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CS151 Complexity Theory

CS151 Complexity Theory. Lecture 12 May 9, 2013. Useful characterization. Recall: L  NP iff expressible as L = { x | 9 y, |y| ≤ |x| k , (x, y)  R } where R  P . Corollary: L  coNP iff expressible as L = { x | 8 y, |y| ≤ |x| k , (x, y)  R } where R  P.

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CS151 Complexity Theory

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  1. CS151Complexity Theory Lecture 12 May 9, 2013

  2. Useful characterization • Recall: L  NP iff expressible as L = { x | 9y, |y| ≤ |x|k, (x, y)  R } where R  P. • Corollary: L  coNP iff expressible as L = { x | 8y, |y| ≤ |x|k, (x, y)  R } where R  P.

  3. Useful characterization Theorem: L  Σi iff expressible as L = { x | 9 y, |y| ≤ |x|k, (x, y)  R } where R  Πi-1. • Corollary: L  Πi iff expressible as L = { x | 8 y, |y| ≤ |x|k, (x, y)  R } where R  Σi-1.

  4. Alternating quantifiers Nicer, more usable version: • LΣi iff expressible as L = { x | 9y1 8y29y3 …Qyi (x, y1,y2,…,yi)R } where Q= 8/9if i even/odd, and RP • LΠi iff expressible as L = { x | 8y19y2 8y3 …Qyi (x, y1,y2,…,yi)R } where Q= 9/8 if i even/odd, and RP

  5. Alternating quantifiers • Proof: • ( )induction on i • base case: true for Σ1=NP and Π1=coNP • consider LΣi: L = {x | 9y1 (x, y1)  R’ }, for R’  Πi-1 L = {x | 9y1 8y2 9y3 …Qyi ((x, y1), y2,…,yi)R} L = {x | 9y1 8y2 9y3 …Qyi (x, y1,y2,…,yi)R} • same argument for L Πi • ( ) exercise.

  6. Alternating quantifiers Pleasing viewpoint: “9898989…” PSPACE const. # of alternations poly(n) alternations PH Δ3 Σ2 Π2 “98” “89” Σi Πi “989…” “898…” Δ2 Σ3 Π3 “898” “989” NP coNP “8” “9” P

  7. Complete problems • three variants of SAT: • QSATi(i odd) = {3-CNFs φ(x1, x2, …, xi) for which 9x18x29x3 … 9xi φ(x1, x2, …, xi) = 1} • QSATi(i even) = {3-DNFs φ(x1, x2, …, xi) for which 9x18x29x3 … 8xi φ(x1, x2, …, xi) = 1} • QSAT = {3-CNFs φ for which 9x18x29x3 … Qxn φ(x1, x2, …, xn) = 1}

  8. QSATi is Σi-complete Theorem: QSATi is Σi-complete. • Proof: (clearly in Σi) • assume i odd; given L  Σi in form { x | 9y18y29y3 … 9yi (x, y1,y2,…,yi)  R } …x… …y1… …y2… …y3… … …yi… C CVAL reduction for R 1 iff (x, y1,y2,…,yi)  R

  9. QSATi is Σi-complete …x… …y1… …y2… …y3… … …yi… • Problem set: can construct 3-CNF φfrom C: 9zφ(x,y1,…,yi, z) = 1  C(x,y1,…,yi) = 1 • we get: 9y18y2…9yi9zφ(x,y1,…,yi, z) = 1 • 9y18y2…9yi C(x,y1,…,yi) = 1 x  L C 1 iff (x, y1,y2,…,yi)  R CVAL reduction for R

  10. QSATi is Σi-complete • Proof (continued) • assume i even; given L  Σi in form { x | 9y18y29y3 … 8yi (x, y1,y2,…,yi)  R } …x… …y1… …y2… …y3… … …yi… C CVAL reduction for R 1 iff (x, y1,y2,…,yi)  R

  11. QSATi is Σi-complete …x… …y1… …y2… …y3… … …yi… • Problem set: can construct 3-DNF φfrom C: 8zφ(x,y1,…,yi, z) = 1  C(x,y1,…,yi) = 1 • we get: 9y18y2… 8yi8zφ(x,y1,y2,…,yi, z) = 1  9y18y2…8yi C(x,y1,y2,…,yi) = 1  x  L C 1 iff (x, y1,y2,…,yi)  R CVAL reduction for R

  12. QSAT is PSPACE-complete Theorem: QSAT is PSPACE-complete. • Proof: 8x19x28x3 … Qxn φ(x1, x2, …, xn)? • in PSPACE: 9x18x29x3 … Qxn φ(x1, x2, …, xn)? • “9x1”: for each x1, recursively solve 8x29x3 … Qxn φ(x1, x2, …, xn)? • if encounter “yes”, return “yes” • “8x1”: for each x1, recursively solve 9x28x3 … Qxn φ(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

  13. QSAT is PSPACE-complete • given TM M deciding L  PSPACE; input x • 2nkpossible configurations • single START configuration • assume single ACCEPT configuration • define: REACH(X, Y, i)  configuration Y reachable from configuration X in at most 2i steps.

  14. QSAT is PSPACE-complete REACH(X, Y, i)  configuration Y reachable from configuration X in at most 2i steps. • Goal: produce 3-CNF φ(w1,w2,w3,…,wm) such that 9w18w2…Qwm φ(w1,…,wm) • REACH(START, ACCEPT, nk)

  15. QSAT is PSPACE-complete • for i = 0, 1, … nk produce quantified Boolean expressionsψi(A, B, W) 9w18w2… ψi(A, B, W)  REACH(A, B, i) • convert ψnkto 3-CNF φ • add variables V • 9w18w2… 9V φ(A, B, W, V)  REACH(A, B, nk) • hardwire A = START, B = ACCEPT 9w18w2… 9Vφ(W, V)  x  L

  16. QSAT is PSPACE-complete • ψo(A, B) = true iff • A = B or • A yields B in one step of M Boolean expression of size O(nk) … config. A STEP STEP STEP STEP config. B …

  17. QSAT is PSPACE-complete • Key observation #1: REACH(A, B, i+1)  9Z[REACH(A, Z, i)  REACH(Z, B, i)] • cannot define ψi+1(A, B) to be 9Z [ψi(A, Z)  ψi(Z, B)] (why?)

  18. QSAT is PSPACE-complete • Key idea #2: use quantifiers • couldn’t do ψi+1(A, B) = 9Z [ψi(A, Z)  ψi(Z, B)] • define ψi+1(A, B) to be 9Z8X8Y[((X=AY=Z)(X=ZY=B))  ψi(X, Y)] • ψi(X, Y) is preceded by quantifiers • move to front (they don’t involve X,Y,Z,A,B)

  19. QSAT is PSPACE-complete ψo(A, B) = true iff A = B or A yields B in 1 step ψi+1(A, B) = 9Z8X8Y[((X=AY=Z)(X=ZY=B))  ψi(X, Y)] • |ψ0| = O(nk) • |ψi+1| = O(nk) + |ψi| • total size of ψnk is O(nk)2 = poly(n) • logspace reduction

  20. PH Σ3 Π3 Δ3 Σ2 Π2 Δ2 NP coNP P PH collapse Theorem: if Σi = Πi then for all j > i Σj = Πj = Δj= Σi “the polynomial hierarchy collapses to the i-th level” • Proof: • sufficient to show Σi = Σi+1 • then Σi+1= Σi = Πi =Πi+1;apply theorem again

  21. PH collapse • recall: L  Σi+1 iff expressible as L = { x | 9 y (x, y)  R } where R  Πi • since Πi = Σi, R expressible as R = { (x,y) | 9z((x, y), z) R’ } where R’  Πi-1 • together: L = { x | 9 (y, z) (x, (y, z))  R’} • conclude L  Σi

  22. 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?

  23. 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 and third level

  24. Natural complete problems in PH • MIN CIRCUIT • good candidate to be 2-complete, still open • MIN DNF: given DNF φ, integer k; is there a DNF φ’ of size at most k computing same function φ does? Theorem (U): MIN DNF is Σ2-complete.

  25. Natural complete problems in PSPACE • General phenomenon: many 2-player games are PSPACE-complete. • 2 players I, II • alternate pick- ing edges • lose when no unvisited choice pasadena auckland athens davis san francisco oakland • GEOGRAPHY = {(G, s) : G is a directed graph and player I can win from node s}

  26. Natural complete problems in PSPACE Theorem: GEOGRAPHY is PSPACE-complete. Proof: • in PSPACE • easily expressed with alternating quantifiers • PSPACE-hard • reduction from QSAT

  27. Natural complete problems in PSPACE 9x18x29x3…(x1x2x3)(x3x1)…(x1x2) I false alternately pick truth assignment true x1 II I II x2 true false I II I I x3 true false pick a clause II II … I I pick a true literal?

  28. Karp-Lipton • 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) • suppose SAT has poly-size circuits • any consequences? • might hope: SAT  P/poly PHcollapses to P, same as if SAT  P

  29. Karp-Lipton Theorem (KL): if SAT has poly-size circuits then PH collapses to the second level. • Proof: • suffices to show Π2 Σ2 • L Π2 implies L expressible as: L = {x : 8y 9z (x, y, z)  R} with R  P.

  30. Karp-Lipton L = {x : 8y 9z (x, y, z)  R} • given (x, y), “9z (x, y, z)  R?” is in NP • pretend C solves SAT, use self-reducibility • Claim: if SAT  P/poly, then L = { x : 9C 8y [use C repeatedly to find some z for which (x, y, z)  R; accept iff (x, y, z)  R] } poly time

  31. Karp-Lipton L = {x : 8y 9z (x, y, z)  R} {x : 9C 8y [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  9C 8y[…] accepts • x  L: • 9y 8z (x, y, z)  R  9C 8y[…] rejects

  32. BPP PH • Recall: don’t know BPP different from EXP Theorem (S,L,GZ): BPP (Π2Σ2) • don’t know Π2Σ2 different from EXP but believe muchweaker

  33. BPP PH • Proof: • BPP language L: p.p.t. TM M: x  L  Pry[M(x,y) accepts] ≥ 2/3 x  L  Pry[M(x,y) rejects] ≥ 2/3 • strong error reduction: p.p.t. TM M’ • use n random bits (|y’| = n) • # strings y’ for which M’(x, y’) incorrect is at most 2n/3 • (can’t achieve with naïve amplification)

  34. BPP PH so many ones, some disk is all ones so few ones, not enough for whole disk • view y’ = (w, z), each of length n/2 • consider output of M’(x, (w, z)): w =000…00000…01000…10 … 111…11 xL 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 … 0 1 1 1 0 1 1 0 1 1 1 1 0 1 0 1 1 1 1 0 0 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 0 1 1 0 1 xL 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 … 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

  35. BPP PH • proof (continued): • strong error reduction: # bad y’ < 2n/3 • y’ = (w, z) with |w| = |z| = n/2 • Claim: L = {x : 9w 8z M’(x, (w, z)) = 1 } • xL: suppose 8w9z M’(x, (w, z)) = 0 • implies  2n/2 0’s; contradiction • xL: suppose 9w8z M’(x, (w, z)) = 1 • implies  2n/2 1’s; contradiction

  36. BPP PH • given BPP language L: p.p.t. TM M: x  L  Pry[M(x,y) accepts] ≥ 2/3 x  L  Pry[M(x,y) rejects] ≥ 2/3 • showed L = {x : 9w 8z M’(x, (w, z)) = 1} • thus BPP  Σ2 • BPP closed under complement BPP  Π2 • conclude: BPP (Π2Σ2)

  37. New Topic The complexity of counting

  38. Counting problems • So far, we have ignored function problems • given x, compute f(x) • important class of function problems: counting problems • e.g. given 3-CNF φ how many satisfying assignments are there?

  39. Counting problems • #P is the class of function problems expressible as: input x f(x) = |{y : (x, y)  R}| where R  P. • compare to NP(decision problem) input x f(x) = y : (x, y)  R ? where R  P.

  40. Counting problems • examples • #SAT: given 3-CNF φ how many satisfying assignments are there? • #CLIQUE: given (G, k) how many cliques of size at least k are there?

  41. Reductions • Reduction from function problem f1 to function problem f2 • two efficiently computable functions Q, A x (prob. 1) Q y (prob. 2) f1 f2 A f2(y) f1(x)

  42. Reductions • problem fis #P-complete if • f is in #P • every problem in #P reduces to f • “parsimonious reduction”: A is identity • many standard NP-completeness reductions are parsimonious • therefore: if #SAT is #P-complete we get lots of #P-complete problems Q x (prob. 1) y (prob. 2) f1 f2 A f2(y) f1(x)

  43. #SAT #SAT: given 3-CNF φ how many satisfying assignments are there? Theorem: #SAT is #P-complete. • Proof: • clearly in #P: (φ, A)  R  A satisfies φ • take any f  #P defined by R  P

  44. #SAT …x… …y… • add new variables z, produce φ such that z φ(x, y, z) = 1  C(x, y) = 1 • for (x, y) such that C(x, y) = 1 this z is unique • hardwire x • # satisfying assignments = |{y : (x, y)  R}| f(x) = |{y : (x, y)  R}| C CVAL reduction for R 1 iff (x, y)  R

  45. Relationship to other classes • To compare to classes of decision problems, usually consider P#P which is a decision class… • easy: NP, coNP  P#P • easy: P#P  PSPACE Toda’s Theorem (homework):PH  P#P.

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