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authorAbigail <abigail@abigail.be>2021-02-24 13:53:35 +0100
committerAbigail <abigail@abigail.be>2021-02-24 13:53:35 +0100
commit4dbfa232abe9797f56c3ba7427dbae9fb6846943 (patch)
tree9413d3d6511ada0970090f1a6693d92632086449 /challenge-101/abigail
parent0dabcd579411c6d1544225d972efe81354416895 (diff)
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Lua solution for week 101, part 2
Diffstat (limited to 'challenge-101/abigail')
-rw-r--r--challenge-101/abigail/README.md1
-rw-r--r--challenge-101/abigail/lua/ch-2.lua52
2 files changed, 53 insertions, 0 deletions
diff --git a/challenge-101/abigail/README.md b/challenge-101/abigail/README.md
index 450cdc1526..e2b747cc38 100644
--- a/challenge-101/abigail/README.md
+++ b/challenge-101/abigail/README.md
@@ -102,6 +102,7 @@ Output: 1 because (0,0) is on the edge connecting B and C.
* [AWK](awk/ch-2.awk)
* [Bash](bash/ch-2.sh)
* [C](c/ch-2.c)
+* [Lua](lua/ch-2.lua)
* [Perl](perl/ch-2.pl)
### Blog
diff --git a/challenge-101/abigail/lua/ch-2.lua b/challenge-101/abigail/lua/ch-2.lua
new file mode 100644
index 0000000000..d7ecaacfb9
--- /dev/null
+++ b/challenge-101/abigail/lua/ch-2.lua
@@ -0,0 +1,52 @@
+#!/opt/local/bin/lua
+
+--
+-- See ../README.md
+--
+
+--
+-- Run as: lua ch-2.lua < input-file
+--
+
+--
+-- See ../README.md
+--
+
+--
+-- 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
+end
+
+for line in io . lines () do
+ --
+ -- Parse input
+ --
+ local _, _, x1, y1, x2, y2, x3, y3 =
+ line : find ("(%S+) (%S+) (%S+) (%S+) (%S+) (%S+)")
+
+ --
+ -- 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).
+ --
+ local s1 = side (x2, y2, x3, y3)
+ local s2 = side (x3, y3, x1, y1)
+ local 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)
+ then print (1)
+ else print (0)
+ end
+end