Create a recursive procedure named (accumulator oddsum next). The procedure will return the sum of the odd numbers entered from
1 answer:
Answer:
Explanation:
The following procedure is written in Python. It takes the next argument, checks if it is an odd number and if so it adds it to oddsum. Then it asks the user for a new number from the keyboard and calls the accumulator procedure/function again using that number. If any even number is passed the function terminates and returns the value of oddsum.
def accumulator(next, oddsum = 0):
if (next % 2) != 0:
oddsum += next
newNext = int(input("Enter new number: "))
return accumulator(newNext, oddsum)
else:
return oddsum
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Answer:
Explanation:
Problem statement:
to simulate a binary search algorithm on an array of random values.
Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
Input/output description
Input:
Size of array: 4
Enter array:10 20 30 40
Enter element to be searched:40
The Output will look like this:
Element is present at index 3
Algorithm and Flowchart:
We basically ignore half of the elements just after one comparison.
Compare x with the middle element.
If x matches with middle element, we return the mid index.
Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
Else (x is smaller) recur for the left half.
The Flowchart can be seen in the first attached image below:
Program listing:
// C++ program to implement recursive Binary Search
#include <bits/stdc++.h>
using namespace std;
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not
// present in array
return -1;
}
int main(void)
{ int n,x;
cout<<"Size of array:\n";
cin >> n;
int arr[n];
cout<<"Enter array:\n";
for (int i = 0; i < n; ++i)
{ cin >> arr[i]; }
cout<<"Enter element to be searched:\n";
cin>>x;
int result = binarySearch(arr, 0, n - 1, x);
(result == -1) ? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}
The Sample test run of the program can be seen in the second attached image below.
Time(sec) :
0
Memory(MB) :
3.3752604177856
The Output:
Size of array:4
Enter array:10 20 30 40
Enter element to be searched:40
Element is present at index 3
Conclusions:
Time Complexity:
The time complexity of Binary Search can be written as
T(n) = T(n/2) + c
The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is Theta(Logn).
Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.
