c# - 获取最少数量的点来创建相同的多边形

Question

我想做什么:

我想从将创建相同多边形的多边形中获取最少量的点:

例如，如果我有这个多边形:

(0, 0), (1, 0), (2, 0), (3, 0), (4, 0), (0,4), (4,4)

它将创建一个位置为 0, 0、宽度为 4、高度为 4 的多边形。

如果我将这个多边形输入到假设算法中，它将返回:

(0, 0), (4, 0), (0, 4), (4, 4)

我为什么要这样做:

我正在创建一个游戏，游戏有动画，每个动画都有自己的图像和多边形(图像的边界)，我已经有了动画的图像，但我没有多边形当然，我可以自己创建多边形，但是手动为 100 多张图像创建多边形会很累，更不用说添加/修改动画了。

我尝试过的:

我的想法是这样的:

逐个像素地扫描图像，检查像素是否为空白，如果不是，则将其添加到列表中，完成后使用某种算法获得最少数量的点来创建相同的多边形。

我做了一些研究，我认为 LLTS(广场万岁)算法是我需要的，所以我使用 cansik 编写了这段代码的 implementation在 C# 中:

    private readonly Bitmap _image;
    private Point[] _result;

    private void Calculate()
    {
        List<Vector2D> points = new List<Vector2D>();
        for (int x = 0; x < _image.Width; x++)
        {
            for (int y = 0; y < _image.Height; y++)
            {
                // Check if the pixel is blank
                if (_image.GetPixel(x, y).ToArgb() != 16777215)
                {
                    // If the pixel isn't blank, add it to the list
                    points.Add(new Vector2D(x, y));
                }

            }
        }
        Vector2D[] resultInVectors = GeoAlgos.MonotoneChainConvexHull(points.ToArray());
        _result = new Point[resultInVectors.Length];
        for (int i = 0; i < resultInVectors.Length; i++)
        {
            _result[i] = new Point((int)resultInVectors[i].X, (int)resultInVectors[i].Y);
        }
    }

我添加了绘画代码:

private void Form_Paint(object sender, PaintEventArgs e)
    {

        e.Graphics.DrawPolygon(Pens.Black, _result);
        e.Graphics.DrawImage(_image, new Point(100, 0));
    }

最后，我运行程序，这是我得到的:

这与我的想法完全不同，至少可以说，我希望它是这样的:

有什么想法吗？

编辑 - 最终解决了它

我用了Wowa对 Trevor Elliott 的回答的 question ，然后，我通过使用我创建的这个函数最小化了结果中的点数:

    private static List<Point> MinimizePoints(List<Point> points)
    {
        if (points.Count < 3)
        {
            return points;
        }
        List<Point> minimumPoints = new List<Point>(points);

        for (int i = minimumPoints.Count - 1; i > 2; i -= 3)
        {
            List<Point> currentPoints = minimumPoints.GetRange(i - 3, 3);
            try
            {
                if ((currentPoints[2].X - currentPoints[0].X) / (currentPoints[1].X - currentPoints[0].X) ==
                    (currentPoints[2].Y - currentPoints[0].Y) / (currentPoints[1].Y - currentPoints[0].Y))
                {
                    minimumPoints.Remove(minimumPoints[i + 1]);
                }
            }
            catch (DivideByZeroException)
            {
                // Ignore
            }
        }
        return minimumPoints;
    }

我用了Oliver Charlesworth对 Prashant C 的回答的 question .

第二次编辑 - 更优化的解决方案

我没有使用我自己的 meh 算法来减少点数，而是使用了 Ramer–Douglas–Peucker 算法并将 ε(公差)设置为 0。这是我使用的实现:

private static class DouglasPeuckerReduction
    {
        public static Point[] ReducePoints(Point[] existingPolygon)
        {
            if (existingPolygon == null || existingPolygon.Length < 3)
                return existingPolygon;

            int firstPoint = 0;
            int lastPoint = existingPolygon.Length - 1;
            List<int> pointIndexsToKeep = new List<int>();

            //Add the first and last index to the keepers
            pointIndexsToKeep.Add(firstPoint);
            pointIndexsToKeep.Add(lastPoint);

            //The first and the last point cannot be the same
            while (existingPolygon[firstPoint].Equals(existingPolygon[lastPoint]))
            {
                lastPoint--;
            }

            ReducePoints(existingPolygon, firstPoint, lastPoint,
                0, ref pointIndexsToKeep);

            pointIndexsToKeep.Sort();
            return pointIndexsToKeep.Select(index => existingPolygon[index]).ToArray();
        }

        /// <summary>
        /// Douglases the peucker reduction.
        /// </summary>
        /// <param name="points">The points.</param>
        /// <param name="firstPoint">The first point.</param>
        /// <param name="lastPoint">The last point.</param>
        /// <param name="tolerance">The tolerance.</param>
        /// <param name="pointIndexesToKeep">The point index to keep.</param>
        private static void ReducePoints(IReadOnlyList<Point> points, int firstPoint, int lastPoint, double tolerance,
            ref List<int> pointIndexesToKeep)
        {
            double maxDistance = 0;
            int indexFarthest = 0;

            for (int index = firstPoint; index < lastPoint; index++)
            {
                double distance = PerpendicularDistance
                    (points[firstPoint], points[lastPoint], points[index]);
                if (distance > maxDistance)
                {
                    maxDistance = distance;
                    indexFarthest = index;
                }
            }

            if (maxDistance > tolerance && indexFarthest != 0)
            {
                //Add the largest point that exceeds the tolerance
                pointIndexesToKeep.Add(indexFarthest);

                ReducePoints(points, firstPoint,
                    indexFarthest, tolerance, ref pointIndexesToKeep);
                ReducePoints(points, indexFarthest,
                    lastPoint, tolerance, ref pointIndexesToKeep);
            }
        }

        /// <summary>
        /// The distance of a point from a line made from point1 and point2.
        /// </summary>
        /// <param name="pt1">The PT1.</param>
        /// <param name="pt2">The PT2.</param>
        /// <param name="p">The p.</param>
        /// <returns></returns>
        private static double PerpendicularDistance
            (Point Point1, Point Point2, Point Point)
        {
            //Area = |(1/2)(x1y2 + x2y3 + x3y1 - x2y1 - x3y2 - x1y3)|   *Area of triangle
            //Base = v((x1-x2)²+(x1-x2)²)                               *Base of Triangle*
            //Area = .5*Base*H                                          *Solve for height
            //Height = Area/.5/Base

            double area = Math.Abs(.5 * (Point1.X * Point2.Y + Point2.X *
                                         Point.Y + Point.X * Point1.Y - Point2.X * Point1.Y - Point.X *
                                         Point2.Y - Point1.X * Point.Y));
            double bottom = Math.Sqrt(Math.Pow(Point1.X - Point2.X, 2) +
                                      Math.Pow(Point1.Y - Point2.Y, 2));
            double height = area / bottom * 2;

            return height;

        }
    }

Answer 1

我认为@ShashwatKumar 的回答误解了您的问题？如果不是，那是我误会了!

在我阅读您的帖子时，您正在寻找图形的多边形轮廓。这称为追踪二值图像的轮廓/轮廓，其第一步， 正如@EugeneKomisarenko 在评论中所说，是“边缘检测”(但随后 是进一步的步骤)。 搜索这些短语的变体会影响许多算法，例如:

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^{图片来自this link .}

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而且你不想去寻找真正最小的兔子洞: “Given a set of 2D vertices, how to create a minimum-area polygon which contains all the given vertices?” 因为这是众所周知的棘手问题(从技术上讲，NP-hard)。