Visual cryptography with extra ability of hiding
confidential data
Wen-Pinn Fang Ja-Chen Lin
National Chiao Tung University
Department of Computer and Information Science 1001 Ta Hsueh Road
Hsinchu, Taiwan E-mail: [email protected]
Abstract. We present a two-in-one visual cryptography (VC) scheme that not only shares an image of moderate confidentiality between two noisy transparencies, but also hides in these two trans-parencies a more confidential text file describing the image. None of the transparencies alone can reveal anything about the image or text. Later, people can view the image by simply stacking the two transparencies; on the other hand, after certain simple computa-tions, the more confidential text data can also be extracted. We also introduce an alternative version in which the decoding of both the image and text requires no computer. © 2006 SPIE and IS&T. 关DOI: 10.1117/1.2193912兴
1 Introduction
Visual cryptography共VC兲 was a sharing technique first pro-posed by Naor and Shamir1 in 1994. According to their design, VC can share an image 共for example, “Lena”兲 be-tween two noisy transparencies 共also known as shares兲. Then stacking these transparencies can unveil the input im-age directly using human vision without any computation. Many other VC systems2–15have been proposed since. On the other hand, there is another kind of tool to deal with sensitive data, namely, data hiding.16–25 People can hide sensitive data in media; and later, the data can be extracted from these host media. Although there are many sophisti-cated VC systems共see Refs. 1–15兲, the reported works are seldom related to a VC design that, besides sharing to the input image共“Lena”兲, also hides a certain amount of data that are more private or more valuable than the input image itself 共for example, Lena’s address and other personal in-formation兲. Therefore, we try to design in this paper a “two-in-one” VC scheme, which shares an image “Lena” of moderate confidentiality using two transparencies; and these two transparencies together can also extract a more confidential text data file describing the image “Lena.” 共More specifically, simply stacking the two transparencies can show the image “Lena” directly; meanwhile, after cer-tain computations, one can also extract the more sensitive text data.兲 Of course, none of the transparencies alone can reveal anything about the image “Lena” or the more sensi-tive text.
There are several kinds of applications for this scheme. Perhaps the most direct application is to store or transmit an image of moderate confidentiality 共for example, the photo of an employee, a suspect, or a VIP member兲 using the two generated transparencies, whose stacking together can yield the image. In addition, the background description of the image 共for example, the employee’s name, address, birth-day, nationality, salary, and professional training兲 can be extracted only from joint information of the two transpar-encies by a high-ranking officer who knows the decoding key discussed in Sec. 4. As a result, if the two shares are collected, both of the lower rank officers and their com-mander can view the image; however, only the comcom-mander can decrypt the background text. Another application is to authenticate the transparencies generated by the VC system so that the combination containing faked transparencies can be identified, as discussed in Sec. 4.
The rest of this paper is organized as follows. The method is stated in Sec. 2; the experimental results are presented in Sec. 3; the application to the authentication of the transparencies is given in Sec. 4; the security concern is discussed in Sec. 5; an alternative version, whose decoding does not require a computer, is given in Sec. 6; finally, the conclusions are presented in Sec. 7. Throughout the paper, “Lena” is an input image of size 256⫻256, whereas “LENA” is the corresponding 512⫻512 stacking result ob-tained from stacking the two transparencies 共both are 512 ⫻512兲 generated by the proposed method.
2 Method
We describe here how to create the two transparencies. Be-fore doing so, we must introduce six fundamental blocks 共type 0 through type 5, see Fig. 1兲 and the combinations to expand a white or a black pixel共see Figs. 2 and 3, respec-tively兲 of a given image “Lena.”
The simplest design is to use 2⫻2 as the expansion rate, i.e., each share is 2⫻2 times greater than the input image “Lena” in size. Then, to define the fundamental blocks, we
Paper 05137R received Jul. 18, 2005; revised manuscript received Sep. 30, 2005; accepted for publication Oct. 25, 2005; published online Apr. 18, 2006.
consider all possible cases in which a 2⫻2 block has two white elements and two black elements. Therefore, we ob-tain C共4,2兲=4⫻3/2=6 types of fundamental blocks, called types 0, 1, 2, 3, 4, and 5, as shown in Fig. 1.
When we stack two shares and obtain the stacking re-sult, the image is 2⫻2 times greater than the input image “Lena;” for each pixel of “Lena” is expanded to a 2⫻2 block of the stacking image “LENA.” To maintain the black-and-white distribution共and hence the visual appear-ance兲 of the input image, the number of black elements in each black block of the stacking result “LENA” 共a 2⫻2 block expanded from a black pixel of “Lena”兲 should be more than that of each white block of the stacking result “LENA” 共a 2⫻2 block expanded from a white pixel of “Lena”兲. For simplicity of the design, we assume that each black block has four black elements, and each white block has two white elements.共This obviously meets the rule that each black block has more black elements than each white block has.兲 Under this assumption, it is easy to see that if we want to share a white pixel of “Lena,” we can use one of the six combinations 兵共0,0兲, 共1,1兲, 共2,2兲, 共3,3兲, 共4,4兲, 共5,5兲其, sketched in Fig. 2, to paint the corresponding posi-tion of the two shares. For example, the combinaposi-tion共3,3兲 means that, when we encode a white pixel, we may paint both shares 1 and 2共at the block position corresponding to the white pixel being processed兲 using the fundamental block of type 3. Then in the decoding process, stacking these two shares will generate a white block共a 2⫻2 block having two white elements and two black elements兲 at the corresponding position. Analogously, we can use Fig. 3 to encode a black pixel of the input image “Lena.”
Also note that to hide the more sensitive text file, we
2.1 Encoding Algorithm
Part I. 关To hide the confidential text data d
=共d1, d2, d3, . . . , d8, d9, . . .兲, where each diis a digit in the range 0 to 5.兴
Step 1. Split each digit di into two parts xi and yi so that if a person only acquires one of the two numbers in兵xi, yi其, he cannot extract di. More precisely, we randomly select a number xi in the range 0 to 5, then find the corresponding yi in the range 0 to 5 so that共xi+ yi兲=dimod 6. For example, if the confidential text data d =共d1, d2, d3, . . . , d8, d9, . . .兲=共1035004234...兲, then we can decompose d as
1 = 3 + 4 mod 6, 0 = 1 + 5 mod 6, 3 = 1 + 2 mod 6, 5 = 0 + 5 mod 6, 0 = 2 + 4 mod 6, 0 = 0 + 0 mod 6, etc.
Step 2. In an interleaved manner, the process sequen-tially generates half of the pixels in each of the two transparencies. For instance, 1 = 3 + 4 mod 6, 0 = 1 + 5 mod 6, and 3 = 1 + 2 mod 6 in the preceding example, so the 2 ⫻2 fundamental blocks of types 3, 1, and 1 are the first, third, and fifth blocks of transpar-ency 1, while the 2⫻2 fundamental blocks of types 4, 5, and 2 are the second, fourth, and sixth blocks of transparency 2. As for the re-maining blocks, they are determined in Part II to share the image “Lena.” In other words, as shown in Fig. 4, right now we have
Transparency 1: 3 ? 1 ? 1 ? 0 ? 2 ? 0 ? . . . , Transparency 2: ? 4 ? 5 ? 2 ? 5 ? 4 ? 0. . .
Part II.共To share the input image “Lena”兲
Step 3. To determine the block types of the remaining blocks 共the question marks written at the end of step 2兲 of the two transparencies, we se-quentially read the pixel values of the input
Fig. 2 Pairs whose stacking results represent white blocks.
image “Lena.” Without the loss of generality, assume that, according to the scanning order, the pixel values are BBWBWB¼. Since the desired visual decoding property requires that, after stacking, the 2⫻2 blocks had better also appear in the sequence BBWBWB¼, thus the block types of the “?” can be determined by looking up the table in Fig. 2 共or Fig. 3兲 if a “W” 共or a “B”兲 is the current pixel value of “Lena.” This results in the two transparencies shown in Fig. 5; in other words, we now have Transparency 1: 3 5គ 1 4គ 1 3គ ... ,
Transparency 2: 2គ 4 1គ 5 1គ 2 ... , Expected stacking: B B W B W B . . . . This ends the algorithm.
In the decoding phase, directly stacking the two trans-parencies can reveal the共enlarged兲 image “LENA” without any computation, while the hidden text file d =共d1, d2, d3, . . . , d8, d9, . . .兲=1035004234... can be ex-tracted by the formula 共xi+ yi兲=dimod 6 from the two transparencies. Here, xi is the block type of the 共2i−1兲’th block of transparency 1, and yi is the block type of the 共2i兲’th block of transparency 2.
3 Experimental Results
The experimental results are shown in Fig. 6. The size of each transparency is 512⫻512, which is 2⫻2 times larger than the 256⫻256 input image “Lena.” The transparencies are noise-like. If we stack them, then we can get the visible image directly, as shown in Fig. 6共d兲. The two
transparen-cies can also cooperate together to extract a sequence of 128⫻256 digits 共each digit is in the range 0 to 5兲 hidden earlier, and this sequence is the confidential text for the purpose of validity verification or background description. The contrast共defined in Ref. 9兲 of our stacked result 关Fig. 6共d兲兴 is always 共4-2兲/共2⫻2兲4=50% because each 2⫻2 block of our stacked result always has four black elements if it is a black block, and two black elements if it is a white block. Notably, according to Ref. 9, the contrast is good if it reaches 50%. Therefore, the visual quality of our stacked result is not worse than those obtained using reported VC methods, while we have the extra advantage of being able to hide the confidential text describing the unveiled image. 4 Application to Transparency Authentication As mentioned in Sec. 1, one of the applications of the pro-posed scheme is to authenticate the transparencies so that a combination of faked transparencies can be identified. This is explained in the following. We only have to transform an authentication message to a sequence d of digits in the range 0 to 5, and then hide this sequence according to part I of the algorithm to obtain shares 1 and 2. Now, assume that a hacker intercepts share 1 关Fig. 6共b兲兴. Using share 1
Fig. 4 Hiding the confidential text file 共see steps 1 and 2 of the encoding algorithm兲.
Fig. 5 Sharing the secret image “Lena”共see step 3 of the encoding algorithm兲 by filling in the undetermined parts of Fig. 4.
Fig. 6 Example illustrating the proposed method:共a兲 original half-tone secret image “Lena,”共b兲 and 共c兲 two transparencies generated by our method,共d兲 image “LENA” obtained by stacking 共b兲 and 共c兲 关note that 共b兲 and 共c兲 can also be used to extract the hidden confi-dential text兴, 共e兲 faked halftone image “Girl” owned by a hacker, 共f兲 faked transparency,共g兲 faked image “GIRL” obtained by stacking 共b兲 and共f兲. 关However, by extracting the hidden text from 共b兲 and 共f兲, our ally can know that共g兲 is faked.兴
sage兲.
5 Techniques to Improve Security
Because shares 1 and 2 are transmitted using two distinct channels or stored in two different places, the chance that both shares are intercepted is very low. However, to reduce the damage that might be caused by a superhacker 共or an internal betrayer of a company兲 who might intercept or ac-cess both shares 1 and 2, the security department can modify our algorithm by using their own manner to reas-sign the locations of the兵xi其 in share 1 and 兵yi其 in share 2. Note that 兵xi其 are not necessarily in the odd positions of share 1, and兵yi其 are not necessarily in the even positions of share 2 共see the final sentence of step 2 of the algorithm兲. For example, to paint the two shares共also called transpar-encies兲, the sequence 共x1, y1, x2, y2, x3, y3, x4, y4, x5, y5, . . .兲 can be permuted by a random-position generator using
a security seed. For instance, it becomes
共x1, x2, x3, y3, y2, y4, x7, y1, y5, x4, x8, x9, . . .兲 after the per-mutation. Then the last two lines of step 2 become Transparency 1: x1x2x3 ? ? ? x7 ? ? x4x8x9. . . , Transparency 2: ? ? ? y3y2y4 ? y1y5 ? ? ? . . . . In the hidden-message extraction phase, the inverse of the random position generator is applied, using the same seed to reverse the sequence 共x1, x2, x3, y3, y2, y4, x7, y1,
y5, x4, x8, x9, . . .兲 back to 共x1, y1, x2, y2, x3, y3,
x4, y4, x5, y5, . . .兲. Then, as usual, 共xi+ yi兲=dimod 6 are uti-lized to generate 共d1, d2, d3, . . .兲. Because the superhacker does not know how the designer distributes the 兵xi其 and 兵yi其, two things can be ensured even if both shares 1 and 2 are intercepted. First, the superhacker cannot extract the confidential text d. Second, the superhacker cannot use the information grabbed from the two intercepted shares 共shares 1 and 2兲 to produce two faked shares 共shares 1
⬘
and 2⬘
兲, which can also pass our authentication test. 关In addi-tion, stacking them yields a faked image similar to the “GIRL” image in Fig. 6共g兲.兴 Of course, if the pixels of the input image “Lena” are also permuted using a random-position generator, then the image “LENA” itself is also free from being viewed by the superhacker intercepting both shares 1 and 2. But this operation共permuting pixels of “Lena”兲 is seldom used unless the image “Lena” itself is also highly confidential, because this operation reduces the advantages of visual cryptography: the image “LENA” can no longer be viewed by simply stacking the shares.共Recall that VC is a tool that balances between security and the benefit of “without-a-computer visual decoding” of images. If we only permute the locations of兵xi, yi其, the disclosure of the image “LENA” still does not require a computer; al-though the disclosure of the confidential text d will then require a computer.兲Then, each team can decode its own moderately confiden-tial image immediately共and begin the discussion about the visually decoded image within the team immediately兲. However, each team leader must wait 共until it is his turn兲 for the computer server to help him in decoding the confi-dential text.
In the following, as suggested by one of the reviewers, we discuss how to modify the design so that the two new transparencies 共generated by the modified design兲 can de-code both “Lena” and the confidential text without using a decoding computer.
Apparently, if the confidential text can also be visually decoded, then there is no need to use a decoding computer. In this kind of approach, we have two images to deal with: the “Lena” image and its background description image. We plan to create two transparencies T1 and T2having the following two properties: 共1兲 stacking T1 with T2 yields “Lena” and共2兲 after the stacking mentioned in 1, if we fix
T1, and then shift T2 by u units horizontally and v units vertically, then the stacking result becomes the confidential text image. Notably, the higher ranking officer keeps the values of u andv. If the higher ranking officer believes that
it is not easy to shift a transparency by u units horizontally and v units vertically in the decoding phase, then he can
use an auxiliary “nonnoisy” transparency T3共prepared ear-lier兲 in which only the boundaries of the transparencies T1 and T2 are sketched.
An encoding algorithm for this modified approach is given next, followed by a descriptive example and an ex-periment. In the algorithm, the image C is either a confi-dential text image describing “Lena” or a logo image for authentication purpose. The coordinate 共i, j兲 indicates a pixel at location共i, j兲 of the image “Lena” 共or the image C兲, or equivalently, the 2⫻2 block at location 共i, j兲 of transpar-ency T1 共or transparency T2兲.
6.1 The Modified Encoding Algorithm (No Decoding Computer Required)
The input to the algorithm are the two natural numbers u andv共u is the horizontal shift amount, and v is the vertical
shift amount兲, a 256⫻256 image “Lena,” and a confiden-tial text image C.
The output is two noise-like transparencies T1 and T2 共both 512⫻512兲, which are useful in visually decoding both “Lena” and C.
The steps are
Step 1. For each coordinate 共i, j兲 in the “easier-to-construct” areas, namely, 兵共i, j兲:0艋i⬍u,0 艋 j艋255其 and 兵共i, j兲:u艋i艋255,0艋 j⬍v其, do the following:
1a. Randomly assign one of the six fundamental block types to T1共i, j兲.
1b. If “Lena”共i, j兲 is a white pixel, then copy the block type of T1共i, j兲 to T2共i, j兲; otherwise, copy the complement of the block type of
T1共i, j兲 to T2共i, j兲.
Step 2. 共For the remaining area where construction is more complicated兲:
2a. Initially, let the value of the counter k be 0. 2b. For each共i, j兲 satisfying u⫻k艋i⬍u⫻共k+1兲
and 0艋 j⬍255−v, do the following:
2b.1. If C共i, j兲 is a white pixel, then copy the block type of T2共i, j兲 to T1共i+u, j+v兲; otherwise, copy the complement of the block type of
T2共i, j兲 to T1共i+u, j+v兲.
2b.2. If “Lena” 共i+u, j+v兲 is a white pixel, then copy the block type of T1共i+u, j+v兲 to T2共i + u , j +v兲; otherwise, copy the complement of
the block type of T1共i+u, j+v兲 to T2共i+u, j +v兲.
2c.
2c.1. Add 1 to the value of k.
2c.2. If k⬍共256/u兲−1, then go to step 2b; other-wise stop the algorithm. This ends the algorithm.
To help the readers understand step 2, we provide an example next. In the example, assume that T1# T2means stacking T1with T2, i.e., “#” is the stacking operator. 6.2 Example
To make the explanation easier, we assume that the hori-zontal shift amount is u = 3, and the vertical shift amount is
v = 0. Without the loss of generality, assume that the pixel
values of the input image “Lena” and confidential text im-age C are
“Lena ” : BWBWBBBWWWBW. . .
Image C: WWBWBWBWB. . .
Since the horizontal shift amount is u = 3, we split “Lena” and C into sectors of 3 pixels each. Therefore, we have “Lena ” : BWB WBB BWW WBW. . .
Image C: WWB WBW BWB . . . .
Iteration 0a. Initially, because u = 3, we randomly assign
the block types to the three blocks of the first sector of T1; for example, assign共0,4,3兲. So we have,
Transparency T1: 0 4 3 ? ? ? ? ? ? ? ? ? . . . , Transparency T2: ? ? ? ? ? ? ? ? ? ? ? ? . . . .
Iteration 0b. For the first three blocks of the stacking
result共T1# T2兲 to be 共B,W,B兲, which are the three blocks in the first sector of “Lena,” the first sector of T2should be
共0˜,4,3˜兲 due to the fact that the first sector of T1is共0,4,3兲. Here, 0˜ is the complement of the block type 0, i.e., 0˜=1. 共See Fig. 3 for explanation; note that stacking block type 0 with block type 1 yields a black block.兲 Similarly, 1˜=0, 2˜ = 3, 3˜ =2, 4˜=5, and 5˜=4. Thus, we now know the first sector of T1, and the first sector of T2, i.e.,
T1: 0 4 3 ? ? ? ? ? ? ? ? ? . . . ,
↓↓↓
T2: 1 4 2 ? ? ? ? ? ? ? ? ? . . . .
Iteration 1a. For the three blocks of the first sector in共T1
# T2
shifted兲 to be 共W,W,B兲, which are the first three blocks of the confidential text image C, the three blocks of the second sector of T1should be of types共1,4,3兲. Again, block type 3 is the complement of block type 2 implies 共1,4,3兲 # 共1,4,2兲=共W,W,B兲. Therefore, we have
Transparency T1: 0 4 3 1 4 3 ? ? ? ? ? ? . . . ,
Transparency T2: 1 4 2 ? ? ? ? ? ? ? ? ? . . . .
Iteration 1b. Since the second sector of “Lena” is
共W,B,B兲, therefore we let the second sector of T2 be 共1,4˜,3˜兲=共1,5,2兲. So we have
Transparency T1: 0 4 3 1 4 3 ? ? ? ? ? ? . . . ,
↓↓↓
Transparency T2: 1 4 2 1 5 2 ? ? ? ? ? ? . . . .
Iteration 2a. Read the next sector, i.e.,共W,B,W兲, of the
confidential text image C. Then, for 共T1 # T2shifted兲 to be 共W,B,W兲 at the current location, the seventh, eighth, and ninth blocks of T1 should be 共1,4,2兲 since 共1,5˜,2兲 =共1,4,2兲. Now we have
Transparency T1: 0 4 3 1 4 3 1 4 2 ? ? ? . . . ,
Transparency T2: 1 4 2 1 5 2 ? ? ? ? ? ? . . . .
Iteration 2b. Since the third sector of the input image
“Lena” is 共B,W,W兲, let the third sector of T2 be 共1˜,4,2兲 =共0,4,2兲. Thus, we have
Transparency T1: 0 4 3 1 4 3 1 4 2 ? ? ? . . . ,
↓↓↓
Transparency T2: 1 4 2 1 5 2 0 4 2 ? ? ? . . . .
Iteration 3a. Read next sector, i.e.,共B,W,B兲, of the
con-fidential text image C. Then, for 共T1 # T2
shifted兲 to be 共B,W,B兲 at the current location, the 10th, 11th, and 12th blocks of T1 should be 共0˜,4,2˜兲=共1,4,3兲. Therefore, we have
Transparency T1: 0 4 3 1 4 3 1 4 2 1 4 3 . . . ,
Transparency T2: 1 4 2 1 5 2 0 4 2 ? ? ? . . . .
Iteration 3b. Since the fourth sector of the input image
“Lena” is 共W,B,W兲, the fourth sector of T2 should be 共1,4˜,3兲=共1,5,3兲. Thus, we have
This ends the example.
An experiment using the modified encoding algorithm was performed and the results are shown in Fig. 7. The decoding process does not require the use of a computer; only the stacking operation is used to get Fig. 7共c兲 and 7共d兲. 7 Conclusions, Suggestions, and Future
Work
We proposed a two-in-one method that not only visually shares an input image of moderate confidentiality, but also hides an extra text file that is more confidential. The extra hiding ability is achieved without giving any artificial ap-pearance in the unveiled image. The unveiled “LENA” is lossless, in the sense that we can sequentially map from the 2⫻2 blocks of “LENA” back to the “exact” pixel values of “Lena.” The method can also be applied to the authentica-tion of the transparencies. We also discussed some other versions with improved security or easier decoding.
Notably, in the design introduced in Sec. 2, if the more confidential text is too long共for example, if “Lena’s” per-sonal information text requires more than 256⫻256/2 dig-its to express it, but the input image “Lena” is only 256 ⫻256兲, then we can enlarge the block size of the transpar-encies to solve the problem.关When we use 2⫻2 blocks 共as we did so far兲, there are only six types of blocks. By using 3⫻3 blocks, there are many more block types that can be used.兴
In future work, we will consider the following two top-ics. First, there is no computer available for decoding, we can still use visual decryption method to decrypt the confi-dential text. However, due to the natural limitations of vi-sual resolution, the hidden text is much shorter 共as com-pared with the data amount carried by the transparencies when a computer is available for decoding兲. Therefore, it will be interesting to discover a solution to increase the length of the text being carried, assuming that no computer can be used for decoding. Second, so far, our method deals with binary images. If we apply the method to the 8-bit planes of gray-value images, there might be some new ap-plications.
Acknowledgments
The authors would like to thank the two reviewers whose comments have greatly improved this paper.
References
1. M. Naor and A. Shamir, “Visual cryptography,” in Cryptology-Eurocrypt’94, Lecture Notes in Computer Science, Vol. 950, pp. 1–12 Springer-Verlag, Berlin共1995兲.
2. M. Naor and A. Shamir, “Improving the contrast via the cover base,” in Visual Cryptograpy II; Proc. Int Workshop on Security Protocols, Amale, Italy, April 1996, M. Lomas Ed., pp. 197–202共1996兲. 3. C. Blundo, A. De Santis, and D. R. Stinson, “On the contrast in visual
cryptography schemes,” J. Cryptology 12, 261–289共1996兲. 4. G. Ateniese, C. Blundo, A. De Santis, and D. R. Stinson, “Visual
cryptography for general access structures,” Inf. Comput. 129, 86– 106共1996兲.
5. M. Naor and B. Pinkas, “Visual authentication and identification,” in Advances in Cryptology—CRYPTO’97, B. Kaliski, Jr., Ed., Lecture Fig. 7 Double-decoding without using a decoding-computer: 共a兲
and共b兲 transparencies T1and T2;共c兲 the result of stacking 共a兲 and
Notes in Computer Science, Vol. 1294, pp. 322–336, Springer, Berlin 共1997兲.
6. D. R. Stinson, “An introduction to visual cryptography,” presented at Public Key Solutions ’97, Toronto, Canada 共Apr. 1997兲; http:// bibd.unl.edu/stinson/VKS-PKS.ps
7. A. Shamir, “Visual cryptanalysis,” Lect. Notes Comput. Sci. 1403, 201–210共1998兲.
8. C. Blundo and A. de Santis, “Visual cryptography schemes with per-fect reconstruction of black pixels,” Comput. Graph. 22, 449–455 共1998兲.
9. T. Hofmeister, M. Kruse, and H. U. Simon, “Contrast-optimal k out of n secret sharing scheme in visual cryptography,” Theor. Comput. Sci. 240, 471–485共2000兲.
10. G. Ateniese, C. Blundo, A. De Santis, and D. R. Stinson, “Extended schemes for visual cryptography,” Theor. Comput. Sci. 250, 137–161 共2001兲.
11. C. C. Chang and J. C. Chuang, “An image intellectual property pro-tection scheme for gray-level images using visual secret sharing strat-egy,” Pattern Recogn. Lett. 23, 931–941共2002兲.
12. C. Blundo, A. D. Santis, and M. Naor, “Visual cryptography for gray level images,” Inf. Process. Lett. 75, 255–259共2002兲.
13. Y. C. Hou, “Visual cryptography for color images,” Pattern Recogn. 36, 1619–1629共2003兲.
14. C. C. Lin and W. H. Tsai, “Visual cryptography for gray-level images by dithering techniques,” Pattern Recogn. Lett. 24, 349–358共2003兲. 15. R. Lukac and K. N. Plataniotis, “Bi-level based secret sharing for
image encryption,” Pattern Recogn. 38, 767–772共2005兲.
16. R. J. Anderson, “Information hiding: 1st international workshop,” Lect. Notes Comput. Sci. 1174, 317–333共1996兲.
17. L. M. Marvel and C. T. Retter, “Hiding information in images,” in Proc. Int. Conf. on Image Processing Vol. 2, pp. 396–398, Boston 共1998兲.
18. L. M. Marvel, C. G. Boncelet, and C. T. Retter, “Spread spectrum image steganography,” IEEE Trans. Image Process. 8共8兲, 1075–1083 共1999兲.
19. T. Jamil, “Steganography: the art of hiding information in plain sight,” IEEE Potentials 18共1兲, 10–12 共1999兲.
20. H. H. Yu and P. Yin, “Multimedia data recovery using information hiding,” in Proc. IEEE GLOBECOM’00, Vol. 3, pp. 1344–1348 共2000兲.
21. Y. C. Tseng, Y. Y. Chen, and H. K. Pan, “A secure data hiding scheme for binary images,” IEEE Trans. Commun. 50共8兲, 1227–1231 共2002兲.
22. C. C. Thien and J. C. Lin, “A simple and high-hiding capacity method for hiding digit-by-digit data in images based on modulus function,” Pattern Recogn. 36共12兲, 2875–2881 共2003兲.
23. M. Wu and B. Liu, “Data hiding in image and video. I. Fundamental issues and solutions,” IEEE Trans. Image Process. 12, 685–695 共2003兲.
24. C. K. Chan and L. M. Cheng, “Hiding data in images by simple LSB substitution,” Pattern Recogn. 37共3兲, 469–474 共2004兲.
25. S. S. Maniccam and N. Bourbakis, “Lossless compression and infor-mation hiding in images,” Pattern Recogn. 37共3兲, 475–486 共2004兲.
Wen-Pinn Fang received his BS degree in mechanical engineering in 1994 from Na-tional Sun-Yet-Sen University and his MS degree in mechanical engineering in 1998 from National Chiao Tung University, where he is currently a PhD candidate in the Computer and Information Science De-partment. His recent research interests in-clude pattern recognition and image pro-cessing.
Ja-Chen Lin received his BS degree in computer science in 1977 and his MS de-gree in applied mathematics in 1979, both from National Chiao Tung University, Tai-wan, and his PhD degree in mathematics from Purdue University, U.S.A., in 1988. In 1981 to 1982 he was an instructor with the National Chiao Tung University and from 1984 to 1988 he was a graduate instructor with Purdue University. In August 1988 he joined the Department of Computer and In-formation Science at National Chiao Tung University, where he is currently a professor. His recent research interests include pattern recognition and image processing. Dr. Lin is a member of the Phi-Tau-Phi Scholastic Honor Society.
