Embedding capacity estimation of reversible watermarking schemes

Abstract

Estimation of the embedding capacity is an important problem specifically in reversible multi-pass watermarking and is required for analysis before any image can be watermarked. In this paper, we propose an efficient method for estimating the embedding capacity of a given cover image under multi-pass embedding, without actually embedding the watermark. We demonstrate this for a class of reversible watermarking schemes which operate on a disjoint group of pixels, specifically for pixel pairs. The proposed algorithm iteratively updates the co-occurrence matrix at every stage to estimate the multi-pass embedding capacity, and is much more efficient vis-a-vis actual watermarking. We also suggest an extremely efficient, pre-computable tree based implementation which is conceptually similar to the co-occurrence based method, but provides the estimates in a single iteration, requiring a complexity akin to that of single pass capacity estimation. We also provide upper bounds on the embedding capacity. We finally evaluate performance of our algorithms on recent watermarking algorithms.

Appendix A

Theorem 3.

Under multi-pass embedding, the embedding capacity estimated by the co-occurrence based algorithm (algorithm 1) is exactly the same as the one estimated by the pixel-pair tree based algorithm. In particular,

$$\begin{array}{@{}rcl@{}} &{\boldsymbol{1)}}& \sum\limits_{\xi \in \mathbb{D}} C_{k}(\xi) b_{\xi} = \sum\limits_{\xi \in \mathbb{D}} C_{0}(\xi) \sum\limits_{s \in \mathbf{S}_{\xi}} \left(~\prod\limits_{m=0}^{k} {p_{s[m]}} \right) {{b}_{s[k]}}\\ &{\boldsymbol{2)}}& \sum\limits_{k = 0}^{P-1} \sum\limits_{\xi \in \mathbb{D} } C_{k}(\xi) b_{\xi} = \sum\limits_{\xi \in \mathbb{D}} C_{0}(\xi) \sum\limits_{s \in \mathbf{S}_{\xi}} \left(~\prod\limits_{k=0}^{P-1} {p_{s[k]}} \right) \sum\limits_{k=0}^{P-1} {{b}_{s[k]}} \end{array} $$

Proof:

For every pixel pair ξ, let N _ξ represent the set of nodes corresponding to the pixel-pair tree of ξ. Let n∈N _ξ refer to a node, i.e., a pixel pair in the pixel-pair tree of ξ. Further, let s _n represent any path of the tree, containing the node n and let d _n represent the depth of node n. Hence s _n [d _n ]=n. Now we prove the first part of the theorem. We start with the LHS and from algorithm 1 it is evident that for any node n∈N _ξ with depth d _n =k, the number of pixel pairs contributed by the pair ξ, to the pixel pair n is $C(\xi ){\prod }_{m = 0}^{k} p_{s_{n}[m]}$. This is clear from the updation step in algorithm 1. We then break down the sum ${\sum }_{\xi \in \mathbb {D}} C_{k}(\xi ) b_{\xi }$, into contributions from every node n∈N _ξ with depth d _n =k, from every pixel pair in $\xi \in \mathbb {D}$. In other words, we can write:

$$ \sum\limits_{\xi \in \mathbb{D}} C_{k}(\xi) b_{\xi} = \sum\limits_{\xi \in \mathbb{D}} \sum\limits_{n \in \mathbf{N}_{\xi} : d_{n} = k} C_{0}(\xi)\prod\limits_{m = 0}^{k} p_{s_{n}[m]} b_{s_{n}[k]}. $$

(42)

Now it is obvious from the definition that: ${\sum }_{n \in \mathbf {N}_{\xi } : d_{n} = k} {\prod }_{m = 0}^{k} p_{s_{n}[m]} b_{s_{n}[k]} = {\sum }_{s \in \mathbf {S}_{\xi }}{\prod }_{m=0}^{k} {p_{s[m]}} {b}_{s[k]}$. Hence using this in Eq. (42) we can prove the first part of the theorem.

We now consider the second part. Using the result from earlier section we can write:

$$\begin{array}{@{}rcl@{}} \sum\limits_{k = 0}^{P-1} \sum\limits_{\xi \in \mathbb{D}} C_{k}(\xi) b_{\xi} &=& \sum\limits_{\xi \in \mathbb{D}} \sum\limits_{k = 0}^{P-1} \sum\limits_{n \in \mathbf{N}_{\xi} : d_{n} = k} C_{0}(\xi)\prod\limits_{m = 0}^{k} p_{s_{n}[m]} b_{s_{n}[m]}\\ &=& \sum\limits_{\xi \in \mathbb{D}} \sum\limits_{n \in \mathbf{N}_{\xi}} C_{0}(\xi)\prod\limits_{k = 0}^{d_{n}} p_{s_{n}[k]} b_{s_{n}[d_{n}]}. \end{array} $$

In the above expression, observe that we are summing over every node in the pixel-pair tree as k goes from 0 to P−1. Hence we re-write the expression, by explicitly summing over every node in the pixel-pair tree. We finally replace the variable m by k, to obtain the above expression. In order to prove the second part of the theorem, it is sufficient to show that for every pixel-pair ξ:

$$ \sum\limits_{n \in \mathbf{N}_{\xi}} C_{0}(\xi).\prod\limits_{k = 0}^{d_{n}}p_{s_{n}[k]} b_{n} = \sum\limits_{s \in \mathbf{S}_{\xi}} C_{0}(\xi) \prod\limits_{k = 0}^{P-1} p_{s[k]} \sum\limits_{k = 0}^{P-1} b_{s[k]}. $$

Note that we used the fact that $b_{s_{n}[d_{n}]} = b_{n}$. We then start with the R.H.S as:

$$\begin{array}{@{}rcl@{}} ~\text{R.H.S} &=& \sum\limits_{s \in \mathbf{S}_{\xi}} C_{0}(\xi) \prod\limits_{k = 0}^{P-1} p_{s[k]} \sum\limits_{k = 0}^{P-1} b_{s[k]} \\ &=& \sum\limits_{n \in \mathbf{N}_{\xi}} \sum\limits_{s \in \mathbf{S}_{\xi}} C_{0}(\xi) \prod\limits_{k = 0}^{P-1} p_{s[k]} {~}_{k = 0}^{P-1} b_{s[k]} I(s[k] = n) \\ &=& \sum\limits_{n \in \mathbf{N}_{\xi}} \sum\limits_{s \in \mathbf{S}_{\xi}} C_{0}(\xi) \prod\limits_{k = 0}^{P-1} p_{s[k]} b_{n} I(s[k] = n). \end{array} $$

In the above equation, we consider only those paths which pass through the node n. Let S _n be a subset of paths which pass through node n. Hence we can write:

$$\begin{array}{@{}rcl@{}} ~\text{R.H.S} &=& \sum\limits_{n \in \mathbf{N}_{\xi}} \sum\limits_{s \in \mathbf{S}_{n}} C_{0}(\xi) \prod\limits_{k = 0}^{d_{n}} p_{s[k]} \prod\limits_{k = d_{n} + 1}^{P} p_{s[k]} b_{n} \\ &=& \sum\limits_{n \in \mathbf{N}_{\xi}} C_{0}(\xi) \prod\limits_{k = 0}^{d_{n}} p_{s_{n}[k]} b_{n} \sum\limits_{s \in \mathbf{S}_{n}} \prod\limits_{k = d_{n} + 1}^{P} p_{s[k]} \\ &=& \sum\limits_{n \in \mathbf{N}_{\xi}} C_{0}(\xi) \prod\limits_{k = 0}^{d_{n}} p_{s_{n}[k]} b_{n} \\ &=& ~\text{L.H.S.} \end{array} $$

Observe that the nodes in the paths s∈S _n with depth greater than d _n , form a complete subtree and hence we have: ${\sum }_{s \in \mathbf {S}_{n}} {\prod }_{k = d_{n} + 1}^{P-1} p_{s[k]} = 1$. Further, also observe that the nodes with a depth ≤d _n occur in every path in s∈S _n . Hence we replace these nodes by s _n [k], since s _n is a path in S _n . Hence proved. Note that in this proof though we have proved the equivalence of the tree based and co-occurrence based algorithms for the estimates of $\omega (\mathcal {B})$ and $\omega (\mathcal {B}_{k})$, the same proof can be used to prove the equivalence of the estimates of $\eta (\mathcal {B})$ and $\eta (\mathcal {B}_{k})$.

Lemma 1.

For any function $f: \mathbb {S} \to [0.5,1]$ , $\underset {\mathbf {s} \in \mathbb {S}}{\min } H_{0}(f(\mathbf {s})) \equiv H_{0}(\underset {\mathbf {s} \in \mathbb {S}}{\max } f(\mathbf {s}))$ . Similarly for any function $g: \mathbb {S} \to [0,0.5]$ , $\underset {\mathbf {s} \in \mathbb {S}}{\min } H_{0}(g(\mathbf {s})) \equiv H_{0}(\underset {\mathbf {s} \in \mathbb {S}}{\min } g(\mathbf {s}))$.

Proof:

First, note that H ₀(z) is a decreasing function for 0.5≤z≤1. Hence H ₀(z ₁)≤H ₀(z ₂) implies z ₁≥z ₂. Using this fact, we now prove this lemma by contradiction. Assume that $\mathbf {s}_{1} = \underset {\mathbf {s} \in \mathbb {S}}{~\text { arg min }} H_{0}(f(\mathbf {s}))$ and $\mathbf {s}_{2} = \underset {\mathbf {s} \in \mathbb {S}}{~\text { arg max }} f(\mathbf {s})$. Further, assume that f(s ₁)≠f(s ₂). Then clearly by definition f(s ₁)≤f(s ₂) and H ₀(f(s ₁))≤H ₀(f(s ₂)). This is a contradiction since H ₀(z) is a decreasing function. Thus $f(\mathbf {s}_{1}) \equiv f(\mathbf {s}_{2}) \Rightarrow \underset {\mathbf {s} \in \mathbb {S}}{\min } H_{0}(f(\mathbf {s})) \equiv H_{0}(\underset {\mathbf {s} \in \mathbb {S}}{\max } f(\mathbf {s}))$. Similarly, we can prove the second part.

Lemma 2.

For any function $h: \mathbb {S} \to [0,1]$ , $\underset {\mathbf {s} \in \mathbb {S}}{\min } H_{0}(h(\mathbf {s})) \ge \min (H_{0}(\underset {\mathbf {s} \in \mathbb {S}}{\max } h(\mathbf {s})), H_{0}(\underset {\mathbf {s} \in \mathbb {S}}{\min } h(\mathbf {s})))$.

Proof:

This lemma directly follows from Lemma 1. If h(s)≤0.5. then using lemma 1, we have: $\underset {\mathbf {s} \in \mathbb {S}}\min H_{0}(h(\mathbf {s})) \equiv H_{0}(\underset {\mathbf {s} \in \mathbb {S}}\min h(\mathbf {s}))$. Similarly if h(s)≥0.5, $\underset {\mathbf {s} \in \mathbb {S}}\min H_{0}(h(\mathbf {s})) \equiv H_{0}(\underset {\mathbf {s} \in \mathbb {S}}\max h(\mathbf {s}))$. Hence $\underset {\mathbf {s} \in \mathbb {S}}\min H_{0}(h(\mathbf {s}))$ is definitely greater than the minimum of these two and hence proved.