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As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′0. Since they cannot be reached from Q′0 there is no path from Q′0 to a state in F′ which passes through them and they can be deleted from the automaton without changing the language it accepts. In practice it is much easier to build Q′ as needed, only including those state sets that actually are needed.
To see how this works, lets carry out an example. For maximum generality, let's start with the NFA with ε-transitions given above, repeated here:
Because it is simpler to write the transition function (δ) out as a table than it is to write out the transition relation (T) as a set of tuples, we will work with the δ representation. When given a transition graph of an NFA with ε-transitions like this there are 6 steps required to reduce it to a DFA:
1. Write out the transition function and set of ?nal states of the NFA.
2. Convert it to an NFA without ε-transitions.
(a) Compute the ε-Closure of each state in the NFA.
(b) Compute the transition function of the equivalent NFA without ε-transitions.
(c) Compute the set of ?nal states of the equivalent NFA without ε- transitions.
The Emptiness Problem is the problem of deciding if a given regular language is empty (= ∅). Theorem 4 (Emptiness) The Emptiness Problem for Regular Languages is decidable. P
I want a proof for any NP complete problem
1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization
a finite automata accepting strings over {a,b} ending in abbbba
DEGENERATE OF THE INITIAL SOLUTION
RESEARCH POSTER FOR MEALY MACHINE
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
how is it important
Can you say that B is decidable? If you somehow know that A is decidable, what can you say about B?
Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
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