1 files changed, 59 insertions, 8 deletions

diff --git a/challenge-113/james-smith/perl/ch-1.pl b/challenge-113/james-smith/perl/ch-1.pl
index d8445c8f47..bdaa3409d9 100644
--- a/challenge-113/james-smith/perl/ch-1.pl
+++ b/challenge-113/james-smith/perl/ch-1.pl
@@ -5,31 +5,82 @@ use strict;
 use warnings;
 use feature qw(say);
 use Test::More;
-use Benchmark qw(timethis);
+use Benchmark qw(timethis cmpthese);
 
 my @ex = ( [25,8,0], [25,7,0], [24,7,1], [24,0,0], [10,0,1], [28,8,1], [26,8,1], [16,8,0], [441,9,1], [431,9,1] );
 
-is( represent( $_->[0], $_->[1]), $_->[2] ) foreach @ex;
+is( represent(          $_->[0], $_->[1]), $_->[2] ) foreach @ex;
+is( represent_unrolled( $_->[0], $_->[1]), $_->[2] ) foreach @ex;
+
+## In this challenge we make the assumption that the numbers
+## that need to be added are all different. This was not made
+## clear in the question itself - but I think that this was
+## implicit.
 
 done_testing();
+
 say '';
-timethis( 1_000_000, sub { represent($_->[0],$_->[1]) for @ex } );
+cmpthese( 1_000_000,{
+  'rolled'   => sub { represent(          $_->[0], $_->[1] ) for @ex },
+  'unrolled' => sub { represent_unrolled( $_->[0], $_->[1] ) for @ex },
+});
 say '';
+
 sub represent {
-  my($t,$n,$d) = (0,@_);
+
+  my( $t, $n, $d ) = ( 0, @_ );
+
   ## If $d is equal to 0
   ##   Any number between 100 & 109 can be represented by itself
   ##   For numbers over 109 we can represent these as 100-109 + a
   ##   number ending in 0...
+  ##   e.g. 534 / 0 = 104 + 430
+  ##
+  ##   So if $d is equal to 0 then all numbers > 100 are possible
+  ##
   ## If $n is between 10*$d and 10*$d+9 then it can be represented as $d$x
   ##   For numbers > than this we can do a similar trick to above
-  ##   We can reprent them as $d$x + a number ending in $d
-  return 1 if $n >= 10 * ($d||10);
+  ##   We can reprent them as a number ending in $d and a number
+  ##   where $d is the penultimate digit
+  ##
+  ##   e.g. 107 & 9 =   **9** +  **9**8
+  ##   e.g. 450 & 8 =  6**8** + 3**8**2
+  ##   e.g. 435 & 2 =  1**2** + 4**2**3
+  ##
+  ##   So if $d is not equal to 0 then all numbers greater than 10x$d
+  ##   are possible
+
+  return 1 if $n >= 10 * ( $d || 10 );
+
   ## Finally we get to the list of numbers less than this - as the only
   ## digit that can contain $d is the last one we just try to see if
   ## we can find a sum of numbers ending in $d which have the same last
-  ## digit as $n and less than or equal to $n.
-  $n>=($t+=$_*10+$d) && ($n%10 == $t%10) && return 1 for 0..9;
+  ## digit as $n and less than or equal to $n. Note as we have already
+  ## removed the numbers greater than 100 we now we only need to loop
+  ## up to 3 - as the next number would be 100 + 4$d....
+
+  ## Return 1 if both conditions hold true...
+
+  $n >= ( $t += $_ * 10 + $d ) &&
+    ( $n % 10 == $t % 10 ) && return 1 for 0..3;
+
+  ## Return 0 if no solution is possible...
+
   0;
 }
 
+sub represent_unrolled {
+  my( $n, $d ) = @_;
+
+## We can speed things up by a factor of 50% by unrolling the
+## for loop and so we can reduce the test to the following -
+## representing the check for the first type of solution and
+## then the 4 checks within the loop...
+
+  $n >= 10 * ( $d || 10 )                      ||
+  $n >= $d                && $n%10 == $d       ||
+  $n >= 2*$d+10           && !( ($n-2*$d)%10 ) ||
+  $n >= 3*$d+30           && !( ($n-3*$d)%10 ) ||
+  $n >= 4*$d+60           && !( ($n-4*$d)%10 ) ? 1 : 0;
+}
+