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I am solving exam question 13.02.2019 problem 6 (d). (

It asks to determinize the FG automata of part (c).

But I am confused in the solution i.e the transition for !a&!b from state P1 should go to another state Q = {{q},{p,q}}, Qf = {} because in the FG automaton in part c from the successor on input !a&!b from {q} is {{q}, {p,q}} and the successor on input !a&!b from {p,q} is {} so union is Q = {{q}, {p,q}} and for Qf the successor on input !a&!b from {p,q} is {} and the intersection with accepting state sets gives {}.

So there should be a new state like Q = {{q}, {p,q}} and Qf = {} (as drawn in my solution).

Could you please correct me what am I doing wrong or misinterpret something ?

in * TF "Emb. Sys. and Rob." by (2.9k points)
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You are right, your result is exactly what is computed also by the teaching tool. However, the example solution is also right since the initial state of your automaton and the state that is missing in the example solution are equivalent to each other so that we can merge them into a single state. Maybe the example solution should be better yours to avoid that confusion.
by (139k points)
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Yes now I can see that the two states are equivalent and hence can be merged.
Thank you for checking and explaining this.
The example solution is fixed meanwhile. You have forgotten the self-loop in the sink state.
Yes I forgot to add self-loop in the sink state.
Thanks for pointing out, I corrected it and updated this problem also.
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