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Consider the regular expression $(0 + 1) (0+1) \dots N$ times. The minimum state finite automaton that recognizes the language represented by this regular expression contains

1. $n$ states
2. $n+1$ states
3. $n+2$ states
4. None of the above
edited | 3.8k views
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Minimal DFA includes dead states also, but not having unreachable or equal states. So here, MFA contains (n+2) states.

As far as for above problem say regular expression for $(0+1)(0+1)...3$ times

$=(0+1)(0+1)(0+1)$

Having DFA:

So, for regular expression $(0+1)(0+1)...N$ times  have $\mathbf{N+2}$ states  in DFA

$N+1$ states in NFA ( we can remove dead state)

When question is about minimum state finite automata (and nothing is mentioned NFA/DFA) then which ever having minimum no.

$N+1$ states.

edited
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sir as stated in this link . should we consider dfa if nothing is given . https://goo.gl/P2jLLi

what u think .

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@Praveen sir then what answer was given in the official key?

should we consider (N+2) as default by DFA
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No,if you put one state then there need to be a self-loop,right? and self loop can contain infinite length of string!

So, if n=1 then you need 2 states,(A-->B)
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@Praveen sir I think Minimum Finite Automata means DFA without ( unreachable and equal states ).

But my above concept contradicts with ur ans

"When question is about minimum state finite automata (and nothing is mentioned NFA/DFA) then which ever having minimum no.  take that"

sir plz can u give any reference regarding ur statement. It would really a great help.

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quite frequently asked question in one or other form
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A finite automata may be

(1) An NFA

(2)DFA

They are asking that finite automata which has minimum number of states.

clearly for any regular language, NFA would have less number of states as compared to DFA because in NFA dead configuration is allowed and through use of dead configuration we can model the reject state of DFA .

So, here they are asking minimum number of states in NFA.
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@Bikram sir

"When question is about minimum state finite automata (and nothing is mentioned NFA/DFA) then which ever having minimum no.  take that"

Any reference for this?

If this is true then an equivalent MINIMAL nfa would always have no. of states less than or equal to the equivalent dfa.

Then, in effect in such situations when nothing is mentioned we should take it to be nfa.

Is my conclusion right?

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@VS

NFA have less state than DFA if we don't consider dead states.

For every n there is a language over {a,b} which is accepted by an NFA having n states, but its minimal DFA has 2n states.

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if an NFA having n states, then

1<= no of states in DFA <= 2^n
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shouldn't the statement be n<= no of states in DFA <= 2^n ??

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i too think the same
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susmita

No

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Praveen Saini Sir please explain this in more detail and expalnation
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Above NFA in comments having 2 states but its corresponding minimized dfa having 1 states.

nfa having n states, can have maximum of $2^n$ states in corresponding DFA.

say $\{q_0,q_1\}$ are states in nfa, then its corresponding DFA have maximum of 4 states, $\{\{\},\{q_1\},\{q_2\},\{q_1,q_2\}\}$
(0+1)(0+1)......... n times

Length will be n..

No of states required for n length is n+1, one more for trap.. total n+2 states..
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Suppose a string of length n+1 is given, it'll reach which state in end?
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Kkk one more state ..

Corrected now !!
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but is it necessary that you will always need a dead state for dfa ,as you are adding 1 dead state every time for the dfa (minimal) asked.
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DFA needs to cover all possibilities so sometimes we need dead or trap ..
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Not sometimes - we need to cover all possible transitions in a DFA- because the definition says "transition function" and not "transition relation" as for a PDA.

https://en.wikipedia.org/wiki/Deterministic_finite_automaton

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But here in question they are asking for minmal state FA means 'nfa' . So, it need not cover all transitions. So, here answer shud be n+1
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@Arjun sir please confirm answer according to me it should be n+2 bcz by default Finite automata is DFA so dead state also need to consider .... Is it? If n+1 then plz explain

See this link: It says- By default, FA is deterministic

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