Lecture 4: Morphological Image Processing

Harvey RhodyChester F. Carlson Center for Imaging Science

Rochester Institute of Technologyrhody@cis.rit.edu

September 15, 2005

AbstractWe will develop morphological image processing techniques that

use combinations and sequences of operations. These techniques areuseful in a wide range of applications.

DIP Lecture 4

Morphological Algorithms

• Boundary Extraction

• Region Filling

• Connected Components

• Hit-or-Miss Transform

• Convex Hull

• Thinning and Thickening

Boundary extraction was done in Lecture 3.

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Region Filling

Develop an algorithm to fill in bounded regions of background color.

1. Let B be a structuring element of type N4, Nd or N8, depending on thedesired connectivity.

2. Select a pixel p inside a background color region.

3. Initialize X0 to be an array of background pixels except X0[p] = 1.

4. Perform the iteration

Xk = (Xk−1 ⊕B)⋂

Ac for k = 1, 2, 3, . . .

until Xk = Xk−1.

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Region Filling Program

Function Region Fill,A,start,filter=B,max iterations=max iterations

IF n params() LT 2 THEN Message,’Too Few Arguments to Function’IF NOT KEYWORD SET(B) THEN B=[[0b,1b,0b],[1b,1b,1b],[0b,1b,0b]]IF NOT KEYWORD SET(max iterations) THEN max iterations=10000L

sa=Size(A,/dim)X=BytArr(sa[0],sa[1]) & X[start]=1bY=BytArr(sa[0],sa[1])Ac=A EQ 0count=0WHILE ((min(X EQ Y) EQ 0) AND (count LT max iterations)) DO BEGIN

count=count+1Y=XX=Dilate(Y,B) AND Ac

ENDWHILERETURN,XEND

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Example: Connected Regions in Maze

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Example: Connected Regions in Maze

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Connected Components

We have used a labeling technique to find connected components. Here we will examine a

morphological technique based on set operations.

1. Let A be a binary image and let Y be a connected component in A. Assume that a

point p ∈ Y is known.

2. Let B be the structuring element

B =

24

1 1 1

1 1 1

1 1 1

35

3. Let X0 = p and perform the iteration

Xk =�Xk−1 ⊕ B

�∩ A for k = 1, 2, 3, . . .

until Xk = Xk−1

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Example: G&W Figure 9.17

Search for the connected component in the figure

to the right starting at the red pixel.

The iteration results are shown in the sequence of

images below.

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Connected Components Program

Function morph cc,A,start,filter=filter,max iterations=max iterations,count=count

IF n params() LT 2 THEN Message,’Too Few Arguments to Function’

IF KEYWORD SET(filter) THEN B=filter ELSE B=[[1b,1b,1b],[1b,1b,1b],[1b,1b,1b]]

IF NOT KEYWORD SET(max iterations) THEN max iterations=10000L

sa=Size(A,/dim)

X=BytArr(sa[0],sa[1]) & X[start]=1b

Y=BytArr(sa[0],sa[1])

count=0

WHILE ((min(X EQ Y) EQ 0) AND (count LT max iterations)) DO BEGIN

count=count+1

Y=X

X=Dilate(Y,B) AND A

ENDWHILE

RETURN,X

END

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Connected Components Search

Let A be a N ×M binary image.

1. Initialization

(a) Set B = A

(b) Set C = 0 (an N ×M array of zeros.)

(c) ib=Where(B NE 0)

(d) k=1

2. Iteration: While Min(ib) GE 0

(a) X=morph cc(B,ib[0])

(b) C=C+kX & k=k+1

(c) B=B AND (X EQ 0)

(d) ib=Where(B NE 0) (returns ib=-1 when B is empty)

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Finding All Connected Components

Shown below is an original image, a color-coded image showing the connected components,

and an index color palette.

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Hit-or-Miss Transform

The morphological hit-or-miss transform is a basic tool for shape detection.

The transform is produced by eroding an image with one structuring element and its

complement with another structuring element and then combining the results.

If A is an image and B = (B1, B2) is a pair of structuring elements, then the H-M

transform is

(A ~ B) = (A B1) ∩ (Ac B2)

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Hit-or-Miss Transform

Shown below left is an image A and below right is a pair of structuring elements designed

to detect the middle-sized box with the H-M transform.

Image with three shapesStructuring elements B1and B2

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Hit-or-Miss Transform

Original Image A (A B1) B1 and B2

(Ac B2) (A ~ B) = (A B1) ∩ (Ac B2)

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Example 1

Use the H-M transform to find the small squares in the image below. Assume that B1 and

B2 are of the exact size to fit the small squares.

Original ImageStructuring elements B1 and B2.(Not to scale)

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Example 1 (cont)

Original Image A C1 = (A B1) Ac

C2 = (Ac B2) (A ~ B) = C1 ∩ C2 Detected objects shown in red.

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Example 2

Find the three different kinds of objects in the image on the left.

To find the small boxes use B = (S1, S2), to find the hollow boxes use C = (S2, S3)

and to find the large boxes use D = (S4, S5).

S1

S2

S3

S4

S5

Original image Search results

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Convex Hull

A set A is said to be convex if any two elements can be joined by a straight path that

does not go outside the set.

The convex hull H of a set S is the smallest convex set such that S ⊆ H.

The set difference H − S = H ∩ Sc is the convex deficiency of S.

The convex set and convex deficiency are needed in many applications.

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Convex Hull Algorithm

The convex hull algorithm uses the H-M transform with four sets of structuring elements,

(Bi, C) as given by the arrays below.

100 111 001 000 000100 000 001 000 010100 000 001 111 000B1 B2 B3 B4 C

1. Initialize four arrays Xi0 = A, i =

1, 2, 3, 4.

2. Compute Xik = (Xi

k−1 ~ Bi) ∪ A

until Xik = Xi

k−1. Let Xiconv be the

“converged” value.

3. C(A) =S4

i=1 Xiconv

4. Trim C(A) to the bounding box of the

component.

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Example: G&W Figure 9.19

The set shown below corresponds to G&W Figure 9.19. The goal is to find the convex hull

H for this set.

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Example (cont)

Shown below is the sequence of arrays X1k. It converges in four steps.

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Example (cont)

Shown below is the sequence of arrays X2k. It converges in two steps.

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Example (cont)

Shown below is the sequence of arrays X3k. It converges in six steps.

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Example (cont)

Shown below is the sequence of arrays X4k. It converges in two steps.

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Example (cont)

The union of the four converged arrays is shown below left. The original set is in white

and the convex set is in red. The convex set may be trimmed to the bounding box of the

original set to form the convex hull H as shown on the right in green and white.

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Morphological Thinning

Thinning removes pixels from a set until only a narrow set remains. It is used to reveal set

structure in recognition applications.

Thinning uses a sequence of structuring elements

B = {B1, B

2, . . . , B

n}

Basic thinning operation

1. Let Xn = A and Y = X0 = [0]N×M

2. Repeat while Xn 6= Y

Y = Xn

X0 = Xn

Xi = Xi−1 ∩ (Xi−1 ~ Bi)c

for i = 1, 2, . . . , n

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Morphological Thinning

Each structuring element Bi consists of a hit element Si and a miss element T i. A system

of eight pairs Bi = (Si, T i) is shown below.

S1 S2 S3 S4 S5 S6 S7 S8000 000 100 110 111 011 001 000010 110 110 110 010 011 011 011111 110 100 000 000 000 001 011

T1 T2 T3 T4 T5 T6 T7 T8111 011 001 000 000 000 100 110000 001 001 001 000 100 100 100000 000 001 011 111 110 100 000

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Example: G&W Figure 9.21

The original image is shown in the upper left. One pass of the algorithm is then made,

with each structuring element used one time. For each Bi, the Si is shown in white and

the T i is shown in red.

A X1 X2

X3 X4 X5

X6 X7 X8

B1 B2

B3 B4 B5

B6 B7 B8

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Example: G&W Figure 9.21 (cont)

the second and third pass with the same algorithm are shown below. There is no change

in X8 from Pass 2 to Pass 3 so the algorithm will stop there.

X0 X1 X2

X3 X4 X5

X6 X7 X8

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M THIN IDL ProgramFUNCTION M_THIN,A,max_iterations=max_iterations,count=count;+;B=m_thin(A);performs morphological thinning on a binary image.;;KEYWORDS:;MAX_ITERATIONS (default=20) Set this value to the maximum; number of iterations to allow.;;COUNT returns the number of iterations actually used.;;REFERENCE:;Gonzalez and Woods, Digital Image Processing (2nd Ed), Sec 9.5.5;;HISTORY:;Written for SIMG-782 class by H. Rhody, Sept. 2002.;-

sa=Size(A)IF sa[0] NE 2 THEN MESSAGE,’A must be 2D’IF N_Elements(max_iterations) LE 0 THEN max_iterations=20

;Pad with zeros to enable the structuring elements to work;to the edge of the set. The padding is removed before;returning the result.

B=BytArr(sa[1]+4,sa[2]+4)B[2:sa[1]+1,2:sa[2]+1]=A NE 0

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M THIN (cont);STRUCTURING ELEMENTSS1=[[0b,0b,0b],[0b,1b,0b],[1b,1b,1b]]T1=[[1b,1b,1b],[0b,0b,0b],[0b,0b,0b]]

S2=[[0b,0b,0b],[1b,1b,0b],[1b,1b,0b]]T2=[[0b,1b,1b],[0b,0b,1b],[0b,0b,0b]]

S3=Reverse(Transpose(S1),1)T3=Reverse(Transpose(T1),1)

S4=Reverse(Transpose(S2),1)T4=Reverse(Transpose(T2),1)

S5=Reverse(S1,2)T5=Reverse(T1,2)

S6=Reverse(S4,1)T6=Reverse(T4,1)

S7=Reverse(S3,1)T7=Reverse(T3,1)

S8=Reverse(S2,1)T8=Reverse(T2,1)

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M THIN (cont)X=BY=BYTARR(sa[1]+4,sa[2]+4)count=0WHILE MAX(X NE Y) GT 0 AND count LT max_iterations DO BEGINY=XX=Y AND NOT Morph_HitOrMiss(X,S1,T1)X=X AND NOT morph_HitOrMiss(X,S2,T2)X=X AND NOT morph_HitOrMiss(X,S3,T3)X=X AND NOT morph_HitOrMiss(X,S4,T4)X=X AND NOT morph_HitOrMiss(X,S5,T5)X=X AND NOT morph_HitOrMiss(X,S6,T6)X=X AND NOT morph_HitOrMiss(X,S7,T7)X=X AND NOT morph_HitOrMiss(X,S8,T8)count=count+1ENDWHILE

RETURN,X[2:sa[1]+1,2:sa[2]+1]END

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Example: Thin the Blobs

The basic structure is captured by the thinned objects (red). The small ends could be removed by further processing to refine theresult.

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Morphological Thickening TransformThickening is the morphological dual of thinning and is defined by

A� B = A ∪ (A ~ B)

The algorithm can be programmed directly from the definition. However, it has been found to be more effective to do thickeningby thinning Ac.

1. Let C = Ac

2. Let D be thinned (using m thin, for example)

3. T = Dc is the thickened object.

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Example: G&W Figure 9.22The steps in the thickening algorithm are illustrated in the figure below.

A C = Ac

D T = Dc

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