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diff --git a/challenge-196/peter-campbell-smith/blog.txt b/challenge-196/peter-campbell-smith/blog.txt
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+https://pjcs-pwc.blogspot.com/2022/12/pattern-132-and-sequential-runs.html
diff --git a/challenge-196/peter-campbell-smith/perl/ch-1.pl b/challenge-196/peter-campbell-smith/perl/ch-1.pl
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+#!/usr/bin/perl
+
+# Peter Campbell Smith - 2022-12-19
+# PWC 196 task 1
+
+use v5.28;
+use utf8;
+use warnings;
+
+# You are given a list of integers, @list. Write a script to find out subsequence that respects Pattern 132. 
+# Pattern 132 is a sequence (a[i], a[j], a[k]) such that i < j < k and a[i] < a[k] < a[j].
+# Return an empty array if no such sequence is found, or the first if several are found.
+
+
+# Blog: https://pjcs-pwc.blogspot.com/2022/12/pattern-132-and-sequential-runs.html
+
+my (@tests, $test, @list, $j, $last, @hard, $i, $k);
+
+# Mohammad's examples
+@tests = ([3, 1, 4, 2], [1, 2, 3, 4], [1, 3, 2, 4, 6, 5], [1, 3, 4, 2]);
+
+# a difficult one
+for $j (1 .. 10000) {
+	push @hard, $j;
+}
+push @hard, 9999;
+push @tests, \@hard;
+
+# loop over tests
+TEST: for $test (@tests) {
+	@list = @$test;
+	$last = scalar @list - 1;
+	
+	# loop over j, which is the largest of the three
+	J: for $j (1 .. $last - 1) {
+		
+		# find a smaller $i to the left of $j
+		for $i (0 .. $j - 1) {
+			if ($list[$i] < $list[$j]) {
+				
+				# one exists so let's see if there's a smaller $k to the right of $j
+				for $k ($j + 1 .. $last) {
+					if ($list[$k] < $list[$j]) {
+						say qq[\nInput:  \@list = (] . join(', ', @list) . qq[)\nOutput: ($list[$i], $list[$j], $list[$k])];
+						next TEST;
+					}
+				}
+				next J;
+			}
+		}
+	}
+	say qq[\nInput:  \@list = (] . join(', ', @list) . qq[)\nOutput: none found];
+	
+}
diff --git a/challenge-196/peter-campbell-smith/perl/ch-2.pl b/challenge-196/peter-campbell-smith/perl/ch-2.pl
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+#!/usr/bin/perl
+
+# Peter Campbell Smith - 2022-12-19
+# PWC 196 task 2
+
+use v5.28;
+use utf8;
+use warnings;
+
+# We are given a sorted unique integer array, @array. We are asked to find all the slices of this array
+# which comprise consecutive integers, and output the first and last element in each such slice.
+
+# Blog: https://pjcs-pwc.blogspot.com/2022/12/pattern-132-and-sequential-runs.html
+
+my (@tests, $test, @array, $start, $in_slice, $output, $j);
+
+# Mohammad's examples
+@tests = ([1,3,4,5,7], [1,2,3,6,7,9], [0,1,2,4,5,6,8,9], [1,3,5,7,9,11]);
+
+# loop over tests
+TEST: for $test (@tests) {
+	@array = @$test;
+	
+	# initialise
+	$in_slice = 0;
+	$output = '';
+	say qq[\nInput:  \@array = (] . join(', ', @array) . ')';
+
+	# add a number at the end of @array which is 2 more than the last one
+	push @array, $array[scalar @array - 1] + 2;
+	
+	# loop over elements in @array
+	for ($j = 0; $j < (scalar @array) - 1; $j ++) {
+		
+		# if we're not already in a sequence and the next element is one more than this, start a sequence
+		unless ($in_slice) {
+			if ($array[$j] == $array[$j + 1] - 1) {
+				$start = $array[$j];
+				$in_slice = 1;
+			}
+		
+		# if we are already in a sequence, end it if the following element isn't one more than this
+		} else {
+			if ($array[$j] != $array[$j + 1] - 1) {
+				$output .= qq[[$start, $array[$j]], ];
+				$in_slice = 0;
+			}
+		}
+	}
+	if ($output) {
+		say qq[Output: ]. substr($output, 0, -2);
+	} else {
+		say qq[Output: no consecutive sequence found];
+	}
+}
+	
+