Discipulus teased me to try solving the problem using the
Marpa parser. It took me a lot more time than I had thought: I didn't find a way how to retrieve the "depth" in a parse tree in order to multiply by it. You can attach 0 as the depth to each number, though, and increment it on each its inclusion into a list:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
use Marpa::R2;
use List::Util qw{ sum };
my $dsl = << '__DSL__';
lexeme default = latm => 1
:default ::= action => ::first
Top ::= List action => resolve
List ::= ('{') Terms ('}')
Terms ::= Term+ separator => comma action => include
Term ::= num action => create
| List
num ~ [0-9]+
comma ~ ','
__DSL__
sub create { [ [ $_[1], 0 ] ] }
sub include { shift; [ map { map [$_->[0], 1 + $_->[1]], @$_ } @_ ] }
+ # do you still love me?
sub resolve { shift; sum map $_->[0] * $_->[1], @{ $_[0] } }
my $grammar = 'Marpa::R2::Scanless::G'->new({ source => \$dsl });
my @strings = ( '{{1,1},2,{1,1}}', # the functio
+n should return 10
'{1,{4,{6}}}', # the functio
+n should return 27
'{{1,1,1,1},{1,1,{2,2}},{1,{2,{3,3}}}}' # Buk 56
);
for my $s (@strings) {
my $v = $grammar->parse(\$s, 'main');
say $$v;
}
($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,