Improved solution #2

1 files changed, 24 insertions, 8 deletions

diff --git a/challenge-046/markus-holzer/raku/ch-2.raku b/challenge-046/markus-holzer/raku/ch-2.raku
index 1fac6c7585..847eca2989 100644
--- a/challenge-046/markus-holzer/raku/ch-2.raku
+++ b/challenge-046/markus-holzer/raku/ch-2.raku
@@ -1,12 +1,28 @@
-say "Open rooms: \n", (1..^500).grep({ 
-    is-open( $_ ) 
-}).join(",");
+# Each door will only be visited by employees whos number is a divisor
+# of the room number. For example room number 12 will be visited
+# by the employees 1, 2, 3, 4, 6 and 12. That is 6 visits. As each
+# pair of visits cancels itself out, the only rooms that will be open
+# at the end are the ones with a number that has an ODD number of divisors.
+# From that we can already tell, doors with prime numbers will never be open
+# because a prime number always has only 2 divisors.
+# 
+# Now divisors always come in pairs, in the case of the 12 these are
+# (1, 12), (2,6) and (3,4) as each of the pairs multiply out to 12.
+#
+# In the case of a square number however, there is always one pair for which both elements are the same.
+# 16 for example, has the divisor pairs are (1,16), (2, 8) and (4,4). 
+# This last pair contains the same number twice. 
+# And that is what makes the total number of divisors odd
+# And that is what tells us the door 16 will be open.
+#
+# Knowing all that we can solve by 
 
-sub is-open( $i )
+say "Open rooms: \n", (1..^500).grep({ .&is-open }) .join(", ");
+
+sub is-open( $room )
 {
-    my $is-open = True;
-    $is-open = !$is-open if $i %% $_ for 2 .. 500;
-    $is-open;
+    my $sqrt = $room.sqrt;
+    $sqrt == $sqrt.Int
 }
 
-# Open rooms: 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484
+# Open rooms: 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484
\ No newline at end of file