Biology Reference
In-Depth Information
x
1
(
t
+
1
)
=
f
x
1
(
x
1
(
t
),
x
2
(
t
),...,
x
n
(
t
))
x
2
(
t
+
1
)
=
f
x
2
(
x
1
(
t
),
x
2
(
t
),...,
x
n
(
t
))
...
x
n
(
t
+
1
)
=
f
x
n
(
x
1
(
t
),
x
2
(
t
),...,
x
n
(
t
)).
(1.9)
A point
(
p
1
,
p
2
,...,
p
n
)
is
a fixed point
for the set of Eqs. (
1.9
), when plugging the
values
(
p
1
,
p
2
,...,
p
n
)
for
x
1
(
t
),
x
2
(
t
),...,
x
n
(
t
)
into the functions
f
x
1
,
f
x
2
,...
f
x
n
,
from Eqs. (
1.9
) returns the same values
(
p
1
,
p
2
,...,
p
n
)
for
x
1
(
t
+
1
),
x
2
(
t
+
1
),...,
x
n
(
t
+
1
)
. In other words,
(
p
1
,
p
2
,...,
p
n
)
is a fixed point for the set of Eqs. (
1.9
)
when
p
1
=
f
x
1
(
p
1
,
p
2
,...,
p
n
)
p
2
=
f
x
2
(
p
1
,
p
2
,...,
p
n
)
...
p
n
=
f
x
n
(
p
1
,
p
2
,...,
p
n
).
(1.10)
We seek a method for determining all points
(
p
1
,
p
2
,...,
p
n
)
that satisfies the con-
dition in Eqs. (
1.10
).
The brute force approach in this context would be to test all possible states for the
fixed-point property defined by Eqs. (
1.10
). However, since the size of the state space
grows exponentially with the number of nodes in the network, we already decided
that going over the entire state space to test all of its elements is not practical (see
Exercise
1.2
). The approach we describe next makes it possible to rephrase the prob-
lem of finding all fixed points
of a Boolean network as a problem of
solving systems of polynomial equations and determining a computationally efficient
way of finding the solutions. This method uses a computational algebra approach and
is based on some fundamental results in abstract algebra and algebraic geometry and
on the use of Groebner bases for solving systems of polynomial equations [
25
].
Equations (
1.10
) state that the fixed points of the model are the solutions
(
p
1
,
p
2
,...,
p
n
)
(
x
1
,
x
2
,...,
x
n
)
=
(
p
1
,
p
2
,...,
p
n
)
of the system of equations
x
1
=
f
x
1
(
x
1
,
x
2
,...,
x
n
)
f
x
1
(
x
1
,
x
2
,...,
x
n
)
−
x
1
=
0
x
2
=
f
x
2
(
x
1
,
x
2
,...,
x
n
)
or
,
equivalently
,
f
x
2
(
x
1
,
x
2
,...,
x
n
)
−
x
2
=
0
...
...
x
n
=
f
x
n
(
x
1
,
x
2
,...,
x
n
)
f
x
n
(
x
1
,
x
2
,...,
x
n
)
−
x
n
=
,
(1.11)
0
where, the functions
f
x
1
,
f
x
2
,...,
f
x
n
, are Boolean functions.
Search WWH ::
Custom Search