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⛄ 内容介绍
近年来，可逆信息隐藏（RDH）是信息安全领域的一个新研究热点，受到了研究人员的越来越多的关注。现有的大多数RDH方案都没有完全考虑到自然图像的纹理会影响嵌入失真。将数据嵌入图像的平滑区域所导致的图像失真要比不平滑区域小得多，这主要是因为在平滑区域中嵌入其他数据对应于直方图移位中的无效移位像素（ISP）更少。因此，我们提出了一种基于图像纹理的RDH方案，以减少直方图移位中像素的无效移位。具体来说，首先，通过棋盘格图案将封面图像分为两个子图像，然后计算每个子图像的波动值。最后，可以将附加数据优先地嵌入到具有较小波动值的子图像区域中。实验结果表明，与现有的一些RDH方案相比，该方法具有更高的容量和更好的隐秘图像质量。
⛄ 部分代码
function varargout = arith07(xC)
% Arith07 Arithmetic encoder or decoder
% Vectors of integers are arithmetic encoded,
% these vectors are collected in a cell array, xC.
% If first argument is a cell array the function do encoding,
% else decoding is done.
%
% [y, Res] = Arith07(xC); % encoding
% y = Arith07(xC); % encoding
% xC = Arith07(y); % decoding
% ------------------------------------------------------------------
% Arguments:
% y a column vector of non-negative integers (bytes) representing
% the code, 0 <= y(i) <= 255.
% Res a matrix that sum up the results, size is (NumOfX+1)x4
% one line for each of the input sequences, the columns are
% Res(:,1) - number of elements in the sequence
% Res(:,2) - unused (=0)
% Res(:,3) - bits needed to code the sequence
% Res(:,4) - bit rate for the sequence, Res(:,3)/Res(:,1)
% Then the last line is total (which include bits needed to store NumOfX)
% xC a cell array of column vectors of integers representing the
% symbol sequences. (should not be to large integers)
% If only one sequence is to be coded, we must make the cell array
% like: xC=cell(2,1); xC{1}=x; % where x is the sequence
% ------------------------------------------------------------------
% Note: this routine is extremely slow on Matlab version 5.x and earlier
% SOME NOTES ON THE FUNCTION
% This function is almost like Arith06, but some important changes have
% been done. Arith06 is buildt almost like Huff06, but this close connection
% is removed in Arith07. This imply that to understand the way Arith06
% works you should read the documentation for Huff06 and especially the
% article on Recursive Huffman Coding. To understand how Arith07 works it is
% only confusing to read about the recursive Huffman coder, Huff06.
%
% Arith07 code each of the input sequences in xC in sequence, the
% method used for each sequence depends on what kind (xType) the
% sequence is. We have
% xType Explanation
% 0 Empty sequence (L=0)
% 1 Only one symbol (L=1) and x>1
% 9 Only one symbol (L=1) and x=0/1
% 11 Only one symbol (L=1) and x<0
% 2 all symbols are equal, L>1, x(1)=x(2)=...=x(L)>1
% 8 all symbols are equal, L>1, x(1)=x(2)=...=x(L)=0/1
% 10 all symbols are equal, L>1, x(1)=x(2)=...=x(L)<0
% 3 non-negative integers, 0<=x(l)<=1000
% 4 not to large integers, -1000<=x(l)<=1000
% 5 large non-negative integers possible, 0<=x(l)
% 6 also possible with large negative numbers
% 7 A binary sequence, x(l)=0/1.
% Many functions are defined in this m-file:
% PutBit, GetBit read and write bit from/to bitstream
% PutABit, GetABit Arithmetic encoding of a bit, P{0}=P{1}=0.5
% PutVLIC, GetVLIC Variable Length Integer Code
% PutS(x,S,C), GetS(S,C) A symbol x, which is in S, is aritmetic coded
% according to the counts given by C.
% Ex: A binary variable with the givene probabilities
% P{0}=0.8 and P{1}=0.2, PutS(x,[0,1],[4,1]).
% PutN(x,N), GetN(N) Arithmetic coding of a number x in range 0<=x<=N where
% we assume equal probability, P{0}=P{1}=...=P{N}=1/(N+1)
%----------------------------------------------------------------------
% Copyright (c) 1999-2001. Karl Skretting. All rights reserved.
% Hogskolen in Stavanger (Stavanger University), Signal Processing Group
% Mail: karl.skretting@tn.his.no Homepage: http://www.ux.his.no/~karlsk/
%
% HISTORY:
% Ver. 1.0 19.05.2001 KS: Function made (based on Arith06 and Huff06)
%----------------------------------------------------------------------
% these global variables are used to read from or write to the compressed sequence
global y Byte BitPos
% and these are used by the subfunctions for arithmetic coding
global high low range ub hc lc sc K code
Mfile='Arith07';
K=24; % number of bits to use in integers (8 <= K <= 24)
Display=1; % display progress and/or results
% check input and output arguments, and assign values to arguments
if (nargin < 1);
error([Mfile,': function must have input arguments, see help.']);
end
if (nargout < 1);
error([Mfile,': function must have output arguments, see help.']);
end
if (~iscell(xC))
Encode=0;Decode=1;
y=xC(:); % first argument is y
y=[y;0;0;0;0]; % add some zeros to always have bits available
else
Encode=1;Decode=0;
NumOfX = length(xC);
end
Byte=0;BitPos=1; % ready to read/write into first position
low=0;high=2^K-1;ub=0; % initialize the coder
code=0;
% we just select some probabilities which we belive will work quite well
TypeS=[ 3, 4, 5, 7, 6, 1, 9, 11, 2, 8, 10, 0];
TypeC=[100, 50, 50, 50, 20, 10, 10, 10, 4, 4, 2, 1];
if Encode
Res=zeros(NumOfX,4);
% initalize the global variables
y=zeros(10,1); % put some zeros into y initially
% start encoding, first write VLIC to give number of sequences
PutVLIC(NumOfX);
% now encode each sequence continuously
Ltot=0;
for num=1:NumOfX
x=xC{num};
x=full(x(:)); % make sure x is a non-sparse column vector
L=length(x);
Ltot=Ltot+L;
y=[y(1:Byte);zeros(50+2*L,1)]; % make more space available in y
StartPos=Byte*8-BitPos+ub; % used for counting bits
% find what kind of sequenqe we have, save this in xType
if (L==0)
xType=0;
elseif L==1
if (x==0) | (x==1) % and it is 0/1
xType=9;
elseif (x>1) % and it is not 0/1 (and positive)
xType=1;
else % and it is not 0/1 (and negative)
xType=11;
end
else
% now find some more info about x
maxx=max(x);
minx=min(x);
rangex=maxx-minx+1;
if (rangex==1) % only one symbol
if (maxx==0) | (maxx==1) % and it is 0/1
xType=8;
elseif (maxx>1) % and it is not 0/1 (and positive)
xType=2;
else % and it is not 0/1 (and negative)
xType=10;
end
elseif (minx == 0) & (maxx == 1) % a binary sequence
xType=7;
elseif (minx >= 0) & (maxx <= 1000)
xType=3;
elseif (minx >= 0)
xType=5;
elseif (minx >= -1000) & (maxx <= 1000)
xType=4;
else
xType=6;
end
end % if (L==0)
if Display >= 2
disp([Mfile,': sequence ',int2str(num),' has xType=',int2str(xType)]);
end
%
if sum(xType==[4,6]) % negative values are present
I=find(x); % non-zero entries in x
Sg=(sign(x(I))+1)/2; % the signs will be needed later, 0/1
x=abs(x);
end
if sum(xType==[5,6]) % we take the logarithms of the values
I=find(x); % non-zero entries in x
xa=x; % additional bits
x(I)=floor(log2(x(I)));
xa(I)=xa(I)-2.^x(I);
x(I)=x(I)+1;
end
% now do the coding of this sequence
PutS(xType,TypeS,TypeC);
if (xType==1) % one symbol and x=x(1)>1
PutVLIC(x-2);
elseif (xType==2) % L>1 but only one symbol, x(1)=x(2)=...=x(L)>1
PutVLIC(L-2);
PutVLIC(x(1)-2);
elseif sum(xType==[3,4,5,6]) % now 'normalized' sequences: 0 <= x(i) <= 1000
PutVLIC(L-2);
M=max(x);
PutN(M,1000); % some bits for M
% initialize model
T=[ones(M+2,1);0];
Tu=flipud((-1:(M+1))'); % (-1) since ESC never is used in Tu context
% and code the symbols in the sequence x
for l=1:L
sc=T(1);
m=x(l);
hc=T(m+1);lc=T(m+2);
if hc==lc % unused symbol, code ESC symbol first
hc=T(M+2);lc=T(M+3);
EncodeSymbol; % code escape with T table
sc=Tu(1);hc=Tu(m+1);lc=Tu(m+2); % symbol with Tu table
Tu(1:(m+1))=Tu(1:(m+1))-1; % update Tu table
end
EncodeSymbol; % code actual symbol with T table (or Tu table)
% update T table, MUST be identical in Encode and Decode
% this avoid very large values in T even if sequence is very long
T(1:(m+1))=T(1:(m+1))+1;
if (rem(l,5000)==0) % l here is letter l (small L)
dT=T(1:(M+2))-T(2:(M+3));
dT=floor(dT*7/8+1/8);
for m=(M+2):(-1):1; T(m)=T(m+1)+dT(m); end;
end
end % for l=1:L
% this end the "elseif sum(xType==[3,4,5,6])"-clause
elseif (xType==7) % L>1 and 0 <= x(i) <= 1
PutVLIC(L-2);
EncodeBin(x,L); % code this sequence a special way
elseif (xType==8) % L>1 and 0 <= x(1)=x(2)=...=x(L) <= 1
PutVLIC(L-2);
PutABit(x(1));
elseif (xType==9) % L=1 and 0 <= x(1) <= 1
PutABit(x(1));
elseif (xType==10) % L>1 and x(1)=x(2)=...=x(L) <= -1
PutVLIC(L-2);
PutVLIC(-1-x(1));
elseif (xType==11) % L=1 and x(1) <= -1
PutVLIC(-1-x);
end % if (xType==1)
% additional information should be coded as well
if 0 % first the way it is not done any more
if sum(xType==[4,6]) % sign must be stored
for i=1:length(Sg); PutABit(Sg(i)); end;
end
if sum(xType==[5,6]) % additional bits must be stored
for i=1:L
for ii=(x(i)-1):(-1):1
PutABit(bitget(xa(i),ii));
end
end
end
else % this is how we do it
if sum(xType==[4,6]) % sign must be stored
EncodeBin(Sg,length(I)); % since length(I)=length(Sg)
end
if sum(xType==[5,6]) % additional bits must be stored
b=zeros(sum(x)-length(I),1); % number of additional bits
bi=0;
for i=1:L
for ii=(x(i)-1):(-1):1
bi=bi+1;
b(bi)=bitget(xa(i),ii);
end
end
if (bi~=(sum(x)-length(I)))
error([Mfile,': logical error, bi~=(sum(x)-length(I)).']);
end
EncodeBin(b,bi); % since bi=(sum(x)-length(I))
end
end
%
EndPos=Byte*8-BitPos+ub; % used for counting bits
bits=EndPos-StartPos;
Res(num,1)=L;
Res(num,2)=0;
Res(num,3)=bits;
if L>0; Res(num,4)=bits/L; else Res(num,4)=bits; end;
if Display
disp([Mfile,': Sequence ',int2str(num),' of ',int2str(L),' symbols ',...
'encoded using ',int2str(bits),' bits.']);
end
end % for num=1:NumOfX
% flush the arithmetic coder'
PutBit(bitget(low,K-1));
ub=ub+1;
while ub>0
PutBit(~bitget(low,K-1));
ub=ub-1;
end
% flush is finished
y=y(1:Byte);
varargout(1) = {y};
if (nargout >= 2)
% now calculate results for the total
Res(NumOfX+1,1)=Ltot;
Res(NumOfX+1,2)=0;
Res(NumOfX+1,3)=Byte*8;
if (Ltot>0); Res(NumOfX+1,4)=Byte*8/Ltot; else Res(NumOfX+1,4)=Byte*8; end;
varargout(2) = {Res};
end
end % if Encode
if Decode
for k=1:K
code=code*2;
code=code+GetBit; % read bits into code
end
NumOfX=GetVLIC; % first read number of sequences
xC=cell(NumOfX,1);
for num=1:NumOfX
% find what kind of sequenqe we have, xType, stored first in sequence
xType=GetS(TypeS,TypeC);
% now decode the different kind of sequences, each the way it was stored
if (xType==0) % empty sequence, no more symbols coded
x=[];
elseif (xType==1) % one symbol and x=x(1)>1
x=GetVLIC+2;
elseif (xType==2) % L>1 but only one symbol, x(1)=x(2)=...=x(L)>1
L=GetVLIC+2;
x=ones(L,1)*(GetVLIC+2);
elseif sum(xType==[3,4,5,6]) % now 'normalized' sequences: 0 <= x(i) <= 1000
L=GetVLIC+2;
x=zeros(L,1);
M=GetN(1000); % M is max(x)
% initialize model
T=[ones(M+2,1);0];
Tu=flipud((-1:(M+1))'); % (-1) since ESC never is used in Tu context
% and decode the symbols in the sequence x
for l=1:L
sc=T(1);
range=high-low+1;
count=floor(( (code-low+1)*sc-1 )/range);
m=2; while (T(m)>count); m=m+1; end;
hc=T(m-1);lc=T(m);m=m-2;
RemoveSymbol;
if (m>M) % decoded ESC symbol, find symbol from Tu table
sc=Tu(1);range=high-low+1;
count=floor(( (code-low+1)*sc-1 )/range);
m=2; while (Tu(m)>count); m=m+1; end;
hc=Tu(m-1);lc=Tu(m);m=m-2;
RemoveSymbol;
Tu(1:(m+1))=Tu(1:(m+1))-1; % update Tu table
end
x(l)=m;
% update T table, MUST be identical in Encode and Decode
% this avoid very large values in T even if sequence is very long
T(1:(m+1))=T(1:(m+1))+1;
if (rem(l,5000)==0)
dT=T(1:(M+2))-T(2:(M+3));
dT=floor(dT*7/8+1/8);
for m=(M+2):(-1):1; T(m)=T(m+1)+dT(m); end;
end
end % for l=1:L
% this end the "elseif sum(xType==[3,4,5,6])"-clause
elseif (xType==7) % L>1 and 0 <= x(i) <= 1
L=GetVLIC+2;
x=DecodeBin(L); % decode this sequence a special way
elseif (xType==8) % L>1 and 0 <= x(1)=x(2)=...=x(L) <= 1
L=GetVLIC+2;
x=ones(L,1)*GetABit;
elseif (xType==9) % L=1 and 0 <= x(1) <= 1
x=GetABit;
elseif (xType==10) % L>1 and x(1)=x(2)=...=x(L) <= -1
L=GetVLIC+2;
x=ones(L,1)*(-1-GetVLIC);
elseif (xType==11) % L=1 and x(1) <= -1
x=(-1-GetVLIC);
end % if (xType==0)
% additional information should be decoded as well
L=length(x);
I=find(x);
if 0
if sum(xType==[4,6]) % sign must be retrieved
Sg=zeros(size(I));
for i=1:length(I); Sg(i)=GetABit; end; % and the signs (0/1)
Sg=Sg*2-1; % (-1/1)
end
if sum(xType==[5,6]) % additional bits must be retrieved
xa=zeros(L,1);
for i=1:L
for ii=2:x(i)
xa(i)=2*xa(i)+GetABit;
end
end
x(I)=2.^(x(I)-1);
x=x+xa;
end
else
if sum(xType==[4,6]) % sign must be retrieved
Sg=DecodeBin(length(I)); % since length(I)=length(Sg)
Sg=Sg*2-1; % (-1/1)
end
if sum(xType==[5,6]) % additional bits must be retrieved
bi=sum(x)-length(I); % number of additional bits
b=DecodeBin(bi);
bi=0;
xa=zeros(L,1);
for i=1:L
for ii=2:x(i)
bi=bi+1;
xa(i)=2*xa(i)+b(bi);
end
end
x(I)=2.^(x(I)-1);
x=x+xa;
end
end
if sum(xType==[4,6]) % sign must be used
x(I)=x(I).*Sg;
end
% now x is the retrieved sequence
xC{num}=x;
end % for num=1:NumOfX
varargout(1) = {xC};
end
return % end of main function, Arith07
%----------------------------------------------------------------------
%----------------------------------------------------------------------
% --- The functions for binary sequences: EncodeBin and DecodeBin -------
% These function may call themselves recursively
function EncodeBin(x,L)
global y Byte BitPos
global high low range ub hc lc sc K code
Display=0;
x=x(:);
if (length(x)~=L); error('EncodeBin: length(x) not equal L.'); end;
% first we try some different coding methods to find out which one
% that might do best. Many more methods could have been tried, for
% example methods to check if x is a 'byte' sequence, or if the bits
% are grouped in other ways. The calling application is the best place
% to detect such dependencies, since it will (might) know the process and
% possible also its statistical (or deterministic) properties.
% Here we just check some few coding methods: direct, split, diff, diff+split
% The main variables used for the different methods are:
% direct: x, I, J, L11, b0
% split: x is split into x1 and x2, L1, L2, b1 is bits needed
% diff: x3 is generated from x, I3, J3 L31, b2
% diff+split: x3 is split into x4 and x5, L4, L5, b3
%
MetS=[0,1,2,3]; % the different methods, direct, split, diff, diff+split
MetC=[9,3,3,1]; % and the counts (which gives the probabilities)
% first set how many bits needed to code the method
b0=log2(sum(MetC))-log2(MetC(1));
b1=log2(sum(MetC))-log2(MetC(2));
b2=log2(sum(MetC))-log2(MetC(3));
b3=log2(sum(MetC))-log2(MetC(4));
I=find(x(1:(L-1))==1); % except last element in x
J=find(x(1:(L-1))==0);
L11=length(I);
% perhaps is 'entropy' interesting
% N1=L11+x(L);
% N0=L-N1;
% b=N1*log2(L/N1)+N0*log2(L/N0);
% disp(['L=',int2str(L),', N1=',int2str(N1),', (e)bits=',num2str(b)]);
b0=b0+BitEst(L,L11+x(L)); % bits needed for the sequence without splitting
if Display
disp(['EncodeBin: x of length ',int2str(L),' and ',int2str(L11+x(L)),...
' ones (p=',num2str((L11+x(L))/L,'%1.3f'),...
') can be coded by ',num2str(b0,'%6.0f'),' bits.']);
end
% diff with a binary sequence is to indicate wether or not a symbol is
% the same as preceeding symbol or not, x(0) is assumed to be '0' (zero).
% This is the DPCM coding scheme on a binary sequence
x3=abs(x-[0;x(1:(L-1))]);
I3=find(x3(1:(L-1))==1); % except last element in x3
J3=find(x3(1:(L-1))==0);
L31=length(I3);
b2=b2+BitEst(L,L31+x3(L)); % bits needed for the sequence without splitting
if Display
disp(['EncodeBin: diff x, L=',int2str(L),' gives ',int2str(L31+x3(L)),...
' ones (p=',num2str((L31+x3(L))/L,'%1.3f'),...
') can be coded by ',num2str(b2,'%6.0f'),' bits.']);
end
%
if (L>40)
% only now we try to split the sequences x and x3
if (L11>3) & ((L-L11)>3)
% try to split x into x1 and x2, depending on previous symbol
if L11<(L/2)
x1=x(I+1);
x2=[x(1);x(J+1)];
L1=L11;L2=L-L11;
else
x1=[x(1);x(I+1)];
x2=x(J+1);
L1=L11+1;L2=L-L11-1;
end
b11=BitEst(L1,length(find(x1))); % bits needed for x1
b12=BitEst(L2,length(find(x2))); % bits needed for x2
% b1 is bits to code: Method, L11, x1 and x2
b1=b1+log2(L)+b11+b12;
if Display
disp(['EncodeBin, x -> x1+x2: lengths are ',int2str(L1),'+',int2str(L2),...
' bits are ',num2str(b11,'%6.0f'),'+',num2str(b12,'%6.0f'),...
'. Total is ',num2str(b1,'%6.0f'),' bits.']);
end
else
b1=b0+1; % just to make this larger
end
if (L31>3) & ((L-L31)>3)
% try to split x3 into x4 and x5, depending on previous symbol
if L31<(L/2)
x4=x3(I3+1);
x5=[x3(1);x3(J3+1)];
L4=L31;L5=L-L31;
else
x4=[x3(1);x3(I3+1)];
x5=x3(J3+1);
L4=L31+1;L5=L-L31-1;
end
b31=BitEst(L4,length(find(x4))); % bits needed for x4
b32=BitEst(L5,length(find(x5))); % bits needed for x5
% b3 is bits to code: Method, L31, x4 and x5
b3=b3+log2(L)+b31+b32;
if Display
disp(['EncodeBin, diff x -> x4+x5: lengths are ',int2str(L4),'+',int2str(L5),...
' bits are ',num2str(b31,'%6.0f'),'+',num2str(b32,'%6.0f'),...
'. Total is ',num2str(b3,'%6.0f'),' bits.']);
end
else
b3=b2+1; % just to make this larger
end
else
b1=b0+1; % just to make this larger
b3=b2+1; % just to make this larger
end
% now code x by the best method of those investigated
[b,MetI]=min([b0,b1,b2,b3]);
MetI=MetI-1;
PutS(MetI,MetS,MetC); % code which method to use
%
if MetI==0
% code the sequence x
N1=L11+x(L);
N0=L-N1;
PutN(N1,L); % code N1, 0<=N1<=L
for n=1:L
if ~(N0*N1); break; end;
PutS(x(n),[0,1],[N0,N1]); % code x(n)
N0=N0-1+x(n);N1=N1-x(n); % update model (of rest of the sequence)
end
elseif MetI==1
% code x1 and x2
clear x4 x5 x3 x I3 J3 I J
PutN(L11,L-1); % 0<=L11<=(L-1)
EncodeBin(x1,L1);
EncodeBin(x2,L2);
elseif MetI==2
% code the sequence x3
N1=L31+x3(L);
N0=L-N1;
PutN(N1,L); % code N1, 0<=N1<=L
for n=1:L
if ~(N0*N1); break; end;
PutS(x3(n),[0,1],[N0,N1]); % code x3(n)
N0=N0-1+x3(n);N1=N1-x3(n); % update model (of rest of the sequence)
end
elseif MetI==3
% code x4 and x5
clear x1 x2 x3 x I3 J3 I J
PutN(L31,L-1); % 0<=L31<=(L-1)
EncodeBin(x4,L4);
EncodeBin(x5,L5);
end
return % end of EncodeBin
function x = DecodeBin(L)
global y Byte BitPos
global high low range ub hc lc sc K code
% these must be as in EncodeBin
MetS=[0,1,2,3]; % the different methods, direct, split, diff, diff+split
MetC=[9,3,3,1]; % and the counts (which gives the probabilities)
MetI=GetS(MetS,MetC); % encode which method to use
if (MetI==1) | (MetI==3) % a split was done
L11=GetN(L-1); % 0<=L11<=(L-1)
if L11<(L/2)
L1=L11;L2=L-L11;
else
L1=L11+1;L2=L-L11-1;
end
x1=DecodeBin(L1);
x2=DecodeBin(L2);
% build sequence x from x1 and x2
x=zeros(L,1);
if L11<(L/2)
x(1)=x2(1);
n1=0;n2=1; % index for the last in x1 and x2
else
x(1)=x1(1);
n1=1;n2=0; % index for the last in x1 and x2
end
for n=2:L
if (x(n-1))
n1=n1+1;
x(n)=x1(n1);
else
n2=n2+1;
x(n)=x2(n2);
end
end
else % no split
N1=GetN(L);
N0=L-N1;
x=zeros(L,1);
for n=1:L
if (N0==0); x(n:L)=1; break; end;
if (N1==0); break; end;
x(n)=GetS([0,1],[N0,N1]); % decode x(n)
N0=N0-1+x(n);N1=N1-x(n); % update model (of rest of the sequence)
end
end
if (MetI==2) | (MetI==3) % x is diff coded
for n=2:L
x(n)=x(n-1)+x(n);
end
x=rem(x,2);
end
return % end of DecodeBin
% ------- Other subroutines ------------------------------------------------
% Functions to write and read a Variable Length Integer Code word
% This is a way of coding non-negative integers that uses fewer
% bits for small integers than for large ones. The scheme is:
% '00' + 4 bit - integers from 0 to 15
% '01' + 8 bit - integers from 16 to 271
% '10' + 12 bit - integers from 272 to 4367
% '110' + 16 bit - integers from 4368 to 69903
% '1110' + 20 bit - integers from 69940 to 1118479
% '1111' + 24 bit - integers from 1118480 to 17895695
% not supported - integers >= 17895696 (=2^4+2^8+2^12+2^16+2^20+2^24)
function PutVLIC(N)
global y Byte BitPos
global high low range ub hc lc sc K code
if (N<0)
error('Arith07-PutVLIC: Number is negative.');
elseif (N<16)
PutABit(0);PutABit(0);
for (i=4:-1:1); PutABit(bitget(N,i)); end;
elseif (N<272)
PutABit(0);PutABit(1);
N=N-16;
for (i=8:-1:1); PutABit(bitget(N,i)); end;
elseif (N<4368)
PutABit(1);PutABit(0);
N=N-272;
for (i=12:-1:1); PutABit(bitget(N,i)); end;
elseif (N<69940)
PutABit(1);PutABit(1);PutABit(0);
N=N-4368;
for (i=16:-1:1); PutABit(bitget(N,i)); end;
elseif (N<1118480)
PutABit(1);PutABit(1);PutABit(1);PutABit(0);
N=N-69940;
for (i=20:-1:1); PutABit(bitget(N,i)); end;
elseif (N<17895696)
PutABit(1);PutABit(1);PutABit(1);PutABit(1);
N=N-1118480;
for (i=24:-1:1); PutABit(bitget(N,i)); end;
else
error('Arith07-PutVLIC: Number is too large.');
end
return
function N=GetVLIC
global y Byte BitPos
global high low range ub hc lc sc K code
N=0;
if GetABit
if GetABit
if GetABit
if GetABit
for (i=1:24); N=N*2+GetABit; end;
N=N+1118480;
else
for (i=1:20); N=N*2+GetABit; end;
N=N+69940;
end
else
for (i=1:16); N=N*2+GetABit; end;
N=N+4368;
end
else
for (i=1:12); N=N*2+GetABit; end;
N=N+272;
end
else
if GetABit
for (i=1:8); N=N*2+GetABit; end;
N=N+16;
else
for (i=1:4); N=N*2+GetABit; end;
end
end
return
% Aritmetic coding of a number or symbol x, the different symbols (numbers)
% are given in the array S, and the counts (which gives the probabilities)
% are given in C, an array of same length as S.
% x must be a number in S.
% example with symbols 0 and 1, where probabilites are P{0}=0.8, P{1}=0.2
% PutS(x,[0,1],[4,1]) and x=GetS([0,1],[4,1])
% An idea: perhaps the array S should be only in the calling function, and
% that it do not need to be passed as an argument at all.
% Hint: it may be best to to the most likely symbols (highest counts) in
% the beginning of the tables S and C.
function PutS(x,S,C)
global y Byte BitPos
global high low range ub hc lc sc K code
N=length(S); % also length(C)
m=find(S==x); % m is a single value, index in S, 1 <= m <= N
sc=sum(C);
lc=sc-sum(C(1:m));
hc=lc+C(m);
% disp(['PutS: lc=',int2str(lc),' hc=',int2str(hc),' sc=',int2str(sc),' m=',int2str(m)]);
EncodeSymbol; % code the bit
return
function x=GetS(S,C)
global y Byte BitPos
global high low range ub hc lc sc K code
range=high-low+1;
sc=sum(C);
count=floor(( (code-low+1)*sc-1 )/range);
m=1;
lc=sc-C(1);
while (lc>count); m=m+1; lc=lc-C(m); end;
hc=lc+C(m);
x=S(m);
% disp(['GetS: lc=',int2str(lc),' hc=',int2str(hc),' sc=',int2str(sc),' m=',int2str(m)]);
RemoveSymbol;
return
% Aritmetic coding of a number x, 0<=x<=N, P{0}=P{1}=...=P{N}=1/(N+1)
function PutN(x,N) % 0<=x<=N
global y Byte BitPos
global high low range ub hc lc sc K code
sc=N+1;
lc=x;
hc=x+1;
EncodeSymbol; % code the bit
return
function x=GetN(N)
global y Byte BitPos
global high low range ub hc lc sc K code
range=high-low+1;
sc=N+1;
x=floor(( (code-low+1)*sc-1 )/range);
hc=x+1;lc=x;
RemoveSymbol;
return
% Aritmetic coding of a bit, probability is 0.5 for both 1 and 0
function PutABit(Bit)
global y Byte BitPos
global high low range ub hc lc sc K code
sc=2;
if Bit
hc=1;lc=0;
else
hc=2;lc=1;
end
EncodeSymbol; % code the bit
return
function Bit=GetABit
global y Byte BitPos
global high low range ub hc lc sc K code
range=high-low+1;
sc=2;
count=floor(( (code-low+1)*sc-1 )/range);
if (1>count)
Bit=1;hc=1;lc=0;
else
Bit=0;hc=2;lc=1;
end
RemoveSymbol;
return;
% The EncodeSymbol function encode a symbol, (correspond to encode_symbol page 149)
function EncodeSymbol
global y Byte BitPos
global high low range ub hc lc sc K code
range=high-low+1;
high=low+floor(((range*hc)/sc)-1);
low=low+floor((range*lc)/sc);
while 1 % for loop on page 149
if bitget(high,K)==bitget(low,K)
PutBit(bitget(high,K));
while ub > 0
PutBit(~bitget(high,K));
ub=ub-1;
end
elseif (bitget(low,K-1) & (~bitget(high,K-1)))
ub=ub+1;
low=bitset(low,K-1,0);
high=bitset(high,K-1,1);
else
break
end
low=bitset(low*2,K+1,0);
high=bitset(high*2+1,K+1,0);
end
return
% The RemoveSymbol function removes (and fill in new) bits from
% file, y, to code
function RemoveSymbol
global y Byte BitPos
global high low range ub hc lc sc K code
range=high-low+1;
high=low+floor(((range*hc)/sc)-1);
low=low+floor((range*lc)/sc);
while 1
if bitget(high,K)==bitget(low,K)
% do nothing (shift bits out)
elseif (bitget(low,K-1) & (~bitget(high,K-1)))
code=bitset(code,K-1,~bitget(code,K-1)); % switch bit K-1
low=bitset(low,K-1,0);
high=bitset(high,K-1,1);
else
break
end
low=bitset(low*2,K+1,0);
high=bitset(high*2+1,K+1,0);
code=bitset(code*2+GetBit,K+1,0);
end
if (low > high); error('low > high'); end;
return
% Functions to write and read a Bit
function PutBit(Bit)
global y Byte BitPos
BitPos=BitPos-1;
if (~BitPos); Byte=Byte+1; BitPos=8; end;
y(Byte) = bitset(y(Byte),BitPos,Bit);
return
function Bit=GetBit
global y Byte BitPos
BitPos=BitPos-1;
if (~BitPos); Byte=Byte+1; BitPos=8; end;
Bit=bitget(y(Byte),BitPos);
return
function b=BitEst(N,N1);
if (N1>(N/2)); N1=N-N1; end;
N0=N-N1;
if (N>1000)
b=(N+3/2)*log2(N)-(N0+1/2)*log2(N0)-(N1+1/2)*log2(N1)-1.3256;
elseif (N1>20)
b=(N+3/2)*log2(N)-(N0+1/2)*log2(N0)-(N1+1/2)*log2(N1)-0.020984*log2(log2(N))-1.25708;
else
b=log2(N+1)+sum(log2(N-(0:(N1-1))))-sum(log2(N1-(0:(N1-1))));
end
return
⛄ 运行结果
⛄ 参考文献
