Bash solution for week 101, part 1

1 files changed, 47 insertions, 0 deletions

diff --git a/challenge-101/abigail/bash/ch-2.sh b/challenge-101/abigail/bash/ch-2.sh
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+#!/bin/sh
+
+#
+# See ../README.md
+#
+
+#
+# Run as: bash ch-2.sh < 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 calc_side () {
+    #
+    # $1: x1; $2: y1; $3 = x2; $4 = y2
+    #
+    side=$((($4 - $2) * $3 - ($3 - $1) * $4))
+}
+
+
+while read x1 y1 x2 y2 x3 y3
+do    #
+      # Determine where the origin is relative to the three lines 
+      # through the vertices of the triangle. Note we have to go
+      # in a specific order through the points. (Either clock wise,
+      # or counter clockwise, as long as we're consistent).
+      #
+      calc_side x2 y2 x3 y3; s1=$side
+      calc_side x3 y3 x1 y1; s2=$side
+      calc_side x1 y1 x2 y2; s3=$side
+
+      #
+      # 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.
+      #
+      echo $(( $s1 <= 0 && $s2 <= 0 && $s3 <= 0 ||
+               $s1 >= 0 && $s2 >= 0 && $s3 >= 0 ? 1 : 0 ))
+done