Question

Consider an ordered array A of size n and the following ternary search algorithm for finding the index i such that A[i] = K. Divide the array into three parts. If A[n/3] > K. the first third of the array is searched recursively, else if A[2n/3] > K then the middle part of the array is searched recursively, else the last thud of the array is searched recursively. Provisions are also made in the algorithm to return n/3 if A[n/3] = K, or return 2n/3 if A[2n/3]=K. Write the recurrence relation for the number of comparisons C(n) in the average case for the binary and ternary searches and solve the ternary average case only. Do the following experiment (need to write a program in the language of your choice): Generate arrays of sizes n = 500, 1000, 2000, 4000 and 8000. Fill the arrays as follows: A[i] = integer(8 times Squareroot i); i=0, 1, 2, ..., n-1. Generate random integer numbers between 0 and 8 times n, and search each array for acquired random number using binary search and ternary search. Record the number of comparison for each array and for each of the two searches. Repeat Steps 2 and 3, ten (10) times. Take the average number of comparisons for the 10 runs, for each array and each search method. Present the results in the form of graph or table, etc. (your choice). Compare the binary search with the ternary search in term of the number of operations needed for the search. How do these result match your findings in (a)? Write one paragraph conclusions.

Answer 1

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The experiment in C

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define n 4000
int bcount=0,tcount=0;
int main()
   {
   int a[n];
   int i,k,pos;
   clrscr();
   for(i=0;i<n;i++)
       {
       a[i]=(int)(8*sqrt(i));
       }
   srand(time(NULL));
   k=rand() % (int)(8*sqrt(n));
   pos=binarysearch(a,0,n-1,k);
   printf("%d is in %d ",k,pos);
   pos=terenarysearch(a,0,n-1,k);
   printf("\n%d is in %d ",k,pos);
   printf("\nTotal comparison in binary search :%d",bcount);
   printf("\nTotal comparison in terenary search : %d",tcount);
   return 0;
   }
int binarysearch(int a[],int p,int r,int k)
   {
   int mid;
   if(p>r) {bcount++;
       return -1;}
   mid=(p+r)/2;
   if (k==a[mid])
       {
       bcount++;
       return mid;
       }
   if(a[mid]>k)
       {
       bcount++;
       return    binarysearch(a,p,mid-1,k);
       }
   else
       {
       bcount++;
       return binarysearch(a,mid+1,r,k);
       }
   }

int terenarysearch(int a[],int p,int r,int k)
   {
   int mid1,mid2;
   if(p>r){tcount++; return -1;}
   mid1=p+(r-p)/3;
   mid2= p+2*(r-p)/3;
   if (k==a[mid1])
       {
       tcount++ ;
       return mid1;
       }
   if (k==a[mid2])
       {
       tcount++ ;
       return mid2;
       }
   if(a[mid1]>k)
       {
       tcount++ ;
   return    terenarysearch(a,p,mid1-1,k);
   }
   else if(a[mid1]<k && a[mid2]>k)
       {
       tcount++ ;
       return terenarysearch(a,mid1+1,mid2-1,k);
       }
   else
       {
       tcount++ ;
       return terenarysearch(a,mid2+1,r,k);
       }
   }

	N=500	1000	2000
	Binary,terenary	Binary,terenary	Binary,terenary
Exp-1	6,5	3,5	7,5
Exp-2	6,5	7,5	7,5
Exp-3	7,2	7,5	8,6
Exp-4	7,5	6,5	12,9
Exp-5	3,3	8,4	8,4

Sorry for only 5 experiment. Time did not permit for all experiment..

how ever you can run the code above 10 times.

From these observation, I can conclude that terenary search is efficient as compared to bany search