Question

我正在使用LSB技术进行视频隐写术。使用traffic.avi和xylophone.mpg作为封面媒体，当我使用licence.txt文件（在附件中）编码成视频时，它运行良好但是，当我使用短消息输入文本时，它显示错误

“ENCODE中的矩阵MSG必须有K列。”有时使用短文时会出现错误“msg太长而无法编码”

我不知道这两组编码意味着什么，以及如何编辑代码以便可以对短信息进行编码......下面是一些我认为与此问题相关的代码

     num2add = 80-length(msg);  % Number of spaces to add to end of MSG.
     if num2add < 0, error('This message is too long to encode.'), end
     newmsg = [msg,repmat(' ',1,num2add)]; % 80 chars always encoded.
     msgmat = dec2bin(newmsg)-48; % Each row is a bin. rep. of an ascii char.

以及此编码

    if m_msg == 1
        type_flag = 2;  % binary vector
        [msg, added] = vec2mat(msg, k);
    elseif m_msg ~= k
        error('comm:encode:InvalidMatrixColumnSize','The matrix MSG in ENCODE must have K columns.');

以下是上述编码的第一部分后继续完整的编码编码！

    B = pic1(:,:,1);   [piclngth pichght] = size(B);  % Choose the first page.
    dim1 = piclngth-2;   dim2 = pichght-3;   keyb = key(end:-1:1);
    rows = cumsum(double(key));   
    columns = cumsum(double(keyb));  % Coord pairs for KEY (rows,columns)
    A = zeros(dim1,dim2); % This matrix will house the hiding points.
    A = crtmtrx(A,rows,columns,dim1,dim2,key);
    idx = find(A==1);  % This same index will be used for pic matrix.

    for vv = 1:80  % This is the encoder.  
    for uu = 1:8 
    if msgmat(vv,uu)==1;
       if rem(B(idx(uu+8*(vv-1))),2)==0
            if(frame==1)
            disp('some pixel value of original frame');
            B(idx(uu+8*(vv-1)))
             end
          B(idx(uu+8*(vv-1))) = B(idx(uu+8*(vv-1)))+1;
            if(frame==1)
            disp('some pixel value of stegno video frame');
            B(idx(uu+8*(vv-1)))
             end
       end
    elseif rem(B(idx(uu+8*(vv-1))),2)==1
            if(frame==1)
            disp('some pixel value of original frame');
            B(idx(uu+8*(vv-1)))
             end
          B(idx(uu+8*(vv-1))) = B(idx(uu+8*(vv-1)))-1;
            if(frame==1)
            disp('some pixel value of stegno video frame');
            B(idx(uu+8*(vv-1)))
             end
         end
     end
   end
    global newpic;
    newpic = pic1;   newpic(:,:,1) = B;
    f(frame) = im2frame(newpic);
  end

    frameRate = get(vidObj,'FrameRate');


    movie2avi(f,'stegano_video.avi','compression','None', 'fps', 20);
    success = 1;

    function A = crtmtrx(A,rows,columns,dim1,dim2,key)
    % Creates the matrix used to find the points to hide the message.

    jj = 1;   idx = 1;
  while 640 > length(idx) %  Need 560 points to hide 80 characters.     
for ii = 1:length(rows)
    if rows(ii) < dim1
       rows(ii) = rem(dim1,rows(ii))+1;
    else
       rows(ii) = rem(rows(ii),dim1)+1;
    end
    if columns(ii) < dim2
       columns(ii) = rem(dim2,columns(ii))+1;
    else
       columns(ii) = rem(columns(ii),dim2)+1;
    end
    A(rows(ii),columns(ii)) = 1;
end
rows = jj*cumsum(double(columns))+round(dim2/2);  % Each pass is diff.
columns = jj*cumsum(double(rows))+round(dim1/2);
if jj > ceil(640/length(key))+2  % Estimate how many iters. needed.
   idx = find(A==1);
 end
   jj = jj+1;
end

this is some of the input text and the right one is the encypted txt

Answer 1

触发错误的代码非常清楚：

num2add = 80-length(msg);  % Number of spaces to add to end of MSG.
if num2add < 0, error('This message is too long to encode.'), end

基本上，只要msg中有超过80个字符，您就会收到错误。我不确定80是否有意义，你可以尝试增加它，但这可能会破坏别的东西。

错误“ENCODE中的矩阵MSG必须有K列”

1 个答案: