digital image processing practice 3

Digital Image ProcessingPractice 3

Atsushi OsaYamaguchi Univ.

How to attend this on-line practice1. Download materials for the practice from the course web

site of Digital Image Processing: – http://ds.cc.yamaguchi-u.ac.jp/~samura/web-login/DIP/

2. Watch a video for this practice (on-demand service) indicated in the course web site.

3. Send me your answers for some exercises by e-mail. They are indicated in the video and the slideshow.

4. When you finish today’s exercises, you may close this Webex meeting.

5. The Webex meeting will be closed at the end of the practice.

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• Let’s watch the video for Practice 3.

• You may close the Webex window as it will interfere with the exercise.

• Visit the Webex meeting when you need it.

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Preparation 1-download materials-

Download materials for today’s practice from the course web site: http://ds.cc.yamaguchi-u.ac.jp/~samura/web-login/DIP/

1.Slideshow: dip-p3.pdf2.Materials: dip-m3.zip

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http://ds.cc.yamaguchi-u.ac.jp/%7Esamura/web-login/DIP/

Preparation 2

1. Open your “Documents” folder, and insideopen the “processing” folder.

2. Decompress the “dip-m3.zip” file.3. Move the “dip-m3” folder inside the

“processing” folder.

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Today’s topics

• Histogram• Contrast change• Automatic gain control

Histogram• The histogram of a digital image with intensity

levels within the range [0,L-1] is a discrete function h(rk) = nk, where rk is the k thintensity value and nk is the number of pixels in the image with intensity rk.

nk

rk0 1 2 3 252 253 254 255

n0

n1n2

n3

・・・

n255

L-1

Exercise 3.1:

Q. What is the summation of nk within the range [0,L-1] ?

A.

Exercise 3.1: Answer

Q. What is the summation of nk within the range [0,L-1] ?

A. The total number of pixels in the image

M rows

N columns

∑−

=

×=1

0

L

kk NMn

Histogram

A histogram shows the basic intensity characteristics of an image: dark, light, low contrast, and high contrast.

Exercise 3.2

• Load the “ex3_2” sketch.• Run the sketch.

Histogram

Input image: f(x,y)

inImage[x][y] in this sketch.

Output image: g(x,y)

outImage[x][y] in this sketch

Histogram of the input image

Histogram of the output image

0rk 255

Attention! Black pixels only in these images

Intensity level: you can select the level

Number of pixels: 200 (width) x 200 (height)

equal

Histogram

nk

rk0 1 2 3 252 253 254 255

n0=40000

n1=0

n2=0・・・

n3=0

nk=0 n255=0

Exercise 3.3

• Change the code as follows:

• Save the sketch with a new name.• Run the sketch.

String fileName=“black.bmp”;

String fileName=“gradation.bmp”;

Histogram of a gradation image

Input image: f(x,y)

inImage[x][y] in this sketch.

Output image: g(x,y)

outImage[x][y] in this sketch

Histogram of the input image

Histogram of the output image

0rk 255

All gray levels have the same number of pixels.

Flat

Exercise 3.4



String fileName=“gradation.bmp”;

String fileName=“sunflower.bmp”;

Histogram of a high-contrast image

0rk 255

Exercise 3.5: Answer the correct combinations between the images and histograms

0 rk255

A B C D

α

β

γ

δ

A

B

C

D

Send your answers by e-mail now

to: [email protected]

Pause the video until you send your answers to me.

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mailto:[email protected]

Exercise 3.6


• Save the sketch as a new name.• Run the sketch.

String fileName=“sunflower.bmp”;

String fileName=“sunflower0.bmp”;

Histogram of a dark image

0rk 255

Exercise 3.7





Histogram of a light image

0rk 255

Exercise 3.8





Histogram of a low-contrast image

0rk 255

Histogram Processing

Histogram manipulation can be used for image enhancement.

Basic intensity transformation functionIdentity transformation

Negative transformation

Logarithmic transformation

Power-Law (Gamma) transformation

f

g

g(x, y) = f(x, y)

g(x, y) = L-1 -f(x, y)

i.e. g(x, y) = 255 -f(x, y)

g(x, y) = C log( f(x, y) )

g(x, y) = C f γ(x, y)

Contrast stretching• Contrast stretching expands the range of

intensity levels in an image so that it spans the full intensity range of the recording medium or display device.

input image: f

outp

ut im

age:

g

0 255

255

0

outp

ut im

age:

g

0 255

255

0

input image: f

Identitytransformation

Contrast stretching

f1 f2

Contrast stretchingrange [0,f1] in input image f

range [f1,f2]

range[f2,255]outp

ut im

age:

g

0 255

255

0

input image: f

Contrast stretching

f1 f2

0=g

( )112

255 ffff

g −−

=

255=g

Exercise 3.9

• Read the sketch for exercise 3.8 (low-contrast). • Change the code as follows:

• Save the sketch as a new name.• Run the sketch.

void imageProcessing(){for (int x = 0; x < imageWidth; x++ ) {

for (int y = 0; y < imageHeight; y++ ) {outImage[x][y] = inImage[x][y];

}}

}next page

Exercise 3.9: Step 1void imageProcessing(){

float f1 = 100;float f2 = 150;for(int y=0;y<imageHeight;y++){for(int x=0;x<imageWidth;x++){

if(inImage[x][y]<f1){

}if(inImage[x][y]>=f1 && inImage[x][y]<=f2){

} if(inImage[x][y]>f2){

}}

}}

outp

ut im

age:

g

0 255

255

0

input image: f

f1 f2

question 1 of step 1

I will ask you in Step2

question 2 of step 1Logic Operation

AND

If statement

if(inImage[x][y]<f1){

}

1.Test thisif trueif false

2.Run codes inside these curly brackets

exit

Exercise 3.9: answer of Step 1void imageProcessing(){


if(inImage[x][y]<f1){outImage[x][y]=0;



outImage[x][y]=255;}

}}

}

I will ask you in Step2

Exercise 3.9: Step 2void imageProcessing(){






}}

}

Answer the correct code

• Send your answer by e-mail

• Pause the video until you send your answers to me.

Answervoid imageProcessing(){




outImage[x][y]=(255/(f2-f1))*(inImage[x][y]-f1);} if(inImage[x][y]>f2){


}}

}

( )112

255 ffff

g −−

=

Exercise 3.10


• Change the values of two variables f1 and f2 in order to obtain a high-contrast image.




float f1 = 100;float f2 = 150;

dark image

Ideal f1 and f2 for this image


14 92

Exercise 3.11


• Change the values of the two variables f1 and f2 in order to obtain a high-contrast image.





light image

Ideal f1 and f2 for this image


180 241

Automatic Gain Control (AGC)

• To find suitable values of f1 and f2 automatically,

nk s are zero in the range [0, min-1] nk s are zero in the range [max+1, 255]min max

Exercise 3.12

• Load the sketch for exercise 3.11. • Change the code as shown on the next slide.• Save the sketch with a new name.• Run the sketch.

void imageProcessing(){float f1 = 100;float f2 = 200;for(int y=0;y<imageHeight;y++){

for(int x=0;x<imageWidth;x++){if(inImage[x][y]<f1){

outImage[x][y]=0;}

if(inImage[x][y]>=f1 && inImage[x][y]<=f2){outImage[x][y]=(255/(f2-f1))*(inImage[x][y]-f1);

} if(inImage[x][y]<f2){


}}}

float f1=0;float f2=255;for(int i=0;i<256;i++){if(inHistogram[i]>0){f1=i;break;

}}for(int i=255;i>=0;i--){if(inHistogram[i]>0){f2=i;break;

}}

Exercise 3.11

Exercise 3.12

Searching the min and max

Exercise 3.13:

• Change the name of the input image to these other files.




• That’s all for today.• You may close today’s Webex meeting.• If you have any questions, please ask me by

the Webex meeting or e-mail.