Computer Science question - Moore Machine

Hello everyone

I am studying for an exam and there is a problem that I don't understand:

Two bit patterns X0 to X7 and Y0 to Y7 should be compared with a sequential network. The two variables Xi and Yi are delivered synchronous with the clock cycle CP and in each case are forwarded into an antivalence circuit. The output of that circuit is the variable V=0 if the digits are equal, and V=1 if not.
This sequential network runs through one cycle with the length 8 and continues to the next one, so that a new comparison can take place.

The question is to make a diagram of the appropriate Moore machine.

The solution is:



but I don't understand it. Why do I need 17 states? What are the numbers in the circles? Basically I know how such a diagram is done, but in this instance, I am lost.

Are there any intelligent people out there, who could please enlighten me and tell me how that damn thing works? :)
 
I dont do computer science, but it looks to me like the numbers in the circle are counts of how many ons and offs there are (top being offs)

I think if X offs and then theres an on, it looks like you end up with 16 - X offs over 1 on.

17th state is the beginning.

I have no idea, I'm just noticing paturns
 
Ok , let's start with the numbers in the circles. The upper number seems to be the number of the state, with 0 being the start state. The lower number should be V.

There are 17 states. The start state is(0/0). Here the first comparison (bit 0) is made. If the first bit are equal is transitions to state (1/0), otherwise to state (16/1). At state 1 the first bit (bit 0) is compared again. If V=0, the next state is (2/0), otherwise it is (15/1). At state 2 the second bit is compared, etc. Since this is a sequential network you need 8 states for V=0, because you need to compare 8 bits. But you need 8 states for V=1 because you cannot skip the rest of the bits. So starting state plus 8 states for V=0 plus 8 states for V=1 = 17 states.

I hope you can understand what I mean. ;-)
 
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