A Projection-Based Method for Shape Measurement

Abstract

This work addresses two main contributions for shape measurement: First, a new circularity measure for planar shapes is introduced based on their geometrical properties in the projection space of Radon transform. Second, a general-purpose evaluation criterion, power of discrimination, for assessing the efficiency of a shape measure is proposed. The new measure ranges over the interval [0, 1] and produces the value 1 if and only if the measured shape is a perfect circle. The proposed measure is invariant with respect to translation, rotation and scaling transformations. Moreover, it is also robust against border distortion of shapes. It is theoretically well founded and can be extended to other problems of shape measurement. Our approach can deal with complex shapes composed of connected components that cannot be handled by classical contour-based methods. Several experiments show its good behavior and demonstrate the efficiency and applicability of our proposed measure. Finally, we also consider our proposed evaluation criterion for assessing different circularity measures.

A Material for the Inverse Problem

The following lemmas establish the relation between convex and non-convex shapes as well as the properties of a convex shape satisfying the conditions in Proposition 2.

Lemma 1

If an arbitrary shape $\mathcal {D}$ satisfies this condition: $\frac{1}{2}\varLambda ^\theta (\mathcal {D})=\varDelta ^\theta _1(\mathcal {D})=\varDelta ^\theta _2(\mathcal {D})=r, \forall \theta \in [0,\pi )$, its convex hull $CV(\mathcal {D})$ satisfies the above condition too.

Proof

First, we will prove that any extremity $E_1^\theta$, $E_2^\theta$ is not in a concave part of the boundary of $\mathcal {D}$ for any direction $\theta$. Suppose that $E_1^\theta$ is in concave part. It means that there exist 2 points A and B on the boundary of $\mathcal {D}$ nearby $E_1^\theta$ at the left and right sides such that $E_1^\theta$ is inside of triangle $E_2^\theta AB$ (see Fig. 18). So, $\max (AE_2^\theta , BE_2^\theta )>E_1^\theta E_2^\theta =2r$. This fact is contradictory to the above condition of $\mathcal {D}$. On the other hand, it is evident that $\mathcal {D}$ and its convex hull $CV(\mathcal {D})$ have a same projected band in any direction. From the two above facts, we can deduce the conclusion of this lemma. $\square$

Fig. 18

Lemma 1

Lemma 2

Let us denote the two extremities in the boundary of $\mathcal {D}$ corresponding to $\rho _0^\theta (\mathcal {D})$ as $E_1^\theta$ and $E_2^\theta$, respectively (see also Sect. 3.1.2 for more details). $E_1^\theta E_2^\theta$ and $E_1^{\theta +\frac{\pi }{2}} E_2^{\theta +\frac{\pi }{2}}$ intersect at the midpoint of each segment $\forall \theta \in [0,\frac{\pi }{2})$.

Proof

Let us consider direction $\theta$, we have the following condition: $E_1^\theta E_2^\theta =2r$ because of $\frac{1}{2}\varLambda ^\theta (\mathcal {D})=r$. Therefore, in direction $\theta +\frac{\pi }{2}$, the length of $\mathcal {D}$ is at least 2r and the minimal value is obtained when $E_1^{\theta }$ and $E_2^{\theta }$ must be on the two supporting lines of $\mathcal {D}$ in this direction, respectively. In Fig. 19, these supporting lines, namely $d_1$ and $d_2$, are perpendicular with $E_1^\theta E_2^\theta$ at $E_1^\theta$ and $E_2^\theta$, respectively. In addition, $\varDelta ^{\theta +\frac{\pi }{2}}_1(\mathcal {D})+\varDelta ^{\theta +\frac{\pi }{2}}_2(\mathcal {D})=2r$. Therefore, in direction $\theta +\frac{\pi }{2}$, the projected band of $\mathcal {D}$ must be sandwiched between two lines $d_1$ and $d_2$ (see Fig. 19).

Because of $\varDelta ^\theta _1(\mathcal {D})=\varDelta ^\theta _2(\mathcal {D})=r, \forall \theta \in [0,\pi )$, $E_1^{\theta +\frac{\pi }{2}} E_2^{\theta +\frac{\pi }{2}}$ must be equidistant from $d_1$ and $d_2$. So, we have $E_1^\theta O =E_2^\theta O$. Similarly, by considering direction $\theta +\frac{\pi }{2}$, we obtain $E_1^{\theta +\frac{\pi }{2}}O= E_2^{\theta +\frac{\pi }{2}}O$. $\square$

Lemma 3

Suppose that O (resp. $O^\theta$) is the intersection between $E_1^0 E_2^0$ and $E_1^{\frac{\pi }{2}} E_2^{\frac{\pi }{2}}$ (resp. $E_1^\theta E_2^\theta$ and $E_1^{\theta +\frac{\pi }{2}} E_2^{\theta +\frac{\pi }{2}}$) (please see Sect. 3.1.2 for the definitions of $E^\theta _1$ and $E^\theta _2$). We have the following property: $OO^\theta \le \sqrt{2}r (\sqrt{5-4\cos (\theta )}-1)$

Fig. 19

Lemma 2

Fig. 20

Lemma 3

Proof

Lemma 2 allows to deduce that the border of $\mathcal {D}$ is decomposed into a set of tuples ($E_1^\theta$, $E_2^\theta$, $E_1^{\theta +\frac{\pi }{2}}$, $E_2^{\theta +\frac{\pi }{2}}$) where the intersection $O^\theta$ is the midpoints of two segments $E_1^\theta E_2^\theta$, $E_1^{\theta +\frac{\pi }{2}} E_2^{\theta +\frac{\pi }{2}}$. Suppose that ${\zeta }(O,r)$ is the circle of center O, of radius r. Without loss of generality, suppose that $O^\theta$ is in the fourth octant (see Fig. 20). In addition, for the simplicity of presentation, we consider $\theta \le \frac{\pi }{4}$ and the other cases can be considered similarly. We then denote $O_1$ (resp. $O_2$) the projection of O on $E_1^\theta E_2^\theta$ (resp. $E_1^{\theta +\frac{\pi }{2}} E_2^{\theta +\frac{\pi }{2}}$); $X^\theta$ and $\overline{E_1^\theta }$ (resp. $X^{\theta +\frac{\pi }{2}}$ and $\overline{E_1^{\theta +\frac{\pi }{2}}}$) are the intersections between $OE_1^\theta$ (resp. $OE_1^{\theta +\frac{\pi }{2}}$) and ${\zeta }(O,r)$ and ${\zeta }(E_2^0,2r)$ (resp. ${\zeta }(O,r)$ and ${\zeta }(E_2^{\frac{\pi }{2}},2r)$).

In this condition, it is evident to deduce that $E_1^\theta$ and $E_1^{\theta +\frac{\pi }{2}}$ are outside or on the boundary of ${\zeta }(O,r)$. In addition, because the length of $\mathcal {D}$ in any direction is a constant (2r), we deduce that $E_2^0 E_1^\theta \le 2r$ and $E_2^{\frac{\pi }{2}}E_1^{\theta +\frac{\pi }{2}}\le 2r$. Therefore, $E_1^\theta$ (resp. $E_1^{\theta +\frac{\pi }{2}}$) must be in the small zone determined by 3 circles: ${\zeta }(O,r)$, ${\zeta }(E_2^0,2r)$, ${\zeta }(E_2^{\frac{\pi }{2}},2r)$ (resp. ${\zeta }(O,r)$, ${\zeta }(E_1^0,2r)$, ${\zeta }(E_2^{\frac{\pi }{2}},2r)$) (see also Fig. 20). So, we have:

$$\begin{aligned} E_1^\theta X^\theta \le \overline{E_1^\theta } X^\theta \text { and } E_1^{\theta +\frac{\pi }{2}} X^{\theta +\frac{\pi }{2}}\le \overline{E_1^{\theta +\frac{\pi }{2}}} X^{\theta +\frac{\pi }{2}} \end{aligned}$$

(11)

Moreover, because $O_1$ and $O_2$ are projections of O, it is evident that $E_1^\theta O_1 \le E_1^\theta O$$\Leftrightarrow E_1^\theta O^\theta + O^\theta O_1 \le E_1^\theta X^\theta +X^\theta O$$\Leftrightarrow$$r+O^\theta O_1\le E_1^\theta X^\theta +r \Leftrightarrow O^\theta O_1\le E_1^\theta X^\theta$ and $E_1^{\theta +\frac{\pi }{2}} O_2 \le E_1^{\theta +\frac{\pi }{2}} O$$\Leftrightarrow E_1^{\theta +\frac{\pi }{2}} O^\theta + O^\theta O_2 \le E_1^{\theta +\frac{\pi }{2}} X^{\theta +\frac{\pi }{2}} +X^{\theta +\frac{\pi }{2}} O$$\Leftrightarrow$$r+O^\theta O_2\le E_1^{\theta +\frac{\pi }{2}} X^{\theta +\frac{\pi }{2}} +r \Leftrightarrow O^\theta O_2\le E_1^{\theta +\frac{\pi }{2}} X^{\theta +\frac{\pi }{2}}$. Briefly, we have:

$$\begin{aligned} O^\theta O_1\le E_1^\theta X^\theta \text { and } O^\theta O_2\le E_1^{\theta +\frac{\pi }{2}} X^{\theta +\frac{\pi }{2}} \end{aligned}$$

(12)

Consider now the triangle $\overline{E_1^\theta } O E_2^0$, we have $\overline{E_1^\theta } O{=}\sqrt{{|\overline{E_1^\theta } E_2^0|}^2{+}{|O E_2^0|}^2{-}2|\overline{E_1^\theta } E_2^0| |O E_2^0| \cos (\angle \overline{E_1^\theta } E_2^0 O) }$$\Leftrightarrow \overline{E_1^\theta }X^\theta {+}r {=}\sqrt{5r^2{-}4r^2\cos (\angle \overline{E_1^\theta } E_2^0 O)}$$\Leftrightarrow \overline{E_1^\theta }X^\theta {=}r(\sqrt{5{-}4\cos (\angle \overline{E_1^\theta } E_2^0 O)}{-}1)$. Because of $\angle \overline{E_1^\theta } E_2^0 O \le \theta$, we deduce that $\overline{E_1^\theta }X^\theta \le r(\sqrt{5{-}4\cos (\theta )}{-}1)$. Similarly, by considering the triangle $\overline{E_1^{\theta {+}\frac{\pi }{2}}} O E_2^{\frac{\pi }{2}}$, the following remarks are obtained.

$$\begin{aligned} \overline{E_1^\theta }X^\theta \text { and } \overline{E_1^{\theta +\frac{\pi }{2}}} X^{\theta +\frac{\pi }{2}} \le r(\sqrt{5-4\cos (\theta )}-1) \end{aligned}$$

(13)

Using the results in Eqs. (11), (12), (13), we have the conclusion of this lemma: $OO^\theta =\sqrt{|OO_1|^2+|OO_2|^2} \le \sqrt{2}r(\sqrt{5-4\cos (\theta )}-1)$. $\square$

Lemma 4

$\lim \limits _{n \rightarrow +\infty } n (\sqrt{5-4\cos (\frac{\pi }{4n})}-1)=0$

Proof

Due to Taylor’s series, we have $1-\frac{x^2}{2}\le cos(x) \le 1$; therefore, $1\le 5-4\cos (x)\le 1+2x^2\le (1+x^2)^2$. So, we have $0\le \sqrt{5-4\cos (x)} -1\le x^2$. Replacing x by $\frac{\pi }{4n}$, we obtain the following result $0\le n (\sqrt{5-4\cos (\frac{\pi }{4n})}-1)\le \frac{\pi ^2}{16n}$. It is evident that $\lim \limits _{n \rightarrow +\infty }\frac{\pi ^2}{16n}=0$. This fact proves our conclusion. $\square$