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7/29/2019 hopfield 1slide

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Neurodynamics

Dynamical System:A dynamical system may be defined as adeterministic mathematical prescription for

evolving the state of a system forward in time

)......,,,(

......

)......,,,(

)......,,,(

21

2122

2111

NNN

N

N

xxxFdt

dx

xxxFdt

dx

xxxFdt

dx

=

=

=

Continuous System: Time is a continuous variable

or = F(x, ) xN

x

With x U N , V Pwhere Uand Vare open sets in N and P


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Neurodynamics

Discrete System: Time is a discrete integer-valued variable

xN+1 = F(xN)

where x is aNdimensional vector. Given an initial state x0,

we can generate an orbit of the discrete time systemx0, x1, x2, .... by iterating the map.

For example: xN+1 = -0.5xN


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Continuous Dynamical Systems

Linearized Stability of Equilibrium Solutions

fixed point(stationary pointor equilibrium point):

is an equilibrium solution of a point xf(t) N such thatF(xf(t)) = 0.

For example (Bonhoeffer-van der Pol neuron model):dx/dt=x - x3/3 - y +A0dy/dt= c(x + a - by)

if b < 1 and 3a+2b 3, only one fixed point


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x - x3/3 - y +A0 = 0

x + a - by = 0

Fixed point equation:

x3/3 -x + (x + a)/b =A0

x3 - qx - p = 0

whereq

= 3(b-1)/

band

p= 3(

A0 -

a/

b).

It has three real roots for 27p2 < 4q3 and one real root and two

complex roots for 27p2

> 4q3

. For the specific choice of theparameters (a = 0.7, b = 0.8, c = 0.1), only one real root and

consequently the system possesses one fixed point only, (xp, yp).


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Action potential characteristics. Current of amplitudeA0 = 0.02 is turned on at

t= 100 and turned off at t= 210, no action potential occurs.A0 = 0.1 is turned

on at t= 300 and turned off at t= 310, no action potential occurs.A0 = 0.2 is

turned on at t= 400 and turned off at t= 410, an action potential.


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y = x - xf(t)

= F(xf(t)) +DF(xf(t))y + (|y|2)

x

+= yxx )( tf

= F(xf(t)),

)(tfx

=DF(xf(t))y + (|y|2)

y


For the dynamical system:

= F(x, ) xN

x (1)

Let (2)

Where xf(t) is a fixed point.

Then,

Using the Taylor expanding about xf(t) gives

Using the factor that we get

(3)

(4)


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=DF(xf(t))y

y

Ayy =


(5)

(6)

Since y is small, so it is reasonable that the stability of the

original system could be answered by studying the associated

linear system:

SinceDF(xf(t)) =DF(xf), then

(7)

0)()( yy x tDF fet =Then,


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( ) ( )

( ) ( )

==xx

xx

xA

n

nn

n

x

f

x

f

x

f

x

f

DF

...

.........

...

)(

1

1

1

1

A is called a Jacobian matrix (partial derivatives):


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=

=N

k

kkk teAt1

)exp()( y

Ae = e

We have an eigenvalue equation:

(8)

(9)

=

=N

k

kkeAy1

0where Akare determined from the initial condition:

k(k = 1, 2, ......, N) are eigenvalues ofA



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Definition 1: (Lyapunov stability) xf(t) is said to be stable or

Lyapunov stable, if given > 0, there exists a = () > 0 such

that, for any other solution, y(t), of the system (2-2), satisfying

|xf(t0) - y(t0)| < , then |xf(t) - y(t)| < for t > t0, t0 .

Definition 2: (asymptotic stability) xf(t) is said to be asymptotic

stable if it is lyapunov stable and if there exists a constant b > 0such that if |xf(t0) - y(t0)| < b, then |xf(t) - y(t)| = 0 when t .


Definition 3: Let x = xfbe a fixed point of = F(x), then, xfis

called a hyperbolic fixed point if none of the eigenvalues ofDF(xf) have zero real part. (a hyperbolic fixed point of a

N-dimensional map is defined as: none of its eigenvalues have

absolute one)


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Stability of Equilibrium State

Theorem 1: Suppose all of the eigenvalues ofA in Eqn.

(7)

have negative real parts. Then the equilibrium solution xf

of the nonlinear flow defined by Eqn.(1) is asymptotically

stable.Theorem 2: if one of the eigenvalues ofA has a positive real

part, the fixed point is unstable


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For a two-dimensional continuous system, the stability of

a fixed point can be classified by its eigenvalues of the

Jacobian matrix evaluated at a fixed point in three types:

The eigenvalues are real and have the same sign. If theeigenvalues are negative, this is a stable point; if the

eigenvalues are positive, this is a unstable point.

The eigenvalues are real and have opposite signs. In this

case, there is a one-dimensional stable manifold and a one-

dimensional unstable manifold. This fixed point is called a

unstable saddle point.

The eigenvalues are complex conjugates. If the real part ofthe eigenvalues is negative, this is a stable spiral or a spiral

sink. If the real part is positive, this is an unstable spiral or a

spiral source. If the real parts are zero, then this is a center.


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y

x

y

x


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y

x


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dx/dt=x - x3/3 - y +I(t)

dy/dt= c(x + a - by)

(a = 0.7, b = 0.8, c = 0.1)

= bccxJ 11

2

1, 2 = -[(bc - 1 +x2) ((x2 -1 + bc)2 - 4c)1/2] /2


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For 0


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Lyapunov Function

According to the definition of stability it would be

sufficient to find a neighborhood Uofxffor which

orbits starting in Uremain in Ufor positive times.

This condition would be satisfied if we could show

that the vector field is either tangent to the boundary

ofUor pointing inward toward xf.


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Lyapunov Function

Theorem: Consider the following continuous dynamical system

= F(x) xN

Let xfbe a fixed point and let V: U be a continuousdifferentiable function defined on some neighborhood Uofxf

such thati) V(xf) = 0 and V(x) > 0 ifx xf.ii) in U-{xf}.

Then xf is stable. Moreover, ifiii) in U-{xf}.

Then xf is asymptotically stable.

0)(

xV

0)( 0in a neighborhood of (xf, yf).


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Lyapunov Function

V = C3

V = C2

(xf, yf)

V = C1


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Lyapunov Function

The gradient ofV, V, is a vector perpendicular to the

tangent vector along each curve ofVwhich points in

the direction of increasing V. So if the vector field were

always to be either tangent or pointing inward for each

of these curves surrounding (xf, yf), we would have

0,xy)V(x,

y


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Lyapunov Function

V

VVV

VV

V

V

(xf, yf)


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Discrete Dynamical Systems


fixed point(stationary pointor equilibrium point):

is an equilibrium solution of a point xf N

such thatxf= F(xf(t)).

Example 1 (logistic map) : xn+1 = f(xn) = xn(1-xn)

There is only one fixed point: x = 0

Example 2 (logistic map):xn+1 = f(xn) = 2xn(1-xn)There are two fixed points: x = 0 and x = 0.5


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Example 3 (Henon map):

There are two fixed points: (x = 1, y = 1) and (x = -1, y = -1)

+=

+

+

n

nn

n

n

xbyxa

yx

2

1

1

a = 1, b = 1


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xN = xf+ yN

yN+1 =DF(xf)yN + (yN2)

yN+1 = AyN

Ae = e

=

+ =N

k

n

kkkn eAy1

1

Conclusion:

directions corresponding to |k| > 1 are unstable;directions corresponding to |k| < 1 is stable.

(10)

(11)

(12)

(13)

(14)


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Example 1 (logistic map) : xn+1 = f(xn) = xn(1-xn)

There is only one fixed point: x = 0

Since f(x)|x = 0 = 1 - 2x|x = 0 = 1, then x = 0 is a center

Example 2 (logistic map): xn+1 = f(xn) = 2xn(1-xn)

There are two fixed points: x = 0 and x = 0.5

Since f(x)|x = 0 = 2 - 4x|x = 0 = 2, then x = 0 is unstable

Since f(x)|x = 0.5 = 2 - 4x|x = 0.5 = 0, then x = 0 stable(super-stable)


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Example 3 (Henon map):

There are two fixed points: (x = 1, y = 1) and (x = -1, y = -1)

+=

+

+

n

nn

n

n

x

byxa

y

x 2

1

1

( )

=

01

12xDF x

For (x=1 , y = 1) ,21,21 21 =+= unstable

For (x=-1 , y = -1) ,21,21 21 =+= unstable

a = 1, b = 1


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Exercises:

1) Let l(x) = ax+b, where a and b are constants. Foe which

values of a and b does l have na attracting fixed point? Arepelling fixed point?

2) Let f(x) = x- x2. Show that x = 0 is a fixed point of f, and

described the dynamical behavior of points near 0.

3) For each of the following linear maps, decide whether

the origin is a sink, source, or saddle.

6.14.0

4.24.0

43

41

211

31

304


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Exercises:

4) Let g(x, y) = (x2-5x+y, x2). Find and classify the fixed

points of g as sink, source, or saddle.

5) Let f(x, y) = (sin(/3)x, y/2). Find all fixed points and

their stability. Where does the orbit of each initial value go?

6) Find

96

5.32

85.4

lim

n

n


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Hopfield Models

General Idea: Artificial Neural Networks Dynamical Systems

Initial Conditions Equilibrium Points


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Continuous Hopfield Model

i

N

jjij

i

iii I

j

txwR

tx

dt

tdxC +

=+=

1))((

)()(

a) the synaptic weight matrix is symmetric, wij = wji, for all i andj.

b) Each neuron has a nonlinear activation of its own, i.e. yi = i(xi).Here, i() is chosen as a sigmoid function;

c) The inverse of the nonlinear activation function exists, so we may

writex = i-1(y).


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Continuous Hopfield Model

== =

=

+=N

i

ii

N

j

N

i

x

ii

i

N

i

jiij yIdxy

R

yywEi

11 10

1

1

)(1

2

1

Lyapunov Function:

= =

+=

N

i

iN

j

i

i

ijij

dt

dyI

R

xyw

dt

dE

1 1

0

)(

)(

1

12

1

1

=

=

=

=

N

i

iiii

iN

i

iii

dtyd

dtdyC

dt

dy

dt

ydC


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Discrete Hopfield Model

Recurrent network

Fully connected

Symmetrically connected (wij

= wji

, or W = WT)

Zero self-feedback (wii = 0)

One layer

Binary States:

xi = 1 firing at maximum valuexi = 0 not firing

or Bipolar

xi = 1 firing at maximum value

xi = -1 not firing


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(Bipole)

=

=

ij

ijiji

ij

ijij

ij

ijij

i

xwx

xw

xw

x

0

01

01

Transfer Function for Neuron i:

x = (x1, x2 ... xN): bipole vector, network state.

i : threshold value of xi.

=

ijijiji xwx sgn ( )= Wxx sgn


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E w x x xij i j i iij ii

= + 1

2

E w x xij i jj ii

=

1

2

For simplicity, we consider all threshold i = 0:

Energy Function:


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Learning Prescription (Hebbian Learning):

Pattern s = (s1, s2, ..., sn), where si take value 1 or -1

==

M

jiij Nw 1,,

1

{ = 1, 2, ..., M}: M memory patterns

In the matrix form:

IW MN

M

T = =11


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Energy function is lowered by this learning rule:

==

i ijji

i ij

jiij

i ij

jiij wxxwE

2

,

2

,

,,

2

1

21

21


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Pattern Association (asynchronous update):

( ) ( )

++=+=

ij

jiji

i ij

jiij

i ij

jiij

k

xwkx

xkxwxkxw

kEkEE

)(

2

11

2

1)()1(

Ek 0


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Example:

Consider a network with three neurons, the weight

matrix is:

=022

202

220

31W


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3

2

3

2

3

2

3

2

32

3

2

1

2 3


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(1, 1, 1)

(1, -1, 1)

stable

(1, -1, -1)(-1, -1, -1)

(-1, 1, -1)stable

(-1, 1, 1)

(-1, -1, 1)

(1, 1, -1)

The model with three neurons has two fundamental

memories (-1, 1, -1)T and (1, -1, 1)T


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=

=44

4

3

1

11

1

022202

220

3

1Wx

[ ] xWx =

=1

1

1

sgn

State (1, -1, 1)T:

A stable state


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=

=

4

4

4

3

1

1

1

1

022

202

220

3

1Wx

[ ] xWx =

=

1

1

1

sgn

State (-1, 1, -1)T:

A stable state


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=

=

0

4

0

3

1

1

1

1

022

202

220

3

1Wx

[ ] xWx

=1

1

1

sgn

State (1, 1, 1)T:

An unstable state. However, it converges to its neareststable state (1, -1, 1)T


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=

=

4

0

0

3

1

1

1

1

022

202

220

3

1Wx

[ ]

=1

1

1

sgn Wx

State (-1, 1, 1)T:

An unstable state. However, it converges to its neareststable state (-1, 1, -1)T


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[ ] [ ]

=

+

=

022

202

220

3

1

100

010

001

32111

1

1

1

31111

1

1

1

31W

Thus, the synaptic weight matrix can be determined by

the two patterns:

C E i


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Computer Experiments


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Significance of Hopfield Model

1) The Hopfield model establishes the bridge between

various disciplines.

2) The velocity of pattern recalling in Hopfield models

is independent on the quantity of patterns stored in

the net.


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Limitations of Hopfield Model

2) Spurious memory;

3) Auto-associative memory;

4) Reinitialization

5) Oversimplification

1) Memory capacity;

The memory capacity is directly dependent on the number

of neurons in the network. A theoretical result is

NNplog2

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