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Levke Bentzien

dblp:16/6302 · DBLP profile ↗
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3ranked-venue papers
1as first author
0since 2021 · last 2000
—ORCID · none

Domains — the database's venue-derived domains; a paper can count in several

Theory of computation · 3 · 1 first-author

Expertise — from the expertise taxonomy: the topics of the expert's papers under the CCF categories. A weight counts papers with recency: 1 for a paper about the topic, 0.3 when the topic is its context, halved every five years.

Theoretical computer science
1 paper
Computational complexity · 100%

Topics — the 2 heaviest of 2, each with the papers that count most for it

TopicWeightPapersLastEvidence papers
Computational complexity › reduction
bounded truth-table reductions
0.011997
Separating NP-Completeness Notions under Strong Hypotheses · CCC 1997
Computational complexity › structural complexity
resource-bounded measure
0.011997
Separating NP-Completeness Notions under Strong Hypotheses · CCC 1997
YearPublicationVenuePosition
2000 Separating NP-Completeness Notions under Strong Hypotheses
Klaus Ambos-Spies, Levke Bentzien
J. Comput. Syst. Sci.2
1998 Positive Turing and Truth-Table Completeness for NEXP Are Incomparable
Levke Bentzien
MFCS1
1997 Separating NP-Completeness Notions under Strong Hypotheses
abstract
J.H. Lutz (1993) proposed the study of the structure of the class NP=NTIME(poly) under the hypothesis that NP does not have p-measure 0 (with respect to Lutz's resource bounded measure. J.H. Lutz and E. Mayordomo (1996) showed that, under this hypothesis, NP-m-completeness and NP-T-completeness differ and they conjectured that further NP-completeness notions can be separated. Here we prove this conjecture for the bounded-query reducibilities. In fact we consider a new weaker hypothesis, namely the assumption that NP is not p-meager with respect to the resource bounded Baire category concept of Ambos-Spies et al.. We show that this category hypothesis is sufficient to get: (i) For every k/spl ges/2, NP-btt(k)-completeness is stronger than NP-btt(k+1)-completeness. (ii) For every k/spl ges/1, NP-bT(k)-completeness and NP-btt(k+1)-completeness are both stronger than NP-bT(k+1)-completeness. (iii) NP-btt-completeness is stronger than NP-tt-completeness.
Klaus Ambos-Spies, Levke Bentzien
CCC2