Image Compression
description
Transcript of Image Compression
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
1 Introduction and background (cont.)
Structral blocks of image compression system. Encoder Decoder
The compression ratio
where n1 and n2 denote the number of information carrying units (usually bits) in the original and encoded images respectively.
2
1
n
nCR
1 Introduction and background (cont.)
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. If F1 and F2 are an original and compressed%image, respectively, CR is the compression ratio.
error(nargchk(2,2,nargin)); %check input argumentscr = bytes(f1) / bytes(f2); %compute the ratio
The Function that finds the compression ratio between two images:
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 (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 (cont.)
Let denote the reconstructed image.
Two types of compression Loseless compression: if Loseless compression: if
),(ˆ and image thedenote ),( yxfyxf
),( of tionrepresentaexact an is ),(ˆ yxfyxf
),( of tionrepresentaexact an is ),(ˆ yxfyxf
1 Introduction and background (cont.)
In lossy compression the error between is defined by root mean square which is given by
),(ˆ and ),( yxfyxf
2/11
0
1
0
2)],(ˆ -),(ˆ[1
M
x
N
yrms yxfyxf
MNe
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 inputs F1 and F2, displays a histogram of the difference, and displays a scaled difference 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 (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); endend
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
1 Introduction and background (cont.)
Error histogram
1 Introduction and background (cont.)
Error image
2 Coding redundancy
Let nk denote the 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
Lkn
nkp k
r ...,,2,1,)(
2 Coding redundancy (cont.)
If the number of bits used to represent each value of rk is
l(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
L
krkavg kprlL
1)()(
2 Coding redundancy (cont.)
When fixed variable length that is l(rk) =m then
m
kpm
kmpL
L
kr
L
kravg
1
1
)(
)(
2 Coding redundancy (cont.)
The average number of bits requred by Code 2 is
8125.18125.1/2
8125.1)1875.0(2)125.0(3)5.0(1)1875.0(3
)()(4
12
r
kkravg
C
rpklL
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 (cont.)
Formulation of generated information
)(log)(
1log)( EP
EPEI
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 (cont.)
Given a source of random events from the discrete set of possible events with associated probabilities naaa ...,,, 21
The average information per source output, called the entropy, is
J
jjj aPaPH
1)(log)(
)(...,),(),( 21 naPaPaP
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
L
kkrk rprpH
1)(log)(
~
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 (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 (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 (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 are widely used and very effective technique for compressing data.
We consider the data to be a sequence of charecters.
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 (cont.)
100
86 14
58 28 14
b:13
a:45
d:16c:12 f:5e:9
100
55
25 30
c:12
a:45
a:45 b:13 14d:16 d:16
f:5 e:9
Fixed-length code
1
0
0
0
0
0
0 0
0
0 0
0
1
1 1 1 1
1
1
1
1
Variable-length code
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.
01011001001010
ebaa 110110100001011101
Constructing Huffman codes
tree theofroot Return theMIN(Q)-EXTRACT9
),(NSERT8
7
MIN(Q)-EXTRACT6
MIN(Q)-EXTRACT5
node new a allocate 4
113
2
1
)HUFFMAN(
return
do
tofor
zQI
yfxfzf
yzright
xzleft
z
ni
CQ
cn
C
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
c:12 b:13e:9f:5 d:16 a:45
14b:13c:12 d:16 a:45
f:5 e:9
14
f:5 e:9
d:16 a:45 25
c:12 b:13
00
000
1
11
Constructing Huffman codes: Example
14
f:5 e:9
d:16
a:45 25
c:12 b:13
30
0 0
0
1
1
1
Constructing Huffman codes: Example
14
f:5 e:9
d:16
a:45
25
c:12 b:13
30
55
0
0 0
0
1
1
1
1
Constructing Huffman codes: Example
14
f:5 e:9
d:16
a:45
25
c:12 b:13
30
55
100
1
1
1
1 1
0
0
0
0
0
Huffman Codes
function CODE = huffman(p)%check the input arguments for reasonablenesserror(nargchk(1,1,nargin));if (ndims(p) ~= 2) | (min(size(p))>1)| ~isreal(p)|~isnumeric(p) error('P must be a real numeric vector');endglobal CODECODE = cell(length(p),1);if length (p) > 1 p=p /sum(p); s=reduce(p); makecode(s,[]);else CODE ={'1'};end;
Huffman Codes(cont.)
%Create a Huffman source reduction tree in a MATLAB cell structure by performing %source symbol reductions until there are only two reduced symbols remaining
function s = reduce(p);s= cell(length(p),1)for i=1:length(p) s{i}=i;endwhile 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(cont.)
%Scan the nodes of a Huffman source reduction tree recursively to generate%the indicated variable length code words.
function makecode(sc,codeword)
global CODEif isa(sc,'cell') makecode(sc{1},[codeword 0]); makecode(sc{2},[codeword 1]);else CODE{sc} = char('0'+codeword);end
Huffman Codes(cont.)
>> p = [0.1875 0.5 0.125 0.1875];>> c = huffman(p)
c =
'011' '1' '010' '00'
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.)
>> 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(cont.)
>> whos('h1f2') Name Size Bytes Class
h1f2 1x16 1018 cell arrayGrand 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 arrayGrand total is 48 elements using 96 bytes
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(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(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 =
10101011010110110000011110011000
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
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(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]; endendx = unravel (y.code',link, m*n); %Decode using C 'unravel'x = x + xmin - 1;x = reshape(x,m,n);
Huffman Encoding(cont.)
(cont.)
#include mex.hvoid 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(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.)
(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);
}
Consider these two images:
•They have virtually identical histograms
•The histograms are three modal indicating the presence of three dominant grey levels
Interpixel redundancy
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.
Random matches Entropy=7.4253 Compression ratio=1.0704
Aligned matches Entropy=7.3505 Compression ratio=1.0821
•Note that the 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 (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 (cont.)
The following figure shows a lossless predictive coding.
Interpixel redundancy (cont.)
The following figure shows a lossless predictive coding.
error. prediction:ˆ
integer.nearest the toroundedpredictor theofoutput The:ˆ
imageinput The:
nnn
n
n
ffe
f
f
The decoder reconstructs the prediction error from the received
variable-length code words and performs the inverse operation
nnn fef ˆ
Interpixel redundancy (cont.)
Various methods can be used to generate . In most cases the prediction is formed by a linearcombination
of m previous pixels.
m
iinin ff
1roundˆ
nf̂
where m is the order of the predictor and
are prediction coefficients. For 1-D linear predictive coding this equation can be rewritten
)...,,2,1(, mii
m
iin iyxff
1),(roundˆ
Interpixel redundancy (cont.)
Example: Consider encoding the Figure 8.7(c) using the simple linear predictor
)1,(roundˆ yxffn
nf̂
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.)
In transform coding, a reversible, linear transform like DFT or DCT
1...,,2,1/2
01
1...,,2,1/2
01
where
2
)12(cos
2
)12(cos)()(),(),(
1
0
1
0
NvN
v/Nα(v)
MuM
u/Mα(u)
N
vy
M
uxvuyxfvuT
M
x
N
y
is used to map an image into a set of transform coefficients which are then quantized and coded.
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 (cont.)
JPEG compression (cont.)
JPEG compression (cont.)
After computing the DCT coefficients, they are normalized in accordance with
),(
),(),(ˆ
vuZ
vuTroundvuT
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)
),(ˆ vuT
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.
),(ˆ vuT
JPEG 2000
JPEG 2000 is 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.
JPEG 2000
Fractal Image Compression
Resim İçindeki Benzer Parçalar
Range : Küçük bloklar Domain: Büyük bloklar
Görüntünün bloklara bölünmesi
Orjinal
görüntü
4x4 Range
Parçaları
Orjinal 8x8
Domain parçaları
4x4’e indirgenmiş
halleri
Parçaların Benzerliği
domainth i':
rangeth i' :
i
i
iii
d
r
edar
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 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 !
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.
Decompression: Iteration-1
Decompression: Iteration-2
Decompression: Iteration-3
Decompression: Iteration-10
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.