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CS221: Computational Complexity Prof. Salil Vadhan
Lecture 10: Alternation
10/11 Scribe: Ethan Abraham
Contents
1 Recap From Last Lecture 1
2 Current Classes 2
3 Alternation - Sections 16.2 & 19.1 2
1 Recap From Last Lecture
Recall the following results from last lecture:
- RCH ∈ P, so NL ⊆ P, since RCH is NL-complete.
- RCH ∈ L^2 by Savitch’s Theorem, thus NL ⊆ L^2 and PSPACE = NPSPACE
- RCH ∈ co-NL by Immerman–Szelepc´enyi, so NSPACE(f (n)) = co-NSPACE(f (n)) ∀proper f (n) ≥ log(n)
A few more things about nondeterministic space are worth pointing out. First, 2SAT is also NL-complete. The proof is in the book. While in previous courses, you have mostly seen completeness as a way to show that problems are possibly intractable, here we have used completeness to prove a positive results, e.g. NL ⊆ P. Although this proof could be accomplished without using completeness, this is not always the case. Lastly, the fact that NL = co-NL was much more surprising than PSPACE = NPSPACE. Some form of the former was first posed around 1950, and, until it was proved, most people believed it was false.
2 Current Classes
The current situation is:
L ⊆ NL = co-NL ⊆ P ⊆ NP?co-NP ⊆ PSPACE = NPSPACE
The following are still open:
- Does NP = co-NP?
- Are any of the above inclusions strict inclusions or equalities? The only separation we know for the above classes is that NL ( PSPACE, which follows from the Space Hierarchy Theorem and Savitch’s Theo- rem.
It is also important to note that even if P 6 = NP, this does not mean that all problems in NP − P are NP-complete. In fact, the following theorem states exactly the opposite:
Theorem 1 (Ladner) If P 6 = NP, there exist languages in NP that are neither in P nor NP-complete.
The proof is by diagonalization, and analogues also probably hold for most pairs of complexity classes (like L and NL). Natural candidates for such languages are Graph Isomorphism, Factoring, Discrete Log, as none of these are known to be NP-complete or in P. (The last two problems are not languages, but still one can ask whether they are NP-hard via Cook reductions or can be solved in polynomial time.)
3 Alternation - Sections 16.2 & 19.
The motivation for this section is to find a PSPACE-complete problem (which will be provable not in L, or even NL, by the Space Hierarchy Theorem).
Definition 2 (Alternation) An alternating TM is a nondeterministic Tur- ing machine in which each state is labelled either existential(∃) or universal(∀).
A configuration of an alternating TM is defined to be accepting if:
- The TM has halted and is in an accepting state.
be an algorithm which simulates the NTM M using the yi’s to determine it’s nondeterministic choices (using the next yi which has the same kind of quantifier as the current state of M ). Note that the nondeterministic choices of ATIME(f (n)) algorithm can indeed be described by at most O(f (n)) bits.
This give us a canonical form for ATMs. We will now prove general inclusion of ATIME and SPACE classes going either way, which will then allow us to conclude the theorem.
Lemma 5 ATIME(f (n)) ⊆ SPACE(O(f (n)))
Proof: Given any ATM in the canonical form and some input, we can evaluate the alternating tree recursively. As we move down the tree, we only need to remember whether each previous node in the tree is a 0 or 1. We know exactly what kind of node each previous one is, as we are assuming the canonical form. Since the depth of the tree is O(f (n)), and calculation of the relation takes time linear in f (n) (and thus use ≤ O(f (n)) space), we use space at most O(f (n)).
Lemma 6 SPACE(f (n)) ⊆ ATIME(f (n)^2 )
Proof: The proof is similar to that of Savitch’s Theorem. Say we are given a TM M : LM ∈ SPACE(f (n)). On input x with length n = |x|, the configuration graph has O(2O(f^ (n))) vertices, and each configuration can be stored in space O(f (n)). Given this graph, we would like to find out if we can get from a start state to an accepting state. Thus we say that P ath(u, v, i) = 1 if ∃ a path with length≤ 2 i^ from configuration u to configuration v. Now, we can evaluate P ath(start, accept, O(f (n))) in ATIME(f (n)^2 ) by using alternating-recursion. For the base, if i = 0, P ath(u, v, i) = 1 if u = v or if u’s child is v. Now, like in Savitch’s Theorem, we recursively guess the midpoints, but here we use alternation.
- First, guess a midpoint w (using ∃ states).
- Next, we can check if both path(u, w, i − 1) and path(w, v, i − 1) are 1 (using a ∀ state).
Because of how we count time in an alternating TM, and second step verifies two conditions in the time of one. Thus, since each vertex w has descrip- tion length O(f (n)), and we are guessing O(f (n)) vertices, we use at most ATIME(f (n)^2 ).
Finally, we can now prove the theorem.
Proof Theorem 3: Since the class of polynomials is closed under the squaring-operation, the two lemmas become inclusion each way AP and PSPACE, thus AP = PSPACE.
How can we think about alternating problems? One simple interpretation is as a game, where the two types of states represent the two players, and R represents a polynomial-time referee. The ∃ player tries to make R = 1, while the ∀ player tries to make R = 0. Then, in general, we can say that x ∈ L ⇔ ∃ a winning strategy for the ∃-player. Since this gives us a good way of thinking about problems in AP, it also gives a good way to think about problems in PSPACE, since we have just shown that these are equal.
