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Image Compression. Chapter 8. 1 Introduction and background. The problem: Reducing the amount of data required to represent a digital image. Compression is achieved by removing the data redundancies: Coding redundancy Interpixel redundancy Psychovisual redundancy.

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1 introduction and background
1 Introduction and background
  • The problem: Reducing the amount of data required to represent a digital image.
  • Compression is achieved by removing the data redundancies:
    • Coding redundancy
    • Interpixel redundancy
    • Psychovisual redundancy
1 introduction and background cont
1 Introduction and background (cont.)
  • Structral blocks of image compression system.
    • Encoder
    • Decoder
  • The compression ratio

where n1 and n2denote the number of information carrying units (usually bits) in the original and encoded images respectively.

1 introduction and background cont2
1 Introduction and background (cont.)

function cr = imratio(f1,f2)

%IMRATIO Computes the ratio of the bytes in two images / variables.

%CR = IMRATIO( F1 , F2 ) returns the ratio of the number of bytes in

%variables / files F1 and F2. IfF1 and F2 are an original and compressed

%image, respectively, CR is the compression ratio.

error(nargchk(2,2,nargin)); %check input arguments

cr = bytes(f1) / bytes(f2); %compute the ratio

The Function that finds the compression ratio between two images:

1 introduction and background cont3
1 Introduction and background (cont.)

(cont.)

% return the number of bytes in input f. If f is a string, assume that it

% is an image filename; if not, it is an image variable.

function b = bytes(f)

if ischar(f)

info = dir(f); b = info.bytes;

elseif isstruct(f)

b = 0;

fields = fieldnames(f);

for k = 1:length(fields)

b = b + bytes(f.(fields{k}));

end

else

info = whos(\'f\'); b = info.bytes;

end

1 introduction and background cont4
1 Introduction and background (cont.)

>> r =imratio( (imread(\'bubbles25.jpg\')), bubbles25.jpg\')

r =

0.9988

>> f = imread(\'bubbles.tif\');

>> imwrite(f,\'bubbles.jpg\',\'jpg\')

>> r = imratio( (imread( \'bubbles.tif\' ) ) , \'bubbles.jpg\')

r =

14.8578

1 introduction and background cont5
1 Introduction and background (cont.)
  • Let denote the reconstructed image.
  • Two types of compression
    • Loseless compression: if
    • Loseless compression: if
1 introduction and background cont6
1 Introduction and background (cont.)
  • In lossy compression the error between is defined by root mean square which is given by
1 introduction and background cont7
1 Introduction and background (cont.)

The M- Function that computes e(rms) and displays both e(x,y) and its histogram

%COMPARE Computes and displays the error between two matrices. RMSE =COMPARE (F1 , F2, SCALE) returns the root-mean-square error between inputsF1 and F2, displays a histogram of the difference, and displays a scaleddifference image. When SCALE ,s omitted, a scale factor of 1 is used

function rmse = compare(f1 , f2 , scale)

error(nargchk(2,3,nargin));

if nargin < 3

scale = 1;

end

%compute the root mean square error

e = double(f1) - double(f2);

[m,n] = size(e);

rmse = sqrt (sum(e(:) .^ 2 ) / (m*n));

1 introduction and background cont8
1 Introduction and background (cont.)

(cont.)

%output error image & histogram if an error

if rmse

%form error histogram.

emax= max(abs(e(:)));

[h,x]=hist(e(:),emax);

if length(h) >= 1

figure; bar(x,h,\'k\');

%scale teh error image symmetrically and display

emax = emax / scale;

e = mat2gray(e,[-emax,emax]);

figure; imshow(e);

end

end

1 introduction and background cont9
1 Introduction and background (cont.)

>> r1 = imread(\'bubbles.tif\');

>> r2 = imread(\'bubbles.jpg\');

>> compare(r1, r2,1)

>> In E:\matlab\toolbox\images\images\truesize.m (Resize1) at line 302

In E:\matlab\toolbox\images\images\truesize.m at line 40

In E:\matlab\toolbox\images\images\imshow.m at line 168

In E:\matlab\work\matlab_code\compare.m at line 32

ans =

1.5660

2 coding redundancy
2 Coding redundancy
  • Let nk denotethe number of times that the kth gray level appears in the image and n denote the total number of pixels in the image. The associated probability for the kth gray level can be expressed as
2 coding redundancy cont
2 Coding redundancy (cont.)
  • If the number of bits used to represent each value of rk isl(rk), then the average number of bits required to represent each pixel is
  • Thus the total number of bits required to code an M×N image is MNLavg
2 coding redundancy cont1
2 Coding redundancy (cont.)
  • When fixed variable length that is l(rk) =m then
2 coding redundancy cont2
2 Coding redundancy (cont.)
  • The average number of bits requred by Code 2 is
2 coding redundancy cont3
2 Coding redundancy (cont.)

How few bits actually are needed to represent the gray levels of an image?

Is there a minimum amount of data that is sufficient to describe completely an image without loss information?

Information theory provides the mathematical framework to answer these questions.

2 coding redundancy cont4
2 Coding redundancy (cont.)

Formulation of generated information

Note that if P(E)=1 that is if the event always occurs then I(E)=0 and no information is attributed to it. No uncertainity associated with the event

2 coding redundancy cont5
2 Coding redundancy (cont.)

Given a source of random events from the discrete set of possible events with associated probabilities

The average information per source output, called the entropy, is

2 coding redundancy cont6
2 Coding redundancy (cont.)

If we assume that the histogram based on gray levels is an estimate of the true probability distribution, the estimate of H can be expressed by

1 introduction and background cont12
1 Introduction and background (cont.)

function h = entropy(x,n)

%ENTROPY computes a first-order estimate of the entropy of a matrix.

%H = ENTROPY(X,N) returns the first-order estimate of matrix X with N

%symbols in bits / symbol. The estimate assumes a statistically independent

%source characterized by the relative frequency of occurence of the

%elements in X

error(nargchk(1,2,nargin));

if nargin < 2

n=256;

end

1 introduction and background cont13
1 Introduction and background (cont.)

(cont)

x = double(x);

xh = hist(x(:),n);

xh = xh/sum(xh(:));

% make mask to eliminate 0\'s since log2(0) = -inf

i = find(xh);

h = -sum(xh(i) .* log2(xh(i))); %compute entropy

1 introduction and background cont14
1 Introduction and background (cont.)

f = [119 123 168 119;123 119 168 168] ;

f = [f; 119 119 107 119 ; 107 107 119 119] ;

f =

119 123 168 119

123 119 168 168

119 119 107 119

107 107 119 119

p=hist(f(:),8)

p =

3 8 2 0 0 0 0 3

1 introduction and background cont15
1 Introduction and background (cont.)

p =p/sum(p)

p =

Columns 1 through 7

0.1875 0.5000 0.1250 0 0 0 0 0.1875

H = entropy(f)

h =

1.7806

huffman codes
Huffman codes
  • Huffman codes are widely used and very effective technique for compressing data.
  • We consider the data to be a sequence of charecters.
huffman codes cont
Huffman codes (cont.)

Consider a binary charecter code wherein each charecter is represented by a unique binary string.

Fixed-length code:

a = 000, b = 001, c = 010, d = 011, e = 100, f = 101

variable-length code:

a = 0, b = 101, c = 100, d = 111, e = 1101, f = 1100

huffman codes cont1
Huffman codes (cont.)

100

100

1

0

1

0

55

a:45

a:45

14

86

0

1

0

0

1

58

28

25

30

14

0

0

1

0

1

0

1

0

1

1

a:45

b:13

d:16

d:16

e:9

f:5

c:12

b:13

14

d:16

c:12

0

1

f:5

e:9

Fixed-length code

Variable-length code

huffman codes cont2
Huffman codes (cont.)

Prefix code:

Codes in which no codeword is also a prefix of some other codeword.

Encoding for binary code:

Example:Variable-length prefix code.

a b c

Decoding for binary code:

Example:Variable-length prefix code.

constructing huffman codes1
Constructing Huffman codes
  • Huffman’s algorithm assumes that Q is implemented as a binary min-heap.
  • Running time:
  • Line 2 : O(n) (uses BUILD-MIN-HEAP)
  • Line 3-8: O(n lg n) (the for loop is executed exactly n-1 times and each heap operation requires time O(lg n) )
constructing huffman codes example
Constructing Huffman codes: Example

f:5

e:9

c:12

b:13

d:16

a:45

c:12

b:13

d:16

a:45

14

1

0

0

f:5

e:9

14

d:16

25

a:45

0

1

0

0

1

f:5

e:9

c:12

b:13

constructing huffman codes example1
Constructing Huffman codes: Example

a:45

25

30

0

1

0

1

c:12

b:13

d:16

14

0

1

f:5

e:9

constructing huffman codes example2
Constructing Huffman codes: Example

55

a:45

1

0

30

25

0

0

1

1

d:16

14

c:12

b:13

0

1

f:5

e:9

constructing huffman codes example3
Constructing Huffman codes: Example

100

1

0

55

a:45

1

0

30

25

0

1

0

1

d:16

14

c:12

b:13

1

0

f:5

e:9

huffman codes1
Huffman Codes

function CODE = huffman(p)

%check the input arguments for reasonableness

error(nargchk(1,1,nargin));

if (ndims(p) ~= 2) | (min(size(p))>1)| ~isreal(p)|~isnumeric(p)

error(\'P must be a real numeric vector\');

end

global CODE

CODE = cell(length(p),1);

if length (p) > 1

p=p /sum(p);

s=reduce(p);

makecode(s,[]);

else

CODE ={\'1\'};

end;

huffman codes cont3
Huffman Codes(cont.)

%Create a Huffman source reduction tree in a MATLAB cell structure byperforming %source symbol reductions until there are only two reducedsymbols remaining

function s = reduce(p);

s= cell(length(p),1)

for i=1:length(p)

s{i}=i;

end

while numel(s) > 2

[p,i] = sort(p);

p(2) = p(1) + p(2);

p(1) = [];

s = s(i);

s{2} = {s{1},s{2}};

s(1) = [];

end

huffman codes cont4
Huffman Codes(cont.)

%Scan the nodes of a Huffman source reduction tree recursively to generate

%the indicated variable length code words.

function makecode(sc,codeword)

global CODE

if isa(sc,\'cell\')

makecode(sc{1},[codeword 0]);

makecode(sc{2},[codeword 1]);

else

CODE{sc} = char(\'0\'+codeword);

end

huffman codes cont5
Huffman Codes(cont.)

>> p = [0.1875 0.5 0.125 0.1875];

>> c = huffman(p)

c =

\'011\'

\'1\'

\'010\'

\'00\'

huffman encoding
Huffman Encoding

>> f2 = uint8 ([2 3 4 2; 3 2 4 4; 2 2 1 2; 1 1 2 2])

f2 =

2 3 4 2

3 2 4 4

2 2 1 2

1 1 2 2

>> whos(\'f2\')

Name Size Bytes Class

f2 4x4 16 uint8 array

Grand total is 16 elements using 16 bytes

huffman encoding cont
Huffman Encoding(cont.)

>> c = huffman(hist(double(f2(:)),4))

c =

\'011\'

\'1\'

\'010\'

\'00\'

>> h1f2 = c(f2(:))\'

h1f2 =

Columns 1 through 9

\'1\' \'010\' \'1\' \'011\' \'010\' \'1\' \'1\' \'011\' \'00\'

Columns 10 through 16

\'00\' \'011\' \'1\' \'1\' \'00\' \'1\' \'1\'

huffman encoding cont1
Huffman Encoding(cont.)

>> whos(\'h1f2\')

Name Size Bytes Class

h1f2 1x16 1018 cell array

Grand total is 45 elements using 1018 bytes

>> h2f2 = char (h1f2)\'

h2f2 =

1010011000011011

1 11 1001 0

0 10 1 1

>> whos(\'h2f2\')

Name Size Bytes Class

h2f2 3x16 96 char array

Grand total is 48 elements using 96 bytes

huffman encoding cont2
Huffman Encoding(cont.)

>> h2f2 = h2f2(:);

>> h2f2(h2f2 == \' \') = [];

>> whos(\'h2f2\')

Name Size Bytes Class

h2f2 29x1 58 char array

Grand total is 29 elements using 58 bytes

huffman encoding cont3
Huffman Encoding(cont.)

>> h3f2 = mat2huff(f2)

h3f2 =

size: [4 4]

min: 32769

hist: [3 8 2 3]

code: [43867 1944]

>> whos(\'h3f2\')

Name Size Bytes Class

h3f2 1x1 518 struct array

Grand total is 13 elements using 518 bytes

huffman encoding cont4
Huffman Encoding(cont.)

>> hcode = h3f2.code;

>> whos(\'hcode\')

Name Size Bytes Class

hcode 1x2 4 uint16 array

Grand total is 2 elements using 4 bytes

>> dec2bin(double(hcode))

ans =

1010101101011011

0000011110011000

huffman encoding cont5
Huffman Encoding(cont.)

>> f = imread(\'Tracy.tif\');

>> c=mat2huff(f);

>> cr1 = imratio(f,c)

cr1 =

1.2191

>> save SqueezeTracy c;

>> cr2 = imratio (\'Tracy.tif\',\'SqueezeTracy.mat\')

cr2 =

1.2627

huffman decoding
Huffman Decoding

function x = huff2mat(y)

% HUFF2MAT decodes a Huffman encoded matrix

if ~isstruct(y) | ~isfield(y,\'min\') | ~isfield(y,\'size\') | ~isfield(y,\'hist\') | ~isfield(y,\'code\')

error(\'The input must be a structure as returned by MAT2HUFF\');

end

sz = double(y.size); m=sz(1); n= sz(2);

xmin = double(y.min) - 32768;

map = huffman(double(y.hist));

code = cellstr(char(\'\',\'0\',\'1\'));

link = [2; 0; 0]; left = [2 3];

found = 0; tofind = length(map);

huffman encoding cont6
Huffman Encoding(cont.)

(cont.)

while length(left) & (found < tofind)

look = find(strcmp(map, code {left(1)}));

if look

link(left(1)) = -look;

left = left (2:end);

found = found + 1;

else

len = length (code);

link(left(1)) = len + 1;

link = [link; 0; 0];

code{end + 1} = strcat(code{left(1)},\'0\');

code{end + 1} = strcat(code{left(1)},\'1\');

left = left(2:end);

left = [left len + 1 len + 2];

end

end

x = unravel (y.code\',link, m*n); %Decode using C \'unravel\'

x = x + xmin - 1;

x = reshape(x,m,n);

huffman encoding cont7
Huffman Encoding(cont.)

(cont.)

#include mex.h

void unravel (unsigned short *hx, double *link, double *x, double xsz, int hxsz)

{

int i = 15, j=0, k=0, n=0;

while (xsz - k)

{

if (*(link + n) > 0)

{

if((*(hx + j) >> i) & 0x0001)

n = *(link + n);

else n = *(link + n) - 1;

if (i) i--;

else {j++; i=15;}

if( j > hxsz )

mexErrMsgTxt("OutOf code bits??");

}

huffman encoding cont8
Huffman Encoding(cont.)

(cont.)

else {

*(x + k++) = -*(link + n);

n = 0;}

}

if ( k == xsz - 1 )

*(x + k++) = -*(link + n);

}

void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])

{

double *link, *x, xsz;

unsigned short *hx;

int hxsz;

if(nrhs != 3)

mexErrMsgTxt("Three inputs required.");

huffman decoding cont
Huffman Decoding(cont.)

(cont.)

else if (nlhs > 1)

mexErrMsgTxt("Too many output arguments");

if( !mxIsDouble(prhs[2]) || mxIsComplex(prhs[2]) || mxGetN(prhs[2]) * mxGetM(prhs[2]) != 1 )

mexErrMsgTxt("Input XSIZE must be a scalar");

hx = mxGetPr(prhs[0]);

link = mxGetPr(prhs[1]);

xsz = mxGetScalar(prhs[2]);

hxsz = mxGetM(prhs[0]);

plhs[0] = mxCreateDoubleMatrix( xsz, 1, mxREAL);

x = mxGetPr(plhs[0]);

unravel(hx, link, x, xsz, hxsz);

}

slide53

Interpixel redundancy

  • Consider these two images:
  • They have virtually identical histograms
  • The histograms are three modal indicating the presence of three dominant grey levels
interpixel redundancy cont
Interpixel redundancy (cont.)
  • Because the gray levels of the image are not equally probable, variable-length coding can be used to reduce the coding redundancy.
  • Note thatthe entropy estimates and the compression ratios of the two images are about the same.
  • The variable length coding does not take advantage of the obvious structral relationship between the aligned matches
interpixel redundancy cont1
Interpixel redundancy (cont.)
  • In order to reduce pixel redundancies, the 2-D pixel array normally used for human viewing and interpretation must be transformed into a more efficient format.
  • For example the difference between the adjacent pixels can be used to represent an image.
  • Transformations of this type are referred to as mapping.
  • If the original image can be reconstructed from such transformed set is referred to as reversible mappings.
interpixel redundancy cont2
Interpixel redundancy (cont.)
  • The following figure showsa lossless predictive coding.
interpixel redundancy cont3
Interpixel redundancy (cont.)
  • The following figure showsa lossless predictive coding.

The decoder reconstructs the prediction error from the received variable-length code words and performs the inverse operation

interpixel redundancy cont4
Interpixel redundancy (cont.)
  • Various methods can be used to generate .
  • In most cases the prediction is formed by a linearcombination of m previous pixels.

wherem is the order of the predictor and

are prediction coefficients. For 1-D linear predictive coding this equation can be rewritten

interpixel redundancy cont5
Interpixel redundancy (cont.)

Example: Consider encoding the Figure 8.7(c) using the simple

linear predictor

psychovisual redundancy
Psychovisual Redundancy
  • Psychovisual Redundancy is associated with real or quantifiable visual information.
  • Its elimination is desirable because the information itself is not essential for normal visual processing.
jpeg compression cont
JPEG compression (cont.)

In transform coding, a reversible, linear transform like DFT or DCT

is used to map an image into a set of transform coefficients which are then quantized and coded.

jpeg compression cont1
JPEG compression (cont.)
  • One of the most popular and comprehensive compression standard is the JPEG standard.
  • In the JPEG baseline coding system, which is based on the discrete cosine transform and is adequate for most compression applications.
  • It is performed in four sequential steps:
jpeg compression cont4
JPEG compression (cont.)
  • After computing the DCT coefficients, they are normalized in accordance with

where , u,v=0,1, ...,7 are the resulting normalized and quantized coefficients, T(u,v) DCT of an 8x8 block of image f(x,y) and Z(u,v) is a transform normalization array like that of Figure 8.12(a)

jpeg compression cont5
JPEG compression (cont.)
  • After each block’s DCT coefficientsare quantized, the elements of

are recorded in accordance with zigzag pattern of Figure 8.12(b).

  • Since the resulting on-dimensionally arrayqualitatively arranged according to increasing spatial frequency,the encoder in Figure 8.11(a) is designed to take advantage of long runs of zero that normally result from the reordering.
jpeg 2000
JPEG 2000
  • JPEG 2000is based on the idea that the coefficients of a transform that decorrelates the pixels of an image can be coded more efficiently than the original pixels themselves. If the transform basis functions-wawelets in the JPEG 2000 case- pack most of the important visual information into a small number of coefficients, the remaining coefficients can be quantized coarsely or truncated to zero with little image distortion.
resim inde ki benzer par alar
Resim İçindeki Benzer Parçalar
  • Range : Küçük bloklar
  • Domain: Büyük bloklar
g r nt n n bloklara b l nmesi
Görüntünün bloklara bölünmesi

Orjinal 8x8

Domain parçaları

4x4’e indirgenmiş halleri

4x4 Range

Parçaları

Orjinal

görüntü

par alar n benzerli i 3
Parçaların Benzerliği-3
  • Benzerini arayacağımız Range üsttedir.
  • Domainler’in orijinal halleri ilk kolonda gösterilmiştir.
  • Regresyon modeline göre düzeltilmiş domainler ise ikinci sütunda gösterilmiştir.
  • Yandaki rakamlar residual mean square(rms) değerlerini göstermektedir.
  • Rms değeri en küçük olan domain eldeki range ile eşleştirilir.
rangeleri domainler cinsinden fade etmek
Rangeleri Domainler Cinsinden İfade Etmek
  • Benzer Range-Domainler bulunduktan sonra bunlar bir dosyada saklanarak resim bilgisini oluştururlar.
  • Örnek resmimiz 320x200x8 = 512,000 bit (64,000 byte) içeriyor.
  • Domain numaraları 0-999 arasında olduğu için 10 bitle ifade edilebilir.
  • Parlaklık ve konrast tamsayıya çevrilip, 0-15 ve 0-31 aralığında kuantize edilirse, 4 ve 5 bitle ifade edilebilir.
  • Dolayısıyla, 4000 Range’e karşılık gelen Domain bilgilerini dosyaya yazarsak : 4000*(Domain Numarası+Parlaklık+Kontrast) = 4000*(10+4+5) = 76,000 bit (9,500 byte) olmaktadır.
  • 512/76 = 6.73:1, (512-76)/512 = %85 sıkıştırmak demektir, üstelik %85 sıkıştıran diğer algoritmalardan çok daha iyi bir kalitede !
g r nt kalitesi
Görüntü Kalitesi
  • Decompression, siyah bir ekrandan başlanarak 8-10 iterasyonla gerçekleştirilir.
  • Resmin kalitesi sonucun orjinale ne kadar yakın olduğuna yani herbir Range-Domain çiftinin ne kadar benzer olduk-larına bağlıdır.
ayn resimden daha fazla domain elde etmek domain transformasyonlar
Aynı Resimden Daha Fazla Domain Elde Etmek : Domain Transformasyonları
  • Range’e uygun Domaini ararken Domainleri olduğu gibi bırakmayıp transformasyon işlemlerine sokabiliriz.
  • Orjinal, 90o, 180o ve 270o derece döndürülür.
  • X eksenine göre yansımaları alınır.
  • Elde edilen 8 ayrı Domainin negatifi alınarak, bir Domain-den hareketle 16 farklı Domain elde edilir.
ad