Counting No of States in the DFA

Question

Minimum number of states required to construct DFA accepting language L={ w | w has even no of 0's and 1's and odd no of 3's }

over alphabet { 0,1,2,3 }

The answer given is 8. Should not the ans be 16? Using the concept of combinations there are 2 possibilities for each character ( even or odd) and there are 4 such characters. Total Combinations - 16. Thus 16 DFA states out of which 2 final states as comb for 0,1,3 is fixed and 2 can take either one.

Is it possible to get 8 states after minimization for the above DFA? Any simpler way of finding that logically?

Comments

it ll b 2*2*2=8

Answer 1

Let

L1={w | w has even no of 0's} -> 2 states

L2={w | w has even no of 1's} -> 2 states

L3={w | w has odd no of 3's} -> 2 states

Using product automata, L={ w | w has even no of 0's and 1's and odd no of 3's } -> L1* L2 * L3 -> 2*2*2 = 8 states

Counting the number of 2's is not at all required. So, not 16 states.

Comments

But it is mentioned dat d string has even number of 0's??

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Why did u do 2 0's? Hmnn. Say.

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Please work out on d DFA part. As in drawing of it to make it more clear.

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do we need dead states for the unncessary transitions ?