Question

我有2600x2600灰色图片。或者它可以被看作是无符号短的矩阵。我想找到最暗的（或通过计算逆图像最亮的）方形具有固定大小的N.N可以被参数化（如果有多个最暗的方形，我想要所有）。

我看了detection-of-rectangular-bright-area-in-a-image-using-opencv 但它需要一个我没有的阈值，而且我搜索一个固定的大小。

有人在c ++或python中找到它吗？

Answer 1

For each row of the image,
    Add up the N consecutive pixels, so you get W - N + 1 pixels.
For each column of the new image,
    For each consecutive sequence of N pixels, (H - N + 1)
        Add them up and compare to the current best.

要添加每个连续的像素序列，您可以减去最后一个像素，然后添加下一个像素。

如果可以修改图像数组，也可以将其重新用作存储。如果没有，则内存优化将仅存储最新的列，并在第一个循环中的每个步骤中进行优化。

运行时：O（ w · h ）

以下是C＃中的一些代码，用于演示（忽略像素格式和任何潜在的溢出）：

List<Point> FindBrightestSquare(int[,] image, int N, out int squareSum)
{
    int width = image.GetLength(0);
    int height = image.GetLength(1);
    if (width < N || height < N)
    {
        return false;
    }

    int currentSum;
    for (int y = 0; y < height; y++)
    {
        currentSum = 0;
        for (int x = 0; x < width; x++)
        {
            currentSum += image[x,y];
            if (x => N)
            {
                currentSum -= image[x-N,y];
                image[x-N,y] = currentSum;
            }
        }
    }

    int? bestSum = null;
    List<Point> bestCandidates = new List<Point>();
    for (int x = 0; x <= width-N; x++)
    {
        currentSum = 0;
        for (int y = 0; y < height; y++)
        {
            currentSum += image[x,y];
            if (y >= N)
            {
                currentSum -= image[x, y-N];
                if (bestSum == null || currentSum > bestSum)
                {
                    bestSum = currentSum;
                    bestCandidates.Clear();
                    bestCandidates.Add(new Point(x, y-N));
                }
                else if (currentSum == bestSum)
                {
                    bestCandidates.Add(new Point(x, y-N));
                }
            }
        }
    }

    squareSum = bestSum.Value;
    return bestCandidates;
}

Answer 2

您可以增加阈值，直到找到正方形，然后使用2D FSM检测正方形。

这将在O(width * height * bpp)中产生匹配（在可能的最低阈值上进行二分搜索，假设两次幂范围）：

- set threshold to its maximum value 
- for every bit of the threshold
  - clear the bit in the threshold
  - if there is a match
    - record the set of matches as a result
  - else
    - set the bit
- if there is no record, then the threshold is its maximum.

to detect a square:
- for every pixel:
  - if the pixel is too bright, set its line-len to 0
  - else if it's the first column, set its line-len to 1
  - else set its line-len to the line-len of the pixel to the left, plus one

  - if the pixel line-len is less than N, set its rect-len to 0
  - else if it's the first row, set its rect-len to 1
  - else set its rect-len to the rect-len of the pixel above, plus one

  - if the rect-len is at least N, record a match.

line-len表示足够暗的连续像素数。

rect-len表示足够长且对齐的连续暗像素行数。

对于视频捕获，请通过线性搜索从前一帧的阈值替换二进制搜索。

显然，你不能比theta(width/N * height/N)最佳情况更好（因为你必须排除每个可能的位置以获得更暗的方块）并且可以假设比特深度是常数，所以这个算法是渐近的对于固定N来说是最优的。对于N来说，它也可能是渐近最优的输入，因为（直觉上）你必须考虑平均情况下的几乎每个像素。

从图片中检测出最暗的固定大小的正方形

2 个答案: