IFT6121 Automne 2008
Complexité du calcul
CS151
Complexity Theory
Partie 2
La hiérarchie polynômiale
Lecture 12
May 3, 2007
Transparents tirés (avec permission)
des séances 12 et 13 du
cours CSC 151 de mai 2007
donné par Chris Umans de la
California Institute of Technology
A motivating question
Oracle Turing Machines
• Oracle Turing Machine (OTM):
• Central problem in logic synthesis:
• is there a circuit C’ of size at most
k that computes the same function
C does?
• Complexity of this problem?
– multitape TM M with special “query” tape
– special states q?, qyes, qno
– on input x, with oracle language A
– MA runs as usual, except…
– when MA enters state q?:
• y = contents of query tape
• y  A Ÿ transition to qyes
• y  A Ÿ transition to qno
š
• given Boolean circuit C, integer k
›
š
x1
š
›
x2 x3
™
… xn
– NP-hard? in NP? in coNP? in PSPACE?
– complete for any of these classes?
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Oracle Turing Machines
Oracle Turing Machines
• Nondeterministic OTM
Shorthand #1:
• applying oracles to entire complexity
classes:
– defined in the same way
– (transition relation, rather than function)
• oracle is like a subroutine, or function in
your favorite programming language
– complexity class C
– language A
– but each call counts as single step
e.g.: given ij1, ij2, …, ijn are even # satisfiable?
– poly-time OTM solves with SAT oracle
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CA = {L decided by OTM M with oracle A with M “in” C}
– example: PSAT
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Oracle Turing Machines
The Polynomial-Time Hierarchy
Shorthand #2:
• using complexity classes as oracles:
• can define lots of complexity classes using
oracles
• the classes on the next slide stand out
– OTM M
– complexity class C
– MC decides language L if for some language
A  C, MA decides L
– they have natural complete problems
– they have a natural interpretation in terms of
alternating quantifiers
– they help us state certain consequences and
containments (more later)
Both together: CD = languages decided by
OTM “in” C with oracle language from D
exercise: show PSAT = PNP
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The Polynomial-Time Hierarchy
The Polynomial-Time Hierarchy
Ȉ0 = Ȇ0 = P
ǻi+1=PȈi
ǻ1=PP
ǻ2=PNP
ǻi+1=P
Ȉi
Ȉ1=NP
Ȇ1=coNP
Ȉ2=NPNP Ȇ2=coNPNP
Ȉi+i=NP
Ȉi
Ȇi+1=coNP
• Example:
– MIN CIRCUIT: given Boolean circuit C,
integer k; is there a circuit C’ of size at most k
that computes the same function C does?
– MIN CIRCUIT  Ȉ2
Ȉi
Polynomial Hierarchy PH = ‰i Ȉi
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The Polynomial-Time Hierarchy
ǻi+1=PȈi
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EXP
The PH
PSPACE: generalized
geography, 2-person
games…
Ȉ0 = Ȇ0 = P
Ȉi+i=NPȈi Ȇi+1=coNPȈi
PSPACE
PH
Ƶ3
– EXACT TSP: given a weighted graph G, and
an integer k; is the k-th bit of the length of the
shortest TSP tour in G a 1?
– EXACT TSP  ǻ2
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2nd level: MIN CIRCUIT,
BPP…
1st level: SAT, UNSAT,
factoring, etc…
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Ƴ3
¨3
3rd level: V-C dimension…
• Example:
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Ȉ0 = Ȇ0 = P
Ȉi+i=NPȈi Ȇi+1=coNPȈi
Ƶ2
Ƴ2
¨2
NP
coNP
P
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Useful characterization
Useful characterization
• Recall: L  NP iff expressible as
Theorem: L  Ȉi iff expressible as
L = { x | y, |y| ” |x|k, (x, y)  R }
L = { x | y, |y| ” |x|k, (x, y)  R }
where R  P.
where R  Ȇi-1.
• Corollary: L  coNP iff expressible as
• Corollary: L  Ȇi iff expressible as
L = { x | y, |y| ” |x|k, (x, y)  R }
L = { x | y, |y| ” |x|k, (x, y)  R }
where R  P.
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where R  Ȉi-1.
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Useful characterization
Useful characterization
Theorem: L  Ȉi iff expressible as
L = { x | y, |y| ” |x|k, (x, y)  R }, where R  Ȇi-1.
Theorem: L  Ȉi iff expressible as
L = { x | y, |y| ” |x|k, (x, y)  R }, where R  Ȇi-1.
( Ÿ)
– given L  Ȉi = NPȈi-1 decided by ONTM M
running in time nk
– try: R = { (x, y) : y describes valid path of M’s
computation leading to qaccept }
– but how to recognize valid computation path
when it depends on result of oracle queries?
• Proof of Theorem:
– induction on i
– base case (i =1) on previous slide
( )
– we know Ȉi = NPȈi-1 = NPȆi-1
– guess y, ask oracle if (x, y)  R
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Useful characterization
Useful characterization
Theorem: L  Ȉi iff expressible as
L = { x | y, |y| ” |x|k, (x, y)  R }, where R  Ȇi-1.
Theorem: L  Ȉi iff expressible as
L = { x | y, |y| ” |x|k, (x, y)  R }, where R  Ȇi-1.
– try: R = { (x, y) : y describes valid path of M’s
computation leading to qaccept }
– valid path = step-by-step description including correct
yes/no answer for each A-oracle query zj (A  Ȉi-1)
– verify “no” queries in Ȇi-1:
e.g: z1 A š z3 A š … š z8 A
– for each “yes” query zj: wj, |wj| ” |zj|k with (zj, wj)  R’
for some R’  Ȇi-2 by induction.
– for each “yes” query zj put wj in description of path y
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– single language R in Ȇi-1 :
(x, y)  R
œ
all “no” zj are not in A and
all “yes” zj have (zj, wj)  R’ and
y is a path leading to qaccept.
– Note: AND of Ȇi-1 predicates is in Ȇi-1.
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Alternating quantifiers
• Proof:
– ( Ÿ) induction on i
– base case: true for Ȉ1=NP and Ȇ1=coNP
– consider LȈi:
L = {x | y1 (x, y1)  R’ }, for R’  Ȇi-1
L = {x | y1y2y3 …Qyi (x, y1,y2,…,yi)R}
– same argument for L Ȇi
– ( ) exercise.
L = { x | y1y2y3 …Qyi (x, y1,y2,…,yi)R }
where Q=/ if i even/odd, and RP
• LȆi iff expressible as
L = { x | y1y2y3 …Qyi (x, y1,y2,…,yi)R }
where Q= / if i even/odd, and RP
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Alternating quantifiers
Nicer, more usable version:
• LȈi iff expressible as
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Alternating quantifiers
Pleasing viewpoint:
const. # of
alternations
Complete problems
• three variants of SAT:
6i-complete
– QSATi (i odd) =
{3-CNFs ij(x1, x2, …, xi) for which
x1x2x3 … xi ij(x1, x2, …, xi) = 1}
– QSATi (i even) =
{3-DNFs ij(x1, x2, …, xi) for which
x1x2x3 … xi ij(x1, x2, …, xi) = 1 }
– QSAT = {3-CNFs ij for which
x1x2x3 … Qxn ij(x1, x2, …, xn) = 1}
“…” PSPACE
poly(n)
alternations
¨3
“” Ƶ2 “” Ƴ2
PH
“…” Ƶi “…”Ƴi
¨2
“” Ƶ3 “”Ƴ3
“” NP “”coNP
P
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QSATi is Ȉi-complete
…x… …y1… …y2… …y3…
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…yi…
CVAL reduction
for R
1 iff (x, y1,y2,…,yi)  R
– Problem set: can construct 3-CNF ij from C:
z ij(x,y1,…,yi,z) = 1 œ C(x,y1,…,yi) = 1
– we get:
y1y2…yi z ij(x,y1,…,yi,z) = 1
œ y1y2…yiC(x,y1,…,yi) = 1 œ x  L
…yi…
C
1 iff (x, y1,y2,…,yi)  R
…
C
– assume i odd; given L  Ȉi in form
{ x | y1y2y3 … yi (x, y1,y2,…,yi)  R }
…
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QSATi is Ȉi-complete
Theorem: QSATi is Ȉi-complete.
• Proof: (clearly in Ȉi)
…x… …y1… …y2… …y3…
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CVAL reduction
for R
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QSATi is Ȉi-complete
QSATi is Ȉi-complete
…x… …y1… …y2… …y3…
• Proof (continued)
…
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– Problem set: can construct 3-DNF ij from C:
z ij(x,y1,…,yi,z) = 1 œ C(x,y1,…,yi) = 1
– we get:
y1y2… yiz ij(x,y1,y2,…,yi,z) = 1
œ y1y2…yiC(x,y1,y2,…,yi) = 1 œ x  L
…yi…
CVAL reduction
for R
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QSAT is PSPACE-complete
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– given TM M deciding L  PSPACE; input x
k
– 2n possible configurations
– single START configuration
– assume single ACCEPT configuration
– in PSPACE: x1x2x3 … Qxn ij(x1, x2, …, xn)?
– “x1”: for each x1, recursively solve
x2x3 … Qxn ij(x1, x2, …, xn)?
• if encounter “yes”, return “yes”
– “x1”: for each x1, recursively solve
x2x3 … Qxn ij(x1, x2, …, xn)?
• if encounter “no”, return “no”
– base case: evaluating a 3-CNF expression
– poly(n) recursion depth
– poly(n) bits of state at each level
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QSAT is PSPACE-complete
Theorem: QSAT is PSPACE-complete.
• Proof:
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CVAL reduction
for R
1 iff (x, y1,y2,…,yi)  R
C
1 iff (x, y1,y2,…,yi)  R
…yi…
C
– assume i even; given L  Ȉi in form
{ x | y1y2y3 … yi (x, y1,y2,…,yi)  R }
…x… …y1… …y2… …y3…
…
– define:
REACH(X, Y, i) œ configuration Y reachable from
configuration X in at most 2i steps.
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QSAT is PSPACE-complete
QSAT is PSPACE-complete
REACH(X, Y, i) œ configuration Y reachable from
configuration X in at most 2i steps.
– Goal: produce 3-CNF ij(w1,w2,w3,…,wm) such
that
– for i = 0, 1, … nk produce quantified Boolean
expressions ȥi(A, B, W)
w1w2… ȥi(A, B, W) œ REACH(A, B, i)
– convert ȥnk to 3-CNF ij
• add variables V
– hardwire A = START, B = ACCEPT
w1w2… V ij(W, V) œ x  L
w1w2…Qwm ij(w1,…,wm)
œ REACH(START, ACCEPT, nk)
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QSAT is PSPACE-complete
QSAT is PSPACE-complete
– ȥo(A, B) = true iff
– Key observation #1:
REACH(A, B, i+1)
œ
• A = B or
• A yields B in one step of M
…
STEP STEP STEP
config.
A
STEP
…
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Boolean expression
of size O(nk)
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Z [REACH(A, Z, i) š REACH(Z, B, i)]
– cannot define ȥi+1(A, B) to be
Z [ȥi(A, Z) š ȥi(Z, B)]
config.
B
(why?)
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QSAT is PSPACE-complete
QSAT is PSPACE-complete
– Key idea #2: use quantifiers
ȥo(A, B) = true iff A = B or A yields B in 1 step
ȥi+1(A, B) =
– couldn’t do ȥi+1(A, B) = Z [ȥi(A, Z) š ȥi(Z, B)]
ZXY [((X=AšY=Z)›(X=ZšY=B)) Ÿ ȥi(X, Y)]
– define ȥi+1(A, B) to be
– |ȥ0| = O(nk)
– |ȥi+1| = O(nk) + |ȥi|
ZXY [((X=AšY=Z)›(X=ZšY=B)) Ÿ ȥi(X, Y)]
– total size of ȥnk is O(nk)2 = poly(n)
– logspace reduction
– ȥi(X, Y) is preceded by quantifiers
– move to front (they don’t involve X,Y,Z,A,B)
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PH collapse
Theorem: if Ȉi = Ȇi then for all j > i
Ȉj = Ȇj = ǻj = Ȉi
“the polynomial hierarchy collapses
to the i-th level”
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PH
¨3
Ƶ2
Ƴ2
¨2
NP
coNP
P
– sufficient to show Ȉi = Ȉi+1
– then Ȉi+1= Ȉi = Ȇi = Ȇi+1; apply theorem again
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PH collapse
Ƴ3
Ƶ3
• Proof:
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– recall: L  Ȉi+1 iff expressible as
L = { x | y (x, y)  R }
where R  Ȇi
– since Ȇi = Ȉi, R expressible as
R = { (x,y) | z ((x, y), z)  R’ }
where R’  Ȇi-1
– together: L = { x | (y, z) (x, (y, z))  R’}
– conclude L  Ȉi
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