# Determine if a polygon is convex

Write a program to determine if the input polygon is convex. The polygon is specified with one line containing N, the number of vertices, then N lines containing the x and y coordinates of each vertex. The vertices will be listed clockwise starting from an arbitrary vertex.

## example 1

### input

4
0 0
0 1
1 1
1 0


### output

convex


## example 2

### input

4
0 0
2 1
1 0
2 -1


### output

concave


## example 3

### input

8
0 0
0 1
0 2
1 2
2 2
2 1
2 0
1 0


### output

convex


x and y are integers, N<1000, and |x|,|y|<1000. You may assume that the input polygon is simple (none of the edges cross, only 2 edges touch each vertex). Shortest program wins.

• "Simple" doesn't include "consecutive edges are non-collinear"?! Also, a couple more test-cases: (0,0) (0,2) (2,2) (2,0) (1,1); and (1,1) (0,0) (0,2) (2,2) (2,0) - to test the cases where finding the concave vertex requires wrapping from the end back to the start. – Peter Taylor Apr 19 '11 at 16:22
• This question is aging, but... Consider adding a concave example with two aligned segments, e.g. a modification of example 2: (0,0),(2,1),(4,2),(1,0)(2,-1). I bring this up because I fudged around example 3 without realizing it. – Jesse Millikan Apr 21 '11 at 19:38

# J, 105

echo>('concave';'convex'){~1=#=(o.1)([:>-.~)(o.2)|3([:-/12 o.-@-/@}.,-/@}:)\(,2&{.)j./"1}.0&".;._2(1!:1)3


Passes all three tests above.

Edit: (111->115) Handle co-linear points by eliminating angles of pi. Gained a few characters elsewhere.

Edit: (115->105) Less dumb.

Explanation for the J-impaired:

• (1!:1)3 read STDIN to EOF. (I think.)
• 0&".;._2 is a nice idiom for parsing this kind of input.
• j./"1}. lop off first line of input (N 0) and convert pairs to complexes.
• (,2&{.) tack first two points onto the end of the list.
• 3(f)\ applies f to sliding window of length 3 (3 points for an angle)
• [:-/12 o.-@-/@}.,-/@}: is a verb that transforms each 3 points into an angle between -pi and pi.
• -@-/@}.,-/@}: produces (p1 - p2),(p3 - p2). (Recall that these are complexes.)
• 12 o. gives an angle for each complex.
• [:-/(...) gives the difference of the two angles.
• (o.1)([:>-.~)(o.2)| mod 2 pi, eliminate angles of pi (straight segments), and compare to pi (greater than, less than, doesn't matter unless the points are supposed to be wound in one direction).
• 1=#= if all those comparison result 1 or 0 (With self-classify. This seems dumb.)
• echo>('concave';'convex'){~ print convex.

## Python - 149 chars

p=[map(int,raw_input().split())for i in[0]*input()]*2
print'ccoonncvaevxe'[all((a-c)*(d-f)<=(b-d)*(c-e)for(a,b),(c,d),(e,f)in zip(p,p[1:],p[2:]))::2]

• I think you need <=, see example 3 I just added. – Keith Randall Apr 19 '11 at 14:33
• dammn, that slice... – st0le Apr 20 '11 at 6:39

## Ruby 1.9, 147133130124 123

gets
puts (\$<.map{|s|s.split.map &:to_i}*2).each_cons(3).any?{|(a,b),(c,d),(e,f)|(e-c)*(d-b)<(d-f)*(a-c)}?:concave: :convex


### scala: 297 chars

object C{class D(val x:Int,val y:Int)
def k(a:D,b:D,c:D)=(b.y-a.y)*(c.x-b.x)>=(c.y-b.y)*(b.x-a.x)
def main(a:Array[String]){val s=new java.util.Scanner(System.in)
def n=s.nextInt

• You can shave of three chars by using def main(a:... instead of def main(args:.... – Gareth Apr 22 '11 at 8:40