Here you can ask questions and find or give answers to organizational, academic and other questions about studying computer science.

862 questions

981 answers

1.4k comments

431 users

0 votes

Below is the Kripke Structure I got after removing initial state with dead-ends :

This answer didn't work in exercise sheet, it is showing incorrect answer.
Answer to part c: ((p&!q&!a& next(p) & !next(q))| (!p&!q&a& next(p) & !next(q)) | (p&q&!a& next(p) & !next(q)) | (p&q&a& next(p) & !next(q))) & next(a) 


Also for part D: I removed states 2,7,8 and wrote as below:
((p&!q&!a& next(p) & !next(q))| (!p&!q&a& next(p) & !next(q)) | (p&q&!a& next(p) & !next(q)) | (p&q&a& next(p) & !next(q))) & next(a) & !(!p & !q & a) & !(p&q&!a) & !(p&q&a)

It is also showing incorrect to me.
Can anyone help me in fixing this ? 


I already referred :https://q2a.cs.uni-kl.de/887/exercise-sheet-5-question-1, tried this approach but answer is being shown incorrect.

in * TF "Emb. Sys. and Rob." by (630 points)
In your drawing, there's a transition from {p} to {p, a} missing.

For reference: This is sheet 5, problem 1, variant 17 in SS2021.

2 Answers

+1 vote
 
Best answer

Even though I should not post the entire solution, I will do so here since there are quite some confusions about this kind of exercise. Using the state encoding s0:{}, s1:{p}, s2:{q}, s3:{p,q}, the FSM has the following states (as can be seen in the state transition diagram you posted): 

  • initial states: {s1,s3} 
  • reachable states: {s1,s3}
  • dead(end) states: {s2}
  • finally dead(end) states: {s2}

The corresponding Kripke structure looks as follows: 

and has

  • initial states: {s2,s3,s6,s7}
  • reachable states: {s2,s3,s6,s7}
  • dead(end) states: {s0,s3,s4,s5}
  • finally dead(end) states: {s0,s3,s4,s5}

Part 1a):

It asks for initial states, but actually it means the initial states that are not finally dead states. Hence, these are states {s2,s3,s6,s7}\{s0,s3,s4,s5} = {s2,s6,s7} represented by p&!q&!a | p&q&!a | p&q&a or simpler p&!a | p&q&a

Part 1b): 

Again, it asks for reachable states, but actually means reachable states that are not finally dead states. Hence, these are states {s2,s3,s6,s7}\{s0,s3,s4,s5} = {s2,s6,s7} which is the same set of states as above. 

Part 1c): 

The (finally) dead states are {s0,s3,s4,s5} where only state s3 has incoming transitions. To remove them, we modify the transition relation as follows:

    (!p&!q& a & !next(q) & next(p) | 
      p& q&     !next(q) & next(p) | 
      p&!q&!a & !next(q) & next(p)
    ) & !(next(p)&!next(q)& next(a))

which is now the following state transition diagram:

If you don't like to inspect the state transition diagram, you may also check that the deadend states are encoded as 

    s0 : !p&!q&!a 
    s3 :  p&!q& a  
    s4 : !p& q&!a 
    s5 : !p& q& a 

so that the set of deadend states is represented by !p&!q&!a | p&!q& a | !p& q or even simpler by !p&!a|p&!q&a|!p&q. The transitions that do not lead to deadend states are therefore R&!next(deadend) which is

    (!p&!q& a & !next(q) & next(p) | 
      p& q&     !next(q) & next(p) | 
      p&!q&!a & !next(q) & next(p)
    ) & !(!next(p)&!next(a)|next(p)&!next(q)&next(a)|!next(p)&next(q))

Part 1d): 

Now we shall remove the transitions from/to unreachable states, and I assume that again, only reachable states without the deadend states is meant here. The reachable states are {s2,s3,s6,s7} and the finally dead states are {s0,s3,s4,s5} so that we have {s2,s3,s6,s7}\{s0,s3,s4,s5} = {s2,s6,s7}. These states are encoded as follows:

    s2:  p&!q&!a
    s6:  p& q&!a
    s7:  p& q& a

So that {s2,s6,s7} is encoded by p&!q&!a | p& q&!a | p& q& a or equivalently by p&q|p&!a. To allow only transitions among these states, we replace the transition relation as follows:

    (!p&!q& a & !next(q) & next(p) | 
      p& q&     !next(q) & next(p) | 
      p&!q&!a & !next(q) & next(p)
    ) 
    & (p&q|p&!a)
    & (next(p)&next(q)|next(p)&!next(a))

This yields the following state transition diagram:

Hence, you can also simplify the transition relation to

      (p&q | p&!q&!a) & !next(q) & next(p)& !next(a)

by (138k points)
selected by
Thank you @KS and @Martin for the help.
Glad I could help. Hope you learned something from my answer.
In 1c., the transition relation for s7 is exempted from R... can you explain why, please?
The transition s7->s2 is still part of the transition relation. Why do you think that it has been removed? State s7 is encoded as p&q&a. It is an initial state, as such also a reachable state, and as it has an outgoing transition to s2, it is not a deadend state. Since s2 has an infinite outgoing path, so does s7 have one, and therefore these states are neither deadend states nor finally deadend states. Hence, these states will not be removed in 1c where we should remove the (finally) deadend states. The transition of s7 which is p&q&a&next(p)&!next(q)&!next(a) is still part of the transition relation.
I get it. Thank you (Prof.) @KS
+1 vote
Not relevant for the answer, but your drawing is inconsistent with respect to transition to the deadend state {p, a}. Most transitions to it are shown, the former self-loop from {p} to {p,a} is not.

I think, you have a typo at the end of your formula for C. At the end, you write “& next(a)” but that's limiting the transitions to the deadend state {p,a}. It should be opposite. You should remove the transitions to the deadend state, not limiting to the transitions to the deadend state. So “& ¬next(a)” is what you are looking for. That would limit the transitions to the ones to {p}

For D, there seems to be a conceptual misunderstanding. Initial states like your states 7 and 8 are reachable.
by (25.2k points)
Imprint | Privacy Policy
...