perform_arith_coding.m.svn-base

signal procesing toolbox

function [y,nbr_bits] = perform_arith_coding(xC, dir)

% perform_arithmetic_coding_slow - perform adaptive arithmetic coding
%
% 	[y,nbr_bits] = perform_arithmetic_coding_slow(x, dir);
%
%   dir=1 for encoding, dir=-1 for decoding.
%
% Based on the code of (c) Karl Skretting
%
%   Copyright (c) 2008 Gabriel Peyre

%
% ------------------------------------------------------------------
% 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)
%  x       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)

if iscell(xC) & length(xC)>1
    nbr_bits = 0;
    for k=1:length(xC)
        [y{k},nb] = perform_arithmetic_coding(xC{k}, dir);
        nbr_bits = nbr_bits + nb;
    end
    return;
end


if dir==1
    xC = {xC};
end

%----------------------------------------------------------------------
% Copyright (c) 1999-2001.  Karl Skretting.  All rights reserved.
% Hogskolen in Stavanger (Stavanger University), Signal Processing Group
% Mail:  [e-mail]   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=0;      % 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;
                counts=floor(( (code-low+1)*sc-1 )/range);
                m=2; while (T(m)>counts); 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;
                    counts=floor(( (code-low+1)*sc-1 )/range);
                    m=2; while (Tu(m)>counts); 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

if dir==1
    nbr_bits = Res(1,3);
else
    % undef for decoding
    nbr_bits = -1;
    y = xC{1};
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;