两条线的交叉算法？

Question

我有2行.两条线都包含它们的2点X和Y.这意味着它们都有长度.

我看到2个公式,一个使用决定因素,一个使用普通代数.哪个是最有效的计算方法,公式是什么样的？

我在代码中使用矩阵很困难.

这是我到目前为止,它是否更有效？

public static Vector3 Intersect(Vector3 line1V1, Vector3 line1V2, Vector3 line2V1, Vector3 line2V2)
{
    //Line1
    float A1 = line1V2.Y - line1V1.Y;
    float B1 = line1V2.X - line1V1.X;
    float C1 = A1*line1V1.X + B1*line1V1.Y;

    //Line2
    float A2 = line2V2.Y - line2V1.Y;
    float B2 = line2V2.X - line2V1.X;
    float C2 = A2 * line2V1.X + B2 * line2V1.Y;

    float det = A1*B2 - A2*B1;
    if (det == 0)
    {
        return null;//parallel lines
    }
    else
    {
        float x = (B2*C1 - B1*C2)/det;
        float y = (A1 * C2 - A2 * C1) / det;
        return new Vector3(x,y,0);
    }
}

Answer 1

假设您有两行表单Ax + By = C,您可以很容易地找到它:

float delta = A1 * B2 - A2 * B1;

if (delta == 0) 
    throw new ArgumentException("Lines are parallel");

float x = (B2 * C1 - B1 * C2) / delta;
float y = (A1 * C2 - A2 * C1) / delta;

从这里拉出来

Answer 2

我最近回到了纸上，用基本代数找到了解决这个问题的方法。我们只需要求解由两条线组成的方程，如果存在有效的解，那么就有一个交集。

您可以查看我的Github 存储库以获取处理潜在精度问题的扩展实现double 和tests。

public struct Line
{
    public double x1 { get; set; }
    public double y1 { get; set; }

    public double x2 { get; set; }
    public double y2 { get; set; }
}

public struct Point
{
    public double x { get; set; }
    public double y { get; set; }
}

public class LineIntersection
{
    //  Returns Point of intersection if do intersect otherwise default Point (null)
    public static Point FindIntersection(Line lineA, Line lineB, double tolerance = 0.001)
    {
        double x1 = lineA.x1, y1 = lineA.y1;
        double x2 = lineA.x2, y2 = lineA.y2;

        double x3 = lineB.x1, y3 = lineB.y1;
        double x4 = lineB.x2, y4 = lineB.y2;

        // equations of the form x = c (two vertical lines)
        if (Math.Abs(x1 - x2) < tolerance && Math.Abs(x3 - x4) < tolerance && Math.Abs(x1 - x3) < tolerance)
        {
            throw new Exception("Both lines overlap vertically, ambiguous intersection points.");
        }

        //equations of the form y=c (two horizontal lines)
        if (Math.Abs(y1 - y2) < tolerance && Math.Abs(y3 - y4) < tolerance && Math.Abs(y1 - y3) < tolerance)
        {
            throw new Exception("Both lines overlap horizontally, ambiguous intersection points.");
        }

        //equations of the form x=c (two vertical parallel lines)
        if (Math.Abs(x1 - x2) < tolerance && Math.Abs(x3 - x4) < tolerance)
        {   
            //return default (no intersection)
            return default(Point);
        }

        //equations of the form y=c (two horizontal parallel lines)
        if (Math.Abs(y1 - y2) < tolerance && Math.Abs(y3 - y4) < tolerance)
        {
            //return default (no intersection)
            return default(Point);
        }

        //general equation of line is y = mx + c where m is the slope
        //assume equation of line 1 as y1 = m1x1 + c1 
        //=> -m1x1 + y1 = c1 ----(1)
        //assume equation of line 2 as y2 = m2x2 + c2
        //=> -m2x2 + y2 = c2 -----(2)
        //if line 1 and 2 intersect then x1=x2=x & y1=y2=y where (x,y) is the intersection point
        //so we will get below two equations 
        //-m1x + y = c1 --------(3)
        //-m2x + y = c2 --------(4)

        double x, y;

        //lineA is vertical x1 = x2
        //slope will be infinity
        //so lets derive another solution
        if (Math.Abs(x1 - x2) < tolerance)
        {
            //compute slope of line 2 (m2) and c2
            double m2 = (y4 - y3) / (x4 - x3);
            double c2 = -m2 * x3 + y3;

            //equation of vertical line is x = c
            //if line 1 and 2 intersect then x1=c1=x
            //subsitute x=x1 in (4) => -m2x1 + y = c2
            // => y = c2 + m2x1 
            x = x1;
            y = c2 + m2 * x1;
        }
        //lineB is vertical x3 = x4
        //slope will be infinity
        //so lets derive another solution
        else if (Math.Abs(x3 - x4) < tolerance)
        {
            //compute slope of line 1 (m1) and c2
            double m1 = (y2 - y1) / (x2 - x1);
            double c1 = -m1 * x1 + y1;

            //equation of vertical line is x = c
            //if line 1 and 2 intersect then x3=c3=x
            //subsitute x=x3 in (3) => -m1x3 + y = c1
            // => y = c1 + m1x3 
            x = x3;
            y = c1 + m1 * x3;
        }
        //lineA & lineB are not vertical 
        //(could be horizontal we can handle it with slope = 0)
        else
        {
            //compute slope of line 1 (m1) and c2
            double m1 = (y2 - y1) / (x2 - x1);
            double c1 = -m1 * x1 + y1;

            //compute slope of line 2 (m2) and c2
            double m2 = (y4 - y3) / (x4 - x3);
            double c2 = -m2 * x3 + y3;

            //solving equations (3) & (4) => x = (c1-c2)/(m2-m1)
            //plugging x value in equation (4) => y = c2 + m2 * x
            x = (c1 - c2) / (m2 - m1);
            y = c2 + m2 * x;

            //verify by plugging intersection point (x, y)
            //in orginal equations (1) & (2) to see if they intersect
            //otherwise x,y values will not be finite and will fail this check
            if (!(Math.Abs(-m1 * x + y - c1) < tolerance
                && Math.Abs(-m2 * x + y - c2) < tolerance))
            {
                //return default (no intersection)
                return default(Point);
            }
        }

        //x,y can intersect outside the line segment since line is infinitely long
        //so finally check if x, y is within both the line segments
        if (IsInsideLine(lineA, x, y) &&
            IsInsideLine(lineB, x, y))
        {
            return new Point { x = x, y = y };
        }

        //return default (no intersection)
        return default(Point);

    }

    // Returns true if given point(x,y) is inside the given line segment
    private static bool IsInsideLine(Line line, double x, double y)
    {
        return (x >= line.x1 && x <= line.x2
                    || x >= line.x2 && x <= line.x1)
               && (y >= line.y1 && y <= line.y2
                    || y >= line.y2 && y <= line.y1);
    }
}