= == Consider a DFA M (Q, E, 8, s, f) with States Q = {s,...

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= == Consider a DFA M (Q, E, 8, s, f) with States Q = {s, q1, 92, f}, where s is the start and f is the final state; Alphabet = {0, 1}; and transition function d. Construct a state transition table for 6 (or you can draw a state transition diagram) that recognizes regular expressions that are binary strings and multiples of 3, for example, the strings would be accepted strings, but 0, 11, 110, 1001, 1100,... 1, 10, 100, 101, ... would not be accepted. (Hint: Think, if n = 3k is a multiple of 3, then the next multiple of 3 is 3k + 3. this could be accomplished by a transition from the current state to a next state by scanning 3 ones.) = == Consider a DFA M (Q, E, 8, s, f) with States Q = {s, q1, 92, f}, where s is the start and f is the final state; Alphabet = {0, 1}; and transition function d. Construct a state transition table for 6 (or you can draw a state transition diagram) that recognizes regular expressions that are binary strings and multiples of 3, for example, the strings would be accepted strings, but 0, 11, 110, 1001, 1100,... 1, 10, 100, 101, ... would not be accepted. (Hint: Think, if n = 3k is a multiple of 3, then the next multiple of 3 is 3k + 3. this could be accomplished by a transition from the current state to a next state by scanning 3 ones.)