**10. Scope for research**

24 Will-be-set-by-IN-TECH

Next, take *<sup>n</sup>* <sup>=</sup> *<sup>k</sup>* <sup>≥</sup> 3 and assume the validity by the above procedure to get all the 2*<sup>k</sup>* binary

**Step-I.** Augment 1 to each of the marking vectors (*a*1, *a*2, ··· , *ak*) at its right most end to get

**Step-II.** In this case, *tk*<sup>+</sup><sup>1</sup> fires and after firing we get the marking vectors (*a*1, *a*2, ··· , *ak*, 0). By the induction hypothesis, we have all the 2*<sup>k</sup>* marking vectors to which Step-I has been applied to obtain the 2*<sup>k</sup>* binary (*k* + 1)-vectors as marking vectors (because of firing at each stage), having 1 in the (*k* + 1)*th* coordinate. Further, each of these 2*<sup>k</sup>* binary (*k* + 1)-vectors has been fired using Step-II to obtain the binary (*k* + 1)-vectors of the form (*a*1, *a*2, ··· , *ak*, 0) which are all distinct from those obtained in Step-I. Together, therefore, we have obtained

The following is a "frustration theorem" due to the negative fact it reveals, to the effect that one cannot hope to have a "forbidden subgraph characterization" of a Boolean Petri net.

**Theorem 8.** *[10]: Every* <sup>1</sup>*-safe Petri net C* = (*P*, *<sup>T</sup>*, *<sup>I</sup>*−, *<sup>I</sup>*+, *<sup>μ</sup>*0)*,* <sup>|</sup>*P*<sup>|</sup> <sup>=</sup> *n with <sup>μ</sup>*0(*p*) = <sup>1</sup> <sup>∀</sup> *<sup>p</sup>* <sup>∈</sup> *<sup>P</sup>*

*Proof.* Let *<sup>C</sup>* = (*P*, *<sup>T</sup>*, *<sup>I</sup>*−, *<sup>I</sup>*+, *<sup>μ</sup>*0), <sup>|</sup>*P*<sup>|</sup> <sup>=</sup> *<sup>n</sup>* be a 1-safe Petri net. If *<sup>C</sup>* is a Boolean Petri net then there is nothing to prove. Hence, assume that *C* is not a Boolean Petri net. Then, we have the following steps to obtain a Boolean Petri net *C*� in which *C* is one of its induced subnets.

**Step-1:** First of all, find those places in *C* each of whose postsets has single distinct sink transition (if the postset of a place has more than one distinct sink transitions then choose only one transition giving *K*2). Suppose such places are *p*1, *p*2, ··· , *pk*, 1 ≤ *k* < *n*. If there is

**Step-2:** Augment *n* − *k* new transitions and join each of them to the remaining *n* − *k* places in

**Step-3:** Thus, in *<sup>C</sup>*� we have *<sup>n</sup>*-copies of *<sup>K</sup>*<sup>2</sup> as its subgraph. Since *<sup>μ</sup>*0(*p*) = <sup>1</sup> <sup>∀</sup>*<sup>p</sup>* <sup>∈</sup> *<sup>P</sup>*, all the transitions are enabled. Firing of *n* transitions forming *n* 'pendant transitions' will produce *nC*<sup>1</sup> distinct binary *n*-vectors whose Hamming distance is 1 from the initial marking vector. At these marking vectors, *n* − 1 transitions out of those *n* transitions are enabled, and after firing give at least *nC*<sup>2</sup> distinct marking vectors, each of whose Hamming distance is 2 from

In general at any stage *<sup>j</sup>*, 3 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> *<sup>n</sup>*, we get a set of at least *nCj* new distinct binary *<sup>n</sup>*-vectors whose Hamming distance is *j* from the initial marking, which are also distinct from the sets of *nCr* distinct marking vectors for all *<sup>r</sup>*, 2 <sup>≤</sup> *<sup>r</sup>* <sup>≤</sup> *<sup>j</sup>* <sup>−</sup> 1. Therefore, at the *<sup>n</sup>th* stage we would have obtained at least *nC*<sup>1</sup> <sup>+</sup>*<sup>n</sup> <sup>C</sup>*<sup>2</sup> <sup>+</sup> ··· <sup>+</sup>*<sup>n</sup> Cn*=2*<sup>n</sup>* <sup>−</sup> 1 distinct binary *<sup>n</sup>*-vectors. Together with the initial marking (1, 1, ··· , 1), we thus see that all the 2*<sup>n</sup>* binary *<sup>n</sup>*-vectors would have been

no sink transition in *C*, then augment one sink transition to each place in *C*.

obtained as marking vectors, possibly with repetitions. Thus *C*� is Boolean.

*C* by an arc from a place to a new transition creating *n* − *k* new active transitions.

. Thus, the proof follows by

*k*-vectors.

PMI.

the initial marking.

Let *n* = *k* + 1. Then, apply the following procedure.

2*<sup>k</sup>* + 2*<sup>k</sup>* = 2*k*+<sup>1</sup> binary (*k* + 1)-vectors as marking vectors from *C*�

**9. An embedding theorem and complexity**

*can be embedded as an induced subnet of a Boolean Petri net.*

the (*k* + 1)-vector (*a*1, *a*2, ··· , *ak*, 1).

Precisely which Petri nets generate all the binary *n*-vectors as their marking vectors, or the so-called Boolean Petri net? This has been a hotly pursued research problems. We have shown in this chapter some necessary and sufficient conditions to characterize a Boolean Petri net, containing an SCC. However, the general problem of characterizing such a 1-safe Petri net *C*, when *C* does not contain an SCC or a strong chain, is still open. A Petri net containing an SCC is strongly connected, in the graph-theoretical sense that any two nodes in it are mutually reachable. However, the converse is not true; that is, if the underlying digraph of a Petri net is strongly connected, it need not contain an SCC. So, even a characterization of strongly connected Boolean Petri net is an open problem. Further, in general, characterizing crisp Boolean Petri nets is open too. If we relax the condition on the depth of the reachability tree in our original definition of minimality of a 'minimal' crisp Boolean Petri net and require instead that the number of enabled transitions be kept at minimum possible, the reachability graphs of such Petri nets may not have their underlying graph structures isomorphic to *<sup>K</sup>*1,2*<sup>n</sup>*−1, whence they would all be trees of the same order 2*<sup>n</sup>*. Since they would be finite in number, determination of the structures of such Petri nets and their enumeration would be of potential practical interest. It involves orienting trees of order 2*<sup>n</sup>* (in general, for theoretical purposes, trees of any order as such) that admit an orientation of their edges to make them the reachability trees of minimal 1−safe crisp Boolean Petri nets.
