1
$\begingroup$

I want to develop a steganography project with using exploiting modification direction method. I started to try and understand how the method works, but I have some questions about decryption.

I hid the secret image in a cover image larger than the secret image by 9 times. So when it comes to decryption I'm not going to use all of the pixels of the cover image: I only need to use those pixels used during encryption. But if the pixel count isn't provided before then how can I know how many pixels should be used to extract the secret image?

If I use all pixels of the cover image the resulting image's size doesn't match the original one and it is quite likely there will be flaws in the image.

Improve this question
$\endgroup$
1
  • $\begingroup$ Hint: keep your sentences short and divide your text into sections to make it more readable. $\endgroup$
    – Maarten Bodewes ♦
    Commented May 6, 2018 at 12:00

1 Answer 1

Reset to default
1
$\begingroup$

The EMD method is a little outdated. The state of the art are Syndrome Tellis Codes. Please, check:

http://dde.binghamton.edu/filler/pdf/Fill10tifs-stc.pdf

In the EMD method, n pixels are used to carry one secret digit. Therefore, to extract this digit you need to use the same n pixels. The receiver needs to know what pixels are used by the sender.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks, so in the extraction stage size of the image we're trying to obtain already have to be known. I'm considering a situation in which in order to carrying each digits of the pixels of secret image, all of the pixels of the cover image didn't be used. $\endgroup$
    – Tolga Karahan
    Commented May 6, 2018 at 12:58
  • $\begingroup$ This is usual not to use all the pixels in the image. In this case is it is appropriate to select those pixels that are more difficult to steganalyze. In this context today researchers are using STC as I told you before. $\endgroup$
    – Daniel Lerch
    Commented May 6, 2018 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.