有了一个点列表,我如何确定它们是否是顺时针顺序的?
例如:
point[0] = (5,0)
point[1] = (6,4)
point[2] = (4,5)
point[3] = (1,5)
point[4] = (1,0)
会说它是逆时针的(对某些人来说是逆时针的)
有了一个点列表,我如何确定它们是否是顺时针顺序的?
例如:
point[0] = (5,0)
point[1] = (6,4)
point[2] = (4,5)
point[3] = (1,5)
point[4] = (1,0)
会说它是逆时针的(对某些人来说是逆时针的)
当前回答
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。
其他回答
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。