Generation of weighting triangles associated with

aggregation functions ∗

T. Calvo1,∗, G. Mayor2, J. Torrens3, J. Suner4, M. Mas5 and M. Carbonell6

Department of Mathematics and Computer Science

University of the Balearic Islands

Ctra. de Valldemossa, km 7.5

07071 Palma de Mallorca, Spain.1dmitcs0@ps.uib.es 2dmigmf0@ps.uib.es 3dmijts0@ps.uib.es4dmijsl0@ps.uib.es 5dmimmg0@ps.uib.es 6dmimch0@ps.uib.es

Abstract

In this work, we present several ways to obtain different types of weighting triangles,due to these types characterize some interesting properties of Extended Ordered WeightedAveraging operators, EOWA, and Extended Quasi–linear Weighted Mean, EQLWM, as wellas of their reverse functions. We show that any quantifier determines an EOWA operatorwhich is also an Extended Aggregation Function, EAF. Moreover, we present different resultsabout generation of these functions by means of sequences and fractal structures. Finally,we introduce a degree of orness of a weighting triangle associated with an EOWA operator.After that, we mention some results on each class of triangle, considering each one of thesetriangles as triangles associated with their corresponding EOWA operator, and we calculatethe orness of some interesting triangles.

Keywords: extended aggregation function, extended Ordered Weighted Averaging op-erator, extended quasi–linear weighted mean, regular, descending and balancing weightingtriangle, quantifier, sequence, negation, fractal structure and orness.

1 Introduction

In the last years, the problem of information aggregation has been a main topic of study for a lotof researchers. Different classes of operators which are very useful in the aggregation field havebeen already introduced, for instance the OWA operators [13, 15], the means [3, 7] and the EAFrecently introduced by G. Mayor and T. Calvo [11]. Furthermore, we have EOWA operators andEQLWM which are respectively a generalization of OWA operators and Quasi–linear Weightedmeans and are also EAF [3, 11, 10]. The extended operators are multidimensional operators,that is to say, we can consider them for n–dimensional lists, for all n > 1. Moreover, we arespecially interested in the comparison of two lists with different dimension.

We foccus on the construction of weighting triangles, but the different classes of these can onlybe understood if we recall these classes close to the corresponding class of aggregation functions.Then, we here present different ways of obtaining EOWA operators and EQLWM which are

∗Corresponding author

1

also EAF and also show the same result for their reverse functions. Each way of obtaining theseextended aggregation functions is determined by the form of defining the weights of the weightingtriangle associated with each function, since the properties studied are characterized by differentconditions on their weighting triangles [3, 4, 5, 6, 10, 11]. According to the characterizationsobtained in the above papers, we analyze several ways to generate : regular, descending andbalancing triangles. The first two classes characterize the EOWA operators and the EQLWMwhich are EAF, and the third class characterizes the self identity property of an EQLWM.

The first way chosen to determine the classes of triangles above is by means of a quantifier,which is equivalent to consider a negation. We show that any quantifier determines a doublyregular triangle and also the conditions when a quantifier determines a doubly descending andbalancing triangle are given. We give some examples of descending and balancing triangles thatcharacterize an EQLWM [3] by means of a negation.

In the second place, the different classes of triangle are built by means of a sequence. Thethird way is more attractive due to the use of fractal structures to define each class of triangle.A family of triangles, called Sierpinski family, is obtained. This name has been chosen becauseits construction is similar to the well–known Sierpinski triangle and Sierpinski carpet fractals.The triangles of this family are regular and descending, which is equivalent to say that we canobtain an EOWA operator and an EQLWM which are EAF with each of them.

Finally, we introduce the definition of degree of orness of a weighting triangle and mentionsome results on each class of triangle. Moreover, we calculate the orness of several interestingtriangles.

2 Preliminaries

2.1 Basic definitions

In this section we present the definitions and basic results that we will use in this work. Most ofthem are related to the concept of an Extended Aggregation Function introduced by G. Mayorand T. Calvo in 1997 (see [11]) and to posterior works developed by other authors like M.Carbonell, M. Mas, J. Suner, et al [10].

Let (L,≤) be a lattice with minimum 0 and maximum 1. For each n ≥ 1, consider the setLn of n–dimensional ordered lists (x1, . . . , xn) with xi ∈ L, i = 1, . . . , n . Finally E =

⋃n≥1

Ln is

the set of all ordered lists that can be constructed from L. In this work we consider L = [0, 1].

Definition 1 Let x = (x1, . . . , xn) , y = (y1, . . . , ym) be elements of E. The following orderingson E can be considered

• x ≤α y if and only if n ≤ m,x1 ≤ y1, . . . , xn ≤ yn, and if n ≤ m then sup(x1, . . . , xn) ≤inf(yn+1, . . . , ym).

• x ≤β y if and only if n ≥ m,x1 ≥ y1, . . . , xn ≥ yn, and if n ≥ m then sup(xm+1, . . . , xn) ≤inf(y1, . . . , ym).

Example 1 If L = [0, 10] (with the usual order), then we have:

(2, 1, 7, 1) <β (2, 1, 7) <α (2, 2, 7, 7) <β (8, 7) <α (8, 7, 8)

2

We now recall the concept of Extended Aggregation Function.

Definition 2 An Extended Aggregation Function, EAF, is a mapping A : E −→ L that satisfiesthe two following conditions:

1. A is monotonic with respect to ≤α and ≤β, i.e., x ≤i y ⇒ A(x) ≤ A(y), with i ∈ {α, β}.

2. A is idempotent, i.e., A(

n times︷ ︸︸ ︷x, . . . , x) = x for all x ∈ L and n ≥ 1.

Well-known operators as the min,max, arithmetic mean or the median are examples ofExtended Aggregation Functions.

According to the idea of Yager [13] the definition of Extended Ordered Weighted Averagingoperator is recalled.

Definition 3 A :⋃n≥1

[0, 1]n −→ [0, 1] is an Extended Ordered Weighted Averaging operator,

EOWA operator, if the restriction of A to each [0, 1]n is an n–dimensional OWA operator. In

other words, if for each n there exists Wn = (wn1 , wn2 , . . . , w

nn) ∈ [0, 1]n with

n∑i=1

wi = 1, such

that A(x1, . . . , xn) =n∑i=1

wiyi for all (x1, . . . , xn) ∈ [0, 1]n and (y1, . . . , yn) being a weighting

list in descending order.

An EOWA operator is idempotent, symmetric and monotonic with respect to the Cartesianproduct order.

Now recall one of the main topics in this paper.

Definition 4 A weighting triangle or triangle of weights is a set of numbers wni ∈ [0, 1], for i =

1, . . . , n, such that:n∑i=1

wni = 1 for each n > 1. It can be represented by

1w21 w2

2

w31 w3

2 w33

w41 w4

2 w43 w4

4

. . .

It will be denoted by 4wni .

We also recall the definition of an Extended Quasi–Linear Weighted Mean.

Definition 5 Let f : [0, 1] −→ [−∞,+∞] be a real continuous and strictly monotonic functionsuch that {f(0), f(1)} 6= {−∞,+∞} and 4wni be a triangle of weights with 0 < wni < 1. AnExtended Quasi–Linear Weighted Mean, EQLWM, A is a mapping from

⋃n≥1

[0, 1]n into [0, 1]

given by

A(x1, . . . , xn) = f−1(

n∑i=1

wni f(xi)

).

3

These weighted means are clearly idempotent and monotonic with respect to the Cartesianproduct order, but they are symmetric only if the weights are identical and they are alwaysbetween the maximum and the minimum. The arithmetic, geometric, harmonic and quadraticmeans are examples of EQLWM.

The reverse of an EOWA operator and an EQLWM can be considered. To do this, it issufficient to replace the weighting triangle 4wni given in each definition for the following one:4unj where unj = wnn−j+1, for n > 1 and all j ∈ {1, . . . , n}.

After recalling the reverse function it is useful to remind the following result.

Lemma 1 Let A be an EOWA operator, then A is monotonic with respect to ≤α if and only ifits reverse operator Ar is monotonic with respect to ≤β.

2.2 Extended aggregation functions and weighting triangles

Recall the main relations between two types of EAF, namely EOWA operator and EQLWM aswell as their reverse functions, and the weighting triangles associated with each one of thesefunctions.

Theorem 1 Let A be an EOWA operator. Then A and Ar are EAF if and only if the weightingtriangle associated with A, 4wni , verifies:

p∑i=1

wn+1i ≤

p∑i=1

wni ≤p+1∑i=1

wn+1i . (1)

for each n > 1 and p = 1, . . . , n and

n+1∑j=q

wn+1j ≤

n+1∑j=q

wnj−1 ≤n+1∑j=q−1

wn+1j . (2)

for each n > 1 and q = n+ 1, . . . , 2.

Proof. For (1) and (2) see [11, 5].

A triangle that satisfies the first condition is called left regular and right regular when itsatisfies the second one. Moreover, it is called doubly regular when the two conditions aresatisfied.

Example 2 The “normalized” triangle of Pascal is doubly regular

11/2 1/2

1/4 2/4 1/41/8 3/8 3/8 1/8

1/16 4/16 6/16 4/16 1/16. . . . . .. . . . . .

. . .(n−1k

)/2n−1 . . .

where k : 0, . . . , n−1 and the weighting lists Wn = 12n−1

((n−10

),(n−1

1

), . . . ,

(n−1n−1

))for each n > 1.

4

Theorem 2 Let A be an EQLWM. Then A and Ar are EAF if and only if the weighting triangleassociated to A, 4wni , verifies:

wn+1i ≤ wni (3)

for each n > 1 and i = 1, . . . , n, andwn+1j+1 ≤ w

nj (4)

for each n > 1 and j = 1, . . . , n.

Proof. The results of (3) and (4) appear in [3, 5].

A triangle satisfying the last two conditions is called doubly descending. If only one conditionis satisfied, it is called left or right descending, respectively.

Example 3 The following triangle is doubly descending and the aggregation operator associatedwit this is the arithmetic mean.

11/4 3/4

1/9 3/9 5/91/16 3/16 5/16 7/16

1/25 3/25 5/25 7/25 9/25. . .

To finish this section, we mention the self identity property, in the position n + 1, of anEQLWM, since it allows us to consider another class of weighting triangle.

Definition 6 An Extended Aggregation Operator A fulfills the self identity property, in theposotion n+ 1, if it holds the following equality

A(x1, . . . , xn, A(x1, . . . , xn)) = A(x1, . . . , xn),

for all n > 1 and all list (x1, . . . , xn) ∈ [0, 1]n.

Theorem 3 Let A be an Extended Quasi–Linear Weighted Mean, EQLWM. Then A and itsreverse function Ar satisfy the self identity property, in the position n + 1, if and only if theweighting triangle associated with A and Ar fulfills:

(i) wn+1i = wni (1− wn+1

n+1), for i = 1, . . . , n. (10)

(ii) wn+1i+1 = wni (1− wn+1

1 ), for i = 1, . . . , n. (11)

Proof. For (10) see [3]. In case of (11), it is enough to use condition (10) on the weightingtriangle associated with Ar and the relation of this with the corresponding weighting triangle ofthe operator A.

5

A weighting triangle that satisfies the two conditions above is called doubly balancing. Thatis, the triangle is left balancing and right balancing.

Remark 1 According to the formulas (10) and (11), the left balancing and right balancingtriangles are respectively determined for the values of wnn and wn1 for all n ≥ 2. Then, thegeneral form of each one of these triangles is as follows: For the left balancing triangle we have:

11− a a

(1− a)(1− b) a(1− b) b(1− a)(1− b)(1− c) a(1− b)(1− c) b(1− c) c

. . .

In the second case, it is obtained:

1a 1− a

b a(1− b) (1− a)(1− b)c b(1− c) a(1− b)(1− c) (1− a)(1− b)(1− c)

. . .

In both cases, a, b, c, . . . ∈ [0, 1] and all of them can be equals.

Proposition 1 (i) If a weighting triangle 4wni is left regular then the triangle 4uni , whereuni = wnn−i+1 is right regular and vice versa.

(ii) If a weighting triangle 4wni is left descending then the triangle 4uni , where uni = wnn−i+1

is right descending and vice versa.

(iii) If a weighting triangle 4wni is left balancing then the triangle 4uni , with uni = wnn−i+1 isright balancing and vice versa.

3 Generating three classes of weighting triangles

In this section we consider different ways to determine weighting triangles, in particular regular,descending and balancing triangles [3, 4, 10, 11]. The first way was introduced by Yager (1988),but other important results are also presented here. In this section we include some remarksabout generation of triangles by means of a negation function. Moreover, we use several typesof sequences in the second way to define the different classes of triangles. However, the mostattractive way presented is the third one due to its geometrical construction. First, we quotesome relationships between balancing and regular triangles.

Proposition 2 Let 4wni be a weighting triangle.

(i) If 4wni is a left (right) regular triangle then the following sequences: {wnn} , ({wn1 }) aredecreasing.

(ii) If 4wni is a right balancing triangle then the triangle is also right regular.

6

(iii) If 4wni is a left balancing triangle and the sequence {wnn} is decreasing then the triangleis also left regular.

Proof. In case of a left regular triangle, for the proof of (i), it is enough to take p = n− 1 in thesecond inequality of left regular condition, then it follows that

1− wnn ≤ 1− wn+1n+1,

which is equivalent town+1n+1 ≤ w

nn.

In a similar way we can obtain that {wn1 } is a decreasing sequence, letting p = 1 in the firstinequality of the regularity condition. Similarly, taking in the first inequality q = n + 1 and inthe second q = 3 the results of a right regular triangle can be obtained.

In the second case, taking into account that the condition of a right balancing triangle canbe expressed by wn+1

i = wni−1(1− wn+11 ), for i = 2, . . . , n+ 1 it follows that wn+1

i ≤ wni−1 hence

n+1∑i=q

wn+1i ≤

n+1∑i=q

wni−1.

For the other inequality, first we show by iteration that

wni = wn−i+11 (1− wn−i+2

1 )(1− wn−i+31 ) . . . (1− wn−11 )(1− wn1 ),

for i = 2, . . . , n. Then we get by induction

n+1∑i=q

wni−1 = (1− wn−q+31 )(1− wn−q+4

1 ) . . . (1− wn−11 )(1− wn1 ).

Then we apply the expression of this sum to prove that the second inequality is always true

n+1∑i=q

wni−1 = (1− wn−q+31 )(1− wn−q+4

1 ) . . . (1− wn−11 )(1− wn1 ) ≤

n+1∑i=q−1

wn+1i = (1− wn−q+4

1 )(1− wn−q+51 ) . . . (1− wn1 )(1− wn+1

1 ),

since this is equivalent to1− wn−q+3

1 ≤ 1

and this inequality is always true. In the third case, suppose that the triangle is left balancing,then it holds

wn+1i = wni (1− wn+1

n+1),

for i = 1, . . . , n. Thus wn+1i ≤ wni , furthermore

p∑i=1

wn+1i ≤

p∑i=1

wni ,

7

for p ≤ n. Now, we need to show that

p∑i=1

wni ≤p+1∑i=1

wn+1i .

For that, using the left balancing condition it can be obtained by iteration that

wni = wii(1− wi+1i+1)(1− wi+2

i+2) . . . (1− wn−1n−1)(1− wnn),

for i = 1, . . . , p and p ≤ n. Consequently, we get by induction

p∑i=1

wni = (1− wp+1p+1)(1− wp+2

p+2) . . . (1− wn−1n−1)(1− wnn).

Furthermore, it can be shown by a simple calculation that the following inequality

p∑i=1

wni ≤p+1∑i=1

wn+1i

is true if and only if wn+1n+1 ≤ wn+1

p+1 for all p ≤ n, which is true by hypothesis. Therefore, thetriangle is left regular. This concludes the proof.

3.1 Generation by a quantifier

Definition 7 A increasing quantifier is a mapping Q : [0, 1]→ [0, 1] such that

i) Q(0) = 0, Q(1) = 1ii) r < t =⇒ Q(r) ≤ Q(t)

This definition appears in [10] and [14], where the increasing quantifiers are used to representthe concepts of “most”, “a lot”, etc. Particularly in [14] it is built an n−dimensional operatorOWA, AQ, by means of a increasing quantifier Q : [0, 1]→ [0, 1], with the weights given by

wj = Q

(j

n

)−Q

(j − 1

n

)for all j = 1, . . . , n.

In this case, it is possible to evaluate the sentence:“Most of the students are poor”

For that we represent the concept of most by a increasing quantifier Q, the term “be poor”by the fuzzy set B, then for a collection of n students x1, . . . , xn, the value of the sentenceshould be given by AQ(b1, . . . , bn) where AQ is the n−dimensional OWA defined by Q and eachbi = B(xi). In the same way we can define an n–dimensional quasi-linear weighted mean givena continuous and monotonic function. This idea can be generalized and applied to extendedoperators and so, given a increasing quantifier Q, we can build a weighting triangle as follows:

wnj = Q(j

n)−Q(

j − 1

n),

for j = 1, . . . , n and n ≥ 1. This triangle can be associated with different classes of ExtendedAggregation Operators.

We see easily that the weights define certainly a weighting triangle:

8

• wnj ∈ [0, 1], for all j = 1, . . . , n, and n > 1,

•n∑j=1

wnj =n∑j=1

[Q( jn)−Q( j−1n )

]= Q(1)−Q(0) = 1.

Another interesting relation is the following:

•k∑j=1

wnj =k∑j=1

[Q( jn)−Q( j−1n )

]= Q( kn).

Example 4 The following increasing quantifiers define the weighting triangles of the Maximum,Minimum and the Median operators. That is,

Q1(x) =

{0 if x = 01 if x ∈ (0, 1]

Q2(x) =

{1 if x = 10 if x ∈ [0, 1)

Q3(x) =

0 if 0 ≤ x < 1/21/2 if x = 1/21 if 1/2 < x ≤ 1

Moreover, in each row of the triangle of maximum appears 1 in the first place and 0 in the othercases, whereas for the minimum 1 is the last element of each row and 0 appears in the otherplaces. The corresponding triangle for the median is

11/2 1/2

0 1 00 1/2 1/2 0

0 0 1 0 0. . .

Example 5 1. If we consider Q(x) = x2, the weights are wnj = 2j−1n2 for j = 1, . . . , n and

n ≥ 1; then the triangle is

11/4 3/4

1/9 3/9 5/91/16 3/16 5/16 7/16

1/25 3/25 5/25 7/25 9/25. . .

2. With Q(x) = x, the corresponding weights are wnj = 1n for j = 1, . . . , n and n ≥ 1. Thus,

the triangle is

11/2 1/2

1/3 1/3 1/31/4 1/4 1/4 1/4

1/5 1/5 1/5 1/5 1/5. . .

9

Proposition 3 Let Q : [0, 1] → [0, 1] be a increasing quantifier. Then the EOWA AQ is anExtended Aggregation Function.

Proof. It is sufficient to prove that the weighting triangle generated by Q is left regular, that isto say,

p∑j=1

wn+1j ≤

p∑j=1

wnj ≤p+1∑j=1

wn+1j ,

for all p = 1, . . . , n, which is equivalent to show that

Q(p

n+ 1) ≤ Q(

p

n) ≤ Q(

p+ 1

n+ 1),

for all p = 1, . . . , n, but these inequalities are true due to the increasingness of Q and thefollowing relation

p

n+ 1≤ p

n≤ p+ 1

n+ 1.

Proposition 4 Let A : ∪n≥1

[0, 1]n → [0, 1] be an EOWA operator with weighting triangle 4wnj .

First assume that A is an EAF. Then there exists a increasing quantifier Q such that A = AQif and only if the following relation holds:

k ·m = n · r =⇒k∑j=1

wnj =r∑j=1

wmj . (5)

Proof. In the first place, we assume that (5) holds. We now define the mapping Q over therational numbers of [0, 1] as follows:

Q(0) = 0, Q(k

n) =

k∑j=1

wnj for all n ≥ 1 and for all k = 1, . . . , n.

If kn = r

m , n 6= m, 0 ≤ k ≤ n, 0 ≤ r ≤ m, then

Q(k

n) =

k∑j=1

wnj =r∑j=1

wmj = Q(r

m).

Thus Q is well defined on Q ∩ [0, 1] and consequently we can extend it over the unit interval.Observe that A = AQ, since

Q(j

n)−Q(

j − 1

n) =

j∑i=1

wni −j−1∑i=1

wni = wnj .

Conversely, suppose that A = AQ, and we want to show that condition (5) is true.We have

k∑j=1

wnj = Q(k

n)

10

andr∑j=1

wmj = Q(r

m).

Hence

k ·m = n · r =⇒ Q(k

n) = Q(

r

m),

since Q is a mapping.

Remark 2 (1) In the first part of the proof above, we have shown that Q is defined onQ ∩ [0, 1]. Now, we can see that if AQ = AQ′ holds then Q = Q′ on Q ∩ [0, 1].

First, observe:

Q(j

n)−Q(

j − 1

n) = wnj = Q′(

j

n)−Q′(j − 1

n).

Then, taking j = 1, it follows that Q(0) = Q′(0) = 1, hence Q( 1n) = Q′( 1

n), and so on untilto obtain Q(1) = Q′(1) = 1.

(2) When, Q is a uniform continuous function on Q∩ [0, 1], there exists a unique extension ofQ in [0, 1] which is also uniform continuous.

Remark 3 Note that “normalized” triangle of Pascal can not be defined by means of a quantifierQ, since Q takes different values on equivalent fractions, for instance 1/3 = 2/6 but Q(1/3) =1/4 6= 6/32 = Q(2/6).

Proposition 5 Any increasing quantifier Q determines a right regular weighting triangle.

Proof. It is sufficient to see that the triangle holds the following relations

n+1∑j=q

wn+1j ≤

n+1∑j=q

wnj−1 ≤n+1∑j=q−1

wn+1j ,

for all q = n+ 1, . . . , 2. But, by means of a simple calculation it follows that

n+1∑j=q

wn+1j = 1−Q(

q − 1

n+ 1).

Then, taking into account the above result the proof of our inequalities is reduced to show that

1−Q(q − 1

n+ 1) ≤ 1−Q(

q − 2

n) ≤ 1−Q(

q − 2

n+ 1),

for all q = n+ 1, . . . , 2, that is

Q(q − 2

n+ 1) ≤ Q(

q − 2

n) ≤ Q(

q − 1

n+ 1).

These inequalities are true, the first one due to the increasingness of Q and the following relation

q − 2

n+ 1≤ q − 2

n

and the second one becauseq − 2

n≤ q − 1

n+ 1

is true for q ≤ n+ 2 and also for the increasingness of Q.

11

Proposition 6 Let Q and Q′ be two increasing quantifiers which define, respectively, the weight-ing triangles 4wni and 4uni . Then uni = wnn−i+1 if and only if

Q(i

n) = 1−Q′(1− i

n),

for i = 1, . . . , n and all n ≥ 1.

Proof. According to the idea of Yager to define a weighting triangle we have

uni = Q′(i

n)−Q′( i− 1

n)

and

wnn−i+1 = Q(n− i+ 1

n)−Q(

n− in

).

Now, considering the relation between both triangles it follows that

Q′(i

n)−Q′( i− 1

n) = Q(

n− i+ 1

n)−Q(

n− in

),

and taking respectively i = n, . . . , 1 in the last equality we get

Q′(1)−Q′(n− 1

n) = Q(

1

n)−Q(0),

Q′(n− 1

n)−Q′(n− 2

n) = Q(

2

n)−Q(

1

n),

Q′(n− 2

n)−Q′(n− 3

n) = Q(

1

n)−Q(

n− 1

n),

. . . . . . ,

. . . . . . ,

Q′(2

n)−Q′( 1

n) = Q(

n− 1

n)−Q(

n− 2

n),

Q′(1

n)−Q′(0) = Q(1)−Q(

n− 1

n).

From the first equality it is obtained

Q(1

n) = 1−Q′(n− 1

n) = 1−Q′(1− 1

n)

and setting this in the second of the above equalities it follows that

Q(2

n) = 1−Q′(1− 2

n),

letting this in the third of the above equalities we get

Q(3

n) = 1−Q′(1− 3

n),

and so on. Therefore,

Q(i

n) = 1−Q′(1− i

n),

12

for i = 1, . . . , n.Conversely, if Q and Q′ are related by

Q(i

n) = 1−Q′(1− i

n),

it is easy to verify the following condition

Q′(i

n)−Q′( i− 1

n) = Q(

n− i+ 1

n)−Q(

n− in

),

which is equivalent touni = wnn−i+1,

for i = 1, . . . , n. This concludes the proof.

Consequently,

Proposition 7 Let Q be a increasing quantifier and let 4wnj be the weighting triangle generatedby Q. Then4wnj is symmetric if and only if Q is symmetric with respect to (1/2, 1/2) on Q∩[0, 1](i. e., Q(1− x) = 1−Q(x) for all x ∈ Q ∩ [0, 1]).

Proof. It is enough to apply the above result to the definition of a symmetric weighting triangle,i. e.,

wnj = wnn−j+1,

for all n > 1, and for all j = 1, . . . , n.

Now, our purpose is to see when a increasing quantifier determines a left descending triangle.

Proposition 8 Let Q be a increasing quantifier on the unit interval. If Q fulfills

Q(i

n)Q(

n

n+ 1) = Q(

i

n+ 1), (6)

for i = 1, . . . , n and all n ≥ 1, then the triangle associated with Q is left descending.

Proof. According to the condition of a left descending triangle we need to prove that

wn+1i = Q(

i

n+ 1)−Q(

i− 1

n+ 1) ≤ Q(

i

n)−Q(

i− 1

n) = wni ,

for i = 1, . . . , n.Firstly, we use that

Q(i

n)−Q(

i− 1

n) ≥ 0

for i = 1, . . . , n and Q( nn+1) ≤ 1. Then(Q(

i

n)−Q(

i− 1

n)

)Q(

n

n+ 1) ≤ Q(

i

n)−Q(

i− 1

n),

or equivalently

Q(i

n)Q(

n

n+ 1)−Q(

i− 1

n)Q(

n

n+ 1) ≤ Q(

i

n)−Q(

i− 1

n),

but taking into account (6), it follows that

Q(i

n+ 1)−Q(

i− 1

n+ 1) ≤ Q(

i

n)−Q(

i− 1

n),

therefore wn+1i ≤ wni for i = 1, . . . , n.

13

According to Proposition 7 and 9, it can be obtained that

Proposition 9 Let Q be a increasing quantifier on the unit interval. If Q fulfills

Q(i

n+ 1) = Q(

i− 1

n) +Q(

1

n+ 1)−Q(

i− 1

n) ·Q(

1

n+ 1), (7)

for i = 1, . . . , n and all n ≥ 1, then the triangle associated with Q is right descending.

For balancing triangles we have:

Proposition 10 Let Q be a increasing quantifier on the unit interval. The quantifier Q deter-mines a left balancing triangle if and only if

Q

((j

n)(

n

n+ 1)

)= Q(

j

n) ·Q(

n

n+ 1),

for j = 1, . . . , n− 1 and all n ≥ 1.

Proof. If a quantifier Q holds the condition of a left balancing triangle, then

Q(j

n+ 1)−Q(

j − 1

n+ 1) =

(Q(

j

n)−Q(

j − 1

n)

)Q(

n

n+ 1), (8)

for j = 1, . . . , n. From here, taking successively j = 1, . . . , n− 1 in (8) we can assure that

Q(j

n+ 1) = Q(

j

n) ·Q(

n

n+ 1),

and using the equality

Q(j

n+ 1) = Q((

j

n)(

n

n+ 1)),

it is obtained that

Q((j

n)(

n

n+ 1)) = Q(

j

n) ·Q(

n

n+ 1),

for j = 1, . . . , n− 1. Consequently, it is easy to obtain

Q(j

n+ 1) = Q(

j

n)Q(

j

j + 1)Q(

j + 1

j + 2) · . . . ·Q(

n

n+ 1), (9)

for j = 1, . . . , n− 1. On the other hand, if a quantifier Q holds the last relation then

wn+1j = Q(

j

n+ 1)−Q(

j − 1

n+ 1) =

=

(Q(

j

j + 1)Q(

j + 1

j + 2) . . . Q(

n− 1

n)−Q(

j − 1

j)Q(

j

j + 1)Q(

j + 1

j + 2) . . . Q(

n− 1

n)

)Q(

n

n+ 1) =

=

(Q(

j

n)−Q(

j − 1

n)

)Q(

n

n+ 1) = wnj (1− wn+1

n+1),

since

wn+1n+1 = Q(

n+ 1

n+ 1)−Q(

n

n+ 1) = 1−Q(

n

n+ 1).

Then the triangle of weights is left balancing.

14

Considering Proposition 7 and 11 it can be deduced that:

Proposition 11 Let Q be a increasing quantifier on the unit interval. The quantifier Q deter-mines a right balancing triangle if and only if

Q(j

n+ 1) = Q(

j − 1

n) +Q(

1

n+ 1)−Q(

j − 1

n) ·Q(

1

n+ 1), (10)

for j = 1, . . . , n− 1 and all n ≥ 1.

3.1.1 Generation by negations

Negations are well–known in the fuzzy set theory, for this reason we consider very interestingto use these functions to generate different classes of Extended Aggregation Operators by meansof their corresponding weighting triangles. We only consider here the left descending, rightdescending, left balancing, right balancing triangles which are related with Extended Quasi–Linear Weighted Mean. These triangles have been already mentioned in the main theorems ofthe above sections. Following the idea of Yager in [14], we can obtain the weights of a weightingtriangle through a negation N : [0, 1] −→ [0, 1] as follows :

wnj = N

(j − 1

n

)−N

(j

n

),

for all n ≥ 1 and j = 1, . . . , n.It is easy to see that these weights define a weighting triangle :

• wnj ∈ [0, 1], for j = 1, . . . , n and n ≥ 1,

•n∑j=1

wnj =n∑j=1

[N( j−1n )−N( jn)

]= N(0)−N(1) = 1.

It is also very useful to take into account the following equality:

k∑j=1

wnj =k∑j=1

[N(

j − 1

n)−N(

j

n)

]= 1−N(

k

n),

for all 1 ≤ k ≤ n.Taking into account the relation between a increasing quantifier and a negation, i.e., N =

1 − Q, all the results mentioned by a increasing quantifier can be rewriten by a negation. Forthis reason, only some examples of triangles generated by well–known negations are mentionedhere.

Example 6 The following well–known negations determine some types of triangles:

1. Intuitionistic negation (Yager, 1980) defines a doubly regular, doubly descending and doublybalancing triangle.

ni(x) =

{1 if x = 00 if x ∈ (0, 1]

15

In this case the weights of the triangle are: wn1 = 1 and wnj = 0, j = 1, . . . , n− 1.

11 0

1 0 01 0 0 0

1 0 0 0 0. . .

2. Dual intuitionistic negation (Ovchinikov, 1983) defines a doubly regular, doubly descendingand doubly balancing triangle.

ndi(x) =

{0 if x = 11 if x ∈ [0, 1)

Here, the weights are : wnn = 1, and wnj = 0, j = 1, . . . , n− 1.

10 1

0 0 10 0 0 1

0 0 0 0 1. . .

3. The Sugeno Negation:

nλ(x) =1− x

1 + λx, λ > −1

defines a doubly regular triangle. For this proof it is enough to consider the followingrelation

k∑j=1

wnj = 1−N(k

n),

for all 1 ≤ k ≤ n, in a similar way to the Proposition 3. However, the triangle generatedby this negation is not always doubly descending, neither doubly balancing as we can seein the following cases:

(a) For λ = 1, n1(x) = 1−x1+x . Then

wn1 =2

n+ 1=

2n

n(n+ 1), wnn =

1

2n− 1=

2n

2n(2n− 1)and wnj =

2n

(n+ j)(n+ j − 1).

Thus, the triangle is

16

14/2 · 3 4/3 · 4

6/3 · 4 6/4 · 5 6/5 · 68/4 · 5 8/5 · 6 8/6 · 7 8/7 · 8

10/5 · 6 10/6 · 7 10/7 · 8 10/8 · 9 10/9 · 10. . .

This triangle is doubly regular and doubly descending triangle but neither left balancing(w5

2 6= w42(1− w5

5)) nor right balancing (w52 6= w4

1(1− w51)).

(b) For λ = 0, n0(x) = 1− x and the weights of the triangle are wnj = 1n , for all j =

1, . . . , n and n ≥ 1.

11/2 1/2

1/3 1/3 1/31/4 1/4 1/4 1/4

1/5 1/5 1/5 1/5 1/5. . .

This triangle is doubly regular, descending and balancing.

4. Circular negation defines a doubly regular, left descending and left balancing, but neitherright descending (w5

2 6= w41(1−w5

1)) nor (w32 6≤ w4

3) right balancing triangle: n(x) = 1−x2then wnj = 2j−1

n2 ; n ≥ 1, j = 1, . . . , n.

11/4 3/4

1/9 3/9 5/91/16 3/16 5/16 7/16

1/25 3/25 5/25 7/25 9/25. . .

The Extended Aggregation Functions associated with the above negations are, respectively:max,min, arithmetic mean (Sugeno negation, with λ = 0). Similarly, choosing a function f asin Definition 5 we obtain different Extended Quasi–Linear Weighted Mean for each triangle.

3.2 Generation by sequences

Let us consider a sequence of positive real numbers λ1, λ2, . . . ≥ 0 such that λ1 > 0. Then wecan define a weighting triangle in the following way :

wni =λi

λ1 + · · ·+ λn,

for i = 1, . . . , n and n ≥ 1. Observe thatn∑i=1

wni =

n∑i=1

λi

λ1+···+λn = 1, therefore we obtain a weighting

17

triangle.

Proposition 12 Every triangle generated by {λn} is left regular if and only if the followingsequence {

Sn+1

Sn

}(11)

is decreasing, with Sn = λ1 + · · ·+ λn.

Proof. Suppose that the triangle is left regular. Then

p∑j=1

wn+1j ≤

p∑j=1

wnj ≤p+1∑j=1

wn+1j ,

for all p = 1, . . . , n.According to the latter condition and the following expression

wnj =λjSn,

we havep∑j=1

λjSn+1

≤p∑j=1

λjSn≤

p+1∑j=1

λjSn+1

,

for all p = 1, . . . , n.Taking p = n− 1 in the second inequality of left regularity condition, it follows that

Sn−1Sn

≤ SnSn+1

.

This shows that{Sn+1

Sn

}is a decreasing sequence.

In the proof of the converse statement, assume that{Sn+1

Sn

}is a decreasing sequence and we try to show that it verifies

p∑j=1

λjSn+1

≤p∑j=1

λjSn≤

p+1∑j=1

λjSn+1

,

for all p = 1, . . . , n, or equivalently

SpSn+1

≤ SpSn≤ Sp+1

Sn+1.

The first inequality is evident, since Sn is increasing, and the second one is also obvious because{Sn+1

Sn

}is decreasing.

18

Remark 4 It can be shown that the same result is obtained for the case of right regular trian-

gles. Therefore, any sequence {λn} defines a doubly regular triangle if and only if the{Sn+1

Sn

}is

decreasing. However, any regular triangle can not be generated by a sequence, as it occurs withthe “normalized” triangle of Pascal.

Example 7 The triangles generated by the following sequences are doubly regular:

a) decreasing sequences,

b) arithmetic progressions,

c) geometric progressions.

d) The Fibonacci sequence: 1, 1, 2, 3, 5, 8, . . . determines the following triangle

11/2 1/2

1/4 1/4 2/41/7 1/7 2/7 3/7

1/12 1/12 2/12 3/12 5/12. . .

It is easy to see by induction that the latter sequence fulfills:

d.1) Sn = λn+2 − 1,

d.2) (λn+3−1)(λn+1−1) ≤ (λn+2−1)2, hence it can be deduced that{Sn+1

Sn

}is decreasing.

According to the Remark 3, this shows that the triangle generated by the Fibonaccisequence is doubly regular.

Remark 5 There exist doubly regular triangles which can be generated by means of a increasingquantifier or via a sequence. However, there are also triangles which only can be determined bya increasing quantifier or a sequence. For this, we quote the following examples.

Example 8 a) Considering Q(x) = x, we easily obtain that the weights defined by meansof this quantifier are wnj = 1/n for all j = 1, . . . , n, and n ≥ 1. Moreover, taking intoaccount the sequence λn = c for all n ≥ 1, c being a constant, the same weights areobtained. These weights determine the weighting triangle associated with the arithmeticmean.

b) The triangle of the median operator can be obtained by means of a increasing quantifier,as we mention at the beginning of this section. However, as w3

1 = 0, the triangle can notbe generated by a sequence.

c) Let us consider the sequence λn = 2n−1 for all n ≥ 1 be a sequence. It is easy to see thatthe triangle generated by this sequence can not be defined by a increasing quantifier.

Proposition 13 1. A triangle generated by a sequence of positive real numbers λn is alwaysleft descending.

19

2. Any sequence of positive real numbers λn defines a right descending triangle, with theweights given by:

wni =λn−i+1

Sn,

for i = 1, . . . , n.

Proof. In the proof of both cases it is enough to consider that Sn ≤ Sn+1.

Remark 6 Not all triangles defined by a sequence of positive real numbers λn as follows

wni =λiSn,

for i = 1, . . . , n are right descending. For example, the triangle generated by the sequence λn = nis not right descending. However, the triangle generated in this way by a decreasing sequenceλn isright descending.

Let us see other sequences that generate right descending triangles with weights given by

wni =λiSn.

As we know, they will be left descending triangles as well.

Example 9 The following sequences λn of positive real numbers generate doubly descendingtriangles with weights given by wni = λi

Sn, for i = 1, . . . , n :

1. Decreasing sequences.

2. Geometric progressions.

3. Constant (different from zero) sequences.

Proposition 14 (i) Any triangle generated by a sequence {λn} is a left balancing triangle.

(ii) Any sequence of positive real numbers {λn} defines a right balancing triangle in the fol-lowing way:

wni =λn−i+1

Sn,

for i = 1, . . . , n.

Proof. The condition of a left balancing triangle is wn+1i = wni (1 − wn+1

n+1) for all i = 1, . . . , n.

But in our case wni = λiSn

, then

wni (1− wn+1n+1) =

λiSn

(1− λn+1

Sn+1) =

λiSn+1

= wn+1i .

The second result can be shown in the same form.

20

The Remark 5 can be rewritten here for the case of balancing instead of descending triangles.In this way, next example is similar to Example 9.

Example 10 The following sequences {λn} of positive real numbers generate doubly balancingtriangles with weights given by wni = λi

Sn, for i = 1, . . . , n :

1. Geometric progressions.

2. Constant (different from zero) sequences.

3.3 Generation by fractals

Before showing the general idea about this section let us consider some examples of fractalstructures (see [2]).

Example 11 Sierpinski triangle : We define the weights of a weighting triangle following thesteps of the construction of this well–known fractal. We obtain

wni = 3i−11

4i, i = 1, . . . , n− 1 and wnn = 3n−1

1

4n−1, for all n ≥ 1.

The sequence 14 ,

142, . . . , 1

4n−1 , in each row n > 1 of the triangle, indicates the different areas ofthe (n − 1)–steps or the iterations in the construction of the fractal. On the other hand, thesequence 1, 3, . . . , 3n−1 shows the number of corresponding smaller triangles with the previousareas. More clearly, the weights wni = 3i−1 1

4i, i = 1, . . . , n − 1 indicate that we have 3i−1

small triangles in the construction of the fractal whose area is equal to 14i

. And the same forthe weight wnn = 3n−1 1

4n−1 . These weights satisfy the conditions :

(i) 0 ≤ wni ≤ 1 for all i = 1, . . . , n,

(ii)n∑i=1

wni = 1,

(iii) wn+1n+1 ≤ wnn,

(iv) wnn = wn+1n + wn+1

n+1 ,

for all n ≥ 1. Thus, the Sierpinski triangle is a weighting triangle according to Definition 4.

1

1.14 3.14

1.14 3. 142

32. 142

1.14 3. 142

32. 143

33. 143

. . . . . . . .

Now, we can consider the Sierpinski carpet to generate in the same way a weighting triangle.We have

21

1

1 · 19 8 · 191 · 19 8 · 1

9282 · 1

92

1 · 19 8 · 192

82 · 193

83 · 193

. . . . . . . .

where wni = 8i−1 · 19i, for i = 1, . . . , n− 1 and wnn = 8n−1 · 1

9n−1 .

This triangle also fulfills the conditions (i), (ii), (iii) and (iv) mentioned in the case of theSierpinski triangle.

After considering these two examples we give a generalization for this kind of fractals whichwe will call Sierpinski’s Family. To obtain this family of triangles we follow the same process asin the previous two examples.

Thus, let us consider an unitary surface and divide it into in two parts, one with an areaequal to a

b < 1 and the second with an area equal to 1 − ab = b−a

b . After that, we take the

surface with area b−ab and we divide it again into two parts: the first one with area a

b ·(b−ab

)and the other one with area ( b−ab )2. In order to build the third row of the weighting trianglewe need to take into account the surface with area a

b , so we obtain

a

b,

a

b·(b− ab

), (

b− ab

)2

Therefore, according to this idea we obtain the following weighting triangle

1

ab

b−ab

ab ( b−ab )(ab ) ( b−ab )2

ab ( b−ab )(ab ) ( b−ab )2(ab ) ( b−ab )3

. . . . . . . .

The weights are

wni =a

b· (b− a

b)i−1, i = 1, . . . , n− 1 and wnn = (

b− ab

)n−1.

They also hold the mentioned properties in (i), (ii), (iii) and (iv). Notice that, if we want toobtain a fractal structure at the end of this process we need to assume that a

b 6=b−ab ; a

b < 1and also that the surface with area a

b should have the same shape as the original surface.

Taking ab = 1

4 and ab = 1

9 we obtain respectively the Sierpinski triangle and the Sierpinskicarpet.

22

Remark 7 The aggregation operators defined by means of a weighting triangle of the Sierpin-ski family can be called “geometric operators”, since the weights in each row of the triangledetermine a geometric progression with reason b−a

b .

It is easy to prove the following results.

Proposition 15 1. Any triangle of the Sierpinski’s family is a doubly regular triangle.

2. Any triangle of the Sierpinski’s family is a doubly descending triangle.

3. Any triangle of the Sierpinski’s family is right balancing but not left balancing.

Proof. In the first case, we show the proof for a left regular triangle, then we need to show that

p∑i=1

wn+1i ≤

p∑i=1

wni ≤p+1∑i=1

wn+1i , (12)

for p ≤ n. For the first inequality, taking into account that

p∑i=1

wn+1i =

p∑i=1

wni ,

for p ≤ n− 1 and also that

wn+1n =

a

b· wnn ≤ wnn,

it can be assured thatp∑i=1

wn+1i =

p∑i=1

wni , for all p ≤ n.

Similarly, the second inequality of (12) can be obtained from the relations

p∑i=1

wni =p∑i=1

wn+1i ≤

p∑i=1

wn+1i + wn+1

p+1 =p+1∑i=1

wni ,

for p ≤ n− 1. The case p = n is trivial.In the second case, it is easy to prove that the triangle is doubly descending; we only explicit

the case of left descending, i.e., wn+1i ≤ wni for all i = 1, . . . , n, since wn+1

i = wni for alli = 1, . . . , n− 1, and wn+1

n = ab ·w

nn ≤ wnn. Finally, we see that this triangle is not left balancing,

i.e.,

wn+1i =

a

b(b− ab

)i−1 6= a

b(b− ab

)i−1(1− (b− ab

)n) = wni (1− wn+1n+1),

for i = 1, . . . , n− 1. The right balancing property is easy to prove.

23

Consequently, note that any EOWA operator generated by a weighting triangle of the Sier-pinski’s family is also an EAF and it can be said the same for an EQLWM generated by aweighting triangle of this family, as well as their reverse functions.

Out of this family, there exist other fractal structures that generate weighting triangles. Oneexample is obtained by the degenerated Pytagorean tree fractal. In this case, we have

1

1 · 14 1 · 341 · 1

16 2 · 316 1 · 9

16

1 · 164 3 · 3

64 3 · 964 1 · 2764

. . . . . . . .

The weights are defined by the following expression

wni =3i−1

4n−1

(n− 1

i− 1

), i = 1, . . . , n and n > 1.

The fraction 3i−1

4n−1 indicates the corresponding area of each square and the number(n−1i−1)

points out the number of squares of each surface. This triangle can also be generated by theBinomial distribution.

It is easy to see that the weighting triangle defined by the degenerated Pytagorean tree fractalis not a left descending triangle, since the condition (3) is only true for n ≤ 4, then the EQLWMdefinided by means of this triangle is not an EAF. This triangle is not balancing either.

The main interest about the weighting triangles generated by fractal structures lies in the factthat this kind of triangles can not be generated by a sequence, nor by a increasing quantifier or anegation. On the other hand, it is an open problem how to find the fractal structure associatedwith a weighting triangle given.

4 Orness of a weighting triangle

First, recall the definition of the degree of orness of an n-dimensional OWA operator

Definition 8 (Yager 1988) Assume An is an n–dimensional OWA operator with weighting listWn = (w1, . . . , wn). The degree of orness of An is defined to be

orness(An) =1

n− 1

n∑i=1

(n− i)wi. (13)

We will denote the degree of orness of An by On.

Now we introduce the degree of orness of a weighting triangle, i. e., the orness of the EOWAoperator defined by the triangle, as follows.

Definition 9 Let 4wni be a weighting triangle. The degree of orness of 4wni , O4 is defined tobe the limit of the sequence of the degree of orness of each row of 4wni , i.e.,

O4 = limn→+∞

On = limn→+∞

1

n− 1

n∑i=1

(n− i)wi. (14)

24

According to the previous definition we want to point out that the orness has a sense onlyif we use the weighting triangle for an EOWA operator.

Example 12 1) The orness of the weighting triangles associated with the Maximum (Mini-mum) is equal to 1, (0), since the orness in each row of the triangle is 1, (0).

2) The orness of any triangle that belongs to the Sierpinski’s family is also equal to 1, as weshow next.The weights of this family of triangles are:

wnj =a

b(b− ab

)j−1, j = 1, . . . , n− 1 and wnn = (b− ab

)n−1.

Then

O4 = limn→+∞

On = limn→+∞

1

n− 1

n−1∑j=1

(n− j) · ab· (b− a

b)j−1

=

= limn→+∞

1

n− 1

n n−1∑j=1

a

b(b− ab

)j−1 −n−1∑j=1

ja

b(b− ab

)j−1

=

= limn→+∞

1

n− 1

[n · a

b·

( b−ab )n−1 − 1b−ab − 1

− a

b·

1− ( b−ab )n−1

(ab )2− (n− 1) ·

( b−ab )n−1

ab

]=

= limn→+∞

[n

n− 1

(1− (

b− ab

)n−1)− 1

n− 1·

1− ( b−ab )n−1

ab

− (b− ab

)n−1]

= 1,

due to b−ab < 1, the first limit is equal to 1 and the other are equals 0.

3) The triangles defined by Sugeno negations have orness 1/2, if λ = 0 or 2(1− ln 2), if λ = 1.We proof the second case, in which the weights of the triangle are:

wnj =2n

(n+ j)(n+ j − 1), j = 1, . . . , n.

Then

O4 = limn→+∞

(On) = limn→+∞

2n

n− 1

n∑j=1

n− j(n+ j)(n+ j − 1)

.

We use the following equality in order to calculate this limit

n− j(n+ j)(n+ j − 1)

=2j

n+ j+

1− 2j

n+ j − 1.

By substituting this in the expression of the orness, a simple calculation gives

O4 = limn→+∞

(On) = limn→+∞

2n

n− 1

n∑j=1

2j

n+ j+

n∑j=1

1− 2j

n+ j − 1

=

= limn→+∞

2n

n− 1

[1−

(1

n+

1

n+ 1+ . . .+

1

2n− 1

)].

25

But

1

n+

1

n+ 1+ . . .+

1

2n− 1= 1 +

1

2+

1

3+ . . .+

1

2n− 1−(

1 +1

2+

1

3+ . . .+

1

n− 1

)=

= ln(2n− 1) + c+ ε2n−1 − ln(n− 1)− c− εn−1 =

= ln(2n− 1

n− 1) + ε2n−1 − εn−1,

where c is the Euler’s constant. Thus,

O4 = limn→+∞

(On) = limn→+∞

2n

n− 1

[1− ln(

2n− 1

n− 1)− ε2n−1 + εn−1

]= 2(1− ln 2),

since the limits of ε2n−1 and εn−1 are equal to 0. The orness of this triangle is below 1.

4) The orness of the triangle associated with the circular negation is equal to 1/3. Now theweights are:

wnj =2j − 1

n2, j = 1, . . . , n.

Then

O4 = limn→+∞

(On) = limn→+∞

1

n− 1

n∑j=1

(n− j)(2j − 1

n2) =

= limn→+∞

1

(n− 1)n2

n2

n∑j=1

j − n

− 2n∑j=1

j2 +n∑j=1

j

=

= limn→+∞

1

(n− 1)n2

[n

(2(1 + n)n

2− n

)− n(n+ 1)(2n+ 1)

3+

(1 + n)n

2

]=

= limn→+∞

1

(n− 1)n2

[2n3 − 3n2 + n

6

]= 1/3.

Remark 8 Note that there are weighting triangles that their orness does not exist, like thefollowing one:

1

0 1

0 0 1

0 1 0 0

0 0 0 0 1

. . . . . . . .

Moreover, this triangle is not regular, nor descending and neither balancing.

Although it is not possible to assure that the orness of the different types of weightingtriangles exists we have found some contions about the ornesses of two consecutive rows, asfollows.

26

Proposition 16 (i) The orness of any left descending triangle satisfies

On+1 ≤ On +1

n,

for all n > 1.

(ii) The orness of any right descending triangle satisfies

On+1 ≤ On + wn+11 ,

for all n > 1.

Proof. We only develop the proof of the second case (ii) since both proves are similar. We have

On+1 =1

n

n+1∑j=1

(n+ 1− j)wn+1j =

1

n

n · wn+11 +

n+1∑j=2

(n+ 1− j)wn+1j

.Replacing j by i+ 1, the last expression can be rewritten as

On+1 =1

n

n · wn+11 +

n+1∑j=2

(n+ 1− j)wn+1j

= wn+11 +

1

n

n∑i=1

(n− i)wn+1i+1 .

On the other hand, we have

On =1

n− 1

n∑i=1

(n− i)wni ,

then due to the right descending condition wn+1i+1 ≤ wni , for i = 1, . . . , n, we get

1

n

n∑i=1

(n− i)wn+1i+1 ≤

1

n− 1

n∑i=1

(n− i)wni .

ThusOn+1 ≤ On + wn+1

1 .

For (i), it is enough to consider the condition (3) and the Definition 9.

Similarly, we can mention the corresponding results of the orness of balancing triangles.

Proposition 17 (i) The orness of any left balancing triangle satisfies

On+1 ≤ On +1

n,

for all n > 1.

(ii) The orness of any right balancing triangle satisfies

On+1 ≤ On + wn+11 ,

for all n > 1.

27

Proof. We here show the proof of the first case. Using the condition (11) and the Definition 9we can easily prove the case (ii). We have

On+1 =1

n

n+1∑j=1

(n+ 1− j)wn+1j =

1

n

n∑j=1

(n+ 1− j)wn+1j =

=1

n

n∑j=1

(n− j)wn+1j +

n∑j=1

wn+1j

,and considering first the condition of left balancing triangles and the following equality

∑nj=1w

n+1j =

(1− wn+1n+1) we can rewrite the last expression as

1

n

n∑j=1

(n− j)wnj (1− wn+1n+1) + (1− wn+1

n+1)

= (1− wn+1n+1)

1

n

n∑j=1

(n− j)wnj +1

n

.On the other hand,

On =1

n− 1

n∑j=1

(n− j)wnj .

From the above expressions we can conclude that

On+1 ≤ On +1

n.

Proposition 18 Every weighting triangle 4wni generated by a continuous, symmetric increas-ing quantifier with respect to 1/2 has orness equal to 1/2.

Proof. Consider 4wni generated by a increasing quantifier Q symmetric with respect to thepoint 1/2. We have that the orness for each row of the triangle is defined as follows

On =1

n− 1

n∑j=1

(n− j)wnj ,

but

wnj = Q(j

n)−Q(

j − 1

n),

then

On =1

n− 1

n∑j=1

(n− j)Q(j

n)−

n∑j=1

(n− j)Q(j − 1

n)

.Developing the last expression we obtain

On =1

n− 1

[Q(

1

n) +Q(

2

n) +Q(

3

n) + . . .+Q(

n− 4

n) +Q(

n− 3

n) +Q(

n− 2

n) +Q(

n− 1

n)

],

and taking into account now that Q is symmetric it follows that

On =1

n− 1

[Q(n− 2

2n) +

n− 2

2)

], if n is even

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and

On =1

n− 1

[n− 1

2

], if n is odd .

Thus, in any case,O4 = lim

n→+∞On = 1/2.

This concludes the proof.

Remark 9 The above result is also true for any symmetric weighting triangle 4wni , i.e., wni =wnn−i+1. In this case, it is sufficient to see that the orness of each even row of the triangle is 1/2and the limit of the orness of odd rows is also 1/2. When we know the expression of its weights,we can calculate the orness of the triangle like in the following example.

Example 13 Consider the “normalized” triangle of Pascal and we calculate its degree of orness.This must be equal to 1/2, since this triangle is symmetric with respect to 1/2. Thus,

O4 = limn→+∞

1

2n−1(n− 1)

n−1∑j=1

(n− j)(n− 1

j − 1

)=

= limn→+∞

1

2n−1(n− 1)

n n−1∑j=1

(n− 1

j − 1

)−n−1∑j=1

j

(n− 1

j − 1

) =

= limn→+∞

1

2n−1(n− 1)

n n−1∑j=1

(n− 1

j − 1

)−n−2∑j=0

(j + 1)

(n− 1

j

) =

= limn→+∞

1

2n−1(n− 1)

n n−1∑j=1

(n− 1

j − 1

)−n−2∑j=0

(n− 1

j

)−n−2∑j=0

j

(n− 1

j

) =

= limn→+∞

1

2n−1(n− 1)

n n−2∑j=0

(n− 1

j

)−n−2∑j=0

(n− 1

j

)−n−2∑j=0

j

(n− 1

j

) =

= limn→+∞

1

2n−1(n− 1)

[(n− 1)(2n−1 − 1)− 2n−2(n− 1)

]=

= limn→+∞

1

2n−1(n− 1)

((n− 1)(2n−2 − 1)

)=

= limn→+∞

(1

2− 1

2n−1

)= 1/2.

5 Conclusions

We have given different ways to determine the main three classes of weighting triangles in orderto define the Extended Aggregation Functions associated with each of these weighting triangles.All of these succesful and useful ways are based on geometical and mathematical processes.In the last section, we introduce the degree of orness of a weighting triangle although it is notpossible to show that if always exists for the mentioned three classes of triangles in this work. Wealso show that a symmetric triangle has orness equal to 1/2 and give an example of a weightingtriangle whose orness does not exist.

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Remark 10 Let 4wni be generated by a quantifier Q. Then the orness of the triangle alwaysexists and it is equal to 1–mean value of a random variable on [0, 1] with distribution functionQ∗ : [0,∞]→ [0, 1], Q∗(x) = sup(Q(t)/t < x, t ∈ [0, 1]).

Moreover, if Q is differentiable then the orness of the triangle is given by 1−1∫0x ·Q′(x)dx,

since

O4 = limn→+∞On = limn→+∞1

n−1n∑i=1

(n− i)wi = limn→+∞nn−1

n∑i=1

(1− in)wi =

= 1− limn→+∞n∑i=1

in(Q( in)−Q( i−1n )) = 1−

1∫0x ·Q′(x)dx.

Hence, under the above conditions the computation of O4 results easier by means of the lastexpression than from the definition, as we show in the following cases:

(1) For a weighting triangle generated by Q(x) = x2, we have

O4 = 1−1∫

0

2x2dx = 1− 2/3 = 1/3.

(2) With Q(x) = x, we obtain

O4 = 1−1∫

0

xdx = 1− 1/2 = 1/2.

(3) If the quantifier is obtained by means of Sugeno negation, with λ = 1, i. e., Q(x) = 2x1+x

then

O4 = 1−1∫

0

2x

(1 + x)2dx = 2− 2ln2.

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