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Sorry for not writing in Chinese, it’s the working computer and my German also not allows me to explain those details.
As respect to the question, I also have no experience on the finding of a pool in the map. But I have an application to find peaks in a 2d or 3d intensity picture. The algorithm is due to find a local maximal area based on 1d theory that a peak in one dimension is a location where the first derivative is zero and second derivative is negative and the full-duration half-maximum of the peak which is the definition of an opening of the peak matches our certain criteria. Expended to a 2d picture like your map here is to find a local maximal (in your case local minimal) by calculating the Jacobian matrix of the data (which in 1d is the 1st derivative) and find the regions which equals to zero (or very small) as well as the corresponding Hessian matrix (which in 1d is the 2nd derivative) and search the regions which equals to a negative value (in your case should be positive), and which define a local protrusion (in your case is a concave). The opening area is also calculated by the area of full-duration half-maximum of the peak (only it is smaller than a certain value can we call it a peak and only it is bigger than a certain value cannot we neglect it like the noise). Sure, the local height of the peak is also counted as a threshold for picking (in your case, it could be the depth of the pool).
The numerical protocol:
1) Low pass filter to cut the peak. This step is to keep a basal level of the data, and later to be used to subtract the background and calculate the height of the peak.
2) Using data from 1) to estimate the local threshold level of the whole data.
3) take the data area which is above the threshold and calculate the Jacobian matrix and Hessian matrix.
4) Gaussian blur those data a little and find zero like area in Jacobian matrix and negative in same area in the corresponding Hessian matrix.
5) Calculate the height of the peak and get the area of full-duration half-maximum of the peak. Mark it according our matching criteria.
Here is how our task looks like:
2D:
3D:
Hope it can give you some help!
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