current state of the art of
steganographic scheme, namely HUGO~\cite{DBLP:conf/ih/PevnyFB10},
WOW~\cite{conf/wifs/HolubF12}, and UNIWARD~\cite{HFD14}.
+Each of these scheme starts with the computation of the distortion cost
+for each pixel switch and is later followed by the STC algorithm.
+Since this last step is shared by all, we do not add its complexity.
+In all the rest of this section, we consider a $n \times n$ square image.
-
-In what follows, we consider a $n \times n$ square image.
First of all, HUGO starts with computing the second order SPAM Features.
This steps is in $O(n^2 + 2\times 343^2)$ due to the calculation
of the difference arrays and next of the 686 features (of size 343).
$O(n^2 +2.343^2 + 2\times 343^2 \times n^2 + 2.n^2 \ln(n))$, \textit{i.e},
$O(2.n^2(343^2 + \ln(n)))$.
+
+
+
+Let us focus now on WOW.
+This scheme starts to compute the residual
+of the cover as a convolution product which is in $O(n^2\ln(n^2))$.
+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 $O(n^2 \times n^2\ln(n^2))$.
+Moreover the suitability is computed for each wavelet level
+detail (HH, HL, LL).
+This distortion computation step is thus in $O(6n^4\ln(n))$.
+Finally a norm of these three values is computed for each pixel
+which adds to this complexity the complexity of $O(n^2)$.
+To summarize, the complixity is in $O(6n^4\ln(n) +n^2)$
+
+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 $O(n^2 \times n^2\ln(n^2))$ and a sum $D$ of
+these $W$ which is in $O(n^2)$.
+This distortion computation step is thus in $O(6n^4\ln(n) + n^2)$.
+
+
Our edge selection is based on a Canny Filter. When applied on a
$n \times n$ square image, the noise reduction step is in $O(5^3 n^2)$.
Next, let $T$ be the size of the canny mask.
are in $O(2Tn)$.
Finally, thresholding with hysteresis is in $O(n^2)$.
The overall complexity is thus in $O((5^3+4T+1)n^2)$.
-To summarize, for the embedding map construction, the complexity of Hugo is
-dramatically larger than our scheme.
+
+
+
+
We are then left to express the complexity of the STC algorithm.
According to~\cite{DBLP:journals/tifs/FillerJF11}, it is
matrix. Its complexity is thus negligible compared with the embedding map
construction.
+To summarize, for the embedding map construction, the complexity of Hugo, WOW
+and UNIWARD are dramatically larger than the one of our scheme:
+STABYLO is in $O(n^2)$
+whereas HUGO is in $O(n^2\ln(n)$, and WOW and UNIWARD are in $O(n^4\ln(n))$.
+Thanks to these complexity results, we claim that STABYLO is lightweight.
+
-Thanks to these complexity results, we claim that STABYLO is lightweight.
