帮助将Python应用程序转换为C#

Question

所有人

下面是一个小python应用程序的链接：

http://en.wikipedia.org/wiki/File:Beta-skeleton.svg

我想我已经正确地转换了它。（来源于帖子底部）

但是数学。Acos总是返回NaN。python版本的acos和数学。Acos？在

    private Random rnd = new Random();
    private double scale = 5;
    private double radius = 10;
    private double beta1 = 1.1;
    private double beta2 = 0.9;
    private double theta1;
    private double theta2;

    private Point[] points = new Point[10];

    public MainWindow()
    {
        InitializeComponent();
        for (int i = 0; i < 100; i++ )
        {
            points[i] = new Point((rnd.NextDouble() * scale), 
                (rnd.NextDouble() * scale));
        }

        theta1 = Math.Asin(1/beta1);
        theta2 = Math.PI - Math.Asin(beta2);
    }

    private double Dot(Point p, Point q, Point r)
    {
        var pr = new Point();
        var qr = new Point();

        //(p[0]-r[0])
        pr.X = p.X-r.X;

        //(p[1]-r[1])
        pr.Y = p.Y-r.Y;

        //(q[0]-r[0])
        qr.X = q.X-r.X;

        //(q[1]-r[1])
        qr.Y = q.Y-r.Y;

        return (pr.X*qr.X) + (pr.Y*qr.Y);
    }

private double Sharp(Point p,Point q)
{
    double theta = 0;

    foreach(var pnt in points)
    {
        if(pnt!=p && pnt!=q)
        {
            var dotpq = Dot(p, q, pnt);
            double t = Math.Acos(dotpq);
            double u = Math.Pow((dotpq * dotpq), 0.5);

            var tempVal = t/u;

            theta = Math.Max(theta, tempVal);
        }
    }
    return theta;

}

    private void DrawPoint(Point p)
    {
        var e = new Ellipse
                    {
                        Width = radius/2,
                        Height = radius/2,
                        Stroke = Brushes.Red,
                        Visibility = Visibility.Visible
                    };

        Canvas.SetTop(e, p.Y + radius);
        Canvas.SetLeft(e, p.X + radius);

        MyCanvas.Children.Add(e);
    }

    private void DrawEdge1(Point p,Point q)
    {
        var l = new Line
                    {
                        X1 = p.X,
                        Y1 = p.Y,
                        X2 = q.X,
                        Y2 = q.Y,
                        Stroke = Brushes.Black,
                        Width = 1,
                        Visibility = Visibility.Visible
                    };

        MyCanvas.Children.Add(l);
    }

    private void DrawEdge2(Point p,Point q)
    {
        var l = new Line
                    {
                        X1 = p.X,
                        Y1 = p.Y,
                        X2 = q.X,
                        Y2 = q.Y,
                        Stroke = Brushes.Blue,
                        Width = 1,
                        Visibility = Visibility.Visible
                    };

        MyCanvas.Children.Add(l);
    }

    private void Window_Loaded(object sender, RoutedEventArgs e)
    {
        foreach (var p in points)
        {
            foreach (var q in points)
            {
                var theta = Sharp(p, q);

                if(theta < theta1) DrawEdge1(p, q);
                else if(theta < theta2) DrawEdge2(p, q);

            }                
        }
    }       

Answer 1

问题是当函数被调用时，dotpq的值是多少。它必须是一个介于-1和1之间的双精度值，如docs中所述。在

Answer 2

你需要做的是从点积得到角度，在aco之前去掉长度。在

python拥有：

prq = acos(dot(p,q,r) / (dot(p,p,r)*dot(q,q,r))**0.5)

你所做的不是在ACO中划分，而是在之后划分。在

所以：

^{pr2}$

当然要改变变量，我只是想让它和python保持类似，以帮助您理解：）

Answer 3

我想这是因为你翻译了Python表达式(dot(p,q,r) / (dot(p,p,r) * dot(q,q,r)) **0.5)。在Python中，求幂有一个最小的运算符，因此取子项dot(p,q,r) / (dot(p,p,r) * dot(q,q,r))的平方根。在C版本中，当计算双“u”的值时，只取最后两项乘积的平方根，即(dotpq * dotpq)。在