2本の線の交差のアルゴリズム？
2行あります。 XとYの2点を含む両方の行。これは、両方とも長さがあることを意味します。
行列式を使用する式と通常の代数を使用する式の2つの式があります。どちらが最も効率的に計算され、式はどのように見えますか？
コードで行列を使用するのに苦労しています。
これは私がこれまで持っていたものですが、より効率的ですか？
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);
}
}
Ax + By = Cという形式の2行があるとすると、かなり簡単に見つけることができます。
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;
here から取得
私は最近、紙に戻り、基本的な代数を使用してこの問題の解決策を見つけました。 2本の線で形成された方程式を解くだけで、有効な解が存在する場合は交点があります。
Githubリポジトリ 拡張 実装 をチェックして、doubleおよび tests で精度の問題を処理する可能性があります。
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);
}
}
2つの線/セグメント/線と長方形の交差点を見つける方法
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 + "]";
}
}
完全なサンプルは[こちら] [1]で説明されています