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Re: Hamming Sequences and Lazy Lists

by Roy Johnson (Monsignor)
on Mar 17, 2005 at 16:40 UTC ( #440428=note: print w/replies, xml ) Need Help??

in reply to Hamming Sequences and Lazy Lists

Here is a lazy-list example of merge, with one of its arguments being a finite list, and the other an infinite list. I may not get around to figuring out the Hamming generator, but I think the techniques I've employed here tell you how it can be done.
use strict; use warnings; # Merge takes two iterators, which can be called # with no arguments, in which case they iterate, returning # their next value; or with a string 'peek', which will # yield the next value without effectively shifting it. # Exhausted iterators return undef on all subsequent calls # # Merge itself is an iterator, returning the next element in # the merged series, plus a continuation coderef sub merge { my ($a, $b) = @_; my ($car_a, $car_b) = ($a->('peek'), $b->('peek')); # Base case: if one of them is empty, return the other defined $car_a or return ($b->(), sub {merge($a, $b)}); defined $car_b or return ($a->(), sub {merge($a, $b)}); # Pull off lesser (both if equal) first element(s) my $low_car; if ($car_a <= $car_b) { $low_car = $a->() } if ($car_b <= $car_a) { $low_car = $b->() } return ($low_car, sub {merge($a, $b)} ); } my $I1 = do { my @arr = (1..10); sub { if (@arr == 0) { return undef; } elsif (@_) { # should be peek, but any other arg works (undocu +mented) return $arr[0]; } else { return shift @arr; } } }; my $I2 = do { my $i = 2; sub { if (@_) { return $i; } else { my $i_was = $i; $i += 2; return $i_was; } } }; my $iterations_to_print = 30; for ( my ($elem, $cont) = merge($I1, $I2); $iterations_to_print-- > 0; ($elem, $cont) = $cont->() ) { print "<$elem>\n"; }

Caution: Contents may have been coded under pressure.

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Re^2: Hamming Sequences and Lazy Lists
by Limbic~Region (Chancellor) on Mar 17, 2005 at 16:55 UTC
    Roy Johnson,
    I don't think this is right. It is my understanding that the results should be the multiples of the factors merged minus duplicates. If you change $iterations_to_print = 23 and @arr = (2,3,5), where is 15 in the results for instance? See Re: Hamming Sequences and Lazy Lists for my understanding and implementation of the problem.

    Cheers - L~R

      I did not implement Hamming here; only merging of two list iterators. But I have just posted a lazy Hamming solution.

      Caution: Contents may have been coded under pressure.
        Roy Johnson,
        Ok - my mistake. In fact, I completely missed the boat on the entire problem. I was so focused on lazy evaluation that I misunderstood "how do you generate the series of numbers composed of a given list of prime factors, where each can be used an unlimited number of times?" To me, that any positive multiple of any factor was valid.

        Cheers - L~R

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