Better Time-Space Lower Bounds for SAT and Related Problems

Transcription

1 Better Time-Space Lower Bounds for SAT and Related Problems Ryan Williams Carnegie Mellon University June 12,

2 Introduction Few super-linear time lower bounds known for natural problems in NP (Existing ones generally use a restricted computational model) 1

3 Introduction Few super-linear time lower bounds known for natural problems in NP (Existing ones generally use a restricted computational model) We will show time lower bounds for the SAT problem, on random-access machines using n o(1) space These lower bounds carry over to MAX-SAT, Hamilton Path, Vertex Cover, etc. [Raz and Van Melkebeek] 1-a

4 Introduction Few super-linear time lower bounds known for natural problems in NP (Existing ones generally use a restricted computational model) We will show time lower bounds for the SAT problem, on random-access machines using n o(1) space These lower bounds carry over to MAX-SAT, Hamilton Path, Vertex Cover, etc. [Raz and Van Melkebeek] Previous time lower bounds for SAT (assuming n o(1) space): ω(n) [Kannan 84] 1-b

5 Introduction Few super-linear time lower bounds known for natural problems in NP (Existing ones generally use a restricted computational model) We will show time lower bounds for the SAT problem, on random-access machines using n o(1) space These lower bounds carry over to MAX-SAT, Hamilton Path, Vertex Cover, etc. [Raz and Van Melkebeek] Previous time lower bounds for SAT (assuming n o(1) space): ω(n) [Kannan 84] Ω(n 1+ε ) for some ε > 0 [Fortnow 97] 1-c

6 Introduction Few super-linear time lower bounds known for natural problems in NP (Existing ones generally use a restricted computational model) We will show time lower bounds for the SAT problem, on random-access machines using n o(1) space These lower bounds carry over to MAX-SAT, Hamilton Path, Vertex Cover, etc. [Raz and Van Melkebeek] Previous time lower bounds for SAT (assuming n o(1) space): ω(n) [Kannan 84] Ω(n 1+ε ) for some ε > 0 [Fortnow 97] Ω(n 2 ε ) for all ε > 0 [Lipton and Viglas 99] 1-d

7 Introduction Few super-linear time lower bounds known for natural problems in NP (Existing ones generally use a restricted computational model) We will show time lower bounds for the SAT problem, on random-access machines using n o(1) space These lower bounds carry over to MAX-SAT, Hamilton Path, Vertex Cover, etc. [Raz and Van Melkebeek] Previous time lower bounds for SAT (assuming n o(1) space): ω(n) [Kannan 84] Ω(n 1+ε ) for some ε > 0 [Fortnow 97] Ω(n 2 ε ) for all ε > 0 [Lipton and Viglas 99] Ω(n φ ε ) where φ = [Fortnow and Van Melkebeek 00] 1-e

8 2 and φ are nice constants... Our Main Result The constant of our work will be (larger, but) not-so-nice. 2

9 2 and φ are nice constants... Our Main Result The constant of our work will be (larger, but) not-so-nice. Theorem: For all k, SAT requires n Υ k time (infinitely often) on a random-access machine using n o(1) workspace, where Υ k is the positive solution in (1, 2) to Υ 3 k (Υ k 1) = k 2 k+3 ( (k 1) 2 k+3 ). 2-a

10 2 and φ are nice constants... Our Main Result The constant of our work will be (larger, but) not-so-nice. Theorem: For all k, SAT requires n Υ k time (infinitely often) on a random-access machine using n o(1) workspace, where Υ k is the positive solution in (1, 2) to Υ 3 k (Υ k 1) = k 2 k+3 ( (k 1) 2 k+3 ). Define Υ := lim k Υ k. Then: n Υ ε lower bound. 2-b

11 2 and φ are nice constants... Our Main Result The constant of our work will be (larger, but) not-so-nice. Theorem: For all k, SAT requires n Υ k time (infinitely often) on a random-access machine using n o(1) workspace, where Υ k is the positive solution in (1, 2) to Υ 3 k (Υ k 1) = k 2 k+3 ( (k 1) 2 k+3 ). Define Υ := lim k Υ k. Then: n Υ ε lower bound. (Note: the Υ stands for Ugly ) 2-c

12 2 and φ are nice constants... Our Main Result The constant of our work will be (larger, but) not-so-nice. Theorem: For all k, SAT requires n Υ k time (infinitely often) on a random-access machine using n o(1) workspace, where Υ k is the positive solution in (1, 2) to Υ 3 k (Υ k 1) = k 2 k+3 ( (k 1) 2 k+3 ). Define Υ := lim k Υ k. Then: n Υ ε lower bound. (Note: the Υ stands for Ugly ) However, Υ , so we ll present the result with 3. 2-d

13 Points About The Method We Use The theorem says for any sufficiently restricted machine, there is an infinite set of SAT instances it cannot solve correctly We will not construct such a set of instances for every machine! Proof is by contradiction: it would be absurd, if such a machine could solve SAT almost everywhere Ours and the above cited methods use artificial computational models (alternating machines) to prove lower bounds for explicit problems in a realistic model 3

14 Outline Preliminaries and Proof Strategy A Speed-Up Theorem (small-space computations can be accelerated using alternation) A Slow-Down Lemma (NTIME can be efficiently simulated implies Σ k TIME can be efficiently simulated with some slow-down) Lipton and Viglas n 2 Lower Bound (the starting point for our approach) Our Inductive Argument (how to derive a better bound from Lipton-Viglas) From n 1.66 to n (a subtle argument that squeezes more from the induction) 4

15 Preliminaries: Two possibly obscure complexity classes DTISP[t, s] is deterministic time t and space s, simultaneously (Note DTISP[t, s] DTIME[t] SPACE[s] in general) We will be looking at DTISP[n k,n o(1) ] for k 1. NQL := c 0 NTIME[n(log n)c ] = NTIME[n poly(log n)] The NQL stands for nondeterministic quasi-linear time 5

16 Preliminaries: SAT Facts Satisfiability (SAT) is not only NP-complete, but also: Theorem: [Cook 85, Gurevich and Shelah 89, Tourlakis 00] SAT is NQL-complete, under reductions doable in O(n poly(log n)) time and O(log n) space (simultaneously). Moreover the ith bit of the reduction can be computed in O(poly(log n)) time. Let D be closed under quasi-linear time, logspace reductions. 6

17 Preliminaries: SAT Facts Satisfiability (SAT) is not only NP-complete, but also: Theorem: [Cook 85, Gurevich and Shelah 89, Tourlakis 00] SAT is NQL-complete, under reductions doable in O(n poly(log n)) time and O(log n) space (simultaneously). Moreover the ith bit of the reduction can be computed in O(poly(log n)) time. Let D be closed under quasi-linear time, logspace reductions. Corollary: If NTIME[n] D, then SAT / D. If one can show NTIME[n] is not contained in some D, then one can name an explicit problem (SAT) not in D (modulo polylog factors) 6-a

18 Preliminaries: Some Hierarchy Theorems For reasonable t(n) n, NTIME[t] contime[o(t)]. Furthermore, for integers k 1, Σ k TIME[t] Π k TIME[o(t)]. So, there s a tight time hierarchy within levels of the polynomial hierarchy. 7

19 Proof Strategy Show if SAT has a sufficiently good algorithm, then one contradicts a hierarchy theorem. 8

20 Proof Strategy Show if SAT has a sufficiently good algorithm, then one contradicts a hierarchy theorem. Strategy of Prior work: 1. Show that DTISP[n c,n o(1) ] can be sped-up when simulated on an alternating machine 2. Show that NTIME[n] DTISP[n c,n o(1) ] allows those alternations to be removed without much slow-down 3. Contradict a hierarchy theorem for small c 8-a

21 Proof Strategy Show if SAT has a sufficiently good algorithm, then one contradicts a hierarchy theorem. Strategy of Prior work: 1. Show that DTISP[n c,n o(1) ] can be sped-up when simulated on an alternating machine 2. Show that NTIME[n] DTISP[n c,n o(1) ] allows those alternations to be removed without much slow-down 3. Contradict a hierarchy theorem for small c Our proof will use the Σ k time versus Π k time hierarchy, for all k 8-b

22 Outline Preliminaries A Speed-Up Theorem A Slow-Down Lemma Lipton and Viglas n 2 Lower Bound Our Inductive Argument From n 1.66 to n

23 A Speed-Up Theorem (Trading Time for Alternations) Let: t(n) = n c for rational c 1, s(n) be n o(1), and k 2 be an integer. Theorem: [Fortnow and Van Melkebeek 00] [Kannan 83] DTISP[t,s] Σ k TIME[t 1/k+o(1) ] Π k TIME[t 1/k+o(1) ]. 10

24 A Speed-Up Theorem (Trading Time for Alternations) Let: t(n) = n c for rational c 1, s(n) be n o(1), and k 2 be an integer. Theorem: [Fortnow and Van Melkebeek 00] [Kannan 83] DTISP[t,s] Σ k TIME[t 1/k+o(1) ] Π k TIME[t 1/k+o(1) ]. That is, for any machine M running in time t and using small workspace, there is an alternating machine M that makes k alternations and takes roughly k t time. Moreover, M can start in either an existential or a universal state 10-a

25 Proof of the speed-up theorem Let x be input, M be the small space machine to simulate Goal: Write a clever sentence in first-order logic with k (alternating) quantifiers that is equivalent to M(x) accepting 11

26 Proof of the speed-up theorem Let x be input, M be the small space machine to simulate Goal: Write a clever sentence in first-order logic with k (alternating) quantifiers that is equivalent to M(x) accepting Let C j denote configuration of M(x) after jth step: bit string encoding head positions, workspace contents, finite control By space assumption on M, C j n o(1) 11-a

27 Proof of the speed-up theorem Let x be input, M be the small space machine to simulate Goal: Write a clever sentence in first-order logic with k (alternating) quantifiers that is equivalent to M(x) accepting Let C j denote configuration of M(x) after jth step: bit string encoding head positions, workspace contents, finite control By space assumption on M, C j n o(1) M(x) accepts iff there is a sequence C 1,C 2,...,C t where C 1 is the initial configuration, C t is in accept state For all i, C i leads to C i+1 in one step of M on input x. 11-b

28 Proof of speed-up theorem: The case k = 2 M(x) accepts iff ( C 0,C t,c 2 t,...,c t) ( i {1,..., t}) [C i t leads to C (i+1) t in t steps, C 0 is initial, C t is accepting] 12

29 Proof of speed-up theorem: The case k = 2 M(x) accepts iff ( C 0,C t,c 2 t,...,c t) ( i {1,..., t}) [C i t leads to C (i+1) t in t steps, C 0 is initial, C t is accepting] Runtime on an alternating machine: takes O( t s) = t 1/2+o(1) time to write down the C j s takes O(log t) time to write down i [ ] takes O( t s) deterministic time to check Two alternations, square root speedup 12-a

30 Proof of speed-up theorem: The k = 3 case, first attempt For k = 2, we had ( C 0,C t,c 2 t,...,c t) ( i {0, 1,..., t}) [C i t leads to C (i+1) t in t steps, C 0 initial, C t accepting] 13

31 Proof of speed-up theorem: The k = 3 case, first attempt For k = 2, we had ( C 0,C t,c 2 t,...,c t) ( i {0, 1,..., t}) [C i t leads to C (i+1) t in t steps, C 0 initial, C t accepting] Observation: The [ ] is an O( t) time and small-space computation, thus we can speed it up by a square root as well 13-a

32 Proof of speed-up theorem: The k = 3 case, first attempt For k = 2, we had ( C 0,C t,c 2 t,...,c t) ( i {0, 1,..., t}) [C i t leads to C (i+1) t in t steps, C 0 initial, C t accepting] Observation: The [ ] is an O( t) time and small-space computation, thus we can speed it up by a square root as well Straightforward way of doing this leads to: ( C 0,C t 2/3,C 2 t 2/3,...,C t )( i {0, 1,...,t 1/3 }) ( C i t 2/3 +t 1/3,C i t 2/3 +2 t 1/3,...,C (i+1) t 2/3)( j {1,..., t}) [C i t 2/3 +j t 1/3 leads to C i t 2/3 +(j+1) t 1/3 in t1/3 steps, C 0 initial, C t accepting] 13-b

33 k = 2 has one stage C 0 C t 1/2 C 2 t 1/2.... Ct

34 k = 2 has one stage C 0 C 0 k = 3 has two stages Ct 2/3 C t 1/2 C C 2/3 2/3 t t + t 1/3 C Ct t 1/2 C.... C t 2 t 2/3 C.... C 2/3 t + 2 t 1/3 2 t 2/3

35 Proof of speed-up theorem: Not quite enough for k = 3 The k = 3 sentence we gave uses four quantifiers, for only t 1/3 time (we want only three quantifier blocks) 15

36 Proof of speed-up theorem: Not quite enough for k = 3 The k = 3 sentence we gave uses four quantifiers, for only t 1/3 time (we want only three quantifier blocks) Idea: Take advantage of the computation s determinism only one possible configuration at any step 15-a

37 Proof of speed-up theorem: Not quite enough for k = 3 The k = 3 sentence we gave uses four quantifiers, for only t 1/3 time (we want only three quantifier blocks) Idea: Take advantage of the computation s determinism only one possible configuration at any step The acceptance condition for M(x) can be complemented: M(x) accepts iff ( C 0,C t,c 2 t,...,c t rejecting) ( i {1,..., t}) [C i t does not lead to C i t+ t in t steps] For all configuration sequences C 1,...,C t where C t is rejecting, there exists a configuration C i that does not lead to C i+1 15-b

38 The k = 3 case We can therefore rewrite the k = 3 case, from ( C 0,C t 2/3,C 2 t 2/3,...,C t accepting)( i {0, 1,...,t 1/3 }) ( C i t 2/3 +t 1/3,C i t 2/3 +2 t 1/3,...,C (i+1) t 2/3)( j {1,..., t}) [C i t 2/3 +j t 1/3 leads to C i t 2/3 +(j+1) t 1/3 in t1/3 steps] to: 16

39 The k = 3 case We can therefore rewrite the k = 3 case, from ( C 0,C t 2/3,C 2 t 2/3,...,C t accepting)( i {0, 1,...,t 1/3 }) ( C i t 2/3 +t 1/3,C i t 2/3 +2 t 1/3,...,C (i+1) t 2/3)( j {1,..., t}) [C i t 2/3 +j t 1/3 leads to C i t 2/3 +(j+1) t 1/3 in t1/3 steps] to: ( C 0,C t 2/3,C 2 t 2/3,...,C t accepting)( i {0, 1,...,t 1/3 }) ( C i t 2/3 +t 1/3,...,C (i+1) t 2/3 t 1/3,C C (i+1) t 2/3 (i+1) t 2/3) ( j {1,..., t}) [C i t 2/3 +j t 1/3 does not lead to C i t 2/3 +(j+1) t 1/3 in t1/3 steps] 16-a

40 The k = 3 case We can therefore rewrite the k = 3 case, from ( C 0,C t 2/3,C 2 t 2/3,...,C t accepting)( i {0, 1,...,t 1/3 }) ( C i t 2/3 +t 1/3,C i t 2/3 +2 t 1/3,...,C (i+1) t 2/3)( j {1,..., t}) [C i t 2/3 +j t 1/3 leads to C i t 2/3 +(j+1) t 1/3 in t1/3 steps] to: ( C 0,C t 2/3,C 2 t 2/3,...,C t accepting)( i {0, 1,...,t 1/3 }) ( C i t 2/3 +t 1/3,...,C (i+1) t 2/3 t 1/3,C C (i+1) t 2/3 (i+1) t 2/3) ( j {1,..., t}) [C i t 2/3 +j t 1/3 does not lead to C i t 2/3 +(j+1) t 1/3 in t1/3 steps] Voila! Three quantifier blocks. This is in Σ 3 TIME[t 1/3+o(1) ] (and similarly one can show it s in Π 3 TIME[t 1/3+o(1) ]) 16-b

41 This can be generalized... For arbitrary k 3, one simply guesses (existentially or universally) t 1/k configurations at each stage 17

42 This can be generalized... For arbitrary k 3, one simply guesses (existentially or universally) t 1/k configurations at each stage Inverting quantifiers means the number of alternations only increases by one for every stage ( ) ( ) ( ) 17-a

43 This can be generalized... For arbitrary k 3, one simply guesses (existentially or universally) t 1/k configurations at each stage Inverting quantifiers means the number of alternations only increases by one for every stage ( ) ( ) ( ) There are k 1 stages of guessing t 1/k configurations, then t 1/k time to deterministically verify configurations 17-b

44 Outline Preliminaries A Speed-Up Theorem A Slow-Down Lemma Lipton and Viglas n 2 Lower Bound Our Inductive Argument From n 1.66 to n

45 A Slow-Down Lemma (Trading Alternations For Time) Main Idea: The assumption NTIME[n] DTIME[n c ] allows one to remove alternations from a computation, with a small time increase 19

46 A Slow-Down Lemma (Trading Alternations For Time) Main Idea: The assumption NTIME[n] DTIME[n c ] allows one to remove alternations from a computation, with a small time increase Let t(n) n be a polynomial, c 1. Lemma: If NTIME[n] DTIME[n c ] then for all k 1, Σ k TIME[t] Σ k 1 TIME[t c ]. We prove the following, which will be very useful in our final proof. 19-a

47 A Slow-Down Lemma (Trading Alternations For Time) Main Idea: The assumption NTIME[n] DTIME[n c ] allows one to remove alternations from a computation, with a small time increase Let t(n) n be a polynomial, c 1. Lemma: If NTIME[n] DTIME[n c ] then for all k 1, Σ k TIME[t] Σ k 1 TIME[t c ]. We prove the following, which will be very useful in our final proof. Theorem: If Σ k TIME[n] Π k TIME[n c ] then Σ k+1 TIME[t] Σ k TIME[t c ]. 19-b

48 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Let M be a Σ k+1 machine running in time t 20

49 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Let M be a Σ k+1 machine running in time t Recall M(x) can be characterized by a first-order sentence: ( x 1, x 1 t( x ))( x 2, x 2 t( x )) (Qz, x k+1 t( x ))[P(x,x 1,x 2,...,x k+1 )] where P runs in time t( x ) 20-a

50 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Let M be a Σ k+1 machine running in time t Recall M(x) can be characterized by a first-order sentence: ( x 1, x 1 t( x ))( x 2, x 2 t( x )) (Qz, x k+1 t( x ))[P(x,x 1,x 2,...,x k+1 )] where P runs in time t( x ) Important Point: input to P is of O(t( x )) length, so P actually runs in linear time with respect to the length of its input 20-b

51 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Define R(x,x 1 ) := ( x 2, x 2 t( x )) (Qz, x k+1 t( x )) [P(x,x 1,x 2,...,x k+1 )] 21

52 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Define R(x,x 1 ) := ( x 2, x 2 t( x )) (Qz, x k+1 t( x )) [P(x,x 1,x 2,...,x k+1 )] So M(x) accepts iff ( x 1, x 1 t( x ))R(x,x 1 ) 21-a

53 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Define R(x,x 1 ) := ( x 2, x 2 t( x )) (Qz, x k+1 t( x )) [P(x,x 1,x 2,...,x k+1 )] So M(x) accepts iff ( x 1, x 1 t( x ))R(x,x 1 ) By definition, R recognized by a Π k machine in time t( x ), i.e. linear time ( x 1 = t( x )). 21-b

54 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Define R(x,x 1 ) := ( x 2, x 2 t( x )) (Qz, x k+1 t( x )) [P(x,x 1,x 2,...,x k+1 )] So M(x) accepts iff ( x 1, x 1 t( x ))R(x,x 1 ) By definition, R recognized by a Π k machine in time t( x ), i.e. linear time ( x 1 = t( x )). By assumption, there is R equivalent to R that starts with an, has k quantifier blocks, is in t( x ) c time 21-c

55 A Slow-Down Lemma: Proof Assume Σ k TIME[n] Π k TIME[n c ] Define R(x,x 1 ) := ( x 2, x 2 t( x )) (Qz, x k+1 t( x )) [P(x,x 1,x 2,...,x k+1 )] So M(x) accepts iff ( x 1, x 1 t( x ))R(x,x 1 ) By definition, R recognized by a Π k machine in time t( x ), i.e. linear time ( x 1 = t( x )). By assumption, there is R equivalent to R that starts with an, has k quantifier blocks, is in t( x ) c time M(x) accepts iff [( x 1, x 1 t( x ))R (x,x 1 )] Σ k TIME[t c ] 21-d

56 Outline Preliminaries A Speed-Up Theorem A Slow-Down Lemma Lipton and Viglas n 2 Lower Bound Our Inductive Argument From n 1.66 to n

57 Lipton and Viglas n 2 Lower Bound (Rephrased) Lemma: If NTIME[n] DTISP[n c,n o(1) ] for some c 1, then for all polynomials t(n) n, NTIME[t] DTISP[t c,t o(1) ] Theorem: NTIME[n] DTISP[n 2 ε,n o(1) ] 23

58 Lipton and Viglas n 2 Lower Bound (Rephrased) Lemma: If NTIME[n] DTISP[n c,n o(1) ] for some c 1, then for all polynomials t(n) n, NTIME[t] DTISP[t c,t o(1) ] Theorem: NTIME[n] DTISP[n 2 ε,n o(1) ] Proof: Assume NTIME[n] DTISP[n c,n o(1) ] (We will find a c that implies a contradiction) 23-a

59 Lipton and Viglas n 2 Lower Bound (Rephrased) Lemma: If NTIME[n] DTISP[n c,n o(1) ] for some c 1, then for all polynomials t(n) n, NTIME[t] DTISP[t c,t o(1) ] Theorem: NTIME[n] DTISP[n 2 ε,n o(1) ] Proof: Assume NTIME[n] DTISP[n c,n o(1) ] (We will find a c that implies a contradiction) Σ 2 TIME[n] NTIME[n c ], by slow-down theorem 23-b

60 Lipton and Viglas n 2 Lower Bound (Rephrased) Lemma: If NTIME[n] DTISP[n c,n o(1) ] for some c 1, then for all polynomials t(n) n, NTIME[t] DTISP[t c,t o(1) ] Theorem: NTIME[n] DTISP[n 2 ε,n o(1) ] Proof: Assume NTIME[n] DTISP[n c,n o(1) ] (We will find a c that implies a contradiction) Σ 2 TIME[n] NTIME[n c ], by slow-down theorem NTIME[n c ] DTISP[n c2,n o(1) ], by assumption and padding 23-c

61 Lipton and Viglas n 2 Lower Bound (Rephrased) Lemma: If NTIME[n] DTISP[n c,n o(1) ] for some c 1, then for all polynomials t(n) n, NTIME[t] DTISP[t c,t o(1) ] Theorem: NTIME[n] DTISP[n 2 ε,n o(1) ] Proof: Assume NTIME[n] DTISP[n c,n o(1) ] (We will find a c that implies a contradiction) Σ 2 TIME[n] NTIME[n c ], by slow-down theorem NTIME[n c ] DTISP[n c2,n o(1) ], by assumption and padding DTISP[n c2,n o(1) ] Π 2 TIME[n c2 /2 ], by speed-up theorem, so c < 2 contradicts the hierarchy theorem 23-d

62 Outline Preliminaries A Speed-Up Theorem A Slow-Down Lemma Lipton and Viglas n 2 Lower Bound Our Inductive Argument From n 1.66 to n

63 Viewing Lipton-Viglas as a Lemma (The Base Case for Our Induction) We deliberately presented Lipton-Viglas s result differently from the original argument. In this way, we get Lemma: NTIME[n] DTISP[n c,n o(1) ] implies Σ 2 TIME[n] Π 2 TIME[n c2 /2 ]. Note if c < 2 then c 2 /2 < c. 25

64 Viewing Lipton-Viglas as a Lemma (The Base Case for Our Induction) We deliberately presented Lipton-Viglas s result differently from the original argument. In this way, we get Lemma: NTIME[n] DTISP[n c,n o(1) ] implies Σ 2 TIME[n] Π 2 TIME[n c2 /2 ]. Note if c < 2 then c 2 /2 < c. Thus, we may not necessarily have a contradiction for larger c, but we can remove one alternation from Σ 3 with only n c2 /2 cost 25-a

65 Viewing Lipton-Viglas as a Lemma (The Base Case for Our Induction) We deliberately presented Lipton-Viglas s result differently from the original argument. In this way, we get Lemma: NTIME[n] DTISP[n c,n o(1) ] implies Σ 2 TIME[n] Π 2 TIME[n c2 /2 ]. Note if c < 2 then c 2 /2 < c. Thus, we may not necessarily have a contradiction for larger c, but we can remove one alternation from Σ 3 with only n c2 /2 cost Slow-down theorem implies Σ 3 TIME[n] Σ 2 TIME[n c2 /2 ] 25-b

66 The Start of the Induction: Σ 3 Assume NTIME[n] DTISP[n c,n o(1) ] and the lemma 26

67 The Start of the Induction: Σ 3 Assume NTIME[n] DTISP[n c,n o(1) ] and the lemma Σ 3 TIME[n] Σ 2 TIME[n c2 /2 ], by slow-down and lemma 26-a

68 The Start of the Induction: Σ 3 Assume NTIME[n] DTISP[n c,n o(1) ] and the lemma Σ 3 TIME[n] Σ 2 TIME[n c2 /2 ], by slow-down and lemma Σ 2 TIME[n c2 /2 ] NTIME[n c3 /2 ], by slow-down 26-b

69 The Start of the Induction: Σ 3 Assume NTIME[n] DTISP[n c,n o(1) ] and the lemma Σ 3 TIME[n] Σ 2 TIME[n c2 /2 ], by slow-down and lemma Σ 2 TIME[n c2 /2 ] NTIME[n c3 /2 ], by slow-down NTIME[n c3 /2 ] DTISP[n c4 /2,n o(1) ], by assumption and padding 26-c

70 The Start of the Induction: Σ 3 Assume NTIME[n] DTISP[n c,n o(1) ] and the lemma Σ 3 TIME[n] Σ 2 TIME[n c2 /2 ], by slow-down and lemma Σ 2 TIME[n c2 /2 ] NTIME[n c3 /2 ], by slow-down NTIME[n c3 /2 ] DTISP[n c4 /2,n o(1) ], by assumption and padding DTISP[n c4 /2,n o(1) ] Π 3 TIME[n c4 /6 ], by speed-up 26-d

71 The Start of the Induction: Σ 3 Assume NTIME[n] DTISP[n c,n o(1) ] and the lemma Σ 3 TIME[n] Σ 2 TIME[n c2 /2 ], by slow-down and lemma Σ 2 TIME[n c2 /2 ] NTIME[n c3 /2 ], by slow-down NTIME[n c3 /2 ] DTISP[n c4 /2,n o(1) ], by assumption and padding DTISP[n c4 /2,n o(1) ] Π 3 TIME[n c4 /6 ], by speed-up Observe: Now c < contradicts time hierarchy for Σ 3 and Π 3 But if c 4 6, then we obtain a new lemma : Σ 3 TIME[n] Π 3 TIME[n c4 /6 ] 26-e

72 Σ 4, Σ 5,... Assume NTIME[n] DTISP[n c,n o(1) ] and lemmas 27

73 Σ 4, Σ 5,... Assume NTIME[n] DTISP[n c,n o(1) ] and lemmas (Here we drop the TIME from Σ k TIME for tidiness) Σ 4 [n] Σ 3 [n c4 6 ] Σ 2 [n c4 6 c2 2 ] NTIME[n c4 6 c2 2 c ], but 27-a

74 Σ 4, Σ 5,... Assume NTIME[n] DTISP[n c,n o(1) ] and lemmas (Here we drop the TIME from Σ k TIME for tidiness) Σ 4 [n] Σ 3 [n c4 6 ] Σ 2 [n c4 6 c2 2 ] NTIME[n c4 6 c2 2 c ], but NTIME[n c4 6 c2 2 c ] DTISP[n c4 6 c2 2 c2,n o(1) ] Π 4 [n c8 /48 ] (c < implies contradiction) 27-b

75 Σ 4, Σ 5,... Assume NTIME[n] DTISP[n c,n o(1) ] and lemmas (Here we drop the TIME from Σ k TIME for tidiness) Σ 4 [n] Σ 3 [n c4 6 ] Σ 2 [n c4 6 c2 2 ] NTIME[n c4 6 c2 2 c ], but NTIME[n c4 6 c2 2 c ] DTISP[n c4 6 c2 2 c2,n o(1) ] Π 4 [n c8 /48 ] (c < implies contradiction) Σ 5 [n] Σ 4 [n c8 48] Σ 3 [n c ] Σ 2 [n c ], and this is in 27-c

76 Σ 4, Σ 5,... Assume NTIME[n] DTISP[n c,n o(1) ] and lemmas (Here we drop the TIME from Σ k TIME for tidiness) Σ 4 [n] Σ 3 [n c4 6 ] Σ 2 [n c4 6 c2 2 ] NTIME[n c4 6 c2 2 c ], but NTIME[n c4 6 c2 2 c ] DTISP[n c4 6 c2 2 c2,n o(1) ] Π 4 [n c8 /48 ] (c < implies contradiction) Σ 5 [n] Σ 4 [n c8 48] Σ 3 [n c ] Σ 2 [n c ], and this is in NTIME[n c ] DTISP[n c ,n o(1) ] Π 5 [n c ] (c < implies contradiction) 27-d

77 An intermediate lower bound, n Υ Assume NTIME[n] DTISP[n c,n o(1) ] Claim: The inductive process of the previous slide converges. The constant derived is Υ := lim k f(k), where f(k) := k 1 j=1 (1 + 1/j)1/2j. Note Υ

78 A Time-Space Tradeoff Corollary: For every c < 1.66 there is d > 0 such that SAT is not in DTISP[n c,n d ]. 29

79 Outline Preliminaries A Speed-Up Theorem A Slow-Down Lemma Lipton and Viglas n 2 Lower Bound Our Inductive Argument From n 1.66 to n

80 From n 1.66 to n DTISP[t,t o(1) ] Π k TISP[t 1/k+o(1) ] is an unconditional result 31

81 From n 1.66 to n DTISP[t,t o(1) ] Π k TISP[t 1/k+o(1) ] is an unconditional result All other derived class inclusions in the above proof actually depend on the assumption that NTIME[n] DTISP[n c,n o(1) ]. 31-a

82 From n 1.66 to n DTISP[t,t o(1) ] Π k TISP[t 1/k+o(1) ] is an unconditional result All other derived class inclusions in the above proof actually depend on the assumption that NTIME[n] DTISP[n c,n o(1) ]. We ll now show how such an assumption can get DTISP[n c,n o(1) ] Π k TISP[n c/(k+ε)+o(1) ] for some ε > 0. This will push the lower bound higher. 31-b

83 From n 1.66 to n DTISP[t,t o(1) ] Π k TISP[t 1/k+o(1) ] is an unconditional result All other derived class inclusions in the above proof actually depend on the assumption that NTIME[n] DTISP[n c,n o(1) ]. We ll now show how such an assumption can get DTISP[n c,n o(1) ] Π k TISP[n c/(k+ε)+o(1) ] for some ε > 0. This will push the lower bound higher. Lemma: Let c 2. Define d 1 := 2, d k := 1 + d k 1 c If NTIME[n 2/c ] DTISP[n 2,n o(1) ], then for all k, DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ].. 31-c

84 From n 1.66 to n DTISP[t,t o(1) ] Π k TISP[t 1/k+o(1) ] is an unconditional result All other derived class inclusions in the above proof actually depend on the assumption that NTIME[n] DTISP[n c,n o(1) ]. We ll now show how such an assumption can get DTISP[n c,n o(1) ] Π k TISP[n c/(k+ε)+o(1) ] for some ε > 0. This will push the lower bound higher. Lemma: Let c 2. Define d 1 := 2, d k := 1 + d k 1 c If NTIME[n 2/c ] DTISP[n 2,n o(1) ], then for all k, DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ]. For c < 2, {d k } is increasing for each k, a bit more of DTISP[n O(1),n o(1) ] is shown to be contained in Π 2 TIME[n 1+o(1) ]. 31-d

85 Proof of Lemma Lemma: Let c < 2. Define d 1 := 2, d k := 1 + d k 1 c If NTIME[n 2/c ] DTISP[n 2,n o(1) ], then for all k N, DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ].. Induction on k. k = 1 case is trivial (speedup theorem). Suppose NTIME[n 2/c ] DTISP[n 2,n o(1) ] and DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ]. 32

86 Proof of Lemma Lemma: Let c < 2. Define d 1 := 2, d k := 1 + d k 1 c If NTIME[n 2/c ] DTISP[n 2,n o(1) ], then for all k N, DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ].. Induction on k. k = 1 case is trivial (speedup theorem). Suppose NTIME[n 2/c ] DTISP[n 2,n o(1) ] and DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ]. Want: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ]. 32-a

87 Proof of Lemma Lemma: Let c < 2. Define d 1 := 2, d k := 1 + d k 1 c If NTIME[n 2/c ] DTISP[n 2,n o(1) ], then for all k N, DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ].. Induction on k. k = 1 case is trivial (speedup theorem). Suppose NTIME[n 2/c ] DTISP[n 2,n o(1) ] and DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ]. Want: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ]. By padding, the purple assumptions imply NTIME[n d k/c ] DTISP[n d k,n o(1) ] Π 2 TIME[n 1+o(1) ]. ( ) 32-b

88 Goal: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ] Consider a Π 2 simulation of DTISP[n 1+d k/c,n o(1) ] with only O(n) bits (n 1 o(1) configurations) in the universal quantifier: 33

89 Goal: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ] Consider a Π 2 simulation of DTISP[n 1+d k/c,n o(1) ] with only O(n) bits (n 1 o(1) configurations) in the universal quantifier: ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( i {1,...,n 1 o(1) 1})[C i does not lead to C i+1 in n d k/c+o(1) time] 33-a

90 Goal: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ] Consider a Π 2 simulation of DTISP[n 1+d k/c,n o(1) ] with only O(n) bits (n 1 o(1) configurations) in the universal quantifier: ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( i {1,...,n 1 o(1) 1})[C i does not lead to C i+1 in n d k/c+o(1) time] Green part is an NTIME computation, input of length O(n), takes n d k/c+o(1) time 33-b

91 Goal: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ] Consider a Π 2 simulation of DTISP[n 1+d k/c,n o(1) ] with only O(n) bits (n 1 o(1) configurations) in the universal quantifier: ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( i {1,...,n 1 o(1) 1})[C i does not lead to C i+1 in n d k/c+o(1) time] Green part is an NTIME computation, input of length O(n), takes n d k/c+o(1) time ( ) = Green can be replaced with Π 2 TIME[n 1+o(1) ] computation, i.e. 33-c

92 Goal: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ] Consider a Π 2 simulation of DTISP[n 1+d k/c,n o(1) ] with only O(n) bits (n 1 o(1) configurations) in the universal quantifier: ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( i {1,...,n 1 o(1) 1})[C i does not lead to C i+1 in n d k/c+o(1) time] Green part is an NTIME computation, input of length O(n), takes n d k/c+o(1) time ( ) = Green can be replaced with Π 2 TIME[n 1+o(1) ] computation, i.e. ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( y, y = c x 1+o(1) ) ( z, z = c z 1+o(1) )[R(C 1,...,C n 1 o(1),x,y,z)], for some deterministic linear time relation R and constant c > d

93 Goal: DTISP[n 1+d k/c,n o(1) ] Π 2 TIME[n 1+o(1) ] Consider a Π 2 simulation of DTISP[n 1+d k/c,n o(1) ] with only O(n) bits (n 1 o(1) configurations) in the universal quantifier: ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( i {1,...,n 1 o(1) 1})[C i does not lead to C i+1 in n d k/c+o(1) time] Green part is an NTIME computation, input of length O(n), takes n d k/c+o(1) time ( ) = Green can be replaced with Π 2 TIME[n 1+o(1) ] computation, i.e. ( configurations C 1,...,C n 1 o(1) of M on x s.t. C n 1 o(1) is rejecting) ( y, y = c x 1+o(1) ) ( z, z = c z 1+o(1) )[R(C 1,...,C n 1 o(1),x,y,z)], for some deterministic linear time relation R and constant c > 0. Therefore, DTISP[n d k+1,n o(1) ] Π 2 TIME[n 1+o(1) ]. 33-e

94 New Lemma Gives Better Bound Corollary 1 Let c (1, 2). If NTIME[n 2/c ] DTISP[n 2,n o(1) ] then c for all ε > 0 such that ε c 1 1, DTISP[n c c 1 ε,n o(1) ] Π 2 TIME[n 1+o(1) ]. 34

95 New Lemma Gives Better Bound Corollary 1 Let c (1, 2). If NTIME[n 2/c ] DTISP[n 2,n o(1) ] then c for all ε > 0 such that ε c 1 1, DTISP[n c c 1 ε,n o(1) ] Π 2 TIME[n 1+o(1) ]. Proof. Recall d 2 = 2, d k = 1 + d k 1 /c. {d k } is monotone non-decreasing for c < 2; converges to d = 1 + d c = d = c/(c 1). (Note c = 2 implies d = 2) It follows that for all ε, there s a K such that d K c c 1 ε. 34-a

96 Now: Apply inductive method from n 1.66 lower bound the base case now resembles Fortnow-Van Melkebeek s n φ lower bound If NTIME[n] DTISP[n c,n o(1) ], Corollary 1 implies ( Σ 2 TIME[n] DTISP[n c2,n o(1) ] DTISP[ n c c2 c 1 Π 2 TIME[n c (c 1)+o(1) ]. φ(φ 1) = 1 ) c/(c 1)+o(1),n o(1) ] 35

97 Now: Apply inductive method from n 1.66 lower bound the base case now resembles Fortnow-Van Melkebeek s n φ lower bound If NTIME[n] DTISP[n c,n o(1) ], Corollary 1 implies ( Σ 2 TIME[n] DTISP[n c2,n o(1) ] DTISP[ n c Inducting as before, we get c2 c 1 Π 2 TIME[n c (c 1)+o(1) ]. φ(φ 1) = 1 ) c/(c 1)+o(1),n o(1) ] 35-a

98 Now: Apply inductive method from n 1.66 lower bound the base case now resembles Fortnow-Van Melkebeek s n φ lower bound If NTIME[n] DTISP[n c,n o(1) ], Corollary 1 implies ( Σ 2 TIME[n] DTISP[n c2,n o(1) ] DTISP[ n c Inducting as before, we get c2 c 1 Π 2 TIME[n c (c 1)+o(1) ]. φ(φ 1) = 1 ) c/(c 1)+o(1),n o(1) ] Σ 3 [n] Σ 2 [n c (c 1) ] DTISP[n c3 (c 1),n o(1) ] Π 3 [n c3 (c 1) 3 ], then 35-b

99 Now: Apply inductive method from n 1.66 lower bound the base case now resembles Fortnow-Van Melkebeek s n φ lower bound If NTIME[n] DTISP[n c,n o(1) ], Corollary 1 implies ( Σ 2 TIME[n] DTISP[n c2,n o(1) ] DTISP[ n c Inducting as before, we get c2 c 1 Π 2 TIME[n c (c 1)+o(1) ]. φ(φ 1) = 1 ) c/(c 1)+o(1),n o(1) ] Σ 3 [n] Σ 2 [n c (c 1) ] DTISP[n c3 (c 1),n o(1) ] Π 3 [n c3 (c 1) 3 ], then Σ 4 [n] Σ 3 [n c3 (c 1) 3 ] Σ 2 [n c4 (c 1) 2 3 ] DTISP[n c6 (c 1) 2 3,n o(1) ] Π 4 [n c6 (c 1) 2 12 ], etc. 35-c

100 Claim: The exponent e k derived for Σ k TIME[n] Π k TIME[n e k ] is e k = c 3 2k 3 (c 1) 2k 3 k (3 2k 4 4 2k 5 5 2k 6 (k 1)). 36

101 Simplifying, e k = ( c 3 2k 3 (c 1) 2k 3 = k (3 2k 4 4 2k 5 5 2k 6 (k 1)) thus e k < 1 Finishing up c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) < 1 ) 2 k 3 37

102 Simplifying, e k = ( c 3 2k 3 (c 1) 2k 3 = k (3 2k 4 4 2k 5 5 2k 6 (k 1)) thus e k < 1 Finishing up c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) < 1 Define f(k) = k 2 k+3 ( (k 1) 2 k+3 ) ) 2 k 3 37-a

103 Simplifying, e k = ( c 3 2k 3 (c 1) 2k 3 = k (3 2k 4 4 2k 5 5 2k 6 (k 1)) thus e k < 1 Finishing up c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) < 1 Define f(k) = k 2 k+3 ( (k 1) 2 k+3 ) f(k) as k ) 2 k 3 37-b

104 Simplifying, e k = ( c 3 2k 3 (c 1) 2k 3 = k (3 2k 4 4 2k 5 5 2k 6 (k 1)) thus e k < 1 Finishing up c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) < 1 Define f(k) = k 2 k+3 ( (k 1) 2 k+3 ) f(k) as k Above task reduces to finding positive root of c 3 (c 1) = ) 2 k 3 37-c

105 Simplifying, e k = ( c 3 2k 3 (c 1) 2k 3 = k (3 2k 4 4 2k 5 5 2k 6 (k 1)) thus e k < 1 Finishing up c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) c 3 (c 1) k 2 k+3 ( (k 1) 2 k+3 ) < 1 Define f(k) = k 2 k+3 ( (k 1) 2 k+3 ) f(k) as k Above task reduces to finding positive root of c 3 (c 1) = ) 2 k 3 = c > yields a contradiction. 37-d

106 The above inductive method can be applied to improve several existing lower bound arguments. Time lower bounds for SAT on off-line one-tape machines Time-space tradeoffs for nondeterminism/co-nondeterminism in RAM model Etc. See the paper! 38