# Why is Binary Search preferred over Ternary Search?

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

 // 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 :

 // 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; }

 import java.io.*;    class GFG { public static void main (String[] args)  {        // A recursive ternary search function.  // It returns location of x in given array // arr[l..r] is present, otherwise -1 static 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; } }

 # A recursive ternary search function. It returns location of x in # given array arr[l..r] is present, otherwise -1 def ternarySearch(arr, l, r, x):     if (r >= l):         mid1 = l + (r - l)//3         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       # This code is contributed by ankush_953

 = \$l)     {         \$mid1 = \$l + (\$r - \$l) / 3;         \$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; }    // This code is contributed by anuj_67 ?>

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.

Exercise:
Why Merge Sort divides input array in two halves, why not in three or more parts?