From 22ed8206ae15baebbaa9732c5124d538135beae6 Mon Sep 17 00:00:00 2001
From: Abigail <abigail@abigail.be>
Date: Wed, 24 Feb 2021 18:23:18 +0100
Subject: Python solution for week 101, part 2

---
 challenge-101/abigail/python/ch-2.py | 48 ++++++++++++++++++++++++++++++++++++
 1 file changed, 48 insertions(+)
 create mode 100644 challenge-101/abigail/python/ch-2.py

(limited to 'challenge-101/abigail/python/ch-2.py')

diff --git a/challenge-101/abigail/python/ch-2.py b/challenge-101/abigail/python/ch-2.py
new file mode 100644
index 0000000000..1ed4dc151f
--- /dev/null
+++ b/challenge-101/abigail/python/ch-2.py
@@ -0,0 +1,48 @@
+#!/opt/local/bin/python
+
+#
+# See ../README.md
+#
+
+#
+# Run as python ch-2.py < input-file
+#
+
+import fileinput
+
+#
+# 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.
+#
+def side (x1, y1, x2, y2):
+    return (y2 - y1) * x2 - (x2 - x1) * y2
+
+
+for line in fileinput . input ():
+    #
+    # Parse input. We need an explicite conversion from strings to floats
+    #
+    x1, y1, x2, y2, x3, y3 = map (lambda f: float (f), line . split ())
+
+    #
+    # 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 (x2, y2, x3, y3)
+    s2 = side (x3, y3, x1, y1)
+    s3 = side (x1, y1, x2, y2)
+
+    #
+    # 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.
+    #
+    if s1 <= 0 and s2 <= 0 and s3 <= 0 or \
+       s1 >= 0 and s2 >= 0 and s3 >= 0:
+        print (1)
+    else:
+        print (0)
-- 
cgit