Number of states in minimal DFA

Question

Let L1  = $0^+1^+$ and L2 = $(01)^+$, $L3 = \frac{L1}{ L2^*}$. The number of state needed for minimal DFA are _____.

Answer 1

we have L1 = { 01,001,011,0011.................}

              L2= {ε, 01,0101,010101.....................}

 now using right quotient : L1/L2 

            = { 01/ε, 001/ε,011/ε ,01/01 , 001/01, 0001/01............, 011/01, 0011/01..........}

            = { 01,001,011,.....ε , 0 , 00 , 000, ................................. ϕ , ϕ    }

            = { 0 * } + 0⁺1⁺

the Dfa will have 4states q0 =  intial and final state both , q1,q3 = final states ,dead state.

answer : 4states.

    

Comments

bro $\Large \frac{L_1}{(L_2)^*}$ and not $\Large \frac{L_1}{L_2}$

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oopsie yes :) thanks a lot :) see now.