Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

It's been a while since I've done this, and I'm embarrassed to ask but here goes: I'm trying to come up with a simple math expression to capture a numeric sequence using the "%" (modulus operator). I'm trying to capture two patterns the first being: 
2,7,12,17...

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and the second being: 
3,8,13,18...

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my thoughts are: 
[(8%3) += 5]
[(8%3) += 6]

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This works fine but I have to look at this expression for every iteration of a while loop so I need some way of retaining the value for each comparison, and it doesn't capture the first 2 or 3 values
Any thoughts?

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Re: What should be an easy math question using %
by Zaxo (Archbishop) on Sep 10, 2003 at 13:12 UTC

    Was the answer something like these? 

    my @first = map { 5 * $_ + 2 } 0..3;
my @second = map { 5 * $_ + 3 } 0..3;

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    You can solve that sort os thing with gcd* of the differences between elements, and then % of any to find the additive part.

    * greatest common divisor

    After Compline,
    Zaxo

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Re: What should be an easy math question using %
by broquaint (Abbot) on Sep 10, 2003 at 13:09 UTC
    You could use a closure 
    sub seq {
  my($v, $i) = ($_[0], $_[1] || 5);
  my $n = 0;

  return sub { $n ? ($n += $i) : ($n = $v) };
}

my $iter = seq 8 % 3, 5;

print join ', ' => map $iter->(), 1 .. 5;

print " ...\n";

$iter = seq 8 % 5;

print join ', ' => map $iter->(), 1 .. 5;

print " ...\n";

__output__

2, 7, 12, 17, 22 ...
3, 8, 13, 18, 23 ...

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    So we start out with the initial value of the first argument and then increment by the second argument or 5 by default.
    HTH

    _________
    broquaint

    update: dropped block as first argument as it was rather unnecessary

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Re: What should be an easy math question using %
by Anonymous Monk on Sep 10, 2003 at 12:47 UTC
    I think I jumped the gun, I just found a solution. Sorry to have wasted your time with this post.
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