p systems simulating oracle computations
DESCRIPTION
TRANSCRIPT
P systems simulatingoracle computations
Antonio E. Porreca Alberto Leporati Giancarlo MauriClaudio Zandron
Dipartimento di Informatica, Sistemistica e ComunicazioneUniversità degli Studi di Milano-Bicocca, Italy
12th Conference on Membrane ComputingFontainebleau, France, 24 August 2011
Summary
I We show how to reuse existing recogniser P systemsas “subroutines”
I This allows us to simulate oraclesI The procedure is quite general
(though technical details may vary)I As an application, we improve the lower bound
on PMCAM(−d,−n) to PPP
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P systems with active membranes
I Known for their ability to solvecomputationally hard problems
I Here we focus on restricted elementary membranes(no nonelementary division, no dissolution)
Object evolution [a→ w]αhCommunication (send-in) a [ ]αh → [b]βhCommunication (send-out) [a]αh → [ ]βh bElementary division [a]αh → [b]βh [c]γh
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Uniform families of recogniser P systemsI For each input length n = |x| we construct a P system Πn
receiving as input a multiset encoding xI Both are constructed by fixed polytime Turing machinesI The resulting P system decides if x ∈ L
M1
1 11
x ∈ Σ?M2
0 10
Y E S
N O
aab
aab
Input multiset
n ∈ N
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The complexity class PMCAM(−d,−n)
It consists of the languages recognised in polytime by uniformfamilies of P systems with restricted elementary membranes
I Contains NP problems [Zandron et al. 2000](semi-uniform solution)
I Contains NP problems [Pérez-Jiménez et al. 2003](first uniform solution)
I Is contained in PSPACE [Sosík, Rodríguez-Patón 2007]I Contains PP problems [Porreca et al. 2010, 2011]
On the other hand, by using nonelementary division(class PMCAM) we obtain exactly PSPACE
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
x2x3x1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
x2x3x1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2x3x1 f2x3x1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2x3x1 f2x3x1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2x3f1 f2 t3x1t2x3t1 f2 f3x1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2x3f1 f2 t3x1t2x3t1 f2 f3x1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t t
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t t
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Solving 3SATIs ϕ(x1, x2, x3) satisfiable?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t tY E S
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The complexity class PP
DefinitionL ∈ PP if it is accepted in polytime by a nondeterministic TMsuch that more than half of its computations are accepting
Solving PP is “essentially the same as” countingthe number of solutions
Problem (T H R E S H O L D -3SAT)Given a Boolean formula ϕ over m variables and an integerk < 2m, do more than k assignments out of 2m satisfy ϕ?
TheoremT H R E S H O L D -3SAT is PP-complete
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
0
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
0 0
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
0 0 0
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t t0 0 0
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t t0 0 0
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t
t− − −
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t
t− − −
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Solving T H R E S H O L D -3SATDoes ϕ(x1, x2, x3) have more than 3 satisfying assignments?
t2 t3f1 f2 t3t1t2 t3t1 f2 f3t1
t2 f3f1 f2 t3f1t2 f3t1 f2 f3f1
t t t
Y E S− − −
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Simulating Turing machines I
q
0 10
S
0− + − 0 0
0 1 2 3 4
H2,q
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Simulating Turing machines II
δ(q1, 0) = (q2, 1, .)
{Hq1,i, [ ]−i → [Hq2,i+1]+i[Hq2,i+1]+i → [ ]+i Hq2,i+1
S
0− + − 0 0
0 1 2 3 4
H2,q1
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Simulating Turing machines II
δ(q1, 0) = (q2, 1, .)
{Hq1,i, [ ]−i → [Hq2,i+1]+i[Hq2,i+1]+i → [ ]+i Hq2,i+1
S
0− + − 0 0
0 1 2 3 4
H2,q1
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Simulating Turing machines II
δ(q1, 0) = (q2, 1, .)
{Hq1,i, [ ]−i → [Hq2,i+1]+i[Hq2,i+1]+i → [ ]+i Hq2,i+1
S
0− + + 0 0
0 1 2 3 4
H3,q2
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Simulating Turing machines II
δ(q1, 0) = (q2, 1, .)
{Hq1,i, [ ]−i → [Hq2,i+1]+i[Hq2,i+1]+i → [ ]+i Hq2,i+1
S
0− + + 0 0
0 1 2 3 4
H3,q2
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Simulating Turing machines II
δ(q1, 0) = (q2, 1, .)
{Hq1,i, [ ]−i → [Hq2,i+1]+i[Hq2,i+1]+i → [ ]+i Hq2,i+1
S
0− + + 0 0
0 1 2 3 4
H3,q2
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 0 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 + 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 + 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 + 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 + 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 + 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 + 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 − 0 0
AY E S
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
0
0 0 − 0 0
AY E S
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
+
0 0 − 0 0
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Using P systems as subroutines
S
0
− + − + 0
0 1 2 3 4
H2,q?
Π Π Π Π Π
A
+
0 0 − 0 0
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Main result
TheoremPPP ⊆ PMCAM(−d,−n)
Proof.I Any polytime TM M with a PP oracle can be simulated
by a polytime TM M′ with an oracle forT H R E S H O L D -3SAT and only one tape
I Just apply a reduction (which always exists,since T H R E S H O L D -3SAT is PP-complete)before querying the oracle
I And we know how to simulate the TM M′
with a polytime AM(−d,−n) uniform family.
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Discussion I
I We can solve QSAT (PSPACE-complete)by using nonelementary divisionand a membrane structure of depth Θ(n)
I QSAT instances have an arbitrary numberof alternations of quantifiers
I By fixing the first quantifier (∀ or ∃) and the numberof alternations, we get complete problems for all levelsof the polynomial hierarchy
I Formulae with k alternations can be solved by P systemsusing nonelementary division and a membrane structureof depth Θ(k)
I Notice that k does not depend on the input size
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Discussion II
I Let PH be the union of the levelsof the polynomial hierarchy
I Toda’s theorem tells us that PH ⊆ PPP
I So we also have PH ⊆ PMCAM(−d,−n)
I This means that all levels of the polynomial hierarchycan be solved by using P systems with only elementarydivision and membrane structure of depth 3
I Does this mean PSPACE ⊆ PMCAM(−d,−n)
and so PSPACE = PMCAM(−d,−n)?I Not immediately: PH is not known neither conjectured
to be PSPACE
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Open problems
I Prove that we can always do the oracle simulationI If we can reset the “oracle P systems”
then we only need a single copy of itI It might still be possible that PMCAM(−d,−n) = PSPACE
even if PH 6= PSPACEI But maybe it would be more interesting if it turns out
that PMCAM(−d,−n) = PPP
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Merci de votre attention!
Thanks for your attention!