Today, another image processing technique is waiting for me to discern. I already dealt with image processing in Fourier space and segmentations which I discussed in my previous blog posts. Now, I will be doing morphological operations. In this technique, binary images are modified based on the operations done with the structuring element. The operations involved uses set theory to modify the images. They can either shrink, expand, ad fill the gaps of a shape depending on the conditions given between the structuring element and the given object.
First of all, I want to discuss some set equations that are essential in understanding morphological operations. Consider two large sets A and B which are both collection of objects like letters, numbers, or points. Eq 1-6 are the basic set notions. Eq 1 stands for a is an element of A and this is true if we can find a in set A. Crossing out the notation signifies negation so Eq 2 means b is not an element of A. Now, if set A is part of a bigger set B, we can write it as Eq 3. There are also notations that yields a new set. If we want to get all the elements of sets A and B, we use the union operator in Eq 4. However, if we are only interested in the intersection of two sets, the intersection operator in Eq 5 is the right notation. Finally, the set notation for saying that the new set has no element, like the intersection of A and D in Eq 6, is to equate the relation to null set.
The next things I want to discuss are some arithmetic operations in set theory. Understanding how these are written and interpreted are very essential in performing basic and complex morphological relations. Believe me. It costed me one quiz before realizing it. Eq 7 says that the compliment of A is a set composed of elements (w) which are not in A. The difference (Eq 8) of sets A and C is a set that contains elements (w) that are in A but not in B. The next two equations are usually used when dealing with set of numerical coordinate values. The reflection (Eq 9) of set A is just negated (-) version of A. For example, if A = {1,2,3,4,5} then the reflection of A is {-1,-2,-3,-4,-5}. The translation (Eq 10) of a set A is done by adding a constant (z) on its elements.
Finally! Let’s do the fun part. I will now apply these set operations to morphological operations. I this activity, two morphological operations are introduced: dilation (Eq 11) and erosion (Eq 12). Set A is composed of points inside the area of a given shape and set B is called the structuring element. The visualization of each set is a binary image in which each coordinate value that is included in the set has value of 1. The structuring element reshapes the larger set A depending on the conditions given for each operation. It is defined from its center. To dilate set A with structuring element B, get the reflection of B first. Then, place the center of the structuring element in each pixel of image of set A and check if the two sets have intersection (translation part). If there is an intersection, mark the current location of the structuring element’s center with 1. After this process is iterated to all the pixels of set A, the result is the dilated image of A. To erode a structure, translate the center of the structuring element (B) to each pixel of image containing the set A and check if the occupied space of the B is a subset of A. Mark the center of B with 1 if this condition is satisfied.
I applied dilation and erosion to various shapes such as square, triangle, hollow square, and cross using structuring elements such as square, vertically and horizontally oriented rectangle, cross and diagonal. The results are in Fig 1 for dilation and Fig 2 for erosion. The topmost rows of both images show the structuring element and the leftmost columns contain the shapes being structured. We can see from the figures that dilation increases the area of the shape while the reverse is observed for erosion. The results also suggests that we need to use a structuring element with the same profile as the shape being structured to preserve the original morphology of the shape.
Fig 1. Dilation results. The area covered of the new image for each shape-structuring element pair is greater than the area of the original shape.
Fig 2. Erosion results. The area covered of the new image for each shape-structuring element pair is smaller than the area of the original shape.
I also did some morphological operations on these shapes with my magnificent free-hand drawing (haha) on a graphing paper. The results are pretty much the same. I posted it in Fig 3.
Fig 3. Hand-drawn dilation (top) and erosion (bottom) of square, triangle, hollow square, and cross.
I want to extend further the application of morphological operations to real world situations. This time, the application is identification of cancer cells. The main problem for isolating cancer cells from the image is the threshold color that segments cells because it is close to the color of the background. What I want to do is to use the combination of erosion and dilation to clean the segmented image so I can clearly estimate the area of normal cells. The mean area is used to separate normal cells from cancerous cells. I asked my professor (Dr. Soriano) about why we are isolating the large cells. According to her, cancerous cells are more active than normal cells therefore they are generally bigger.
To do mean area estimation, we are tasked to divide the test image into 256×256 subimages, perform morphological operations to clean up the unnecessary artifacts of thresholding, and estimate the area of the cells. The result of thresholding is in Fig 4. This shows that we need additional steps to clean the image before area estimation.
Fig 4. After thresholding, artifacts are found in some subimages. These must be removed before proceeding to area estimation.
Fig 5. Subimages after close-open morphological operation was performed. The circles with red outlines are used to estimate the mean area.
After performing morphological operations to clean the images and area estimation, the mean area obtained was 528 and the standard deviation was 42. This was used to isolate the cancerous cells to normal cells. We were given an image with cancer cells. I performed the morphological operation I used in the segmentation part and then I searched the circles with areas greater than the mean area plus the standard deviation. My results are in Fig 6.
Fig 6. Image with identified cancer cells (outlined with red).
Finally I’m done! This task is both rewarding and exhausting. I think I clearly provided the necessary steps to perform this tasks and obtained good results. For this, my self rating is 10.
References:
[1] M. Soriano, AP 186 A8- Morphological operations 2014. 2016
[2] “GpuArray.” Morphologically Close Image – MATLAB Imclose. N.p., n.d. Web. 06 Dec. 2016.
[3] “GpuArray.” Morphologically Open Image – MATLAB Imopen. N.p., n.d. Web. 06 Dec. 2016.
Applied Physics 186 is life 