Re: Challenge: Another Infinite Lazy List

[kvale]

Here is a simple iterative solution:

my @factors = (2,3,5);  # Assmue increasing sequence
my $seq_len = 10;

my $index = 0;
foreach my $num (1..$seq_len*$factors[0]) {
   foreach my $factor (@factors) {
      unless ($num % $factor) {
         print "$num ";
         $index++;
         last;
      }
   }
   last if $index >= $seq_len;
}

[tall_man]

If you want a pure iterative solution, you could compute a factor wheel and print everything but the usual trial divisors. The code looks messier, but it requires no mod calculations in the loop, just adds. I computed the wheel using Math::Big::Factors (so for people following the original articles, we have come full circle...).
Read more... (1117 Bytes)

[tlm]

I am confused. Where is the "infinite lazy list"?

In the implementation I posted, a variable like

my $lazy235 = challenge( 2, 3, 5 );

is, effectively, an infinite lazy list. One can repeatedly extract an arbitrarily long prefix from it, using take, or repeatedly find its n-th element with something like:

sub nth {
  my $ll = shift;
  my $n = shift;
  die "Invalid argument: $n\n" if $n < 0;
  return $n == 0 ? head( $ll ) : nth( tail( $ll ), $n - 1 );
}

For example, the following:

my @first_50 = take( 50, $lazy235 );
my @again    = take( 50, $lazy235 );

print "@first_50\n\n@again\n\n";

for ( map int( rand( 1000 ) ), ( 1 .. 5 ) ) {
  printf "%3d %d\n", $_, nth( $lazy235, $_ );
}

prints out something like

0 2 3 4 5 6 8 9 10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 
[phone] 45 46 48 50 51 52 54 55 56 57 58 60 62 63 64 65 
+66

0 2 3 4 5 6 8 9 10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 
[phone] 45 46 48 50 51 52 54 55 56 57 58 60 62 63 64 65 
+66

347 474
857 1168
530 723
994 1355
579 789

[tall_man]

My reply was not meant to be an "infinite lazy list". It was a reply to another poster who had an interative solution, giving another way to do it. I perfer the lazy list solutions myself.