Reversible video watermarking through recursive histogram modification

Abstract

In this study, a new reversible video watermarking method is developed. Unlike the existing reversible video watermarking algorithms that are based on motion compensated prediction or interpolation error expansion, motion compensated interpolation errors in the proposed method are watermarked in a totally different manner with the aid of recursive histogram modification algorithm developed recently for the purpose of reversible image watermarking. However,the recursive histogram modification can not be used directly for reversible video watermarking since the important problems of ensuring reversibility for each frame and distribution of total capacity among frames are encountered. In this study, novel ideas are proposed to solve the aforementioned two problems. The proposed method is tested on the video sequences commonly used in the literature. It is shown to give better performance than the existing reversible video watermarking algorithms in terms of capacity and distortion by means of computer simulations. Also, a numerical example is provided.

Appendix A: Numerical example

1.1 A.1 Watermark insertion

In the following, watermarking of a single block is explained in an example in which the block is assumed to contain 16 elements that take on values from the set {1, 2, 3, 4}. ${e_{i}^{k}}$ and Q _Y|X and Q _X|Y are assumed to be given by

$$ {e_{i}^{k}} =\left[\begin{array}{llllllllllllllll} 1 & 2 & 4 & 3 & 4 & 2 & 4 & 3 & 3 & 4 & 4 & 1 & 4 & 3 & 4 & 4 \end{array}\right] $$

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$$ Q_{Y\vert{}X} = \left[\begin{array}{llll} 1/2 & 0 & 0 & 0\\ 1/2 & 1 & 1/4 & 0\\ 0 & 0 & 3/4 & 1/8\\ 0 & 0 & 0 & 7/8 \end{array}\right] $$

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$$ Q_{X\vert{}Y} = \left[\begin{array}{llll} 1 & 1/4 & 0 & 0 \\ 0 & 2/4 & 0 & 0 \\ 0 & 1/4 & 3/4 & 0 \\ 0 & 0 & 1/4 & 1 \end{array}\right] $$

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Furthermore, let the watermark bit sequence be given as

$$ M_{i} = [ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \ 0 \ 1 \ 0 \ 1 \ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \ {\dots} ] $$

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Since ${e_{i}^{k}}$ contains two 1 (that is, h ₁ = 2), M _i is subjected to arithmetic decompression algorithm until a DNS of length 2 is obtained. The first column of Q _Y|X is used as a parameter in the decompression process. This step produces the DNS $y_{i,1}^{k} = [2 \quad 1]$ from the first three bits (1 1 0) in M _i . Then, 1’s in ${e_{i}^{k}}$ are replaced by the DNS $y_{i,1}^{k}$. This process is repeated for 3 and 4 in ${e_{i}^{k}}$. The following 6 bits (0 1 0 1 1 0) in M _i is applied to decompression by using the third column of Q _Y|X to get the DNS $y_{i,3}^{k} = [3 \ 3 \ 3 \ 2]$. Similarly, applying the next 13 bits in M _i to the decompression process by using the fourth column of Q _Y|X results in the DNS $y_{i,4}^{k} = [4 \ 4 \ 4 \ 3 \ 4 \ 4 \ 4 \ 4]$. 3’s and 4’s in ${e_{i}^{k}}$ are replaced by the DNS $y_{i,3}^{k}$ and $y_{i,4}^{k}$, respectively. Since the entropy of the distribution in the second column of Q _Y|X is zero, 2’s in ${e_{i}^{k}}$ are kept as they are. Hence, the watermarked block is given as

$$ {y_{i}^{k}} = \left[ \begin{array}{cccccccccccccccc} 2 & 2 & 4 & 3 & 4 & 2 & 4 & 3 & 3 & 3 & 4 & 1 & 4 &2 & 4 & 4 \end{array}\right] $$

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The above process is illustrated pictorially in Fig. 10 for the 4 in ${e_{i}^{k}}$. Next, the SI required for watermark decoding and the original block restoration has to be generated. For this purpose, locations of 1,2,3 and 4 in ${y_{i}^{k}}$ are stored in the respective sequences given by I Y ₁ = [1 2], I Y ₂ = [1 2 6 14], I Y ₃ = [4 8 9 10] and I Y ₄ = [3 5 7 11 13 15 16]. Since the entropies computed from the first and last columns of Q _X|Y are zero, locations 1 and 4 do not need to be provided the watermark decoder (1’s and 4’s in the original and watermarked blocks are the same). Subjecting the sequences ${e_{i}^{k}}({IY}_{2}) = [1 \ 2 \ 2 \ 3]$ and ${e_{i}^{k}}({IY}_{3} ) = [3 \ 3 \ 3 \ 4]$ to arithmetic compression by using the corresponding columns in Q _X|Y results in the bit sequences given by $O\left ({e_{i}^{k}},2\right ) = [1 \ 1 \ 0 \ 1 \ 0 \ 0 \ 1]$ and $O\left ({e_{i}^{k}},3\right ) = [0 \ 0 \ 1 \ 0 \ 1]$. Concatenating $O\left ({e_{i}^{k}},2\right )$ and $O\left ({e_{i}^{k}},3\right )$ leads to the $O\left ({e_{i}^{k}}\right )$ given by

$$ O\left( {e_{i}^{k}}\right) = O\left( {e_{i}^{k}},2\right) \parallel{} O\left( {e_{i}^{k}},3\right) = \left[ \begin{array}{cccccccccccccccc} 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \end{array}\right] $$

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$O\left ({e_{i}^{k}}\right )$ is appended to the beginning of the remaining bits in M _i to get the watermark bit sequence for the next block given by

$$ M_{i+1} = \left[ \begin{array}{cccccccccccccccc} 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & \dots \end{array}\right] $$

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SI construction for 2 in ${e_{i}^{k}}$ is shown in Fig. 11.

Fig. 10

Watermarking of 4 in ${e_{i}^{k}}$

Fig. 11

SI construction for 2 in ${e_{i}^{k}}$

1.2 A.2 Watermark decoding and block restoration

First, the original block ${e_{i}^{k}}$ is restored from ${y_{i}^{k}}$ with the aid of Q _X|Y . From the first and last columns of Q _X|Y , we conclude that 1 and 4 in the watermarked block are equal to the corresponding elements in the original block. Since 2 is repeated four times in ${y_{i}^{k}}$ (that is, l ₂ = 4), the watermark bit sequence for the (i+1)-th block is decompressed until a DNS of length 4 is obtained by using the second column of Q _X|Y . This step results in the DNS [1 2 2 3] determined from the first 7 bits [1 1 0 1 0 0 1] in M _i+1. 2’s in ${y_{i}^{k}}$ are replaced by the [1 2 2 3]. The same step is repeated for 3 in ${y_{i}^{k}}$ to restore the original block ${e_{i}^{k}}$. Restoring the original value for 2 in ${y_{i}^{k}}$ is shown in Fig. 12.

Fig. 12

Original value restoration for 2 in ${y_{i}^{k}}$

Once the original block is restored, the watermark is decoded from ${e_{i}^{k}}$, ${y_{i}^{k}}$ and Q _Y|X . First locations of 1, 2, 3 and 4 in ${e_{i}^{k}}$ are stored in the respective sequences I X ₁ = [1 12], I X ₂ = [2 6], I X ₃ = [4 8 9 14] and I X ₄ = [3 5 7 10 11 13 15 16]. From the second column of Q _Y|X , we conclude that 2 in the original block is not watermarked. For the remaining elements, subjecting the sequences ${y_{i}^{k}}({IX}_{1}) = [2 \ 1]$, ${y_{i}^{k}}({IX}_{3} ) = [3 \ 3 \ 3 \ 2]$ and ${y_{i}^{k}}({IX}_{4} ) = [4 \ 4 \ 4 \ 3 \ 4 \ 4 \ 4 \ 4]$ to arithmetic compression under the corresponding columns in Q _X|Y results in the bit sequences given by [1 1 0], [0 1 0 1 1 0] and [1 0 1 1 1 0 0 1 0 1 0 0 1] that are concatenated to obtain the watermark for the i-th block given by

$$ M_{i}=[1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \ 0 \ 1 \ 0 \ 1 \ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \ ...] $$

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Watermark decoding for 4 in ${e_{i}^{k}}$ is shown in Fig. 13.

Fig. 13

Watermark decoding for 4 in ${e_{i}^{k}}$