Fill in the blanks to write a recurrence relation called RecSearch find a specific value x in a list of size n. You may assume that the list in held in an array called A. Efficiency is not a concern. You must use recursion. The function should return an index (position) of the desired item. Do not make any assumptions about the nature of the list.
RecSearch(A,n,x) = ______________ if ________________= ________________ // _________________ >= 1 (can also index from zero)
RecSearch(A,n,x) = ______________ // otherwise
Solution
The given recurrence relation can be shown in the form of algorithm as follows:
int RecSearch(int A[], int n, int x)
{
if(n <= 0)
{
return - 1;
}
else if(A[n-1] == x)
{
return (n - 1);
}
return RecSearch(A, n - 1, x);
}
The recurrence relation RecSearch(A, n, x) will return -1 if the value of n is less than or equal to 0. The recurrence relation RecSearch(A, n, x) will return the index of the element if the condition if(A[n – 1] == x) is satisfied.
The RecSearch(A, n, x) will return the index returned by the multiple calls to the RecSearch(A, n - 1, x).
The blanks given in the problem can be filled as follows:
RecSearch(A, n, x) = -1 if n < = 0 // if n >= 1 (can also index from zero) RecSearch(A, n, x) = RecSearch(A, n – 1, x) // otherwise.
.
Fill in the blanks to write a recurrence relation called RecSearch fin.docx
1. Fill in the blanks to write a recurrence relation called RecSearch find a specific value x in a list
of size n. You may assume that the list in held in an array called A. Efficiency is not a concern.
You must use recursion. The function should return an index (position) of the desired item. Do
not make any assumptions about the nature of the list.
RecSearch(A,n,x) = ______________ if ________________= ________________ //
_________________ >= 1 (can also index from zero)
RecSearch(A,n,x) = ______________ // otherwise
Solution
The given recurrence relation can be shown in the form of algorithm as follows:
int RecSearch(int A[], int n, int x)
{
if(n <= 0)
{
return - 1;
}
else if(A[n-1] == x)
{
return (n - 1);
}
return RecSearch(A, n - 1, x);
}
2. The recurrence relation RecSearch(A, n, x) will return -1 if the value of n is less than or equal to
0. The recurrence relation RecSearch(A, n, x) will return the index of the element if the condition
if(A[n – 1] == x) is satisfied.
The RecSearch(A, n, x) will return the index returned by the multiple calls to the RecSearch(A, n
- 1, x).
The blanks given in the problem can be filled as follows:
RecSearch(A, n, x) = -1 if n < = 0 // if n >= 1 (can also index from zero) RecSearch(A, n, x) =
RecSearch(A, n – 1, x) // otherwise.
![Fill in the blanks to write a recurrence relation called RecSearch find a specific value x in a list
of size n. You may assume that the list in held in an array called A. Efficiency is not a concern.
You must use recursion. The function should return an index (position) of the desired item. Do
not make any assumptions about the nature of the list.
RecSearch(A,n,x) = ______________ if ________________= ________________ //
_________________ >= 1 (can also index from zero)
RecSearch(A,n,x) = ______________ // otherwise
Solution
The given recurrence relation can be shown in the form of algorithm as follows:
int RecSearch(int A[], int n, int x)
{
if(n <= 0)
{
return - 1;
}
else if(A[n-1] == x)
{
return (n - 1);
}
return RecSearch(A, n - 1, x);
}](https://image.slidesharecdn.com/fillintheblankstowritearecurrencerelationcalledrecsearchfin-230131193350-fe03097e/85/Fill-in-the-blanks-to-write-a-recurrence-relation-called-RecSearch-fin-docx-1-320.jpg)
![The recurrence relation RecSearch(A, n, x) will return -1 if the value of n is less than or equal to
0. The recurrence relation RecSearch(A, n, x) will return the index of the element if the condition
if(A[n – 1] == x) is satisfied.
The RecSearch(A, n, x) will return the index returned by the multiple calls to the RecSearch(A, n
- 1, x).
The blanks given in the problem can be filled as follows:
RecSearch(A, n, x) = -1 if n < = 0 // if n >= 1 (can also index from zero) RecSearch(A, n, x) =
RecSearch(A, n – 1, x) // otherwise.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)