This section aims at justifying the lightweight attribute of our approach.
-To be more precise, we compare the complexity of our schemes to the
-state of the art steganography, namely HUGO~\cite{DBLP:conf/ih/PevnyFB10}.
+To be more precise, we compare the complexity of our schemes to some of
+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 an $n \times n$ square image.
First of all, HUGO starts with computing the second order SPAM Features.
-This steps is in $O(n^2 + 2.343^2)$ due to the calculation
+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).
Next for each pixel, the distortion measure is calculated by +1/-1 modifying
its value and computing again the SPAM
features. Pixels are thus selected according to their ability to provide
-an image whose SPAM features are close to the original one.
-The algorithm is thus computing a distance between each computed feature,
-andthe original ones
-which is at least in $O(343)$ and an overall distance between these
-metrics which is in $O(686)$. Computing the distance is thus in
+an image whose SPAM features are close to the original ones.
+The algorithm thus computes a distance between each feature
+and the original ones,
+which is at least in $O(343)$, and an overall distance between these
+metrics, which is in $O(686)$. Computing the distance is thus in
$O(2\times 343^2)$ and this modification
is thus in $O(2\times 343^2 \times n^2)$.
-Ranking these results may be achieved with a insertion sort which is in
+Ranking these results may be achieved with an insertion sort, which is in
$2.n^2 \ln(n)$.
-The overall complexity of the pixel selection is thus
-$O(n^2 +2.343^2 + 2\times 343^2 \times n^2 + 2.n^2 \ln(n))$, \textit{i.e}
+The overall complexity of the pixel selection is finally
+$O(n^2 +2.343^2 + 2\times 343^2 \times n^2 + 2.n^2 \ln(n))$, \textit{i.e},
$O(2.n^2(343^2 + \ln(n)))$.
-Our edge selection is based on a Canny Filter,
-whose complexity is in $O(2n^2.\ln(n))$ thanks to the convolution step
-which can be implemented with FFT.
-To summarize, for the embedding map construction, the complexity of Hugo is
-at least $343^2/\ln{n}$ times higher than
-our scheme. For instance, for a squared image with 4M pixel per slide,
-this part of our algorithm is more than 14100 faster than Hugo.
-We are then left to express the complexity of the STC algorithm .
+
+
+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.
+Computing gradients is in $O(4Tn)$ since derivatives of each direction (vertical or horizontal)
+are in $O(2Tn)$.
+Finally, thresholding with hysteresis is in $O(n^2)$.
+The overall complexity is thus in $O((5^3+4T+1)n^2)$.
+
+
+
+
+
+We are then left to express the complexity of the STC algorithm.
According to~\cite{DBLP:journals/tifs/FillerJF11}, it is
in $O(2^h.n)$ where $h$ is the size of the duplicated
-matrix. Its complexity is thus negligeable compared with the embedding map
+matrix. Its complexity is thus negligible compared with the embedding map
construction.
-Thanks to these complexity result, we claim that STABYLO is lightweight.
+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.
+
+
+
+
+
