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in * TF "Emb. Sys. and Rob." by (250 points)

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As you can see on slide 48 of the chapter on omega-automata, we just make the conjunctions of the initial conditions and transition relations. That is the same as we do for the product automaton on slide 46 for the disjunction of totally defined automata, and on slide 44 for the conjunction of all automata.

Why is that so? To accept an infinite word, we have to have an infinite run through the automaton that satisfies the acceptance condition. To check this, we may consider the input word as a transition system so that we can compute the product with the automaton to check whether the acceptance condition holds in the product structure for at least one path (for existential automata) or for all paths (for universal automata). For a nested automaton the same reasoning applies now for a path of the product with the outer automaton that needs to be accepted by the inner automaton.
by (166k points)
Thanks for the answer professor. But then, for the first nested automaton i.e. 2020 Feb q5, after resolving the nested automaton, shouldn't the initial condition be !q&p&!r ?
Yes, true! That is a typo. Thanks for pointing this out!
thanks, that's what i was confused about.
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