+Let us focus now on WOW.
+This scheme starts to compute the residual
+of the cover as a convolution product which is in $\theta(n^2\ln(n^2))$.
+The embedding suitability $\eta_{ij}$ is then computed for each pixel
+$1 \le i,j \le n$ thanks to a convolution product again.
+We thus have a complexity of $\theta(n^2 \times n^2\ln(n^2))$.
+Moreover the suitability is computed for each wavelet level
+detail (HH, HL, LL).
+This distortion computation step is thus in $\theta(6n^4\ln(n))$.
+Finally a norm of these three values is computed for each pixel
+which adds to this complexity the complexity of $\theta(n^2)$.
+To summarize, the complixity is in $\theta(6n^4\ln(n) +n^2)$
+
+What follows details the complexity of the distortion evaluation of the
+UNIWARD scheme. This one is based to a convolution product $W$ of two elements
+of size $n$ and is again in $\theta(n^2 \times n^2\ln(n^2))$,
+ and a sum $D$ of
+these $W$ which is in $\theta(n^2)$.
+This distortion computation step is thus in $\theta(6n^4\ln(n) + n^2)$.
+
+
+Our edge selection is based on a Canny filter. When applied on a
+$n \times n$ square image, the noise reduction step is in $\theta(5^3 n^2)$.
+Next, let $T$ be the size of the Canny mask.
+Computing gradients is in $\theta(4Tn^2)$ since derivatives of each direction (vertical or horizontal)
+are in $\theta(2Tn^2)$.
+Finally, thresholding with hysteresis is in $\theta(n^2)$.
+The overall complexity is thus in $\theta((5^3+4T+1)n^2)$.
+
+
+
+
+
+We are then left to express the complexity of the STC algorithm.
+According to~\cite{DBLP:journals/tifs/FillerJF11}, it is
+in $\theta(2^h.n)$ where $h$ is the size of the duplicated
+matrix. Its complexity is thus negligible compared with the embedding map
+construction.