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| -rw-r--r-- | challenge-114/james-smith/perl/ch-2.pl | 22 |
1 files changed, 22 insertions, 0 deletions
diff --git a/challenge-114/james-smith/perl/ch-2.pl b/challenge-114/james-smith/perl/ch-2.pl index 4f46a237a0..773f79b0cb 100644 --- a/challenge-114/james-smith/perl/ch-2.pl +++ b/challenge-114/james-smith/perl/ch-2.pl @@ -37,10 +37,32 @@ sub next_bin { } } +## All numbers can be written in the binary form as +## ^[01]*(01)1*0*$ +## This we can match with the regexp.. +## /01(1*)(0*)$/ +## The next highest number with the same number of bits +## flips the 01 to 10 and switches the 1s with the 0s +## The regex replace is then: +## /01(1*)(0*)$/10$2$1/ + sub next_bin_rex { return oct '0b'.sprintf('0%b',shift) =~ s{01(1*)(0*)$}{10$2$1}r; } +## We further note we can find the "01" with rindex +## rather than having to use a regex {regex's are expensive} +## +## We also note that to flip 1111000 to 0001111 we don't need to +## know how many 1s there are or 0s we just reverse the string. +## +## This gives us the following similar function which DOES NOT +## use regexs +## +## Usually avoiding regexs leads to more performant code (unless the +## replacement for the regex is particularly complex - which in this +## case it isn't!) + sub next_bin_rrev { my $t = rindex my $s = sprintf('0%b',shift),'01'; return oct '0b'.substr($s,0,$t).'10'.reverse substr $s,$t+2; |