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| -rw-r--r-- | challenge-101/abigail/README.md | 1 | ||||
| -rw-r--r-- | challenge-101/abigail/python/ch-2.py | 48 |
2 files changed, 49 insertions, 0 deletions
diff --git a/challenge-101/abigail/README.md b/challenge-101/abigail/README.md index adcc052059..a6488bf50a 100644 --- a/challenge-101/abigail/README.md +++ b/challenge-101/abigail/README.md @@ -105,5 +105,6 @@ Output: 1 because (0,0) is on the edge connecting B and C. * [Lua](lua/ch-2.lua) * [Node.js](node/ch-2.js) * [Perl](perl/ch-2.pl) +* [Python](python/ch-2.py) ### Blog 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) |