Fast and robust watermarking in still images based on QR decomposition

Abstract

A blind watermarking technique based on QR decomposition is proposed on still images. The method is presented in spatial as well as transform domains and its robustness against some well-known image processing attacks is evaluated. It is shown that the QR decomposition offers capacity and robustness comparable to or better than similar watermarking based on the DCT and SVD transformations. Also, an interesting property of QR decomposition, which for the first time has been introduced by our previous paper, is proved and put in practice in more details.

Appendix

In Sections 3.1, it was stated that the QR decomposition of a matrix A has an interesting property, which we have used for watermark embedding. The property is the following: when the columns of A are correlated (as in the case of an image), the absolute values of elements of the first row of R tend to be larger than the rest of the rows. We know from (2) that the i ^th row of R is the projection of columns of A onto the i ^th column of Q matrix. Also from Gram-Schmidt process and QR decomposition, we know that the first column of Q is the normalized form of the first column of A. Also the other columns of Q are orthogonal to the first column of Q and to each other. Therefore the more the columns of A look similar or correlated with its first column (as in our case), their projection onto first column of Q (which is the normalized form of the first column of A) will be larger than their projection to other columns of Q. This fact is now shown for a 2 × 2 matrix and it can be generalized to higher dimensions. The QR decomposition for a 2 × 2 matrix can be derived as follow:

$$ \label{eq.10} \mathbf{A}=\left[ \begin{matrix} a & \quad c \\ b & \quad d \\ \end{matrix} \right] $$

(10)

$$ \label{eq.11} ad-bc>0 \overset{QR}{\Rightarrow} \left\{\begin{array}{l} \mathbf{Q}=\dfrac{1}{\sqrt{a^2+b^2}} \left[\begin{matrix} a &-b \\ b & a \\ \end{matrix} \right] \\[6pt] \mathbf{R}=\dfrac{1}{\sqrt{a^2+b^2}} \left[\begin{matrix} a ^2+b^2 & ac+bd \\ 0 & ad-bc \\ \end{matrix} \right] \end{array}\right. $$

(11)

$$ \label{eq.12} ad-bc<0 \overset{QR}{\Rightarrow} \left\{\begin{array}{l} \mathbf{Q}=\dfrac{1}{\sqrt{a^2+b^2}} \left[\begin{matrix} a & b \\ b & -a \\ \end{matrix} \right] \\ \mathbf{R}=\dfrac{1}{\sqrt{a^2+b^2}} \left[\begin{matrix} a ^2+b^2 & ac+bd \\ 0 & bc-ad \\ \end{matrix} \right] \end{array}\right. $$

(12)

As seen from (11), (12) in all cases r ₁₁ is larger than r ₂₁. From matrix algebra we know that a 2 × 2 matrix such as R can be shown as:

$$ \label{eq.13} \mathbf{R}_{2\times 2}=\left[ \begin{matrix} r_{11} & \quad r_{12} \\ r_{21} & \quad r_{22} \\ \end{matrix} \right] $$

(13)

Now we are going to derive the condition in which r ₁₂ is larger than r ₂₂. To do this we consider three cases based on the values of a, b and then derive the locus of c, d in which r ₁₂ is larger than r ₂₂. In our case (in spatial domain), a, b, c, d belong to the set {0, 1, 2, ..., 255}. For simplicity we first derive the area in which r ₂₂ is larger than r ₁₂ and then the remaining area will be the required one.

• Case1: a = b

  In this case for both (11) and (12), and for all value of c and d belong to {0, 1, 2, ..., 255} the value of r ₁₂ is larger than r ₂₂. The reader can easily verify it.

• Case2: a > b

  In this case, as depicted in Fig. 8a, we derive the locus of c, d from (11) by replacing them respectively with x and y for r ₁₂ to be larger than r ₂₂. The complementary for this is r ₁₂ to be smaller than r ₂₂ and it is depicted by shaded area in Fig. 8a. Thus we have:

  $$ ad-bc>0 \Rightarrow y>\frac{b}{a}x $$

  (14)
  $$ \begin{array}{rll} r_{12}<r_{22} &\Rightarrow \dfrac{ac+bd}{\sqrt{a^2+b^2}}<\dfrac{ad-bc}{\sqrt{a^2+b^2}} \\ &\Rightarrow y>\dfrac{a+b}{a-b}x \end{array} $$

  (15)

• Case3: a < b

  In this case, as depicted in Fig. 8b, we derive the locus of c, d from (12) by replacing them respectively with x and y for r ₁₂ to be larger than r ₂₂. The complementary for this is r ₁₂ to be smaller than r ₂₂ and it is depicted by shaded area in Fig. 8b. Thus we have:

  $$ ad-bc<0 \Rightarrow y<\frac{b}{a}x \\ $$

  (16)
  $$ \begin{array}{rll} r_{12}<r_{22} &\Rightarrow \dfrac{ac+bd}{\sqrt{a^2+b^2}}<\dfrac{bc-ad}{\sqrt{a^2+b^2}} \\ &\Rightarrow y<\dfrac{b-a}{a+b}x \end{array} $$

  (17)

Fig. 8

a The shaded area S ₁ is the locus of c, d in which r ₂₂ > r ₁₂ and a > b. b The shaded area S ₂ is the locus of c, d in which r ₂₂ > r ₁₂ and a < b

In Fig. 8, X′ and Y′, are the new axes which are produced from QR decomposition and the columns of A are projected to them, and L ₁ is the bisector of X′ and Y′. Thus in both pictures if the pair (c, d) from A matrix covers the shaded area then r ₁₂ will be smaller than r ₂₂. The probability that the pair (c, d) sit in the shaded area can be calculated from the law of total probability as:

$$ \begin{array}{rll} \label{eq.18} P(r_{22}>r_{12})&=&P(r_{22}>r_{12}|a>b).P(a>b) \\ &&+P(r_{22}>r_{12}|a<b).P(a<b)\nonumber\\ &=&0.5\times P(r_{22}>r_{12}|a>b)\nonumber\\ &&+0.5\times P(r_{22}>r_{12}|a<b)\nonumber\\ &=&0.5\times P(S_{1})+0.5\times P(S_{2})\nonumber\\ &=&0.5\times \dfrac{a-b}{2(a+b)}+0.5\times \dfrac{b-a}{2(a+b)}\nonumber\\ &=&0.5\times \dfrac{|a-b|}{a+b} \end{array} $$

(18)

Thus Equation (18) shows that how much a, b close to each other (this means that the pair must be around the line y = x), the probability that r ₂₂ > r ₁₂ is smaller.

To verify this result, we divide the test images (Fig. 9) into 2 × 1 and 3 × 1 blocks and depict them by spot in 2D and 3D charts respectively in Fig. 10. Also we do this in DWT domain for LL sub-band which is a scaled version of the original image and depict the result in Fig. 11. As seen from these figures, most of the dots are around the y = x line and this confirms our statement: In QR decomposition of 8 × 8 blocks of an image, the absolute values of the elements of the first row of R matrix statistically tend to be larger than the elements of the other rows.

Fig. 9

Test images

Fig. 10

a Charts of 2 × 1 blocks of test images. b Charts of 3 × 1 blocks of test images

Fig. 11

a Charts of 2 × 1 blocks for LL sub-band of test images. b Charts of 3 × 1 blocks for LL sub-band of test images