Solution to task 2

1 files changed, 73 insertions, 0 deletions

diff --git a/challenge-211/jo-37/perl/ch-2.pl b/challenge-211/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(forcomb lastfor);
+use List::Util qw(sum);
+use experimental qw(signatures);
+
+our ($tests, $examples);
+
+run_tests() if $tests || $examples;	# does not return
+
+die <<EOS unless @ARGV;
+usage: $0 [-examples] [-tests] [--] [N...]
+
+-examples
+    run the examples from the challenge
+ 
+-tests
+    run some tests
+
+N...
+    list of numbers
+
+EOS
+
+
+### Input and Output
+
+say same_average(@ARGV) ? 'true' : 'false';
+
+
+### Implementation
+
+# It can easily be shown that if two partitions of the list have the
+# same average, this common average equals the average of the whole
+# list.  Thus all we need to do is find a nonempty true subset of the
+# list having the same average as the list.  The complement will have
+# the same average.
+
+sub same_average (@list) {
+    state $delta = 1e-6;
+    my $avg = sum(@list) / @list;
+    my $res;
+    # Loop over all subsets.
+    forcomb {
+        # Skip non true subsets.
+        return unless @_ && @_ != @list;
+        $res = 1, lastfor if abs($avg - sum(@list[@_]) / @_) < $delta;
+    } @list;
+
+    $res;
+}
+
+
+### Examples and tests
+
+sub run_tests {
+    SKIP: {
+        skip "examples" unless $examples;
+
+        ok same_average(1, 2, 3, 4, 5, 6, 7, 8), 'example 1';
+        ok !same_average(1, 3), 'example 2';
+    }
+
+    SKIP: {
+        skip "tests" unless $tests;
+        ok same_average(1, 1, 1, 1, 5, 9,), '(1 5), (1 1 1 9)';
+	}
+
+    done_testing;
+    exit;
+}