CSE596, Fall 2015
Problem Set 8
Due Fri. Nov. 13
Reading. For next week, first read Homer-Selman section 5.6 in tandem with what the ALR
notes say about padding and translation, and skim the rest of the chapter to note the statement
of the Immerman-Szelepcsényi Theorem. The most particular thing to understand about both
it and Savitch’s Theorem is that by padding-and-translation they follow from the statements
in the case s(n) = log n: NL = co-NL and NL ⊆ DSPACE[log2 n]. Hence I will come back to
them in the last weeks when we reach Chapter 7 which covers logspace.
In the ALR notes, finish section 2 but skip subsection 2.8 on alternating TMs (also the
NTIME Hierarchy Theorem in section 2.6 is a ‘skim’) and skim section 3. We will take Theorem 3.1 from ALR for granted with the slogan “Software can be efficiently burned into hardware”
and use it to set up the proof of the Cook-Levin Theorem in the ALR notes. This shortcut
enables Cook-Levin to be proved in half a page, so the Homer-Selman’s longer “traditional”
proof is a skim. That is enough to read for next week—NP-hard problem from both the text
and the ALR notes will be covered the week after that.
Again because I am away Monday and Tuesday, this homework is due Friday, Nov. 13;
I will have my regular Wednesday hours 11am–1pm and substitute hours on Thursday 11am–
12:30pm. Also a reminder that Prelim II is set on the Monday before Thanksgiving, Monday
Nov. 23, in class period. It will cover assignments up through this one and readings up
through the Cook-Levin theorem and variations of it, but not reductions from SAT to graphtype problems.
(a) Discussion Problem: Suppose we try to “improve” the DSPACE hierarchy theorem as
follows: Take a fixed TM M2 that space-constructs the function s2 (n) in the theorem as a
“clock.” Let MU on inputs w = hz, vi first attempt to decode z as a TM M with 1 worktape
plus M2 as an attached “space clock.” If this fails—in particular if M does not include M2 as
an attached space clock metering its main worktape—then MU rejects w. Else it runs M (w)
to completion—since M is space-clocking itself—and accepts w iff M rejects w. Then L(MU )
belongs to DSPACE[s2 (n)] but is different from every language L(Me ) in DSPACE[s2 (n)]. Thus
we have a “poof” by diagonalization that DSPACE[s2 (n)] 6= DSPACE[s2 (n)].
What has gone wrong?? Find the “lie.” Note also some key similarities to the discussion
problem with “P, Q” on Problem Set 5: MU is “P ” with the slight difference that its input is
not exactly “z” but a padded input w = hz, vi that contains z. Q is the step of MU above
that verifies the presence of the M2 space clock which incidentally guarantees that the machine
M decoded from z is total. The last part of MU simulates M (w) and inverts the answer the
same way P does. Does all of complexity crumble into contradiction? (3 pts. checkoff credit
for giving a short summary of the flaw.)
(1) For each of the following stated relationships between complexity classes, say whether
it is known to be true or not. In all cases where it is “known,” you can prove it by applying
theorems in Homer-Selman, section 5.4. Where you say “not known,” explain why the theorem
comparing the two complexity measures involved fails to yield the stated relationship. Note
O(1)
that E = DTIME[2O(n) ] and EXP = DTIME[2n ]. (5 × 6 = 30 pts., items overleaf)
(a) NP ⊆ EXP.
(b) PSPACE ⊆ E.
(c) NSPACE[(log n)2 ] ⊆ P.
(d) P ⊆ DSPACE[n2 ].
(e) NTIME[n + 1] ⊆ E.
(2) A collection { Ci } of complexity classes forms a proper hierarchy if given any Ci and Cj
with i 6= j, one of them is properly contained in the other. Which of the following collections
are proper hierarchies? Justify your answers, mainly by verifying the relevant “little-o” or “Θ”
relations between time bounds. Here Q+ stands for the positive rational numbers.
(a) { DTIME[nc ] : c ∈ Q+ , c ≥ 1 }.
(b) { DTIME[(n + c)3 ] : c ∈ Q+ , c ≥ 1 }.
(c) { DTIME[2cn ] : c ∈ Q+ }.
1/c
(d) { DTIME[2n ] : c ∈ Q+ , c ≥ 1 }.
(24 pts. total)
(3) First, use padding-and-translation to show that if NLBA is contained in NP then
PSPACE = NP. Then, after quoting theorems to explain why NLBA ⊂ PSPACE (proper
containment), show how the same if-then implication follows from PSPACE being the closure
of NLBA under downward ≤pm -reductions. (12 + 9 = 21 pts., for 78 on the set including the 3
pts. for A).