Binary Search vs. Brute Force Search Algorithms for Finding a Local Minimum in a
ID: 3601702 • Letter: B
Question
Binary Search vs. Brute Force Search Algorithms for Finding a Local Minimum in a TwoDimensional Array
In this project, you will implement the binary search-based (n) algorithm to determine the local minimum in an 'n' x 'n' two-dimensional array (as discussed in Module 2) and compare its run-time performance with that of a brute force (n2 ) algorithm that searches for the local minimum element by element until one is found.
Note that both the binary search and the brute force search algorithms should stop once a local minimum is found.
You should create random two-dimensional arrays (with numRows = numCols) with unique elements in the range [1... numRows * numCols] for the following values of numRows (numCols) and determine the average execution time of the binary search and the brute force search algorithms by running 100 trials for each of the numRows (numCols) values. Determine the running times in nano seconds or milli seconds, as appropriate.
numRows (numCols) values: 4, 6, 8, 10, 15, 20, 25, 30, 35, 40, 50, 75, 100, 150, 200, 250, 300, 350
Plot the results with numRows in X-axis and the average execution times (in nanoseconds or milliseconds) of the binary search and the brute force search algorithms in the Y-axis.
Submission Submit a hardcopy of your codes for the binary search and brute force search algorithms, the Excel plots of the numRows vs. average execution times of the two algorithms as well as your interpretation of the results.
Explanation / Answer
Dear student ,
Please find solution below,if having any confusion in solution feel free to comment.
Note:Code for brute force search and binary search is in C.
For comparision between brute force na d binary search,lets take a sorted matrix given below and search for elelment 25 in both case.
5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25 26
Bruteforce search with time complexity O(n2)
#include <stdio.h>
#include <limits.h>
#include <time.h>
#include <sys/time.h>
//BruteForce
int search(int mat[5][5], int n, int k)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(mat[i][j]==k)
{
// printf("localMinindex=(%d,%d)",i,j);
break;
}
}
}
}
int main() {
// your code goes here
int i,j,r,c;
printf("Enter number of rows ");
scanf("%d",&r);
printf("Enter number of column ");
scanf("%d",&c);
int mat[r][c];
for(i=0;i<r;i++)
{
for(j=0;j<c;j++)
{
scanf("%d",&mat[i][j]);
}
}
int locmin=INT_MAX;
struct timeval timer_usec;
long long int timestamp_usec; /* timestamp in microsecond */
if (!gettimeofday(&timer_usec, NULL)) {
timestamp_usec = ((long long int) timer_usec.tv_sec) * 1000000ll +
(long long int) timer_usec.tv_usec;
}
else {
timestamp_usec = -1;
}
printf("%lld microseconds since epoch ", timestamp_usec);
int loop;
for(loop=1;loop<=100000;loop++)
{
search(mat, r, 25);
}
if (!gettimeofday(&timer_usec, NULL)) {
timestamp_usec = ((long long int) timer_usec.tv_sec) * 1000000ll +
(long long int) timer_usec.tv_usec;
}
else {
timestamp_usec = -1;
}
printf("%lld microseconds since epoch ", timestamp_usec);
return 0;
}
Binary search with time complexity O(nlogn)
#include <stdio.h>
#include <limits.h>
#include <time.h>
#include <sys/time.h>
//Binary Search
int bsearch(int mat[5][5], int n, int k)
{
int i = 0, j = n-1; //set indexes for top right element
while ( i < n && j >= 0 )
{
if ( mat[i][j] == k )
{
// printf("n Found at %d, %d", i, j);
return 1;
}
if ( mat[i][j] > k )
j--;
else // if mat[i][j] < k
i++;
}
printf("n Element not found");
return 0; // if ( i==n || j== -1 )
}
int main() {
// your code goes here
int i,j,r;
printf("Enter number of rows & column ");
scanf("%d",&r);
int mat[r][r];
for(i=0;i<r;i++)
{
for(j=0;j<r;j++)
{
scanf("%d",&mat[i][j]);
}
}
// int locmin=INT_MAX;
struct timeval timer_usec;
long long int timestamp_usec; /* timestamp in microsecond */
if (!gettimeofday(&timer_usec, NULL)) {
timestamp_usec = ((long long int) timer_usec.tv_sec) * 1000000ll +
(long long int) timer_usec.tv_usec;
}
else {
timestamp_usec = -1;
}
printf("%lld microseconds since epoch ", timestamp_usec);
int loop;
for(loop=1;loop<=100000;loop++)
{
bsearch(mat, 5, 25);
}
if (!gettimeofday(&timer_usec, NULL)) {
timestamp_usec = ((long long int) timer_usec.tv_sec) * 1000000ll +
(long long int) timer_usec.tv_usec;
}
else {
timestamp_usec = -1;
}
printf("%lld microseconds since epoch ", timestamp_usec);
return 0;
}
Note:It is apllicable for a sorted matrix only
Run both codes in your compiler compute time diffrence in both case by substracting second epoc time from first
Then you can compare results and plot graph in Excel.
Thanks !!!
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