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diff --git a/challenge-191/jo-37/perl/ch-1.pl b/challenge-191/jo-37/perl/ch-1.pl
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+++ b/challenge-191/jo-37/perl/ch-1.pl
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+#!/usr/bin/perl -s
+
+use v5.16;
+use Test2::V0 '!float';
+use PDL;
+
+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 integers
+
+EOS
+
+
+### Input and Output
+
+say twice_largest(@ARGV);
+
+
+### Implementation
+#
+# The task requires careful reading: Operating on *items*, not on
+# *values*.  This differentiation becomes crucial if the maximum is not
+# taken on a unique item.  Interpreting this task as: Take the largest
+# two items from the list and check if the first is at least twice the
+# second.
+
+sub twice_largest {
+    state $w = long(1, -2)->dummy(0);
+
+    my $n = long(@_);
+    # Find the indices of the two largest items.
+    maximum_n_ind($n, my $max2 = zeroes(indx, 2));
+
+    # Check if the maximum item is at least twice the next.  As items
+    # are integers, we may compare the scalar product of the maximum two
+    # items with the weight vector against -0.5 to gain the requested
+    # -1/1 result.
+    ($n->index($max2) x $w)->sclr <=> -0.5;
+}
+
+
+### Examples and tests
+
+sub run_tests {
+    SKIP: {
+        skip "examples" unless $examples;
+
+        is twice_largest(1, 2, 3, 4), -1, 'Example 1';
+        is twice_largest(1, 2, 0, 5),  1, 'Example 2';
+        is twice_largest(2, 6, 3, 1),  1, 'Example 3';
+        is twice_largest(4, 5, 2, 3), -1, 'Example 4';
+    }
+
+    SKIP: {
+        skip "tests" unless $tests;
+
+        is twice_largest(1, 3, 3), -1, 'non-unique max';
+        is twice_largest(-10, 1, -20), 1, 'negative items';
+	}
+
+    done_testing;
+    exit;
+}