How to determine if a list of polygon points are in clockwise order

Question

Having a list of points  how do I find if they are in clockwise order   For example   point 0     5 0  point 1     6 4  point 2     4 5  point 3     1 5  point 4     1 0    would say that it is anti-clockwise  or counter-clockwise  for some people

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This is my solution using the explanations in the other answers   def segments poly          A sequence of  x y  numeric coordinates pairs         return zip poly  poly 1      poly 0     def check clockwise poly       clockwise   False     if  sum x0 y1 - x1 y0 for   x0  y0    x1  y1   in segments poly     lt  0          clockwise   not clockwise     return clockwise  poly     2 2   6 2   6 6   2 6   check clockwise poly  False  poly     2  6    6  6    6  2    2  2   check clockwise poly  True

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The cross product measures the degree of perpendicular-ness of two vectors  Imagine that each edge of your polygon is a vector in the x-y plane of a three-dimensional  3-D  xyz space  Then the cross product of two successive edges is a vector in the z-direction   positive z-direction if the second segment is clockwise  minus z-direction if it s counter-clockwise   The magnitude of this vector is proportional to the sine of the angle between the two original edges  so it reaches a maximum when they are perpendicular  and tapers off to disappear when the edges are collinear  parallel    So  for each vertex  point  of the polygon  calculate the cross-product magnitude of the two adjoining edges    Using your data  point 0     5  0  point 1     6  4  point 2     4  5  point 3     1  5  point 4     1  0    So Label the edges consecutively as  edgeA is the segment from point0 to point1 and  edgeB between point1 to point2       edgeE is between point4 and point0      Then Vertex A  point0   is between    edgeE  From  point4 to point0     edgeA  From  point0 to  point1    These two edges are themselves vectors  whose x and y coordinates can be determined by subtracting the coordinates of their start and end points   edgeE   point0 - point4    1  0  -  5  0     -4  0    and    edgeA   point1 - point0    6  4  -  1  0     5  4    and   And the cross product of these two adjoining edges is calculated using the determinant of the following matrix  which is constructed by putting the coordinates of the two vectors below the symbols representing the three coordinate axis  i  j   amp  k   The third  zero -valued coordinate is there because the cross product concept is a 3-D construct  and so we extend these 2-D vectors into 3-D in order to apply the cross-product    i    j    k  -4    0    0  1    4    0       Given that all cross-products produce a vector perpendicular to the plane of two vectors being multiplied  the determinant of the matrix above only has a k   or z-axis  component  The formula for calculating the magnitude of the k or z-axis component is   a1 b2 - a2 b1   -4  4 - 0  1          -16  The magnitude of this value  -16   is a measure of the sine of the angle between the 2 original vectors  multiplied by the product of the magnitudes of the 2 vectors  Actually  another formula for its value is   A X B  Cross Product     A     B    sin AB      So  to get back to just a measure of the angle you need to divide this value   -16   by the product of the magnitudes of the two vectors     A     B     4   Sqrt 17     16 4924     So the measure of sin AB    -16   16 4924   - 97014     This is a measure of whether the next segment after the vertex has bent to the left or right  and by how much  There is no need to take arc-sine  All we will care about is its magnitude  and of course its sign  positive or negative    Do this for each of the other 4 points around the closed path  and add up the values from this calculation at each vertex     If final sum is positive  you went clockwise  negative  counterclockwise

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Here s a simple Python 3 implementation based on this answer  which  in turn  is based on the solution proposed in the accepted answer   def is clockwise points         points is your list  or array  of 2d points      assert len points   gt  0     s   0 0     for p1  p2 in zip points  points 1      points 0             s     p2 0  - p1 0      p2 1    p1 1       return s  gt  0 0

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As also explained in this Wikipedia article Curve orientation  given 3 points p  q and r on the plane  i e  with x and y coordinates   you can calculate the sign of the following determinant    If the determinant is negative  i e  Orient p  q  r   lt  0   then the polygon is oriented clockwise  CW   If the determinant is positive  i e  Orient p  q  r   gt  0   the polygon is oriented counterclockwise  CCW   The determinant is zero  i e  Orient p  q  r     0  if points p  q and r are collinear   In the formula above  we prepend the ones in front of the coordinates of p  q  and r because we are using homogeneous coordinates

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For what it is worth  I used this mixin to calculate the winding order for Google Maps API v3 apps   The code leverages the side effect of polygon areas  a clockwise winding order of vertexes yields a positive area  while a counter-clockwise winding order of the same vertexes produces the same area as a negative value   The code also uses a sort of private API in the Google Maps geometry library   I felt comfortable using it - use at your own risk   Sample usage   var myPolygon   new google maps Polygon    options      var isCW   myPolygon isPathClockwise      Full example with unit tests   http   jsfiddle net stevejansen bq2ec       Mixin to extend the behavior of the Google Maps JS API Polygon type     to determine if a polygon path has clockwise of counter-clockwise winding order           Tested against v3 14 of the GMaps API          author  stevejansen github icloud com         license http   opensource org licenses MIT         version 1 0         mixin           param   number Array google maps MVCArray    path  - an optional polygon path  defaults to the first path of the polygon      returns  boolean  true if the path is clockwise  false if the path is counter-clockwise      function       var category    google maps Polygon isPathClockwise           check that the GMaps API was already loaded   if  null    google    null    google maps    null    google maps Polygon        console error category   Google Maps API not found        return        if  typeof google maps geometry spherical computeArea       function         console error category   Google Maps geometry library not found        return         if  typeof google maps geometry spherical computeSignedArea       function         console error category   Google Maps geometry library private function computeSignedArea   is missing  this may break this mixin           function isPathClockwise path        var self   this          isCounterClockwise       if  null     path        throw new Error  Path is optional  but cannot be null            default to the first path     if  arguments length     0          path   self getPath            support for passing an index number to a path     if  typeof path       number           path   self getPaths   getAt path        if   path instanceof Array  amp  amp   path instanceof google maps MVCArray        throw new Error  Path must be an Array or MVCArray            negative polygon areas have counter-clockwise paths     isCounterClockwise    google maps geometry spherical computeSignedArea path   lt  0        return   isCounterClockwise          if  typeof google maps Polygon prototype isPathClockwise       function         google maps Polygon prototype isPathClockwise   isPathClockwise

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After testing several unreliable implementations  the algorithm that provided satisfactory results regarding the CW CCW orientation out of the box was the one  posted by OP in this thread  shoelace formula 3     As always  a positive number represents a CW orientation  whereas a negative number CCW

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Here is a simple C  implementation of the algorithm based on this answer   Let s assume that we have a Vector type having X and Y properties of type double   public bool IsClockwise IList lt Vector gt  vertices        double sum   0 0      for  int i   0  i  lt  vertices Count  i              Vector v1   vertices i           Vector v2   vertices  i   1    vertices Count           sum     v2 X - v1 X     v2 Y   v1 Y             return sum  gt  0 0        is the modulo or remainder operator performing the modulo operation which  according to Wikipedia  finds the remainder after division of one number by another

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Another solution for this   const isClockwise    vertices       gt        const len   vertices length      const sum   vertices map   x  y   index    gt            let nextIndex   index   1          if  nextIndex     len  nextIndex   0           return               x1  x              x2  vertices nextIndex  x              y1  x              y2  vertices nextIndex  x                  map    x1  x2  y1  y2     gt    x2 - x1     y1   y2    reduce  a  b    gt  a   b        if  sum  gt  -1  return true      if  sum  lt  0  return false      Take all the vertices as an array like this   const vertices     x  5  y  0    x  6  y  4    x  4  y  5    x  1  y  5    x  1  y  0    isClockwise vertices

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An implementation of Sean s answer in JavaScript    x000D   x000D  function calcArea poly    x000D      if  poly    poly length  lt  3  return null  x000D      let end   poly length - 1  x000D      let sum   poly end  0  poly 0  1  - poly 0  0  poly end  1   x000D      for let i 0  i lt end    i    x000D          const n i 1  x000D          sum    poly i  0  poly n  1  - poly n  0  poly i  1   x000D        x000D      return sum  x000D    x000D   x000D  function isClockwise poly    x000D      return calcArea poly   gt  0  x000D    x000D   x000D  let poly     352 168   305 208   312 256   366 287   434 248   416 186    x000D   x000D  console log isClockwise poly    x000D   x000D  let poly2     618 186   650 170   701 179   716 207   708 247   666 259   637 246   615 219    x000D   x000D  console log isClockwise poly2    x000D   x000D   x000D    Pretty sure this is right  It seems to be working  -   Those polygons look like this  if you re wondering

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Start at one of the vertices  and compute the angle subtended by each side     The first and the last will be zero  so skip those   for the rest  the sine of the angle will be given by the cross product of the normalizations to unit length of  point n -point 0   and  point n-1 -point 0     If the sum of the values is positive  then your polygon is drawn in the anti-clockwise sense

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This is the implemented function for OpenLayers 2  The condition for having a clockwise polygon is area  lt  0  it confirmed by this reference   function IsClockwise feature        if feature geometry    null          return -1       var vertices   feature geometry getVertices        var area   0       for  var i   0  i  lt   vertices length   i              j    i   1    vertices length           area    vertices i  x   vertices j  y          area -  vertices j  x   vertices i  y             console log area              return  area  lt  0

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I think in order for some points to be given clockwise all edges need to be positive not only the sum of edges  If one edge is negative than at least 3 points are given counter-clockwise

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find the center of mass of these points    suppose there are lines from this point to your points   find the angle between two lines for line0 line1  than do it for line1 and line2            if this angle is monotonically increasing than it is counterclockwise    else if monotonically decreasing it is clockwise  else  it is not monotonical   you cant decide  so it is not wise

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Solution for R to determine direction and reverse if clockwise  found it necessary for owin objects    coords  lt - cbind x   c 5 6 4 1 1  y   c 0 4 5 5 0   a  lt - numeric   for  i in 1 dim coords  1       print i    q  lt - i   1   if  i     dim coords  1    q  lt - 1   out  lt -   coords q 1   -  coords i 1        coords q 2      coords i 2      a q   lt - out   rm q out     end i loop  rm i   a  lt - sum a   -ve is anti-clockwise  b  lt - cbind x   rev coords  1    y   rev coords  2     if  a gt 0  coords  lt - b  reverses coords if polygon not traced in anti-clockwise direction

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A much computationally simpler method  if you already know a point inside the polygon    Choose any line segment from the original polygon  points and their coordinates in that order  Add a known  inside  point  and form a triangle  Calculate CW or CCW as suggested here with those three points

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My C    LINQ solution is based on the cross product advice of  charlesbretana is below  You can specify a reference normal for the winding  It should work as long as  the curve is mostly in the plane defined by the up vector   using System Collections Generic  using System Linq  using System Numerics   namespace SolidworksAddinFramework Geometry       public static class PlanePolygon                    lt summary gt              Assumes that polygon is closed  ie first and last points are the same              lt  summary gt         public static bool Orientation             this IEnumerable lt Vector3 gt  polygon  Vector3 up                        var sum   polygon                  Buffer 2  1     from Interactive Extensions Nuget Pkg                  Where b   gt  b Count    2                   Aggregate                     Vector3 Zero                      p  b    gt  p   Vector3 Cross b 0   b 1                                      b 0  Length   b 1  Length                  return Vector3 Dot up  sum   gt  0                         with a unit test  namespace SolidworksAddinFramework Spec Geometry       public class PlanePolygonSpec                Fact          public void OrientationShouldWork                          var points   Sequences LinSpace 0  Math PI 2  100                   Select t   gt  new Vector3  float  Math Cos t    float  Math Sin t   0                    ToList                 points Orientation Vector3 UnitZ  Should   BeTrue                points Reverse                points Orientation Vector3 UnitZ  Should   BeFalse

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Find the vertex with smallest y  and largest x if there are ties   Let the vertex be A and the previous vertex in the list be B and the next vertex in the list be C  Now compute the sign of the cross product of AB and AC     References    How do I find the orientation of a simple polygon  in  Frequently Asked Questions  comp graphics algorithms  Curve orientation at Wikipedia

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I m going to throw out another solution because it s straightforward and not mathematically intensive - it just uses basic algebra  Calculate the signed area of the polygon  If it s negative the points are in clockwise order  if it s positive they are counterclockwise   This is very similar to Beta s solution   Calculate the signed area  A   1 2    x1 y2 - x2 y1   x2 y3 - x3 y2         xn y1 - x1 yn  Or in pseudo-code  signedArea   0 for each point in points      x1   point 0      y1   point 1      if point is last point         x2   firstPoint 0          y2   firstPoint 1      else         x2   nextPoint 0          y2   nextPoint 1      end if      signedArea     x1   y2 - x2   y1  end for return signedArea   2  Note that if you are only checking the ordering  you don t need to bother dividing by 2  Sources  http   mathworld wolfram com PolygonArea html

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Some of the suggested methods will fail in the case of a non-convex polygon  such as a crescent  Here s a simple one that will work with non-convex polygons  it ll even work with a self-intersecting polygon like a figure-eight  telling you whether it s mostly clockwise    Sum over the edges   x2 - x1  y2   y1   If the result is positive the curve is clockwise  if it s negative the curve is counter-clockwise   The result is twice the enclosed area  with a   - convention    point 0     5 0    edge 0    6-5  4 0      4 point 1     6 4    edge 1    4-6  5 4    -18 point 2     4 5    edge 2    1-4  5 5    -30 point 3     1 5    edge 3    1-1  0 5      0 point 4     1 0    edge 4    5-1  0 0      0                                          ---                                          -44  counter-clockwise

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If you use Matlab  the function ispolycw returns true if the polygon vertices are in clockwise order

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Here s swift 3 0 solution based on answers above       for  i  point  in allPoints enumerated             let nextPoint   i    allPoints count - 1   allPoints 0    allPoints i 1          signedArea     point x   nextPoint y - nextPoint x   point y             let clockwise    signedArea  lt  0

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C  code to implement lhf s answer      https   en wikipedia org wiki Curve orientation Orientation of a simple polygon public static WindingOrder DetermineWindingOrder IList lt Vector2 gt  vertices        int nVerts   vertices Count         If vertices duplicates first as last to represent closed polygon         skip last      Vector2 lastV   vertices nVerts - 1       if  lastV Equals vertices 0            nVerts -  1      int iMinVertex   FindCornerVertex vertices          Orientation matrix               1  xa  ya          O     1  xb  yb                1  xc  yc       Vector2 a   vertices WrapAt iMinVertex - 1  nVerts        Vector2 b   vertices iMinVertex       Vector2 c   vertices WrapAt iMinVertex   1  nVerts           determinant O     xb yc   xa yb   ya xc  -  ya xb   yb xc   xa yc      double detOrient    b X   c Y   a X   b Y   a Y   c X  -  a Y   b X   b Y   c X   a X   c Y           TBD  check for    0   in which case is not defined         Can that happen   Do we need to check other vertices   eliminate duplicate vertices      WindingOrder result   detOrient  gt  0               WindingOrder Clockwise               WindingOrder CounterClockwise      return result     public enum WindingOrder       Clockwise      CounterClockwise       Find vertex along one edge of bounding box     In this case  we find smallest y  in case of tie also smallest x  private static int FindCornerVertex IList lt Vector2 gt  vertices        int iMinVertex   -1      float minY   float MaxValue      float minXAtMinY   float MaxValue      for  int i   0  i  lt  vertices Count  i                  Vector2 vert   vertices i           float y   vert Y          if  y  gt  minY              continue          if  y    minY              if  vert X  gt   minXAtMinY                  continue              Minimum so far          iMinVertex   i          minY   y          minXAtMinY   vert X             return iMinVertex        Return value in  0  n-1      Works for i in  -n   infinity      If need to allow more negative values  need more complex formula  private static int WrapAt int i  int n             n   Moves  -n    up to  0         return  i   n    n

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While these answers are correct  they are more mathematically intense than necessary  Assume map coordinates  where the most north point is the highest point on the map  Find the most north point  and if 2 points tie  it is the most north then the most east  this is the point that lhf uses in his answer   In your points   point 0     5 0   point 1     6 4   point 2     4 5   point 3     1 5   point 4     1 0   If we assume that P2 is the most north then east point either the previous or next point determine clockwise  CW  or CCW  Since the most north point is on the north face  if the P1  previous  to P2 moves east the direction is CW  In this case  it moves west  so the direction is CCW as the accepted answer says  If the previous point has no horizontal movement  then the same system applies to the next point  P3  If P3 is west of P2  it is  then the movement is CCW  If the P2 to P3 movement is east  it s west in this case  the movement is CW  Assume that nte  P2 in your data  is the most north then east point and the prv is the previous point  P1 in your data  and nxt is the next point  P3 in your data  and  0  is horizontal or east west where west is less than east  and  1  is vertical   if  nte 0   gt   prv 0   amp  amp  nxt 0   gt   nte 0   return CW   if  nte 0   lt   prv 0   amp  amp  nxt 0   lt   nte 0   return CCW      Okay  it s not easy-peasy  so now  do the math if  nte 0    nxt 1  - nte 1    nxt 0  - prv 0     nxt 1  - crt 1     prv 1     nxt 0  - nte 0    gt   0  return CCW      For quadrant 3 return CW  return CW     For quadrant 3 return  CCW

[math] How to determine if a list of polygon points are in clockwise order?

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