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+34 votes
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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 (115k points)
edited by | 4.1k views
0
??
+1
how to create DFA for all possible substrings of a string ?

6 Answers

+47 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 $n+2$).
answered by Veteran (395k points)
edited by
+3
Why answer is not n+2? for string of length n we need n+1 states and one extra state to reject the string of length greater than n.
+1
Sir its an NFA, so here i dont think we need a reject state, we only need n+1 states for counting the length. Please clarify.
+3
In NFA also we need dead state. @Omesh-Pandita can you confirm?
0
Sir a minimized NFA that accepts only "1" will have two states, 1 final and 1 non-final(start state) and a transition from non-final to final state on "1" and no other transitions... am I right ?
+6

i don't think we need a dead state, the  transition function is Δ : Q × Σ → P(Q). P(Q) also includes empty set.

+11

In NFA we do not need dead state. 

http://www.eecs.wsu.edu/~ananth/CptS317/Lectures/FiniteAutomata.pdf

Page no 18.

For each state, not all symbols necessarily have to be defined in the transition function

 answer should be n.

+3
Yes. I have corrected. But still we need n+1 states, for length 0, 1, ...., n.
0
Yes. You are right
+1
for DFA n+2 states ?
0
@arjun sir how can u  say that  for dfa it would take n+2 states .    try this containing substring '11' it would take 3 states. for containing substring of length dfa always takes n+1 states. bcz here no trap states comes to be in picture . but if they ask starting wid then it will take 'n+2' states bcz here dead state is involved. am i correct plz verify @praveen saini sir @habib khan also
0
@Arjun sir, and all these n+1 states are final states right??
0
@Arjun Sir how n+2 for dfa ?
+1

 set2018 To reject other strings, one more state and self loop transitiion.

+1

sir question is asking substring and not prefix of string so how n+1 states and also it is asked for NFA and not epsilon NFA .

0
does substring include string itself .

e.g. let w=101

then substring may be e,0,1,00,11,01,10 rt ?
0
We cannot generalize that we need n+2 states for DFA.

eg: w=010  ; n=3

L={ ε , 0 , 1 , 01 , 10 , 101 }

Try drawing a DFA, it will have 7 states , which is not equal to n+2=5
0
^^ I dont think there will be 7 states for above
0

@Praveen Saini sir

How many then ? May be I am doing some mistake :)

0
It should be n + 1 where n is length of string
0
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.
+1

 VS .we can make dfa with 4 state.

+2
BEWARE: THE PICTURE WITH N+1 STATE IS NOT DFA. SEE (START STATE , 0 ) TRANSITION.
0

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

0
101 is not a substring of 010
0
Same doubt....how can 101 be a substring oo 010?
0
hii arjun please me mail id to discuss it
0
This DFA doesn’t accept the substrings of 110
0

@balraj_allam

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

 

+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 (12.6k points)
+1

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

+1
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 (291 points)
edited by
0
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'
0
Hi,

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

Thanks
+1
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
+1
yea 0, epsilon are also substrings of this string(updated in answer too)
0

@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:https://gateoverflow.in/1660/gate1998-1-23

+1 vote
  1. here substring requires all states makes as initial and final
answered by Junior (759 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 (1.2k points)
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