141 add more performant method; still buggy

1 files changed, 50 insertions, 21 deletions

diff --git a/challenge-141/jake/perl/ch-1.pl b/challenge-141/jake/perl/ch-1.pl
index e03638a082..ce8c9da777 100755
--- a/challenge-141/jake/perl/ch-1.pl
+++ b/challenge-141/jake/perl/ch-1.pl
@@ -6,11 +6,12 @@ use feature 'say';
 
 # number attributes
 # num specifies the lowest number to start analysis
-# div_required specifies for how many divisors we aiming at
+# div_required specifies for how many divisors we're aiming at
 # output specifies how many numbers will be displayed
+# div_required and output_required can be adjusted as desired
 my $num_atr = {
     num => '1',
-    div_required => '8',
+    div_required => '6',
     output_required => '10'
 };
 
@@ -18,39 +19,67 @@ my $num_atr = {
 my $res = collect_numbers();
 say join (' ', @$res);
 
-
-
 # aggregate required amount of numbers according to output_required
 sub collect_numbers {
     my @result = ();
 
     while ( $#result+1 < $num_atr->{output_required} ) {
-        push @result, count_divisors( $num_atr, 0, 0 );
+        push @result, count_divisors( $num_atr, 1 );
         $num_atr->{num}++;
     }
     return \@result;
 }
 
+# starting at 1 we aggregate all divisors in an array
+# we also aggregate all corresponding reciprocal values in another array
+# if these arrays are symmetric to each other at the position of half our's div_required, we know they have the required amount of divisors
+# p.e. num = 24; div_required = 8; the divisor arrays are cross-symmetric at div_required / 2
+#         <==\\==>
+# 24 12  8  6\\4  3  2  1
+#  1  2  3  4\\6  8 12 24
+#         <==\\==>
 
 sub count_divisors {
-    my ( $spec_num, $div_cntr, $subtractor) = @_;
+    my ( $spec_num, $divisor) = @_;
 
-# divide num through all numbers <= num and count every time modulo is 0
-# each time modulo is 0 we know it's a divisor
-    while ( $subtractor != $num_atr->{num} ) {
-        $div_cntr++ if $num_atr->{num} % ( $num_atr->{num} - $subtractor ) == 0;
-        $subtractor++;
-    }
+    # each iteration needs a fresh set of arrays
+    my @divisors = ();
+    my @reciprocal = ();
 
-# only return to sub collect_numbers if we hit the required amount of divisors
-    if ( $div_cntr == $num_atr->{div_required} ) {
-        return $num_atr->{num};
-    }
+    # in case our divisor gets as large as our number, we stop aggregating our arrays
+    # because then we have arrived at the largest and smallest divisors
+    while ( $num_atr->{num} / $divisor >= 1 ) {
+        
+        # finding our divisors
+        if ( $num_atr->{num} % $divisor == 0 ) {
+            push @divisors, $divisor;
+            push @reciprocal, ( $num_atr->{num} / $divisor );
+        }
 
-# when ever the current num did not meet the required divisors we need to start over counting them for the next num
-# in this case we must not return num to sub collect_numbers
-    else {
-        $num_atr->{num}++;
-        return count_divisors( $num_atr, 0, 0 );
+        # if div_required is even
+        if ( $num_atr->{div_required} % 2 == 0 ) {
+            return $num_atr->{num} if $divisors[$num_atr->{div_required} / 2] && $divisors[$num_atr->{div_required} / 2] == $reciprocal[$num_atr->{div_required} / 2 - 1];
+        }
+
+        # if div_required is odd; CAPUTTED: ref massacre
+        #if ( $num_atr->{div_required} % 2 == 1 ) {
+        #    return $num_atr->{num} if $divisors[int($num_atr->{div_required} / 2) + 1]
+        #                            && $divisors[int($num_atr->{div_required} / 2) + 1] == $reciprocal[int($num_atr->{div_required} / 2) + 1];
+        #}
+        if ( $num_atr->{div_required} % 2 == 1 ) {
+            return $num_atr->{num} if $divisors[3]
+                                    && $divisors[3] == $reciprocal[3];
+        }
+        #say ("line $num_atr->{div_required}");
+        #say[int($num_atr->{div_required} / 2) + 1];
+        #say $divisors[int($num_atr->{div_required} / 2) + 1];
+        #say $reciprocal[int($num_atr->{div_required} / 2) + 1];
+        #die if $num_atr->{num}==15;
+        $divisor++;
     }
+
+    # when ever the current num did not meet the required divisors we need to start over counting them for the next num
+    # in this case we must not return num to sub collect_numbers
+    $num_atr->{num}++;
+    return count_divisors( $num_atr, 1 );
 }
\ No newline at end of file