CS151 Complexity Theory Lecture 12 May 6, 2004

CS151 Lecture 122 Outline The Polynomial-Time Hierarachy (PH) Complete problems for classes in PH, PSPACE BPP and the PH non-uniformity and the PH

May 6, 2004CS151 Lecture 123 The Polynomial-Time Hierarchy Σ 0 = Π 0 = P Δ 1 =P P Σ 1 =NPΠ 1 =coNP Δ 2 =P NP Σ 2 =NP NP Π 2 =coNP NP Δ i+1 =P Σ i Σ i+i =NP Σ i Π i+1 =coNP Σ i Polynomial Hierarchy PH =  i Σ i

May 6, 2004CS151 Lecture 124 The Polynomial-Time Hierarchy Σ 0 = Π 0 = P Δ i+1 =P Σ i Σ i+i =NP Σ i Π i+1 =coNP Σ i 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  Σ 2

May 6, 2004CS151 Lecture 125 The Polynomial-Time Hierarchy Σ 0 = Π 0 = P Δ i+1 =P Σ i Σ i+i =NP Σ i Π i+1 =coNP Σ i Example: –EXACT TSP: given a weighted graph G, and in integer k; is the k-th bit of the length of the shortest TSP tour in G a 1? –EXACT TSP  Δ 2

May 6, 2004CS151 Lecture 126 The PH PSPACE: generalized geography, 2-person games… 3rd level: V-C dimension… 2nd level: MIN CIRCUIT, Succinct Set Cover, BPP… 1st level: SAT, UNSAT, factoring, etc… P NP coNP Σ3Σ3 Π3Π3 Δ3Δ3 PSPACE EXP PH Σ2Σ2 Π2Π2 Δ2Δ2

May 6, 2004CS151 Lecture 127 Useful characterization Recall: L  NP iff expressible as L = { x |  y, |y| ≤ |x| k, (x, y)  R } where R  P. Corollary: L  coNP iff expressible as L = { x |  y, |y| ≤ |x| k, (x, y)  R } where R  P.

May 6, 2004CS151 Lecture 128 Useful characterization Theorem: L  Σ i iff expressible as L = { x |  y, |y| ≤ |x| k, (x, y)  R } where R  Π i-1. Corollary: L  Π i iff expressible as L = { x |  y, |y| ≤ |x| k, (x, y)  R } where R  Σ i-1.

May 6, 2004CS151 Lecture 129 Useful characterization Proof of Theorem: –induction on i –base case on previous slide (  ) –we know Σ i = NP Σ i-1 = NP Π i-1 –guess y, ask oracle if (x, y)  R

May 6, 2004CS151 Lecture 1210 Useful characterization Proof (continued): (  ) –given L  Σ i = NP Σ i-1 decided by ONTM M running in time n k –try: R = { (x, y) : y describes valid path of M’s computation leading to q accept } –but how to recognize valid computation path when it depends on result of oracle queries?

May 6, 2004CS151 Lecture 1211 Useful characterization Proof (continued): –try: R = { (x, y) : y describes valid path of M’s computation leading to q accept } –valid path = step-by-step description including correct yes/no answer for each A-oracle query z j (A  Σ i-1 ) –verify “no” queries in Π i-1 : e.g: z 1  A  z 3  A  …  z 8  A –for each “yes” query z j :  w j, |w j | ≤ |z j |k with (z j, w j )  R’ for some R’  Π i-2 by induction. –for each “yes” query z j put w j in description of path y

May 6, 2004CS151 Lecture 1212 Useful characterization Proof (continued): –single language R in Π i-1 : (x, y)  R  all “no” z j  A and all “yes” z j have (z j, w j )  R’ and y is a path leading to q accept. –Note: AND of Π i-1 predicates is in Π i-1.

May 6, 2004CS151 Lecture 1213 Alternating quantifiers Nicer, more usable version: L  Σ i iff expressible as L = { x |  y 1  y 2  y 3 …Qy i (x, y 1,y 2,…,y i )  R } where Q=  /  if i even/odd, and R  P L  Π i iff expressible as L = { x |  y 1  y 2  y 3 …Qy i (x, y 1,y 2,…,y i )  R } where Q=  /  if i even/odd, and R  P

May 6, 2004CS151 Lecture 1214 Alternating quantifiers Proof: –(  ) induction on i –base case: true for Σ 1 =NP and Π 1 =coNP –consider L  Σ i : L = {x |  y 1 (x, y 1 )  R’ }, for R’  Π i-1 L = {x |  y 1  y 2  y 3 …Qy i (x, y 1,y 2,…,y i )  R} –same argument for L  Π i –(  ) exercise.

May 6, 2004CS151 Lecture 1215 Alternating quantifiers Pleasing viewpoint: P NP coNP Δ3Δ3 Σ2Σ2 Π2Π2 Δ2Δ2 “”“” “”“” “  ”“  ” Σ3Σ3 Π3Π3 PSPACE PH “  ” “  ” “  …” ΣiΣi ΠiΠi “  …” “  …” const. # of alternations poly(n) alternations

May 6, 2004CS151 Lecture 1216 Complete problems Recall: 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? { (C, k) |  C’  x (|C’| ≤ k and C’(x) = C(x)) } –Conclude: in Σ 2 –(open whether it is complete for Σ 2 )

May 6, 2004CS151 Lecture 1217 Complete problems three variants of SAT: –QSAT i (i odd) = {3-CNFs φ(x 1, x 2, …, x i ) for which  x 1  x 2  x 3 …  x i φ(x 1, x 2, …, x i ) = 1} –QSAT i (i even) = {3-DNFs φ(x 1, x 2, …, x i ) for which  x 1  x 2  x 3 …  x i φ(x 1, x 2, …, x i ) = 1 } –QSAT = {3-CNFs φ for which  x 1  x 2  x 3 … Qx n φ(x 1, x 2, …, x n ) = 1}

May 6, 2004CS151 Lecture 1218 QSAT i is Σ i -complete Theorem: QSAT i is Σ i -complete. Proof: (clearly in Σ i ) –assume i odd; given L  Σ i in form { x |  y 1  y 2  y 3 …  y i (x, y 1,y 2,…,y i )  R } …x… …y 1 … …y 2 … …y 3 … … …y i … C CVAL reduction for R 1 iff (x, y 1,y 2,…,y i )  R

May 6, 2004CS151 Lecture 1219 QSAT i is Σ i -complete –Problem set: can construct 3-CNF φ from C:  z φ(x,y 1,…,y i,z) = 1  C(x,y 1,…,y i ) = 1 –we get:  y 1  y 2 …  y i  z φ(x,y 1,…,y i,z) = 1   y 1  y 2 …  y i C(x,y 1,…,y i ) = 1  x  L …x… …y 1 … …y 2 … …y 3 … … …y i … C CVAL reduction for R 1 iff (x, y 1,y 2,…,y i )  R

May 6, 2004CS151 Lecture 1220 QSAT i is Σ i -complete Proof (continued) –assume i even; given L  Σ i in form { x |  y 1  y 2  y 3 …  y i (x, y 1,y 2,…,y i )  R } …x… …y 1 … …y 2 … …y 3 … … …y i … C CVAL reduction for R 1 iff (x, y 1,y 2,…,y i )  R

May 6, 2004CS151 Lecture 1221 QSAT i is Σ i -complete –Problem set: can construct 3-DNF φ from C:  z φ(x,y 1,…,y i,z) = 1  C(x,y 1,…,y i ) = 1 –we get:  y 1  y 2 …  y i  z φ(x,y 1,y 2,…,y i,z) = 1   y 1  y 2 …  y i C(x,y 1,y 2,…,y i ) = 1  x  L …x… …y 1 … …y 2 … …y 3 … … …y i … C CVAL reduction for R 1 iff (x, y 1,y 2,…,y i )  R

May 6, 2004CS151 Lecture 1222 QSAT is PSPACE-complete Theorem: QSAT is PSPACE-complete. Proof: –in PSPACE:  x 1  x 2  x 3 … Qx n φ(x 1, x 2, …, x n )? –“  x 1 ”: for each x 1, recursively solve  x 2  x 3 … Qx n φ(x 1, x 2, …, x n )? if encounter “yes”, return “yes” –“  x 1 ”: for each x 1, recursively solve  x 2  x 3 … Qx n φ(x 1, x 2, …, x n )? if encounter “no”, return “no” –base case: evaluating a 3-CNF expression –poly(n) recursion depth –poly(n) bits of state at each level

May 6, 2004CS151 Lecture 1223 QSAT is PSPACE-complete Proof (continued): –given TM M deciding L  PSPACE; input x –configuration graph has 2 n k nodes –recall: PATH(X, Y, i)  path from X to Y of length at most 2 i –goal: 3-CNF φ(w 1,w 2,w 3,…,w m )  w 1  w 2 …Qw m φ(w 1,…,w m )  PATH(START, ACCEPT, n k )

May 6, 2004CS151 Lecture 1224 QSAT is PSPACE-complete –for i = 0, 1, … n k produce quantified Boolean expressions ψ i (A, B)  w 1  w 2 … ψ i (A, B, W)  PATH(A, B, i) –convert ψ n k to 3-CNF φ add variables V –hardwire START, ACCEPT  w 1  w 2 …  V φ(W, V)  x  L

May 6, 2004CS151 Lecture 1225 QSAT is PSPACE-complete Proof (continued): –ψ o (A, B) = 1 iff A = B or A yields B in one step of M … … STEP config. A config. B Boolean expression of size O(n k )

May 6, 2004CS151 Lecture 1226 QSAT is PSPACE-complete –recall Savitch’s algorithm: PATH(A, B, i+1)   Z [ PATH(A, Z, i)  PATH(Z, B, i) ] –cannot define ψ i+1 (A, B) to be  Z [ ψ i (A, Z)  ψ i (Z, B) ] (why?)

May 6, 2004CS151 Lecture 1227 QSAT is PSPACE-complete Proof (continued): –Key: reuse expressions just as Savitch reuses stack records… –define ψ i+1 (A, B) to be  Z  X  Y [ ((X=A  Y=Z)  (X=Z  Y=B))  ψ i (X, Y) ] –ψ i (X, Y) is preceded by quantifiers –move to front (they don’t involve X,Y,Z,A,B)

May 6, 2004CS151 Lecture 1228 QSAT is PSPACE-complete ψ o (A, B) = 1 iff A = B or A yields B in 1 step  Z  X  Y [ ((X=A  Y=Z)  (X=Z  Y=B))  ψ i (X, Y) ] –|ψ 0 | = O(n k ) –|ψ i+1 | = O(n k ) + |ψ i | –total size of ψ n k is O(n k ) 2 = poly(n) –logspace reduction

May 6, 2004CS151 Lecture 1229 PH collapse Theorem: if Σ i = Π i then for all j > i Σ j = Π j = Δ j = Σ i “the polynomial hierarchy collapses to the i-th level” Proof: –sufficient to show Σ i = Σ i+1 –then Σ i+1 = Σ i = Π i = Π i+1 ; apply theorem again

May 6, 2004CS151 Lecture 1230 PH collapse –recall: L  Σ i+1 iff expressible as L = { x |  y (x, y)  R } where R  Π i –since Π i = Σ i, R expressible as R = { (x,y) |  z ((x, y), z)  R’ } where R’  Π i-1 –together: L = { x |  (y,z) (x, (y,z))  R’} –conclude L  Σ i

May 6, 2004CS151 Lecture 1231 Oracles vs. Algorithms A point to ponder: given poly-time algorithm for SAT –can you solve MIN CIRCUIT efficiently? –what other problems? Entire complexity classes? given SAT oracle –same input/output behavior –can you solve MIN CIRCUIT efficiently?

May 6, 2004CS151 Lecture 1232 Natural complete problems We now have versions of SAT complete for levels in PH, PSPACE Natural complete problems? –PSPACE: games –PH: almost all natural problems lie in the second level

May 6, 2004CS151 Lecture 1233 Natural complete problems –MIN CIRCUIT good candidate, still open –MIN DNF: given DNF φ, integer k; is there a DNF φ’ of size at most k computing same function φ does? –example: x 1 x 2 x 3  x 1 x 2  x 3  x 4

May 6, 2004CS151 Lecture 1234 Natural complete problems –MIN CIRCUIT good candidate, still open –MIN DNF: given DNF φ, integer k; is there a DNF φ’ of size at most k computing same function φ does? –example: x 1 x 2 x 3  x 1 x 2  x 3  x 4  x 1 x 2  x 4

May 6, 2004CS151 Lecture 1235 Simpler version of MIN DNF Theorem (U): MIN DNF is Σ 2 -complete. we’ll consider a simpler variant: –IRREDUNDANT: given DNF φ, integer k; is there a DNF φ’ consisting of at most k terms of φ computing same function φ does?

May 6, 2004CS151 Lecture 1236 Simpler version of MIN DNF analogy with an NP-complete problem: –SET COVER: given subsets S 1,S 2,...,S m  U, integer k, is there a collection of at most k sets that cover U. U sets S i {0,1} n φ -1 (0) φ -1 (1) terms of φ