Lecture Notes 08: DFA-NFA equivalence and RE-FA equivalence

Outline

DFAs Vs NFAs
REs Vs FAs

A Slideshow:

GUIDED NOTES (Optional)

DFA-NFA equivalence

Last lecture we asked the question: Are NFAs stronger than DFAs?

Let’s make a claim:

NFAs and DFAs are equally powerful!

DFA VS NFA

The proof approach is by construction (in two steps):

What we know: $M_{NFA}=(Q,\Sigma,\delta, q_0, F)$

Need to construct: $M_{DFA}=(Q^\prime,\Sigma,\delta^\prime, q_0^\prime, F^\prime)$ that recognizes L(M)
Prove that $ L(M_{DFA}) = L(M_{NFA}) $

Step 1: Building the equivalent $M_{DFA}$

We need to make the 5-tuple for the DFA equivalent to out NFA:

From NFA to DFA: the states $Q \rightarrow Q^\prime $

It’s possible that in an NFA, we could be in multiple states at each step. If we want to emulate the NFA’s behavior, we’ll have to have a state for each combination.

DFA VS NFA

From NFA to DFA: the Start states $q_0 \rightarrow q_0^\prime$

DFA VS NFA

(In this case, only $A$)

From NFA to DFA: the Accepting states $F \rightarrow F^\prime$

DFA VS NFA

(In this case, $ \{ C, AC, BC, ABC \} $)

From NFA to DFA: the Transitions $\delta \rightarrow \delta^\prime$

At each symbol reading, if there are multiple paths to follow, we want to jump the the single state (in the DFA) that is the Union of all the possible destination states (in the NFA) for that symbol.

This can be expressed like this: \[ \begin{alignat}{2} &\delta^\prime ( q, \mathit{a} ) = \bigcup\limits_{p\in q} \delta(\mathit{p}, a) \\ &\qquad \text{ where $q \in Q^\prime$; $a \in \Sigma$} \\ &\qquad \text{ and $q$ can be a set of states from $Q$} \end{alignat} \]

Let’s see it in action.

Example: building the DFA for this NFA

DFA VS NFA

Solution:

DFA VS NFA

Step2: Prove $L(M_{DFA}) = L(M_{NFA}) $:

DFA VS NFA

Hint:

To prove that L(DFA) = L(NFA), use induction to show that the result of running the DFA on a word $w$ (denoted $ \hat{\delta}^\prime(q_0^{\prime} , w))$ is equivalent to the result of running the NFA on the same word [denoted $\hat{\delta}(q_0, w)$] for every word $w \in \Sigma^*$.

A summary of recommended steps is:

Base Case: Prove that $ \hat{\delta}^\prime(q_0^{\prime} , w)) = \hat{\delta}(q_0, w)$ for the base case of $w = \epsilon$, which is when $ \vert w \vert = 0 $
Induction Hypothesis: Asume that $ \hat{\delta}^\prime(q_0^{\prime} , w)) = \hat{\delta}(q_0, w)$ for when $ \vert w \vert = k $
Induction Step: Prove that, for a word $z$ that has a length $ \vert z \vert = k+1 $, (or which we can represent as a concatenation of a word $w$ of size $k$ and one more symbol $a$), that $ \hat{\delta}^\prime(q_0^{\prime} , z)) = \hat{\delta}(q_0, z)$