A novel data hiding for color images based on pixel value difference and modulus function

Abstract

This paper proposes a novel data hiding method using pixel-value difference and modulus function for color image with the large embedding capacity(hiding 810757 bits in a 512×512 host image at least) and a high-visual-quality of the cover image. The proposed method has fully taken into account the correlation of the R, G and B plane of a color image. The amount of information embedded the R plane and the B plane determined by the difference of the corresponding pixel value between the G plane and the median of G pixel value in each pixel block. Furthermore, two sophisticated pixel value adjustment processes are provided to maintain the division consistency and to solve underflow and overflow problems. The most importance is that the secret data are completely extracted through the mathematical theoretical proof.

Appendix A. Proof of Theorem 1

Proof

Without loss of generality, we only prove Theorem 1 for the B plane. Now we discuss three cases for t _i −B _i mod 2ⁿ¹.

Case 1

If

$$\lceil\frac{2^{n_{1}}-1}{2}\rceil<t_{i}-B_{i}\mod 2^{n_{1}}<2^{n_{1}},$$

By the first formula of (4), we obtain

$$\overline{B_{i}}-B_{i}=t_{i}-B_{i}\mod 2^{n_{1}}.$$

The

$$\lceil\frac{2^{n_{1}}-1}{2}\rceil<\overline{B_{i}}-B_{i}<2^{n_{1}}.$$

That is to say

$$ \overline{B_{i}}-B_{i}\in (\lceil\frac{2^{n_{1}}-1}{2}\rceil,2^{n_{1}}). $$

(13)

And because

$$(\lceil\frac{2^{n_{1}}-1}{2}\rceil,2^{n_{1}})\subset[0,\lceil\frac{2^{n_{1}+1}-1}{2}\rceil),$$

therefore, it follows from the second expression of the equation(6) that

$$B^{\prime}_{i}=\overline{B_{i}}+2^{n_{1}}.$$

According to the adjustment process, we have

$$B^{\prime}_{i}-B_{i}=B^{'}_{i}-\overline{B_{i}}+\overline{B_{i}}-B_{i}=2^{n_{1}}+\overline{B_{i}}-B_{i}.$$

From (13), we can derive

$$B^{\prime}_{i}-B_{i}\in(\lceil\frac{2^{n_{1}}-1}{2}\rceil+2^{n_{1}},2^{n_{1}}+2^{n_{1}}) $$

$$=(\lceil\frac{2^{n_{1}}-1}{2}\rceil+2^{n_{1}},2^{n_{1}+1})\subseteq(2^{n_{1}},2^{n_{1}+1}).$$

Case 2

Suppose that the following conditions hold

$$-\lfloor\frac{2^{n_{1}}-1}{2}\rfloor\leq t_{i}-B_{i}\mod 2^{n_{1}}\leq\lceil\frac{2^{n_{1}}-1}{2}\rceil. $$

By the second formula of (4), we get

$$\overline{B_{i}}-B_{i}=t_{i}-B_{i}\mod2^{n_{1}}+2^{n_{1}}.$$

It’s not difficult to obtain

$$\overline{B_{i}}-B_{i}\in[2^{n_{1}}-\lfloor\frac{2^{n_{1}}-1}{2}\rfloor,\lceil\frac{2^{n_{1}}-1}{2}\rceil+2^{n_{1}}] $$

$$=\left[\lfloor\frac{2^{n_{1}+1}-2^{n_{1}}+1}{2}\rfloor, \lceil\frac{2^{n_{1}}+2^{n_{1}+1}-1}{2}\rceil\right]$$

$$=[\lfloor\frac{2^{n_{1}}+1}{2}\rfloor,\lceil\frac{3\times2^{n_{1}} -1}{2}\rceil]. $$

Therefor

$$\overline{B_{i}}-B_{i} \in\left[\lfloor\frac{2^{n_{1}}+1}{2}\rfloor,\lceil\frac{3\times2^{n_{1}}-1}{2}\rceil\right]. $$

In the following, we shall discuss two cases for$\overline {B_{i}}-B_{i}$

• 1) When

  $$ \overline{B_{i}}-B_{i} \in[\lfloor\frac{2^{n_{1}}+1}{2}\rfloor,\lceil\frac{2^{n_{1}+1}-1}{2}\rceil] $$

  (14)

  So we conclude from the second formula of(6)

  $$B^{\prime}_{i}=\overline{B_{i}}+2^{n_{1}}.$$

  Then

  $$B^{\prime}_{i}-B_{i}=B^{'}_{i}-\overline{B_{i}}+\overline{B_{i}}-B_{i}=2^{n_{1}}+\overline{B_{i}}-B_{i}.$$

  As a result, from (14), we can derive

  $$B^{\prime}_{i}-B_{i}\in[\lfloor\frac{2^{n_{1}}+1}{2}\rfloor+2^{n_{1}},\lceil\frac{2^{n_{1}+1}-1}{2}\rceil+2^{n_{1}})$$
  $$= [\lfloor\frac{3\times2^{n_{1}}+1}{2}\rfloor, \lceil\frac{2^{n_{1}+2}-1}{2}\rceil)=[2^{n_{1}},2^{n_{1}+1})$$

• 2) Whe

  $$\overline{B_{i}}-B_{i} \in[\lceil\frac{2^{n_{1}+1}-1}{2}\rceil,\lceil\frac{3\times2^{n_{1}}-1}{2}\rceil].$$

  According to the first formula of (6), we know

  $$B^{\prime}_{i}=\overline{B_{i}}.$$

  So

  $$B^{\prime}_{i}-B_{i}\in[\lceil\frac{2^{n_{1}+1}-1}{2}\rceil,\lceil\frac{3\times2^{n_{1}}-1}{2}\rceil]\subset[2^{n_{1}},2^{n_{1}+1}).$$

Case 3

If

$$-2^{n_{1}}+1\leq t_{i}-B_{i}\mod 2^{n_{1}}< -\lfloor\frac{2^{n_{1}}-1}{2}\rfloor.$$

In other word

$$ t_{i}-B_{i}\mod 2^{n_{1}}\in[-2^{n_{1}}+1,-\lfloor\frac{2^{n_{1}}-1}{2}\rfloor) $$

(15)

an

$$\overline{B_{i}}-B_{i}=t_{i}-B_{i}\mod 2^{n_{1}}+2^{n_{1}+1}.$$

Consequently, from (15), we have

$$\overline{B_{i}}-B_{i}\in\left[2^{n_{1}+1}-2^{n_{1}}+1,2^{n_{1}+1}-\lfloor\frac{2^{n_{1}}-1}{2}\rfloor)\right. $$

$$\subset[2^{n_{1}}+1,2^{n_{1}}+\frac{2^{n_{1}}+1}{2}). $$

From the result $B^{\prime }_{i}=\overline {B_{i}}$, we can reduce to

$$B^{\prime}_{i}-B_{i}=\overline{B_{i}}-B_{i}\in[2^{n_{1}}+1,2^{n_{1}}+\frac{2^{n_{1}}+1}{2})\subset[2^{n_{1}},2^{n_{1}+1}).$$

If $B^{\prime }_{i}-B_{i}\subset [2^{n_{1}},2^{n_{1}+1}),$ and $B^{\prime }_{i} <0,or B^{\prime }_{i}>255$, according to Step 7 of the embedding process, we have

$$|B^{*}_{i}-B_{i}|=|2B_{i}-B^{\prime}_{i}-B_{i}|=|B_{i}-B^{\prime}_{i}|\subset[2^{n_{1}},2^{n_{1}+1}).$$

Similarly, we obtain

$$|R^{*}_{i}-R_{i}|=|2R_{i}-R^{\prime}_{i}-R_{i}|=|R_{i}-R^{\prime}_{i}|\subset[2^{n_{2}},2^{n_{2}+1}).$$