Java Ternary Search

Question

I am trying to perfrom a ternary search on a array of strings. I have got most of code down and I think I am going on the right track but can't seem to get any results other then -1. Below is the code that I have genearated thus far. I know the problem is in the search algorithm. I do not want to use any built is as I am learning.

public static void main(String[] args) {
    //declare a string array with initial size
    String[] songs =  {"Ace", "Space", "Diamond"};
    System.out.println("\nTest binary (String):");
    System.out.println(search(songs,"Ace"));

}  

public static int search(String[] x, String target) {     

   int start=0, end=x.length;

   while (start > end) {
      int midpoint1 = start+(end - start)/3;
      int midpoint2 = start +2*(end-start)/3;
      if ( target.compareTo(x[midpoint1]) == 0 ) 
          return midpoint1;
      else if ( target.compareTo(x[midpoint2]) == 0 ) 
          return midpoint2;   
      else if ( target.compareTo(x[midpoint1]) < 0 ) 
          return end = midpoint1-1;
      else if ( target.compareTo(x[midpoint2]) > 0 ) 
          return start = midpoint2+1;
  }

   return -1;
}

You should learn how to debug these kinds of problems on your own. You'll get stuck with things like this all the time, especially in the beginning. In this case a few prints would've sufficed. — keyser, Apr 20 '15 at 20:24

score 3 · Answer 1 · answered Apr 20 '15 at 20:18

3

You never get into the loop.

int start=0, end=x.length;
while (start > end)

answered Apr 20 '15 at 20:18

Thomas

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Can you expanded more please. – M Lill Apr 20 '15 at 20:24
x.length is always 0 or larger, therefore `start`(0) can never be larger than `end` – Thomas Apr 20 '15 at 20:25
Thank you very much. I can't belive I over looked that – M Lill Apr 20 '15 at 20:40

score 1 · Answer 2 · answered Apr 20 '15 at 20:22

1

You're while statement is wrong, it should contain start < end. I recomend learning the debug settings that are on most IDEs because if you're going to stick around it makes it so much easier to view the states of vars.

answered Apr 20 '15 at 20:22

Feek

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xerx593 · Answer 3 · 2017-05-18T15:55:35.770

1

Try this corrected version, additionally to the bug identified by the Thomas and user2789574, you also have a bug in the recursion:

public static void main(String[] args) {
    //declare a string array with initial size

    String[] songs = {"Ace", "Space", "Diamond"};

    System.out.println("\nTest binary (String):");
    System.out.println(search(songs, "Ace", 0, 3));

}

public static int search(String[] x, String target, int start, int end) {

    if (start < end) {
        int midpoint1 = start + (end - start) / 3;
        int midpoint2 = start + 2 * (end - start) / 3;
        if (target.compareTo(x[midpoint1]) == 0) {
            return midpoint1;
        } else if (target.compareTo(x[midpoint2]) == 0) {
            return midpoint2;
        } else if (x[midpoint1].compareTo(x[midpoint2]) < 0) {
            return search(x, target, midpoint1, end);
        } else {
            return search(x, target, start, midpoint2);
        }
    }

    return -1;
}

edited May 18 '17 at 15:55

answered Apr 20 '15 at 20:27

xerx593

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1

Thanks for showing my what I was doing wrong. I have an array larger then the one that I showed that contains 50 strings, I know to change "Ace", 0, 3 to Ace, 0 , 50 but when I run seach another string within the array it just hangs up – M Lill Apr 20 '15 at 20:43
"A ternary search algorithm is a technique in computer science for finding the **minimum or maximum** of an **increasing or decreasing** function", (from wikipedia), of course this raises problems on "unsorted" data. – xerx593 Apr 20 '15 at 20:58
What you most likely want to implement is not a ternary search, but maybe a [Lookup in a ternary Tree](http://en.wikipedia.org/wiki/Ternary_search_tree#Look_up) ...for this you have to convert your String[] into another structure (or at least ensure a distinct structure/ordder...it is possible to store a k-tree as an array) – xerx593 Apr 20 '15 at 21:04
..this is nor trivial, nor very efficient, the most effecient datastructure for your use case would be the `HashSet`(/`table`)! – xerx593 Apr 20 '15 at 21:25
Also the `while` statement makes no sense if you are going to return anyway. It has some overhead on the `if` statement – Sam Segers May 12 '17 at 20:31

score 0 · Answer 4 · answered Jan 23 '17 at 22:47

0

end = x.length,will always return numeric value greater than zero for not null string and it's comparison with start =0 , it will never enter into the loop.

answered Jan 23 '17 at 22:47

vipin

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score 0 · Answer 5 · answered Nov 02 '20 at 07:45

I am not sure if its still relevant now, but here is the ternary search done with loops in java(assuming that the array is pre-sorted):

  public static int ternSearch(String[] x, String target) {
        
        if((target.toLowerCase()).compareTo(x[0].toLowerCase()) == 0)
        {
            return 0;
        }
        if((target.toLowerCase()).compareTo(x[x.length - 1].toLowerCase()) == 0)
        {
            return x.length-1;
        }
        int mid1 = (int) Math.ceil( x.length/3);
        if((target.toLowerCase()).compareTo(x[mid1].toLowerCase()) == 0)
        {
            return mid1;
        }
        int mid2 = (int) Math.ceil( x.length*2/3);
        if((target.toLowerCase()).compareTo(x[mid2].toLowerCase()) == 0)
        {
            return mid2;
        }
        if((target.toLowerCase()).compareTo(x[0].toLowerCase()) > 0&& (target.toLowerCase()).compareTo(x[mid1].toLowerCase()) < 0)
        {
            for (int i = 1; i < mid1; i++)
            {
                if((target.toLowerCase()).compareTo(x[i].toLowerCase()) == 0) 
                {
                    return i;
                }
            }
        }
        else if((target.toLowerCase()).compareTo(x[mid1].toLowerCase()) > 0&& (target.toLowerCase()).compareTo(x[mid2].toLowerCase()) < 0)
        {
            for (int i = mid1+1; i < mid2; i++)
            {
                if((target.toLowerCase()).compareTo(x[i].toLowerCase()) == 0) 
                {
                    return i;
                }
            }
        }
        else if((target.toLowerCase()).compareTo(x[mid2].toLowerCase()) > 0&& (target.toLowerCase()).compareTo(x[x.length-1].toLowerCase()) < 0)
        {
            for (int i = mid2+1; i < x.length-2; i++)
            {
                if((target.toLowerCase()).compareTo(x[i].toLowerCase()) == 0) 
                {
                    return i;
                }
            }
        }
        return -1;
    }

Java Ternary Search

5 Answers5