1.9k views

Consider the following C program that attempts to locate an element $x$ in an array $Y[ \ ]$ using binary search. The program is erroneous.

f (int Y[10] , int x) {
int u, j, k;
i= 0; j = 9;
do {
k = (i+ j) / 2;
if( Y[k] < x) i = k;else j = k;
} while (Y[k] != x) && (i < j)) ;
if(Y[k] == x) printf(" x is in the array ") ;
else printf(" x is not in the array ") ;
}

The correction needed in the program to make it work properly is

1. Change line 6 to: if $(Y[k] < x) i = k + 1$; else $j = k-1$;
2. Change line 6 to: if $(Y[k] < x) i = k - 1$; else $j = k +1$;
3. Change line 6 to: if $(Y[k] < x) i = k$; else $j = k$;
4. Change line 7 to: } while $((Y[k] == x) \&\& (i < j))$ ;

edited | 1.9k views
+3
Given progeam is errorneous bcz it is doing malfunction when we are serching for an element that not present in the array.

I.e. 10 20 30 40 50 60 70 80 90 100

Now search 65 it will go into infinite loop

And that problem occured due to improper index (i,j) updation.

and the correct updation is Option A which is followed by Binary Search.
+1

if( Y[k] < x) then i = k + 1;

if given element that  we are searching is greater, then searching will be continued in the upper half of array

otherwise $\mathbf{j = k - 1}$;

in the lower half.

Take few case in consideration i.e.

1. All elements are same
2. Increasing order with no repetition
3. Increasing order with  repetition.
by Boss (38.7k points)
edited
0
manoj could u explain a bit more
+9

@asu ... see .. suppose our array is - 1, 2, 3, 4,  5, 6

and we want to search x=7, then check what happens here-

In the beginning , i=0, and j=5.

k= (0 + 5)/2 = 2.

A[2] <7, so  i=k=2.

now while (A[2] != x&& i <j) - true

2nd time - i=2, j=5.

k= (2+5)/2= 3

A[3] < 7, so i=k=3

while(A[3] !=x && i<j) - true

3rd time - i=3, j=5.

k= (3+5)/2= 4.

A[4] < 7, so i= k=4.

while(A[4] !=x && i<j) - true

3rd time - i=4, j=5.

k= (4+5)/2= 4.

A[4] < 7, so i= k=4.

while(A[4] !=x && i<j) - true

3rd time - i=4, j=5.

k= (4+5)/2= 4.

A[4] < 7, so i= k=4.

.

.

.

will go into infinite loop.

So to get rid of this infinite loop if we put- i = k+1. instead of i=k then it would not fall into infinite loop.

..

.and you can put j= k-1 and even j=k is also okay.

.

0
@vijay...thank u for explaniation ....but to check each option it will take much time..any suggestion for these kind of question
+1

Sorry @asu .. I am also working on it  ... but I think .. while reading BST .or any topic... if we consider every cases and especially edge case then we can solve most of the question related to any topic ..

+5

Is there any case in which::  if we update i=k+1 and keep j=k , then the given program will give erroneous answer??

As far as I understood. I think even i=k+1 and j=k will work fine.

but if we change

do{

}while(Y[k] !=x && i<=j)

then j=k-1 is necessary.

Correct me if I am wrong !!!

0
@VS

I think it will work fine for even j=k, no need to change to j=k-1
0
if array is {2,2,2,2,2,2,2,2,2,2} and we wat to search 3 now in this case option 1 will not work
+1
if(Y[k]<x) then i=k+1 else j=k-1

if(Y[k]<x) then i=k+1 else j=k

Both will work fine.

Second one will work fine because loop terminates at $i \not \leq j$.

Just try to search Last element : it will go into infinite loop because of interger division.

i=k+1 will solve the problem.

lets elemets are : 10 20 30 40 50 60 70 80 90 100

now try to search 25 or 35 or 45 or 55 on above array so it will run into infinite loop.

so we need both the conditions=i=k+1 and j=k-1

P.S: This program is errorenous because of unsuccessful search.
by Boss (25.7k points)

There are 2 changes to be made:

1. updation conditions : i=k+1 and j=k-1 //reason can be found in other ans

2.while loop condition : while (Y[k] != x) && (i < = j) //equal to added here , other ans missed this

let a[]={1,2,3,4,5,6,7,8,9,10};

Without 2nd condition f(a,10) will fail

by (67 points)