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Getting a random sign

Posted on 2004-08-03
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Last Modified: 2006-11-17
Hi all,

I want to get a randomly generated binary choice:  either -1 or 1.  This function does it, but is there is a 'nicer' way?

  private int getRandomSign() {
    if (Math.random() <= 0.5) return -1;
      else return 1;
    }
  }
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Question by:Dawkins
  • 3
  • 2
  • 2
  • +2
10 Comments
 
LVL 14

Assisted Solution

by:sudhakar_koundinya
sudhakar_koundinya earned 100 total points
ID: 11706737
That is the nicer way :)
0
 
LVL 36

Expert Comment

by:girionis
ID: 11707086
Can you define "nicer"? Your code looks fine.
0
 
LVL 36

Accepted Solution

by:
girionis earned 100 total points
ID: 11707122
Maybe you could do

 private int getRandomSign() {
    if (Math.random() < 0.5) return -1;
      else return 1;
    }
  }

since it can return a value equal to zero but never one.
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LVL 14

Expert Comment

by:sudhakar_koundinya
ID: 11707291
I haven't understand the difference between if (Math.random() < 0.5) return -1; and if (Math.random() <= 0.5) return -1;

both results gives the value less than one only.

Regards
Sudhakar
0
 

Author Comment

by:Dawkins
ID: 11707310
>>>  Maybe you could do...

Good point, it should be < rather than <=. I'll change that

>>> Can you define "nicer"? Your code looks fine.

well basically I am teaching myself how to program in Java and know no Java programers in real life so have no one to ask "is this completely ridiculous", so I come here instead :)  By nicer I mean "more elegant, simpler, less code, there's already a java library function that does that" etc

Seems like I've already found the best way so I'll give it another 30 minutes to make sure and then give you two the points.
0
 
LVL 6

Expert Comment

by:Roger Neyman
ID: 11707336

Here's another approach that may be closer to what you want:

private int getRandomSigh() {
    return (int) (Math.signum(0.5-Math.random())) ;
}

An additional note:  using "<= 0.5" favors -1  VERY slightly  since Math.random() returns values between 0 and .999...
0
 
LVL 6

Expert Comment

by:Roger Neyman
ID: 11707408
There's a problem with my posted solution: there's a very slight chance it will return 0. If you can't live with that, here's an amendment:


private int getRandomSign() {
    int sign = (int) (Math.signum(0.5-Math.random())) ;
    return sign = 0 ? 1 : sign ;
}
0
 
LVL 15

Expert Comment

by:JakobA
ID: 11707712
nice & short:

private int getRandomSign() {
    return (Math.random()<0.5) ? 1 : -1;                           // ?: operator same as in C
}

regards JakobA
0
 

Author Comment

by:Dawkins
ID: 11708880
Thanks everyone, I'll keep it as I originally had it but put < 0.5 instead to ensure the split is exactly 50/50
0
 
LVL 36

Expert Comment

by:girionis
ID: 11712677
:)
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