Image Edge Detection: A New Approach Based on Fuzzy Entropy and Fuzzy Divergence

Abstract

In image pre-processing, edge detection is a non-trivial task. Sometimes, images are affected by vagueness so that the edges of objects are difficult to distinguish. Hence, the usual edge-detecting operators can give unreliable results, thus necessitating the use of fuzzy procedures. In literature, Chaira and Ray approach is a popular technique for fuzzy edge detection in which fuzzy divergence formulation is exploited. However, this approach does not specify the threshold technique must be applied. Then, in this work, starting from Chairy and Ray procedure, we present a new fuzzy edge detector based on both fuzzy divergence (thought and proved to be a distance) and fuzzy entropy minimization for the thresholding sub-step in gray-scale images. Eddy currents, thermal infrared, and electrospinning images were used to test the proposed procedure after their fuzzification by a suitable adaptive S-shaped fuzzy membership function. Moreover, the fuzziness content of each image has been quantified by new specific indices proposed here and formulated in terms of fuzzy divergence. The results have been evaluated by suitable assessment metrics here formulated and are considered to be encouraging when qualitatively and quantitatively compared with those obtained by some well-known I- and II-order edge detectors.

A Proof of Theorem 1

Remark 9

Let us consider three fuzzy images, $F({\bf{I}}_{norm})_j$, $j=1,2,3$, whose gray levels are $\hat{a_{ij}}$, $\hat{b_{ij}}$, and $\hat{c_{ij}}$ $\in X$, respectively. Let us say, for simplicity, that

$$\begin{aligned} m_{F({\bf{I}}_{norm})_1}(\hat{a_{ij}})-m_{F({\bf{I}}_{norm})_2}(\hat{b_{ij}})= & {} \alpha ,\nonumber \\ m_{F({\bf{I}}_{norm})_2}(\hat{b_{ij}})-m_{F({\bf{I}}_{norm})_3}(\hat{c_{ij}})= & {} \beta ,\nonumber \\ m_{F({\bf{I}}_{norm})_3}(\hat{c_{ij}})-m_{F({\bf{I}}_{norm})_1}(\hat{a_{ij}})= & {} \gamma , \end{aligned}$$

(26)

such that $\alpha +\beta +\gamma =0.$ Further, with (26) taken into account, $d({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})$, $d({F({\bf{I}}_{norm})_2},{F({\bf{I}}_{norm})_3})$, and

$d({F({\bf{I}}_{norm})_3},{F({\bf{I}}_{norm})_1})$, as the generic addends of $D({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})$, $D({F({\bf{I}}_{norm})_2},{F({\bf{I}}_{norm})_3})$, and $D({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_1})$, can be written as follows:

$d({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})=2- (1-\alpha ))\rm{{e}}^{\alpha }- (1+\alpha )\rm{{e}}^{-\alpha }$, $d({F({\bf{I}}_{norm})_2},{F({\bf{I}}_{norm})_3})=2- (1-\beta ))\rm{{e}}^{\beta }- (1+\beta ) )\rm{{e}}^{-\gamma }$, $d({F({\bf{I}}_{norm})_3},{F({\bf{I}}_{norm})_1})=2- (1-\gamma ))\rm{{e}}^{\gamma }- (1+\gamma ) )\rm{{e}}^{-\gamma }$. If they have the values

$$\begin{aligned} d({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})\ge & {} 0, \end{aligned}$$

(27)

$$\begin{aligned} d({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})= & {} 0\;\Leftrightarrow A=B, \end{aligned}$$

(28)

$$\begin{aligned} d({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})& = d({F({\bf{I}}_{norm})_2},{F({\bf{I}}_{norm})_1}), \end{aligned}$$

(29)

$$\begin{aligned} d({F({\bf{I}}_{norm})_1},{F({\bf{I}}_{norm})_2})& \le d({F({\bf{I}}_{norm})_2},{F({\bf{I}}_{norm})_3}) + d({F({\bf{I}}_{norm})_3},{F({\bf{I}}_{norm})_1}), \end{aligned}$$

(30)

then, with the double summation operator applied to them, (8), (9), (10), and (11) apply.

We prove (27).

If (27) is true, then $d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)=2-(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha }\ge 0$, from which $2\ge (1-\alpha )\rm{{e}}^{\alpha }+(1+\alpha )\rm{{e}}^{-\alpha }=\rm{{e}}^{\alpha }-\alpha \rm{{e}}^{\alpha }+\rm{{e}}^{-\alpha }\alpha \rm{{e}}^{-\alpha }=(\rm{{e}}^{\alpha }+\rm{{e}}^{-\alpha })-\alpha (\rm{{e}}^{\alpha }-\rm{{e}}^{-\alpha }).$ Then, $2\ge 2\cosh (\alpha )-2\alpha \sinh (\alpha )$ and, again, $1\ge \cosh (\alpha )-\alpha \sinh (\alpha ).$ Now, we set $f(\alpha )=\cosh (\alpha )-(\alpha )\sinh (\alpha )$. However, aiming to search for the minimum value of $f(\alpha )$, we impose $f'(\alpha )=\sinh (\alpha )-\sinh (\alpha )-\alpha \cosh (\alpha )=0$ to achieve $\alpha \cosh (\alpha )=0.$ $\cosh (\alpha )=0$ is never null, hence the stationary value of $f(\alpha )$ that one has for $\alpha =0$. Again, if $\alpha =0$, $f(\alpha )=1$, while, if $\alpha =1$, $f(\alpha )=\frac{1}{e}<1$. From this, $\alpha$ is a point of maximum for $f(\alpha )$. Thus, $1\ge \cosh (\alpha )-\alpha \sinh (\alpha )$ is always true. Then, given Remark 9, inequality (8) is also verified.

We prove (28).

The sufficient condition is easy to prove. In fact, if $F({\bf{I}}_{norm})_1=F({\bf{I}}_{norm})_2$, then $\alpha =0$. Thus, $d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)=2-(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha }=0$. Vice versa, to prove the necessary condition, we impose $d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)=0$, obtaining $2=(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha }=0.$ Further, it is reasonable to write the following chain of equalities: $2=(1-\alpha )\rm{{e}}^{\alpha }+(1+\alpha )\rm{{e}}^{-\alpha }=\rm{{e}}^{\alpha }-\alpha \rm{{e}}^{\alpha }+\rm{{e}}^{-\alpha }+\alpha \rm{{e}}^{-\alpha }=\rm{{e}}^{\alpha }+\rm{{e}}^{-\alpha }-\alpha (\rm{{e}}^{\alpha }-\rm{{e}}^{-\alpha })-\alpha (\rm{{e}}^{\alpha }-\rm{{e}}^{-\alpha })=\rm{{e}}^{\alpha }+\rm{{e}}^{-\alpha }-\alpha (\rm{{e}}^{\alpha }-\rm{{e}}^{-\alpha })=2\cosh (\alpha )-2(\alpha )\sinh (\alpha )$ to achieve $1=\cosh (\alpha )-\alpha \sinh (\alpha )$ verified iff $\alpha =0$. Thus, (28) is verified. Finally, by Remark 9, (9) is verified.

We prove (29).

$d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)=2-(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha }=2-(1-\alpha )\rm{{e}}^{-\alpha }-(1+\alpha )\rm{{e}}^{-\alpha }=d(F({\bf{I}}_{norm})_2,F({\bf{I}}_{norm})_1)$.

We prove (30).

$d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)=2-(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha }\le 2-(1-\beta )\rm{{e}}^{\beta }-(1+\beta )\rm{{e}}^{-\beta }+2-(1-\gamma )\rm{{e}}^{\gamma }-(1+\gamma )\rm{{e}}^{-\gamma }$, from which

$$\begin{aligned} &d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2) \le 2-(1+\alpha +\gamma )\rm{{e}}^{-\alpha -\gamma }\nonumber \\&\quad -\,(1-\alpha -\gamma )\rm{{e}}^{\alpha +\gamma }+\nonumber \\&\quad \times\,\gamma \rm{{e}}^{\alpha +\gamma }+2-(1-\gamma )\rm{{e}}^{\gamma }-(1+\gamma )\rm{{e}}^{-\gamma }. \end{aligned}$$

(31)

By (31), we can write $d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)\le 2-(1+\alpha )^{-\alpha -\gamma }-\gamma \rm{{e}}^{-\alpha -\gamma }-(1-\alpha )\rm{{e}}^{\alpha +\gamma }+\gamma \rm{{e}}^{\alpha +\gamma }+2-(1-\gamma )\rm{{e}}^{\gamma }-(1+\alpha )\rm{{e}}^{-\alpha }.$ Further, by adding and subtracting $(1+\alpha )\rm{{e}}^{-\alpha }+(1-\alpha )\rm{{e}}^{\alpha }$, we obtain

$$\begin{aligned}&d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)\le 2-(1+\alpha )\rm{{e}}^{-\alpha }-(1-\alpha )\rm{{e}}^{\alpha }+(1+\alpha )\rm{{e}}^{-\alpha }+(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha )-\gamma }-\gamma \rm{{e}}^{-\alpha -\gamma }-(1-\gamma )\rm{{e}}^{\alpha +\gamma }+\gamma \rm{{e}}^{\alpha +\gamma }+2-(1-\gamma )\rm{{e}}^{\gamma }-(1+\gamma )\rm{{e}}^{-\gamma }. \end{aligned}$$

(32)

In (32), $2-(1+\alpha )\rm{{e}}^{-\alpha }-(1-\alpha )\rm{{e}}^{\alpha }=d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)$, while $2-(1-\gamma )\rm{{e}}^{\gamma }-(1+\gamma )\rm{{e}}^{-\gamma }=d(C,A).$ Then, (32) becomes

$$\begin{aligned} d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)&\le\,d(F({\bf{I}}_{norm})_1,F({\bf{I}}_{norm})_2)+(1+\alpha )\rm{{e}}^{-\alpha }\\&\quad+\,(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha -\gamma }- \gamma \rm{{e}}^{-\alpha -\gamma }\\&\quad-\,(1-\alpha )\rm{{e}}^{\alpha +\gamma }+d(F({\bf{I}}_{norm})_3,F({\bf{I}}_{norm})_1), \end{aligned}$$

(33)

thereby reducing the problem to show that $(1+\alpha )\rm{{e}}^{-\alpha }+(1-\alpha )\rm{{e}}^{\alpha }-(1+\alpha )\rm{{e}}^{-\alpha -\gamma }-\gamma \rm{{e}}^{-\alpha -\gamma }-(1-\alpha )\rm{{e}}^{\alpha +\gamma }$ in (33) is not negative. However, $(1+\alpha )\rm{{e}}^{-\alpha }\ge 0$ and $(1-\alpha )\rm{{e}}^{\alpha }\ge 0.$. Thus, it remains to be shown that $-(1+\alpha )\rm{{e}}^{-\alpha -\gamma }-\gamma )\rm{{e}}^{-\alpha -\gamma }-(1-\alpha )\rm{{e}}^{\alpha +\gamma }+\gamma \rm{{e}}^{\alpha +\gamma }\ge 0$. If, absurdly, it were $-(1+\alpha )\rm{{e}}^{-\alpha -\gamma }-\gamma \rm{{e}}^{-\alpha -\gamma }-(1-\alpha )\rm{{e}}^{\alpha +\gamma }+\gamma \rm{{e}}^{\alpha +\gamma }< 0,$ we would get

$$\begin{aligned} \gamma \rm{{e}}^{\alpha +\gamma } < (1+\alpha )\rm{{e}}^{-\alpha -\gamma }+\gamma \rm{{e}}^{-\alpha -\gamma }+(1-\alpha )\rm{{e}}^{\alpha +\gamma }=(1+\alpha +\gamma )\rm{{e}}^{-\alpha -\gamma }+(1-\alpha )\rm{{e}}^{\alpha +\gamma }, \end{aligned}$$

(34)

from which $\gamma <(1+\alpha +\gamma )\frac{\rm{{e}}^{-\alpha -\gamma }}{\rm{{e}}^{\alpha +\gamma }}+(1-\alpha ).$ If (34) is always true, it is necessary that $\sup \{\gamma \}<\inf \{ (1+\alpha +\gamma )\frac{\rm{{e}}^{-\alpha -\gamma }}{\rm{{e}}^{\alpha +\gamma }}+(1-\alpha )\}$. However, $\inf \{ 1-m_B(b_{ij})-m_A(a_{ij}) \}=0$ for which $\inf \{1+\alpha +\gamma \}=0$ and $\sup \{\gamma \}=1$ which results in a false inequality. Thus, (30) is true and, by Remark 9, (11) holds.