Local Turn-Boundedness: A Curvature Control for Continuous Curves with Application to Digitization

Abstract

This article focuses on the classical problem of the control of information loss during the digitization step. The properties proposed in the literature rely on smoothness hypotheses that are not satisfied by the curves including angular points. The notion of turn introduced by Milnor in the article On the Total Curvature of Knots generalizes the notion of integral curvature to continuous curves. Thanks to the turn, we are able to define the local turn-boundedness. This promising property of curves does not require smoothness hypotheses and shares several properties with the par(r)-regularity, in particular well-composed digitizations. Besides, the local turn-boundedness enables to constrain spatially the continuous curve as a function of its digitization.

Appendices

Proof of Corollary 1

Lemma 7

The boundary of a convex shape with nonempty interior has no cusp.

Proof

Let $\mathcal C$ be a Jordan curve whose interior is convex. Let $p\in \mathcal C$. Since $\mathcal C$ is a Jordan curve, there exist in $\mathcal C$ two points q, r not colinear with p. Let P, Q, R be three straight lines passing, respectively, by the points p, q, r and separating $\mathcal C$ from one of the half planes they delimit (see Fig. 18). The interior of the convex hull of $\{p,q,r\}$ is included in the interior of $\mathcal C$. Then, the arc from q to r passing through p is included in the half plane delimited by P and containing pqr deprived of the triangle pqr. Then, if $\mathcal C$ has semi-tangent vectors $\mathbf {u}$ and $\mathbf {v}$ in p, the angle $\angle (\mathbf {u}, \mathbf {v})$ is bounded from above by $\pi -\alpha$ where $\alpha$ is the interior angle of the triangle [p, q, r] at p. $\square$

Fig. 18

Bounding a convex curve (in blue) passing through three non-colinear points (Color figure online)

Proof of Lemma 1

Lemma 1

Let $\mathcal C$ be a $(\theta , \delta )$-LTB curve with $\theta <2\pi /3$. Then,

$$\begin{aligned} \delta \le {\text {diam}}(\mathcal C), \end{aligned}$$

where ${\text {diam}}$ denotes the diameter.

Proof

By contradiction, we assume $\delta > {\text {diam}}(\mathcal C)$. Let $\ell$ be the length of $\mathcal C$, a be a non-angular point of $\mathcal C$ and $k<1/2$. We prove by induction that, for any n, there exists an arc $\mathcal C_n$ of $\mathcal C$ whose ends are non-angular points and containing a whose turn is less than $\theta$ and whose length is greater than $(1-(1-k)^n)\ell$. Furthermore, the sequence $(\mathcal C_i)$ is increasing for the inclusion. We initialize the induction by taking a smooth point b of $\mathcal C$ such that the geodesic distance from a to b is greater than $k\ell$ (recall that the set of angular points of a LTBcurve is countable). Since $\delta > {\text {diam}}(\mathcal C)$ and $\mathcal C$ is LTB, one of the arcs from a to b, that we denote by $\mathcal C_1$, has a turn less than or equal to $\theta$. Assuming the property is true for some $i\ge 1$, we denote by $a_i$ and $b_i$ the end points of $\mathcal C_i$. There exists a smooth point $c\in \mathcal C\setminus \mathcal C_i$ such that the geodesic distance from c to both $a_i$ and $b_i$ is greater than or equal to $k(\ell - \mathcal {L}(\mathcal C_i))$. If the arc from $a_i$ to c not passing through $b_i$ has a turn less than $\theta$, we set $a_{i+1}=c$, $b_{i+1}=b_i$ and $\mathcal C_{i+1}$ is the arc from $a_{i+1}$ to $b_{i+1}$ (that is from c to $b_i$) passing through $a_i$. Indeed, the other arc from $a_{i+1}$ to $b_{i+1}$ has a turn greater than $2\pi -2\theta$ by Fenchel’s Theorem (Property 3) and Property 7. So, since $\mathcal C$ is $(\theta ,\delta )$-LTB and $\mathrm d(a_{i+1}, b_{i+1})\le {\text {diam}}(\mathcal C)<\delta$, the turn of $\mathcal C_{i+1}$ is less than $\theta$. Moreover, $\mathcal C_i\subseteq \mathcal C_{i+1}$. If the arc from $a_i$ to c not passing through $b_i$ has a turn greater than $\theta$, we define $\mathcal C_{i+1}$ as the arc from $a_i$ to c passing through $b_i$ since it has a turn less than $\theta$ and we set $a_{i+1}=a_i$, $b_{i+1}=c$. In both cases, we have $\mathcal C_i\subseteq \mathcal C_{i+1}$ and $\mathcal {L}(\mathcal C_{i+1})\ge \mathcal {L}(\mathcal C_i) + k(\ell -\mathcal {L}(\mathcal C_i))$, and, since $\mathcal {L}(\mathcal C_i)\ge (1-(1-k)^i)\ell$ by induction hypothesis, we obtain $\mathcal {L}(\mathcal C_{i+1})\ge (1-(1-k)^{i+1})\ell$. This completes the induction. Now, on the one hand, considering the arc $\mathcal C_{\infty }=\bigcup \mathcal C_{i}$, we claim that $\mathcal C_{\infty }$ has a length greater than $(1-(1-k)^i)\ell$ for any positive integer i. Thus, $\mathcal {L}(\mathcal C_{\infty })=\ell$. Then, $\mathcal C\setminus \mathcal C_{\infty }$ is reduced to a point. On the other hand, since $\mathcal C_{\infty }$ is the supremum of an increasing sequence of arcs whose turns are bounded from above by $\theta$, it also has a turn-bounded from above by $\theta$. This contradicts Corollary 1. $\square$

Local Increase of the Distance for $(\theta ,\delta )$-LTB Curves

We now show that, when $\theta \le \pi /2$, the Euclidean distance d(p, q) between two points p and q of a parameterized LTB curve is locally monotonic in function of the parameter of one of the two points p and q. Visually, Proposition 12 states that ($\theta , \delta )$-LTB have no local U-turns (see Fig. 19).

Fig. 19

This configuration cannot occur in a $(\theta , \delta )$-LTB since the distance to the point c is not locally increasing ($d_1 <d_2$)

Proposition 12

Let $\theta \in (0,\pi /2]$ and ${\mathcal {C}}$ be a $(\theta ,\delta )$-LTB curve. Let $\gamma :[0, t_{M}] \rightarrow \mathcal C$ be an injective parametrization of the curve $\mathcal C$ and $t_{m}\in (0,t_M)$ be such that the arc $\gamma ([0,t_{m}])$ is included in $B(\gamma (0), \frac{\delta }{2})$. Then, the restriction of the function $t \mapsto \Vert \gamma (t)-\gamma (0)\Vert$ on $[0,t_{m}]$ is increasing.

Proof

We prove a contrapositive statement. Let $\phi :t\in [0,t_{m}] \mapsto \Vert \gamma (t)-\gamma (0)\Vert$. Suppose that $\phi$ is not monotonic. Then, there exists $t_1$, $t_2$ in $(0,t_m)$ such that $t_1<t_2$ and $\phi (t_1)>\phi (t_2)$. Therefore, the turn of the polygonal line $[\gamma (0),\gamma (t_1),\gamma (t_2)]$ is strictly greater than $\pi /2$. Hence, the turn of the arc $\gamma ([0, t_2])$ is a fortiori greater than $\pi /2$. Since $\mathcal C$ is $(\theta ,\delta )$-LTB for some $\theta <\pi /2$, the turn of the arc $\gamma ([t_2,t_M])$ is strictly less than $\pi /2$ and, according to Proposition 4, the arc $\gamma ([t_2,t_M])$ is therefore included in the disk of diameter $[\gamma (t_2),\gamma (t_M)]$ which is itself included in the ball

$B(\gamma (0),\delta /2)$ (for $\gamma (t_M)=\gamma (0)$). We conclude that the whole curve $\mathcal C$ is included in the ball $B(\gamma (0),\frac{\delta }{2})$. Then, the diameter of $\mathcal C$ is strictly less than $\delta$ which contradicts Lemma 1. $\square$

Proof of Proposition 7

Proposition 7 1

Let $\theta \in (0,\pi /2]$ and $\mathcal C$ be a $(\theta ,\delta )$-LTB Jordan curve. Let d be the diameter of $\mathcal C$ . Let T be a closed set included in an open disk B(c, r) with r less than or equal to $\min (\frac{1}{2}\delta , \frac{\sqrt{2}}{4} d).$ Then, the arc $\mathcal C_T$ passing through T is the unique arc of $\mathcal C$ of turn less than or equal to $\frac{\pi }{2}$ having its end points in T and such that the straightest arc between any two points of T is included in $\mathcal C_T$. Moreover,

$$\begin{aligned} (\mathcal C\setminus \mathcal C_T) \cap T= \emptyset . \end{aligned}$$

Proof

Observe that, obviously, any straightest arc between two points of $\mathcal C\cap T$ is included in $\mathcal C_T$. Furthermore,

$$\begin{aligned} \mathcal C\cap T = \bigcup _{a \in \mathcal C\cap T}{\mathcal C_{a}^{a}} \subset \mathcal C_T, \end{aligned}$$

where $\mathcal C_a^a=\{a\}$ is the straightest arc from a to a. Thus,

$$\begin{aligned} (\mathcal C\setminus \mathcal C_T) \cap T = \emptyset . \end{aligned}$$

Fig. 20

Step 4 of the proof of Proposition 7: the points $q_1$ and $q_2$ are in the ball of center c and of radius r. Then, the ball of diameter $[q_1,q_2]$ is included in the ball $B(c,\sqrt{2}r)$ for $\sin \frac{\phi }{2} + \cos \frac{\phi }{2}\le \sqrt{2}$

The proof is divided into six steps. In the first step, we show that $\mathcal C_T$ is an arc of $\mathcal C$. In the second step, we show that the ends of $\mathcal C_T$, $q_1$ and $q_2$ are in T. In the third step, we show that $\mathcal C_T$ is the straightest arc between $q_1$ and $q_2$. In the fourth step, we show that $\mathcal C_T$ is included in the open disk $B(c,\sqrt{2}r)$ (see Fig. 20). In the fifth step, we show that $\mathcal C_T$ is not the whole curve $\mathcal C$. In the sixth step, we show that $\mathcal C_T$ is the unique arc of $\mathcal C$ with turn less than or equal to $\pi /2$ and including any straightest arc of $\mathcal C$ between points of T.

Step 1. Let $p_1$ and $p_2$ be two points of $\mathcal C_T$. By Definition 6, there exist four points $a_1$, $b_1$, $a_2$, $b_2$ in T such that $p_1$ lies in the straightest arc $\mathcal C_{a_1}^{b_1}$ and $p_2$ lies in the straightest arc $\mathcal C_{a_2}^{b_2}$. Hence, the straightest arc between $a_1$ and $a_2$, which is included in $\mathcal C_T$, connects the arcs $\mathcal C_{a_1}^{b_1}$ and $\mathcal C_{a_2}^{b_2}$. Thereby, $p_1$ and $p_2$ are connected by a path in $\mathcal C_T$. We derive that $\mathcal C_T$ is path-connected: $\mathcal C_T$ is an arc of $\mathcal C$.

Step 2. The ends of $\mathcal C_T$, $q_1$ and $q_2$ are limits of points that are the ends of straightest arcs between two points in T. Indeed, for any $\epsilon >0$, there exists a point $q_1'$ of $\mathcal C_T$ such that the length of the subarc of $\mathcal C_T$, $\mathcal C_{q_1}^{q_1'}$ is less than $\epsilon$. The point $q_1'$ belongs to a straightest arc between two points in T, one of these two points is on the arc $\mathcal C_{q_1}^{q_1'}$, hence at geodesic distance from $q_1$ less than $\epsilon$ and a fortiori at Euclidean distance from $q_1$ less than $\epsilon$. The same holds for $q_2$. Since T is a closed set, $\mathcal C_T$ has its end points $q_1$ and $q_2$ in T.

Step 3. The straightest arc between $q_1$ and $q_2$ is included in $\mathcal C_T$ (by definition of $\mathcal C_T$). Then, $\mathcal C_T$ is the straightest arc between $q_1$ and $q_2$.

Step 4. (Figure 20) From Step 3 and Proposition 4, we derive that $\mathcal C_T$ is included in the disk of diameter $[q_1, q_2]$. By the hypotheses, the segment $[q_1, q_2]$ is included in B(c, r). Then, the arc $\mathcal C_T$ is included in the open ball $B(c, (\sin \frac{\phi }{2} + \cos \frac{\phi }{2} )r)\subseteq B(c, \sqrt{2}r)$ where $\phi$ is the geometric angle $\widehat{q_1 c q_2}$.

Step 5. By hypothesis, the diameter of $\mathcal C$ is greater than or equal to $2\sqrt{2}r$. Since $\mathcal C_T$ is included in the open disk ${B}(c,\sqrt{2}r)$, $\mathcal C_T$ cannot be the whole curve $\mathcal C$.

Step 6. If there exits another arc $\mathcal C'$ of curvature less than or equal to $\frac{\pi }{2}$ having its ends in T such that each other arc having its ends in T and of turn less than or equal to $\frac{\pi }{2}$ is included in $\mathcal C'$, then $\mathcal C' \subset \mathcal C_T$ and $\mathcal C_T \subset \mathcal C'$ hence $\mathcal C'= \mathcal C_T$. $\square$

Fig. 21

Thick: the polygonal line $P=[a_0,a_1,a_2,a_3,a_4,a_5]$. The four angles $\theta _1$$\theta _2$, $\theta _3$ and $\theta _4$ are subtended by the circle arcs defined, respectively, by the triples $(a_0,a_1,a_2)$, $(a_1,a_2,a_3)$, $(a_2,a_3,a_4)$ and $(a_3,a_4,a_5)$. Then, the turn of P is half the sum of the $\theta _i$: $\kappa (P)=\tfrac{1}{2}\sum _{i=1}^4\theta _i$

Turn of a Polygonal Line Inscribed in a Par-regular Curve

We establish two results allowing us to control the turn of a polygonal line inscribed in a par(r)-regular curve (Lemma 8 and Lemma 9). The first of these two lemmas is an easy consequence of the inscribed angle theorem. It is illustrated in Fig. 21. Its proof is left to the reader.

Lemma 8

Let P be a polygonal line $[a_i]_{i=0}^N$. Then,

$$\begin{aligned} \kappa (P)&= \frac{1}{2} \sum _{i=1}^{N-1}\theta _i\\&= \sum _{i=1}^{N-1} \arcsin \left( \frac{\Vert a_i-a_{i-1}\Vert }{2r_i}\right) \\&\quad +\arcsin \left( \frac{\Vert a_{i+1}-a_{i} \Vert }{2r_i}\right) , \end{aligned}$$

where, for any $i\in \llbracket 1,N-1 \rrbracket$, $r_i$ is the radius of the circumcircle $C_i$ of the triangle $a_{i-1}a_{i}a_{i+1}$ and $\theta _i$ is the central angle of $C_i$ subtended by the arc $[a_{i-1},a_i,a_{i+1}]$.

Lemma 9

Let $P=[a_i]_{i=0}^N$ be a polygonal line inscribed in a par(r)-regular curve. If the maximal edge length of P is less than 2r, then the turn of P is less than the turn of a polygonal line which has the same edge length sequence and is inscribed in a circle of radius r. In other words,

$$\begin{aligned} \kappa (P) \le&\arcsin \left( \frac{\Vert a_1-a_{0}\Vert }{2r}\right) \\&+ 2 \sum _{i=1}^{N-2} \arcsin \left( \frac{\Vert a_{i+1}-a_{i} \Vert }{2r}\right) \\&+\arcsin \left( \frac{\Vert a_N-a_{N-1} \Vert }{2r}\right) . \end{aligned}$$

Fig. 22

Thick, black: a polygonal line $[a_{i-1},a_i,a_{i+1}]$ inscribed in a par(r)-regular curve (not depicted). The circles of par(r)-regularity are drawn in red. Thin, orange: a polygonal line $[a'_{i-1},a'_i,a'_{i+1}]$ inscribed in a circle with radius r (the circle is the osculating circle that is on the side of the turn of the oriented polygonal line $[a_{i-1},a_i,a_{i+1}]$). The polygonal lines have the same edge length sequence (Color figure online)

Proof

Figure 22 illustrates the main argument of the proof. Let ${\mathcal {C}}$ be a par(r)-regular curve and $P=[a_i]_{i=0}^N$ be a polygonal line inscribed in ${{\mathcal {C}}}$. For any $i \in \llbracket 0, N-1 \rrbracket$, we set $\ell _i= \Vert a_{i+1} -a_i\Vert$ and we assume $\ell _i \le 2r$. Let $P'=[a'_i]_{i=0}^N$ be a polygonal line inscribed in a circle with radius 2r such that $\ell _i= \Vert a'_{i+1} -a'_i\Vert$ for any $i\in \llbracket 0, N-1 \rrbracket$.

For each $i \in \llbracket 1, N-1\rrbracket$, let us denote, respectively, by $\kappa _{i}$ and $\kappa '_{i}$ the turn of the polygonal lines $[a_{i-1}, a_{i}, a_{i+1}]$ and $[a'_{i-1}, a'_{i}, a'_{i+1}]$. On the one hand, from Lemma 8 (“Appendix E”, we have $\kappa '_{i}=\arcsin \left( \frac{\ell _{i-1}}{2r}\right) +\arcsin \left( \frac{\ell _{i+1}}{2r}\right)$. On the other hand, the turns $\kappa ([a_{i-1},a_i,a_{i+1}])$ and $\kappa ([a'_{i-1},a'_i,a'_{i+1}])$ are, respectively, the supplementary angles of $\widehat{a_{i-1}a_ia_{i+1}}$ and $\widehat{a'_{i-1}a'_ia'_{i+1}}$. Then, from the definition 10 of the par-regularity, $\kappa ([a_{i-1},a_i,a_{i+1}]) \le \kappa ([a'_{i-1},a'_i,a'_{i+1}])$ (see Fig. 22). As $\kappa (P)=\sum _{i=1}^N\kappa _i$ and $\kappa (P')=\sum _{i=1}^N\kappa '_i$, we conclude straightforwardly. $\square$