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| -rw-r--r-- | challenge-013/dave-jacoby/c2.pl | 69 |
1 files changed, 69 insertions, 0 deletions
diff --git a/challenge-013/dave-jacoby/c2.pl b/challenge-013/dave-jacoby/c2.pl new file mode 100644 index 0000000000..8a7c0663a4 --- /dev/null +++ b/challenge-013/dave-jacoby/c2.pl @@ -0,0 +1,69 @@ +#!/usr/bin/env perl + +# Perl Weekly Challenge 013-2 + +# Write a script to demonstrate Mutually Recursive methods. +# Two methods are mutually recursive if the first method calls +# the second and the second calls first in turn. Using the +# mutually recursive methods, generate Hofstadter Female +# and Male sequences. + +## F ( 0 ) = 1 ; M ( 0 ) = 0 +## F ( n ) = n − M ( F ( n − 1 ) ) , n > 0 +## M ( n ) = n − F ( M ( n − 1 ) ) , n > 0. + +# Thinking through this problem + +# ff(1) = 1 - mm( ff( 0 ) ) +# ff(1) = 1 - mm( 1 ) +# mm(1) = 1 - ff( mm( 0 ) ) +# mm(1) = 1 - ff( 0 ) +# mm(1) = 1 - 1 +# mm(1) = 0 +# ff(1) = 1 - 0 +# ff(1) = 1 + +use strict; +use warnings; +use utf8; +use feature qw{ postderef say signatures state switch fc }; +no warnings + qw{ experimental::postderef experimental::smartmatch experimental::signatures }; + +for my $n ( 0 .. 3 ) { + my $f = ff($n); + my $m = mm($n); + # say ''; + say qq{ f( $n ) = $f \t m( $n ) = $m }; +} + +exit; + +sub ff( $n ) { + # print qq{ ff($n) }; + return 1 if $n == 0 ; + return $n - mm( ff( $n-1 )); +} + +# using mm() because m() is a match operator, and using ff() to +# keep consistent, even though there isn't an f() operator. +sub mm( $n ) { + # print qq{ mm($n) }; + return 0 if $n == 0 ; + return $n - ff( mm( $n-1 )); +} + +__DATA__ + ff(0) mm(0) + f( 0 ) = 1 m( 0 ) = 0 + + ff(1) ff(0) mm(1) mm(0) ff(0) mm(1) mm(0) ff(0) + f( 1 ) = 1 m( 1 ) = 0 + + ^^^ Verified + + ff(2) ff(1) ff(0) mm(1) mm(0) ff(0) mm(1) mm(0) ff(0) mm(2) mm(1) mm(0) ff(0) ff(0) + f( 2 ) = 2 m( 2 ) = 1 + + ff(3) ff(2) ff(1) ff(0) mm(1) mm(0) ff(0) mm(1) mm(0) ff(0) mm(2) mm(1) mm(0) ff(0) ff(0) mm(3) mm(2) mm(1) mm(0) ff(0) ff(0) ff(1) ff(0) mm(1) mm(0) ff(0) + f( 3 ) = 2 m( 3 ) = 2 |