CS–172 Computability & Complexity, Spring 2009
Feb. 19, 2009
Homework 2 Solutions
Note: These solutions are not necessarily model answers. Rather, they are designed to be tutorial in nature,
and sometimes contain a little more explanation than an ideal solution. Also, bear in mind that there may
be more than one correct solution.
1 First, we observe that any DFA is an all-paths NFA, so all-paths NFAs accept all regular languages.
For the other direction, we need to show that any language accepted by an all-paths NFA is regular.
Let N= (Q, Σ, δ, q0, F )be an arbitrary all-paths NFA. We will construct a standard NFA M=
(Q′,Σ, δ′, q′
0, F ′)that recognizes the same language as N. The construction will be similar to the
conversion from NFAs to DFAs discussed in class (and in Theorem 1.19 of Sipser). The resulting NFA
Mwill be almost a DFA, in that there will be at most one possible path in each computation. However,
unlike in a DFA,every time we see a “dying” computation in N, the corresponding computation in M
will die as well. (Note that this is where our construction differs from the NFA-to-DFA construction!)
Q′=P(Q)
q′
0={q0}
δ′(R, a) = (∅if for some r∈R,δ(r, a) = ∅;
{q∈Q|q∈E(δ(r, a)) for some r∈R}otherwise.
(Recall that E(R)is the set of states reachable from Rusing zero or more ǫtransitions.) Finally, we
want Mto be accepting if its final state contains only accepting states of N, so we define F′=P(F).
An alternative argument is the following: First, augment the all-paths NFA Ninto an all-paths NFA
N1such that N1accepts the same language asN, but no computation of N1dies. (One way to do this
is to add a rejecting state qdie to N; then for all states q∈Q∪ {qdie}, add a transition from qto qdie
labeled by all symbols in Σunused by other outgoing arrows at q.) Now, reverse the accepting and
rejecting states of N1; call the resulting machine N2. Think of N2as a standard NFA: The language
accepted by N2is then the complement of the language Laccepted by the all-paths NFA N1. It
follows that the complement of Lis regular. Since regular languages are closed under complement
(see Problem 3 of Homework 1), Lmust be regular as well.
3 (Coming)
4 For each of these, there are many possible valid regular expressions.
(b) a∪e∪i∪o∪uΣ∗∪ǫing
(d) 0∗10∗10∗10∗∗
