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Chair Professor Chin-Chen Chang (張真誠) National Tsing Hua University

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Presentation on theme: "Chair Professor Chin-Chen Chang (張真誠) National Tsing Hua University"— Presentation transcript:

1 Some Steganographic Methods for Delivering Secret Messages Using Cover Media
Chair Professor Chin-Chen Chang (張真誠) National Tsing Hua University National Chung Cheng University Feng Chia University

2 Introduction Information Hiding Hiding system Stego image Cover image
Secret message

3 Introduction (Cont.) Cover Carriers Image Video Sound Text

4 Kim et al.’s Method 9 1 2 3 4 5 6 7 8

5 Kim et al.’s Method (Embedding)
5 8 3 4 7 6 1 2 9 1 2 3 4 5 6 7 8 Cover Image Cover Image 6 9 7 3 8 1 2 5 4 Stego Image Stego Image

6 Kim et al.’s Method (Embedding)
6 9 7 3 8 1 2 5 4 9 1 2 3 4 5 6 7 8 Stego Image Stego Image

7 Zhang and Wang’s Method (Embedding)
Extracting function: 8 7 9 4 79 54 55 11 20 21 12 24 10 Secret data: … p2 255 1 2 3 4 1 2 3 4 1 : : : : : : : : : : : : : 10002 1 35 11 2 3 4 1 2 3 4 1 2 3 2 10 1 2 3 4 1 2 3 4 1 Cover image 9 3 4 1 2 3 4 1 2 3 4 3 8 1 2 3 4 1 2 3 4 1 2 1 7 4 1 2 3 4 1 2 3 4 4 6 2 3 4 1 2 3 4 1 2 3 2 5 1 2 3 4 1 2 3 4 1 7 10 4 4 3 4 1 2 3 4 1 2 3 4 3 3 1 2 3 4 1 2 3 4 1 2 1 2 4 1 2 3 4 1 2 3 4 4 1 2 3 4 1 2 3 4 1 2 3 2 1 2 3 4 1 2 3 4 1 Stego image 1 2 3 4 5 6 7 8 9 10 11 255 p1 Magic Matrix

8 Zhang and Wang’s Method (Extracting)
p2 7 10 4 255 1 2 3 4 1 2 3 4 1 : : : : : : : : : : : : : 11 2 3 4 1 2 3 4 1 2 3 2 10 1 2 3 4 1 2 3 4 1 9 3 4 1 2 3 4 1 2 3 4 3 8 1 2 3 4 1 2 3 4 1 2 1 Stego image 7 4 1 2 3 4 1 2 3 4 4 6 2 3 4 1 2 3 4 1 2 3 2 5 1 2 3 4 1 2 3 4 1 4 3 4 1 2 3 4 1 2 3 4 3 3 1 2 3 4 1 2 3 4 1 2 1 2 4 1 2 3 4 1 2 3 4 4 1 35 1 2 3 4 1 2 3 4 1 2 3 2 1 2 3 4 1 2 3 4 1 p1 1 2 3 4 5 6 7 8 9 10 11 255 Extracted secret data: 10002 Magic Matrix

9 Sudoku A logic-based number placement puzzle

10 Sudoku (Cont.) Property
A Sudoku grid contains nine 3 × 3 matrices, each contains different digits from 1 to 9. Each row and each column of a Sudoku grid also contain different digits from 1 to 9. Possible solutions: 6,670,903,752,021,072,936,960 (i.e. ≈ 6.671×1021)

11 Sudoku (Cont.) - 1 Reference Matrix M 6 5 1 7 4 8 2 3 9 5 4 6 3 7 1 2
6 3 7 1 2 8 Reference Matrix M

12 Sudoku (Cont.) 279 Cover Image Stego Image d( , ) = d((8,4) , (8,7)) =
11 12 79 54 55 20 21 24 10 9 Secret data: … 279 Cover Image 9 7 Stego Image d( , ) = d((8,4) , (8,7)) = d( , ) = d((9,7) , (8,7)) = d( , ) = d((6,8) , (8,7)) = min.

13 Sudoku (Cont.) 279 Cover Image d( , ) = min. Stego Image
8 7 11 12 79 54 55 20 21 24 10 9 Secret data: … 279 Cover Image d( , ) = 9 7 14 min. Stego Image

14 Sudoku (Cont.) 9 7 14 Stego Image Extracted data: 279 =

15 Turtle Shell Based Matrix
2 3 ( 7 1 2 3 4 5 2 ( 4 5 6 7 1 2 3 3 ( 1 2 3 4 5 6 7 2 ( 1 2 3 4 5 6 First, like this, we draw a turtle shell with the number between 0 and 7 on it. So what we are going to do now, is to continuously write down 0 to 7 in every row. And according to this turtle shell we’ve constructed, we found that the difference between the bottom row and the upper row is 2, and the next difference is 3, then it is 2 again. So we apply this rule, alternately add 2 and 3 to every row, to complete the entire matrix.

16 255 9 8 7 6 5 4 3 2 1 6 7 1 2 3 4 5 2 3 ( Pixel b 2 ( 3 As we can see: digits on every turtle shell include 0 to 7. To me, I think it is simple but fabulous. And on vertical and horizontal axis are values of a pixel pair, where values are from 0 to 255. What does this mean and how does this matrix work? Let’s see the following examples. 1 2 3 4 5 6 7 8 9 255 Pixel a

17 9 8 7 6 5 4 3 2 1 6 7 1 2 3 4 5 3 4 4 3 4 1 9 7 71 80 55 2 12 30 21 64 73 28 95 91 76 83 23 202 42 19 3 57 11 234 39 27 40 18 215 89 255 107 59 16 99 200 185 97 243 25 101 235 79 33 198 211 3 5 4 Pixel b Here we’ve got an original image, which we called a cover image. And we’ve got a secret as well. And the turtle shell based matrix must be prepared. We show the image in the pixel form. And divide this binary secret into 3 bits patterns. As a result, the secret can be expressed in decimal form where every digit is between 0 and 7. We can find out the purpose in the next step. To embed the secret into pixels, for a start, we take the 1st pixel pair in the image. And match these two pixels in coordinates respectively. Then we can map a point on the matrix. As we can see, the point is on the back of a turtle shell. So now we define this as the 1st situation. The embedding procedure is quite easy. Find the secret digit within this turtle shell and refer back to the coordinates. And then change our pixel values into this coordinates. So the secret is successfully embedded into pixels by modifying them very slightly. No matter how it changed, it still within a turtle shell. And now we can understand why we want our secret becoming numbers between 0 and 7. Because every turtle shell only includes the number between 0 and 7. 110 010 000 111 1 2 3 4 5 6 7 8 9 6 2 7 Pixel a

18 9 8 7 6 5 4 3 2 1 6 7 1 2 3 4 5 4 1 9 7 71 80 55 2 12 30 21 64 73 28 95 91 76 83 23 202 42 19 3 57 11 234 39 27 40 18 215 89 255 107 59 16 99 200 185 97 243 25 101 235 79 33 198 211 3 5 3 6 3 5 4 Pixel b But now you might notice a problem: how about the points on the edge of the turtle shells? Like this example, we can see that actually the point is at the intersection of three different turtle shells. So this time, we find out the secret digits within all these three turtle shells. And choose the nearest one, because we do care about the image quality. To find the nearest one, means to minimize the image distortion. 110 010 000 111 1 2 3 4 5 6 7 8 9 6 2 7 Pixel a

19 9 8 7 6 5 4 3 2 1 6 7 1 2 3 4 5 4 1 9 7 71 80 55 2 12 30 21 64 73 28 95 91 76 83 23 202 42 19 3 57 11 234 39 27 40 18 215 89 255 107 59 16 99 200 185 97 243 25 101 235 79 33 198 211 3 6 4 4 4 5 Pixel b How about the point on this side? In fact, this is the totally same situation. It’s just a matter of perspective. When the related turtle shells come out, it become the same situation. 110 010 000 111 1 2 3 4 5 6 7 8 9 6 2 7 Pixel a

20 9 8 7 6 5 4 3 2 1 6 7 1 2 3 4 5 4 9 7 71 80 55 2 12 30 21 64 73 28 95 91 76 83 23 202 42 19 3 57 11 234 39 27 40 18 215 89 255 107 59 16 99 200 185 97 243 25 101 235 79 33 198 211 3 6 4 5 2 1 1 Pixel b Through all efforts, there is still a tiny flaw in this scheme. That is, how to deal with the points on the edge of the matrix. They do not be involved in any turtle shells. Well… if we take a detailed observation on the matrix, we can discover another interesting feature of the matrix. That is, wherever we draw a 3 by 3 block, we find the numbers between 0 and 7 are all within it. Then apparently, as what we did before, find the secret digit within the block. 110 010 000 111 1 2 3 4 5 6 7 8 9 6 2 7 Pixel a

21 9 8 7 6 5 4 3 2 1 6 7 1 2 3 4 5 4 9 7 71 80 55 2 12 30 21 64 73 28 95 91 76 83 23 202 42 19 3 57 11 234 39 27 40 18 215 89 255 107 59 16 99 200 185 97 243 25 101 235 79 33 198 211 3 6 4 5 2 Pixel b For the extracting phrase, I think it is much easier than the embedding procedures. Whatever the embedding strategies had been used, in the extracting phrase, we just easily extract the secret by drawing out pixel pairs sequentially. And use them to map the secret in our matrix. 110 010 000 111 1 2 3 4 5 6 7 8 9 6 2 7 Pixel a

22 Experimental Results Here comes our experimental results. These pictures above are generally used in many related research. After using our scheme to hide secret into these images, they still remain the high qualities.

23 Conclusions Simple embedding & extracting Design Philosophy: Goal:
So here is my conclusions for our proposed scheme and the future objective of researching. For our proposed scheme, I think the most remarkable property is that the embedding and extracting procedure are both quite simple. For the future objective of researching, well… here comes a design philosophy: with higher embedding capacity, comes lower image quality. So this will be our goal to pursue high embedding capacity; in the meantime, remain high image quality. Embedding Capacity Image Quality

24 Questions? Thank you! Here is my presentation for today. Are there any questions?


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