**Appendix A**

This section contains numerical examples related to the previous five sections (excluding Section 6) to increase their legibility.

Example A1 relates to Section 3.2, explains the notations of matrices and shows the detailed computation of the F-transform components which will be omitted in other examples. It illustrates the preimage problem and shows one vector that solves it.

**Example A1.** *Let a set of nodes Y* = {3, 4, 6} *be a selected subset of X* = {1, 2, 3, 4, 5, 6, 7, <sup>8</sup>}*, the universe, and*

$$\forall t \in \mathcal{Y} \colon A\_t(\mathbf{x}) = \begin{cases} \frac{\mathbf{x} - t}{\mathcal{S}} + \mathbf{1} & \mathbf{x} \in [t - 3, t] \cap X \\ \frac{t - \mathbf{x}}{\mathcal{S}} + \mathbf{1} & \mathbf{x} \in [t, t + 3] \cap X \\ 0 & \mathbf{x} \in X \backslash [t - 3, t + 3] \end{cases}$$

*be triangular basic functions on X associated with elements of Y that determine the closenessdescribing (weight) matrix*

$$
\mathcal{W} = \begin{bmatrix}
\frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\
0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} & 0 & 0 \\
0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3}
\end{bmatrix},
$$

*and the scaling (node degree) matrix*

$$D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$

.

.

.

*Note that these two matrices, W and the derived D, are independent of the functions acting on X. Let u* = *x*2 *be a particular function on X which has a vector form*

> *u* = 1 4 9 16 25 36 49 64

*The function u is described by its values* (*ui*)*Li*=<sup>1</sup> = (*u*(*xi*))*xi*∈*<sup>X</sup> that represent registered values of a certain variable (physical, chemical, etc.) measured at all points (on all objects) of the universe (data set) X. Another possible interpretation of u consists in assuming that there is a continuous signal that spreads over a certain medium during a time frame described by an interval* [*y*0, *yl*+<sup>1</sup>] ⊃ *X and we are able to register that signal only in discrete time steps* (*xi*)*Li*=1*. The components of the direct F-transform of the function u corresponding to all At's (t* ∈ *Y) are:*

$$F[u]\_1 = \left(1 \cdot \frac{1}{3} + 4 \cdot \frac{2}{3} + 9 \cdot 1 + 16 \cdot \frac{2}{3} + 25 \cdot \frac{1}{3}\right) : \left(\frac{1}{3} + \frac{2}{3} + 1 + \frac{2}{3} + \frac{1}{3}\right) = \frac{31}{3}$$

$$F[u]\_2 = \left(4 \cdot \frac{1}{3} + 9 \cdot \frac{2}{3} + 16 \cdot 1 + 25 \cdot \frac{2}{3} + 36 \cdot \frac{1}{3}\right) : \left(\frac{1}{3} + \frac{2}{3} + 1 + \frac{2}{3} + \frac{1}{3}\right) = \frac{52}{3}$$

$$F[u]\_3 = \left(16 \cdot \frac{1}{3} + 25 \cdot \frac{2}{3} + 36 \cdot 1 + 49 \cdot \frac{2}{3} + 64 \cdot \frac{1}{3}\right) : \left(\frac{1}{3} + \frac{2}{3} + 1 + \frac{2}{3} + \frac{1}{3}\right) = \frac{112}{3}$$

*Writing them in the vector form, we obtain*

$$F[\mu] = \begin{bmatrix} \frac{31}{3} & \frac{52}{3} & \frac{112}{3} \end{bmatrix}^\top.$$

*Obviously, it holds that DF*[*u*] = *Wu, i.e.,*

$$
\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} \frac{31}{5} \\ \frac{52}{5} \\ \frac{112}{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} & 1 & \frac{2}{5} & \frac{1}{5} & 0 & 0 & 0 \\ 0 & \frac{1}{5} & \frac{2}{5} & 1 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{5} & \frac{2}{5} & 1 & \frac{2}{5} & \frac{1}{5} \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 4 \\ 16 \\ 25 \\ 36 \\ 36 \\ 64 \end{bmatrix} = \begin{bmatrix} 31 \\ 52 \\ 112 \end{bmatrix}.
$$

*The preimage problem given by the vector we ended up with (this ensures that b* ∈ *B) then has the form*

$$b = DF[u] = \begin{bmatrix} 31 \\ 52 \\ 112 \end{bmatrix} \quad = \quad \begin{bmatrix} \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} \end{bmatrix} \cdot v = Wv \cdot T$$

*We will use SVD of W to find one of the right inverse matrices that constitute a solution. For one particular SVD decomposition, we obtain an explicit solution*

$$v = ZS^{+}P^{\top}b \doteq \begin{bmatrix} 10.563025 \\ 8.92437 \\ 7.285714 \\ 6.403361 \\ 29.92437 \\ 53.445378 \\ 43.764706 \\ 21.882353 \end{bmatrix}$$

.

*To verify it, we compute*

$$\mathcal{W}\upsilon \doteq \begin{bmatrix} \mathbf{31} & \mathbf{52} & \mathbf{112} \end{bmatrix}^\top \ .$$

*and comparing it to b we see that we found a solution to the preimage problem.*

In Example A1, we obtained the same particular solution while using the explicit formula from Theorem 1 as well as while using the right inverse given by SVD of *W* (Corollary 1) and the right inverse given by Lemma 8. As SVD is widely used to compute pseudoinverses, this was not too surprising and our software is no exception.

Examples A2 and A3 relate to Section 4 and show that under certain conditions, the inverse F-transform is one of the possible solutions to the preimage problem. The former example illustrates the degenerated case where not all points of the universe are covered by basic functions (are not connected with at least one node), the latter example uses non-convex basic functions.

**Example A2.** *Let Y* = {3, 4, 6} *be a selected subset of nodes of the set X* = {1, 2, 3, 4, 5, 6, 7, <sup>8</sup>}*,*

$$\forall t \in \mathcal{Y} \colon A\_t(x) = \begin{cases} 1 & t = x \\ 0 & t \neq x \end{cases}$$

*be singleton basic functions on X associated with elements of Y that determine the closeness matrix*

$$
\mathcal{W} = \begin{bmatrix}
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0
\end{bmatrix},
$$

*and the scaling matrix*

$$D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot$$

*Let u be a particular function on X with a vector form*

$$
\mu = \begin{bmatrix} 0 & 0 & 9 & 16 & 0 & 36 & 0 & 0 \end{bmatrix}^{\top}.
$$

*The components of the direct F-transform of u corresponding to all At's (t* ∈ *Y) form the vector:*

$$F[\mu] = \begin{bmatrix} 9 & 16 & 36 \end{bmatrix}^\perp.$$

*Obviously, it holds that DF*[*u*] = *Wu, i.e.,*

$$
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 9 \\ 16 \\ 36 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 16 \\ 0 \\ 36 \\ 36 \\ 0 \end{bmatrix} = \begin{bmatrix} 9 \\ 16 \\ 36 \end{bmatrix}.
$$

*The preimage problem given by the vector we ended up with then has the form*

$$b = DF[u] = \begin{bmatrix} 9 \\ 16 \\ 36 \end{bmatrix} \quad = \quad \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot v = Wv \cdot u$$

*The inverse F-transform of u with respect to W has the form*

$$\mathcal{F}[u] = \mathcal{W}^{\top} F[u] = \begin{bmatrix} 0 & 0 & 9 & 16 & 0 & 36 & 0 & 0 \end{bmatrix}^{\top}.$$

*Since this vector is equal to u, we obtain that WF*<sup>ˆ</sup>[*u*] = *Wu, so <sup>F</sup>*<sup>ˆ</sup>[*u*] *is a solution to the preimage problem. Note that W is such that WW* = *I* = *D.*

**Example A3.** *Let Y* = {3, 4, 6} *be a selected subset of nodes of the set X* = {1, 2, 3, 4, 5, 6, 7, 8} *and let the closeness be determined by the matrix*

> *W*= ⎡⎣11100000 00011000 00000111⎤⎦ ,

*and the corresponding scaling matrix*

> *D* = ⎡⎣300 020 003⎤⎦ .

*Let u be a particular function on X with a vector form*

$$
\mu = \begin{bmatrix} 0 & 0 & 9 & 16 & 0 & 36 & 0 & 0 \end{bmatrix}^{\top}.
$$

*The components of the direct F-transform of u corresponding to all At's (t* ∈ *Y) form the vector:*

.

$$F[u] = \begin{bmatrix} 3 & 8 & 12 \end{bmatrix}^\top.$$

*Obviously, it holds that DF*[*u*] = *Wu, i.e.,*

$$
\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} 3 \\ 8 \\ 12 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 9 \\ 16 \\ 0 \\ 36 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 9 \\ 16 \\ 36 \end{bmatrix}.
$$

*The preimage problem given by the vector we ended up with then has the form*

$$b = DF[u] = \begin{bmatrix} 9 \\ 16 \\ 36 \end{bmatrix} \quad = \quad \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \end{bmatrix} \cdot v = Wv \dots$$

*The inverse F-transform of u with respect to W has the form*

$$\hat{F}[u] = \mathcal{W}^{\top} F[u] = \begin{bmatrix} 3 & 3 & 3 & 8 & 8 & 12 & 12 & 12 \end{bmatrix}^{\top} .$$

*We see that*

$$\mathcal{MF}[u] = \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 3\\3\\3\\8\\8\\12\\12\\12 \end{bmatrix} = \begin{bmatrix} 9\\16\\36 \end{bmatrix} = \mathcal{W}u\_{\prime \prime}$$

*so F* <sup>ˆ</sup>[*u*] *is a solution to the preimage problem. Note that again W is such that WW* = *D.*

Examples A4 and A5 relate to Section 5 and illustrate that creating a compatible set of basic functions leads to a solution to the preimage problem. The latter example shows a drawback of requiring reflexivity of closeness.

**Example A4.** *Let X* = {1, 2, 3, 4, 5, 6, 7, 8, 9} *be the universe with a selected subset Y* = {2, 5, 8} *of nodes and let the closeness be determined by the matrix*


,

.

.

*and the corresponding scaling matrix*

$$D = \begin{bmatrix} \frac{7}{4} & 0 & 0\\ 0 & \frac{7}{4} & 0\\ 0 & 0 & \frac{7}{4} \end{bmatrix} \cdot$$

*Let u* = *x*2 *be a particular function on X which has a vector form*

*u* = 1 4 9 16 25 36 49 64 81

*The components of the direct F-transform of the function u corresponding to all At's (t* ∈ *Y) form the vector:*

$$\begin{aligned} |F[\mu] &= \begin{bmatrix} \frac{32}{7} & \frac{179}{7} & \frac{452}{7} \end{bmatrix}^\top . \\ | &= W \mu\_\epsilon \text{ i.e.} \end{aligned} $$

*Obviously, it holds that DF*[*u*] = *Wu, i.e.,*

$$
\begin{bmatrix}
\frac{7}{4} & 0 & 0\\ 0 & \frac{7}{4} & 0\\ 0 & 0 & \frac{7}{4}
\end{bmatrix} \cdot \begin{bmatrix}
\frac{32}{7}\\ \frac{179}{7}\\ \frac{452}{7}
\end{bmatrix} = \begin{bmatrix}
\frac{1}{2} & \frac{3}{4} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2} & \frac{3}{4} & \frac{1}{2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{3}{4} & \frac{1}{2}
\end{bmatrix} \cdot \begin{bmatrix}
1\\ 4\\ 9\\ 16\\ 25\\ 36\\ 49\\ 9\\ 8
\end{bmatrix} = \begin{bmatrix}
8\\ 179\\ 113
\end{bmatrix}.
$$

*The preimage problem given by the vector we ended up with then has the form*

$$b = \begin{bmatrix} 8 \\ \frac{179}{4} \\ 113 \end{bmatrix} \quad = \quad \begin{bmatrix} \frac{1}{2} & \frac{3}{4} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{3}{4} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{3}{4} & \frac{1}{2} \end{bmatrix} \cdot v = Wv \text{ .} $$

*Let us consider, e.g., the following matrix M satisfying WM* = *I:*

$$M = \begin{bmatrix} \frac{1}{2} & \frac{2}{3} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2} & \frac{2}{3} & \frac{1}{2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{2}{3} & \frac{1}{2} \end{bmatrix}$$

*Then the vector*

$$w = M^\top b = \begin{bmatrix} \frac{1}{2} & \frac{2}{3} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2} & \frac{2}{3} & \frac{1}{2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{2}{3} & \frac{1}{2} \end{bmatrix}^\top \cdot \begin{bmatrix} 8\\ \frac{8}{4}\\ \frac{179}{4}\\ 113 \end{bmatrix} = \begin{bmatrix} \frac{4}{3} \\ \frac{16}{3}\\ \frac{179}{3}\\ \frac{179}{3}\\ \frac{179}{3}\\ \frac{256}{3}\\ \frac{256}{3}\\ \frac{113}{2} \end{bmatrix}$$

2

,

*is a solution to the preimage problem that can be verified by*

$$\mathcal{W}v = \begin{bmatrix} \frac{1}{2} & \frac{3}{4} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2} & \frac{3}{4} & \frac{1}{2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{3}{4} & \frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} 4\\ \frac{16}{3} \\ \frac{4}{3} \\ \frac{179}{3} \\ \frac{179}{3} \\ \frac{179}{3} \\ \frac{113}{3} \\ \frac{226}{3} \\ \frac{113}{2} \end{bmatrix} = \begin{bmatrix} 8\\ 8\\ 113 \end{bmatrix} = b \cdot A$$

*Moreover, it demonstrates the fact that the sets of basic functions At's (given by the rows of W) and Bt's (given by the rows of M), t* ∈ *Y, are compatible.*

If we assumed that for each *t* ∈ *Y* it holds *At*(*t*) = 1 (reflexive closeness), the compatible set of convex basic functions satisfying that also for each *t* ∈ *Y* it holds *Bt*(*t*) = 1, would be uniquely formed by singletons (degenerated case). This is illustrated by Example A5.

**Example A5.** *Let X* = {1, 2, 3, 4, 5, 6, 7, 8, 9} *be the universe with the selected subset Y* = {2, 5, 8} *of nodes and let the closeness be determined by the matrix*

$$
\mathcal{W} = \begin{bmatrix}
\frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2}
\end{bmatrix} \mathcal{I}
$$

*and the corresponding scaling matrix*

$$D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}.$$

*Let u* = *x*2 *be a particular function on X which has a vector form*

$$
\mu = \begin{bmatrix} 1 & 4 & 9 & 16 & 25 & 36 & 49 & 64 & 81 \end{bmatrix}^\top \dots
$$

*The components of the direct F-transform of the function u corresponding to all Ats (t* ∈ *Y) form the vector:*

$$F[\mu] = \begin{bmatrix} \frac{9}{2} & \frac{51}{2} & \frac{129}{2} \end{bmatrix}^{\top}.$$

*Obviously, it holds that DF*[*u*] = *Wu, i.e.,*

$$
\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \cdot \begin{bmatrix} \frac{9}{7} \\ \frac{81}{7} \\ \frac{129}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 4 \\ 9 \\ 25 \\ 36 \\ 49 \\ 64 \\ 81 \end{bmatrix} = \begin{bmatrix} 9 \\ 51 \\ 129 \end{bmatrix}.
$$

*The preimage problem given by the vector we ended up with then has the form*

$$\begin{array}{rcl} b = \begin{bmatrix} 9\\51\\129 \end{bmatrix} &=& \begin{bmatrix} \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \end{bmatrix} \cdot v = Wv \end{array}$$

*Let us consider the following matrix M satisfying WM* = *I (if we want each of its rows to form a convex fuzzy set with M<sup>t</sup>* [*t*] = 1*, then this is the only option):*

$$M = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

*Note that if for any s* ∈ *Y* \ {*t*} *it holds that At*(*s*) > 0*, then no compatible matrix M exists. The vector*

.

$$w = M^\top b = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}^\top \cdot \begin{bmatrix} 9 \\ 51 \\ 51 \\ 129 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 51 \\ 0 \\ 0 \\ 0 \\ 129 \\ 0 \end{bmatrix},$$

*is a solution to the preimage problem that can be verified by*

$$\mathcal{W}\upsilon = \begin{bmatrix} \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} 0\\ 9\\ 0\\ 0\\ 51\\ 0\\ 0\\ 0\\ 129\\ 0 \end{bmatrix} = \begin{bmatrix} 9\\ 51\\ 51\\ 129\\ 0 \end{bmatrix} = b \cdot A$$

*This again demonstrates the fact that the sets of basic functions At's (given by the rows of W) and Bt's (given by the rows of M), t* ∈ *Y, are compatible. More importantly, it demonstrates that requiring At*(*t*) = 1 *and Bt*(*t*) = 1 *for all node indices t* ∈ *Y can lead to an empty set of compatible convex basic functions.*
