in reply to 2 bit Math puzzle

You don't want to add 1 here: 
sub Get_Target{
   ###Gets two random primes a creates the goal.
   my $i = @primes;
   $i++;
   my $value1 = int(rand($i));
   my $value2 = int(rand($i));
   $goal = $primes[$value1] * $primes[$value2];   
}

[download]
Remove the $i++ line. With it in there, you can end up attempt to fetch $primes[@primes], which returns undef.

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

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Re: Re: 2 bit Math puzzle
by sparkyichi (Deacon) on Dec 13, 2001 at 23:54 UTC
    I do because I want to include the last prime number in the randomization. If I don't it will go from 0 to lest numb but not including last numb.

    Sparky
[reply]
      No, that's not true. 
      @a = qw( a e i o u );
# indices are 0, 1, 2, 3, 4

$x = int rand @a;
# $x is a random integer
# st 0 <= $x < @a
# where @a is 5
# thus, 0 <= $x < 5
# thus, $x is from 0 to 4

      [download]
      If you add one, you run the risk of accessing an element that doesn't exist. Trust me.

      _____________________________________________________
      Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
      s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

[reply]
[d/l]
        In the Perl Cook Book it states In chapter 2.7. Generating Random Numbers. "The rand function returns a fractional number, from (and including) 0 up to (but not including) its argument. We give it an argument of 51 to get a number that can be 0 or more, but never 51 or more." I take it that i meens if there are 25 elements to the ararry then the posible numbers are from 0 to 24.999999... the remainder get cut off.

        I see your right because the elements are 0..24 thanks Japhy. I will make the correction.

        Sparky
[reply]

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