Better Time-Space Lower Bounds for SAT and Related Problems Ryan - PowerPoint PPT Presentation

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

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 j th step: bit string encoding head positions, workspace contents, finite control By space assumption on M , | C j | ∈ n o (1) 11-a

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 j th 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

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 steps, C 0 is initial, C t is accepting] √ √ t in 12

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 steps, C 0 is initial, C t is accepting] √ √ t in Runtime on an alternating machine: √ t · s ) = t 1 / 2+ o (1) time to write down the C j ’s • ∃ takes O ( • ∀ takes O (log t ) time to write down i √ • [ · · · ] takes O ( t · s ) deterministic time to check Two alternations, square root speedup 12-a

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 steps, C 0 initial, C t accepting] t in 13

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 steps, C 0 initial, C t accepting] t in √ 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

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 steps, C 0 initial, C t accepting] t in √ 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 t 1 / 3 steps, C 0 initial, C t accepting] 13-b

k = 2 has one “stage” C 0 C t 1/2 C t 1/2 2 . . . . C t

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

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

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

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 steps] t in t + “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

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 t 1 / 3 steps] to: 16

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 t 1 / 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 ′ ( i +1) · t 2 / 3 � = C ( 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 t 1 / 3 steps] 16-a

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 t 1 / 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 ′ ( i +1) · t 2 / 3 � = C ( 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 t 1 / 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

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

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

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

Outline • Preliminaries • A Speed-Up Theorem • A Slow-Down Lemma √ 2 Lower Bound • Lipton and Viglas’ n • Our Inductive Argument • From n 1 . 66 to n 1 . 732 18

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

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

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

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

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

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

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

A Slow-Down Lemma: Proof Assume Σ k TIME [ n ] ⊆ Π k TIME [ n c ] R ( x, x 1 ) := ( ∀ x 2 , | x 2 | ≤ t ( | x | )) · · · Define ( 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

A Slow-Down Lemma: Proof Assume Σ k TIME [ n ] ⊆ Π k TIME [ n c ] R ( x, x 1 ) := ( ∀ x 2 , | x 2 | ≤ t ( | x | )) · · · Define ( 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

A Slow-Down Lemma: Proof Assume Σ k TIME [ n ] ⊆ Π k TIME [ n c ] R ( x, x 1 ) := ( ∀ x 2 , | x 2 | ≤ t ( | x | )) · · · Define ( 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

A Slow-Down Lemma: Proof Assume Σ k TIME [ n ] ⊆ Π k TIME [ n c ] R ( x, x 1 ) := ( ∀ x 2 , | x 2 | ≤ t ( | x | )) · · · Define ( 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 − Σ k TIME [ t c ] M ( x ) accepts iff [( ∃ x 1 , | x 1 | ≤ t ( | x | )) R ′ ( x, x 1 )] ← 21-d

Outline • Preliminaries • A Speed-Up Theorem • A Slow-Down Lemma √ 2 Lower Bound • Lipton and Viglas’ n • Our Inductive Argument • From n 1 . 66 to n 1 . 732 22

√ 2 Lower Bound (Rephrased) Lipton and Viglas’ n 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) ] √ 2 − ε , n o (1) ] Theorem: NTIME [ n ] � DTISP [ n 23

√ 2 Lower Bound (Rephrased) Lipton and Viglas’ n 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) ] √ 2 − ε , n o (1) ] Theorem: NTIME [ n ] � DTISP [ n Proof: Assume NTIME [ n ] ⊆ DTISP [ n c , n o (1) ] (We will find a c that implies a contradiction) 23-a

√ 2 Lower Bound (Rephrased) Lipton and Viglas’ n 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) ] √ 2 − ε , n o (1) ] Theorem: NTIME [ n ] � DTISP [ n 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

√ 2 Lower Bound (Rephrased) Lipton and Viglas’ n 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) ] √ 2 − ε , n o (1) ] Theorem: NTIME [ n ] � DTISP [ n 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 c 2 , n o (1) ] , by assumption and padding 23-c

√ 2 Lower Bound (Rephrased) Lipton and Viglas’ n 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) ] √ 2 − ε , n o (1) ] Theorem: NTIME [ n ] � DTISP [ n 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 c 2 , n o (1) ] , by assumption and padding • DTISP [ n c 2 , n o (1) ] ⊆ Π 2 TIME [ n c 2 / 2 ] , by speed-up theorem, so √ c < 2 contradicts the hierarchy theorem � 23-d

Outline • Preliminaries • A Speed-Up Theorem • A Slow-Down Lemma √ 2 Lower Bound • Lipton and Viglas’ n • Our Inductive Argument • From n 1 . 66 to n 1 . 732 24

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 c 2 / 2 ] . Note if c < 2 then c 2 / 2 < c . 25

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 c 2 / 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 c 2 / 2 cost 25-a

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 c 2 / 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 c 2 / 2 cost • Slow-down theorem implies Σ 3 TIME [ n ] ⊆ Σ 2 TIME [ n c 2 / 2 ] 25-b

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

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 c 2 / 2 ] , by slow-down and lemma 26-a

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 c 2 / 2 ] , by slow-down and lemma • Σ 2 TIME [ n c 2 / 2 ] ⊆ NTIME [ n c 3 / 2 ] , by slow-down 26-b

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 c 2 / 2 ] , by slow-down and lemma • Σ 2 TIME [ n c 2 / 2 ] ⊆ NTIME [ n c 3 / 2 ] , by slow-down • NTIME [ n c 3 / 2 ] ⊆ DTISP [ n c 4 / 2 , n o (1) ] , by assumption and padding 26-c

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 c 2 / 2 ] , by slow-down and lemma • Σ 2 TIME [ n c 2 / 2 ] ⊆ NTIME [ n c 3 / 2 ] , by slow-down • NTIME [ n c 3 / 2 ] ⊆ DTISP [ n c 4 / 2 , n o (1) ] , by assumption and padding • DTISP [ n c 4 / 2 , n o (1) ] ⊆ Π 3 TIME [ n c 4 / 6 ] , by speed-up 26-d

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 c 2 / 2 ] , by slow-down and lemma • Σ 2 TIME [ n c 2 / 2 ] ⊆ NTIME [ n c 3 / 2 ] , by slow-down • NTIME [ n c 3 / 2 ] ⊆ DTISP [ n c 4 / 2 , n o (1) ] , by assumption and padding • DTISP [ n c 4 / 2 , n o (1) ] ⊆ Π 3 TIME [ n c 4 / 6 ] , by speed-up Observe: √ 4 • Now c < 6 ≈ 1 . 565 contradicts time hierarchy for Σ 3 and Π 3 √ 4 • But if c ≥ 6 , then we obtain a new “lemma”: Σ 3 TIME [ n ] ⊆ Π 3 TIME [ n c 4 / 6 ] 26-e

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

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) c 4 c 4 6 · c 2 c 4 6 · c 2 6 ] ⊆ Σ 2 [ n 2 ] ⊆ NTIME [ n 2 · c ] , but Σ 4 [ n ] ⊆ Σ 3 [ n 27-a

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) c 4 c 4 6 · c 2 c 4 6 · c 2 6 ] ⊆ Σ 2 [ n 2 ] ⊆ NTIME [ n 2 · c ] , but Σ 4 [ n ] ⊆ Σ 3 [ n c 4 6 · c 2 c 4 6 · c 2 2 · c 2 , n o (1) ] ⊆ Π 4 [ n c 8 / 48 ] 2 · c ] ⊆ DTISP [ n NTIME [ n √ 8 ( c < 48 ≈ 1 . 622 implies contradiction) 27-b

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) c 4 c 4 6 · c 2 c 4 6 · c 2 6 ] ⊆ Σ 2 [ n 2 ] ⊆ NTIME [ n 2 · c ] , but Σ 4 [ n ] ⊆ Σ 3 [ n c 4 6 · c 2 c 4 6 · c 2 2 · c 2 , n o (1) ] ⊆ Π 4 [ n c 8 / 48 ] 2 · c ] ⊆ DTISP [ n NTIME [ n √ 8 ( c < 48 ≈ 1 . 622 implies contradiction) c 8 c 12 c 14 Σ 5 [ n ] ⊆ Σ 4 [ n 48 ] ⊆ Σ 3 [ n 48 · 6 ] ⊆ Σ 2 [ n 48 · 6 · 2 ] , and this is in 27-c

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) c 4 c 4 6 · c 2 c 4 6 · c 2 6 ] ⊆ Σ 2 [ n 2 ] ⊆ NTIME [ n 2 · c ] , but Σ 4 [ n ] ⊆ Σ 3 [ n c 4 6 · c 2 c 4 6 · c 2 2 · c 2 , n o (1) ] ⊆ Π 4 [ n c 8 / 48 ] 2 · c ] ⊆ DTISP [ n NTIME [ n √ 8 ( c < 48 ≈ 1 . 622 implies contradiction) c 8 c 12 c 14 Σ 5 [ n ] ⊆ Σ 4 [ n 48 ] ⊆ Σ 3 [ n 48 · 6 ] ⊆ Σ 2 [ n 48 · 6 · 2 ] , and this is in c 15 c 16 c 16 48 · 12 , n o (1) ] ⊆ Π 5 [ n NTIME [ n 48 · 12 ] ⊆ DTISP [ n 48 · 60 ] √ 16 ( c < 2880 ≈ 1 . 645 implies contradiction) 27-d

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 / 2 j . Note Υ ′ ≈ 1 . 66 . 28

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

Outline • Preliminaries • A Speed-Up Theorem • A Slow-Down Lemma √ 2 Lower Bound • Lipton and Viglas’ n • Our Inductive Argument • From n 1 . 66 to n 1 . 732 30

From n 1 . 66 to n 1 . 732 DTISP [ t, t o (1) ] ⊆ Π k TISP [ t 1 /k + o (1) ] is an unconditional result 31

From n 1 . 66 to n 1 . 732 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

From n 1 . 66 to n 1 . 732 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

From n 1 . 66 to n 1 . 732 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

From n 1 . 66 to n 1 . 732 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

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

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

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

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

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

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

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

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 . 33-d

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

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 c − 1 − ε ≥ 1 , c c − 1 − ε , n o (1) ] ⊆ Π 2 TIME [ n 1+ o (1) ] . DTISP [ n 34

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 c − 1 − ε ≥ 1 , c c − 1 − ε , n o (1) ] ⊆ Π 2 TIME [ n 1+ o (1) ] . DTISP [ n 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 ) c It follows that for all ε , there’s a K such that d K ≥ c − 1 − ε . � 34-a

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 � c/ ( c − 1)+ o (1) � n c 2 · c − 1 Σ 2 TIME [ n ] ⊆ DTISP [ n c 2 , n o (1) ] ⊆ DTISP [ , n o (1) ] c ⊆ Π 2 TIME [ n c · ( c − 1)+ o (1) ] . φ ( φ − 1) = 1 35

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 � c/ ( c − 1)+ o (1) � n c 2 · c − 1 Σ 2 TIME [ n ] ⊆ DTISP [ n c 2 , n o (1) ] ⊆ DTISP [ , n o (1) ] c ⊆ Π 2 TIME [ n c · ( c − 1)+ o (1) ] . φ ( φ − 1) = 1 Inducting as before, we get 35-a

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 � c/ ( c − 1)+ o (1) � n c 2 · c − 1 Σ 2 TIME [ n ] ⊆ DTISP [ n c 2 , n o (1) ] ⊆ DTISP [ , n o (1) ] c ⊆ Π 2 TIME [ n c · ( c − 1)+ o (1) ] . φ ( φ − 1) = 1 Inducting as before, we get c 3 · ( c − 1) Σ 3 [ n ] ⊆ Σ 2 [ n c · ( c − 1) ] ⊆ DTISP [ n c 3 · ( c − 1) , n o (1) ] ⊆ Π 3 [ n ] , then 3 35-b

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 � c/ ( c − 1)+ o (1) � n c 2 · c − 1 Σ 2 TIME [ n ] ⊆ DTISP [ n c 2 , n o (1) ] ⊆ DTISP [ , n o (1) ] c ⊆ Π 2 TIME [ n c · ( c − 1)+ o (1) ] . φ ( φ − 1) = 1 Inducting as before, we get c 3 · ( c − 1) Σ 3 [ n ] ⊆ Σ 2 [ n c · ( c − 1) ] ⊆ DTISP [ n c 3 · ( c − 1) , n o (1) ] ⊆ Π 3 [ n ] , then 3 c 3 · ( c − 1) c 4 · ( c − 1)2 c 6 · ( c − 1)2 , n o (1) ] Σ 4 [ n ] ⊆ Σ 3 [ n ] ⊆ Σ 2 [ n ] ⊆ DTISP [ n 3 3 3 c 6 · ( c − 1)2 ⊆ Π 4 [ n ] , etc. 12 35-c

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