如何分辨圆和矩形在二维欧几里得空间中是否相交?(即经典二维几何)
当前回答
我为处理形状创建了类 希望你喜欢
public class Geomethry {
public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectCenterX = rectangleX + rectHalfWidth;
float rectCenterY = rectangleY + rectHalfHeight;
float deltax = Math.abs(rectCenterX - circleX);
float deltay = Math.abs(rectCenterY - circleY);
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectHalfWidth2 = rectangleWidth2/2.0f;
float rectHalfHeight2 = rectangleHeight2/2.0f;
float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle
if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
}
其他回答
假设你有矩形的四条边，检查从这些边到圆心的距离，如果小于半径，那么这些形状是相交的。
if sqrt((rectangleRight.x - circleCenter.x)^2 +
(rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect
if sqrt((rectangleRight.x - circleCenter.x)^2 +
(rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect
if sqrt((rectangleLeft.x - circleCenter.x)^2 +
(rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect
if sqrt((rectangleLeft.x - circleCenter.x)^2 +
(rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect
我为处理形状创建了类 希望你喜欢
public class Geomethry {
public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectCenterX = rectangleX + rectHalfWidth;
float rectCenterY = rectangleY + rectHalfHeight;
float deltax = Math.abs(rectCenterX - circleX);
float deltay = Math.abs(rectCenterY - circleY);
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectHalfWidth2 = rectangleWidth2/2.0f;
float rectHalfHeight2 = rectangleHeight2/2.0f;
float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle
if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
}
为了可视化，拿你的键盘的numpad。如果键“5”代表你的矩形，那么所有的键1-9代表空间的9个象限除以构成矩形的线(5是里面的线)。
1)如果圆的中心在象限5(即在矩形内)，则两个形状相交。
这里有两种可能的情况: a)圆与矩形的两条或多条相邻边相交。 b)圆与矩形的一条边相交。
第一种情况很简单。如果圆与矩形的两条相邻边相交，则它必须包含连接这两条边的角。(或者说它的中心在象限5，我们已经讲过了。还要注意，圆只与矩形的两条相对边相交的情况也被覆盖了。)
2)如果矩形的任意角A、B、C、D在圆内，则这两个形状相交。
第二种情况比较棘手。我们应该注意到，只有当圆的中心位于2、4、6或8象限中的一个象限时，才会发生这种情况。(事实上，如果中心在1、3、7、8象限中的任何一个象限上，则相应的角将是离它最近的点。)
现在我们有了圆的中心在一个“边”象限内的情况，它只与相应的边相交。那么，边缘上最接近圆中心的点必须在圆内。
3)对于每条直线AB, BC, CD, DA，构造经过圆中心p的垂直线p(AB, p)， p(BC, p)， p(CD, p)， p(DA, p)，对于每条垂直线，如果与原边的交点在圆内，则两个图形相交。
最后一步有一个捷径。如果圆的圆心在象限8，边AB是上边，交点的y坐标是A和B, x坐标是P。
你可以构造四条线的交点并检查它们是否在相应的边上，或者找出P在哪个象限并检查相应的交点。两者都应该化简为相同的布尔方程。要注意的是，上面的步骤2并没有排除P位于“角落”象限之一;它只是在寻找一个十字路口。
编辑:事实证明，我忽略了一个简单的事实，即#2是#3的子情况。毕竟，角也是边缘上的点。请看下面@ShreevatsaR的回答，你会得到很好的解释。与此同时，忘记上面的第二条，除非你想要一个快速但冗余的检查。
球面和矩形相交于IIF 圆心和矩形的一个顶点之间的距离小于球体的半径 或 圆心与矩形的一条边之间的距离小于球面的半径([点线距离]) 或 圆的中心在矩形的内部 一点上距离:
P1 = [x1,y1] P2 = [x2,y2] Distance = sqrt(abs(x1 - x2)+abs(y1-y2))
点线路距离:
L1 = [x1,y1],L2 = [x2,y2] (two points of your line, ie the vertex points) P1 = [px,py] some point Distance d = abs( (x2-x1)(y1-py)-(x1-px)(y2-y1) ) / Distance(L1,L2)
矩形内圆中心: 采用分离轴的方法:如果存在一个投影到一条直线上，将矩形与点分开，它们就不相交
您将点投影在平行于矩形边的直线上，然后可以很容易地确定它们是否相交。如果它们不在所有4个投影上相交，它们(点和矩形)就不能相交。
你只需要内积(x= [x1,x2]，y = [y1,y2]，x *y = x1*y1 + x2*y2)
你的测试应该是这样的:
//rectangle edges: TL (top left), TR (top right), BL (bottom left), BR (bottom right)
//point to test: POI
seperated = false
for egde in { {TL,TR}, {BL,BR}, {TL,BL},{TR-BR} }: // the edges
D = edge[0] - edge[1]
innerProd = D * POI
Interval_min = min(D*edge[0],D*edge[1])
Interval_max = max(D*edge[0],D*edge[1])
if not ( Interval_min ≤ innerProd ≤ Interval_max )
seperated = true
break // end for loop
end if
end for
if (seperated is true)
return "no intersection"
else
return "intersection"
end if
它没有假设一个轴对齐的矩形，并且很容易扩展用于测试凸集之间的交集。
为我工作(只工作时，矩形的角度是180)
function intersects(circle, rect) {
let left = rect.x + rect.width > circle.x - circle.radius;
let right = rect.x < circle.x + circle.radius;
let top = rect.y < circle.y + circle.radius;
let bottom = rect.y + rect.height > circle.y - circle.radius;
return left && right && bottom && top;
}
