concerning the attached machine:

each arrow may represent multiple arrows (for different inputs), it is unknown exactly which inputs produce which outputs
and lead to which state. the number of inputs and outputs is finite but unknown.

which of the following can be said for certain about the minimization of the machine
(using moore algorithm)?
a. If the initial state is H, "I" is the only state to be minimized
b. If the initial state is A I is the only state to be minimized
c. If the initial state is C only 4 states will be minimized
d. If the initial state is B the machine cannot be minimized
e. if the initial state is I the machine cannot be minimized

the correct answer is a, I couldn't figure out why though.

your help will be very appreciated.

 

Hint: If a state is not reachable from the initial state via any possible set of moves, what does that mean about that state?

 

Thank you for the hint

I think this is the idea:

since the machine can never realy reach that position it will never create a contradiction in the algorithm
(nevermind which position it leads to) meaning I could put him in whatever division group I want at p1 and it will stay in this group until the end of the minimization process
along with at least one more state, meaning it will always be minimized.


I'm not sure this explanation works though because it may be that "I" is "1 distinguishable" with any other state in the first place
and therefore will be on its own division group in p1 already.

It does make sense though that if you can't get to this state it's not a part of the machine and can be minimized,
but is there any formal way to state that?

 

Unreachable states are generally eliminated before any attempt at state reduction is performed.