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I had my own idea of how to do it, and I came up with restartable iterators:
#!/usr/bin/perl use strict; use warnings; sub restartable_iter { my ($start, $end) = @_; sub { $start = shift if @_; return if $start > $end; $start++; } } sub choose_m_of_n_iter { my ($m, $n) = @_; my @iter; for my $i (0..($m-1)) { push @iter, restartable_iter($i, $n-$m+$i); } join_iter(@iter); } sub join_iter { my $it = shift; while ( my $tmp = shift ) { $it = append_iter( $it, $tmp ); } $it; } sub append_iter { my ($it1, $it2) = @_; my (@ret1, @ret2) = $it1->(); sub { return @ret1, @ret2 if @ret2 = $it2->(); return unless @ret1 = $it1->(); return @ret1, $it2->($ret1[-1]+1); } } my $iter = choose_m_of_n_iter(4, 10); while (my @arr = $iter->()) { print "@arr\n"; }

In the append iterator, when the second iterator is exhausted, it causes the first iterator to iterate, and then restarts the second iterator at the appropriate starting point.

Update:Unlike the other solutions, the solution above returns an array of indices instead of a set of elements from a list, but that's easy to adjust:

sub choose_n { my $n = shift; my @set = @_; my $iter = choose_m_of_n_iter($n, scalar(@set)); sub { @set[$iter->()]; } }

Update: Simplified code. Which may or may not be a good thing :-)


In reply to Re: Recursively-generated Iterators by runrig
in thread Recursively-generated Iterators by Roy Johnson

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