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### Paper:

TR05-111 | 3rd October 2005 00:00

#### A Generic Time Hierarchy for Semantic Models With One Bit of Advice

TR05-111
Authors: Dieter van Melkebeek, Konstantin Pervyshev
Publication: 8th October 2005 19:21
Keywords:

Abstract:

We show that for any reasonable semantic model of computation and for
any positive integer $a$ and rationals $1 \leq c < d$, there exists a language
computable in time $n^d$ with $a$ bits of advice but not in time $n^c$
with $a$ bits of advice. A semantic model is one for which there exists a
computable enumeration that contains all machines in the model but may
also contain others. We call such a model reasonable if it has an
efficient universal machine that can be complemented within the model
in exponential time and if it is efficiently closed under deterministic
transducers.

Our result implies the first such hierarchy theorem for randomized
machines with zero-sided error, quantum machines with one- or zero-sided
error, unambiguous machines, symmetric alternation,
Arthur-Merlin games of any signature,
interactive proof protocols with one or multiple provers, etc. Our
argument yields considerably simpler proofs of known hierarchy
theorems with one bit of advice for randomized or quantum machines with
two-sided error and randomized machines with one-sided error.

Our paradigm also allows us to derive stronger separation results in a
unified way. For models that have an efficient universal machine that can be
simulated deterministically in exponential time and that are efficiently
closed under randomized reductions with two-sided error, we establish the
following: For any constants $a$ and $c$, there exists a language computable
in polynomial time with one bit of advice but not in time $n^c$ with
$a \log n$ bits of advice.
In particular, we obtain such separation for randomized and quantum
machines with two-sided error.
For randomized machines with one-sided error, we get that for any
constants $a$ and $c$ there exists a language computable in polynomial
time with one bit of advice but not in time $n^c$ with $a (\log n)^{1/c}$