我如何确定两条直线是否相交,如果相交,在x,y点处?
当前回答
我从《多视图几何》这本书里读到了这些算法
以下文本使用
'作为转置符号
*作为点积
当用作算子时,X作为叉乘
1. 线的定义
点x_vec = (x, y)'在直线ax + by + c = 0上
标记L = (a, b, c)',点为(x, y, 1)'为齐次坐标
直线方程可以写成
(x, y, 1)(a, b, c)' = 0或x' * L = 0
2. 直线交点
我们有两条直线L1=(a1, b1, c1)', L2=(a2, b2, c2)'
假设x是一个点,一个向量,x = L1 x L2 (L1叉乘L2)。
注意,x始终是一个二维点,如果你对(L1xL2)是一个三元素向量,x是一个二维坐标感到困惑,请阅读齐次坐标。
根据三重积,我们知道
L1 * (L1 x L2) = 0, L2 * (L1 x L2) = 0,因为L1,L2共平面
我们用向量x代替L1*x,那么L1*x=0, L2*x=0,这意味着x在L1和L2上,x是交点。
注意,这里x是齐次坐标,如果x的最后一个元素是零,这意味着L1和L2是平行的。
其他回答
有一个很好的方法来解决这个问题就是用向量叉乘。定义二维向量叉乘v × w为vx wy−vy wx。
假设这两条线段从p到p + r,从q到q + s。那么第一行上的任意点都可以表示为p + t r(对于标量参数t),第二行上的任意点可以表示为q + u s(对于标量参数u)。
如果t和u满足以下条件,两条直线相交:
P + t r = q + u s
两边叉乘s,得到
(p + r) × s = (q + u s) × s
由于s × s = 0,这意味着
T (r × s) = (q−p) × s
因此,求解t:
T = (q−p) × s / (r × s)
同样地,我们可以解出u:
(p + r) × r = (q + u s) × r U (s × r) = (p−q) × r U = (p−q) × r / (s × r)
为了减少计算步骤,可以方便地将其重写为以下形式(记住s × r =−r × s):
U = q−p × r / (r × s)
现在有四种情况:
If r × s = 0 and (q − p) × r = 0, then the two lines are collinear. In this case, express the endpoints of the second segment (q and q + s) in terms of the equation of the first line segment (p + t r): t0 = (q − p) · r / (r · r) t1 = (q + s − p) · r / (r · r) = t0 + s · r / (r · r) If the interval between t0 and t1 intersects the interval [0, 1] then the line segments are collinear and overlapping; otherwise they are collinear and disjoint. Note that if s and r point in opposite directions, then s · r < 0 and so the interval to be checked is [t1, t0] rather than [t0, t1]. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting. If r × s ≠ 0 and 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1, the two line segments meet at the point p + t r = q + u s. Otherwise, the two line segments are not parallel but do not intersect.
来源:该方法是3D线相交算法的2维专门化,来自Ronald Goldman发表在Graphics Gems,第304页的文章“三条线在三维空间中的相交”。在三维空间中,通常的情况是直线是倾斜的(既不平行也不相交),在这种情况下,该方法给出了两条直线最接近的点。
许多答案把所有的计算都打包成一个函数。如果您需要计算直线斜率、y轴截距或x轴截距,以便在代码的其他地方使用,那么这些计算将是冗余的。我分离出了各自的函数,使用了明显的变量名,并注释了我的代码以使其更易于理解。我需要知道直线是否无限超出它们的端点,所以在JavaScript中:
http://jsfiddle.net/skibulk/evmqq00u/
var point_a = {x:0, y:10},
point_b = {x:12, y:12},
point_c = {x:10, y:0},
point_d = {x:0, y:0},
slope_ab = slope(point_a, point_b),
slope_bc = slope(point_b, point_c),
slope_cd = slope(point_c, point_d),
slope_da = slope(point_d, point_a),
yint_ab = y_intercept(point_a, slope_ab),
yint_bc = y_intercept(point_b, slope_bc),
yint_cd = y_intercept(point_c, slope_cd),
yint_da = y_intercept(point_d, slope_da),
xint_ab = x_intercept(point_a, slope_ab, yint_ab),
xint_bc = x_intercept(point_b, slope_bc, yint_bc),
xint_cd = x_intercept(point_c, slope_cd, yint_cd),
xint_da = x_intercept(point_d, slope_da, yint_da),
point_aa = intersect(slope_da, yint_da, xint_da, slope_ab, yint_ab, xint_ab),
point_bb = intersect(slope_ab, yint_ab, xint_ab, slope_bc, yint_bc, xint_bc),
point_cc = intersect(slope_bc, yint_bc, xint_bc, slope_cd, yint_cd, xint_cd),
point_dd = intersect(slope_cd, yint_cd, xint_cd, slope_da, yint_da, xint_da);
console.log(point_a, point_b, point_c, point_d);
console.log(slope_ab, slope_bc, slope_cd, slope_da);
console.log(yint_ab, yint_bc, yint_cd, yint_da);
console.log(xint_ab, xint_bc, xint_cd, xint_da);
console.log(point_aa, point_bb, point_cc, point_dd);
function slope(point_a, point_b) {
var i = (point_b.y - point_a.y) / (point_b.x - point_a.x);
if (i === -Infinity) return Infinity;
if (i === -0) return 0;
return i;
}
function y_intercept(point, slope) {
// Horizontal Line
if (slope == 0) return point.y;
// Vertical Line
if (slope == Infinity)
{
// THE Y-Axis
if (point.x == 0) return Infinity;
// No Intercept
return null;
}
// Angled Line
return point.y - (slope * point.x);
}
function x_intercept(point, slope, yint) {
// Vertical Line
if (slope == Infinity) return point.x;
// Horizontal Line
if (slope == 0)
{
// THE X-Axis
if (point.y == 0) return Infinity;
// No Intercept
return null;
}
// Angled Line
return -yint / slope;
}
// Intersection of two infinite lines
function intersect(slope_a, yint_a, xint_a, slope_b, yint_b, xint_b) {
if (slope_a == slope_b)
{
// Equal Lines
if (yint_a == yint_b && xint_a == xint_b) return Infinity;
// Parallel Lines
return null;
}
// First Line Vertical
if (slope_a == Infinity)
{
return {
x: xint_a,
y: (slope_b * xint_a) + yint_b
};
}
// Second Line Vertical
if (slope_b == Infinity)
{
return {
x: xint_b,
y: (slope_a * xint_b) + yint_a
};
}
// Not Equal, Not Parallel, Not Vertical
var i = (yint_b - yint_a) / (slope_a - slope_b);
return {
x: i,
y: (slope_a * i) + yint_a
};
}
根据t3chb0t的答案:
int intersezione_linee(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
//L1: estremi (x1,y1)(x2,y2) L2: estremi (x3,y3)(x3,y3)
int d;
d = (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4);
if(!d)
return 0;
p_x = ((x1*y2-y1*x2)*(x3-x4) - (x1-x2)*(x3*y4-y3*x4))/d;
p_y = ((x1*y2-y1*x2)*(y3-y4) - (y1-y2)*(x3*y4-y3*x4))/d;
return 1;
}
int in_bounding_box(int x1, int y1, int x2, int y2, int p_x, int p_y)
{
return p_x>=x1 && p_x<=x2 && p_y>=y1 && p_y<=y2;
}
int intersezione_segmenti(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
if (!intersezione_linee(x1,y1,x2,y2,x3,y3,x4,y4,p_x,p_y))
return 0;
return in_bounding_box(x1,y1,x2,y2,p_x,p_y) && in_bounding_box(x3,y3,x4,y4,p_x,p_y);
}
iMalc回答的Python版本:
def find_intersection( p0, p1, p2, p3 ) :
s10_x = p1[0] - p0[0]
s10_y = p1[1] - p0[1]
s32_x = p3[0] - p2[0]
s32_y = p3[1] - p2[1]
denom = s10_x * s32_y - s32_x * s10_y
if denom == 0 : return None # collinear
denom_is_positive = denom > 0
s02_x = p0[0] - p2[0]
s02_y = p0[1] - p2[1]
s_numer = s10_x * s02_y - s10_y * s02_x
if (s_numer < 0) == denom_is_positive : return None # no collision
t_numer = s32_x * s02_y - s32_y * s02_x
if (t_numer < 0) == denom_is_positive : return None # no collision
if (s_numer > denom) == denom_is_positive or (t_numer > denom) == denom_is_positive : return None # no collision
# collision detected
t = t_numer / denom
intersection_point = [ p0[0] + (t * s10_x), p0[1] + (t * s10_y) ]
return intersection_point
只是想提一下,一个很好的解释和明确的解决方案可以在数字食谱系列中找到。我有这本书的第三版,答案在1117页21.4节。另一种不同命名的解决方案可以在玛丽娜·加夫里洛娃(Marina Gavrilova)的论文中找到。在我看来,她的解决办法要简单一些。
我的实现如下:
bool NuGeometry::IsBetween(const double& x0, const double& x, const double& x1){
return (x >= x0) && (x <= x1);
}
bool NuGeometry::FindIntersection(const double& x0, const double& y0,
const double& x1, const double& y1,
const double& a0, const double& b0,
const double& a1, const double& b1,
double& xy, double& ab) {
// four endpoints are x0, y0 & x1,y1 & a0,b0 & a1,b1
// returned values xy and ab are the fractional distance along xy and ab
// and are only defined when the result is true
bool partial = false;
double denom = (b0 - b1) * (x0 - x1) - (y0 - y1) * (a0 - a1);
if (denom == 0) {
xy = -1;
ab = -1;
} else {
xy = (a0 * (y1 - b1) + a1 * (b0 - y1) + x1 * (b1 - b0)) / denom;
partial = NuGeometry::IsBetween(0, xy, 1);
if (partial) {
// no point calculating this unless xy is between 0 & 1
ab = (y1 * (x0 - a1) + b1 * (x1 - x0) + y0 * (a1 - x1)) / denom;
}
}
if ( partial && NuGeometry::IsBetween(0, ab, 1)) {
ab = 1-ab;
xy = 1-xy;
return true;
} else return false;
}