نتایج جستجو برای: quasi p
تعداد نتایج: 1343077 فیلتر نتایج به سال:
Let F be a class of finite groups, p a prime and π a set of some primes. A finite group G is called a p-quasi-F-group (respectively, by π-quasi-F-group) provided that for every F-eccentric G-chief factor H/K of order divisible by p (respectively, by at least one prime in π), the automorphisms of H/K induced by all elements of G are inner. In this paper, we obtain the characterizations of p-quas...
The notion of h-hypertree was deened in 9] in order to improve the Bonferroni inequalities. We prove that the problem of nding the optimal spanning h-hypertree of a complete hypergraph is NP complete, for h 3. Moreover, no polynomial time approximate algorithm even with poor approximation ratio seems to exist for this problem. The existence of such an algorithm for the minimum version achieving...
We present a computational analysis of de re, de dicto, and de se belief and knowledge reports. Our analysis solves a problem rst observed by Hector-Neri Casta~ neda, namely, that the simple rulè(A knows that P) implies P' apparently does not hold if P contains a quasi-indexical. We present a single rule, in the context of a knowledge-representation and reasoning system, that holds for all P, i...
Dualization problems have been intensively studied in combinatorics, AI and pattern mining for years. Roughly speaking, for a partial order (P, ) and some monotonic predicate Q over P , the dualization consists in identifying all maximal elements of P verifying Q from all minimal elements of P not verifying Q, and vice versa. The dualization is equivalent to the enumeration of minimal transvers...
1. Introduction. Combining the work of many people yields a complete quasi-isometry classification of irreducible lattices in semisimple Lie groups (see [F] for an overview of these results). One of the first general results in this classification is the complete description, up to quasi-isometry, of all nonuniform lattices in semisimple Lie groups of rank 1, proved by R. Schwartz [S1]. He show...
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