[CS@TUK] Questions and Answers - Recent questions and answers in FBI (Department of CS / Fachbereich Informatik)
https://q2a.cs.uni-kl.de/qa/general-questions/fbi-department-of-cs-fachbereich-informatik
Powered by Question2AnswerAnswered: Product of two Kripk
https://q2a.cs.uni-kl.de/2658/product-of-two-kripk?show=2660#a2660
<p>I'm not exactly sure if this is what you mean, but the state SQ<sub>0</sub> that is mentioned on page 15 is the resulting state that you get by combining S<sub>0</sub> and Q<sub>0</sub> in the product structure where S<sub>0</sub> is a state of S<sub>K1</sub> and Q<sub>0</sub> is a state of S<sub>K2</sub>. As it is a new state in a new Kripke (product) structure it simply has a new name, as it is the 0th state in the SQ-product structure (the same holds for SQ<sub>1</sub> and SQ<sub>2</sub>).</p>FBI (Department of CS / Fachbereich Informatik)https://q2a.cs.uni-kl.de/2658/product-of-two-kripk?show=2660#a2660Fri, 27 May 2022 07:53:54 +0000Answered: State Transition Diagram (Exercise)
https://q2a.cs.uni-kl.de/2393/state-transition-diagram-exercise?show=2395#a2395
<p>To see all transitions, you would have to compute a canonical DNF, i.e., fully expanded DNF, e.g., as follows extneding your computation to a DNF:</p><pre> ¬(next(q)∧p∧q ∨ next(p) ∨ (o->a))
<-> ¬(next(q)∧p∧q ∨ next(p) ∨ (¬o ∨ a))
<-> (¬next(q) ∨ ¬p ∨ ¬q)∧¬next(p)∧o∧¬a
<-> ¬next(q)∧¬next(p)∧o∧¬a ∨ ¬p∧¬next(p)∧o∧¬a ∨ ¬q∧¬next(p)∧o∧¬a
<-> ¬p∧¬q∧¬a∧o∧¬next(p)∧¬next(q) ∨
¬p∧¬q∧¬a∧o∧¬next(p)∧ next(q) ∨
¬p∧ q∧¬a∧o∧¬next(p)∧¬next(q) ∨
¬p∧ q∧¬a∧o∧¬next(p)∧ next(q) ∨
p∧¬q∧¬a∧o∧¬next(p)∧ next(q) ∨
p∧¬q∧¬a∧o∧¬next(p)∧¬next(q) ∨
p∧ q∧¬a∧o∧¬next(p)∧¬next(q)
</pre><p>However, that is usually a lot of work and is not recommended. You can better compute the transitions as follows: With state variable p and q, we have states {},{p},{q},{p,q}. To compute their transitions, we instantiate the transition relation with their possible values:</p><pre> {} : ¬(next(p) ∨ (o->a))
{p} : ¬(next(p) ∨ (o->a))
{q} : ¬(next(p) ∨ (o->a))
{p,q} : ¬(next(q) ∨ next(p) ∨ (o->a))
</pre><p>Now you see that states {},{p},{q} have the same transitions, namely two transitions both labeled with ¬(o->a) and leading to {} and {q}, respectively. This is so, since ¬(next(p) ∨ (o->a)) is equivalent to ¬next(p) ∧ ¬(o->a)).</p><p>The transitions of the remaining state {p,q} are given by ¬(next(q) ∨ next(p) ∨ (o->a)) which is equivalent to ¬next(q) ∧ ¬next(p) ∧ ¬(o->a), so that there is only one transition labeled with ¬(o->a) and leading to {}.</p><p>Hence, this is your FSM:</p><p><img alt="" src="https://q2a.cs.uni-kl.de/?qa=blob&qa_blobid=15745874633331569891" style="height:275px; width:250px"></p>FBI (Department of CS / Fachbereich Informatik)https://q2a.cs.uni-kl.de/2393/state-transition-diagram-exercise?show=2395#a2395Sat, 29 May 2021 09:12:23 +0000