arith07.m

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 (maxx>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('Arith06-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('Arith06-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