Question

我可以检测到两条线的交点，但是如果我的线条没有我的屏幕长度，它会检测到它不应该的位置。

预览： Intersection 因此，它不应该检测到这个交叉点，因为水平线不是那么长。

代码：

- (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;
}

Answer 1

如果我正确理解您的问题，您需要确定两个线段的交叉点。这应该使用以下方法：

- (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];
}

Answer 2

这是Hayden Holligan's answer的略微修改版本，可与Swift 3配合使用：

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))
}

Answer 3

Swift版

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))
    }

Answer 4

这是正确的等式：

+(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;  
}

此外，您可以检查此库以计算线交叉点： http://www.cprogramdevelop.com/5045485/

Answer 5

这是Swift 4.2中的另一个解决方案。这在功能上与MartinR的解决方案相同，但是使用simd向量和矩阵对其进行清理。

/// 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)
    }
}

该算法之所以有效，是因为如果您有由两个点 p ₁ 和 p _{2定义的线段 a} 和由 p ₃ 和 p ₄定义的线段 b a 和 b 上的点分别由

定义

p ₁ + t _a（ p ₂ - p ₁ ）
p ₃ + t _b（ p ₄ - p ₃ ）

所以交点将在

p ₁ + t _a（ p ₂ - p ₁ ）= p ₃ + t _b（ p ₄ - p ₃ ）

可以重新排列为

p ₁ - p ₃ = t _b（ p ₄ - p ₃ ）+ t _a（ p ₁ - p ₂ ）

加上一点点扑克游戏，您可以得到以下等价物

p ₁ - p ₃ = A 。 t

其中 t 是向量（t _b，t _a）， A 是其列的矩阵是 p ₄ - p ₃ 和 p ₁ >- p ₂

方程式可以重新排列为

A ^-1（ p ₁ - p ₃ ）= t

左侧的所有信息都是已知的，或者可以通过计算得出 t 。可以将 t 的任意一个分量插入相应的原始方程式以获得交点（NB浮点数舍入误差将意味着两个答案可能并不完全相同，但非常接近）

请注意，如果线是平行的，则 A 的行列式将为零。另外，如果任一分量超出0 ... 1范围，则需要扩展一个或两个线段才能到达交点。

Answer 6

我知道答案已经给出，而且所有答案都是正确的，我觉得可以回答这个问题。就在这里。

$(document).ready(function () {
jQuery(function ($) {
    var ft = FooTable.init('.table', {
        "columns": $.get('/js/mycols.json'),
    });
    $('.table').on('ready.ft.table', function(e) {
      $.ajax({
          url: "/api/GetEvents",
          dataType: "json",
          type: "GET"
      }).done(function (e) {
          console.log(e);
          ft.rows.load(e.value);
      })             
    });
  });
});

我是从this site那里得到的。

这也是非常简单和容易的。

快乐编码：）

坐标中两条线之间的交点

6 个答案: