Advanced Methods for Inconsistent Knowledge Management by Ngoc Thanh Nguyen

By Ngoc Thanh Nguyen

This ebook is a primary. It fills a massive hole out there and offers a large photograph of clever applied sciences for inconsistency solution. the necessity for this solution of data inconsistency arises in lots of useful functions of computers. this sort of inconsistency effects from using a variety of assets of information in understanding useful projects. those assets are frequently independent and use various mechanisms for processing wisdom concerning the similar actual international. this may bring about compatibility difficulties.

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2. Consensus Theory – A Case Study • • • 51 Neutrality (N): The consensus for situation (A, P) should not be changed despite changing the indexes of elements of set A; that is, all votes should be treated in the same way. Symmetry (S): A consensus function should be symmetrical if conditions (N) and (ND) are satisfied; that is, it should justly treat the voters and the votes. Condorcet consistency (Cc): A consensus function C should satisfy condition Cc, if for any profiles P1 and P2 the following dependency is true, C ( P1 ) ∩ C ( P2 ) ≠ ∅ ⇒ C ( P1 ∪ P2 ) = C ( P1 ) ∩ C ( P2 ) ; that is, if two groups of voters voting separately have common preferences then they should have the same preferences voting together.

In fact some of them may be identical because of repetitions. • The total average distance in profile X: ∑ dt_mean(X) = d ( x, y ) x , y∈ X M ( M + 1) . This value serves to represent the average distance of all distances between elements of profile X. The sum of these distances is expressed by the numerator of the quotient. However, one can ask a question: why is k2 not in the denominator, but k(k + 1) is? The answer is: in the numerator each distance d(x,y), where x ≠ y, occurs exactly twice, whereas each distance d(x,y), where x = y, occurs exactly only once.

Let Y = X − {y} and card(X) = n > 1, then card(Y) = n − 1. Let x′ be such an element of universe U that 34 2. Model of Knowledge Conflict d(x′,Y) = min {d(t,Y): t ∈ U}. We have then d(x′,Y) ≤ d(x,Y). Also, from d(x,y) = max d (x, z ) z∈ X it follows that (n − 1) ⋅ d(x, y) ≥ d(x,Y). Thus min {d (t , X ) : t ∈ U } d ( x, X ) d ( x, Y ) + d ( x, y ) = = n n card ( X ) ≥ min {d (t , Y ) : t ∈U } d ( x, Y ) d ( x ' , Y ) = . ≥ n −1 n −1 card (Y ) Because function c satisfies postulate P6, then there should be c(X) ≤ c(Y).

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