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Suppose an NFA which accepts language of the form L(N) = {w| w has 1 in n$^t$$^h$ from last symbol.} Then the corresponding DFA would have 2$^n$ states(worst case of subset construction). If we are to prove that equivalent DFA has 2$^n$ states, then we take 2 string as a$_1$a$_2$a$_3$.....a$_n$ and b$_1$b$_2$b$_3$....b$_n$ and consider two cases:

1) a$_i$ $\neq$ b$_1$, i=1. this case is clear to me....

2) a$_i$ $\neq$ b$_i$ , i>1. I am having trouble in understanding this case. If both a$_i$ and b$_i$ has same value for n$^t$$^h$ symbol from last, then automaton would be in accepted state after reading a$_n$ and b$_n$. So why do we need to remember every possible string sequence for this case.

Thanx!

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  • $\begingroup$ The language of words that have a $1$ in the $n$-th position can be recognized by a DFA with $n+2$ states. $\endgroup$ – Brian M. Scott Apr 12 '13 at 18:28
  • $\begingroup$ @BrianM.Scott It's n-th from last, so for equivalent DFA it must have 2$^n$ states. $\endgroup$ – greendragons Apr 12 '13 at 18:36
  • $\begingroup$ Ah, okay, but then you need to change the description: ‘last $n$-th’ does not mean ‘$n$-th from last’. $\endgroup$ – Brian M. Scott Apr 12 '13 at 18:42
  • $\begingroup$ @BrianM.Scott Sorry mistake is mine. $\endgroup$ – greendragons Apr 12 '13 at 18:46
  • $\begingroup$ No problem! ${}{}{}{}$ $\endgroup$ – Brian M. Scott Apr 12 '13 at 18:48
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The difficulty is that you don't know until you get to the end which symbol is the $n$-th from last. To understand what is going on, you need to think about how the automaton should process strings of more than $n$ symbols, rather than of exactly $n$ symbols as you seem to be doing.

As a hint to hopefully help you come up with a formal proof: Think about what information you need to remember as you are processing a string. You need to remember every $1$ you see, and how far 'back' in the string you saw it, in case it turns out to have been in the $n$th-from-last position. Except you can forget any information from before the most recent $n$ characters of the string, since any $1$ read before then can't possibly be the $n$th-from-last symbol.

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  • $\begingroup$ I understand that from subset construction, it'll have 2$^n$ states, how to prove it taking string w, |w| > n. $\endgroup$ – greendragons Apr 13 '13 at 17:29
  • $\begingroup$ Sorry, I don't understand your question (if it is a question). $\endgroup$ – Tara B Apr 13 '13 at 18:43
  • $\begingroup$ I am asking how to prove(formally not intuitively) it that for string greater than n, requires 2$^n$ states if n-th from last symbol is 1. $\endgroup$ – greendragons Apr 13 '13 at 18:54
  • $\begingroup$ Do you understand it intuitively, for a start? (In your original question, you didn't seem to believe it was true, so I guess at least at that point you hadn't understood it intuitively.) $\endgroup$ – Tara B Apr 13 '13 at 19:21
  • $\begingroup$ your answer was useful so did i mark it. It gave me insight to case length greater than n. $\endgroup$ – greendragons Apr 13 '13 at 19:39

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