Here's another iterator solution that avoids the recursion altogether. There are three differences:

- It will return undef instead of an empty array ref if the number of items is less than the amount to choose (ie 3 choose 5).
- The amount to choose is passed before the list of items instead of afterwards because I thought it was weird, although your way does make sense if one thinks of it like "Here are 50 things, choose 5".
- The output order is a bit different.

All three of these can certainly be fixed, but I wanted to keep the code clear.

`sub choose_n_iter {
my @todo = [ shift, [ @_ ], [] ];
return sub {
while ( @todo ) {
my ( $n, $pool, $tally ) = @{ shift @todo };
return $tally if $n == 0;
next if @$pool == 0;
my ( $first, @rest ) = @$pool;
push @todo, [ $n - 1, \@rest, [ @$tally, $first ]],
[ $n , \@rest, $tally ];
}
return;
};
}
`

And here is a more effecient version:

`sub choose_n_iter {
my @todo = [ shift, [ @_ ], [] ];
return sub {
while ( @todo ) {
my ( $n, $pool, $tally ) = @{ shift @todo };
return $tally if $n == 0;
return [ @$tally, @$pool ] if $n == @$pool;
next if @$pool == 0;
my ( $first, @rest ) = @$pool;
push @todo, [ $n - 1, \@rest, [ @$tally, $first ]];
push @todo, [ $n , \@rest, $tally ] if @res
+t;
}
return;
};
}
`

Thanks goes out to

HOP for the recursion-to-iterator help.

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