Boyer-Moore Algorithm
Boyer-Moore Algorithm Code in C#
And my running application looks like this:
Wednesday, November 12, 2014
Tuesday, November 4, 2014
Image Processing: Matlab code - Local Histogram equalization 3x3 window
Letus begin by considering following 64x64 image.
Letus apply local window processing by taking 3x3 window and move certral pixel of local window x(2,2) to output image J. Store above image to your harddisk folder and provide name in path variable below (line#2).
clear all
path = 'E:\xxx\woman_1.gif'
I = imread(path)
[row, col]=size(I);
eI = histeq(I);
h = imhist(I);
for i=2:row-1
for j=2:col-1
a = i
b = j
win = I(i-1:i+1, j-1:j+1);
ewin = histeq(win)
J(i,j) = ewin(2,2)
end
end
imshow(I);
imhist(I);
imshow(J);
imshow(J);
Result of the local window processing is as follows:
And
to
clear all
path = 'E:\xxx\woman_1.gif'
I = imread(path)
[row, col]=size(I);
eI = histeq(I);
h = imhist(I);
for i=2:row-1
for j=2:col-1
a = i
b = j
win = I(i-1:i+1, j-1:j+1);
ewin = histeq(win)
J(i,j) = ewin(2,2)
end
end
imshow(I);
imhist(I);
imshow(J);
imshow(J);
Result of the local window processing is as follows:
And
Note: it may take certain time :)
Image Processing: Matlab Introduction with Histogram Equalization on image
>>>>>>>>>>>>
Some basic command and mathematics>>>>>>>>>>>>>
Desc.
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Command
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Results
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Clear all variables
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>> clear
>> clear all
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Clear screen
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>> clc
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Array single dimension
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x=1:5
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X = 1 2
3 4 5
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Array with specific
interval
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t=0:0.1:1
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T = 0 0.1000
0.2000 0.3000 0.4000
0.5000 0.6000 0.7000
0.8000 0.9000 1.0000
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2D array, matrix with
random values
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x= rand(3:3)
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0.6258 0.8439
0.1939
0.2507 0.3974
0.3631
0.2630 0.1046
0.8745
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Another 2D array,
matrix
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y= rand(3:3)
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0.6258 0.8439
0.1939
0.2507 0.3974
0.3631
0.2630 0.1046
0.8745
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Show graph
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plot(x,y)
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List all files in
current directory
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dirlist = dir('.');
for i = 1:length(dirlist)
dirlist(i)
end
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How to find help
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>> help imhist
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Generate single dimensional
and array with zeros
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>> h=zeros(1,300);
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Variable names
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Avoide giving reserved names
like pi, i= =sqrt(-1)
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Matrix
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>> v=[1, 2, 3]
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1 2 3
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>> v=[1 2 3]
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1 2 3
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>> v=[1; 2; 3]
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1
2
3
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>> w=[1+ 5 3 4]
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6 3 4
|
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>> v=[1 +2 3]
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1 2 3
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Multiplication of
matrices
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>> v=[4 5 6]
>> u=[1; 2; 3]
>> v*u
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4 5
6
8 10
12
12 15
18
3x3 matrix
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>> u*v
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32
1x1 matrix
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Plot Graphs
>> a=0
>> b=1
>> n=10
>> x=linspace(a,b,n+1) à h
= b-a/(n+1)-1 = 1/10=0.1
>> y = sin(x)
>> z = length(x)
Answer:
x = 0
0.1000 0.2000 0.3000
0.4000 0.5000 0.6000
0.7000 0.8000 0.9000
1.0000
y = 0
0.0998 0.1987 0.2955
0.3894 0.4794 0.5646
0.6442 0.7174 0.7833
0.8415
z = 11
>> plot(x,y)
>> plot(x,y, 'r') à r=red, g=green, b=blue, y=yellow, c=cyan, m=magenta, k=black
>> plot(x,y, 'red')
>> plot(x,y, 'red--') à -- dashed, *, x, <, >, . , .- , -.
>> plot(x,y, 'red*')
>> plot(x,y, 'red.')
>> plot(x,y, 'red*',
'linewidth', 3, 'markersize', 20)
plot(x,y, 'red*', 'linewidth',
3, 'markersize', 10)
Quadretic
equation
Imagine an equateion Ax^2 +
Bx + C = 0
x = -B +/- Sqrt(B^2 – 4AC)/2A
Program:
>> A=1, B=2, C=3
>> x(1)=(-B+sqrt(B^2-4*A*C))/(2*A)
à ans: x = -1.0000 + 1.4142i
>> x(2)=(-B-sqrt(B^2-4*A*C))/(2*A) à
ans: x = -1 + 1.4142i -1 - 1.4142i
X now contains two values in
an array.
>>>>>>>>>>>>
Image Processing >>>>>>>>>>>>>
Read an
image
>> path = E:\MS_CS\Semester_1\3.
Digtal.Img.Processing\Matlab\images\a.jpg'
>> A =
imread(path)
Read built-in
image
>>
I=imread('cameraman.tif');
>> imshow(I)
display an
image
>> imshow(A)
Negative of
an image
>> B =
imcomplement(A)
>> imshow(B)
Blue
Filtering of an image
>> B = A
>> B =
double(A);
B(:,:,3) =
3*B(:,:,3);
B = uint8(B);
>> imshow(B)
r
= A(:,:,1);
g
= A(:,:,2);
b
= A(:,:,3);
Color and
sizing of an image
>> [m,n,k] =
size(A)
m = h, n = w, k =
colors/graylevels
Display
histogram of an image
Method 1:
>> [count,bin]
= hist(A(:), 0:255);
>>
figure,plot(bin, count);
image >>
Method 2:
>> imhist(A);
Error
using imhist
Expected
input number 1, I or X, to be two-dimensional.
Reason: input image
is RGB – change to gray scale.
>> imhist(rgb2gray(A));
Display
histogram equalization of an image
>> B =
rgb2gray(A)
>> J =
histeq(B);
>> figure,
imshow(B), figure, imshow(J)
And histogram is now
stretched:
>> imhist(J)
Number of
Pixels in an image
>> numel(A)
Just copy paste in matlab,
it should work
It is an alternative program not using built-in histeq function.
It is an alternative program not using built-in histeq function.
clear all
clc
I=imread('cameraman.tif');
I=double(I);
maximum_value=max((max(I)));
[row col]=size(I);
c=row*col;
h=zeros(1,300);
z=zeros(1,300);
for n=1:row
for m=1:col
if I(n,m) == 0
I(n,m)=1;
end
end
end
for n=1:row
for m=1:col
t = I(n,m);
h(t) = h(t) + 1;
end
end
pdf = h/c;
cdf(1) = pdf(1);
for x=2:maximum_value
cdf(x) = pdf(x) + cdf(x-1);
end
new = round(cdf *
maximum_value);
new= new + 1;
for p=1:row
for q=1:col
temp=I(p,q);
b(p,q)=new(temp);
t=b(p,q);
z(t)=z(t)+1;
end
end
b=b-1;
subplot(2,2,1),
imshow(uint8(I)) , title(' Image1');
subplot(2,2,2), bar(h) ,
title('Histogram of d Orig. Image');
subplot(2,2,3),
imshow(uint8(b)) , title('Image2');
subplot(2,2,4),bar(z) ,
title('Histogram Equalisation of image2');
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