# Thread: Machine Of Mealy And Moore

1. Originally Posted by Apprentice123
You make (q1|b).
The correct not is (q1|0) ???

() is the circle that represents the state
Highlighted in red:

Maybe you can't read my handwriting? That says $\displaystyle q_1/0$.

2. Ok. Thank you. And if a have the Moore in the picture. I make the Mealy in the picture is correct ?

3. Originally Posted by Apprentice123
Ok. Thank you. And if a have the Moore in the picture. I make the Mealy in the picture is correct ?
Well I'm confused by some things. First of all your Mealy machine has an accepting state, whereas the Moore does not. Second of all, does your professor use some convention regarding the first character of output in a Moore machine? Because my interpretation of the Moore machine you posted is that it always prints a 1 at the beginning no matter what, because that's the initial state (assuming q0 is the initial state). I used a lambda in my Moore machine because it seemed like a universal convention. Lastly, in your Mealy diagram, the upper-rightmost b has no output associated with it, which I suspect is a typo.

4. There is no convention for the first character. I picked up this exercise from a book
The upper-rightmost is b|0 sorry.
What I did is correct?

5. Originally Posted by Apprentice123
There is no convention for the first character. I picked up this exercise from a book
The upper-rightmost is b|0 sorry.
What I did is correct?
You didn't address all my questions.. but I'll answer as best I can.

So you are given the Moore diagram and want to make an equivalent Mealy diagram. I'd say your Mealy diagram is equivalent provided that you turn the double circle around q1 into a single circle in your Mealy diagram, and also that we're assuming the Moore machine starts in state q0 without printing any output. Since there is no accepting state in the Moore diagram, making an accepting state in the Mealy diagram adds functionality and results in a non-equivalent machine. Also, note that the Moore diagram could be drawn with only two states, since q1 and q2 could be combined to make an equivalent Moore machine (just remove q2 and have the "b" from q1 loop back to q1).

6. OK. Thank you

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