state of the art steganography, namely HUGO~\cite{DBLP:conf/ih/PevnyFB10}.
-In what follows, we consider an $n \times n$ 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 $O(n^2 + 2.343^2)$ due to the calculation
of the difference arrays and next of the 686 features (of size 343).
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
+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}
$O(2.n^2(343^2 + \ln(n)))$.
-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.
+Our edge selection is based on a Canny Filter. When applied on a
+$n \times n$ square image the Noise reduction steps is in $O(5^3 n^2)n$.
+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
-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.
+dramatically higher than our scheme.
-We are then left to express the complexity of the STC algorithm .
+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
construction.
+
+
+
+
+
Thanks to these complexity result, we claim that STABYLO is lightweight.
