Why is Binary Search preferred over Ternary Search?

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The following is a simple recursive Binary Search function in C++.

// 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;
}

The following is a simple recursive Ternary Search function in C++.

// A recursive ternary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
int ternarySearch(int arr[], int l, int r, int x)
{
   if (r >= l)
   {
        int mid1 = l + (r - l)/3;
        int mid2 = mid1 + (r - l)/3;
        // If x is present at the mid1
        if (arr[mid1] == x)  return mid1;
        // If x is present at the mid2
        if (arr[mid2] == x)  return mid2;
        // If x is present in left one-third
        if (arr[mid1] > x) return ternarySearch(arr, l, mid1-1, x);
        // If x is present in right one-third
        if (arr[mid2] < x) return ternarySearch(arr, mid2+1, r, x);
        // If x is present in middle one-third
        return ternarySearch(arr, mid1+1, mid2-1, x);
   }
   // We reach here when element is not present in array
   return -1;
}

Which of the above two does less comparisons in worst case?
From the first look, it seems the ternary search does less number of comparisons as it makes Log3n recursive calls, but binary search makes Log2n recursive calls. Let us take a closer look.
The following is recursive formula for counting comparisons in worst case of Binary Search.

   T(n) = T(n/2) + 2,  T(1) = 1

The following is recursive formula for counting comparisons in worst case of Ternary Search.

   T(n) = T(n/3) + 4, T(1) = 1

In binary search, there are 2Log2n + 1 comparisons in worst case. In ternary search, there are 4Log3n + 1 comparisons in worst case.

Time Complexity for Binary search = 2clog2n + O(1)
Time Complexity for Ternary search = 4clog3n + O(1)

Therefore, the comparison of Ternary and Binary Searches boils down the comparison of expressions 2Log3n and Log2n . The value of 2Log3n can be written as (2 / Log23) * Log2n . Since the value of (2 / Log23) is more than one, Ternary Search does more comparisons than Binary Search in worst case.

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source and credits.
source and credits: http://www.geeksforgeeks.org
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rakesh

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