Biology Reference
In-Depth Information
F p that describes moving
one space forward on an n -by- n grid, we need to show that
In order to show that Eq. ( 5.2 ) is the polynomial over
f
(
x
) =
x
+
1 for all
x
=
n
=
1 and f
(
x
) =
0for x
=
n
1. Note that p
=
0 and
(
p
1
) !=−
1in
F p .
￿ Case: x
=
n
1
:
f
(
x
) =
1
+
x
+
n
(
0
) =
1
+
x .
￿ Case: x
=
n
1:
p
1
f
(
n
1
) =
1
+
n
1
+
n
i
n
+
1
i = 0 , i = n 1
=
n
+
n
(
1
n
)(
2
n
) ··· (
n
2
n
+
1
)
× (
n
n
+
1
)(
n
+
1
n
+
1
) ··· (
p
1
n
+
1
)
=
n
+
n
(
p
+
1
n
)(
p
+
2
n
) ··· (
p
1
)(
1
)(
2
) ··· (
p
n
)
=
n
+
n
(
1
)(
2
) ··· (
p
n
)(
p
+
1
n
) ··· (
p
1
)
=
n
+
n
(
p
1
) !=
n
+
n
(
1
) =
n
n
=
0
.
Exercise 5.30.
Show that Eqs. ( 5.3 )-( 5.6 ) indeed describe the movements as stated
above.
5.9.2 Uphill and Downhill Movement
In many applications, agents can scan their close environment and move towards a
desired resource. Netlogo has this behavior implemented as “uphill” and “downhill.”
“Uphill” moves an agent to the neighboring patch with the highest value of the desired
variable. If no neighboring patch has a higher value than the current patch, the agent
stays put. Since the variables are discrete, there may exist a tie between neighboring
patches for highest (or lowest) concentration, in which case the agent moves toward
the lowest arbitrarily numbered neighboring patch i . The polynomials describing the
movement of agent x are the following.
￿ Uphill:
8
i
1
(
p
1
p
1
f
(
x
) =
1
(
C
I i )
)
l i
0 (
C
I j )
i
=
1
j
=
p
1
C
1
C
8
p
1
+ (
1
(
C
I 0 )
) ·
l 0 +
i (
k
I i )
m
=
0
k
=
m
+
1
j
=
8
i
1
p
p
1
1
·
1 (
1
(
m
I i )
)
l i
0 (
m
I j )
i
=
j
=
p
1
,
+ (
1
(
m
I 0 )
)
l 0
(5.7)
 
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