Just for the sake of it, and I apologise profusely for the top posting..
Here is a summary of the posted prime calculators for testing your
results against.
>(1) Ohad
#-------------------------------------------------------
perl -e 'print"@{[grep{(1x$_)!~/^(11+?)\1+$/}2..shift||1e2]}\n"'
<number>
#-------------------------------------------------------
>(2) Joseph
#-------------------------------------------------------
my @primes = (2);
my $candidate = 3;
while(1) {
my $limit = $candidate ** 0.5;
my $has_factor = 0;
foreach (@primes) {
if ($_ > $limit) {
push @primes, $candidate;
last;
}
unless ($candidate % $_) {
$has_factor = 1;
last;
}
}
unless ($has_factor) {
print "$_ " foreach @primes;
print "\n";
}
$candidate++;
}
#-------------------------------------------------------
>(3) me (Hey, it's a Mee Too!)
#-------------------------------------------------------
$max=shift||17393;
%notprime=();
@prime=();
for $n (2..$max){
unless($notprime{$n}){
push @prime,$n;
$notprime{$_*$n}=1 for 2..$max/$n;
}
}
print "Found $#prime prime numbers..".$/;
print "@prime\n";
#-------------------------------------------------------
Brief explanation..
(1) Ohad posted an obfuscated calculator which runs the array
(2..<number>) through
grep(). The pattern in grep is '(1 x $_) !~ /^(11+?)\1+$/' which is
a tricky
way of getting perls enormously powerful regex exgine to determine
if there is
a way of dividing a string of length int($_) characters into a
number of same length
substrings. Probably memory hungry, but definitely cpu hungry for
the regexp.
Note that for any value '$x', "(1 x $x) !~ /^(11=?)\1+$/" will
return true if the
number is prime. Excellent way to perform a (quick?) once off test
against a
known number without calculating every other prime. It uses the
obscure perl
@{[ expr ]} to magically interpolate code within quotes.
(2) Joseph posted a calculator which uses the feature of prime numbers
that the
largest number you need to divide it by is the square root of itself
($_**0.5)
and that you only need to test if the number you are examining is
divisible
(ie mod==0) by another prime you have already found. Hence, very
quick and
very cpu/memory cheap. Why not test against every number? If you
test against
a non-prime number, you are simply testing against a multiple of a
prime of
course.
(3) A very simple implementation of an erogenous sieve (iud? did I say
that wrong?)
which keeps all primes in '@primes' and all 'eliminated' entries in
'%notprime'.
quick, but likely more memory intensive than (2).
Cheers.
David
--
Senior Technical Specialist
I d e n t i t y S o l u t i o n s
Level 1, 369 Camberwell Road, Melbourne, Vic 3124
Ph 03 9813 1388 Fax 03 9813 1688 Mobile 0417 595 550
Email [EMAIL PROTECTED]
> -----Original Message-----
> From: Stuart White [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, 2 March 2004 11:14 AM
> To: James Edward Gray II
> Cc: [EMAIL PROTECTED]
> Subject: Re: listing prime numbers in a range (Beginning Perl
> exercise)
>
>
> > That's not really accurate. What was said to that:
> >
> > @list = "@list";
> >
> > Creates a joined string of elements and replaces the
> > list with that one
> > item. It's perfectly reasonable to stringify an
> > array, without
> > altering the array's contents:
> >
> > print "@list\n";
> >
>
> I'm not sure I understand.
>
> >
> > That can just be if ($_ % 2), if you prefer. Zero
>
> That makes perfect sense. Thanks.
>
> > > {
> > > unshift(@primelist, $_);
> > > @primelist\n";
> >
> > The above line is not a valid Perl statement.
> >
>
> Oh, the "@primelist\n"; ? I'm not sure how that got
> there...
>
> > > Now what I'd like to do to test whether or not I
> > have
> > > a prime number is to get at each element in
> > > @primelist, and use the modulus operator on each
> > > element against each element in @list. For
> > example:
> > > $primelist[0] % $list[0]
> > > $primelist[0] % $list[1]
> > > $primelist[0] % $list[2]
> > > $primelist[1] % $list[0]
> > > $primelist[1] % $list[1]
> > > $primelist[1] % $list[2]
> > > $primelist[2] % $list[0]
> > > $primelist[2] % $list[1]
> > > $primelist[2] % $list[2]
> > >
> > > and if the result doesn't equal 0 for every test
> > > except for against itself, then I want to unshift
> > it
> > > onto @primelist.
> > >
> > > I thought I'd do this with nested foreach loops,
> > but
> > > that didn't seem to work. Any ideas?
> >
> > Define "didn't seem to work"? On second thought,
> > just post that nested
> > foreach and let us help you fix it.
> >
> > James
> >
>
> Ok, I went with for loops. As above, I want to get
> the modulus of numbers in one array by comparing the
> numbers of another array. It's not doing what I
> expect it to. I'm not sure if the first loop is
> really running either. What am I doing wrong? (Does
> that question make sense?)
>
> Also, I'm getting an error: Use of uninitialized value
> in modulus (%) at primeNumbers.pl line 40, <STDIN>
> line 1.
> Illegal modulus zero at primeNumbers.pl line 40,
> <STDIN> line 1.
> What do those errors mean exactly? Thanks.
>
>
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