CS 4510: Automata and Complexity
Problem Set 1 Solutions
Problem 4 Give a regular expression for each of the following languages.
1. {w: The length of wis a multiple of 3}.
Solution: (ΣΣΣ)∗
2. {w:weither starts with 01 or ends with 10}.
Solution: (01Σ∗)∪(Σ∗10)
3. {w:wdoes not contain the substring 001}.
Solution: (1 ∪01)∗0∗
Problem 6
The procedure for converting an NFA to an equivalent DFA given in class yields an exponential
blowup in the number of states. That is, if the original NFA has nstates, then the resulting DFA
has 2nstates. In this problem, you will show that such an exponential blowup is necessary in the
worst case.
Define Ln={w: The nth symbol from the right is 1}.
1. Give an NFA with n+ 1 states that recognizes Ln.
2. Prove that any DFA with fewer than 2nstates cannot recognize Ln. (Hint: Let Mbe any
DFA with fewer than 2nstates. Start by showing that there exist two different strings of
length nthat drive Mto the same state.)
Proof: Let Mbe any DFA with fewer than 2nstates. We will show that Mcannot recognize
Ln. Since there are 2nstrings of length n, by the Pigeonhole Principle, there are two different
strings x=x1x2· · · xnand y=y1y2· · · ynthat drive Mto the same state. Since x6=y, there
is some isuch that xi6=yi. Without loss of generality, say that xi= 1 and yi= 0. Let
x0=x0i−1and y0=y0i−1. It is easy to see that x0∈Lnand y06∈ Ln. However, since xand
ydrive Mto the same state, it is clear that x0=x0i−1and y0=y0i−1also drive Mto the
same state, yet x0∈Lnand y06∈ Ln. Therefore Mcannot recognize Ln.
1