Re: Math.random()

Lew wrote:
maya wrote:
oh brother... I think get gist of what you're saying.. will try it....;) thank you very much.. (boolean array.... hmmm.. not sure if I've ever even seen code for a boolean array...;)

package testit;
import java.util.Random;
import java.util.Set;
import java.util.HashSet;

public class Scrambler
{
private final Random rand = new Random();

public Set <Integer> scrambleEggs( int upper, int count )
{
Set <Integer> chosen = new HashSet <Integer> ( count * 4 / 3 + 1 );
boolean [] selected = new boolean [ upper ];

while ( count > 0 )
{
int indx = rand.nextInt( upper );
if ( ! selected [ indx ] )
{
if ( chosen.add( indx ))
{
--count;
}
selected [ indx ] = true;
}
}
return chosen;
}
}

Untried, untested. (Pseudo-)Non-deterministic run time.

The bizarre reference to chosen.add()'s return value is a hint that one doesn't really need the boolean array if one is using a Set, in that Set guarantees uniqueness of its values with respect to equals(), and add() tells you if the item was not already in the Set. This could really help if 'upper' has a large value.

Things get trickier if you want multi-threaded use or a deterministic number of times through the loop.


Why have the boolean array at all? The technique only makes sense if
upper is very large compared with count, so that the collision risk is
very low, and in that case it is bad to have to create an array with
upper elements. Here's a simplified, though still untested, version.

package testit;
import java.util.Random;
import java.util.Set;
import java.util.HashSet;

public class Scrambler
{
private final Random rand = new Random();

public Set <Integer> scrambleEggs( int upper, int count )
{
Set <Integer> chosen = new HashSet <Integer> ( count * 4 / 3 + 1 );

while ( chosen.size() < count )
{
chosen.add(Integer.valueOf(rand.nextInt( upper ));
}
return chosen;
}
}

Testing chosen.size(), rather than counting add attempts, automatically
deals with duplicates, because adding a duplicate does not increase the
size of the Set.

Patricia
.

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