正如这篇维基百科文章中所解释的曲线方向，给定平面上的3个点p, q和r(即x和y坐标)，您可以计算以下行列式的符号

如果行列式为负(即定向(p, q, r) < 0)，则多边形是顺时针方向(CW)。如果行列式为正(即定向(p, q, r) > 0)，则多边形是逆时针方向(CCW)。如果点p, q和r共线，行列式为零(即定向(p, q, r) == 0)。

在上面的公式中，由于我们使用的是齐次坐标，我们将1放在p, q和r的坐标前面。

解决方案R确定方向和反向如果顺时针(发现这是必要的owin对象):

coords <- cbind(x = c(5,6,4,1,1),y = c(0,4,5,5,0))
a <- numeric()
for (i in 1:dim(coords)[1]){
  #print(i)
  q <- i + 1
  if (i == (dim(coords)[1])) q <- 1
  out <- ((coords[q,1]) - (coords[i,1])) * ((coords[q,2]) + (coords[i,2]))
  a[q] <- out
  rm(q,out)
} #end i loop

rm(i)

a <- sum(a) #-ve is anti-clockwise

b <- cbind(x = rev(coords[,1]), y = rev(coords[,2]))

if (a>0) coords <- b #reverses coords if polygon not traced in anti-clockwise direction

另一个解决方案是;

const isClockwise = (vertices=[]) => {
    const len = vertices.length;
    const sum = vertices.map(({x, y}, index) => {
        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}) => ((x2 - x1) * (y1 + y2))).reduce((a, b) => a + b);

    if (sum > -1) return true;
    if (sum < 0) return false;
}

把所有的顶点作为一个数组;

const vertices = [{x: 5, y: 0}, {x: 6, y: 4}, {x: 4, y: 5}, {x: 1, y: 5}, {x: 1, y: 0}];
isClockwise(vertices);

c#代码实现lhf的答案:

// https://en.wikipedia.org/wiki/Curve_orientation#Orientation_of_a_simple_polygon
public static WindingOrder DetermineWindingOrder(IList<Vector2> 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 > 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<Vector2> vertices)
{
    int iMinVertex = -1;
    float minY = float.MaxValue;
    float minXAtMinY = float.MaxValue;
    for (int i = 0; i < vertices.Count; i++)
    {
        Vector2 vert = vertices[i];
        float y = vert.Y;
        if (y > minY)
            continue;
        if (y == minY)
            if (vert.X >= 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;
}

这是我使用其他答案中的解释的解决方案:

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))) < 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

找出y最小的顶点(如果有平手，则x最大)。假设顶点是A，列表中的前一个顶点是B，列表中的下一个顶点是c。现在计算AB和AC的叉乘的符号。

引用:

如何确定一个简单多边形的方向?在 常见问题:计算机。图形。算法。 维基百科的曲线定位。