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decay. However, a fundamental difference exists between both models. In the present
case interaction lists have a particular order that establishes a global frustration state
in the system. On the contrary, in [16] lists were random which allowed a much
greater diversity of interactions lifetimes. Hence, in that work, power law (scale free)
behavior reflected the absence of a typical lifetime.
N
W
10
0
5
4
3
2
1
0
10
10
10
10
10
10
100000
-2
N=21
10000
10
1000
N=20
-4
100
10
10
-6
10
1
0
5
10
15
20
4
6
8
10
12
14
W
N
N
Fig. 2.
Left:
Frequency of interactions that lasted
iterations in populations with
21
or
20
elements. These histograms were calculated using one population for 10
6
iterations. The first
10
3
where not considered to avoid including transient effects.
Right:
The number of iterations
required to reach the stable configuration for a population with N even. There is an exponential
growth of the number of iterations
N
it
required:
N
it
~ exp(0.8N)
. For a population with
N=20
,
around 10
7
iterations would be required.
τ
In order to better understand this result consider a conjugate formed between cells
i
and
r
. At each time-step each cell has a probability respectively
p
and
q
to find a
higher ranked cell to interact with, and to terminate the former
i-r
conjugate. Hence,
the probability that the
i-r
conjugate terminates is:
P=p+q-pq .
(2)
The probability that a conjugate lives for exactly
τ
time-steps is then:
P
τ
=(1-P)
τ−1
P.
(3)
This equation implies that
any
conjugate displays a typical exponential lifetime decay
behavior:
P
τ
=(1-P)
τ−1
P
≅
In the particular case of the
IL in (1), equation (2) is simplified because
q=1-p,
which leads to
P/(1-P) exp(-P
τ) ∼
exp(-P
τ).
2
.
Hence, in this case
P
varies between 3/4 and 1, whereas in the most general case of
random ILs,
P
can vary between 0 and 1. This is fundamentally different because it
implies that in the former case interactions are short-lived, whereas in the last case
there are interactions that never terminate. In order to calculate
N
τ
,
P
=
1
−
p
+
p
a sum over the
possible interactions has to be considered. Assuming for simplicity that all conjugates
occur with an equal frequency
f
P
, then we get:
b
b
τ
−
1
τ
−
1
N
=
f
(
−
P
)
P
dP
~
(
−
P
)
P
dP
.
(4)
∫
∫
τ
P
a
a
The integral can be integrated by parts. The difference between the two cases is now
in the correct choice of the limits of integration
a
and
b
. In the frustrated case
a=3/4
