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CS221: Computational Complexity Prof. Salil Vadhan
Lecture 6: Nondeterministic Time
Oct. 2, 2002 Scribe: Arthur Rudolph
Contents
1 Nondeterministic Time 1
2 Nondeterministic Time Hierarchy 2
3 Relationships between Resources 3
1 Nondeterministic Time
- NTIME(f (n)) leads to NP, NEXP
- NSPACE(f (n)) leads to NL, NPSPACE ←− next week
For a complexity class C: co-C = { L¯ : L ∈ C}( 6 = C¯)
It is clear that P = co-P, L = co-L, etc. (deterministic algorithms can flip their answer), but
NP =? co-NP, NL =? co-NL, NPSPACE =? co-NPSPACE are intriguing questions. Problems in co-NP:
Unsatisfiability = {ϕ : ¬∃a s.t. ϕ(¯a) = 1} Tautology = {ϕ : ∀¯aϕ(¯a) = 1}
If NP = co-NP, then every tautology has a short proof. (This observation has led to an entire area, called Proof Complexity, that studies the lengths of proofs.)
Proposition 1 If NP 6 = co-NP then P 6 = NP
1.1 What about NEXP?
Exp-Proof = {(ϕ, (^) ︸︷︷︸n binary
) : ϕ is math. thm. with a proof of length ≤ n} ∈ NEXP
NEXP is like NP but with succinct inputs (equivalently, with exponentially long certificates).
Proposition 2 EXP 6 = NEXP ⇒ P 6 = NP
We prove this using a simple but useful technique called padding.
Proof: We prove the contrapositive: P = NP ⇒ EXP = NEXP. Assume P = NP. Let L ∈ NEXP. Then it is decided by a NTM M running in time 2n c
. Define
Lpad = {(x, 12 |x|c ) : x ∈ L}
Claim 3 Lpad ∈ NP
Proof of claim: On input (x, 1 k), check k = 2|x|c , if not reject. Otherwise, simulate M on x and behave the same. §
By our assumption, we then have Lpad ∈ P, with poly-time alg A.
Claim 4 L ∈ EXP
Proof of claim: Given x, run A((x, 12 |x|c )) ←time poly (|x|, 2 |x|
c ) ≤ 2 |x| c′ §
More generally, with padding we can prove:
Proposition 5 If f (n), g(n) ≥ n proper, then
TIME(f (n)) = NTIME(f (n)) ⇒ TIME(f (g(n))) = NTIME(f (g(n)))
Padding is even more general:
- Applies to any pair of resources, bounds need not be same.
- “Equality (or inclusion) translates upwards, inequality (or separation) translates downwards.”
2 Nondeterministic Time Hierarchy
Such a thing exists even though we can’t flip output (don’t know if closed under complement).
Theorem 6 If f (n), g(n) ≥ n proper, g(n) = ω(f (n)) then NTIME(f (n)) ( NTIME(g(n))
Proof Sketch: For f (n) = n^2 , g(n) = n^3. The technique we’ll use is called “Lazy Diagonaliza- tion.” Divide N into huge intervals, define n 1 = 1 and ni+1 = 2n (^4) i for i > 1. Define Ii = [ni, ni+1]. Construct n^3 time NTM M , diagonalize against Mi somewhere in Ii.
M (x):
- If x 6 ∈ 1 ∗, reject.
- If x = 1n, say n ∈ Ii:
- If n < ni+1, simulate M i on 1n+1^ for (n + 1)^2 steps and do the same.
- If n = ni+1, compute whether Mi accepts or rejects 1ni^ (within n^2 i steps) and do opposite. Do this by enumerating over all 2O(n (^2) i ) ≤ O(ni+1) computations of length n^2 i.
