Improvements on Audio and Optical Cryptography - Semantic Scholar

JOURNAL OF INFORMATION SCIENCE AND ENGINEERING 18, 381-391 (2002)

Short Paper_________________________________________________ Improvements on Audio and Optical Cryptography CHING-NUNG YANG Department of Computer Science and Information Engineering National Dong Hwa University Hualien, 974 Taiwan E-mail: [email protected]

Audio and optical cryptography schemes in [1] are secret sharing schemes that protect audio and visual secrets by using sounds and images, respectively. However, 2-out-of-n schemes in [1] are constructed by log2 n independent 2-out-of-2 schemes, meaning so that a 2-out-of-n scheme needs log2 n different cover sounds or images. More cover sounds (images) will be more difficult for the human hearing(vision) system to hear(see) the secret. In this paper new constructions for the 2-out-of-n audio and optical secret sharing schemes are proposed. Our schemes only need one cover sound or image. Keywords: visual cryptography, audio cryptography, secret sharing scheme, optical cryptography, optical interferometer

1. INTRODUCTION In 1998 Desmedt proposed an audio cryptography scheme. It is a secret sharing scheme that high (low) volume represents the secret bit “1”(“0”) embedded into an audio signal, and the decoder for this audio cryptography scheme is the “ears” of listener. Due to constructive and destructive interference of sound waves, one can hear the volume changing. The authors also generalized the 2-out-of-2 to a 2-out-of-n scheme. However, the 2-out-of-n scheme needs log2 n different pieces of sounds as the cover sounds, where the word “cover” represents the sound used to embed or hide the shared secret message. More cover sounds will cause the human hearing system to overload, and one may not interpret the volume changing correctly. In [1] an optical cryptography scheme is also proposed. This 2-out-of-2 optical cryptography scheme can be considered as a Visual Secret Sharing (VSS) scheme [2-7] and does not need the aid of a computer to decode the shared secret. The approach of Desmedt’s optical cryptography scheme is completely different from Shamir’s VSS scheme. The former uses the destructive interference property of optical waves to decode the shared secret image by using a Mach-Zehnder optical interferometer, and the shadow images are high quality images rather than the conventional VSS schemes. Similar to the

Received June 14, 2000; revised February 5, 2001; accepted October 8, 2001. Communicated by Wen-Hsiang Tsai.

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audio cryptography scheme, the 2-out-of-n optical cryptography scheme can be constructed using log2 n different 2-out-of-2 schemes with log2 n different cover images. Here, new constructions for the 2-out-of-n audio and optical secret sharing schemes are proposed, and our proposed schemes need only one cover sound or image.

2. THE BASIC AUDIO AND OPTICAL CRYPTOGRAPHY SCHEMES 2.1 Desmedt’s Audio Cryptography Scheme In this section parameters S : a plaintext message, L : the length of the embedded message, T : the length of secret bit(seconds) are defined in section 3.2 of [1]. Then we use the following figure, a harmonic sound, to briefly describe the key concept of a 2-out-of-2 audio cryptography scheme. To encrypt the plaintext message S, when deciding the high(bit 1) or low(bit 0) volume, we randomly select a phase (0 or π) in the first share and select the opposite phase in the second share as shown in Fig. 1(a). The phase diagram is illustrated in Fig. 1(b). We can see that S = (0 0 1 1), L = 4, T = Tb, and the length of the cover sound = 4 × Tb. Each share will have phases “0” and “π”, half and half, in every time slot Tb, and fixed amplitude A. Thus we cannot hear the volume change for each share. When share 1 “adds” to share 2, where “adds” means playing different sounds from two speakers, we will get the amplitude 2A(resp. 0) in time slot where the plaintext message is 1(resp. 0). Finally, we can hear the volume change and get the secret S. Note that one cannot hear the complete silence, due to the incomplete destruction of the signal. Tb

Volume Shares S1 S2

H 0 0

π π

L 1

0 π π 0

0 A

π

π

π

0

π

Embedded secret Share 1

(a) The choice of phase for each share H (L): High (Low) volume. 2A

1

0

π

0 Share 2

2A Share 1 + Share 2

A (b) Phase diagram.

Cover harmonic sound

(c) A 2-out-of-2 scheme with simple harmonic sound.

Fig. 1. A 2-out-of-2 audio cryptography scheme illustration.

2.2 Desmedt’s Optical Cryptography Scheme Desmedt’s 2-out-of-2 optical cryptography scheme uses destructive interference of optical waves to decode the black and white secret image by using the Mach-Zehnder optical interferometer as shown in Fig. 2(a). We use a cover image with only four pixels,

2-OUT-OF-n AUDIO AND OPTICAL SECRET SHARING SCHEMES

383

where li is the gray level for the i-th pixel and the binary form of li is (bi8, bi7, bi6, bi5, bi4, bi3, bi2, bi1), i = 1, 2, 3, 4. Based on the resolution, we may choose which j-th bit, j = 1, 2, …, 8, is randomly flipped. For example we use j = 4, l1 = 100, l2 = 20, l3 = 210, l4 = 72, secret S = (1, 0, 0, 1), and the random flipping sequence for share 1 is (0, 1, 1, 0). Thus the flipping sequence for share 2 is (0Ƨ1, 1Ƨ0, 1Ƨ0, 0Ƨ1) = (1, 1, 1, 1) , then the gray level of four pixels in share 1 is 100, 28, 218, 72, and the gray level of four pixels in share 2 is 108, 28, 218, 80. Share 1: 100(01100100) → 100 20 (00010100) → 28 210(11010010) → 218 72 (01000100) → 72

(01100100), (00011100), (11011010), (01000100).

Share 2 : 100 (01100100) → 108 20 (00010100) → 28 210 (11010010) → 218 72 (01000100) → 80

(01101100), (00011100), (11011010), (01001100).

If the light beams meet out of phase, they will interfere destructively. So, when the two light beams pass through share 1 and share 2 and have 180o difference in phase, we will get the four pixels in the reconstructed image (8, 0, 8, 0). Referring to the figures in Fig. 2(b), we see the original image, cover image, share 1, share 2, and the reconstructed image.

Original image

Share 1 180o shift

From laser light source

l1

l2

l3

l4

Cover image

100

28

108

28

8

0

218

72

218

80

0

8

To Observer

Share 2

Share 1 (a)

Share 2 (b)

Reconstructed image

Fig. 2. A 2-out-of-2 optical cryptography scheme.

In [1], the authors had generalized previous 2-out-of-2 audio and optical cryptography schemes to 2-out-of-n schemes using log2 n different cover sounds or images. If S is a plaintext message, one has S = r0i Ƨr1i , where 1 ≤ i ≤ log2 n and Ƨ is the exclusive or. When numbering the participants from 0 to n-1, participant j receives shares r1i, if the ith bit of the binary representation of the integer j is 1, else receives shares r0i, where r0i and r1i are the two shraes of 2-out-of-2 scheme for the i-th cover sound or music.

3. CONSTRUCTION OF 2-OUT-OF-n AUDIO CRYPTOGRAPHY SCHEME In this section, two different methods for constructing a 2-out-of-n audio cryptography scheme with only one cover sound is developed. One (Construction 1) is wave-based and the other (Construction 2) is digitized sampling-based.

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In the first construction method is for a different time slot T we change the original sound using multiple phases instead of only two phases (0, π) as in the 2-out-of-2 scheme. For example, a 2-out-of-3 scheme uses three phases: 0, 2π/3, and 4π/3. When decoding, if they are in-phase they will cause high volume, they otherwise, will cause low volume. In fact, for this scheme the low volume amplitude is half of the high volume. Construction 1: Let the two 2 × 1 matrices in BH and BL be the representation for Fig. 1(a), and be called as the high matrix, and low matrix for 2-out-of-2 scheme, respectively. 0 π 

 0  π 

BH = {   ,   }, and BL = {  ,   }. π   0  0 π  Then, a 2-out-of-n scheme with one cover sound, where n ≥ 3, has the n × 1 high matrices CH and low matrices CL. θ 1  θ 2   θn  θ n  θ 1  θ 2  θ  θ  θ  θ  θ  θ   2  3  1   n  1  2  CH = { θ 1 , θ 2 , ..., θ n  }, and CL = { θ 3 , θ 4 , ...,  θ 2  },             M M  M  M M M  θ n  θ 1  θ n −1  θ n  θ 1  θ 2 

where θi = 2π/n × (i − 1), 1 ƙ i ƙ n.

Theorem 1: The scheme from Construction 1 is a 2-out-of-n audio cryptography scheme with one cover sound. When decoding any two shares (participants i1 and i2), the amplitudes of high and low volume are 2A and 2A × cos((i2 − i1) × π/n), respectively, if the amplitude of the cover sound is A. The minimum difference between high and low volume when adding any two shares is 2A(1−cosπ/n), and the maximum difference is 2A(1−cos(n/2 × π/n) ). Each share has no information about the secret, i.e., the scheme holds the perfect privacy. Proof: From any two rows in CH, it is observed that for a certain time slot if one chooses θj in share 1, then he has θj in share 2. So, in this time slot, the amplitude of volume will be Acosθj + Acosθj = 2Acosθj, and the amplitude is 2A. Any two rows in CL, for example row i1 and row i2, and i2 Ɨ i1, for a certain time slot, if one chooses θj in share 1, then he has θj + (i2 − i1) × 2π/n in share 2. So, in this time slot, the amplitude of volume will be Acosθj + Acos(θj + (i2 − i1) × 2π/n). If the angle between two vectors is small, then the amplitude of the sum of these two vectors is high; this situation will occur when i2 − i1=1. On the contrary, the phase difference between two angles will be at most n/2 × 2π/n. It is obvious that the difference between the high and low volumes in Theorem 1 is correct. For a proof of the condition “security” of this secret sharing scheme, it is obvious that the amplitude of all shares are “A”, and each share chosen from CH and CL has the phase θi, i∈{1, 2, …, n} with the same probability. Thus the “security” of 2-out-of-n scheme holds.


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From Theorem 1, it is easily seen that when n = 2, CH and CL can be reduced to BH and BL, and the amplitude of high and low volume is 2A and 2A × cos((i2 − i1) × π/n) = 2A × cos(π/2) = 0. The results are in agreement with Fig. 1(c). Example 1. Considering the harmonic sound case in Fig. 1, design a 2-out-of-3 scheme using Construction 1. When the amplitude of cover sound is A, the high volume is 2A and the low volume is A. The 3 × 3 high and low matrices CH, CL for 2-out-of-3 scheme are constructed as

 0  2π 3  4π 3  0 2π 3  4π 3      2π 4π   , }, and CL = { 2π 3 , 4π 3 ,  0  }. CH = { 0 , 4π     0 2π 3  4π 3   3   0  2π 3     3   3  We show the result of 2-out-of-3 scheme in the following figure. Our proposed scheme needs only one cover sound. Desmedt’ scheme needs log2 3 = 2 cover sounds. In fact, we pay price of a degradation of the difference of high and low volume, shown in Fig. 3, to get the benefit “one cover sound”. Share 1 + Share 2 Cover harmonic sound Embedded secret

2A A

A 1

1

0

0

0

2π/3

0

4π/

2π/3

4π/3

0

4π/

4π/

0

0

4π/

Share 1 + Share 3

Share 1

Share 2

Share 2 + Share 3

Share 3

Fig. 3. A 2-out-of-3 audio cryptography scheme with simple harmonic sound.

The second method is analogous to the optical cryptography scheme using only destructive interference. We change two, three, … bits in a sound sample, for example using PCM coding 8 bits per sample, to construct (2, 4), (2, 8), … schemes, respectively. Construction 2: Let the sample in the cover sound be digitized to m bits, and its binary form be (bm, bm-1, …, b1). We will randomly flip any s bits i1, i2, …, is (from LSB to MSB) of each sound sample. Then, a 2-out-of-n scheme with one cover sound, where n = 2s, has high matrices CH and low matrices CL, shown as follows. One constructs n participants according the matrices in CH and CL which will cause the high and low volume,

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respectively. The minimum difference between high and low volume when “adding” any two shares is 2 i1 −1 and the maximum difference is 2 i1 −1 + 2 i2 −1 + … + 2 is −1 . Note that the decoding method will need the two shares (sounds) meeting 180o difference in phase. 0 

0 

1

CH = {all the matrices obtained by permuting the rows of   }, and CL = {   ,   } 0 1 1  for s = 1, i.e., the 2-out-of-2 scheme. CH = {all the matrices obtained by permuting the 0 0 rows of 1  1

0 0 0  }, and C = { L 0   1  0 0

1

0

0 , 0  0

0 0 0  0

1

1 0 1 1 0   , 1 0 , 1 

1

  1 0

1 1 1  1

1 1 } 1  1

for s = 2, i.e., the 2-out-of-4

scheme. CH and CL for s ≥ 3 can be produced using the same procedure as 2-out-of-2 and 2-out-of-4. The correctness of Construction 2 is evident. We now give an example to clarify the idea of this construction and also show the correctness. Example 2. Consider a 2-out-of-4 scheme using Construction 2. The volume sample has 256 quantization levels, or 8 bits per symbol. For example, the four samples in the cover sound are 60 (0 0 1 1 1 1 0 0), 24 (0 0 0 1 1 0 0 0), 112 (0 1 1 1 0 0 0 0), and 200 (1 1 0 0 1 0 0 0). Assume the embedded message is (1 0 1 0), and choose the 3-th and 4-th LSB bits to randomly flip. Use CH and CL for s = 2 to construct a 2-out-of-4 scheme. Denote any two bits in one sample from share 1 to share 4 and plaintext as (s11, s12), (s21, s22), (s31, s32), (s41, s42), and (s1, s2), respectively. Then they are related by  s1 = si1 ⊕ sj1, and s2 = si2 ⊕ sj2, where i, j∈{1, 2, 3, 4}, and i ≠ j.   If the volume is low, then (s1, s2) = (0, 0).  If the volume is high, then (s1, s2) = (1, 0), or (0, 1), or (1, 1). 

(1-1) (1-2) (1-3)

The first embedded bit is “1”(high volume). One randomly flips a coin to choose the flipping sequence (s11, s12) for share 1. If (s11, s12) is (0, 1), then (s21, s22) could be (1, 0), (1, 1), or (0, 0) according (1−1) and (1−3). Similarly, if one chooses (s21, s22) = (1, 0), then (s31, s32) could be (1, 1), or (0, 0). If one chooses (s31, s32) = (1, 1), then (s41, s42) is (0, 0). So, the first sample of share 1 to share 4 will be 60 (0 0 1 1 1 1 0 0) → 52 (0 0 1 1 60 (0 0 1 1 1 1 0 0) → 56 (0 0 1 1 60 (0 0 1 1 1 1 0 0) → 60 (0 0 1 1 60 (0 0 1 1 1 1 0 0) → 48 (0 0 1 1

01 10 11 00

0 0), 0 0), 0 0), 0 0).

Similarly, the four samples of share 1 to share 4 are (52, 28, 112, 204), (56, 28, 116, 204), (60, 28, 120, 204), and (48, 28, 124, 204). When decoding, use the destructive interference, i.e., let the two shares meet out of phase. Then combine any two shares with 180o difference in phase, giving share 1 + share 2 = (4, 0, 4, 0), share 1 + share 3 = (8, 0, 8, 0), share 1 + share 4 = (4, 0, 12, 0), share 2 + share 3 = (4, 0, 4, 0), share 2 + share 4 = (8, 0, 8, 0), and share 3 + share 4 = (12, 0, 4, 0). In this situation, the human hearing system


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cares only about is “contrast”, not whether the high volume is a little higher or much higher than the low volume. For example, from the difference of share 1 and share 4, (4, 0, 12, 0), we still get the embedded message (1 0 1 0). Desmedt’s 2-out-of-4 scheme uses a wave-based approach, and needs two 2-out-of-4 schemes and two cover sounds. Let r01 and r11 be two shares of a 2-out-of-2 Desmedt’s scheme for the first cover sound, and. r02 and r12 be two shares for the second cover sound. Then four participants of Desmedt’s 2-out-of-4 scheme hold (r01, r02), (r01, r12), (r11, r02), and (r11, r12) as their shares. If the number n is not of the type 2s, it is obvious that we can shorten the 2-out-of-n1 scheme to get a 2-out-of-n2 scheme by removing the n1-n2 shares, where n1>n2.

4. CONSTRUCTION FOR 2-OUT-OF-N OPTICAL CRYPTOGRAPHY SCHEME Here, we propose a method for constructing a 2-out-of-n optical cryptography scheme with only one cover image. The crux of the new scheme is that the human visual system can get the secret, if the level of the secret image is different from the level of background, no matter whether the image is higher or much higher than the background. We herein change two, three, … bits instead of one bit in a pixel of each image(share) to construct a 2-out-of-n scheme. Construction 3: Let the sample in the cover image be digitized into the m bits, and its binary form be (bm, bm-1, …, b1). We will randomly flip any s bits i1, i2, …, is of each image sample. Therefore, a 2-out-of-n scheme with one cover image, where n = 2s, has the black matrices CB = CH and white matrices CW = CL, where CH and CL are defined in Construction 2, for different s. One constructs n participants according the CB and CW matrices which will cause the black and white pixels, respectively. Note that the decoding method uses the Mach-Zehnder optical interferometer described in [1]. In fact, Construction 2 and Construction 3 use the same approach. Thus a scheme using Construction 3 is indeed a 2-out-of-n scheme with only one cover image. The correctness is easy to demonstrate and shown in Example 3. Example 3. For a 2-out-of-4 optical cryptography scheme, the cover image is Lenna image of size 512 × 512 pixels. The secret image text “LENNA” is black & white. (Note that Desmedt’s scheme is only used to share the black & white (i.e., binary) image.) Figs. 4(a) to 4(d) are the four shares, when choosing i1 = 4 and i2 = 5. The level from the matrices in CB may be 0 × 24 + 0 × 23 = 0, 0 × 24 + 1 × 23 = 8, 1 × 2 + 0 × 23 = 16, and 1 × 24 + 1 × 23 = 24. So, the difference of level between any two shares will be 8(8−0, 16−8, 24−16), 16(16−0, 24−8), and 24(24−0). The minimum(maximum) contrast of the recovered image is 8(24), and the probability of “8”, “16”, “24” is 1/2, 1/3, 1/6, respectively. Thus the difference in contrast will not cause difficulty in seeing the recovered secret since the probability of “8” is highest. For example, the number of different pixels between share 3 and share 4 in Figs. 4(c) and 4(d) is 8612, 5816, 2936 for “8”, “16”, “24”, and the probability is 0.496, 0.335, 4

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(a) Share 1

(b) Share 2

(c) Share 3

LENNA (d) Share 4

(e) Original secret image

(f) Share 3 “+” Share 4

Fig. 4. A 2-out-of-4 scheme, using Lenna image as the cover image, for i1 = 4 and i2 = 5.

0.169, respectively. The original secret image is shown in Fig. 4(e) and the recovered image of share 3 “+” share 4 using the destructive property to decode, is shown in Fig. 4(f). Although the contrast of the recovered image is not very good, we did improve Desmedt’s 2-out-of-n scheme with only cover image. The noise in recovered image Fig. 4(f) is a compromise by our improved “one cover image”. How does one choose the s bits (i1, i2, …, is) from m-tuple (bm, bm-1, …, b1) of each pixel so that the different contrast will not result in difficulty for the human visual system to see the recovered secret? We use the following theorem to formally describe the optimum choice for m = 8, and s = 2 in Construction 3. Theorem 2. For a 2-out-of-4 scheme, using Construction 3, the optimum choices for i1 and i2, where i2 > i1 ≥ 1, from 8-tuple of each pixel in the covered image is that i2 = i1 + 1. The dominant contrast level of the recovered image is 2 i1 −1 , and the probability is 0.5. Proof: There are four possible random flippings for 2-tuple (i1, i2), i.e., (0, 0), (0, 1), (1, 0), and (1, 1). There are six possible differences of level between any two shares since  4   = 6 with values 2 i1 −1 ,2 i1 −1 ,2 i2 −1 ,2 i2 −1 ,2 i2 −1 − 2 i1 −1 ,2 i2 −1 + 2 i1 −1. Since (2 i2 −1 + 2 i1 −1 ) >  2 2 i2 −1 > (2 i2 −1 + 2 i1 −1 ) ≥ 2 i1 −1 , if we let 2 i2 −1 − 2 i1 −1 = 2 i1 −1 , then we will have the difference level “ 2 i1 −1 ” with probability 3/6 = 0.5. When 2 i2 −1 − 2 i1 −1 = 2 i1 −1 , we get i2 = i1 + 1.


389

 8  2

Example 4. For a 2-out-of-4 scheme, using Construction 3 and m = 8, there are   = 28 possible 2-tuple (i1, i2) choices from an 8-tuple. Seven optimum (i1, i2) choices are (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), and (7, 8). Table 1 shows the probability distribution of the different contrast levels. The entities in the table are the contrast level. We can see that although the level of the secret image is a little higher or much higher than the background level, we can easily get the embedded message because of the difference level with 50% probability shown in the first row of Table 1. Table 1. Probability distribution of different contrast levels for optimum (i1, i2). (i1, i2) Probability 1/2 1/3 1/6

(1, 2)

(2, 3)

(3, 4)

(4, 5)

(5, 6)

(6, 7)

(7, 8)

1 2 3

2 4 6

4 8 12

8 16 24

16 32 48

32 64 96

64 128 192

5. COMPARISON OF RESULTS Comparisons for 2-out-of-n audio and optical cryptography schemes are given in Table 2 and Table 3. A major benefit of the proposed schemes is that only one cover sound (image) is needed. In fact, we compromise the contrast of the sound and image to achieve this benefit. In Table 2 “cs” denotes the number of cover sound, and δ1(δ2) indicates the minimum(maximum) difference between high and low volume. For Construction 1, δ1 = 2A(1-cosπ/n) and δ2 = 2A(1−cos(n/2 × π/n)), where A is the amplitude of the original harmonic sound. For Construction 2, δ1 = 2 i1 −1 and δ2 = 2 i1 −1 + 2 i2 −1 + … + 2 is −1 , where we choose the third bit for s = 1, third and fourth bits for s = 2, third, fourth and fifth bits for s = 3 in this table. Table 2. 2-out-of-n audio cryptography schemes. Construction Method

n=2

n=3 cs

δ1 ( ) δ2

2A (2A)

2

1

2A (2A)

1

4 (4)

cs

δ1 ( ) δ2

Desmedt [1]

1

Construction 1

Construction 2

n=4 cs

δ1 ( ) δ2

2A (2A)

2

1

A (A)

1

4 (12)

n=5 cs

δ1 ( ) δ2

2A (2A)

3

2A (2A)

1

0.59 A (2A)

1

1

4 (12)

1

0.38A (1.38A)

4 (28)

n=6 cs

δ1 ( ) δ2

3

n=7

n=8

cs

δ1 ( ) δ2

cs

2A (2A)

3

2A (2A)

3

2A (2A)

1

0.27A (2A)

1

1

0.15 A (2A)

1

4 (28)

1

1

4 (28)

0.20A (1.55A)

4 (28)

(

δ1 δ2 )

In Table 3 “ci” denotes the number of cover images, and ∆1(∆2) indicates the minimum(maximum) contrast of the recovered image. For Desmedt’s 2-out-of-n scheme [1], ∆1 = ∆2 = 2 i1 −1 , where the i1 bit of each pixel in the cover image is randomly flipped.

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Table 3. 2-out-of-n optical cryptography schemes. n=2 Construction Method

i1=3 ci

(

∆1 ∆2)

i1=4

(

∆1 ∆2)

n=4 i1=3, i2=4

i1=5

(

∆1 ∆2)

ci

(

∆1 ∆2)

i1=4, i2=5 (

∆1 ∆2)

n=8 i1=3, i2=5 (

i1=2, i1=3, i1=4, i2=3, i3=4 i2=4, i3=5 i2=5, i3=6

∆1 ∆2)

ci

(

∆1 ∆2)

(

∆1 ∆2)

(

∆1 ∆2)

Desmedt [1]

1

4 (4)

8 (8)

16 (16)

2

*

*

*

2

*

*

*

Construction 3

1

4 (4)

8 (8)

16 (16)

1

4 (12)

8 (24)

4 (20)

1

2 (14)

4 (28)

8 (56)

6. CONCLUSIONS In this paper, we have proposed two methods (Construction 1, Construction 2) to construct the 2-out-of-n audio cryptography scheme with only one cover sound, and one method (Construction 3) to construct the 2-out-of-n optical cryptography scheme with only one cover image. In fact, the proposed Construction 1 and Construction 3 are generalized methods of Desmedt’s 2-out-of-2 schemes. It is easily seen that when n = 2, Construction 1 reduces to Desmedt’s 2-out-of-2 audio cryptography scheme and when s = 1, Construction 3 to Desmedt’s 2-out-of-2 optical cryptography scheme.

ACKNOWLEDGMENT This research was supported by the National Science Council, Republic of China, under contract NSC 89-2218-E-259-003.

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7. C. N. Yang and C. S. Laih, “New colored visual secret sharing schemes,” Designs, Codes and Cryptography, Vol. 20, 2000, pp. 325-336.

Ching-Nung Yang () was born on May 9, 1961 in Kaohsiung, Taiwan. He received the B.S. degree in 1983 and the M.S. degree in 1985, both from the Department of Telecommunication Engineering at National Chiao Tung University. He received Ph.D. degree in Electrical Engineering from National Cheng Kung University in 1997. During 1987-1989 and 1990-1999, he worked at Telecommunication Lab., and Tainning Institute Kaohsiung Center, Chunghwa Telecom Co., Ltd., respectively. He is presently an assistant professor in the Department of Computer Science and Information Engineering at National Dung Hwa University. His research interests include coding theory, information security and cryptography.