Polynomial-Time Hierarchy
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Transcript of Polynomial-Time Hierarchy
Polynomial-Time Hierarchy
1. Stockmeyer2. Wrathall
Definitions
Let A Θ+ and B Δ+ for finite alphabets Θ and Δ. A transforms to B within logspace via f (A B via f) iff f is a transformation, f:Θ+→Δ+, such that f є logspace and xєA↔f(x)єB for all x є Θ+
1
( )k
k
P DTIME n
1
( )k
k
NP NTIME n
1
( )k
k
PSPACE DSPACE n
log
The Hierarchy The polynomial time hierarchy is
where:
and for k≥0
Also define
0 0 0p p p P
1
1
1
pk
pk
pk
pk
pk
pk
NP
coNP
P
{ , , , 0}p p pk k k k
0
pkk
PH
2 notes Note that and . Since obviously BєPB
and for any set B, the P-hierarchy has the following structure:
Also
1p NP 1
p coNP B B BP NP coNP
1 1 1p p p p pk k k k k
1pkp
k NP
Lemmas Let L a language and i≥1. L in Σk
P iff there is a poly-balanced relation R s.t. the language{x,y: (x,y)єR} is in Πk-1
P and L={x: Эy s.t. (x,y) єR}
Let L a language and i≥1. L in ΠkP iff
there is a poly-balanced relation R s.t. the language{x,y: (x,y)єR} is in Σk-1
P and L={x: for all y with |y|≤|x|k, (x,y) єR}
Proof
ΠkP =co Σk
P so it suffices to prove it for Σk
P .
For i=1 it holds. Let i>1 and R exists. NDTM M choses a y nondet. And with
a Σi-1P oracle decides if (x,y) not in R
(since R in Πi-1P)
Proof continues Let L in Σk
P we will show that a proper R exists. L is decided by NDTM M with oracle for KєΣi-1
P. By induction Э relation S s.t. zєK iff Эw : (z,w)єS,
SєΠi-2P.
R poly-balanced and poly decidable for L. xєL iff Э acc. comput. of MK on x. y records computation of MK.
Some steps are queries to K. For each yes query (zi) y will contain the certificate wi s.t.
(zi,wi)єS. (x,y)єR iff y records an acc computation of M with a
certificate wi for each yes querry zi in computation.\ (x,y)єR can be checked in Πi-1
P
Main Theorem
Let L S+ be a language. For any k≥1, Lє if and only if there exist polynomials p1,…,pk and a language L’ є P such that for all x є S+,
x є L iffDually, L є if and only if
x є L ifffor some L’ є P and polynomials p1,…,pk
pk
pk
1 21 2 1( ) ( ) ...( ) [ , ,..., ']kp p k p ky y Qy x y y L
1 21 2 1( ) ( ) ...( ) [ , ,..., ']kp p k p ky y Qy x y y L
2 propositions1. For any k ≥ 1, a language L S+ is in iff there exist
a homomorphism h:S*→T*, a language L’ T+ in and a polynomial p(n) such that L=h(L’) and for any x є L’, |x|≤p(|h(x)|), That is ={h(L’): L’ є , h a homomorphism that performs poly-bounded erasing on L’}
2. For each k ≥ 1, is closed under poly-bounded existential quantification and is closed under poly-time bounded universal quantification.
pk
pk
1pk
1pk
pk
pk
If for some k≥1 then for all j≥K Assume for some k≥1 By induction on j we will prove it For j=k it stands Assume that for some j>k we will
show that
From previous theorem: There is a 2-ary relation R and a polynomial p such that for all x, xєA iff
By induction we have R . A because for k≥1, is closed under the operation of poly-bounded existential quantification over variables of relations (prop 2).
Thus and by definition
p pk k
p pk k
p p pj j k
1 1p p pj j k
p pj k
1pj
( )[ (| |) ( , )]y y p x R x y pk p
k pk
p pj k p p
j k
1. If for some k ≥ 1, then P ≠ NP2. If contains infinitely main distinct classes,
then for all k ≥ 0.
Baker points out that NPPSPACE=PSPACE is an immediate consequence from Savitch’s theorem NSPACE(S(n)) DSPACE(S2(N)). By induction on k we have for all k.
0p p
k { : 0}p
k k
pk PSPACE
PH PSPACE
1p pk k
If for all k, then Let k≥1
Bk={F(X1,…,Xk)|F(X1,…,Xk) is a boolean formula, and
} 1. Bω is log-complete in PSPACE.
2. Suppose A B and B є NPC. Then also A є NPC.
PH PSPACE. If PSPACE PH then for some j, Since is closed under logspace reductions, implies that and then .
pm
1p pk k PH PSPACE
1k
k
B B
1 2 1( )( )...( )[ ( ,..., ) 1]k kX X QX F X X
pjB
pjB
pj
pjPSPACE 1
p pj j