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 since it is all Matlab code
% 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 dokumentation 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 (4 <= 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)
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
