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+34 votes

Let $w$ be any string of length $n$ in $\{0,1\}^*$. Let $L$ be the set of all substrings of $w$. What is the minimum number of states in non-deterministic finite automation that accepts $L$?

  1. $n-1$
  2. $n$
  3. $n+1$
  4. $2^{n-1}$
asked in Theory of Computation by Veteran (97.4k points)
edited by | 4.3k views
how to create DFA for all possible substrings of a string ?

6 Answers

+48 votes
Best answer
We need a state for counting the length. So, for length n we need $n+1$ states (one for length zero). We don't need a reject state for larger strings as we have NFA and not DFA. So, totally $n+1$ states are required. (For DFA it would be $n+2$).

Correct Answer: $C$
answered by Veteran (413k points)
edited by
for nfa answer is n+1 states

But all dfa does not have dead state. for some of languages we have dead state for dfa.

so for dfa it can be n+1 or n+2 states

can anybody give perfect example.

 VS .we can make dfa with 4 state.


@abhishekmehta4u in your diagram edge from start state to finish state is not needed , if possible for you please change image

101 is not a substring of 010
Same can 101 be a substring oo 010?
hii arjun please me mail id to discuss it
This DFA doesn’t accept the substrings of 110


Language here is substrings of a any specific string and not substrings of all the strings.. 



What would be the answer had it been given maximum number of [email protected] sir

+12 votes

Suppose a string w from (0+1)* is 001 {n=3} 
Step1. Draw a NFA that accept 001 so it requires (n+1=4) states 

step2. Now our requirement is to draw a NFA for "L be the set of all substrings of w". so From NFA given in step1 attach an branch containing ε from initial state to all other states. This  ε-NFA accepts L.
--- but there may be a confusion that it is a ε-NFA as i know that for an ε-NFA an equivalent NFA(w/o ε) contains same no of states as in ε-NFA 
Hence this NFA also contains "n+1" states

Step3. now someone ask that how many minimum states DFA it requires to accept the L so i want to tell u that there is no any standard result for it and also it is wrong to say that DFA for L contains "n+2" states it is variying {u can check it also by taking some examples}.

answered by Boss (12.2k points)

Sir, why it is wrong to say that DFA for L contains "n+2" states

well explained :)
+5 votes
For n length string, the nfa needs n+1 states where all the states are final states and there are epsilon transitions from the initial state to all final states. So, answer is n+1.
answered by Active (1.7k points)
+3 votes
For example - string is 101 (n=3) then possible sub-strings are : (epsilon, 1,0,10,101,01,1) , so max length sub-string will be string itself.

And for accepting 'n' length string atleast 'n+1' states will be required ,either in DFA or NDFA (may require more states for acceptance of other sub-strings too)

So Answer is C)  n+1
answered by (301 points)
edited by
Also can you help me with the substring of 'pratik' (assume input alphabet = a to z)

I guess you are just taking set of prefixes Union set of suffixes

According to me there are 2^6 substrings for the string 'pratik'

Please tell me why '0' alone is not a substring. Also why empty string not a substring.

substrings are simply possible factors of the string. In this question we needed to find the states, so we need is only maximum possible substring of string, and maximum possible substring of any string is string itself
yea 0, epsilon are also substrings of this string(updated in answer too)

@Pratikkumar Bulani  26 doesn't seems to be correct option to find all possible strings of length 6. I think for any given string w of length |W|=n, the number of possible strings (including null)= (n(n+1))/2 +1 where +1 is for epsilon.

Please check:

+1 vote
  1. here substring requires all states makes as initial and final
answered by Junior (667 points)
0 votes
I thought about this question as follows

Question is asking for set of all substrings to be accepted by nfa where a is of size n

So substrings possible are {€, 0,1,00,01,10,11} now draw nfa for this

q0---0,1-->q1--0,1-->q2 where all states are accepting state hence it's 3 for n=2 so n+1
answered by Active (1.3k points)

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