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.
+Since this last step is shared by all,
+we separately eavluate this complexity.
In all the rest of this section, 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
+This steps is in $\theta(n^2 + 2\times 343^2)$ 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
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 $O(2\times 343^2 \times n^2)$.
+which is at least in $\theta(343)$, and an overall distance between these
+metrics, which is in $\theta(686)$. Computing the distance is thus in
+$\theta(2\times 343^2)$ and this modification
+is thus in $\theta(2\times 343^2 \times n^2)$.
Ranking these results may be achieved with an insertion sort, which is in
-$2.n^2 \ln(n)$.
+$2 \times n^2 \ln(n)$.
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},
-$O(2.n^2(343^2 + \ln(n)))$.
+$\theta(n^2 +2 \times 343^2 + 2\times 343^2 \times n^2 + 2 \times n^2 \ln(n))$, \textit{i.e},
+$\theta(2 \times n^2(343^2 + \ln(n)))$.
Let us focus now on WOW.
This scheme starts to compute the residual
-of the cover as a convolution product which is in $O(n^2\ln(n^2))$.
+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 $O(n^2 \times n^2\ln(n^2))$.
+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 $O(6n^4\ln(n))$.
+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 $O(n^2)$.
-To summarize, the complixity is in $O(6n^4\ln(n) +n^2)$
+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 $O(n^2 \times n^2\ln(n^2))$ and a sum $D$ of
-these $W$ which is in $O(n^2)$.
-This distortion computation step is thus in $O(6n^4\ln(n) + n^2)$.
+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 $O(5^3 n^2)$.
+$n \times n$ 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 $O(4Tn)$ 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 $O((5^3+4T+1)n^2)$.
-To summarize, for the embedding map construction, the complexity of Hugo is
-dramatically larger than our scheme.
+Computing gradients is in $\theta(4Tn)$ since derivatives of each direction (vertical or horizontal)
+are in $\theta(2Tn)$.
+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 $O(2^h.n)$ where $h$ is the size of the duplicated
+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.
+The Fig.~\ref{fig:compared}
+summarizes the complexity of the embedding map construction, for Hugo, Wow
+and Uniward. It deals with square images
+of size $n \times n$ when $n$ ranges from
+512 to 4096. The $y$-coordinate is expressed in a logarithm scale.
+It shows that the complexity of all algorithms
+is dramatically larger than the one of the STABYLO scheme.
+Thanks to these complexity results, we claim that STABYLO is lightweight.
+\begin{figure}
+\begin{center}
+\includegraphics[scale=0.4]{complexity}
+\end{center}
+\caption{Complexity evaluation of Wow/Uniward, Hugo, and Stabylo}
+\label{fig:compared}
+\end{figure}
-Thanks to these complexity results, we claim that STABYLO is lightweight.
-
-
