Intersection between two lines in coordinates

问题:

I can detect the intersection point of two lines, but if my line don't has the length of my screen, it detects the point, where it shouldn't be.

Here a preview: So, it shouldn't detect this intersection because the horizontal line isn't that long.

Code:

- (NSMutableArray *) intersectWithLines:(CGPoint)startPoint andEnd:(CGPoint)endPoint {
    NSMutableArray *intersects = [[NSMutableArray alloc] init];

    for(GameLine *line in [_lineBackground getLines]) {

        double lineStartX = line.startPos.x;
        double lineStartY = line.startPos.y;
        double tempEndX = line.endPos.x;
        double tempEndY = line.endPos.y;

        double d = ((startPoint.x - endPoint.x)*(lineStartY - tempEndY)) - ((startPoint.y - endPoint.y) * (lineStartX - tempEndX));

        if(d != 0) {            
            double sX = ((lineStartX - tempEndX) * (startPoint.x * endPoint.y - startPoint.y * endPoint.x) - (startPoint.x - endPoint.x) * (lineStartX * tempEndY - lineStartY * tempEndX)) / d;
            double sY = ((lineStartY - tempEndY) * (startPoint.x * endPoint.y - startPoint.y * endPoint.x) - (startPoint.y - endPoint.y) * (lineStartX * tempEndY - lineStartY * tempEndX)) / d;


            if([self isValidCGPoint:CGPointMake(sX, sY)]) {
                [intersects addObject:[NSValue valueWithCGPoint:CGPointMake(sX, sY)]];
            }            
        }
    }

    return intersects;
}

回答1:

If I understand your question correctly, you need to determine the intersection point of two line segments. This should work with the following method:

- (NSValue *)intersectionOfLineFrom:(CGPoint)p1 to:(CGPoint)p2 withLineFrom:(CGPoint)p3 to:(CGPoint)p4
{
    CGFloat d = (p2.x - p1.x)*(p4.y - p3.y) - (p2.y - p1.y)*(p4.x - p3.x);
    if (d == 0)
        return nil; // parallel lines
    CGFloat u = ((p3.x - p1.x)*(p4.y - p3.y) - (p3.y - p1.y)*(p4.x - p3.x))/d;
    CGFloat v = ((p3.x - p1.x)*(p2.y - p1.y) - (p3.y - p1.y)*(p2.x - p1.x))/d;
    if (u < 0.0 || u > 1.0)
        return nil; // intersection point not between p1 and p2
    if (v < 0.0 || v > 1.0)
        return nil; // intersection point not between p3 and p4
    CGPoint intersection;
    intersection.x = p1.x + u * (p2.x - p1.x);
    intersection.y = p1.y + u * (p2.y - p1.y);

    return [NSValue valueWithCGPoint:intersection];
}

回答2:

This is a slightly modified version of Hayden Holligan's answer to work with Swift 3:

func getIntersectionOfLines(line1: (a: CGPoint, b: CGPoint), line2: (a: CGPoint, b: CGPoint)) -> CGPoint {

    let distance = (line1.b.x - line1.a.x) * (line2.b.y - line2.a.y) - (line1.b.y - line1.a.y) * (line2.b.x - line2.a.x)
    if distance == 0 {
        print("error, parallel lines")
        return CGPoint.zero
    }

    let u = ((line2.a.x - line1.a.x) * (line2.b.y - line2.a.y) - (line2.a.y - line1.a.y) * (line2.b.x - line2.a.x)) / distance
    let v = ((line2.a.x - line1.a.x) * (line1.b.y - line1.a.y) - (line2.a.y - line1.a.y) * (line1.b.x - line1.a.x)) / distance

    if (u < 0.0 || u > 1.0) {
        print("error, intersection not inside line1")
        return CGPoint.zero
    }
    if (v < 0.0 || v > 1.0) {
        print("error, intersection not inside line2")
        return CGPoint.zero
    }

    return CGPoint(x: line1.a.x + u * (line1.b.x - line1.a.x), y: line1.a.y + u * (line1.b.y - line1.a.y))
}

回答3:

Swift version

func getIntersectionOfLines(line1: (a: CGPoint, b: CGPoint), line2: (a: CGPoint, b: CGPoint)) -> CGPoint {
        let distance = (line1.b.x - line1.a.x) * (line2.b.y - line2.a.y) - (line1.b.y - line1.a.y) * (line2.b.x - line2.a.x)
        if distance == 0 {
            print("error, parallel lines")
            return CGPointZero
        }

        let u = ((line2.a.x - line1.a.x) * (line2.b.y - line2.a.y) - (line2.a.y - line1.a.y) * (line2.b.x - line2.a.x)) / distance
        let v = ((line2.a.x - line1.a.x) * (line1.b.y - line1.a.y) - (line2.a.y - line1.a.y) * (line1.b.x - line1.a.x)) / distance

        if (u < 0.0 || u > 1.0) {
            print("error, intersection not inside line1")
            return CGPointZero
        }
        if (v < 0.0 || v > 1.0) {
            print("error, intersection not inside line2")
            return CGPointZero
        }

        return CGPointMake(line1.a.x + u * (line1.b.x - line1.a.x), line1.a.y + u * (line1.b.y - line1.a.y))
    }

回答4:

That's the correct equation:

+(CGPoint) intersection2:(CGPoint)u1 u2:(CGPoint)u2 v1:(CGPoint)v1 v2:(CGPoint)v2 {  
    CGPoint ret=u1;  
    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))  
    /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));  
    ret.x+=(u2.x-u1.x)*t;  
    ret.y+=(u2.y-u1.y)*t;  
    return ret;  
}  

Also, you can check this library to calculate line intersections: http://www.cprogramdevelop.com/5045485/

回答5:

Here is another solution in Swift 4.2. This is functionally identical to MartinR's solution but uses simd vectors and matrices to clean it up.

/// Protocol adoped by any type that models a line segment.
protocol LineSegment
{
    /// Point defining an end of a line segment.
    var p1: simd_double2 { get }
    /// Point defining an end of a line segment.
    var p2: simd_double2 { get }
}

extension LineSegment
{
    /// Calcualte the intersection between this line segment and another line
    /// segment.
    ///
    /// Algorithm from here:
    /// http://www.cs.swan.ac.uk/~cssimon/line_intersection.html
    ///
    /// - Parameter other: The other line segment.
    /// - Returns: The intersection point, or `nil` if the two line segments are
    ///            parallel or the intersection point would be off the end of
    ///            one of the line segments.
    func intersection(lineSegment other: LineSegment) -> simd_double2?
    {
        let p3 = other.p1 // Name the points so they are consistent with the explanation below
        let p4 = other.p2
        let matrix = simd_double2x2(p4 - p3, p1 - p2)
        guard matrix.determinant != 0 else { return nil } // Determinent == 0 => parallel lines
        let multipliers = matrix.inverse * (p1 - p3)
        // If either of the multipliers is outside the range 0 ... 1, then the
        // intersection would be off the end of one of the line segments.
        guard (0.0 ... 1.0).contains(multipliers.x) && (0.0 ... 1.0).contains(multipliers.y)
            else { return nil }
        return p1 + multipliers.y * (p2 - p1)
    }
}

The algorithm works because, if you have line segment a defined by two points p₁ and p₂ and line segment b defined by p₃ and p₄ the points on a and b are respectively defined by

• p₁ + t_a(p₂ - p₁)
• p₃ + t_b(p₄ - p₃)

so the point of intersection would be where

p₁ + t_a(p₂ - p₁) = p₃ + t_b(p₄ - p₃)

This can be rearranged as

p₁ - p₃ = t_b(p₄ - p₃) + t_a(p₁ - p₂)

and with a bit of jiggery pokery you can get to the following equivalent

p₁ - p₃ = A.t

where t is the vector (t_b, t_a) and A is the matrix whose columns are p₄ - p₃ and p₁ - p₂

The equation can be rearranged as

A^-1(p₁ - p₃) = t

Everything on the left hand side is already known or can be calculated to get us t. Either of the components of t can be plugged into the respective original equation to get the intersection point (NB floating point rounding errors will mean that the two answers probably aren't exactly the same but are very close).

Note that, if the lines are parallel, the determinant of A will be zero. Also, if either component is outside the range 0 ... 1, then one or both line segments needs to be extended to get to the point of intersection.