tan. If a cycle has length one (i.e. the element appearing in the cycle is a fixed point of
the corresponding permutation) we omit that cycle from writing. For example, and ∗
are the fixed points of the permutation (+, −). The corresponding permutation groups
are:
G(cs(v
1
),v
1
)=(+, −), (∗, ),
G(cs(v
2
),v
2
)=(ln, sin, cos)(tan, cot), (tan, cot)(sin, cos), (ln, sin),
G(cs(v
3
),v
3
)=(+, , ∗), (−, ∗),
G(cs(v
4
),v
4
)=(x, w)(y, z), (+, ∗)(−, ),
G(cs(v
5
),v
5
)=(x, y, z), (+, −, ∗, ), and
G(cs(v
6
),v
6
)=(x, w)(y, z), (x, w)(sin, cos, tan, cot), (+, − ).
The cycle decomposition makes it easy to compute the corresponding orbits:
O(cs(v
1
),v
1
, +) = {+, −}, O (cs(v
2
),v
2
, ln) = {ln, sin, cos},
O(cs(v
3
),v
3
, )={+, −, ∗, }, O (cs(v
4
),v
4
,x)={x, w},
O(cs(v
5
),v
5
,x)={x, y, z} and O(cs(v
6
),v
6
,y)={y, z} .
Now suppose the initial population is the same as in example 4.7.2. In ord er to apply
theorem 4.9.2, for every node v of u we need to compute the number |os(v, o)(P )|.
Recall that the schema os(v, o) is obtained from the schema os(v) by attaching the
operation o at the node v and labelling its children nodes b y the # signs. For the pop-
ulation P in example 4.7.2 it was already computed that |os(v
1
, +)| = |os(v
1
)| =2.
There are exactly 2 individuals (namely x
3
and x
4
) fitting the schema os(v
1
, ∗) and
so |os(v
1
, ∗)| =2. Exactly one individual, namely x
2
, and one individual, namely
x
6
, fit the schemata os(v
1
, sin) and os(v
2
, cos) respectively and so |os(v
1
, sin)| =
|os(v
1
, cos)| =1. For all the other operations o ∈ (Σ
0
∪ Σ
1
∪ Σ
2
) −{+, ∗, sin, cos}
there are no individuals in P fi tting the schema os(v
1
,o) and so we have |os(v
1
,o)| =
0. There are no individuals in P fitting the schema os(v
2
, ln) = os(v
2
) for the
individual u of the current example, and no individuals fitting the schemata of the
form os(v
2
,o) wher e o/∈{∗, sin, cos} so that for such o we have |os(v
2
,o)(P )| =
0. Moreover, there is exactly one individual, namely x
1
fitting the schema os(v
2
, ∗)
and exactly one, namely x
4
fitting the schema os(v
2
, cos) so that |os(v
2
, ∗)(P )| =
|os(v
2
, cos)(P )| =1; exactly two individuals, namely x
3
and x
5
fit the schema
os(v
2
, sin) so that |os(v
2
, sin)(P )| =2. Continuing in this manner with the rest
of the nodes of u we obtain |os(v
3
,o)(P )| =0for o/∈{+, ∗}; x
1
and x
3
fit
os(v
3
, +) while x
4
and x
5
fit os(v
3
, +) and so |os(v
3
, +)(P )| = |os(v
3
, ∗)(P )| =2.
|os(v
4
,o)(P )| =0for o/∈{x, y, +}; x
3
is the only individual fitting the schema
os(v
4
,y), x
4
is the only individual fi tting the schema os(v
4
,x) and x
5
is the only indi-
vidual fitting the schema os(v
4
, +) and so we have |os(v
4
,x)(P )| = |os(v
4
,y)(P )| =
|os(v
4
, +)(P )| =1. |os(v
5
,o)(P )| =0for o/∈{∗,x,y}; x
1
is the only individual
fitting the schema os(v
5
, ∗) and x
5
is the only individual fitting the schema os(v
5
,y)
while the individuals x
3
and x
4
are the only two which fit the schema os(v
5
,x) so that
we have |os(v
5
, ∗)(P )| = |os(v
5
,y)(P )| =1and |os(v
5
,x)(P)| =2. |os(v
6
,o)(P )| =
43 / 66
Del 8.3 Report on Population dynamics in EvE
DBE Project (Contract n° 507953)