Solution to task 2

1 files changed, 73 insertions, 0 deletions

diff --git a/challenge-175/jo-37/perl/ch-2.pl b/challenge-175/jo-37/perl/ch-2.pl
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+#!/usr/bin/perl -s
+
+use v5.16;
+use Test2::V0;
+use Math::Prime::Util qw(euler_phi);
+use List::Gen qw(:iterate :while);
+
+our ($tests, $examples);
+
+run_tests() if $tests || $examples;	# does not return
+
+die <<EOS unless @ARGV == 1;
+usage: $0 [-examples] [-tests] [N]
+
+-examples
+    run the examples from the challenge
+ 
+-tests
+    run some tests
+
+N
+    Print the first N perfect totient numbers.
+
+EOS
+
+
+### Input and Output
+
+perfect_totient()->say(shift);
+
+
+### Implementation
+
+# Another exercise in using List::Gen.
+
+sub perfect_totient {
+    # Build a generator for perfect totient numbers.  Here we use two
+    # nested generators to accomplish the task.
+
+    # Build a non-caching generator for odd numbers N starting with 3
+    # and filter for perfect totient numbers. 
+    iterate_stream {$_ + 2}->from(3)->filter_stream(sub {
+        # Build a generator for the iterative sequence of totients for N
+        # and sum over the sequence's elements.  The generated sequence
+        # will start with N because of "from($_)" and will not include 1
+        # according to the chained "until" method.  Therefore the
+        # expected sum needs to be adjusted: There is an extra "N" and 1
+        # is missing. Thus we expect a sum of 2 * N - 1 for a perfect
+        # totient number.
+        iterate {euler_phi($_)}->from($_)->until('== 1')->sum == 2 * $_ - 1;
+    });
+}
+
+### Examples and tests
+
+sub run_tests {
+    SKIP: {
+        skip "examples" unless $examples;
+
+        is perfect_totient()->take(20), [3, 9, 15, 27, 39, 81, 111, 183,
+            243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375,
+            5571], 'task 2';
+    }
+
+    SKIP: {
+        skip "tests" unless $tests;
+
+        is perfect_totient()->(25), 46791, 'from A082897';
+	}
+
+    done_testing;
+    exit;
+}