1 / 47

Time-Space Lower Bounds for the Polynomial-Time Hierarchy on Randomized Machines

Time-Space Lower Bounds for the Polynomial-Time Hierarchy on Randomized Machines. Scott Diehl Dieter van Melkebeek University of Wisconsin-Madison. Complexity of SAT. Conjecture: requires exponential time. Best lower bound: Ω (n). [FvM00]If machines use small space:

deo
Download Presentation

Time-Space Lower Bounds for the Polynomial-Time Hierarchy on Randomized Machines

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Time-Space Lower Bounds for the Polynomial-Time Hierarchy on Randomized Machines Scott Diehl Dieter van Melkebeek University of Wisconsin-Madison

  2. Complexity of SAT • Conjecture: requires exponential time. • Best lower bound: Ω(n). • [FvM00]If machines use small space: SAT cannot be solved deterministically in time nc & space no(1) for c<φ1.618.

  3. Complexity of SAT • Conjecture: requires exponential time. • Best lower bound: Ω(n). • [Wil05] If machines use small space: SAT cannot be solved deterministically in time nc & space no(1) for c < 1.732.

  4. Complexity of SAT • Conjecture: requires exponential time. • Best lower bound: Ω(n). • If machines use small space: SAT cannot be solved deterministically in time nc & space no(1) for c < 1.759.

  5. Complexity of QSATk • Generalization of SAT to Σk:QSATk. • i.e., QSAT2: Is xyψ(x,y) valid? • Conjecture: requires exponential time. • Theorem[FvM00]: (k ≥ 2, c < k) QSATk can’t be solved by deterministic machines in time nc & space no(1).

  6. Complexity of QSATk • Generalization of SAT to Σk:QSATk. • i.e., QSAT2: Is xyψ(x,y) valid? • Conjecture: requires exponential time even on randomized machines.

  7. Complexity of QSATk • Generalization of SAT to Σk:QSATk. • i.e., QSAT2: Is xyψ(x,y) valid? • Conjecture: requires exponential time even on randomized machines. • Main Theorem: (k ≥ 2, c < k) QSATk can’t be solved by randomized machines in time nc & space no(1).

  8. Outline • Ingredients. • Derivation for QSAT2. • Other Results. • Open Problems.

  9. QSATk vs. ΣkTIME[n] • Strong version of Cook-Levin: QSATkDTISP[nc, no(1)]  ΣkTIME[n] DTISP[nc, no(1)] (up to polylog factors)

  10. QSATk vs. ΣkTIME[n] • Strong version of Cook-Levin: QSATkBPTISP[nc, no(1)]  ΣkTIME[n] BPTISP[nc, no(1)] (up to polylog factors)

  11. Proof Techniques Prove ΣkTIME[n]  BPTISP[nc,no(1)] by indirect diagonalization. • Assume ΣkTIME[n]  BPTISP[nc,no(1)]. • Derive more unlikely inclusions. • Contradict known diagonalization.

  12. Warmup To prove Σ2TIME[n]  DTISP[nc,no(1)]. • Assume Σ2TIME[n]  DTISP[nc,no(1)]. • a) Speed up at the cost of alternations. b) Remove alternations at cost of time. • Contradict known diagonalization Σ2TIME[Ta] Π2TIME[Tb] for b < a.

  13. DTISP[nc,no(1) ] speedup no(1) Start config C nc Accept config C’

  14. DTISP[nc,no(1) ] speedup no(1) Start config C0 xLC1,…,Cb-10ib-1 CiCi+1 in nc/b steps nc/b C1 nc/b DTISP[nc,no(1)] Σ2TIME[bno(1)+nc/b]. C2 … … Choosing b=nc/2, DTISP[nc,no(1)] Σ2TIME[nc/2+o(1)]. Cb-1 nc/b Accept config Cb

  15. Warmup (cont’d) • Assume Σ2TIME[n]  DTISP[nc,no(1)].

  16. Warmup (cont’d) • Assume Σ2TIME[n]  DTISP[nc,no(1)]. • Then Σ2TIME[n]  DTISP[nc,no(1)] Π2TIME[nc/2+o(1)].

  17. Warmup (cont’d) • Assume Σ2TIME[n]  DTISP[nc,no(1)]. • Then Σ2TIME[n]  DTISP[nc,no(1)] Π2TIME[nc/2+o(1)]. • Contradiction for c/2 < 1.

  18. Warmup (cont’d) • Assume Σ2TIME[n]  DTISP[nc,no(1)]. • Then Σ2TIME[n]  DTISP[nc,no(1)] Π2TIME[nc/2+o(1)]. • Contradiction for c/2 < 1. (c < 2) QSAT2 DTISP[nc,no(1)].

  19. QSAT2 vs. BPTISP[nc,no(1)] • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Space-bounded speedup applicable? • Allows elimination of alternations? • Since BPP Π2, yields complementation Σ2TIME[n] Π2TIME[ f (n,c)]. • Is f small enough?

  20. Lautemann’s Proof BPP Π2 • Consider randomized algorithm A(x,r). • Time: T, Random bits: R, Error: < min(2-R/v, 1/v). • x  L all sets of v shifts of accepting set has some overlap. • Expressed as Π2 predicate: s1,...,sv{0,1}Ry{0,1}R1≤i≤vA(x,ysi) • Time: vR+ vT.

  21. Lautemann’s Proof BPP Π2 • Typical setting: • Error: < min(2-R/v, 1/v) satisfied by  < 1/R and v = R. • Easily guaranteed via O(log T) trials. s1,...,sR{0,1}Ry{0,1}R1≤i≤RA’(x,ysi) • Time: R2 + RT1+o(1)

  22. Improving the Simulation • Restriction  < min(2-R/v, 1/v) • v = O(1) shifts work when  is 2-(R). • Naïve amplification does not work. s1,...,sR{0,1}Ry{0,1}R1≤i≤RA’(x,ysi) • Time: R2 + RT1+o(1)

  23. Improving the Simulation • Restriction  < min(2-R/v, 1/v) • v = O(1) shifts work when  is 2-(R). • Deterministic amplification does! • Theorem [CW, IZ] Amplification via length O(R) walk on expander uses R’ = O(R) bits, gives error 2-R = 2-(R’). s1,...,sR{0,1}Ry{0,1}R1≤i≤RA’(x,ysi) • Time: R2 + RT1+o(1)

  24. Improving the Simulation • Restriction  < min(2-R/v, 1/v) • v = O(1) shifts work when  is 2-(R). • Deterministic amplification does! • Theorem [CW, IZ] Amplification via length O(R) walk on expander uses R’ = O(R) bits, gives error 2-R = 2-(R’). s1,...,sv{0,1}Ry{0,1}R1≤i≤vA’(x,ysi) • Time: O(RT)

  25. Nisan’s Derandomization • Theorem [Nisan ’92]: Every L  BPTISP[T,S] simulated in: • Time: T1+o(1) • Space: O(S log T) • Random bits: R’ = O(S log T) s1,...,sv{0,1}Ry{0,1}R1≤i≤vA’(x,ysi) • Time: O(RT)

  26. Nisan’s Derandomization • Theorem [Nisan ’92]: Every L  BPTISP[T,S] simulated in: • Time: T1+o(1) • Space: O(S log T) • Random bits: R’ = O(S log T) s1,...,sv{0,1}R’y{0,1}R’1≤i≤vA’(x,ysi) • Time: ST1+o(1).

  27. Complementation • BPTISP[T, To(1)] simulation in Π2: To(1) To(1) DTISP[T1+o(1),To(1)] • Assuming Σ2TIME[n]  BPTISP[nc, no(1)], Σ2TIME[n]Π2TIME[nc+o(1)].

  28. Speedup Σ2TIME[T1/2+o(1)] • BPTISP[T, To(1)] simulation in Π2: To(1) To(1) DTISP[T1+o(1),To(1)] T1/2+o(1) T1/2+o(1)TIME[T1/2+o(1)]

  29. Speedup Σ2TIME[T1/2+o(1)] • BPTISP[T, To(1)] simulation in Π2: To(1) To(1)DTISP[T1+o(1),To(1)] T1/2+o(1) T1/2+o(1)TIME[T1/2+o(1)]

  30. Speedup Σ2TIME[T1/2+o(1)] • BPTISP[T, To(1)] simulation in Π2: To(1) To(1)DTISP[T1+o(1),To(1)] • Speedup of final space-bounded stage BPTISP[T, To(1)] Π3TIME[T1/2+o(1)]. T1/2+o(1) T1/2+o(1)TIME[T1/2+o(1)]

  31. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T]  BPTISP[Tc, To(1)]

  32. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T]  BPTISP[Tc, To(1)] Π3TIME[Tc/2+o(1)]

  33. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T]  BPTISP[Tc, To(1)] Π3TIME[Tc/2+o(1)] = To(1).Σ2TIME[Tc/2+o(1)]

  34. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T]  BPTISP[Tc, To(1)] Π3TIME[Tc/2+o(1)] = To(1).Σ2TIME[Tc/2+o(1)] Π2TIME[Tc2/2+o(1)]

  35. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T]  BPTISP[Tc, To(1)] Π3TIME[Tc/2+o(1)] = To(1).Σ2TIME[Tc/2+o(1)] Π2TIME[Tc2/2+o(1)]. Π2TIME[Tc2/2+o(1)]

  36. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T] BPTISP[Tc, To(1)] Π3TIME[Tc/2+o(1)] = To(1).Σ2TIME[Tc/2+o(1)] Π2TIME[Tc2/2+o(1)]. Π2TIME[Tc2/2+o(1)]

  37. Putting It Together • Assume Σ2TIME[n]  BPTISP[nc, no(1)]. • Σ2TIME[T] BPTISP[Tc, To(1)] Π3TIME[Tc/2+o(1)] = To(1).Σ2TIME[Tc/2+o(1)] Π2TIME[Tc2/2+o(1)]. • Contradiction for c2/2 < 1. (c < √2) QSAT2  BPTISP[nc, no(1)]. Π2TIME[Tc2/2+o(1)]

  38. Inductive Argument Initially: Σ2TIME[T] Π2TIME[Tc+o(1)].

  39. Inductive Argument Initially: Σ2TIME[T] Π2TIME[Tc+o(1)]. New: Σ2TIME[T] Π2TIME[Tc2/2+o(1)]

  40. Inductive Argument Initially: Σ2TIME[T] Π2TIME[Tc+o(1)]. New: Σ2TIME[T] Π2TIME[Tc2/2+o(1)] • Allows for stronger complementations Σ2TIME[T] Π2TIME[Tc*f(k)], for f(k) = (c/2)k.

  41. Inductive Argument Initially: Σ2TIME[T] Π2TIME[Tc+o(1)]. New: Σ2TIME[T] Π2TIME[Tc2/2+o(1)] • Allows for stronger complementations Σ2TIME[T] Π2TIME[Tc*f(k)], for f(k) = (c/2)k. • (c < 2)(k ≥1) s.t. c*f(k) < 1

  42. Result Theorem: (c<2) QSAT2BPTISP[nc,no(1)]

  43. Result Theorem: (c<2)(d>0) QSAT2BPTISP[nc,nd], d  ½ from below as c 1 from above

  44. Result Theorem: (k2, c<k)(d>0) QSATkBPTISP[nc,nd] d 1 from below as c 1 from above for k3.

  45. Other Results • [FvM00] Tautologies, one-sided error: (c<21.414)(d>0) TAUT RTISP[nc, nd].

  46. Other Results • Tautologies, one-sided error: (c < 1.759 )(d>0) TAUT RTISP[nc, nd].

  47. Open Problems • Time-Space lower bounds for SAT on two-sided error randomized machines. • Match new deterministic results[Wil05]. QSAT2 DTISP[n2.78,no(1)].

More Related