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Randomise An Array of integers

Posted on 2002-03-11
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Last Modified: 2010-03-31
Hi, My problem is that, first of all I have an array that contains integer values. Now, what I want is to shuffle the contents in a random manner.

I tried using the random class and failed by far cause it appears that it allows a certain range that it can select from. I dont need to select any range. I just want to randomise the values that I already have in the array. Hope it is clear and understandable.

Thanx...!!
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Question by:q_bic
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9 Comments
 
LVL 9

Accepted Solution

by:
Venci75 earned 300 total points
ID: 6855726
then - try this:
for (int i = 0; i < arr.length; i++) {
  j = (new java.util.Random()).nextInt(); // random position
  int buff = arr[i];
  arr[i] = arr[j];
  arr[j] = buff;
}
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LVL 28

Expert Comment

by:rrz
ID: 6855744
How about?

import java.io.*;
import java.util.*;
public class Tryrandom
{
public static void main(String args[])
{
 Object numbers[] = {"0","1","2","3","4","5","6","7"};
 ArrayList as = new ArrayList(Arrays.asList(numbers));
 Collections.shuffle(as);
 numbers = as.toArray();
 for (int i=0;i<8;i++)
        System.out.println("result["+i+"]= "+ numbers[i]);
}
}
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LVL 18

Expert Comment

by:bobbit31
ID: 6855745
something like this would work:

          Vector v1 = new Vector();

          for (int i=0;i<100;i++) {
               v1.add(new Integer(i));
          }
         
          Collections.shuffle(v1);
         
          for (int i=0;i<100;i++) {
               out.println(v1.elementAt(i));
          }

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LVL 18

Expert Comment

by:bobbit31
ID: 6855746
damn, too slow ;(
0
 
LVL 7

Expert Comment

by:Igor Bazarny
ID: 6855753
Hi,

> j = (new java.util.Random()).nextInt();
There is no need to create new Random instance each time. And don't let index to go out of range:

Before loop: Random random = new java.util.Random();

Inside loop: j = random.nextInt() % arr.length;

Igor
0
 
LVL 6

Expert Comment

by:gadio
ID: 6855968
Heres an option. This is pseudo code:

java.util.Random r = new java.util.Random();
Object numbers[] = {"0","1","2","3","4","5","6","7"};
for( int cnt = 0; cnt < number.length; cnt++ ) {
   int index = (int)((numbers.length - cnt - 1)* r.nextDouble() + 0.5);
   // do the swap
   int tmp = numbers[cnt];
   numbers[cnt] = numbers[cnt+1+index];
   numbers[cnt+1+index] = tmp;
}

This promis you the following things:
1. Time complexity of o(n).
2. Space complexity of o(n).
3. You will touch each cell of the array at least once.
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LVL 9

Expert Comment

by:Venci75
ID: 6856005
bazarny -> yes - you are right - I forgot this:
j = random.nextInt() % arr.length;
but regarding the new Random() - I used it only for demonstration.

rrz - great suggestion!
0
 

Author Comment

by:q_bic
ID: 6859999
thanx for the great Idea.
0
 
LVL 6

Expert Comment

by:gadio
ID: 6863845
I am sorry to say that after you had accepted the answer, however, the solution by Venci75 is wrong in my opinion.

This:
j = random.nextInt() % arr.length;

is not so good since you may touch any cell, also the ones that you already shuffled. The probability that you will undo shuffles is not zero. This is why in my solution I used:

int index = (int)((numbers.length - cnt - 1)* r.nextDouble() + 0.5);
 
And the range is getting smaller each time - to the ones that were not touched yet.
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