C solution for week 101, part 2

2 files changed, 69 insertions, 0 deletions

diff --git a/challenge-101/abigail/README.md b/challenge-101/abigail/README.md
index 8fa0cc8907..450cdc1526 100644
--- a/challenge-101/abigail/README.md
+++ b/challenge-101/abigail/README.md
@@ -100,6 +100,8 @@ Output: 1 because (0,0) is on the edge connecting B and C.
 
 ### Solutions
 * [AWK](awk/ch-2.awk)
+* [Bash](bash/ch-2.sh)
+* [C](c/ch-2.c)
 * [Perl](perl/ch-2.pl)
 
 ### Blog
diff --git a/challenge-101/abigail/c/ch-2.c b/challenge-101/abigail/c/ch-2.c
new file mode 100644
index 0000000000..bfeba9c679
--- /dev/null
+++ b/challenge-101/abigail/c/ch-2.c
@@ -0,0 +1,67 @@
+# include <stdlib.h>
+# include <stdio.h>
+# include <string.h>
+
+/*
+ * See ../README.md
+ */
+
+/*
+ * Run as: cc -o ch-2.o ch-2.c; ./ch-2.o < input-file
+ */
+
+/*
+ * See https://stackoverflow.com/questions/2049582/
+ */
+
+typedef long double num;
+
+/*
+ * 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.
+ */
+
+num side (num x1, num y1, num x2, num y2) {
+    return (y2 - y1) * x2 - (x2 - x1) * y2;
+}
+
+int main (void) {
+    char *  line = NULL;
+    size_t  len  = 0;
+
+    while (getline (&line, &len, stdin) != -1) {
+        num x1, y1, x2, y2, x3, y3;
+        num s1, s2, s3;
+        /*
+         * Parse input
+         */
+        if (sscanf (line, "%Lf %Lf %Lf %Lf %Lf %Lf",
+                           &x1, &y1, &x2, &y2, &x3, &y3) != 6) {
+            fprintf (stderr, "Could not parse input\n");
+            exit (1);
+        }
+
+        /*
+         * 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.
+         */
+        printf ("%d\n", (s1 <= 0 && s2 <= 0 && s3 <= 0) ||
+                        (s1 >= 0 && s2 >= 0 && s3 >= 0) ? 1 : 0);
+    }
+    free (line);
+
+    return (0);
+}