The ternary search algorithm locates an element in a list of increasing integers by successively splitting the list into three sub-lists of equal (or as close to equal as possible) size, and restricting the search to the appropriate piece. Specify the steps of this algorithm.

Algorithm

The steps involved in this algorithm are:

(The list must be in sorted order)

•Step 1: Divide the search space (initially, the list) in three parts (with two mid-points: mid1 and mid2)

•Step 2: The target element is compared with the edge elements that is elements at location mid1, mid2 and the end of the search space. If element matches, go to step 3 else predict in which section the target element lies. The search space is reduced to 1/3rd. If the element is not in the list, go to step 4 or to step 1.

•Step 3: Element found. Return index and exit.

•Step 4: Element not found. Exit.

Complexity

•Worst case time complexity: O(log N)

•Average case time complexity: O(log N)

•Best case time complexity: O(1)

•Space complexity: O(1)

Algorithm

The steps involved in this algorithm are:

(The list must be in sorted order)

•Step 1: Divide the search space (initially, the list) in three parts (with two mid-points: mid1 and mid2)

•Step 2: The target element is compared with the edge elements that is elements at location mid1, mid2 and the end of the search space. If element matches, go to step 3 else predict in which section the target element lies. The search space is reduced to 1/3rd. If the element is not in the list, go to step 4 or to step 1.

•Step 3: Element found. Return index and exit.

•Step 4: Element not found. Exit.

Complexity

•Worst case time complexity: O(log N)

•Average case time complexity: O(log N)

•Best case time complexity: O(1)

•Space complexity: O(1)

#include <stdio.h>

int ternarySearch(int array[], int left, int right, int x)

{

  if (right >= left) {

    int intvl = (right - left) / 3;

    int leftmid = left + intvl;

    int rightmid = leftmid + intvl;

    if (array[leftmid] == x)

       return leftmid;

    if (array[rightmid] == x)

       return rightmid;

    if (x < array[leftmid]) {

      return ternarySearch(array, left, leftmid, x);

    }

    else if (x > array[leftmid] && x < array[rightmid]) {

      return ternarySearch(array, leftmid, rightmid, x);

    }

    else {

      return ternarySearch(array, rightmid, right, x);

    }

  }

  return -1;

}

int main(void)

{

  int array[] = {1, 2, 3, 5};

  int size = sizeof(array)/ sizeof(array[0]);

  int find = 3;

  printf("Position of %d is %d\n", find, ternarySearch(array, 0, size-1, find));

  return 0;

}

Step-by-step explanation:

the answer is b.)

< 1=< 2

where are the choices?