Umans Complexity Theory Lectures

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1 Umans Complexity Theory Lectures Lecture 12: The Polynomial-Time Hierarchy Oracle Turing Machines Oracle Turing Machine (OTM): Deterministic multitape TM M with special query tape special states q?, q yes, q no on input x, with oracle language A M A runs as usual, except when M A enters state q? : y = contents of query tape y A transition to q yes y A transition to q no 2 Oracle Turing Machines Nondeterministic Oracle TM (NOTM) defined in the same way (transition relation, rather than function) oracle is like a subroutine, or function in your favorite programming language but each call counts as single step e.g.: given φ 1, φ 2,, φ n are even # satisfiable? poly-time OTM solves with SAT oracle Oracle Turing Machines Shorthand #1: applying oracles to entire complexity classes: complexity class C language A C A = {L decided by OTM M with oracle A with M in C} example: P SAT 3 4

2 Oracle Turing Machines Shorthand #2: using complexity classes as oracles: OTM M complexity class C M C decides language L if for some language A C, M A decides L Both together: C D = languages decided by OTM in C with oracle language from D exercise: show P SAT = P NP The Polynomial-Time Hierarchy can define lots of complexity classes using oracles the classes on the next slide stand out 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) 5 6 The Polynomial-Time Hierarchy Σ 0 = Π 0 = P Δ 1 =P P Σ 1 =NP Π 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 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: MN CRCUT: given Boolean circuit C, integer k; is there a circuit C of size at most k that computes the same function C does? MN CRCUT Σ 2 7 8

3 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 an integer k; is the k-th bit of the length of the shortest TSP tour in G a 1? EXACT TSP Δ 2 The PH PSPACE: generalized geography, 2-person games 3rd level: V-C dimension 2nd level: MN CRCUT, BPP 1st level: SAT, UNSAT, factoring, etc EXP PSPACE PH Σ 3 Π 3 Δ 3 Σ 2 Π 2 Δ 2 NP conp 9 P 10 Useful characterization Recall: L NP iff expressible as L = { x y, y x k, (x, y) R } where R P. 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 conp iff expressible as L = { x y, y x k, (x, y) R } where R P. Corollary: L Π i iff expressible as L = { x y, y x k, (x, y) R } where R Σ i

4 Useful characterization Theorem: L Σ i iff expressible as L = { x y, y x k, (x, y) R }, where R Π i-1. Proof of Theorem: induction on i base case (i =1) on previous slide ( ) we know Σ i = NP Σ i-1 = NP Π i-1 guess y, ask oracle if (x, y) R Useful characterization Theorem: L Σ i iff expressible as L = { x y, y x k, (x, y) R }, where R Π i-1. ( ) 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? Useful characterization Theorem: L Σ i iff expressible as L = { x y, y x k, (x, y) R }, where R Π i-1. 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 15 Useful characterization Theorem: L Σ i iff expressible as L = { x y, y x k, (x, y) R }, where R Π i-1. single language R in Π i-1 : (x, y) R all no z j are not in A and all yes z j have (z j, w j ) R and y is a path leading to q accept. Note: AND of polynomially-many Π i-1 predicates is in Π i-1. 16

5 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 17 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} 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 same argument for L Π i ( ) exercise. 18 Alternating quantifiers Complete problems Pleasing viewpoint: const. # of alternations NP Δ 3 Σ 2 Π 2 Δ 2 P conp poly(n) alternations PSPACE PH Σ i Π i Σ 3 Π 3 19 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} 20

6 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 Boolean Circuit C 1 iff (x, y 1,y 2,,y i ) R CVAL reduction for R QSAT i is Σ i -complete x y 1 y 2 y 3 y i 1 iff (x, y 1,y 2,,y i ) R Boolean Circuit C CVAL reduction for R 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 Where CVAL = Boolean Circuit evaluation problem 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 Boolean Circuit C 1 iff (x, y 1,y 2,,y i ) R CVAL reduction for R QSAT i is Σ i -complete x y 1 y 2 y 3 y i 1 iff (x, y 1,y 2,,y i ) R Boolean Circuit C CVAL reduction for R 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 23 24

7 QSAT is PSPACE-complete Theorem: QSAT is PSPACE-complete. Proof: x 1 x 2 x 3 Qx n φ(x 1, x 2,, x n )? 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 QSAT is PSPACE-complete given TM M deciding L PSPACE; input x 2 nk possible configurations single START configuration assume single ACCEPT configuration define: REACH(X, Y, i) configuration Y reachable from configuration X in at most 2 i steps QSAT is PSPACE-complete QSAT is PSPACE-complete REACH(X, Y, i) configuration Y reachable from configuration X in at most 2 i steps. Goal: produce 3-CNF φ(w 1,w 2,w 3,,w m ) such that w 1 w 2 Qw m φ(w 1,,w m ) REACH(START, ACCEPT, n k ) for i = 0, 1, n k produce quantified Boolean expressions ψ i (A, B, W) w 1 w 2 ψ i (A, B, W) REACH(A, B, i) convert ψ n k to 3-CNF φ add variables V w 1 w 2 V φ(a, B, W, V) REACH(A, B, n k ) hardwire A = START, B = ACCEPT w 1 w 2 V φ(w, V) x L 27 28

8 QSAT is PSPACE-complete QSAT is PSPACE-complete ψ o (A, B) = true iff A = B or A yields B in one step of M Boolean expression of size O(n k ) Key observation #1: REACH(A, B, i+1) Z [REACH(A, Z, i) REACH(Z, B, i)] STEP STEP STEP STEP config. A config. B cannot define ψ i+1 (A, B) to be Z [ψ i (A, Z) ψ i (Z, B)] why: then formula grows exponentially large) QSAT is PSPACE-complete QSAT is PSPACE-complete Key idea #2: use quantifiers Couldn t do ψ i+1 (A, B) = Z [ψ i (A, Z) ψ i (Z, B)] Key dea: define ψ i+1 (A, B) to be ψ o (A, B) = true iff A = B or A yields B in 1 step ψ i+1 (A, B) = Z X Y[((X=A Y=Z) (X=Z Y=B)) ψ i (X, Y)] 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 ψ i (X, Y) is preceded by quantifiers move to front (they don t involve X,Y,Z,A,B) 31 total size of ψ n k is O(n k ) 2 = poly(n) logspace reduction 32

9 PH collapse PH Σ 3 Π 3 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 NP Δ 3 Σ 2 Π 2 Δ 2 P conp 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 Oracles vs. Algorithms Natural complete problems A point to ponder: given poly-time algorithm for SAT can you solve MN CRCUT efficiently? what other problems? Entire complexity classes? given SAT oracle same input/output behavior can you solve MN CRCUT efficiently? 35 We now have quantified versions of SAT complete for levels in PH, PSPACE Natural complete problems? PSPACE: QSAT and games PH: bounded quantifier QSAT & games with bounded alternations almost all other natural problems lie in the second and third level of PH 36

10 Natural complete problems in PH MN CRCUT good candidate to be Σ 2 -complete, still open MN DNF: given DNF φ, integer k; is there a DNF φ of size at most k computing same function φ does? Theorem (U): MN DNF is Σ 2 -complete. Natural complete problems in PSPACE General phenomenon: many 2-player games are PSPACE-complete. 2 players, alternate picking edges lose when no unvisited choice auckland san francisco pasadena athens davis oakland GEOGRAPHY = {(G, s) : G is a directed graph and player can win from node s} Natural complete problems in PSPACE Theorem: GEOGRAPHY is PSPACEcomplete. Proof: in PSPACE easily expressed with alternating quantifiers PSPACE-hard reduction from QSAT 39 Proof Sketch: GEOGRAPHY is PSPACE-complete. x 1 x 2 x 3 ( x 1 x 2 x 3 ) ( x 3 x 1 ) (x 1 x 2 ) true true true x 1 x 2 x 3 false false false alternately pick truth assignment pick a true literal? pick a clause 40

11 Karp-Lipton Theorem we know that P = NP implies SAT has polynomial-size circuits. (showing SAT does not have poly-size circuits is one route to proving P NP) P/poly = Family of languages recognized by polynomial size boolean circuits suppose SAT has poly-size circuits, so: SAT P/poly any consequences? might hope: SAT P/poly PH collapses to Karp-Lipton Theorem Theorem (KL): if SAT has poly-size circuits then PH collapses to the second level. Proof: suffices to show Π 2 Σ 2 L Π 2 implies L expressible as: L = { x : y z (x, y, z) R} with R P. P, same as if SAT P Proof of Karp-Lipton Theorem L = { x : y z (x, y, z) R} given (x, y), z (x, y, z) R? is in NP pretend C solves SAT, use self-reducibility Claim: if SAT P/poly, then L = poly time { x : C y [use C repeatedly to find some z for which (x, y, z) R; accept iff (x, y, z) R] } Proof of Karp-Lipton Theorem L = { x : y z (x, y, z) R} {x : C y [use C repeatedly to find some z for which (x,y,z) R; accept iff (x,y,z) R] } x L: some C decides SAT C y [ ] accepts x L: y z (x, y, z) R C y [ ] rejects 43 44

12 Theorem BPP PH BPP language L: probabilistic TM M: x L Pr y [M(x,y) accepts] 2/3 x L Pr y [M(x,y) rejects] 2/3 Don t know BPP different from EXP Also don t know if Π 2 Σ 2 is different from EXP but believe much weaker Theorem (S,L,GZ): BPP (Π 2 Σ 2 ) 45 Proof of Theorem BPP PH Proof: BPP language L: p.p.t. TM M: x L Pr y [M(x,y) accepts] 2/3 x L Pr y [M(x,y) rejects] 2/3 Known Strong error reduction: There exists a probabilistic TM M : use n random bits ( y = n) # strings y for which M (x, y ) incorrect is at most 2 n/3 (can t achieve with naïve amplification) 46 Proof of Theorem BPP PH view y = (w, z), each of length n/2 consider output of M (x, (w, z)): w = x L x L so so many few ones, ones, not some enough disk fo is whole all ones disk Proof of Theorem BPP PH proof (continued): strong error reduction: # bad y < 2 n/3 y = (w, z) with w = z = n/2 Claim: L = {x : w z M (x, (w, z)) = 1 } x L: suppose w z M (x, (w, z)) = 0 implies 2 n/2 number of 0 s in output; contradiction x L: suppose w z M (x, (w, z)) = 1 implies 2 n/2 number of 1 s in output; contradiction 48

13 Proof of Theorem BPP PH given BPP language L: p.p.t. TM M: x L Pr y [M(x,y) accepts] 2/3 x L Pr y [M(x,y) rejects] 2/3 showed L = {x : w z M (x, (w, z)) = 1} thus BPP Σ 2 BPP closed under complement BPP Π 2 conclude: BPP (Π 2 Σ 2 ) 49

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