Biology Reference
In-Depth Information
Example 3.6. Consider a 3-node system having the following states in
Z 5 :
200
431
314
043
which we interpret as
s 1 = 2
0
t 1 = 4
1 ,
,
0
,
,
3
,
s 2 = 4
1
t 2 = 3
4 ,
,
3
,
,
1
,
s 3 = 3
4
t 3 = 0
3 .
,
1
,
,
4
,
The corresponding ideals are
I
(
s 1 ) =
x 1
2
,
x 2 ,
x 3 ,
I
(
s 2 ) =
x 1
4
,
x 2
3
,
x 3
1
,
.
To compute the intersection I of the ideals, we can use the computer algebra system
Macaulay 2 [ 26 ] with the following code:
I
(
s 3 ) =
x 1
3
,
x 2
1
,
x 3
4
R=ZZ/5[x1,x2,x3]
I1=ideal(x1-2,x2,x3)
I2=ideal(x1-4,x2-3,x3-1)
I3=ideal(x1-3,x2-1,x3-4)
I=intersect{I1,I2,I3} .
Then I is computed as
2 x 3 +
x 3 +
x 1 +
2 x 2 +
x 3 +
2
,
2 x 2 x 3
x 3 ,
x 2 x 3
x 2 +
x 3 ,
x 2 +
2 x 2
x 2 x 3 +
2 x 2 +
2 x 3 ,
2 x 2 x 3 +
x 2 ,
x 1 x 2 +
x 1 x 3 +
x 2 +
x 3 ,
2
2 x 1 x 2 +
2 x 1 x 3
x 1
x 2 +
x 3 +
.
Note that negative coefficients can be written as positive numbers:
1
4mod5
,
2
3 mod 5, etc.
To find the function f 1 (
x 1 ,
x 2 ,
x 3 )
for node x 1 such that
f 1 2
0 =
,
0
,
4
,
f 1 4
1 =
,
3
,
3
,
(3.1)
f 1 3
4 =
,
1
,
0
,
we compute the r polynomials:
r 1 x 1 ,
x 3 = 2
4 3 x 1
4 2
3 3 x 1
3 =
3 x 1 +
x 2 ,
4 x 1 +
1
,
r 2 x 1 ,
x 3 = 4
2 3 x 1
2 4
3 3 x 1
3 =
3 x 1 +
x 2 ,
3
,
r 3 x 1 ,
x 3 = 3
2 3 x 1
2 3
4 3 x 1
4 =
4 x 1 +
x 2 ,
x 1 +
2
.
 
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