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The number of states in the minimum sized DFA that accepts the language defined by the regular expression.

$(0+1)^{*} (0+1) (0+1)^{*}$

is ________.
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5 Answers

Best answer
95 votes
95 votes

All strings over $\{0,1\}$ having length $\geq 1$

$(0+1)^*(0+1)(0+1)^* = (0+1)(0+1)^*=(0+1)^*(0+1)= (0+1)^+$

Having DFA:

Number of states in minimal DFA $=2$.

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3 votes
3 votes

The regular expression generates the min string “0” or “1” and then any number of 0’s and 1’s .
So, the DFA has two states.

1 votes
1 votes
The given regular expression is  (0+1)*(0+1) (0+1)*

After applying R*. R*=R* it becomes (0+1)(0+1)*

in (0+1)(0+1)* there is atleast one appearance of (0+1) hence we can write it as (0+1)^+

A={0,1,00,01,10,11,000,001,011,........} if draw DFA we need atleast 2 states so the answer is 2.
Answer:

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