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let herbivore start with a positive score at each step (while carnivores has to meet
food to survive) [12].
Sexual reproduction implies some modication in the reproductive phase. One
can simulate haploid or diploid individuals, with explicit sex determination of just
with a recombination among the genomes of the two parents.
Also the displacement may depend on the phenotypic distribution near a given
location: preys may try to escape predators, who are pursuing the rst. Dierent
sexes may nd convenient to stay nearby, and so on.
With the assumption of \well mixed" population, i.e., disregarding spatial cor-
relation, and in the limit of very large population the dynamics is described by the
mutation-selection equation [13] (mean-eld approach)
X
1
N
K
n(x; t + 1) =
A(x; n(t))
M(xjy)n(y; t + 1) ;
(15.3)
y
where n(x; t) is the number of individuals with genome x at time t, M is the
mutation matrix (
P
x
M(xjy) = 1) and A(x; n(t)) if the tness of genotype x (or
better: of the relative phenotype), given the rest of the population n(t). In the
following, we shall neglect to indicate the time dependence, and use a prime to
indicate quantities computed one time step further.
The quantity K denotes the carrying capacity, and N =
P
x
n(x) is the total
number of individuals. The logistic term 1N=K implies that there is no repro-
duction in the absence of free space, and therefore models the competition among
all individual for space. In the limit of populations whose size is articially kept
constant, this term may be included into A.
In vector terms, Eq. (15.3) can be written as
1
N
K
n
0
=
AMn ;
where A is a diagonal matrix.
By summing over x, we obtain
1
N
K
N
0
=hAi
N ;
(15.4)
P
x
A(x; n)n(x)=N
i.e., the logistic equation, where the reproduction ratehAi=
depends on the population and therefore on time.
By dividing Eq. (15.3) by Eq. (15.4), and introducing the frequencies p(x) =
n(x)=N, we get
X
p
0
(x) =
A(x; n(t))
hAi
M(xjy)p(y) :
(15.5)
y
or
A
hAi
Mp
p
0
=
