Image watermarking using LSB

Question

Elysi Cochin 2018-3-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/387023-image-watermarking-using-lsb

编辑： Guillaume 2018-3-8

Below are few line from LSB image watermarking. Please can someone explain why we divide by 32 in the first code line. Why do we choose 224 and 248 for image2hide and coverImage and not same number or other number

im2hide = bitand(floor(im2hide), 224) / 32;
coverimage = bitand(floor(coverimage), 248);

same way why 7 * 32 is done in decryption line

im2hide = bitand(floor(watermarkedIm), 7) * 32;

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Christoph F. 2018-3-8

It looks like the code is doing some bit manipulation on the raw pixel data. What image format is used for im2hide?

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Answer 1

Guillaume 2018-3-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/387023-image-watermarking-using-lsb#answer_309006

编辑：Guillaume 2018-3-8

在 MATLAB Online 中打开

This is simple bit masking and shifting. Whenever you encounter this sort of thing, if you're not familiar with powers of 2, then convert the decimal values to binary to see which bits are actually kept or discarded.

anding with 224 (11100000b) keep the 3 highest bits. Dividing by 32 then shift these bits to the 3 lowest, hence that first line is equivalent to:

im2hide = bitshit(floor(im2hide), -5)

The second line is another masking operation, this time with 11111000b, hence it keeps the 5 highest bit.

The third line is the reverse of the first, it keeps the lowest 3 bits. Then multipliying by 32 shifts these 3 bits to the highest 3 bits, and it is thus equivalent to:

im2hide = bitshift(floor(watermarkedIm), 5);

I'm not sure why the floor are there. If you're dealing with integers (as you should since we're talking about integer bits), they serve no purpose.