CSE 311 Lecture 26:
Limitations of DFAs, NFAs, and Regular Expressions

Topics

From NFAs to DFAs

A quick wrap-up of Lecture 25.

DFAs $\equiv$ NFAs $\equiv$ regular expressions

They are all the same and represent regular languages.

Languages and representations

Regular, context-free, and other languages.

Proving irregularity

A proof template for showing that a language is not regular.

NFAs and DFAs

Every DFA is an NFA.

A DFA is an NFA that satisfies more constraints.

Proof (and algorithm) idea:

The DFA constructed for an NFA keeps track of all the states that a prefix of an input string can reach in the NFA.

So there will be one state in the DFA for each subset of the states of the NFA that can be reached by some string.

We’ll see how to construct the start state, remaining states and transitions, and the final states of the DFA.

NFAs to DFAs: the start state

The start state of the DFA represents the following set of states in the NFA:

All states reachable from the start state of the NFA using only $\varepsilon$ edges.

NFAs to DFAs: states and transitions

Repeat until fixed point:

Let $D_Q$ be a state of the DFA corresponding to a set $Q$ of the NFA states.

Let $a\in\Sigma$ be a symbol for which $D_Q$ has no outgoing edge.

Let $T$ be the (possibly empty) set of NFA states reachable from some state in $Q$ by following one $a$ edge and zero or more $\varepsilon$ edges.

Add a state $D_T$ to the DFA, if not included, that represents the set $T$.

Add an edge labeled $a$ from $D_Q$ to $D_T$.

NFAs to DFAs: final states

The final states of the DFA:

Every DFA state that represents a set of NFA states containing a final state.

Example: NFA to DFA conversion

Exponential blow-up in simulating nondeterminism

In general the DFA might need a state for every subset of states of the NFA.

Power set of the set of states of the NFA.

$n$-state NFA yields DFA with up to $2^n$ states.

We saw an example of this worst case outcome.

The famous “P=NP?” question asks whether a similar blow-up is always necessary to get rid of nondeterminism for polynomial-time algorithms.

Equivalence of DFAs, NFAs, and regular expressions

We have shown how to build an optimal DFA for every regular expression.

Build an NFA.

Convert the NFA to a DFA using the subset construction.

Minimize the resulting DFA.

Languages represented by NFAs, DFAs, and regular expressions are called regular languages.

A hierarchy of languages and representations

In HW8, you’ll (almost :) prove that regular languages are context-free. How do we prove that all finite languages are regular? By showing how to construct a DFA/NFA/RegEx for every finite language!

Converting a finite language to a regular expression

Back to languages and representations …

We saw in Lecture 21 that the language $B$ of all binary palindromes can be represented by a CFG. We also said that $B$ can’t be represented by any regular expression. How would you prove that?

Why isn’t $B$ (binary palindromes) a regular language?

If $B$ were regular, we could express it as a DFA/NFA/RegEx.

Let’s choose a DFA as our hypothetical representation.

This is the intuition for why $B$ is not regular. Let’s see how to turn this intuition into a formal proof.

A strategy for proving that $B$ is not regular

Proving that $B$ is not regular

Suppose that some DFA $M$ recognizes $B$. We show $M$ accepts or rejects a string it shouldn’t. Consider the set $S = \{0^n1 : n \geq 0 \}$.

Since there are finitely many states in $M$ and infinitely many strings in $S$, there exist strings $0^a1 \in S$ and $0^b1 \in S$ with $a\neq b$ that end in the same state of $M$.

Important: We don’t get to choose $a$ and $b$! We just know they exist.

Now, consider appending $0^a$ to both strings.

Since $0^a1$ and $0^b1$ end in the same state, so do $0^a10^a$ and $0^b10^a$, call it $q$. But then $M$ must make a mistake: $q$ needs to be an accept state since $0^a10^a \in B$, but then $M$ would accept $0^b10^a \not\in B$, which is an error. $\square$

A template for proving that a language $L$ is not regular

Example: prove that $L = \{0^n1^n : n \geq 0\}$ is not regular

Suppose for contradiction that some DFA $M$ recognizes $L$.

Consider the set $S = $ $\{0^n : n\geq 0 \}$.

Since $S$ is infinite and $M$ has finitely many states, there must be two strings $0^a, 0^b$ $\in S$ with $a \neq b$ that both end in the same state of $M$.

Consider appending $1^a$ to both $0^a$ and $0^b$.

Since $0^a$ and $0^b$ end in the same state of $M$, then $0^a1^a$ and $0^b1^a$ also end in the same state $q$ of $M$. Since $0^a1^a\in L$ and $0^b1^a\not\in L$, $M$ does not recognize $L$.

Since $M$ was arbitrary, no DFA recognizes $L$.

Example: prove that $L = \{(^n)^n : n \geq 0\}$ is not regular

Suppose for contradiction that some DFA $M$ recognizes $L$.

Consider the set $S = $ $\{(^n : n\geq 0 \}$.

Since $S$ is infinite and $M$ has finitely many states, there must be two strings $(^a, (^b$ $\in S$ with $a \neq b$ that both end in the same state of $M$.

Consider appending $)^a$ to both $(^a$ and $(^b$.

Since $(^a$ and $(^b$ end in the same state of $M$, then $(^a)^a$ and $(^b)^a$ also end in the same state $q$ of $M$. Since $(^a)^a\in L$ and $(^b)^a\not\in L$, $M$ does not recognize $L$.

Since $M$ was arbitrary, no DFA recognizes $L$.

A fun fact about this proof method

Suppose that for a language $L$, the set $S$ is a largest set of prefix strings with the property that for every pair $s_a \neq s_b \in S$, there is some string $t$ such that one of $s_at, s_bt$ is in $L$ but the other isn’t.

If $S$ is infinite, then $L$ is not regular.

If $S$ is finite, then the minimal DFA for $L$ has precisely $|S|$ states, one reached by each member of $S$.