Biology Reference
In-Depth Information
Example 1.9.
Repeat Example
1.8
with the following system of polynomial equa-
tions.
x
2
y
z
3
−
=
0
2
xy
−
4
z
−
1
=
0
y
2
z
−
=
0
x
3
−
4
zy
=
0
.
After entering the appropriate commands as in Example
1.8
for this system of equa-
tions,
SAGE
returns the Groebner basis [
1
] for the functions
f
1
x
2
y
z
3
=
−
;
f
2
=
y
2
x
3
2
xy
4
zy
.
This means that the solution set of the system of equations above is equivalent to
the solution set of the equation 1 = 0, indicating that the solution set is empty. The
system does not have a real-valued solution.
Example 1.10.
Consider again Exercise
1.19
where you wrote the polynomial form
of the Boolean model of the
lac
operon given by Eqs. (
1.7
). If in those equations we
rename the variables
M
−
4
z
−
1
;
f
3
=
z
−
;
f
4
=
−
x
9
(in this order) and the parameters
L
e
and
G
e
to
a
and
g
, respectively, and rewrite the
equations in a form where the right-hand side is zero, we will obtain
3
,
P
,
B
,
C
,
R
,
A
,
A
l
,
L
,
L
l
to
x
1
,
x
2
,
x
3
,
x
4
,
x
5
,
x
6
,
x
7
,
x
8
,
x
1
+
x
4
∗
x
5
+
x
4
=
0
x
1
+
x
2
=
0
,
x
1
+
x
3
=
0
x
4
+
(
g
+
1
)
=
0
x
5
+
x
6
∗
x
7
+
x
6
+
x
7
+
1
=
0
x
6
+
x
3
∗
x
8
=
0
(1.13)
x
6
+
x
7
+
x
8
+
x
9
+
x
8
∗
x
9
+
x
6
∗
x
8
+
x
6
∗
x
9
+
x
6
∗
x
8
∗
x
9
=
0
x
8
+
(
g
+
1
)
∗
a
∗
x
2
=
0
x
9
+
(
g
+
1
)
∗
(
x
8
+
a
∗
x
8
+
a
)
=
0
.
We now need to solve this system four times, using all four combinations of
a
=
0,1
and
g
0,1 for the parameter values. Using
SAGE
we need to “evaluate” the following
command lines:
P.<x1,x2,x3,x4,x5,x6,x7,x8,x9>=PolynomialRing(GF(2),
9, order = 'lex').
We use GF(2) here since we want to find the solutions of the system of equations
from Eqs. (
1.13
) over the field
=
, often referred to as the Galois field of two
elements (hence the notation). We also use nine to indicate the number of variables.
After setting
a
{
0
,
1
}
=
g
=
0 in the polynomials from Eqs. (
1.13
), we enter the set of
functions:
3
To get the system in this form we used the facts that, over the field {0,1}, 1 + 1 = 0,
−
1
=
1, and
a
−
b=a+b.
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