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Computational Complexity
Lecture 4 in which Diagonalization takes on itself, and we enter Space Complexity (But first Ladner’ s Theorem)
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Ladner’ s Theorem
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Computational Complexity Lecture 4 in which Diagonalization takes - - PowerPoint PPT Presentation
Computational Complexity Lecture 4 in which Diagonalization takes - - PowerPoint PPT Presentation
Computational Complexity Lecture 4 in which Diagonalization takes on itself, and we enter Space Complexity (But first Ladner s Theorem) 1 Ladner s Theorem 2 Ladner s Theorem If P ! NP, then are all non-P NP languages equally
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Ladner’ s Theorem
If P!NP, then are all non-P NP languages equally hard? (Are all NP-complete?)
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Ladner’ s Theorem
If P!NP, then are all non-P NP languages equally hard? (Are all NP-complete?) No!
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Ladner’ s Theorem
If P!NP, then are all non-P NP languages equally hard? (Are all NP-complete?) No! Can show an NP language which is neither in P, nor NP complete (unless P = NP)
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Ladner’ s Theorem: Proof
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)}
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP.
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP.
Padding maps problem to a lower complexity class
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP. If SATH in P and H(|x|) bounded by const. then SAT in P!
Padding maps problem to a lower complexity class
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP. If SATH in P and H(|x|) bounded by const. then SAT in P! |pad| < |x|i* implies SAT !p SATH
Padding maps problem to a lower complexity class
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP. If SATH in P and H(|x|) bounded by const. then SAT in P! |pad| < |x|i* implies SAT !p SATH If SATH is NPC (⇒ SATH not in P) and H(|x|) goes to infinity, then SAT in P!
Padding maps problem to a lower complexity class
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP. If SATH in P and H(|x|) bounded by const. then SAT in P! |pad| < |x|i* implies SAT !p SATH If SATH is NPC (⇒ SATH not in P) and H(|x|) goes to infinity, then SAT in P! Suppose f(x) = (x’,pad), |(x’,pad)| ! c|x|c. If |x’|>|x|/2, then |pad| = |x’|H(|x’|) > c|x|c (for long enough x). So |x’| is at most |x|/2. Repeat to solve SAT
Padding maps problem to a lower complexity class
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Ladner’ s Theorem: Proof
SATH = { (x,pad) | x ∈ SAT and |pad|=|x|H(|x|)} H(|x|) will be computable in poly(|x|) time. SATH in NP. If SATH in P and H(|x|) bounded by const. then SAT in P! |pad| < |x|i* implies SAT !p SATH If SATH is NPC (⇒ SATH not in P) and H(|x|) goes to infinity, then SAT in P! Suppose f(x) = (x’,pad), |(x’,pad)| ! c|x|c. If |x’|>|x|/2, then |pad| = |x’|H(|x’|) > c|x|c (for long enough x). So |x’| is at most |x|/2. Repeat to solve SAT To define H s.t. H(n) bounded by const. iff SATH in P
Padding maps problem to a lower complexity class
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Proof (ctd.)
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Proof (ctd.)
Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Proof (ctd.)
Mi|Ti be Mi restricted to Ti Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put H(n) be least i < log log n s.t. Mi|Ti correct for all |z|<log n Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put H(n) be least i < log log n s.t. Mi|Ti correct for all |z|<log n
log log n log n
Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put H(n) be least i < log log n s.t. Mi|Ti correct for all |z|<log n
log log n log n
Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put H(n) be least i < log log n s.t. Mi|Ti correct for all |z|<log n H is poly-time computable
log log n log n
Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put H(n) be least i < log log n s.t. Mi|Ti correct for all |z|<log n H is poly-time computable SATH in P iff H(n) < i*
log log n log n
Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Mi|Ti |z|
Proof (ctd.)
Mi|Ti be Mi restricted to Ti Put at (i,t) if Mi|Ti agrees with SATH on all z, |z|=t; else put H(n) be least i < log log n s.t. Mi|Ti correct for all |z|<log n H is poly-time computable SATH in P iff H(n) < i* Both equivalent to having a row of all
log log n log n
Mi be ith TM. Ti be ith polynomial (i.e., Ti(t)=i.ti)
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Meta-Questions
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Meta-Questions
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Meta-Questions
“Real” Questions
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Meta-Questions
“Real” Questions “Meta” Questions
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? “Meta” Questions
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? Is my problem NP-complete? “Meta” Questions
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? Is my problem NP-complete? Results non-specialists would care about “Meta” Questions
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? Is my problem NP-complete? Results non-specialists would care about “Meta” Questions What can we do with an
- racle for SAT?
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? Is my problem NP-complete? Results non-specialists would care about “Meta” Questions What can we do with an
- racle for SAT?
Will this proof technique work?
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? Is my problem NP-complete? Results non-specialists would care about “Meta” Questions What can we do with an
- racle for SAT?
Will this proof technique work? Tools & Techniques, intermediate results
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Meta-Questions
“Real” Questions SAT in DTIME(n2)? Is my problem NP-complete? Results non-specialists would care about “Meta” Questions What can we do with an
- racle for SAT?
Will this proof technique work? Tools & Techniques, intermediate results Under-the-hood stuff
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Oracles
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Oracles
What if we had an oracle for language A
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if L decided by a TM MA, in poly time
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if L decided by a TM MA, in poly time Turing reduction: L !T A
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if L decided by a TM MA, in poly time Turing reduction: L !T A Class NPA: L ∈ NPA if
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if L decided by a TM MA, in poly time Turing reduction: L !T A Class NPA: L ∈ NPA if L decided by an NTM MA, in poly time
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if L decided by a TM MA, in poly time Turing reduction: L !T A Class NPA: L ∈ NPA if L decided by an NTM MA, in poly time Equivalently, L = {x| ∃w, |w| < poly(|x|) s.t. (x,w) ∈ L ’ }, where L ’ is in PA
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Oracles
What if we had an oracle for language A Class PA: L ∈ PA if L decided by a TM MA, in poly time Turing reduction: L !T A Class NPA: L ∈ NPA if L decided by an NTM MA, in poly time Equivalently, L = {x| ∃w, |w| < poly(|x|) s.t. (x,w) ∈ L ’ }, where L ’ is in PA Equivalence carries over!
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Proofs that Relativize
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Proofs that Relativize
Often entire theorems/proofs carry over, with the
- racle tagging along
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Proofs that Relativize
Often entire theorems/proofs carry over, with the
- racle tagging along
e.g. Time hierarchy theorems (and proofs!) hold for machines with access to any given oracle A
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Proofs that Relativize
Often entire theorems/proofs carry over, with the
- racle tagging along
e.g. Time hierarchy theorems (and proofs!) hold for machines with access to any given oracle A Said to “relativize”
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P vs. NP with oracles
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P vs. NP with oracles
How does P vs. NP fare relative to different oracles?
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P vs. NP with oracles
How does P vs. NP fare relative to different oracles? Does their relation (equality or not) relativize?
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P vs. NP with oracles
How does P vs. NP fare relative to different oracles? Does their relation (equality or not) relativize? No! Different in different worlds!
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P vs. NP with oracles
How does P vs. NP fare relative to different oracles? Does their relation (equality or not) relativize? No! Different in different worlds! There exist languages A, B such that PA = NPA, but PB ! NPB!
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A s.t. PA = NPA
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A s.t. PA = NPA
If A is EXP-complete (w.r.t !Cook or !P), PA = NPA = EXP
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A s.t. PA = NPA
If A is EXP-complete (w.r.t !Cook or !P), PA = NPA = EXP A EXP-hard ⇒ EXP ⊆ PA ⊆ NPA
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A s.t. PA = NPA
If A is EXP-complete (w.r.t !Cook or !P), PA = NPA = EXP A EXP-hard ⇒ EXP ⊆ PA ⊆ NPA A in EXP ⇒ NPA ⊆ EXP (note: to decide a language in NPA can try all possible witnesses, and carry out PA computation in exponential time)
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A s.t. PA = NPA
If A is EXP-complete (w.r.t !Cook or !P), PA = NPA = EXP A EXP-hard ⇒ EXP ⊆ PA ⊆ NPA A in EXP ⇒ NPA ⊆ EXP (note: to decide a language in NPA can try all possible witnesses, and carry out PA computation in exponential time) A simple EXP-complete language:
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A s.t. PA = NPA
If A is EXP-complete (w.r.t !Cook or !P), PA = NPA = EXP A EXP-hard ⇒ EXP ⊆ PA ⊆ NPA A in EXP ⇒ NPA ⊆ EXP (note: to decide a language in NPA can try all possible witnesses, and carry out PA computation in exponential time) A simple EXP-complete language: EXPTM = { (M,x,1n) | TM represented by M accepts x within time 2n }
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B s.t. PB ! NPB
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B s.t. PB ! NPB
Building B and L, s.t. L in NPB\PB
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B}
Building B and L, s.t. L in NPB\PB
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} B
Building B and L, s.t. L in NPB\PB
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B}
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B}
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n)
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. When Mi queries B on x > 1n-1, set B(X)=0 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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B s.t. PB ! NPB
L={1n| ∃w, |w|=n and w∈B} L in NPB. To do: L not in PB For each i, ensure MiB in 2n-1 time gets L(1n) wrong (for some new n) Pick n s.t. B not yet set beyond 1n-1. Run Mi on 1n for 2n-1 steps. When Mi queries B on x > 1n-1, set B(X)=0 After Mi finished set B up to x=1n s.t. L(1n) ! MiB(1n) 1n Mi
1n 11 1
L B
Building B and L, s.t. L in NPB\PB
00 01 10
...
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Meta-Result of the Day
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Meta-Result of the Day
P vs. NP cannot be resolved using a relativizing proof
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Meta-Result of the Day
P vs. NP cannot be resolved using a relativizing proof “Diagonalization proofs” relativize
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Meta-Result of the Day
P vs. NP cannot be resolved using a relativizing proof “Diagonalization proofs” relativize Just need a way to enumerate/ encode machines, and to simulate one without much
- verhead given its encoding
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Meta-Result of the Day
P vs. NP cannot be resolved using a relativizing proof “Diagonalization proofs” relativize Just need a way to enumerate/ encode machines, and to simulate one without much
- verhead given its encoding
Do not further depend on internals of computation
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Meta-Result of the Day
P vs. NP cannot be resolved using a relativizing proof “Diagonalization proofs” relativize Just need a way to enumerate/ encode machines, and to simulate one without much
- verhead given its encoding
Do not further depend on internals of computation e.g. of non-relativizing proof: that of Cook-Levin theorem
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Space Complexity
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Space Complexity
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Space Complexity
Natural complexity question
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Space Complexity
Natural complexity question How much memory is needed
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Space Complexity
Natural complexity question How much memory is needed More pressing than time:
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Space Complexity
Natural complexity question How much memory is needed More pressing than time: Can’ t generate memory on the fly
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Space Complexity
Natural complexity question How much memory is needed More pressing than time: Can’ t generate memory on the fly Or maybe less pressing:
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Space Complexity
Natural complexity question How much memory is needed More pressing than time: Can’ t generate memory on the fly Or maybe less pressing: Turns out, often a little memory can go a long way (if we can spare the time)
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DSPACE and NSPACE
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DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape
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DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape Model allows o(n) memory usage
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DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape Model allows o(n) memory usage DSPACE(n) may already be inefficient in terms of time
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DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape Model allows o(n) memory usage DSPACE(n) may already be inefficient in terms of time We shall stick to "(log n)
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DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape Model allows o(n) memory usage DSPACE(n) may already be inefficient in terms of time We shall stick to "(log n) Less than log is too little space to remember locations in the input
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SLIDE 100
DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape Model allows o(n) memory usage DSPACE(n) may already be inefficient in terms of time We shall stick to "(log n) Less than log is too little space to remember locations in the input DSPACE/NSPACE more robust across models
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SLIDE 101
DSPACE and NSPACE
Measure of working memory (work-tape) used by a TM/NTM: input kept in a read-only tape Model allows o(n) memory usage DSPACE(n) may already be inefficient in terms of time We shall stick to "(log n) Less than log is too little space to remember locations in the input DSPACE/NSPACE more robust across models Constant factor (+O(log n)) simulation overhead
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L ∈ NSPACE(S): Two Equivalent views
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space Deterministic M’
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space Deterministic M’ input: x and read-once w
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space Deterministic M’ input: x and read-once w reads bits from w (certificate)
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space Deterministic M’ input: x and read-once w reads bits from w (certificate) x ∈ L iff for some cert. w, M’ accepts
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space Deterministic M’ input: x and read-once w reads bits from w (certificate) x ∈ L iff for some cert. w, M’ accepts in at most S(|x|) space
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L ∈ NSPACE(S): Two Equivalent views
Non-deterministic M input: x makes non-det choices x ∈ L iff some thread of M accepts in at most S(|x|) space Deterministic M’ input: x and read-once w reads bits from w (certificate) x ∈ L iff for some cert. w, M’ accepts in at most S(|x|) space
Equivalent
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L and NL
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L and NL
L = DSPACE(O(log n))
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SLIDE 116
L and NL
L = DSPACE(O(log n)) L = ∪a,b > 0 DSPACE(a.log n+b)
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SLIDE 117
L and NL
L = DSPACE(O(log n)) L = ∪a,b > 0 DSPACE(a.log n+b) NL = NSPACE(O(log n))
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SLIDE 118
L and NL
L = DSPACE(O(log n)) L = ∪a,b > 0 DSPACE(a.log n+b) NL = NSPACE(O(log n)) NL = ∪a,b > 0 NSPACE(a.log n+b)
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SLIDE 119
L and NL
L = DSPACE(O(log n)) L = ∪a,b > 0 DSPACE(a.log n+b) NL = NSPACE(O(log n)) NL = ∪a,b > 0 NSPACE(a.log n+b) “L and NL are to space what P and NP are to time”
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SLIDE 120
Space Hierarchy
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SLIDE 121
Space Hierarchy
UTM space-overhead is only a constant factor
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SLIDE 122
Space Hierarchy
UTM space-overhead is only a constant factor Tight hierarchy: if T(n) = o(T’(n)) (no log slack) then DSPACE(T(n)) ⊊ DSPACE(T’(n))
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SLIDE 123
Space Hierarchy
UTM space-overhead is only a constant factor Tight hierarchy: if T(n) = o(T’(n)) (no log slack) then DSPACE(T(n)) ⊊ DSPACE(T’(n)) Same for NSPACE
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SLIDE 124
Space Hierarchy
UTM space-overhead is only a constant factor Tight hierarchy: if T(n) = o(T’(n)) (no log slack) then DSPACE(T(n)) ⊊ DSPACE(T’(n)) Same for NSPACE Again, tighter than for NTIME (where in fact, we needed T(n+1) = o(T’(n) )
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SLIDE 125
Space Hierarchy
UTM space-overhead is only a constant factor Tight hierarchy: if T(n) = o(T’(n)) (no log slack) then DSPACE(T(n)) ⊊ DSPACE(T’(n)) Same for NSPACE Again, tighter than for NTIME (where in fact, we needed T(n+1) = o(T’(n) ) No “delayed flip,” because, as we will see later, NSPACE(O(S)) = co-NSPACE(O(S))!
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SLIDE 126
Space, Today
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SLIDE 127
Space, Today
DSPACE, NSPACE
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SLIDE 128
Space, Today
DSPACE, NSPACE Tight hierarchy.
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SLIDE 129
Space, Today
DSPACE, NSPACE Tight hierarchy. Coming up:
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SLIDE 130
Space, Today
DSPACE, NSPACE Tight hierarchy. Coming up: Connections with DTIME/NTIME
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SLIDE 131
Space, Today
DSPACE, NSPACE Tight hierarchy. Coming up: Connections with DTIME/NTIME Savitch’ s theorem: NSPACE(S) ⊆ DSPACE(S2)
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SLIDE 132
Space, Today
DSPACE, NSPACE Tight hierarchy. Coming up: Connections with DTIME/NTIME Savitch’ s theorem: NSPACE(S) ⊆ DSPACE(S2) Hence PSPACE = NPSPACE
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SLIDE 133
Space, Today
DSPACE, NSPACE Tight hierarchy. Coming up: Connections with DTIME/NTIME Savitch’ s theorem: NSPACE(S) ⊆ DSPACE(S2) Hence PSPACE = NPSPACE PSPACE-completeness and NL-completeness
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SLIDE 134
Space, Today
DSPACE, NSPACE Tight hierarchy. Coming up: Connections with DTIME/NTIME Savitch’ s theorem: NSPACE(S) ⊆ DSPACE(S2) Hence PSPACE = NPSPACE PSPACE-completeness and NL-completeness NSPACE = co-NSPACE
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SLIDE 135
Space, Today
P
PSPACE
EXP NP NEXP L NL
NPSPACE
DSPACE, NSPACE Tight hierarchy. Coming up: Connections with DTIME/NTIME Savitch’ s theorem: NSPACE(S) ⊆ DSPACE(S2) Hence PSPACE = NPSPACE PSPACE-completeness and NL-completeness NSPACE = co-NSPACE
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