C++ Neural Networks and Fuzzy Logic: Preface

C++ Neural Networks and Fuzzy Logic

by Valluru B. Rao

MTBooks, IDG Books Worldwide, Inc.

ISBN: 1558515526   Pub Date: 06/01/95

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Intersection and Complement of Two Fuzzy Sets

Analogously, the degree of membership of an element in the intersection of two fuzzy sets is the minimum, or

the smaller value of its degree of membership individually in the two sets forming the intersection. For

example, if today has 0.8 for degree of membership in the set of rainy days and 0.5 for degree of membership

in the set of days of work completion, then today belongs to the set of rainy days on which work is completed

to a degree of 0.5, the smaller of 0.5 and 0.8.

Recall the fuzzy sets A and B in the previous example. A = (0.9, 0.4, 0.5, 0) and B = (0.7, 0.6, 0.3, 0.8).

A[cap]B, which is the intersection of the fuzzy sets A and B, is obtained by taking, in each component, the

smaller of the values found in that component in A and in B. Thus A[cap]B = (0.7, 0.4, 0.3, 0).

The idea of a universal set is implicit in dealing with traditional sets. For example, if you talk of the set of

married persons, the universal set is the set of all persons. Every other set you consider in that context is a

subset of the universal set. We bring up this matter of universal set because when you make the complement

of a traditional set A, you need to put in every element in the universal set that is not in A. The complement of

a fuzzy set, however, is obtained as follows. In the case of fuzzy sets, if the degree of membership is 0.8 for a

member, then that member is not in that set to a degree of 1.0 – 0.8 = 0.2. So you can set the degree of

membership in the complement fuzzy set to the complement with respect to 1. If we return to the scenario of

having a degree of 0.8 in the set of rainy days, then today has to have 0.2 membership degree in the set of

nonrainy or clear days.

Continuing with our example of fuzzy sets A and B, and denoting the complement of A by A’, we have A’ =

(0.1, 0.6, 0.5, 1) and B’ = (0.3, 0.4, 0.7, 0.2). Note that A’ [cup] B’ = (0.3, 0.6, 0.7, 1), which is also the

complement of A [cap] B. You can similarly verify that the complement of A [cup] B is the same as A’ [cap]

B’. Furthermore, A [cup] A’ = (0.9, 0.6, 0.5, 1) and A [cap] A’ = (0.1, 0.4, 0.5, 0), which is not a vector of

zeros only, as would be the case in conventional sets. In fact, A and A’ will be equal in the sense that their fit

vectors are the same, if each component in the fit vector is equal to 0.5.