NFA
Question 1 |
Regarding the power of recognition of languages, which of the following statements is false?
The non-deterministic finite-state automata are equivalent to deterministic finite-state automata. | |
Non-deterministic Push-down automata are equivalent to deterministic Push- down automata. | |
Non-deterministic Turing machines are equivalent to deterministic Push-down automata. | |
Both B and C |
Question 1 Explanation:
B: No conversion possible from NPDA to DPDA.
C: Power (TM) > NPDA > DPDA.
C: Power (TM) > NPDA > DPDA.
Question 2 |
Let δ denote the transition function and 
denote the extended transition function of the ε-NFA whose transition table is given below:
Then 
is
∅ | |
{q0,q1,q3} | |
{q0,q1,q2} | |
{q0,q2,q3} |
Question 2 Explanation:
Extended transition function describes, what happens when we start in any state and follow any sequence of inputs.
If δ is our transition function, then the extended transition function is denoted by δ.
The extended transition function is a function that takes a state q and a string w and returns a state p (the state that the automaton reaches when starting in state q and processing the sequence of inputs w).
The starting state is q2, from q2, transition with input “a” is dead so we have to use epsilon transition to go to other state.
With epsilon transition we reach to q0, at q0 we have a transition with input symbol “a” so we reach to state q1.
From q1, we can take transition with symbol “b” and reach state q2 but from q2, again we have no further transition with symbol “a” as input, so we have to take another transition from state q1, that is, the epsilon transition which goes to state q2.
From q2 we reach to state q0 and read input “b” and then read input “a” and reach state q1. So q1 is one of the state of extended transition function.
From q1 we can reach q2 by using epsilon transition and from q2 we can reach q0 with epsilon move so state q2 and q0 are also part of extended transition function.
So state q0,q1,q2.
If δ is our transition function, then the extended transition function is denoted by δ.
The extended transition function is a function that takes a state q and a string w and returns a state p (the state that the automaton reaches when starting in state q and processing the sequence of inputs w).
The starting state is q2, from q2, transition with input “a” is dead so we have to use epsilon transition to go to other state.
With epsilon transition we reach to q0, at q0 we have a transition with input symbol “a” so we reach to state q1.
From q1, we can take transition with symbol “b” and reach state q2 but from q2, again we have no further transition with symbol “a” as input, so we have to take another transition from state q1, that is, the epsilon transition which goes to state q2.
From q2 we reach to state q0 and read input “b” and then read input “a” and reach state q1. So q1 is one of the state of extended transition function.
From q1 we can reach q2 by using epsilon transition and from q2 we can reach q0 with epsilon move so state q2 and q0 are also part of extended transition function.
So state q0,q1,q2.
There are 2 questions to complete.