C# 获取零星结果

Question

所以我实际上已经在这个问题上坚持了几个小时了，这非常令人沮丧。我得到零星的结果。该程序有时可以工作，但有时它找不到我正在寻找的“切线”之一，即使从数学上来说肯定存在一个。

我对 c# 真的很陌生，所以如果有人有时间帮助菜鸟，我将不胜感激。

private List<PointF> SplitAndSolve(Graphics g, List<PointF> pointList)
    {
        if (pointList.Count < 4)
            return pointList;
        List<List<PointF>> leftAndRight = SplitByX(pointList);
        List<PointF> left = leftAndRight[0];
        List<PointF> right = leftAndRight[1];
        drawPolygon(g, left);// just a way to visually assess the correctness
        drawPolygon(g, right);// same
        left = SplitAndSolve(g, left);
        right = SplitAndSolve(g, right);
        Combine(g, left, right);
        return pointList;
    }
    private List<PointF> Combine(Graphics g, List<PointF> left, List<PointF> right)
    {
        //find tangents
        List<PointF> topTangents = GetTangents(g, left, right, TOP);
        drawPoint(g, topTangents[0]);//visual debug
        drawPoint(g, topTangents[1]);//visual debug
        List<PointF> botTangents = GetTangents(g, left, right, BOT);
        drawPoint(g, botTangents[0]);//visual debug
        drawPoint(g, botTangents[1]);//""
        // get a new polygon

        return left;// just a place holder so I don't get errors for the time being
    }
    private List<PointF> GetTangents(Graphics g, List<PointF> left, List<PointF> right, float topOrBot)
    {
        List<PointF> tangents = new List<PointF>();
        foreach (PointF leftAnchor in left)
        {
            foreach (PointF rightAnchor in right)
            {
                double lax = leftAnchor.X;
                double lay = leftAnchor.Y;
                double rax = rightAnchor.X;
                double ray = rightAnchor.Y;
                double m = (lay - ray) / (lax - rax);
                double b = (-1 * m * lax) + lay;
                bool isTangent = true;
                foreach (PointF lpoi in left)
                {
                    if ((topOrBot == TOP) && (Test(m, b, lpoi) > 0))
                    {
                        isTangent = false;

                    }
                    if ((topOrBot == BOT) && (Test(m, b, lpoi) < 0))
                    {
                        isTangent = false;

                    }
                }

                foreach (PointF rpoi in right)
                {
                    if ((topOrBot == TOP) && (Test(m, b, rpoi) > 0))
                    {
                        isTangent = false;

                    }
                    if ((topOrBot == BOT) && (Test(m, b, rpoi) < 0))
                    {
                        isTangent = false;

                    }
                }
                if (isTangent)
                {
                    tangents.Add(leftAnchor);
                    tangents.Add(rightAnchor);
                    return tangents;
                }

            }

        }
        return null;
    }

    /*  Test, test to see the location of a point in relation to a line
     *  @float m slope of the line
     *  @float b the constast of the y intercept form
     *  @Pointf r is the point to be tested against the line
     * 
     *  returns some k > 0 if point is below the line
     *  returns some k < 0 if point is above the line
     *  returns 0 if point is found along the line
     */
    private double Test(double m, double b, PointF r)
    {
        return (m*(double)r.X) + b - (double)r.Y;
    }

所以我有点相信这是一个编程错误，因为我已经一遍遍地重复它，尽管我可能是错的。

如果这是一个不恰当的帖子，我很抱歉，我真的只是被困住了，我无意打扰任何人，如果我滥用论坛，请告诉我。

我找到了穿过两个顶点的线，每个顶点都有一个，所有其他线要么在下面（顶部切线），要么在上面（底部切线）。错误在于，使用该算法，它应该始终迭代循环永远不会返回 null。但有时确实如此。我猜这是一个精度错误。

Answer 1

这可能只是 double 的精度问题吗？也许你应该让你的测试函数“模糊”。

像这样：

  double result = (m*(double)r.X) + b - (double)r.Y;

  if ((result > -0.0001) && (result < 0.0001))
     return 0.0;
  else
     return result;