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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:

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

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  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)].

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