Mutual Functions

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diff --git a/challenge-013/dave-jacoby/c2.pl b/challenge-013/dave-jacoby/c2.pl
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+#!/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