C# 两条直线相交的算法？
我有两行。包含X和Y两个点的两条线。这意味着它们都有长度 我看到两个公式，一个使用行列式，一个使用正规代数。哪一种计算效率最高，公式是什么样的 我很难在代码中使用矩阵 这就是我目前所拥有的，它能更有效吗C# 两条直线相交的算法？,c#,algorithm,linear-algebra,C#,Algorithm,Linear Algebra,我有两行。包含X和Y两个点的两条线。这意味着它们都有长度 我看到两个公式，一个使用行列式，一个使用正规代数。哪一种计算效率最高，公式是什么样的 我很难在代码中使用矩阵 这就是我目前所拥有的，它能更有效吗 public static Vector3 Intersect(Vector3 line1V1, Vector3 line1V2, Vector3 line2V1, Vector3 line2V2) { //Line1 float A1 = line1V2.Y - line1V1
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);
}
}
假设您有两行
Ax+By=Cfloat 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;
从如何找到两条直线/线段/光线与矩形的交点
public class LineEquation{
public LineEquation(Point start, Point end){
Start = start;
End = end;
IsVertical = Math.Abs(End.X - start.X) < 0.00001f;
M = (End.Y - Start.Y)/(End.X - Start.X);
A = -M;
B = 1;
C = Start.Y - M*Start.X;
}
public bool IsVertical { get; private set; }
public double M { get; private set; }
public Point Start { get; private set; }
public Point End { get; private set; }
public double A { get; private set; }
public double B { get; private set; }
public double C { get; private set; }
public bool IntersectsWithLine(LineEquation otherLine, out Point intersectionPoint){
intersectionPoint = new Point(0, 0);
if (IsVertical && otherLine.IsVertical)
return false;
if (IsVertical || otherLine.IsVertical){
intersectionPoint = GetIntersectionPointIfOneIsVertical(otherLine, this);
return true;
}
double delta = A*otherLine.B - otherLine.A*B;
bool hasIntersection = Math.Abs(delta - 0) > 0.0001f;
if (hasIntersection){
double x = (otherLine.B*C - B*otherLine.C)/delta;
double y = (A*otherLine.C - otherLine.A*C)/delta;
intersectionPoint = new Point(x, y);
}
return hasIntersection;
}
private static Point GetIntersectionPointIfOneIsVertical(LineEquation line1, LineEquation line2){
LineEquation verticalLine = line2.IsVertical ? line2 : line1;
LineEquation nonVerticalLine = line2.IsVertical ? line1 : line2;
double y = (verticalLine.Start.X - nonVerticalLine.Start.X)*
(nonVerticalLine.End.Y - nonVerticalLine.Start.Y)/
((nonVerticalLine.End.X - nonVerticalLine.Start.X)) +
nonVerticalLine.Start.Y;
double x = line1.IsVertical ? line1.Start.X : line2.Start.X;
return new Point(x, y);
}
public bool IntersectWithSegementOfLine(LineEquation otherLine, out Point intersectionPoint){
bool hasIntersection = IntersectsWithLine(otherLine, out intersectionPoint);
if (hasIntersection)
return intersectionPoint.IsBetweenTwoPoints(otherLine.Start, otherLine.End);
return false;
}
public bool GetIntersectionLineForRay(Rect rectangle, out LineEquation intersectionLine){
if (Start == End){
intersectionLine = null;
return false;
}
IEnumerable<LineEquation> lines = rectangle.GetLinesForRectangle();
intersectionLine = new LineEquation(new Point(0, 0), new Point(0, 0));
var intersections = new Dictionary<LineEquation, Point>();
foreach (LineEquation equation in lines){
Point point;
if (IntersectWithSegementOfLine(equation, out point))
intersections[equation] = point;
}
if (!intersections.Any())
return false;
var intersectionPoints = new SortedDictionary<double, Point>();
foreach (var intersection in intersections){
if (End.IsBetweenTwoPoints(Start, intersection.Value) ||
intersection.Value.IsBetweenTwoPoints(Start, End)){
double distanceToPoint = Start.DistanceToPoint(intersection.Value);
intersectionPoints[distanceToPoint] = intersection.Value;
}
}
if (intersectionPoints.Count == 1){
Point endPoint = intersectionPoints.First().Value;
intersectionLine = new LineEquation(Start, endPoint);
return true;
}
if (intersectionPoints.Count == 2){
Point start = intersectionPoints.First().Value;
Point end = intersectionPoints.Last().Value;
intersectionLine = new LineEquation(start, end);
return true;
}
return false;
}
public override string ToString(){
return "[" + Start + "], [" + End + "]";
}
}
公共类线性方程{
公共直线方程（点起点、点终点）{
开始=开始；
结束=结束；
IsVertical=数学绝对值（End.X-start.X）<0.00001f；
M=（End.Y-Start.Y）/（End.X-Start.X）；
A=-M；
B=1；
C=Start.Y-M*Start.X；
}
公共bool是垂直的{get；private set；}
公共双M{get；私有集；}
公共点开始{get；private set；}
公共点结束{get；private set；}
公共双A{get；私有集；}
公共双B{get；私有集；}
公共双C{get；私有集；}
公共边界相交线（直线、外点相交点）{
相交点=新点（0,0）；
if（IsVertical&&otherLine.IsVertical）
返回false；
if（IsVertical | | otherLine.IsVertical）{
intersectionPoint=GetIntersectionPointIfOneIsVertical（另一条线，此线）；
返回true；
}
双三角=A*otherLine.B-otherLine.A*B；
布尔交叉点=数学绝对值（δ-0）>0.0001f；
if（交叉口）{
双x=（otherLine.B*C-B*otherLine.C）/delta；
双y=（A*otherLine.C-otherLine.A*C）/delta；
相交点=新点（x，y）；
}
返回交叉口；
}
私有静态点GetIntersectionPointIfOneIsVertical（直线方程line1，直线方程line2）{
LineEquation verticalLine=line2.Is垂直？line2:line1；
LineEquation NonVerticaline=line2.Is垂直？line1:line2；
双y=（verticalLine.Start.X-nonVerticalLine.Start.X）*
（nonVerticalLine.End.Y-nonVerticalLine.Start.Y）/
（（nonVerticalLine.End.X-nonVerticalLine.Start.X））+
非维特卡林；
double x=line1.IsVertical？line1.Start.x:line2.Start.x；
返回新点（x，y）；
}
公共布尔线与边线相交（边线、外点相交点）{
bool hasIntersection=相交线（其他线，外相交点）；
if（交叉口）
返回两个点之间的交点.Is（otherLine.Start，otherLine.End）；
返回false；
}
公共bool GetIntersectionLineForRay（矩形、外线方程intersectionLine）{
如果（开始==结束）{
相交线=空；
返回false；
}
IEnumerable lines=rectangle.GetLinesForRectangle（）；
相交线=新线方程（新点（0，0），新点（0，0））；
var=newdictionary（）；
foreach（直线方程中的直线方程）{
点-点；
if（与直线（方程式，外点）相交）
交点[方程]=点；
}
如果（！crossions.Any（））
返回false；
var intersectionPoints=new SortedDictionary（）；
foreach（交叉点中的变量交叉点）{
两个点（起点、交点、值）之间的if（End.Is）||
两个点之间的交点.Value.Is（起点、终点））{
double distanceToPoint=Start.distanceToPoint（intersection.Value）；
相交点[distanceToPoint]=相交点的值；
}
}
if（intersectionPoints.Count==1）{
点端点=intersectionPoints.First（）.值；
相交直线=新直线方程（起点、终点）；
返回true；
}
if（intersectionPoints.Count==2）{
点开始=相交点.First（）.值；
点端点=相交点.Last（）.值；
相交直线=新直线方程（起点、终点）；
返回true；
}
返回false；
}
公共重写字符串ToString（）{
返回“[”+开始+”]，[“+结束+”]；
}
}
完整的示例在[这里][1]我最近回到纸上，用基本代数找到了这个问题的解决方案。只要解出由两条线组成的方程，如果存在有效解，则存在交点 检查my是否存在双精度和双精度的扩展处理潜在精度问题
公共结构行
{
公共双x1{get；set；}
公共双y1{get；set；}
公共双x2{get；set；}
公共双y2{get；set；}
}
公共结构点
{
公共双x{get；set；}
公共双y{get；set；}
}
公共类线路交叉口
{
//如果不相交，则返回交点，否则返回默认点（null）
公共静态点FindIntersection（直线A，直线B，双公差=0.001）
{
双x1=线性A.x1，y1=线性A.y1；
双x2=lineA.x2，y2=lineA.y2；
双x3=线路B.x1，y3=线路B.y1；
双x4=lineB.x2，y4=lineB.y2；
//形式为x=c的方程式（两条垂直线）
如果（数学Abs（x1-x2）<公差和数学Abs（x3-x4）<公差和数学Abs（x1-x3）<公差）
{
抛出新异常（“两条线垂直重叠，交点不明确”）；
}
//形式为y=c的方程式（两条水平线）
如果（数学Abs（y1-y2）<公差和数学Abs（y3-y4）<公差和数学Abs（y1-y3
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 lines)
if (Math.Abs(x1 - x2) < tolerance && Math.Abs(x3 - x4) < tolerance)
{
return default(Point);
}
//equations of the form y=c (two horizontal lines)
if (Math.Abs(y1 - y2) < tolerance && Math.Abs(y3 - y4) < tolerance)
{
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(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 null (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);
}
}