\section{Theoretical background} \label{sec:notions} \subsection{Blurred segment} This work relies on the notion of digital straight line as classically defined in the digital geometry literature \cite{KletteRosenfeld04}. Only the 2D case is considered here. \begin{definition} A digital line $\mathcal{L}(a,b,c,\nu)$, with $(a,b,c,\nu) \in \mathbb{Z}^4$, is the set of points $P(x,y)$ of $\mathbb{Z}^2$ that satisfy : $0 \leq ax + by - c < \nu$. \end{definition} In the following, we note $\delta(\mathcal{L}) = b/a$ the slope of digital line $\mathcal{L}$, $w(\mathcal{L}) = \nu$ its arithmetical width, $h(\mathcal{L}) = c$ its {\it shift} to origin, and $p(\mathcal{L}) = max(|a|,|b|)$ its period (i.e. the length of its periodic pattern). When $\nu = p(\mathcal{L})$, then $\mathcal{L}$ is the narrowest 8-connected line and is called a naive line. \begin{definition} A blurred segment $\mathcal{B}$ of assigned width $\varepsilon$ is a set of points in $\mathbb{Z}^2$ that all belong to a digital line $\mathcal{L}$ of arithmetical width $w(\mathcal{L}) = \varepsilon$. \end{definition} A linear-time algorithm to recognize a blurred segment of assigned width $\varepsilon$ \cite{DebledAl05} is used in the work. It is based on an incremental growth of the convex hull of the blurred segment when adding each point $P_i$ successively. The minimal width $\mu$ of the blurred segment $\mathcal{B}$ is the arithmetical width of the narrowest digital straight line that contains $\mathcal{B}$. %It is also the minimal width of the convex hull of $\mathcal{B}$, %that can be computed by Melkman's algorithm \cite{Melkman87}. The enclosing digital segment $E(\mathcal{B})$ is the section of this optimal digital straight line bounded by the end points of $\mathcal{B}$. As depicted on \RefFig{fig:bs}, the extension of the blurred segment $\mathcal{B}_{i-1}$ of assigned width $\varepsilon$ and minimal width $\mu_{i-1}$ at step $i-1$ with a new input point $P_i$ is thus controlled by the recognition test $\mu_i < \varepsilon$. \begin{figure}[h] \center \input{Fig_notions/bswidth} \caption{A growing blurred segment $\mathcal{B}_i$ : when adding the new point $P_i$, the blurred segment minimal width augments from $\mu_{i-1}$ to $\mu_i$; if the new width $\mu_i$ exceeds the assigned width $\varepsilon$, then the new input point is rejected and $\mathcal{B}_i = \mathcal{B}_{i-1}$.} \label{fig:bs} \end{figure} Associated to this primitive, the following definition of a directional scan also based on digital straight lines is also used in this work. \subsection{Directional scan} \begin{definition} A directional scan $DS$ is an ordered partition restricted to the image domain $\mathcal{I}$ of a digital straight line $\mathcal{D}$, called the {\it scan strip}, into scans $S_i$, each of them being a segment of a naive line $\mathcal{N}_i$, called a {\it scan line}, orthogonal to $\mathcal{D}$. \end{definition} \begin{equation} DS = \left\{ S_i = \mathcal{D} \cap \mathcal{N}_i \cap \mathcal{I} \left| \begin{array}{l} \delta(\mathcal{N}_i) = - \delta^{-1}(\mathcal{D}) \\ \wedge~ h(\mathcal{N}_i) = h(\mathcal{N}_{i-1}) + p(\mathcal{D}) \end{array} \right. \right\} %S_i = \mathcal{D} \cap \mathcal{N}_i, \mathcal{N}_i \perp \mathcal{D} \end{equation} In this definition, the clause $\delta(\mathcal{N}_i) = - \delta^{-1}(\mathcal{D})$ expresses the othogonality constraint between the scan lines $\mathcal{N}_i$ and the scan strip $\mathcal{D}$. Then the shift of the period $p(\mathcal{D})$ between successive scans guarantees that all points of the scan strip are travelled one and only one time. The scans $S_i$ are developed on each side of a start scan $S_0$, and ordered by their distance to the start line $\mathcal{N}_0$ with a positive (resp. negative) sign if they are on the left (resp. right) side of $\mathcal{N}_0$ (\RefFig{fig:ds}). The directional scan is iterately parsed from the start scan to both ends. At each iteration $i$, the scans $S_i$ and $S_{-i}$ are successively processed. \begin{figure}[h] \center % \input{Fig_notions/fig} \includegraphics[width=0.8\textwidth]{Fig_notions/scanstrip.eps} \begin{picture}(1,1)(0,0) \thicklines \put(-176,112){\vector(2,-1){30}} \put(-90,19){\vector(-2,1){30}} {\color{dwhite}{ \put(-181,114.5){\circle*{10}} \put(-84,16.5){\circle*{10}} \put(-16,102.5){\circle*{10}} \put(-132,66.5){\circle*{12}} \put(-72,96.5){\circle*{12}} \put(-175.5,65.5){\circle*{20}} \put(-117,10.5){\circle*{14}} \put(-54,32.5){\circle*{14}} \put(-161,10.5){\circle*{20}} }} \put(-88,13.5){$A$} \put(-185,111.5){$B$} \put(-20,98){$\mathcal{D}$} \put(-137,64){\color{blue}{$S_0$}} \put(-77,94){\color{red}{$S_8$}} \put(-183,64){\color{dgreen}{$S_{-5}$}} \put(-123,8){\color{blue}{$\mathcal{N}_0$}} \put(-60,30){\color{red}{$\mathcal{N}_8$}} \put(-169,8){\color{dgreen}{$\mathcal{N}_{-5}$}} \end{picture} \caption{A directional scan. The start scan $S_0$ is drawn in blue, odd scans in green, even scans in red, the bounds of scan lines $\mathcal{N}_i$ with plain lines and the bounds of scan strip $\mathcal{D}$ with dotted lines.} \label{fig:ds} \end{figure} A directional scan can be defined by its start scan $S_0$. If $A(x_A,y_A)$ and $B(x_B,y_B)$ are the end points of $S_0$, and if we note $\delta_x = x_B - x_A$, $\delta_y = y_B - y_A$, $c_1 = \delta_x\cdot x_A + \delta_y\cdot y_A$, $c_2 = \delta_x\cdot x_B + \delta_y\cdot y_B$ and $\nu_{AB} = max (|\delta_x|, |\delta_y|)$, it is then defined by the following scan strip $\mathcal{D}^{A,B}$ and scan lines $\mathcal{N}_i^{A,B}$: \begin{equation} \left\{ \begin{array}{l} \mathcal{D}^{A,B} = \mathcal{L}(\delta_x,~ \delta_y,~ min (c1,c2),~ 1 + |c_1-c_2|) \\ \mathcal{N}_i^{A,B} = \mathcal{L}(\delta_y,~ -\delta_x,~ \delta_y\cdot x_A - \delta_x\cdot y_A + i\cdot \nu_{AB},~ \nu_{AB}) \end{array} \right. \end{equation} %The scan lines length is $d_\infty(AB)$ or $d_\infty(AB)-1$, where $d_\infty$ %is the chessboard distance ($d_\infty = max (|d_x|,|d_y|)$). %In practice, this difference of length between scan lines is not a drawback, %as the image bounds should also be processed anyway. A directional scan can also be defined by its central point $C(x_C,y_C)$, its direction $\vec{D}(X_D,Y_D)$ and its width $w$. If we note $c_3 = x_C\cdot Y_D - y_C\cdot X_D$, $c_4 = X_D\cdot x_C + Y_D\cdot y_C$, $\nu_{\vec{D}} = max (|X_D|,|Y_D|)$, it is then defined by the following scan strip $\mathcal{D}^{C,\vec{D},w}$ and scan lines $\mathcal{N}_i^{C,\vec{D},w}$: \begin{equation} \left\{ \begin{array}{l} \mathcal{D}^{C,\vec{D},w} = \mathcal{L}(Y_D,~ -X_D,~ c_3 - w / 2,~ w) \\ \mathcal{N}_i^{C,\vec{D},w} = \mathcal{L}(X_D,~ Y_D,~ c_4 - w / 2 + i\cdot w,~ \nu_{\vec{D}} \end{array} \right. \end{equation}