CSci 5403 Lecture 6

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CSci 5403
COMPLEXITY THEORY
LECTURE VI ALTERNATING QUANTIFIERS
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ORACLE TM
An oracle Turing Machine is a TM with a query
Tape and three states q?, qyes, qno.
S
With an oracle for set S, it goes in one step
from state q? with x on the query tape to qx2S.
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ORACLE COMPLEXITY
We denote M running with oracle S by MS.
For fixed O, we can define complexity
classes NPO, PO, PSPACEO, etc
S 2 PO iff 9 poly(n)-time P . L(PO) S
S 2 NPO iff 9 poly(n)-time V, poly q. S x
9y20,1q(x). VO(x,y) accepts
We can also define complexity classes relative to
other classes
S 2 PNP iff 9 A 2 NP, poly(n)-time P. L(PA) S.
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ORACLE ARITHMETIC
9 M, 9 A 2 NP. MA decides L in polynomial time.
PNP L
9 M. MSAT decides L in polynomial time.
PSAT L
Claim. PNP PSAT. (why?)
Quiz. We know that PPSPACE NPPSPACE. Does PNP
NPNP?
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PNP vs NPNP
Does PNP NP?
Notice that coNP µ PNP.
Consider the following problems
C C has a smaller, equivalent circuit
C.
NMIN-CIRCUIT
9X20,1n. ?(X,y1,,yn) is a tautology
?(x1xn,y1yn)
QSAT2
Claim. NMIN-CIRCUIT, QSAT2 2 NPNP.
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THE POLYNOMIAL HIERARCHY
Definition. Let ?0P ?0P ?0P P. Define the
hierarchy of complexity classes ?1 P ?1
NP ?1 coNP ?2 PNP ?2 NPNP ?2
coNPNP ? ?i P?i-1 Si NP?i-1 Pi
coNP?i-1 ? PH i ?i
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THE POLYNOMIAL HIERARCHY
?4
?4
?4
?3
?2
?3
?3
?2
?2
P
NP
coNP
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NMIN-CIRCUIT 2 ?2
MIN-CIRCUIT 2 ?2
for all C lt C accept if 9x .(C(x) ? C(x))
guess a circuit C accept if 8x. C(x)C(x).
Each path is a circuit Each node is an OR Each
leaf is an oracle call to CIRCUIT-TAUT
Each path is a circuit Each node is an AND Each
leaf is an oracle call to CIRCUIT-SAT
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Another way to express these languages
NMIN-CIRCUIT C 9 C20,1C.8x 2 0,1n.
CltC Æ C(x)C(x)
C 9 C20,1C.8x 2 0,1n. R(C,C,x)
QSAT2 ? 9 X 2 0,1n . 8 Y20,1n . ?(X,Y)

? 9 X 2 0,1n . 8 Y20,1n . R(?,X,Y)
MIN-CIRCUIT C 8 C20,1C. 9x2 0,1n.
CgtC Ç C(x)?C(x)
C 8 C20,1C. 9x2 0,1n. R(C,C,x)
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Notation we will write x 9Y1.8Y2.9Y3QnYn.
R(x,Y1Yn) as shorthand for x
9Y120,1p1(x).8Y220,1p2(x)QnYn20,1pn(x
) R(x,Y1,Yn) where p1,p2,,pn are
polynomials and R is a poly(n) time computable
relation.
Then NP L L coNP L L

x 9Y. R(x,Y)
x 8 Y. R(x,Y)
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Theorem. L 2 ?i , L x 9 Y1 8 Y2
QiYi . R(x,Y1,,Yi)
Proof. By induction on i. Base case i 1
?1 NP L 2 NP iff L x 9Y.V(x,Y) .
Induction step (()
Let N (x,Y1) 9Y2. 8Y3 QiYi. R(x,Y1,,Yi)
By I.H., N 2 Si-1. But then L 2 NPN
guess Y1, accept if oracle call (x,Y1) returns no.
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Theorem. L 2 ?i , L x 9 Y1 8 Y2
QiYi . R(x,Y1,,Yi)
Induction step ())
Let L 2 NPN, for some N 2 ?i-1. So NTM MN decides
L.
MN(x) accepts if 9 guesses c1cm 9 answers
a1ak 9 queries q1qk M(xc1cm,a1ak) accepts
and ai (qi 2 N)
c1
q ? a
c2
c3
accept
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MN(x) accepts if 9 guesses c1cm 9 answers
a1ak 9 queries q1qk M(xc1cm,a1ak) accepts
and aj (qj 2 N)
c1
q ? a
c2
c3
accept
qj 2 N 9 Y18 Y2 Qi-1Yi-1. RN(qj,Y1Yi-1)
qj ? N 8Y19Y2 QiYi-1.RN(qj,Y1Yi-1)
9c,a1,q1,Y1 8 Y1,Y2 9 Y2,Y3QiYi-1 .
R(M,x,c,a1,q1) Æ (a1 0 Æ RN(q1,Y1,,Yi-1)) Ç
(a11 Æ RN(q1,Y1,Yi-1))
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COLLAPSING PH
Conjecture 8 i, ?i ( ?i1. Why?
- The arithmetic hierarchy is proper.
- Problems higher up seem harder.
Theorem. If for some i, ?i ?i then PH ?i.
(The hierarchy collapses to the ith level)
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Theorem. If for some i, ?i ?i then PH ?i.
Proof. Suppose ?i ?i. We will show that ?i1
µ ?i.
Choose any L 2 ?i1. Then L has the form
x 9Y1 . R(x,Y1)
Where R(x,Y) is a ?i relation (x,Y1)
8Y2 9Y3 R?(x,Y1Yi1)
Since ?i ?i, R(x,Y) is also a ?i relation
(x,Y1) 9 Y28 Y3 R?(x,Y1Yi1)
Thus L x 9 Y1Y28 Y3 R?(x,Y1Y2Yi1)
only i alternations
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Corollary. NP coNP ) NPPH.
Corollary. PNP ) PHP.
Given a polytime algorithm for SAT, we
can construct a polytime algorithm for any
problem in PH!
Given an oracle for SAT, we only know how
to build polytime algorithms for ?1.
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Let QSATi ? 9X18X2 QiXi ?(X1,,Xi)
Where each Xi is a string of bits and ? is 3CNF.
Theorem. QSATi is ?i-complete.
Proof. Let L 2 ?i. Then L has the form x
9 Y1 8 Y2 Qi Yi R(x,Y1,,Yi) From Cook-Levin
we know how to produce a ?R(x,Y1Yi,Z) such that
R(x,Y1,,Yi) accepts iff 9Z. ?R(x,Y1Yi,Z).
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Is there a complete problem for all of PH?
Theorem. Suppose L is PH-complete. Then PH
collapses to the ith level, for some i.
Proof.
Let L be PH-complete. Then L 2 ?i, for some
i L x 9 Y18 Y2 QiYi R(x,Y1Yi)
Let L 2 ?i1. By hypothesis, L P L, so L
x 9 Y18 Y2 QiYi R((x),Y1Yi) where (x)
2 L , x 2 L.
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Theorem. Unless PH collapses, PH ( PSPACE.
Proof.
PH µ PSPACE, since we can decide QSATi using a
similar procedure as TQBF.
Suppose PH PSPACE. Then TQBF is PH-complete
and the hierarchy collapses.
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ALTERNATING TMs
A nondeterministic TM has a transition relation
and accepts if some path leads to accept.
A conondeterministic TM has a transition relation
and accepts if all paths lead to accept.
An alternating TM has a transition relation and
a label 2 8, 9 for each state. A 8
configuration accepts if all of its branches
accept, and a 9 configuration accepts if one
branch accepts.
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ALTERNATING TMs
An alternating TM has a label 2 8, 9 for each
state. 8 configurations accept if all branches
do 9 configurations accept if one branch does.
The time complexity is the length of the longest
path from C0 to accept or reject.
The space complexity is the maximum space used
by any Ci reachable from C0.
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Definition. ATIME(t(n)) L L is decided by
an ATM in time O(t(n))
Definition. ASPACE(s(n)) L L is decided by
an ATM in space O(s(n))
Definition. ?iTIME(t(n)) L L is decided by
an ATM in time O(t(n)), where label(q0) 9
and every path has at most i alterations
Theorem. For every i, ?i c ?iTIME(nc)
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Theorem. ATIME(t(n)) µSPACE(t(n)) µATIME(t(n)2)
Proof. SPACE(t(n)) µ ATIME(t(n)2) by
using recursive search of the configuration graph.
M accepts if C0 !2t(n) Caccept. Ci !m Cj if 9 Ck
8 (D,E) 2 (Ci,Ck (Ck,Cj) D!m/2 E.
ATIME(t(n)) µ SPACE(t(n)) by using
space- efficient recursion enumerate all
length-t(n) paths and reuse the O(t(n)) space
for each one.
Corollary. AP PSPACE.
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