13th International Conference on Membrane Computing - MTA Sztaki

F.G.C. Cabarle, H.N. Adorna
OR-split is synchronized by an AND-join, it is possible to have a deadlock. The
free-choice property also avoids the possibility of deadlocks (see Fig. 2) and has
been studied extensively in literature. See for example [5] and [21,19] to name a
few. These nets are also known as extended free-choice nets in [15] and [4]. Next
Fig. 2. (a) A net that is not well-handled. (b) A non-free-choice net.
we provide the definition of an SNP system, slightly modified from [18].
Definition 4 (SNP system). An SNP system without delay of a finite degree
m ≥ 1, is a construct of the form
where:
Π = (O, σ 1 , . . . , σ m , syn, out),
1. O = {a} is the singleton alphabet (a is called spike).
2. σ 1 , . . . , σ m are neurons of the form σ i = (α i , R i ), 1 ≤ i ≤ m,
where:
(a) α i ≥ 0 is the number of spikes in σ i
(b) R i is a finite set of rules of the general form
E/a c → a b
where E is a regular expression over O, c ≥ 1, b ≥ 0, with c ≥ b.
3. syn ⊆ {1, 2, . . . , m} × {1, 2, . . . , m}, (i, i) /∈ syn for 1 ≤ i ≤ m, are synapses
between neurons.
4. out ∈ {1, 2, . . . , m} is the index of the output neuron.
A spiking rule is a rule E/a c → a b where b ≥ 1. A forgetting rule is a rule
where b = 0 is written as E/a c → λ. If L(E) = {a c } then spiking and forgetting
rules are simply written as a c → a b and a c → λ, respectively. Applications of
rules are as follows: if neuron σ i contains k spikes, a k ∈ L(E) and k ≥ c, then the
rule E/a c → a b ∈ R i is enabled and the rule can be fired or applied. If b ≥ 1, the
application of this rule removes c spikes from σ i , so that only k −c spikes remain
in σ i . The neuron fires b number of spikes to every σ j such that (i, j) ∈ syn. If
b = 0 then no spikes are produced. SNP systems assume a global clock, so the
application of rules and the sending of spikes by neurons are all synchronized.
The nondeterminism in SNP systems occurs when, given two rules E 1 /a c1 → a b1
and E 2 /a c2 → a b2 , it is possible to have L(E 1 ) ∩ L(E 2 ) ≠ ∅. In this situation,
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