Hello, o wise ones.

Disclaimer in advance: This is homework. The task is to determine whether a list is a sublist of another (sub in_list()). Order matters, these are real lists, not sets. I coded a solution, see below.

Can you think of any other approaches, possibly simpler and without using LCS? 

use Test::More tests => 6;
use List::MoreUtils qw(each_arrayref);
use Algorithm::Diff qw(LCS);

sub _identic {
    # Returns true is two lists are identic.
    my ($lsa, $lsb) = @_;
    return unless scalar @{$lsa} == scalar @{$lsb};    # bail if unequ
+al length
    my $iterator = each_arrayref($lsa, $lsb);
    while (my ($first, $second) = $iterator->()) {
        return unless $first eq $second;
        # bail if two elements at the same position differ
    }

    # at here, all elements are pairwise identic
    return 1;
}

sub in_list {
    # Returns true if lsa is a proper subsequence of lsb
    my ($lsa, $lsb) = @_;
    my $LCS = LCS($lsa, $lsb);
    return _identic($LCS, $lsa);
}

my (@lsa, @lsb);

@lsa = (1, 2, 3);
@lsb = (1, 2, 3, 4, 5);
ok in_list(\@lsa, \@lsb), 'partial list at start';

@lsa = (1, 2, 3);
@lsb = (2, 1, 2, 3);
ok in_list(\@lsa, \@lsb), 'partial list at end';

@lsa = (1, 2, 3);
@lsb = (3, 2, 1);
ok !in_list(\@lsa, \@lsb), 'same elements, but nothing in common due t
+o order';

@lsa = ();
@lsb = ();
ok in_list(\@lsa, \@lsb), 'null list, identic';

@lsa = (1);
@lsb = (2);
ok !in_list(\@lsa, \@lsb), 'single element, different';

@lsa = (1);
@lsb = (1);
ok in_list(\@lsa, \@lsb), 'single element, identic';

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In reply to sublist of a list by Anonymous Monk

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