2
$\begingroup$

I'm a bit confused about the definition of a Petri-Net. From Wikipedia (EN) I have this defintion:

A Petri-Net is a 5-Tupel (P, T, F, W, s)

P and T are disjoint finite sets of places and transitions, respectively.

So the defintion says: $P \cap T = \emptyset $ & $P \cup T \neq \emptyset $

Let's say T is $ \emptyset $ and P = { p } so both prerequisites are given. I think that is okay. We have a system which stays allways in one state and can not change to another so we do not need any other state or transition.

Now think about the other case: T = { t } and P is $ \emptyset $. Now we have a system which does not have a state. The transition can't do anything because we do not have any states and it does not have any meaningful meaning in this case.

So does any Petri-Net which has minimum one transition need two places? Or have I missed out something?

$\endgroup$
  • $\begingroup$ In a net with 1 transition and 0 places, the transition can fire indefinitely. $\endgroup$ – reinierpost Apr 25 at 12:59
3
$\begingroup$

A net without place does have "state", or a marking in proper terminilogy; as that is a mapping from the set of places into $\Bbb N$, technically this does exist even when the set of places is empty (as a kind of empty relation).

A net with one place and one transition can even change state. If the transition has an outgoing arrow to the place, the transition can add tokens into the place. (Tokens can be created, they aren't physically moved from one place to another.) Effectively the behaviour will be that of a upward counter.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.