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Re: Hamming Sequences and Lazy Lists

by kvale (Monsignor)
on Mar 17, 2005 at 08:56 UTC ( #440306=note: print w/ replies, xml ) Need Help??


in reply to Hamming Sequences and Lazy Lists

Here is a solution that is not lazy and not quite as efficient as the Haskell version, but is simple:

use Algorithm::Loops qw(NestedLoops); use List::Util qw(reduce); my @factors = (2,3,5); # Assmue increasing sequence my $seq_len = 10; my $depth = int( $seq_len ** (1/@factors)) + 1; my @seq; my @list= NestedLoops( [ ( [ 0..$depth ] ) x @factors ], sub { push @seq, reduce {$a * $b} map {$factors[$_]**$_[$_]} 0..$#_;}, ); my @sorted = sort {$a <=> $b} @seq; print "$sorted[$_] " foreach 0..$seq_len-1;
Update: Improved the depth bound.

-Mark


Comment on Re: Hamming Sequences and Lazy Lists
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Re^2: Hamming Sequences and Lazy Lists
by Limbic~Region (Chancellor) on Mar 17, 2005 at 16:48 UTC
    kvale,
    I don't think this is right. Shouldn't the result include all factors of all 3 lists merged minus duplicates? If you change $seq_len = 23 for instance, why is 16 for instance missing from the results? See Re: Hamming Sequences and Lazy Lists for my understanding and implementation of the problem.

    Cheers - L~R

      Given the OP's reference to the factors of a composite number thread, I interpreted 'use an unlimited number of times" to mean create numbers of the form
      2**$i * 3**$j * 5**$k
      with $i, $j, and $k as integers >= 0. The program I wrote generalizes this by handling an arbitrary number of arbitrary factors.

      I don't know what you mean by 'factors of 3 lists', but if I guess that each list is a multiple of each factor, then I think that must not be right. The example given by the OP had 1 as the first member, but 1 is not any multiple 2, 3, or 5.

      That said, there is an error my program :) 16 should be in the list even in my understanding of the problem. The mistake with is that the bound on the $depth I set was too low. In the harsh light of the morning, a safe bound is

      my $depth = $seq_len;
      But I am sure this bound can be made tighter, right after I have some tea :)

      -Mark

        kvale,
        I understand the need for caffeine. I plan on cleaning my code up too. Here is a visual representation of my understanding of the Hamming sequence.
        2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, ... 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ... 5 = 5, 10, 15, 20, 25, 30, ... H = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 16, 18, 20, 21, 22, 24, 25, 26 +, 27, 28, 30, ...

        Cheers - L~R

        Update: This is wrong. 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 meant any positive multiple of any factor was valid.

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