest element and exchange it with the element in the first position, find the second smallest element and exchange it with the one in the second position, continue in this way until the entire array is sorted. Selection sort: l x e a m e p a x e l m e p a e x l m e p a e e l m x p a e e l m p x Program 1 Selection sort (sort array a[1..n]) Procedure seletion(a[1..n]) { for (i=1; i < n; i++) { min = i; for ( j=i+1; j <=n; j++) if (a[j] < a[min]) min = j; exch(a[i], a[min]); } Section sort is a stable sort

3
Running Time of Section Sort case (n-1) comparisons (n-2) comparisons (n-3) comparisons case (n-1) comparisons (n-2) comparisons (n-3) comparisons Running time does not depend on the state of the input.

4
(2) Insertion Sort a m x e l e p a e m x l e p a e m l x e p a e e l m x p a e e l m p x Sort Consider the elements one at a time, inserting each into its proper place among these already considered (keeping them sorted). Program 2 Insertion sort (sort array a[1..n]) Procedure insertion(a[1..n]) { for (i = 2; i<=n; i++) { j = i; v = a[i]; while (v < a[j-1]) { a[j] = a[j-1]; j--; } a[j] = v; } m a x e l e p Insertion Sort is a stable sort. a m x e l e p

5
Running Time of Insertion Sort case 1 comparison case 1 comparisons 2 comparisons 1 comparison (n-1) comparisons 3comparisons Running time depends on the state of the input. In the worst case, the running time is Better Algorithms can solve sorting problem in O(nlogn) time.

6
Example 4 Problems using exhaustive search (1) Traveling sales man problem Given a graph G, find the shortest simple circuit (Hamiltonian circuit) in G which passes through all the vertices. a cd b 2 3 1 5 87 Tour length abcda 18 abdca 11 … Totally there are P(4,4)=4! tours. Generally, there may have n! tours in a graph of n vertices and if we check all the tours the computing time is O(n!). People couldn’t find a good algorithm (using polynomial time) to solve it!

7
(2) Knapsack problem knapsack: W=10 item 1: w=7, v=$42 item 2: w=3, v=$12, item 3: w=4, v= $40, item 4: w=5, v=$25 People couldn’t find a good algorithm (using polynomial time) to solve it ! The problems which people could not find polynomial algorithms for are called as NP-hard problems.

8
Chapter 4 Divide-and-Conquer Divide-and-conquer: one of the most important technique used for designing algorithms (1)Divide a large problem into subproblems of about the same size (2)Solve each of subproblems (3)Merge the solutions of the subproblems to that of original one. Example 1 Quick sort Quick sort n elements in array a[1..n] (1)Partition array a[1..n] to two parts: one whose elements is smaller than a[n] and one whose elements are larger than a[n]. (2)Quick sort the first part and then quick sort the second part. (3)Output the first part, a[n], then output the second part.

9
Program 1 Quicksort Sort a[l..r], where v=a[r]. Function partition(a,l,r) return a value i such that after partitioning a[i]=v, the elements a[l..i-1] are smaller than v, and the elements a[i+1..r] are larger than v. void quicksort(int a[], int l, int r) { if (r <= l) return; int i = partition(a, l, r); quicksort(a,l,i-1); quicksort(a,i+1,r); } Program 2 Partitioning Function partition(a,l,r) return a value i such that after partitioning a[i]=v, the elements a[l..i-1] are smaller than v, and the elements a[i+1..r] are larger than v. void partition(int a[], int l, int r) { int i = l-1, j = r; Item v = a[r]; for (; ;) { while (a[++i] < v); while (v < a[--j]) if (j == 1) break; if (i >= j) break; exch(a[i],a[j]); } exch(a[i],a[r]); return i; } the number for comparisons is n when partition a[1..n]

10
Program 2 Partitioning Function partition(a,l,r) return a value i such that after partitioning a[i]=v, the elements a[l..i-1] are smaller than v, and the elements a[i+1..r] are larger than v. void partition(int a[], int l, int r) { int i = l-1, j = r; Item v = a[r]; for (; ;) { while (a[++i] < v); while (v < a[--j])if (j == l) break; if (i >= j) break; exch(a[i],a[j]); } exch(a[i],a[r]); return i; } Less than or equal to vGreater than or equal to v ijl r Example Partitioning a[l..r]=[11,6,15,7,9,18,9,5,17,14] 11, 6, 15, 7, 9, 18, 9, 5, 17, 14 v ij ij ij 11, 6, 5, 7, 9, 18, 9, 15, 17, 14 ij ij ij 11, 6, 5, 7, 9, 9, 18, 15, 17, 14 ij ji

11
Program 1 Quicksort Sort a[l..r]. Let v=a[r]. Function partition(a,l,r) return a value i such that after partitioning a[i]=v, the elements a[l..i-1] are smaller than v, and the elements a[i+1..r] are larger than v. void quicksort(int a[], int l, int r) { if (r <= l) return; int i = partition(a, l, r); quicksort(a,l,i-1); quicksort(a,i+1,r); } How recursive algorithms work? Example quicksort a[1..8] partition(a,1,8) quicksort(a,1,4) quicksort(a,5,8) partition(a,1,4) quicksort(a,1,2) quicksort(a,3,4) partition(a,5,8) quicksort(a,5,6) quicksort(a,7,8) partition(a,1,2) quicksort(a,1,1) quicksort(a,2,2) partition(a,3,4) quicksort(a,3,3) quicksort(a,4,4) partition(a,5,6) quicksort(a,5,5) quicksort(a,6,6) partition(a,7,8) quicksort(a,7,7) quicksort(a,8,8)

12
Performance Characteristics of Quicksort Property 1 Quicksort uses about comparisons in the worst case. In the case that a file is already in order, items are always partitioned into two parts, in which one part contains only one item. Therefore, the time for partitioning in Quicksort(a,1,n) is n+(n-1)+(n-2)+…+2+1=n(n+1)/2, which is. In the best case, a file is always divided exactly in half. The time for partitioning in Quicksort(a,1,n) is Property 2 Quicksort uses about comparisons on the average case.

13
An improvement of Quick-Sort: Median-of-three partitioning An improvement to quicksort is to use a partition element that is more likely to divide the file near the middle. Use a random element from the array for a partitioning element. The worst case will happen with negligibly small probability. take a sample of three elements from the file, then use the median of three for the partitioning element. Another improvement is to cut off the recursion for small subfiles –sort a file directly by insertion sort when it is small. void Quicksort(int a[], int l, int r) If r-l <= M { insertion(a,l,r); return } find the median from a[l], a[(l+r)/2], a[r] and exchange it with a[r]; i := partition(a, l,r); quicksort(a,l,i-1); quicksort(a,i,r) Three merits (1)The worst case much more unlikely to occur. (2)It eliminates the need for a sentinel key for partitioning. (3)It reduces the average time of the algorithm by 5%.

14
Program 3 Improved quicksort static const int M = 10; void quicksort( int a[], int l, int r) { if (r-l < M) return; exch(a[(l+r)/2], a[r-1]); compexch(a[l],a[r-1]); compexch(a[l],a[r]); compexch(a[r-1],a[r]); int i = partition(a, l+1, r-1); quicksort(a, l, i-1); quicksort(a, i+1,r); } template void hybridsort(Item a[], int l, int r) { quicksort(a, l,r), insertion(a, l, r); } For randomly ordered files, Program 3 is superfluous. Quicksort is widely used because it runs well in a variety of situations. Other methods might be more appropriate for particular cases, but quicksort handles more types of sorting problems than are handled by many other methods, and it is often significantly faster than alternative approaches.

15
Example 2 Merge Sort Two-Way Merging Given two ordered input files, combine then into one ordered output file. 11, 12, 25, 56, 57 12, 14, 26, 58, 59, 62, 73, 81 12 1425265657115859 62 73 81 Program 4 Merging void mergeAB( int c[], int b[], int a[], int l, int m, int r) { for (int i = 0, j = 0, k = l; k < =r; k++) { if ( i == m) { c[k] = b[j++]; continue; if ( j == r) { c[k] = a[i++]; continue; c[k] = ( a[i] < b[j]) ? a[i++] : b[j++]; }; } a[l..m]a[m+1..r] a[l..r] How to merge a[l..m] and a[m+1..r] inside array a[l..m]? Mergesort sorts a file of n elements in

16
Abstract In-Place Merge a[l..m]a[m+1..r] a[l..r] How to merge a[l..m] and a[m+1..r] inside array a[l..m]? 12,14,21,33,4511,21,32,34,35,46 lmm+1r a[l..r] 12,14,21,33,4546,35,34,32,21,11 aux[l..r] ij 11 j 12 i 14 ii 21 ij a[l..r] Program 5 Abstract in-place merge void merge( int a[], int l, int m, int r) { int i, j; static int aux[maxN] for (i = l; i <=m; i++) aux[i] = a[i]; for (j = m+1; j <= r; j++) aux[j]=a[r+m-j+1]; for (i=l; j=r; int k = l; k <= r; k++) if (aux[j] < aux[i]) a[k] = aux[j--]; else a[k] = aux[i++]; }

17
Mergesort based divide-and-conquer technique. Mergesort a[l..r] (1)Divide array a[l..r] to two parts: a[l..m] and a[m+1..r]. (2)Mergesort a[l..m] and a[m+1..r], respectively. (3)Merge the sorted a[l..m] and a[m+1..r].

18
Program 6 mergesort void mergesort (int a[], int l, int r) {if (r <= l) return; int m = (r+l)/2; mergesort(a,l,m); mergesort(a,m+1,r); merge(a,l,m,r); } How mergesort algorithm works? Example mergesort a[1..8] mergesort(a,1,4) mergesort(a,5,8) merge(a,1,4,8) mergesort(a,1,2) mergesort(a,3,4) merge(a,1,2,4) mergesort(a,5,6) mergesort(a,7,8) merge(a,5,6,8) mergesort(a,1,1) mergesort(a,2,2) merge(a,1,1,2) mergesort(a,3,3) mergesort(a,4,4) merge(a,3,3,4) mergesort(a,5,5) mergesort(a,6,6) merge(a,5,5,6) mergesort(a,7,7) mergesort(a,8,8) merge(a,7,7,8) T(n) = 2T(n/2) + n T(1) = 0 T(n) = O(nlgn) Property 8.1 Mergesort requires about O(nlgn) comparisions to sort any file of n elements.

19
Project 1 Task: Using Divide-and-conquer to design algorithms for a Binary Search Tree which support: (1) Search, (2) Insertion, (3) Transversal, (3) Finding the height of a BST, and (4) Finding the size of a BST Requirements First, build a BST with 100 nodes. To simplify the problem, the data in each node contains only one key which is an integer between -10000 and 10000. The key is generated randomly. Each of the algorithms for search, insert, transversal, finding the height and size of a BST has to designed by Divide-and-Conquer technique. Implement the algorithms using C or C++, or any other programming language. A simple user interface has to be contained. That is, user can select operations and read the results of the operations. Submission Project description, Algorithms, Algorithm analysis, Experiment output, Code.

Similar presentations

Presentation is loading. Please wait....

OK

Searching and Sorting Arrays

Searching and Sorting Arrays

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on code switching and code mixing Ppt on 9/11 conspiracy theory facts Ppt on content development manager Ppt on hydrogen fuel cell vehicles cost Ppt on buddhism in india Ppt on obesity prevention in children Ppt on bluetooth technology download Ppt on cell the fundamental unit of life Ppt on dubai in hindi Ppt on amplitude shift keying pdf