Problem Solving in Automata, Languages, and Complexity
dana p.
(Dana P.)
#1
6.5 Context-Sensitive Languages 319
for all a E C, where
generates exactly the
Lh is a new nonterminal symbol. Then, grammar G2
nonempty strings u) E L as follows:
s$Bh#BlL l B$Bw cl s#B jBWLh&LhWjw.
I W I
Finally, the only rule in this grammar G:! that violates the context-sensitive
requirement is the rule BLh + &. So, we can convert Gz to an equivalent
context-sensitive grammar G by the technique of Solution 2 of Example 6.33.
That is, we can simply attach the leftmost and rightmost blanks to their
neighboring symbols and make every rule satisfying the context-sensitive re-
quirement. We leave the details to the reader as an exercise. Cl
Whether NSPACE(n) = DSPACE( n ) is, like the question of whether
NP = P, a major open question in complexity theory. (Savitch’s theorem
only gives a weaker result: NSPA CE (n) C DSPA CE (n2) .) We note, however,
that, unlike the class NP, the class NSFACE (n) is known to be closed un-
der complementation. In the following, we prove a more general result that
all classes NSPACE(s(n)) is closed under complementation, if s(n) is fully
space-constructible and s(n) 2 logn.
In order to prove this result, we first define the notion of an NTM computing
a function. We say that an NTM A4 with a write-only output tape computes
a partial function f : C + C, if the following conditions hold:
(i) For each x E Domain(f), there exists at least one accepting path of A4
on input 2.
(ii) For each x E Domain(f), every accepting path of A4 on input II: has the
same output y = f(z).
(iii) For each x # Domain(f), th ere is no accepting path in the computation
tree of A4 on input z.
In other words, an NTM A4 is considered as a transducer only if all of its
accepting paths on an input x have the same output value.
* Theorem 6.36 For any fully space-constructible function s(n) > logn,
NSPACE(s(n)) = co-NSPACE(s(n)).
Proof. Let A4 be a one-worktape NTM with the space bound s(n). Recall
the predicate read-@, p, Ic) defined in the proof of Savitch’s theorem. In this
proof, we further explore the concept of reachable configurations. Consider a
fixed input II: of length n. Let C& be the class of all configurations of A4 on z
that is of length s(n). That is,
where j indicates the position of the input
configuration of the work tape. Note that
tape head, and ybz -
each configuration
is the tape
in Cz is of