I have a few issues with this code. Or at least observations of how it
differs from the classic "SEND + MORE = MONEY" problem. see below.
>> #!perl6
>> use v6;
>> my $s; my $e; my $n; my $d; my $m; my $o; my $r;
>> my $y;
>> $s = any(0..10) & none(0);
>> $e = any(0..10);
>> $n = any(0..10);
>> $d = any(0..10);
>> $m = any(0..10) & none(0);
>> $o = any(0..10);
>> $r = any(0..10);
>> $n = any(0..10);
>> $y = any(0..10);
> I think these should be any(0..9).
Indeed they should. I will assume they are written as such in my
discussion below.
>> my $send := construct($s,$e,$n,$d);
>> my $more := construct($m,$o,$r,$e);
>> my $money := construct($m,$o,$n,$e,$y);
>> if ($send + $more == $money) {
>> say " send = $send";
>> say "+more = $more";
>> say "-------------"
>> say "money = $money";
>> }
>> sub foldl(Code &op, Any $initial, *@values) returns Any {
>> if (+@values == 0) {
>> return $initial;
>> } else {
>> return &op(shift @values, &?SUB(&op, $initial, @values));
>> }
>> }
>> sub add(Int $x, Int $y) returns Int {
>> return $x + $y;
>> }
>> sub construct(*@values) returns Junction {
>> return foldl( sub ($x, $y) { $x * 10 + $y}, 0, @values);
>> }
> How would the if (...) {...} work if there were more than one possible
> match to this equation?
As written, this generates a rather large number of solutions. I do not
see any test to make sure that the individual letters are different. Nor
is there any check to see if all the "e"'s assume the same value.
So what you reach the:
if ($send + $more == $money) {...}
stage is $send being something equivalent to any(0000..9999) &
none(0000..0999), only no where near as simplified. Similar values can
be found in $more and $money. Therefore, you're asking something like:
"Are there any two numbers between 1000 and 9999 that together total
between 10000..99999?" The answer is yes, and then you get an error as
you attempt to print a raw junction, if you're lucky, or 9000 "send"
lines, followed by 9000 "more" lines, followed by 90000 "money" lines.
Junctions are _not_ the same as unbound variables in Prolog. They do not
"widdle away" inconsistent values as those inconsistencies are found.
Most of them (all except all()) do not have to use the same value each
time they are evaluated.
If I'm wrong about this interpretation of this code, I apologize. But it
certainly fits my understanding of junctions, which has grown by leaps
and bounds over the last few weeks.
HTH,
-- Rod Adams
