### comment on

 Need Help??

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.

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

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