How to show the encrypted share for the Jarvis halftone algorithm in following code?

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Mangai s 2017-2-20

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https://ww2.mathworks.cn/matlabcentral/answers/325913-how-to-show-the-encrypted-share-for-the-jarvis-halftone-algorithm-in-following-code

评论： Walter Roberson 2017-2-22

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How to show the encrypted share for the Jarvis halftone algorithm?

[Merged from duplicate]

function outImg = jarvisHalftone(inImg)
a = imread('D:\our project\crypt\rose.jpg');
figure
imshow(a);
I = rgb2gray(a);
inImg = double(I);
figure
imshow(inImg);
[M,N] = size(inImg);
T = 127.5;
y = inImg;
error = 0;
y= [127.5*ones(M,2) y 127.5*ones(M,2) ; 127.5*ones(2,N+4)];
z = y;
M
N
for rows = 1:M  
  for cols = 3:N+2
    z(rows,cols) =255*(y(rows,cols)>=T);
    error = -z(rows,cols) + y(rows,cols);
      y(rows,cols+2) = 5/48 * error + y(rows,cols+2);
      y(rows,cols+1) = 7/48 * error + y(rows,cols+1);
      y(rows+1,cols+2) = 3/48 * error + y(rows+1,cols+2);
      y(rows+1,cols+1) = 5/48 * error + y(rows+1,cols+1);
      y(rows+1,cols+0) = 7/48 * error + y(rows+1,cols+0);
      y(rows+1,cols-1) = 5/48 * error + y(rows+1,cols-1);
      y(rows+1,cols-2) = 3/48 * error + y(rows+1,cols-2);
      y(rows+2,cols+2) = 1/48 * error + y(rows+2,cols+2);
      y(rows+2,cols+1) = 3/48 * error + y(rows+2,cols+1);
      y(rows+2,cols+0) = 5/48 * error + y(rows+2,cols+0);
      y(rows+2,cols-1) = 3/48 * error + y(rows+2,cols-1);
      y(rows+2,cols-2) = 1/48 * error + y(rows+2,cols-2);
    end
  end
  figure
  imshow(z);
outImg = z(1:M,3:N+2);
figure
imshow(outImg);
outImg = im2bw(uint8(outImg));
figure
imshow(outImg)

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Mangai s 2017-2-22

The actual encrypted image should be blurred one. But I am not able to extract that from the code.

Walter Roberson 2017-2-22

It would help if you posted a copy of your rose.jpg image along with a copy of the image we should be looking to extract (so we would be able to recognize it.)

Is your rose.jpg created in the rather uncommon JPEG lossless compression? If not then the JPEG compression process would have damaged the payload. Though perhaps that is part of the challenge, that perhaps all of the shares together should be able to reconstruct the data even in the face of damage?

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