Complexity Classes

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Tutorial 2 : Complexity classes Computational complexity theory, 5th semester. 2017 Exercise 1 — Belonging to complexity classes For each of the problems in exercice 1 of the previous tutorial, tell to which of the following complexity class that problem belongs: P, NP, Co-NP, PSPACE, EXPTIME, NEXPTIME or EXPSPACE. Exercise 2 — Démonstrations simples 1. Show that DTIME(f(n)) ⊂ NTIME(f(n)). 2. Show that DTIME(f(n)) ⊂ DSPACE(f(n)). 3. Show that, if NP = P then Co-NP = NP. Is the reciprocate true? 4. Show that P = Co-P. 5. Using the Savitch theorem, saying that NSPACE(f(n)) ⊂ DSPACE(f(n)2), prove that PSPACE = NPSPACE and EXPSPACE = NEXPSPACE. Deduce that NPSPACE = Co- NPSPACE and that NEXPSPACE = Co-NEXPSPACE. 6. Using the deterministic time hierarchy theorem saying that DTIME(f(n)) ( DTIME(f(n)2) (among other results), prove that P ( EXPTIME. Exercise 3 — PSPACE ⊂ EXPTIME 1. Let Π be a PSPACE problem, how much cells of the tape can use a Turing machine that solve Π in polynomial space? 2. In how many configurations that machine can be? A configuration is defined by the position of the head, the state pointed by the state register and what is written on the tape. 3. Can the machine be twice in the same configuration during the computation? 4. Deduce that the complexity of the machine is at most exponential. Exercise 4 — P = NP ) EXPTIME = NEXPTIME We assume that P = NP. 1. Let Π be a NEXPTIME problem. What is the complexity of a non-deterministic Turing machine solving Π in exponential time? c 2jxj 2. Let c be a constant, let Π2 be a decision problem for which the instances are {x · π nx ⊂ f0; 1g∗g, where πd is the symbol π repeated d times. The positive instances are the ones where x encodes a positive instance of Π. Build a non deterministic algorithm that solves Π2 in polynomial time. To obtain such a complexity, choose well the constant c. 3. Deduce that Π 2 EXPTIME. 4. Deduce that EXPTIME = NEXPTIME. 1.

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COMPLEXITY THEORY Review Lecture 13: Space Hierarchy and Gaps
COMPLEXITY THEORY Review Lecture 13: Space Hierarchy and Gaps Markus Krotzsch¨ Knowledge-Based Systems TU Dresden, 26th Dec 2018 Markus Krötzsch, 26th Dec 2018 Complexity Theory slide 2 of 19 Review: Time Hierarchy Theorems Time Hierarchy Theorem 12.12 If f , g : N N are such that f is time- → constructible, and g log g o(f ), then · ∈ DTime (g) ( DTime (f ) ∗ ∗ Nondeterministic Time Hierarchy Theorem 12.14 If f , g : N N are such that f → is time-constructible, and g(n + 1) o(f (n)), then A Hierarchy for Space ∈ NTime (g) ( NTime (f ) ∗ ∗ In particular, we ﬁnd that P , ExpTime and NP , NExpTime: , L NL P NP PSpace ExpTime NExpTime ExpSpace ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ , Markus Krötzsch, 26th Dec 2018 Complexity Theory slide 3 of 19 Markus Krötzsch, 26th Dec 2018 Complexity Theory slide 4 of 19 Space Hierarchy Proving the Space Hierarchy Theorem (1) Space Hierarchy Theorem 13.1: If f , g : N N are such that f is space- → constructible, and g o(f ), then For space, we can always assume a single working tape: ∈ Tape reduction leads to a constant-factor increase in space • DSpace(g) ( DSpace(f ) Constant factors can be eliminated by space compression • Proof: Again, we construct a diagonalisation machine . We deﬁne a multi-tape TM Therefore, DSpacek(f ) = DSpace1(f ). D D for inputs of the form , w (other cases do not matter), assuming that , w = n hM i |hM i| Compute f (n) in unary to mark the available space on the working tape Space turns out to be easier to separate – we get: • Space Hierarchy Theorem 13.1: If f , g : N N are such that f is space- Initialise a separate countdown tape with the largest binary number that can be → • constructible, and g o(f ), then written in f (n) space ∈ Simulate on , w , making sure that only previously marked tape cells are • M hM i DSpace(g) ( DSpace(f ) used Time-bound the simulation using the content of the countdown tape by • Challenge: TMs can run forever even within bounded space.
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