Thanks a lot, this works well.
This new function is even able to partially decode the watermarked
picture I made with your previous sample code, using:
image.stegano(WatermarkImage)
It's obvious the algorithm is not the same, because I get a highly
modified watermark, but I get it! More precisely, I see there is
something looking vaguely like a MBS Logo.
Thanks for having done this job. Greatly appreciated.
Now, It would be great to know why the Stegano module does not work, as in:
call n.read("stegano:/tmp/test.png")
And, of course, to make it work.
Thanks again.
Jean-Luc Arnaud
Le 23/03/2016 13:37, Christian Schmitz a écrit :
>> Am 23.03.2016 um 11:08 schrieb Jean-Luc Arnaud <[email protected]>:
>>
>> Ok, so let try another way...
>> What kind of algorithm do you use for this Stegano function ? Your own
>> or a known one (i.e. LSB one)?
> This GraphicsMagick plugin uses the code from GraphicsMagick library.
>
>> Do you know a tool able to decode it? I just need to know if the
>> watermark is in the image and to decode it.
> Well, I just added my own here:
>
>
> const AllChannels = &h111
>
> dim p as new Picture(500, 500, 32)
>
> dim w as new window1
> w.Title = "white"
> w.Backdrop = p
>
> dim l as Picture = LogoMBS(500)
>
> w = new window1
> w.Title = "Logo"
> w.Backdrop = l
>
> // adds picture. You will normally not see the modification
> dim x as Picture = p.AddSteganographyPictureMBS(AllChannels, l)
>
> w = new window1
> w.Title = "Logo hidden in white picture"
> w.Backdrop = x
>
> // as we store in lowest bit, this picture will look strange
> dim y as Picture = x.SteganographyPictureMBS(AllChannels)
>
> w = new window1
> w.Title = "Logo extracted"
> w.Backdrop = y
>
>
>
> Sincerely
> Christian
>
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