From 0160915bd273e56b782576ce40a6c202883afb7b Mon Sep 17 00:00:00 2001
From: Abigail <abigail@abigail.be>
Date: Wed, 24 Feb 2021 01:57:22 +0100
Subject: AWK solution from week 101, part 2

---
 challenge-101/abigail/README.md    |  3 ++-
 challenge-101/abigail/awk/ch-2.awk | 45 ++++++++++++++++++++++++++++++++++++++
 2 files changed, 47 insertions(+), 1 deletion(-)
 create mode 100644 challenge-101/abigail/awk/ch-2.awk

diff --git a/challenge-101/abigail/README.md b/challenge-101/abigail/README.md
index da150e755c..8fa0cc8907 100644
--- a/challenge-101/abigail/README.md
+++ b/challenge-101/abigail/README.md
@@ -99,6 +99,7 @@ Output: 1 because (0,0) is on the edge connecting B and C.
 ~~~~
 
 ### Solutions
-* [Perl](perl/ch-1.pl)
+* [AWK](awk/ch-2.awk)
+* [Perl](perl/ch-2.pl)
 
 ### Blog
diff --git a/challenge-101/abigail/awk/ch-2.awk b/challenge-101/abigail/awk/ch-2.awk
new file mode 100644
index 0000000000..d8d3bc561b
--- /dev/null
+++ b/challenge-101/abigail/awk/ch-2.awk
@@ -0,0 +1,45 @@
+#!/usr/bin/awk
+
+#
+# See ../README.md
+#
+
+#
+# Run as: awk -f ch-2.awk < input-file
+#
+
+#
+# See https://stackoverflow.com/questions/2049582/
+#
+
+#
+# This determines on which side of the line through ($x1, $y1) and
+# ($x2, $y2) the origin lies. If > 0, then the origin lies to the left
+# of the line, if < 0, the origin lies to the right of the line, if
+# = 0, the origin lies on the line.
+#
+function side (x1, y1, x2, y2) {
+    return (y2 - y1) * x2 - (x2 - x1) * y2
+}
+
+{
+    # x1 y1 x2 y2 x3 y3
+    # $1 $2 $3 $4 $5 $6
+    #
+    # Determine where the origin is relative to the three lines 
+    # through the vertices of the triangle. Note we have to go
+    # in a specific order through the points. (Either clock wise,
+    # or counter clockwise, as long as we're consistent).
+    #
+    s1 = side($3, $4, $5, $6)
+    s2 = side($5, $6, $1, $2)
+    s3 = side($1, $2, $3, $4)
+
+    #
+    # If the origin either lies to the left (or on) each of the
+    # lines, or to the right (or on) each of the lines, the origin
+    # lies inside the triangle. If not, it does not.
+    #
+    print (s1 <= 0 && s2 <= 0 && s3 <= 0 ||
+           s1 >= 0 && s2 >= 0 && s3 >= 0 ? 1 : 0)
+}
-- 
cgit