CSE200: Computability and Complexity Fall 2008, Homework 6

Instructor: Daniele Micciancio Due Tuesday November 18, 2008

Problem 1: Log-space, Polylog-space

(a) Recall the WHILE language we were using in the first homeworks. The input to any WHILE program is encoded as a single list (which has as elements other lists and so on). For instance the string [[][[]][][[[]]]] is a valid input to a WHILE program whereas [][[][]] (two separate lists) and []][ (right parenthesis not preceded by a left one) aren’t. Consider the following set of Valid WHILE Inputs

∗ VWI = {x ∈ {[, ]} x is a Valid WHILE Input}

Show that VWI ∈ L. (b) Consider the polylogarithmic space complexity class defined in class as follows [ polyL = SP ACE(logkn) k>0

We say that a language C is polyL-complete if: • C ∈ polyL and

• A ≤L C for every A ∈ polyL. (every polylogarithmic space language is log-space (map) reducible to C).

(i) Show that NL ⊂ polyL

(ii) Show that there exist no polyL-complete problems under the above definition of completeness.

(iii) Show that polyL 6= P.

(iv) Part (iii) implies that either (a) polyL ⊂ P, or (b) P ⊂ polyL or (c) polyL ⊂ polyL∪P and P ⊂ polyL∪P (none of them is strictly contained in the other). Note that (a), (b), (c) are mutually exclusive. So, part (iii) shows that one and only one of (a), (b) or (c) holds true. Still which of the three cases holds true is not currently known. Justify why deterining which of (a), (b) or (c) holds true might be pretty hard question, by pointing out the implications each of these three relations would have. Note: You can take as given (no need to prove it) that P has complete languages under log-space reductions. Hint 1: Use the Space Hierarchy Theorem. Hint 2: The solution for part (b) is not trivial but is fairly short (for all four subquestions). A solution longer than a page or so, might be an indication that you are on the wrong track.

1 Problem 2: PSPACE, PSPACE-Completeness

(a) Consider the following definition for PSPACE-Completeness We say that a language C is P SP ACE-complete if: • C ∈ P SP ACE and • A ≤P SP ACE C for every A ∈ P SP ACE (every PSPACE language is PSPACE (map) reducible to C). Show that under the above definition of completeness every problem in P SP ACE is P SP ACE-Complete.

(b) P1 and P2 are playing the following game: they both get as an input a graph G = (V,E), an integer vector 1 ~w = (w1, ..., wn) (with wi > 0) that corresponds to the weights of the nodes and a positive number T. P1 and P2 alternately select nodes with P1 playing first. At each stage of the game, the player that moves can pick a node only if this node doesn’t have any neighbors in the set of previously selected nodes (in other words, at each stage all the previously selected nodes must form an independent set in G). Player P2 wins if at the end of the game (when there are no independent nodes left) the sum of the weights of the nodes he has selected is at least T. Based on the above game, we can define the following language

L = {(G, ~w, T )|P2 has a winning strategy no matter how P1 plays }

Show that L can be decided in P SP ACE 2.

1 You may assume that the nodes are labeled 1, ..., n and that node i was weight wi. 2It turns out that L is P SP ACE-Complete but here you are not required to prove the P SP ACE-hardness.

2