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Exponential Blowup of Subset Construction Definition: $L_n = \{w ∈ \{0, 1\}^\star\,|\,w_{|w|-n}=1\}$. In other words, $L_n$ is the set of binary words having a $1$ in the nth position from the end. Here $|w|$ is the length of the word $w$.

Note: an NFA with only $n+1$ states can be designed to recognized $L_n$.

Attempt: Construct a NFA $B$ with $n+1$ states for the reverse of $w$. Change the start state of $B$ to be the last state in $B$.

Theorem: A DFA that recognizes $L_n$ can not have less than $2^n$ states

Why not less than $n+2$? All we need to do is to add one more branch to $B$ to make it DFA.

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    Can you explain what you have tried so far and what difficulty are you having? – 2017-02-23
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    Can you explain overall what exponential blowup of subset construction means? – 2017-02-23
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    maybe cross post to cs stackexchange? – 2017-02-23
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    I guess you want to complete the automaton for the reversal of $L_n$. That requires just one more transition, but the automaton you get by reversing the transitions is still nondeterministic. – 2017-02-23

2 Answers 2

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A simple proof that no DFA with fewer than $2^n$ states accepts $L_n$ is by invocation of the Myhill-Nerode theorem. The idea is to count the number of distinct language quotients, because each of them must correspond to a distinct state of the minimal DFA for $L_n$. Let $\Sigma = \{0,1\}$. Let's recall that the quotient of $L$ w.r.t. word $x$ is defined by

$$ x^{-1}L = \{w \mid xw \in L \} \enspace. $$

To keep notation at a minimum, let's examine the case of $n=3$. This means that words whose ends match $1\Sigma\Sigma$ are accepted, while the other words---those whose ends match $0\Sigma\Sigma$ or are shorter than three letters---are rejected.

Let $x$ be a word that ends in $000$. Then $x^{-1}L_n = 1\Sigma\Sigma$. The same is true for $|x| < 3$ as long as $x$ contains no $1$. On the other hand, if $x$ ends in $001$, then $x^{-1}L_n = \Sigma\Sigma \cup 1\Sigma\Sigma$. Continuing, we get eight different quotients all the way to $(\Sigma^*111)^{-1}L_n = \epsilon \cup \Sigma \cap \Sigma\Sigma \cup 1\Sigma\Sigma$.

This argument generalizes to arbitrary $n$ and shows that the minimal DFA for $L_n$ needs $2^n$ states.

Incidentally, an implementation of the DFA that accepts $L_n$ is an $n$-bit shift register, started from the all-zero state, and with the output taken from the "oldest" bit.

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Completing the excellent answer of @Fabio Somenzi, $L_n^R$ the reversal is easier to build because you exactly know how to count from the initial state. Hence you need just $n+1$ states.

When turning to $L_n$, you know the beginning of the word but you don't know where the end is, therefore you must check in some sense all the possible subsets of $\{1,2,\ldots,n\}$ (the possible situations before end) and this set has size $2^n$.