Fuzzy histograms, weak fuzzification and accumulation of periodic quantities

Abstract

The influence of the scale of a fuzzy membership function used to fuzzify a histogram is analysed. It is shown that for a class of fuzzifying functions it is possible to indicate the limit for fuzzification, at which the mode of the histogram equals the mean of the data accumulated in it. The fuzzification functions for which this appears are: the quadratic function for aperiodic histograms and the cosine square function for periodic ones. The scaled and clipped versions of these functions can be used to control the degree of fuzzification belonging to the interval [0,1]. While the quadratic function is related to the widely known Huber-type clipped mean or the kernel function derived from the Epanechnikov kernel, the clipped cosine square seems to be less known. The indications for using strong or weak fuzzification, according to the value of the fuzzification degree, are justified by examples in two applications: classic Hough transform-based image registration and novel accumulation-based line detection. Typically, the weak fuzzification is recommended. The images used are related to simulation images from teleradiotherapy and to mammographic images.

Proofs of the properties

The proofs of the properties presented in the paper are simple and consist solely in straightforward transformations; nevertheless, they seem necessary to justify the statements made.

Proof (of Property 1: Quadratic fuzzifying function) The fuzzifying function according to (9) is convex in the real interval [i _min,i _max], so (11) has a unique maximum when

$$ \frac{\partial H_{{\rm f}2}(x)}{\partial x} = \sum\limits_{i\in I} H(i) \frac{\partial \mu_2(x-i)}{\partial x} = 0. $$

(38)

Substituting (9) and expanding, one gets

$$ -\frac{2a}{s^2} \left[ x\sum\limits_{i\in I} H(i) - \sum\limits_{i\in I} i H(i) \right] = 0, $$

which holds when

$$ x = x_{\rm m} = \frac{\sum_{i\in I} iH(i)}{\sum_{i\in I} H(i)}.$$

(39)

This is the formula for the mean of the histogram. □

Proof (of Property 2: Symmetrical fuzzifying function) If the function μ(x), x∈X, X ≡ [i _min  −  i _max, i _max  −  i _min] can be expanded into a Maclaurin series (see, e.g. [57]), then

$$ f(x/s) = f(0) + \frac{x^2}{2! s^2} f''(0) + \frac{x^4}{4! s^4} f^{(4)}(0) + \cdots $$

(40)

or, using an explicit form of the remainder and substituting f(0) = 1, f′(0) = 0 due to the presence of a maximum

$$\begin{aligned} f(x/s) &= 1 + \frac{x^2}{2! s^2} f''(0) + R_4, \; \hbox{where}\\ R_4 &= \frac{x^4}{4! s^4}f^{(4)} (\theta x) \;\hbox{and}\; \theta \in [0,1].\\ \end{aligned}$$

(41)

All the even-order derivatives are zero for x = 0 due to symmetry. Note (Fig. 4) that in application to the fuzzification of a histogram, X is the domain of f(x) and also the interval to which the real expression θ x belongs. If the derivative f ⁽⁴⁾(·) is bounded, i.e.

$$ \forall_{\kappa \in X} \; \exists_{c_4 > 0} \; |f^{(4)}(\kappa)| < c_4, $$

(42)

where κ = θ x, then it is possible to indicate the value of the scale parameter s = s ₄ for which the remainder R ₄ is arbitrarily small:

$$ \forall_{\varepsilon > 0} \; \forall_{x,\kappa \in X} \; \exists_{s_4} \left|\frac{x^4}{4! s_4^4}f^{(4)}(\kappa) \right| <\varepsilon. $$

(43)

After elementary transformations the necessary condition is received:

$$ s > s_4, \;\hbox{where}\; s_4 = x_{\max} \sqrt[4]{\frac{c_4}{4! \varepsilon}}, \; x_{\max} = i_{\max} - i_{\min}. $$

(44)

Then, the membership function (15) is arbitrarily close to the quadratic function (9), with a =  − f′′(0)/2!.

Similarly, it can be shown that the derivative of the function (15) is arbitrarily close to the derivative of the quadratic function for a sufficiently large value of s, if the fifth derivative f ⁽⁵⁾(·) is bounded, i.e.

$$ \forall_{\kappa \in X} \; \exists_{c_5 > 0} \; |f^{(5)}(\kappa)| < c_5. $$

(45)

To do this it is necessary to differentiate the expression (41) with respect to x. The problem resolves to showing that it is possible to indicate a value of the parameter s = s ₅ for which the derivative of the remainder ∂R ₄/ ∂x is arbitrarily small:

$$ \forall_{\varepsilon > 0} \; \forall_{x,\kappa \in X} \; \exists_{s_5} \; \left| \frac{4 x^3}{4! s_5^4} f^{(4)}(\kappa) + \frac{x^4\theta}{4! s_5^4} f^{(5)}(\kappa) \right| < \varepsilon. $$

(46)

The upper bound for x is x _max, for θ is 1, for f ⁽⁴⁾(κ) is c ₄ by virtue of (42), and for f ⁽⁵⁾(κ) is c ₅ by virtue of (45). The moduli can be replaced by their arguments, as the estimations of both addends are positive. It can then be written

$$ \frac{4 c_4 x^3_{\max} + c_5 x^4_{\max}}{4! s_5^4} < \varepsilon. $$

Finally, combining with condition (44), one gets

$$\begin{aligned} s &> \max(s_4,s_5),\; \hbox{where} \\ s_4 &= x_{\max} \sqrt[4]{\frac{c_4}{4! \varepsilon}},\\ s_5 &= x_{\max} \sqrt[4]{\frac{4c_4/x_{\max} + c_5}{4! \varepsilon}},\\ \end{aligned}$$

(47)

which ends the considerations. Boundedness of the fourth and fifth derivatives is equivalent to the continuity of the function up to the fourth derivative. □

Proof (of Properties 3: Cosine square fuzzifying function, and 4: Intensity of the fuzzy periodic histogram) In the proof the basic properties of the harmonics will be used.

Let us recall the formula (22) for the fuzzy histogram:

$$ H_{\rm fc}(\xi)=\sum\limits_{i\in I} H(i)\mu_{\rm c}(\xi-i) $$

(48)

and for the fuzzifying function (20):

$$ \mu_{\rm c}(\xi) = \cos^2(\pi \phi(\xi)/T). $$

(49)

It holds that $\cos^2(\beta) = (1/2) \cos (2\beta) + (1/2),$ so the function (49) can be rewritten as

$$ \mu_{\rm c}(\xi) = \frac{1}{2} \cos (2\pi \phi(\xi)/T) + \frac{1}{2}.$$

(50)

In the process of accumulation, the multiplicative and additive constants change the values of the histogram elements so that these values are systematically affine transformed, which does not change the relations between them. At this point the requirement that the fuzzifying function as well as the histogram values must be non-negative is postponed. What has been called fuzzification can now be treated as mere convolution. In place of (49) the following function will be used:

$$ \nu_T(\xi) = \cos(2\pi \phi(\xi)/T) $$

(51)

and the convolution will be

$$ H_{\rm cc}(\xi) = \sum\limits_{i\in I} H(i) \nu_T(\xi-i) = \sum\limits_{i\in I} H(i)\cos(2\pi \phi(\xi-i)/T). $$

(52)

By using the period ratio τ defined by (23) and recalling that ϕ(ξ) is linear in Ξ, one gets

$$ H_{\rm cc}(\xi)=\sum\limits_{i\in I} H(i)\cos(\tau\phi(\xi)-\tau\phi(i)). $$

(53)

This formulation is equivalent to the summation of harmonics, where H(i) are amplitudes, τ represents the constant frequency, and τϕ(i) represents phases, different for each summand. The sum of harmonics is a harmonic, so (53) can be written as

$$ H_{\rm cc} (\xi) = \sum\limits_{i\in I} H(i)\cos(\tau\phi(\xi)-\beta_i) = H \cos(\tau\phi(\xi)-\beta), $$

(54)

where the phases are

$$ \beta_i=\tau\phi(i) $$

(55)

and the known formulae for addition of harmonics are

$$\begin{aligned} H &= \sqrt{H_x^2+H_y^2},\\ \sin\beta &= H_x/H, \\ \cos\beta &= H_y/H, \\ H_x &= \sum\limits_{i\in I} H(i)\cos\beta_i = \sum\limits_{i\in I} H_{xi},\\ H_y &= \sum\limits_{i\in I} H(i)\sin\beta_i = \sum\limits_{i\in I} H_{yi}.\\ \end{aligned}$$

(56)

Having in mind (54), with the cosine function having the maximum at τϕ(ξ) = β, it can be easily seen that β is the angle measure of the mode of the histogram, and

$$\xi_{\rm m} = \xi(\beta/\tau) $$

(57)

where ξ(·) = ϕ⁻¹(·).

Returning to the full form of the fuzzifying function (50), the formula (48) can be rewritten as

$$\begin{aligned} H_{\rm fc}(\xi) &= \frac{1}{2} \sum\limits_{i\in I} H(i)\cos(2\pi \phi(\xi-i)/T) + \frac{1}{2} \sum\limits_{i\in I} H(i)\\ &= \frac{1}{2} \sum\limits_{i\in I} H(i) \cos(\tau\phi(\xi) - \tau\phi(i)) + \frac{1}{2} \sum\limits_{i\in I} H(i)\\ &= \frac{1}{2} H_{\rm cc}(\xi) + \frac{1}{2} \sum\limits_{i\in I} H(i) \\ \end{aligned}.$$

(58)

Finally, H _fc(ξ) can be rewritten as

$$ H_{\rm fc}(\xi) = \frac{1}{2} H \cos(\tau\phi(\xi)-\beta) + \frac{1}{2} \sum_{i\in I} H(i). $$

(59)

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