P systems simulating oracle computations Antonio E. Porreca - PowerPoint PPT Presentation

P systems simulating oracle computations Antonio E. Porreca Alberto Leporati Giancarlo Mauri Claudio Zandron Dipartimento di Informatica, Sistemistica e Comunicazione Università degli Studi di Milano-Bicocca, Italy 12th Conference on Membrane Computing Fontainebleau, France, 24 August 2011

Summary ◮ We show how to reuse existing recogniser P systems as “subroutines” ◮ This allows us to simulate oracles ◮ The procedure is quite general (though technical details may vary) ◮ As an application, we improve the lower bound on PMC AM ( − d , − n ) to P PP 2/16

P systems with active membranes ◮ Known for their ability to solve computationally hard problems ◮ Here we focus on restricted elementary membranes (no nonelementary division, no dissolution) [ a → w ] α Object evolution h h → [ b ] β a [ ] α Communication (send-in) h h → [ ] β [ a ] α Communication (send-out) h b h → [ b ] β h [ c ] γ [ a ] α Elementary division h 3/16

Uniform families of recogniser P systems ◮ For each input length n = | x | we construct a P system Π n receiving as input a multiset encoding x ◮ Both are constructed by fixed polytime Turing machines ◮ The resulting P system decides if x ∈ L M 1 n ∈ N 1 1 Y E S 1 aab M 2 aab x ∈ Σ ⋆ 0 1 N O 0 Input multiset 4/16

The complexity class PMC AM ( − d , − n ) It consists of the languages recognised in polytime by uniform families of P systems with restricted elementary membranes ◮ Contains NP problems [Zandron et al. 2000] (semi-uniform solution) ◮ Contains NP problems [Pérez-Jiménez et al. 2003] (first uniform solution) ◮ Is contained in PSPACE [Sosík, Rodríguez-Patón 2007] ◮ Contains PP problems [Porreca et al. 2010, 2011] On the other hand, by using nonelementary division (class PMC AM ) we obtain exactly PSPACE 5/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? x 1 x 2 x 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? x 1 x 2 x 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? x 1 t 2 x 3 x 1 f 2 x 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? x 1 t 2 x 3 x 1 f 2 x 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 2 x 3 t 2 x 3 x 1 x 1 t 1 f 2 t 3 f 1 f 2 f 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 2 x 3 t 2 x 3 x 1 x 1 t 1 f 2 t 3 f 1 f 2 f 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t t t t 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t t t t 6/16

Solving 3SAT Is ϕ ( x 1 , x 2 , x 3 ) satisfiable? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t t t Y E S 6/16

The complexity class PP Definition L ∈ PP if it is accepted in polytime by a nondeterministic TM such that more than half of its computations are accepting Solving PP is “essentially the same as” counting the number of solutions Problem (T H R E S H O L D -3SAT) Given a Boolean formula ϕ over m variables and an integer k < 2 m , do more than k assignments out of 2 m satisfy ϕ ? Theorem T H R E S H O L D -3SAT is PP -complete 7/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 0 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 0 0 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 0 0 0 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t t t t 0 0 0 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t t t t 0 0 0 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t − − − t t t 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 t − − − t t t 8/16

Solving T H R E S H O L D -3SAT Does ϕ ( x 1 , x 2 , x 3 ) have more than 3 satisfying assignments? t 1 t 2 t 3 t 2 t 3 t 1 f 2 t 3 t 1 f 1 f 2 f 3 t 1 t 2 f 3 t 2 f 3 f 2 t 3 f 1 f 1 f 1 f 2 f 3 Y E S − − − t t t 8/16

Simulating Turing machines I q 0 1 0 0 − − + 0 0 0 1 2 3 4 H 2, q S 9/16

Simulating Turing machines II � H q 1 , i , [ ] − i → [ H q 2 , i + 1 ] + i δ ( q 1 , 0 ) = ( q 2 , 1 , ⊲ ) [ H q 2 , i + 1 ] + i → [ ] + H q 2 , i + 1 i 0 − − + 0 0 0 1 2 3 4 H 2, q 1 S 10/16

Simulating Turing machines II � H q 1 , i , [ ] − i → [ H q 2 , i + 1 ] + i δ ( q 1 , 0 ) = ( q 2 , 1 , ⊲ ) [ H q 2 , i + 1 ] + i → [ ] + H q 2 , i + 1 i 0 − − + 0 0 0 1 2 3 4 H 2, q 1 S 10/16

Simulating Turing machines II � H q 1 , i , [ ] − i → [ H q 2 , i + 1 ] + i δ ( q 1 , 0 ) = ( q 2 , 1 , ⊲ ) [ H q 2 , i + 1 ] + i → [ ] + H q 2 , i + 1 i 0 − + + 0 0 H 3, q 2 0 1 2 3 4 S 10/16

Simulating Turing machines II � H q 1 , i , [ ] − i → [ H q 2 , i + 1 ] + i δ ( q 1 , 0 ) = ( q 2 , 1 , ⊲ ) [ H q 2 , i + 1 ] + i → [ ] + H q 2 , i + 1 i 0 − + + 0 0 H 3, q 2 0 1 2 3 4 S 10/16

Simulating Turing machines II � H q 1 , i , [ ] − i → [ H q 2 , i + 1 ] + i δ ( q 1 , 0 ) = ( q 2 , 1 , ⊲ ) [ H q 2 , i + 1 ] + i → [ ] + H q 2 , i + 1 i 0 − + + 0 0 0 1 2 3 4 H 3, q 2 S 10/16

Using P systems as subroutines 0 H 2, q ? − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 0 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 + 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 + 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 + 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 + 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 + 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 0 0 + 0 0 Π Π Π Π Π 0 H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 − 0 0 0 0 Π Π Π Π Π 0 H 2, q ? A Y E S A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 − 0 0 0 0 Π Π Π Π Π 0 H 2, q ? A Y E S A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 − 0 0 0 0 Π Π Π Π Π + H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Using P systems as subroutines 0 − 0 0 0 0 Π Π Π Π Π + H 2, q ? A − − + + 0 0 1 2 3 4 S 11/16

Main result Theorem P PP ⊆ PMC AM ( − d , − n ) Proof. ◮ Any polytime TM M with a PP oracle can be simulated by a polytime TM M ′ with an oracle for T H R E S H O L D -3SAT and only one tape ◮ 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 ◮ And we know how to simulate the TM M ′ with a polytime AM ( − d , − n ) uniform family. 12/16

Discussion I ◮ We can solve QSAT ( PSPACE -complete) by using nonelementary division and a membrane structure of depth Θ( n ) ◮ QSAT instances have an arbitrary number of alternations of quantifiers ◮ By fixing the first quantifier ( ∀ or ∃ ) and the number of alternations, we get complete problems for all levels of the polynomial hierarchy ◮ Formulae with k alternations can be solved by P systems using nonelementary division and a membrane structure of depth Θ( k ) ◮ Notice that k does not depend on the input size 13/16