$O(n^2 +2.343^2 + 2\times 343^2 \times n^2 + 2.n^2 \ln(n))$, \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.
According to~\cite{DBLP:journals/tifs/FillerJF11}, it is
