Mathematical Morphology (Chapters 9, 10) PDF

Chapter 9 Mathematical morphology (1) 9.1 Introduction Morphology, or morphology for short, is a branch of image processing which is particularly useful for analyzing shapes in images. We shall develop basic morphological tools for investigation of binary images, and then show how to ext...

Chapter 9 Mathematical morphology (1) 9.1 Introduction Morphology, or morphology for short, is a branch of image processing which is particularly useful for analyzing shapes in images. We shall develop basic morphological tools for investigation of binary images, and then show how to extend these tools to greyscale images. Matlab has many tools for binary morphology in the image processing toolbox; most of which can be used for greyscale morphology as well. 9.2 Basic ideas The theory of mathematical morphology can be developed in many different ways. We shall adopt one standard method which uses operations on sets of points. A very solid and detailed account can be found in Haralick and Shapiro. Translation




Suppose that is a set of pixels in a binary image, and Then
is the set “translated” in direction 
. That is is a particular coordinate point.



 


 
 For example, in figure 9.1, is the cross shaped set, and 

. The set has been shifted in

the and directions by the values given in. Note that here we are using matrix coordinates, rather than Cartesian coordinates, so that the origin is at the top left, goes down and goes across. Reflection If
is set of pixels, then its reflection, denoted
 , is obtained by reflecting
in the origin:




 For examples, in figure 9.2, the open and closed circles form sets which are reflections of each other. 163 164 CHAPTER 9. MATHEMATICAL MORPHOLOGY (1) 0 1 2 3 4 0 1 2 3 4 0 0 1 1 2 2 3 3 4 4 5 5

Figure 9.1: Translation    0 1 2 3  0 1 2 3 Figure 9.2: Reflection 9.3. DILATION AND EROSION 165 9.3 Dilation and erosion These are the basic operations of morphology, in the sense that all other operations are built from a combination of these two. 9.3.1 Dilation Suppose
and  are sets of pixels. Then the dilation of
by  , denoted

 , is defined as
 

  
What this means is that for every point   , we translate
by those coordinates. Then we take the union of all these translations. An equivalent definition is that
 
 







   From this last definition, dilation is shown to be commutative; that
 

 An example of a dilation is given in figure 9.3. In the translation diagrams, the grey squares show the original position of the object. Note that 

 is of course just itself. In this example, we

have 









 
In general,
 and those these are the coordinates by which we translate. can be obtained by replacing every point 
 in with a copy of , placing

at
. Equivalently, we can replace every point

the point of of with a copy of  
. Dilation is also known as Minkowski addition; see Haralick and Shapiro for more information. As you see in figure 9.3, dilation has the effect of increasing the size of an object. However, it is


 not necessarily true that the original object will lie within its dilation. Depending on the

coordinates of , may end up quite a long way from. Figure 9.4 gives an example of this:   is the same as in figure 9.3; has the same shape but a different position. In this figure, we have  
 (

%

%




  so that
 

 "    
    
    
    
     For dilation, we generally assume that is the image being processed, and
 is a small set of pixels. In this case is referred to as a structuring element or as a kernel.  Dilation in Matlab is performed with the command >> imdilate(image,kernel) To see an example of dilation, consider the commands: 166 CHAPTER 9. MATHEMATICAL MORPHOLOGY (1) 1 2 3 4 5 1 2 3 4 5 1 1 2 2 3 3 4  0 1 4 5  5 6 0 6 7 1 7

   1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7
 !  
 ! 
 ! !  1 2 3 4 5 1 2 3 4 5 6 7
  Figure 9.3: Dilation 9.3. DILATION AND EROSION 167 1 2 3 4 5 6 7 8 1 2 3 4 1 2 3 4 5 1 6 2 7 3 8 4 9 5 10 6 12 7 13 8 14 
   Figure 9.4: A dilation for which 
 >> t=imread(’text.tif’); >> sq=ones(3,3); >> td=imdilate(t,sq); >> subplot(1,2,1),imshow(t) >> subplot(1,2,2),imshow(td) The result is shown in figure 9.5. Notice how the image has been “thickened”. This is really what dilation does; hence its name. 9.3.2 Erosion Given sets
and  , the erosion of
by  , written
  , is defined as:
 
   


for which 

In other words the erosion of by consists of all points
is in. To   perform an erosion, we can move over , and find all the places it will fit, and for each such place  mark down the corresponding
point of. The set of all such points will form the erosion.  An example of erosion is given in figures 9.6. Note that in the example, the erosion
 was a subset of. This is not necessarily the case; 
it depends on the position of the origin in. If contains the origin (as it did in figure 9.6), then   the erosion will be a subset of the original object. Figure 9.7 shows an example where does not contain the origin. In this figure, the open circles  in the right hand figure form the erosion. 168 CHAPTER 9. MATHEMATICAL MORPHOLOGY (1) Figure 9.5: Dilation of a binary image position is different. Since the origin of  in figure 9.7 is translated by  Note that in figure 9.7, the shape of the erosion is the same as that in figure 9.6; however its 
from its position in figure 9.6, we can assume that the erosion will be translated by the same amount. And if we compare figures 9.6 and 9.7, we can see that the second erosion has indeed been shifted by 
 from the first.
For erosion, as for dilation, we generally assume that is the image being processed, and is  a small set of pixels: the structuring element or kernel.
Erosion is related to Minkowski subtraction: the Minkowski subtraction of from is defined  as



  
Erosion in Matlab is performed with the command >> imerode(image,kernel) We shall give an example; using a different binary image: >> c=imread(’circbw.tif’); >> ce=imerode(c,sq); >> subplot(1,2,1),imshow(c) >> subplot(1,2,2),imshow(ce) The result is shown in figure 9.8. Notice how the image has been “thinned”. This is the expected result of an erosion; hence its name. If we kept on eroding the image, we would end up with a completely black result. Relationship between erosion and dilation It can be shown that erosion and dilation are “inverses” of each other; more precisely, the complement of an erosion is equal to the dilation of the complement. Thus:
 

  9.3. DILATION AND EROSION 169 1 2 3 4 5 6 1 2 3  0 1 4  5 0 6 1
 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 1 2 3 4 5 6 1 2 3 4 5 6
  Figure 9.6: Erosion with a cross-shaped structuring element 170 CHAPTER 9. MATHEMATICAL MORPHOLOGY (1) 0 1 2 3 4 5 6 0 1 2 3 4  0 0 1 1 2 2 3 3 4 4 5 5 6 6 
  Figure 9.7: Erosion with a structuring element not containing the origin Figure 9.8: Erosion of a binary image 9.3. DILATION AND EROSION 171 A proof of this can be found in Haralick and Shapiro. It can be similarly shown that the same relationship holds if erosion and dilation are inter- changed; that
 

   We can demonstrate the truth of these using Matlab commands; all we need to know is that the complement of a binary image b is obtained using >> ~b and that given two images a and b; their equality is determined with >> all(a(:)==b(:)) To demonstrate the equality
 

  pick a binary image, say the text image, and a structuring element. Then the left hand side of this equation is produced with >> lhs=~imerode(t,sq); and the right hand side with >> rhs=imdilate(~t,sq); Finally, the command >> all(lhs(:)==rhs(:)) should return 1, for true. 9.3.3 An application: boundary detection

If is an image, and a small structuring element consisting of point symmetrically places about  the origin, then we can define the boundary of by any of the following methods:

 (i)

“ internal boundary”  (ii)

 “external boundary”  (iii) “morphological gradient”  
In each definition the minus refers to set difference. For some examples, see figure 9.9. Note that the
internal boundary consists of those pixels in which are at its edge; the external boundary consists of pixels outside which are just next to it, and that the morphological gradient is a combination of both the internal and external boundaries. To see some examples, choose the image rice.tif, and threshold it to obtain a binary image: >> rice=imread(’rice.tif’); >> r=rice>110; 172 CHAPTER 9. MATHEMATICAL MORPHOLOGY (1)  0 1 2 3 4 5 6  0 1 2 3 4 5 0 0 1 1 2 2 3 3 4 4 5 5 6

   0 1 2 3 4 5 6  0 1 2  0 1 3  4 0 5 1 6 
 
 0 1 2 3 4 5 6  0 1 2 3 4 5 0 0 1 1 2 2 3 3 4 4 5 5 6
 
 
  Figure 9.9: Boundaries 9.3. DILATION AND EROSION 173 Then the internal boundary is obtained with: >> re=imerode(r,sq); >> r_int=r&~re; >> subplot(1,2,1),imshow(r) >> subplot(1,2,2),imshow(r_int) The result is shown in figure 9.10. Figure 9.10: “Internal boundary” of a binary image The external boundary and morphological gradients can be obtained similarly: >> rd=imdilate(r,sq); >> r_ext=rd&~r; >> r_grad=rd&~re; >> subplot(1,2,1),imshow(r_ext) >> subplot(1,2,2),imshow(r_grad) The results are shown in figure 9.11. Note that the external boundaries are larger than the internal boundaries. This is because the internal boundaries show the outer edge of the image components; whereas the external boundaries show the pixels just outside the components. The morphological gradient is thicker than either, and is in fact the union of both. Exercises 1. For each of the following images
and structuring elements :

 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 0 174 CHAPTER 9. MATHEMATICAL MORPHOLOGY (1) Figure 9.11: “External boundary” and the morphological gradient of a binary image 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 1 0 1 1 1 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 calculate the erosion
  , the dilation
  , the opening

 and the closing
. Check your answers with Matlab. 2. Suppose a square object was eroded by a circle whose radius was about one quarter the side of the square. Draw the result. 3. Repeat the previous question with dilation. 4. Using the binary images circbw.tif, circles.tif, circlesm.tif, logo.tif and testpat2.tif, view the erosion and dilation with both the square and the cross structuring elements. Can you see any differences? Chapter 10 Mathematical morphology (2) 10.1 Opening and closing These operations may be considered as “second level” operations; in that they build on the basic operations of dilation and erosion. They are also, as we shall see, better behaved mathematically. 10.1.1 Opening Given
and a structuring element  , the opening of
by  , denoted
 , is defined as:


    So an opening consists of an erosion followed by a dilation. An equivalent definition is

    


That is, is the union of all translations of which fit inside. Note the difference with
  erosion: the erosion consists only of the
point of for those translations which fit inside ;  the opening consists of all of. An example of opening is given in figure 10.1.  0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5

 

 Figure 10.1: Opening The opening operation satisfies the following properties: 1.

. Note that this is not the case with erosion; as we have seen, an erosion may not  necessarily be a subset. 175 176 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) 2.


. That is, an opening can never be done more than once. This property   is called idempotence. Again, this is not the case with erosion; you can keep on applying a sequence of erosions to an image until nothing is left. 3. If
 , then

  . 4. Opening tends to “smooth” an image, to break narrow joins, and to remove thin protrusions. 10.1.2 Closing erosion, and is denoted :
Analogous to opening we can define closing, which may be considered as a dilation followed by an 


     Another definition of closing is that 
if all translations
which contain have non-empty   intersections with. An example of closing is given in figure 10.2. The closing operation satisfies 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6

 
 Figure 10.2: Closing the following properties: 1.

. 2.
 

 ; that is, closing, like opening, is idempotent. 3. If
 , then
  . 4. Closing tends also to smooth an image, but it fuses narrow breaks and thin gulfs, and eliminates small holes. Opening and closing are implemented by the imopen and imclose functions respectively. We can see the effects on a simple image using the square and cross structuring elements. >> cr=[0 1 0;1 1 1;0 1 0]; >> >> test=zeros(10,10);test(2:6,2:4)=1;test(3:5,6:9)=1;test(8:9,4:8)=1;test(4,5)=1 test = 10.1. OPENING AND CLOSING 177 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 >> imopen(test,sq) ans = 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 >> imopen(test,cr) ans = 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Note that in each case the image has been separated into distinct components, and the lower part has been removed completely. >> imclose(test,sq) ans = 178 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 >> imclose(test,cr) ans = 0 0 1 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 With closing, the image is now fully “joined up”. We can obtain a joining-up effect with the text image, using a diagonal structuring element. >> diag=[0 0 1;0 1 0;1 0 0] diag = 0 0 1 0 1 0 1 0 0 >> tc=imclose(t,diag); >> imshow(tc) The result is shown in figure 10.3. An application: noise removal
some of the white pixels are back. An example is given in figure 10.4. Then
 Suppose is a binary image corrupted by impulse noise—some of the black pixels are white, and  will remove the single black pixels, but will enlarge the holes. We can fill the holes by dilating twice:
     10.1. OPENING AND CLOSING 179 Figure 10.3: An example of closing The first dilation returns the holes to their original size; the second dilation removes them. But this will enlarge the objects in the image. To reduce them to their correct size, perform a final erosion:
       The inner two operations constitute an opening; the outer two operations a closing. Thus this noise removal method is in fact an opening followed by a closing:
   This is called morphological filtering.
Suppose we take an image and apply  shot noise to it: >> c=imread(’circles.tif’); >> x=rand(size(c)); >> d1=find(x> d2=find(x>=0.95); >> c(d1)=0; >> c(d2)=1; >> imshow(c) The result is shown as figure 10.4(a). The filtering process can be implemented with >> cf1=imclose(imopen(c,sq),sq); >> figure,imshow(cf1) >> cf2=imclose(imopen(c,cr),cr); >> figure,imshow(cf2) and the results are shown as figures 10.4(b) and (c). The results are rather “blocky”; although less so with the cross structuring element. 180 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) (a) (b) (c) Figure 10.4: A noisy binary image and results after morphological filtering with different structuring elements. Relationship between opening and closing Opening and closing share a relationship very similar to that of erosion and dilation: the complement of an opening is equal to the closing of a complement, and complement of an closing is equal to the opening of a complement. Specifically:



 and


  Again see Haralick and Shapiro for a formal proof. 10.2 The hit-or-miss transform This is a powerful method for finding shapes in images. As with all other morphological algorithms, Suppose we wish to locate    it can be defined entirely in terms of dilation and erosion; in this case, erosion only. square shapes, such as is in the centre of the image in
figure 10.5. Figure 10.5: An image containing a shape to be found If we performed an erosion
  with  being the square structuring element, we would obtain the result given in figure 10.6. 10.2. THE HIT-OR-MISS TRANSFORM 181 Figure 10.6: The erosion 

The result contains two pixels, as there are exactly two places in where will fit. Now suppose     
we also erode the complement of with a structuring element which fits exactly around the  square; and are shown in figure 10.7. (We assume that
is at the centre of.) 
:  : Figure 10.7: The complement and the second structuring element If we now perform the erosion
  we would obtain the result shown in figure 10.8. Figure 10.8: The erosion 
 
The intersection of the two erosion operations would produce just one pixel at the position of the centre of the  square in , which is just what we want. If had contained more than one
square, the final result would have been single pixels at the positions of the centres of each. This combination of erosions forms the hit-or-miss transform. In general, if we are looking for a particular shape in an image, we design two structuring elements: which is the same shape, and  which fits around the shape. We then write  

and



     
  for the hit-or-miss transform.  182 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) As an example, we shall attempt to find the hyphen in “Cross-Correlation” in the text image shown in figure 9.5. This is in fact a line of pixels of length six. We thus can create our two structuring elements as: >> b1=ones(1,6); >> b2=[1 1 1 1 1 1 1 1;1 0 0 0 0 0 0 1; 1 1 1 1 1 1 1 1]; >> tb1=erode(t,b1); >> tb2=erode(~t,b2); >> hit_or_miss=tb1&tb2; >> [x,y]=find(hit_or_miss==1) 
 ( % and this returns a coordinate of , which is right in the middle of the hyphen. Note that the command >> tb1=erode(t,b1); is not sufficient, as there are quite a few lines of length six in this image. We can see this by viewing the image tb1, which is given in figure 10.9. Figure 10.9: Text eroded by a hyphen-shaped structuring element 10.3 Some morphological algorithms In this section we shall investigate some simple algorithms which use some of the morphological techniques we have discussed in previous sections. 10.3.1 Region filling  Suppose in an image we have a region bounded by an -connected boundary, as shown in figure 10.10.  Given a pixel within the region, we wish to fill up the entire region. To do this, we start with  , and dilate as many times as necessary with the cross-shaped structuring element (as used in  10.3. SOME MORPHOLOGICAL ALGORITHMS 183 Figure 10.10: An -connected boundary of a region to be filled figure 9.6), each time taking an intersection with
before continuing. We thus create a sequence of sets:   







  





for which


 
!  Finally 


is the filled region. Figure 10.11 shows how this is done. %    
:    Figure 10.11: The process of filling a region In the right hand grid, we have

 










Note that the use of the cross-shaped structuring element means that we never cross the boundary. 184 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) 10.3.2 Connected components We use a very similar algorithm to fill a connected component; we use the cross-shaped structuring   element for -connected components, and the square structuring element for -connected compo-  nents. Starting with a pixel , we fill up the rest of the component by creating a sequence of sets

 





such that


 !  
until




!. Figure 10.12 shows an example.                       Using the cross Using the square Figure 10.12: Filling connected components In each case we are starting in the centre of the square in the lower left. As this square is itself a  -connected component, the cross structuring element cannot go beyond it. Both of these algorithms can be very easily implemented by Matlab functions. To implement region filling, we keep track of two images: current and previous, and stop when there is no  current the dilation 
difference between them. We start with previous being the single point in the region, and . At the next step we set  
   




   

  
 Given , we can implement the last step in Matlab by  imdilate(current,B)&~A. The function is shown in figure 10.13 We can use this to fill a particular region delineated by a boundary. >> n=imread(’nicework.tif’); >> imshow(n),pixval on >> nb=n&~imerode(n,sq); >> figure,imshow(nb) >> nf=regfill(nb,[74,52],sq); >> figure,imshow(nf) 10.3. SOME MORPHOLOGICAL ALGORITHMS 185 function out=regfill(im,pos,kernel) % REGFILL(IM,POS,KERNEL) performs region filling of binary image IMAGE, % with kernel KERNEL, starting at point with coordinates given by POS. % % Example: % n=imread(’nicework.tif’); % nb=n&~imerode(n,ones(3,3)); % nr=regfill(nb,[74,52],ones(3,3)); % current=zeros(size(im)); last=zeros(size(im)); last(pos(1),pos(2))=1; current=imdilate(last,kernel)&~im; while any(current(:)~=last(:)), last=current; current=imdilate(last,kernel)&~im; end; out=current; Figure 10.13: A simple program for filling regions The results are shown in figure 10.14. Image (a) is the original; (b) the boundary, and (c) the result of a region fill. Figure (d) shows a variation on the region filling, we just include all boundaries. This was obtained with >> figure,imshow(nf|nb) (a) (b) (c) (d) Figure 10.14: Region filling The function for connected components is almost exactly the same as that for region filling, except that whereas for region filling we took an intersection with the complement of our image, for connected components we take the intersection with the image itself. Thus we need only change one line, and the resulting function is shown in 10.15 We can experiment with this function with the “nice work” image. We shall use the square structuring element, and also a larger structuring    element of size. >> sq2=ones(11,11); >> nc=components(n,[57,97],sq); 186 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) function out=components(im,pos,kernel) % COMPONENTS(IM,POS,KERNEL) produces the connected component of binary image % IMAGE which nicludes the point with coordinates given by POS, using % kernel KERNEL. % % Example: % n=imread(’nicework.tif’); % nc=components(nb,[74,52],ones(3,3)); % current=zeros(size(im)); last=zeros(size(im)); last(pos(1),pos(2))=1; current=imdilate(last,kernel)&im; while any(current(:)~=last(:)), last=current; current=imdilate(last,kernel)&im; end; out=current; Figure 10.15: A simple program for connected components >> imshow(nc) >> nc2=components(n,[57,97],sq2); >> figure,imshow(nc2)       and the results are shown in figure 10.16. Image (a) uses the square; image (b) uses the square. (a) (b) Figure 10.16: Connected components 10.3.3 Skeletonization Recall that the skeleton of an object can be defined by the “medial axis transform”; we may imagine fires burning in along all edges of the object. The places where the lines of fire meet form the skeleton. The skeleton may be produced by morphological methods. 10.3. SOME MORPHOLOGICAL ALGORITHMS 187 Consider the table of operations as shown in table 10.1. Erosions Openings Set differences






 
  
 
  
 
  
 
  
  
   
  
   .........
 
  
 
   Table 10.1: Operations used to construct the skeleton 
 Here we use the convention that a sequence of erosions using the same structuring element is denoted. We continue the table until  is empty. The skeleton is then
   obtained by taking the unions of all the set differences. An example is given in figure 10.17, using
 the cross structuring element. Since  is empty, we stop here. The skeleton is the union of all the sets in the third  column; it is shown in figure 10.18. This method of skeletonization is called Lantuéjoul’s method; for details see Serra. This algorithm again can be implemented very easily; a function to do so is shown in figure 10.19. We shall experiment with the nice work image. >> nk=imskel(n,sq); >> imshow(nk) >> nk2=imskel(n,cr); >> figure,imshow(nk2) The result is shown in figure 10.20. Image (a) is the result using the square structuring element; Image (b) is the result using the cross structuring element. Exercises 1. Read in the image circlesm.tif. (a) Erode with squares of increasing size until the image starts to split into disconnected components. (b) Using pixval on, find the coordinates of a pixel in one of the components. (c) Use the components function to isolate that particular component. 2. (a) With your disconnected image from the previous question, compute its boundary. (b) Again with pixval on, find a pixel inside one of the boundaries. (c) Use the regfill function to fill that region. (d) Display the image as a boundary with one of the regions filled in. 188 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2)





 
  
 
  
 
  
 
   Figure 10.17: Skeletonization Figure 10.18: The final skeleton 10.3. SOME MORPHOLOGICAL ALGORITHMS 189 function skel = imskel(image,str) % IMSKEL(IMAGE,STR) - Calculates the skeleton of binary image IMAGE using % structuring element STR. This function uses Lantejoul’s algorithm. % skel=zeros(size(image)); e=image; while (any(e(:))), o=imopen(e,str); skel=skel | (e&~o); e=imerode(e,str); end Figure 10.19: A simple program for computing skeletons (a) (b) Figure 10.20: Skeletonization of a binary image 190 CHAPTER 10. MATHEMATICAL MORPHOLOGY (2) 3. Using the    square structuring element, compute the skeletons of ( (a) a square, )  (b) a rectangle, (c) an L shaped figure formed from an    square with a    square taken from a corner,         (d) an H shaped figure formed from a square with squares taken from the centres of the top and bottom, )     (e) a cross formed from an square with squares taken from each corner. In each case check your answer with Matlab 4. Repeat the above question but use the cross structuring element. 5. For the images listed in question 4, obtain their skeletons by both the bwmorph function, and by using the function given in figure 10.19. Which seems to provide the best result? 6. Use the hit-or-miss transform with appropriate structuring elements to find the dot on the “ i ” in the word “ in ” in the image text.tif.