Download Activation Models

NEURONAL DYNAMICS 2:
ACTIVATION MODELS
2002.10.8
Chapter 3. Neuronal Dynamics 2 :Activation Models
3.1 Neuronal dynamical system
Neuronal activations change with time. The way
they change depends on the dynamical equations
as following:

x  g（FX ，FY ，
）
(3-1)

y  h（FX ，FY ，
）
(3-2)
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3.1 ADDITIVE NEURONAL DYNAMICS
first-order passive decay model
In the absence of external or neuronal stimuli,
the simplest activation dynamics model is:

x i   xi

y
j
 yj
(3-3)
(3-4)
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3.1 ADDITIVE NEURONAL DYNAMICS
since for any finite initial condition
xi (t )  xi (0)e
t
The membrane potential decays
exponentially quickly to its zero potential.
Passive Membrane Decay
Passive-decay rate
Ai  0 scales the rate to
the membrane’s resting potential.

xi   Ai xi
i t） x（
i 0）e
x（
 Ai t
solution :
Passive-decay rate measures: the cell membrane’s
resistance or “friction” to current flow.
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property
Pay attention to Ai property
The larger the passive-decay rate,the faster the
decay--the less the resistance to current flow.
Membrane Time Constants
The membrane time constant C i scales the time
variable of the activation dynamical system.
The multiplicative constant model:

C i x i  -A i x i
（3－8）
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Solution and property
solution
x i (t )  xi (0)e

Ai
t
Ci
property
The smaller the capacitance ,the faster things change
As the membrane capacitance increases toward
positive infinity,membrane fluctuation slows to stop.
Membrane Resting Potentials
Definition
Define resting Potential Pi as the activation value
to which the membrane potential equilibrates in the
absence of external or neuronal inputs:

C i x i  -A i x i  Pi
(3-11)
Solutions
x i (t)  x i (0)e
-
Ai
t
Ci
Pi

(1 - e
Ai
-
Ai
t
Ci
)
(3-12)
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Note
The capacitance appear in the index of the
solution,it is called time-scaling capacitance.
It does no affect the steady-state solution and
does not depend on the finite initial condition.
In resting case,we can find the solution quickly.
Additive External Input
Add input
Apply a relatively constant numeral input to a neuron.

x i  -x i  I i
(3-13)
solution
x i (t)  x i (0)e-t  I i (1- e -t )
(3-14)
Meaning of the input
Input can represent the magnitude of directly
experiment sensory information or directly apply
control information.
The input changes slowly,and can be assumed
constant value.
3.2 ADDITIVE NEURONAL FEEDBACK
Neurons do not compute alone. Neuron modify their
state activations with external input and with the feedback
from one another.
This feedback takes the form of path-weighted signals
from synaptically connected neurons.
Synaptic Connection Matrices
n neurons in field FX
p neurons in field FY
The ith neuron axon in FX
jth neurons in FY
a synapse m ij
m ij is constant,can be positive,negative or zero.
Meaning of connection matrix
The synaptic matrix or connection matrix M is an
n-by-p matrix of real number whose entries are the
synaptic efficacies m ij.the ijth synapse is excitatory
if
m ij  0
m ij  0 inhibitory if
The matrix M describes the forward projections from
neuron field FX to neuron field FY
The matrix N describes the feedforward projections
from neuron field FY to neuron field F
X
Bidirectional and Unidirectional connection
Topologies
Bidirectional networks
M and N have the same or approximately the same
structure. N  M T
M  NT
Unidirectional network
A neuron field synaptically intraconnects to itself.
BAM
M is symmetric,
network is BAM
M  MT
the unidirectional
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Augmented field and augmented matrix
Augmented field
FX
FY
FZ  

FX

FY 
M connects FX to FY ,N connects FY to FX then the
augmented field FZ intraconnects to itself by the square
block matrix B
0
B  
N
M

0
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Augmented field and augmented matrix
In the BAM case,when N  M then B  BT hence
a BAM symmetries an arbitrary rectangular matrix M.
T
In the general case,
P
C  
N
M

Q
P is n-by-n matrix.
Q is p-by-p matrix.
T
If and only if, N  M T P  PT Q  Q the neurons in FZ
are symmetrically intraconnected
C  CT
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3.3 ADDITIVE ACTIVATION MODELS
Define additive activation model
n+p coupled first-order differential equations defines the
additive activation model

y j  -A j y j 

x i  -A i x i 
p
 S ( x )m
i
i
ij
Ij
（3－15）
 Ii
（3－16）
j 1
p
 S ( y )n
j
j
ji
j 1
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additive activation model define
The additive autoassociative model correspond to a system
of n coupled first-order differential equations

p
x i  -Ai x i   S j ( x j )m ji  I i
（3－17）
j 1
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additive activation model define
A special case of the additive autoassociative
model

where Ri'
x j  xi
 r I
 S j ( x j )mij  I i
xi
Ci x i   
Ri
xi
 ' 
Ri
(3-18)
i
ij
j
j
is
n
1
1
1
 
'
Ri Ri
j rij
(3-19)
（3-20）
rij
measures the cytoplasmic resistance between
neurons i and j.
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Hopfield circuit and continuous additive bidirectional
associative memories
Hopfield circuit arises from if each neuron has a strictly
increasing signal function and if the synaptic connection
matrix is symmetric

xi
Ci xi   '   S j ( x j )mij  I i
Ri
j
(3-21)
continuous additive bidirectional associative memories

p

j 1
p
x i  -Ai x i   S j ( y j )mij  I i
y j  -Aj y j   Si ( xi )mij  I j
(3-22)
(3-23)
i 1
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3.4 ADDITIVE BIVALENT FEEDBACK
Discrete additive activation models correspond to neurons
with threshold signal function
The neurons can assume only two value: ON and OFF.
ON represents the signal value +1. OFF represents 0 or –1.
Bivalent models can represent asynchronous and stochastic
behavior.
Bivalent Additive BAM
BAM-bidirectional associative memory
Define a discrete additive BAM with threshold signal
functions, arbitrary thresholds and inputs,an arbitrary but
constant synaptic connection matrix M,and discrete time
steps k.
p
xik 1   S j ( y kj )mij  I i
(3-24)
j 1
p
y kj 1   Si ( xik )mij  I j
(3-25)
i 1
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Bivalent Additive BAM
Threshold binary signal functions
 1

k
Si ( xi )  Si ( xik 1 )
 0

 1

k
S j ( y j )  S j ( y kj 1 )
 0

if
if
xik  U i
xik  U i
if
xik  U i
if
if
y kj  V j
y kj  V j
if
y kj  V j
(3-26)
(3-27)
For arbitrary real-value thresholds U  U1 , , U n 
for neurons FX V  V1 , , V p for neurons FY


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A example for BAM model
Example
A 4-by-3 matrix M represents the forward synaptic
projections from FX to FY .
A 3-by-4 matrix MT represents the backward synaptic
projections from FY to FX .
2
 3 0


 1 2 0 
M 
0
3
2


  2 1  1


  3 1 0  2


T
M  0 2 3 1 
 2

0
2

1


A example for BAM model
Suppose at initial time k all the neurons in FY are ON.
So the signal state vector S (Yk ) at time k corresponds to
S (Yk )  (1 1 1)
Input
X k  ( x , x , x , x ,)  (5  2 3 1)
k
1
k
2
k
3
k
4
Suppose
Ui  V j  0
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A example for BAM model
is:
first:at time k+1 through synchronous operation,the result
S ( X k )  (1 0 1 1)
next:at time k+1 ,theseFX signals pass “forward” through the
filter M to affect the activations of the FY neurons.
The three neurons compute three dot products,or correlations.
The signal state vector S ( X k ) multiplies each of the three
columns of M.
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A example for BAM model
the result is:
4
S ( X k ) M  (
i 1
Si ( xik )mi1 ,
 (5
 ( y1k 1
4
3)
y2k 1
4

i 1
Si ( xik )mi 2 ,
4
k
S
(
x
 i i )mi3 )
i 1
y3k 1 )
 Yk 1
synchronously compute the new signal state vector S (Yk 1 ):
S (Yk 1 )  (0 1 1)
A example for BAM model
the signal vector passes “backward” through the synaptic
filter S (Yk 1 ) at time k+2:
S (Yk 1 )M T  (2  2 5 0)
 ( x1k 2
 X k 2
x2k 2
x3k 2
x4k 2 )
synchronously compute the new signal state vector
S ( X k  2 )  (1 0 1 1)  S ( X k )
:
A example for BAM model
since S ( X k 2 )  S ( X k ) then
S (Yk 3 )  S (Yk 1 )
conclusion
These same two signal state vectors will pass back and
forth in bidirectional equilibrium forever-or until new
inputs perturb the system out of equilibrium.
A example for BAM model
asynchronous state changes may lead to different
bidirectional equilibrium
keep the first FY neurons ON,only update the second
and third FY neurons. At k,all neurons are ON.
Yk 1  S ( X k ) M  ( 5 4 3)
new signal state vector at time k+1 equals:
S (Yk 1 )  (1 1 1)
A example for BAM model
new FX
activation state vector equals:
X k 2  S (Yk 1 ) M T  ( 1  1 5  2)
synchronously thresholds
S ( X k 2 )  (0 0 1 0)
passing this vector forward to FY gives
Yk 3  S ( X k 2 ) M  (0 3 2)
S (Yk 3 )  (1 1 1)
 S (Yk 1 )
A example for BAM model
similarly,
S ( X k 4 )  S ( X k 2 )  (0 0 1 0)
for any asynchronous state change policy we apply to the
neurons FX
the system has reached a new equilibrium,the binary pair
(0
0 1 0), (1 1 1)represents a fixed point of the system.
conclusion
conclusion
Different subset asynchronous state change policies applied
to the same data need not product the same fixed-point
equilibrium. They tend to produce the same equilibria.
All BAM state changes lead to fixed-point stability.
Bidirectional Stability
definition
A BAM system ( Fx , F y , M ) is Bidirectional stable if all
inputs converge to fixed-point equilibria.
A denotes a binary n-vector in
B denotes a binary p-vector in
0,1n
0,1p
Bidirectional Stability
Represent a BAM system equilibrates to bidirectional fixed
point
Af
) as
 M
 MT
 M
 MT

 M
Af

( Af , B f
A
A'
A'
A ''
MT





B
B
B'
B'

Bf

Bf
Lyapunov Functions
Lyapunov Functions L maps system state variables to real
numbers and decreases with time. In BAM case,L maps the
Bivalent product space to real numbers.
Suppose L is sufficiently differentiable to apply the chain
rule:

L
n

i
L dxi

xi dt

i
L 
xi
xi
(3-28)
Lyapunov Functions
The quadratic choice of L
1
1
T
L  xIx 
2
2

x i2
（3-29）
i
Suppose the dynamical system describes the passive
decay system.

xi   xi
（3-30）
The solution
xi (t )  xi (0)e t
（3-31）
Lyapunov Functions
The partial derivative of the quadratic L:
L
 xi
xi

L

i
（3-32）

xi2 （3-33） or L  


i
2
xi
(3-34)
(3-35)
In either case L  0

(3-36)
At equilibrium
L0
This occurs if and only if all velocities equal zero

xi  0
conclusion
A dynamical system is stable if some Lyapunov Functions L
decreases along trajectories.
A dynamical system is asymptotically stable if it strictly
decreases along trajectories
Monotonicity of a Lyapunov Function provides a sufficient
condition for stability and asymptotic stability.
Linear system stability
For symmetric matrix A and square matrix B,the quadratic
T
L

xAx
form
behaves as a strictly decreasing Lyapunov

function for any linear dynamical system x  xB if and
only if the matrix ABT  BA is negative definite.
T

L  xA x  x Ax T
 xAB T x T  xBAx T
 x[ AB T  BA]x T
The relations between convergence rate
and eigenvalue sign
A general theorem in dynamical system theory relates
convergence rate and ergenvalue sign:
A nonlinear dynamical system converges exponetially
quickly if its system Jacobian has eigenvalues with negative
real parts. Locally such nonlinear system behave as linearly.
A Lyapunov Function summarizes total system behavuor.
A Lyapunov Function often measures the energy of a
physical sysem.
Potential energy function represented by quadratic form
Consider a system of n variables and its potential-energy
function E. Suppose the coordinate x i measures the
displacement from equilibrium of ith unit.The energy depends
on only coordinate x i ,so E  E ( x1 ,  xn )
since E is a physical quantity,we assume it is sufficiently
smooth to permit a multivariable Taylor-series expansion
about the origin:
Potential energy function represented by quadratic form
E  E (0, ,
0) 

i
1

3!
1

2

i
j

i
j
k
2
E
1
xi 
x i
2

i
j
3E
xi x j x k  
xi  j  k
 E
xi x j
xi x j
1
 xAx T
Where2A is symmetric,since
2E
2E
aij 

 a ji
xi x j x j xi
2E
xi x j
x i x j
The reason of (3-42)follows
First,we defined the origin as an equilibrium of zero potential
energy;so E (0, , 0)  0
Second,the origin is an equilibrium only if all first partial
derivatives equal zero.
Third,we can neglect higher-order terms for small
displacement,since we assume the higher-order products are
smaller than the quadratic products.
Bivalent BAM theorem
The average signal energy L of the forward pass of theFX
Signal state vector S ( X ) through M,and the backward pass
Of the FY signal state vector S (Y ) through M T :
S ( X ) MS (Y ) T  S (Y ) MS ( X ) T
L
2
since S (Y ) M T S ( X ) T  [S (Y ) M T S ( X )T ]T
 S ( X ) M T S (Y )T
L  S ( X )M T S (Y )T  
n
p
 S ( x )S ( y )m
i
i
j
i
j
j
ij
Lower bound of Lyapunov function
The signal is Lyapunov function clearly bounded below.
For binary or bipolar,the matrix coefficients define the
attainable bound:
L
mij

i
j
The attainable upper bound is the negative of this expression.
Lyapunov function for the general BAM system
The signal-energy Lyapunov function for the general BAM
system takes the form
L  S ( X ) MS(Y )T  S ( X )[I  U ]T  S (Y )[J  V ]T
Inputs I  [ I1 , , I N ] and J  [ J 1 , , J P ] and
constant vectors of thresholds U  [U1 , , U N ] V  [V1 , , VN ]
the attainable bound of this function is.
L
 m  [ I
ij
i
j
i
i
 Ui ] 
[ J
j
j
V j ]
Bivalent BAM theorem
Bivalent BAM theorem.every matrix is bidrectionally stable
for synchronous or asynchronous state changes.
Proof consider the signal state changes that occur from
time k to time k+1,define the vectors of signal state
changes as:
S (Y )  S (Yk 1 )  S (Yk )
 S1 ( y1 ), , S p ( y p ) ,
S ( X )  S ( X k 1 )  S ( X k )
 S1 ( x1 ), , S n ( x n ) ,
Bivalent BAM theorem
define the individual state changes as:
S j ( y j )  S j ( y j k 1 )  S j ( y j k )
Si ( xi )  Si ( xi
k 1
)  S i ( xi )
k
We assume at least one neuron changes state from k
to time k+1.
Any subset of neurons in a field can change state,but in
only one field at a time.
For binary threshold signal functions if a state change
is nonzero,
Bivalent BAM theorem
Si ( xi )  1  0  1 Si ( xi )  0  1  1
For bipolar threshold signal functions
S i ( xi )  2
Si ( xi )  2
The “energy”change
L  Lk 1  Lk
L
Differs from zero because of changes in field FX or in
field FY
Bivalent BAM theorem
L  S ( X ) MS(Yk )T  S ( X )[I  U ]T
 S ( X )[S (Yk ) M T  [ I  U ]]T

 S ( x ) I   S ( x )U
  S ( x ) S ( y ) m   S ( x ) I   S ( x )U
   S ( x )[  S ( y ) m  I  U ]
  Si ( xi )[ xik 1  U i ]

S i ( xi ) S j ( y kj ) T mij 
i
j
i
i
i
i
i
i
i
k T
i
i
i
ij
j
j
i
i
0
j
i
k T
i
j
j
j
i
i
i
i
i
i
ij
i
i
i
i
i
Bivalent BAM theorem
Suppose S i ( xi )  0
Then Si ( xi )  Si ( xi
k 1
)  Si ( xi k )
 1 0
k 1
This implies xi  U i so the product is positive:
Si ( xi )[xik 1  U i ]  0
Another case suppose S i ( xi )  0
Si ( xi )  Si ( xi
k 1
 0 1
)  Si ( xi )
k
Bivalent BAM theorem
k 1
This implies xi
 Ui
so the product is positive:
Si ( xi )[xik 1  U i ]  0
So Lk 1  Lk  0 for every state change.
Since L is bounded,L behaves as a Lyapunov function for
the additive BAM dynamical system defined by before.
Since the matrix M was arbitrary,every matrix is
bidirectionally stable. The bivalent Bam theorem is proved.
Property of globally stable dynamical system
Two insights about the rate of convergence
First,the individual energies decrease nontrivially.the BAM
system does not creep arbitrary slowly down the toward the
nearest local minimum.the system takes definite hops into
the basin of attraction of the fixed point.
Second,a synchronous BAM tends to converge faster
than an asynchronous BAM.In another word, asynchronous
updating should take more iterations to converge.