Grayscale Morphology

What is grayscale morphology?
Implementation using the van Herk/Gil-Werman algorithm
The Tophat and H-dome transforms
Generalization: rank order filters

What is grayscale morphology?

If you are not familiar with binary morphology, you may want to read about it first.

As mentioned in the section on binary morphology, grayscale morphology is simply a generalization from 1 bpp (binary) images to images with multiple bits/pixel, where the Max and Min operations are used in place of the OR and AND operations, respectively, of binary morphology. The following should be noted:

• These operations, as with binary morphology, are nonlinear. The significance of nonlinear operations, as opposed to linear operations, is that nonlinear operations can be used directly for making decisions about regions of the image. Nonlinear operations are therefore of particular use in image analysis.
• The generalization can be seen in reverse: binary morphology is a specialization of grayscale morphology. The Max of a set of values {0, 1} is equivalent to an OR, and the Min is equivalent to an AND.
• The generalization only applies to flat structuring elements. With grayscale images, dilation and erosion can be defined with non-uniform structuring elements. These are rarely used, and will not be discussed further.
• With flat structuring elements, grayscale morphology is itself a specialization of rank-order filters. When a rank-order filter acts on a source image, it generates a dest image where each pixel is computed as follows: place the structuring element with its origin on the corresponding source pixel, and choose the rank (ordered) pixel in the set of source pixels that is covered by the structuring element. Erosion is thus a rank-order filter with rank = 0.0; dilation has rank = 1.0; and the median filter, which is useful for removing noise, has rank 0.5. (To be completely accurate, when doing a dilation as described here, invert the structuring element about its origin.)
• Because the standard photometry of binary and grayscale images is opposite (white is min val in binary and max val in grayscale), grayscale dilation lightens a grayscale image and erosion darkens it, visually.
• Without special hardware registers that do Min and Max comparisons on 4 or 8 bytes simultaneously, grayscale morphology must be done by comparing integers, one pixel at a time. It is thus much more than 8 times slower than implementations of binary morphology that pack 32 pixels into each 32-bit word.
• This inefficiency is somewhat offset by the existence of a simple algorithm that computes grayscale morphological operations in a time independent of the size of the structuring element.

For the general case of an arbitrary SE of size A, where A, the "area" of the SE, is the number of hits, the computation of the Max or Min requires A pixel comparisons on the src image for each pixel of the dest. However, things get more efficient when the SE is a brick, by which we mean it is a solid rectangle of hits. For a brick SE, dilation (or erosion) can be decomposed into a sequence of 2 dilations (or erosions) on 1-D horizontal and vertical SEs. Each 1-D grayscale morphological operation can be done with less comparisons than the "brute-force" method. Luc Vincent described one such method, that uses an ordered queue of pixels, given by the set of pixels currently under the SE. The queue is ordered so that the Min or Max can be read off from the end. When the SE is moved one position, one pixel is removed from the queue, and a new pixel is placed into it, in the proper location. Most of the work goes into sorting the queue to maintain the order.

More recently, van Herk, Gil and Werman have described an algorithm that uses a maximum of 3 comparisons for a linear SE, regardless of the size of the SE. This method is described below, and it is the method we have implemented here.

Some useful image transforms are composed of a sequence of grayscale operations. For example, the tophat operator is often used to find local maxima as seeds for some filling operation. It is defined as the difference between an image and the opening of the image by a structuring element of given size. Another local maximum finder is called the h-dome operator, and it uses a grayscale reconstruction method for seed filling. (Yes, with h-domes, you do a seed-filling operation to find seeds for another operation!). These are discussed in detail in the section on tophats and h-domes below.

Implementation using the van Herk/Gil-Werman algorithm

The Tophat and H-dome transforms

Binary segmentation most simply is carried out by selecting connected components in the image, using a seed image, a mask image (which is typically the actual image), and filling selected connected components in the mask that have seeds in them. Here, the mask is used to clip the filling process. In a more complicated situation, typified by a binary image of touching coffee beans, you need to split binary connected components. This can be done using the distance function and looking for low saddle points that are most easily cut through. How do you do that? Put the distance function upside-down. Then the seeds, which are the local maxima of the distance function (the points that are farthest from the boundaries) are at the bottom of the inverted function. The boundaries are at the highest point, and the "low passes" through the saddle are on the boundaries of the catchment basins. Segmentation is achieved by filling the basins to define the watersheds, which are all the points that drain into the same basin.

Similarly, grayscale segmentation usually proceeds by finding markers, or "seeds," and an image of catchment basins surrounded by walls at the boundaries. The catchment filling "mask" can be computed in various ways; a popular one is to use a properly smoothed morphological gradient, which has large peaks at places where the image intensity is rapidly varying -- a likely position to find a segmentation boundary. Filling then proceeds from the seeds into these basins.

There are various ways to get the seeds. Two popular ones are tophat and h-dome transforms. The tophat is simpler, and you can envision its operation as follows. Suppose you have a dark grayscale image with some small bright regions. To identify those regions, apply the tophat, using a SE that is larger than the extent of the regions. The opening is a Min operation that removes those bright regions that are smaller in dimension than the SE used in the opening. Then, subtracting this image with the thin peaks cut off from the original image gives you just those peaks, plus some low amplitude noise. The tophat is typically followed by a thresholding operation on the peaks. We've just described the white tophat. There is a black tophat that is a dual to the white tophat, and it subtracts the original image from the closing with a SE. The black tophat gives large pixel values in the result where there were small dark regions in the original image.

The h-domes are another method for finding local maxima. They use a grayscale reconstruction (seed filling) method, where the original image is the mask and the seed is derived from the mask by subtracting a constant value "h" from each pixel value. Reconstruction expands the seed into the original image (mask). This is visualized as having each local maximum of the seed expand horizontally until it hits a value of the mask that is of equal or greater value, at which point it can expand no further. To complete the h-dome transform, the filled seed is then subtracted from the original image, resulting in an image composed of the "domes" of local maxima of the original, none of which can exceed the value h in size.

You can find implementations of the tophat transform pixTophat() and the hdome transform pixHDome() in morphapp.c. We have implemented both black and white tophat transforms, even though they are trivially related. Intuitively, applying a black tophat to find small dark regions should be equivalent to applying a white tophat to the inverse of the image. (The inverse of an image is found by replacing each pixel by its reflection in the light-dark axis; specifically, if the pixel has value v, it is replaced by the value 255 - v.) And, in fact, this is correct. These operations have exactly that simple duality relation: the white tophat on an image is equal to the black tophat on the inverse of the image, and v.v. The proof, which uses the duality of the opening and closing operations, is very simple. Let the opening of an image I with a structuring element S be given by O_S(I), and the closing of I by S be C_S(I). From here on, due to a typographic limitation of HTML, we'll suppress the "S". Denote the inverse of an image I by I, where I = 255 - I. The opening and closing are dual, in the sense that O(I) = C(I). (Note that this is a different sense of "dual" than for the black and white tophats, which we are in the process of showing.) Denote the white tophat by Tw(I) = I - O(I) and the black tophat by Tb(I) = C(I) - I. Then the black tophat on the inverse of I is given by

Tb(I) = C(I) - I = O(I) - I = 255 - O(I) - 255 + I = I - O(I) = Tw(I)

which is the result that was to be proven. By the way, the duality relations for opening and closing, O(I) = C(I), or C(I) = O(I), can be stated in words in several ways. Here's one for the second of the pair: The closing of an image can be equivalently found by inverting the opening of the inverted image. This holds for binary as well as grayscale, of course.

High frequency noise, which is not well filtered by the white tophat, can be greatly reduced by doing a closing first. Likewise, an opening is often necessary before a black tophat. The size of the SE in the opening or closing can be comparable to that used in the tophat.

(Parenthetical note. For seeing the results clearly, we provide a function pixMaxDynamicRange() that expands the dynamic range to cover the full set of 256 values for an 8 bpp image. We let you do this with either a linear or log transform. The latter is useful for images that cover a large dynamic range to begin with and have small values that you wish to see and which appear black on a linear scale.)

Generalization: rank order filters

How is a brick rank order filter implemented efficiently?

The brute force method is to sort all the neighboring pixels, and pick the value of the r^th one. The best sorting algorithms are O(n*logn), where n, the area of the filter, is the number of values to be sorted. For a 50x50 filter, the number of operations at each pixel is over 10,000, which is not practical. (These are not machine operations.)

However, we only need the r-th largest pixel. So our problem is really a "selection" one (e.g., quickselect), not a sorting problem. Now it turns out that using a variation of quicksort, selection of any specific rank value is O(n). This is because for selection you only need to sort the appropriate segment at each bifurcation in the quicksort method, rather than the entire tree. (For details, see Numerical Recipes in C, 2nd edition, 1992, p. 355ff.) There is a constant 2 in front of the n, relative to quicksort, so this brings the count down to 5,000 for our example, which is still ridiculously large.

Actually, we only need to do an incremental selection or sorting, because moving the filter down by one pixel causes one filter-width of pixels to be added and another to be removed. Can we do this incrementally in an efficient manner? Perhaps, but I don't know how. The sorted values will be in an array. Even if the filter width is 1, we can expect to have to move O(n) pixels, because insertion and deletion can happen anywhere in the array. By comparison, heapsort is excellent for incremental sorting, where the cost for insertion or deletion is O(logn), because the array itself doesn't need to be sorted into strictly increasing order. The heap, which is represented by an array, only needs to be in "heap" order, so just a few elements must be rearranged. However, heapsort only gives the max (or min) value, not the general rank value.

The conclusion is that sorting and selection are far too slow. All is not lost, because we can still use a histogram of pixel values. But how is this to be represented?

Suppose we represent the histogram as an array of 256 bins, which is the usual situation. At each new filter location, the histogram must be updated and the rank value computed. The problem we face is that, in general, to find the rank value, a significant fraction of the entire histogram must be summed. Suppose the filter size is 5x5. There are at most 25 different bins occupied, and we will mostly be adding zeros to the sum of bin occupancies. That is painful!

We can instead use a linked list, so that there won't be any unoccupied cells and the sum can be quickly done. However, we lose random access for insertion and deletion, so those operations become unacceptably slow.

However, we can mostly overcome the empty bin problem and retain random access by using two histograms represented as arrays, with bin sizes 1 and 16. Each of these histograms is updated for each pixel added or removed when the filter is moved. To find the rank value, proceed from coarse to fine, first locating the coarse bin for the given rank value, of which there are only 16. Then for the fine histogram with bin size 1 and 256 entries, we need look only at a maximum of 16 bins. On average, there will be a total of about 16 sums.

Points of interest in the implementation

• Two histograms are maintained throughout: a 16-bin coarse histogram and a 256-bin fine histogram, as mentioned above.
• The histograms are updated incrementally, using either row major order or column major order, depending on the aspect ratio of the filter. For row major order, which is used when the filter height hf is larger than the width wf, the fast scan is down the column. For that case, for each column, generate the histogram for the entire filter at the first location (the top of the column). Then as the filter is moved down, remove the values for the old top row and add the values for the new bottom row.
• Avoid special-case processing for pixels near the boundary by expanding the image by half the filter dimension on each side (specifically, half the horizontal dimension on left and right sides, and half the vertical dimension on top and bottom).
• To avoid bias as far as possible in the added boundary pixels, we mirror the pixel values across the boundary.
• For convenience, choose the filter "origin" implicitly at the UL corner. Then we can simply march through dimensions of the width and height of the original (un-bordered) image to generate the destination (rank-order) image.
• If both filter dimensions are odd and the rank is either 0.0 or 1.0, we use instead grayscale erosion or dilation, respectively. The reason both dimensions have to be odd is that grayscale erosion and dilation are only computed using filters of odd width and height.
• If the input rank is 0.0 or 1.0, with at least one filter dimension being even, it is necessary to use the slower rank order function. However, we move the rank value slightly off the input value (e.g., move 0.0 to 0.0001) to allow use of a simpler comparison in the inner loop that does the sum over histogram bins.

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