1. KANCHANA RANI G
MTECH R2
ROLL No: 08

2. Hopfield Nets
 Hopfield has developed a number of neural networks based on
fixed weights and adaptive activations.
 These nets can serve as associative memory nets and can be used
to solve constraint satisfaction problems such as the "Travelling
Salesman Problem.“
 Two types:
 Discrete Hopfield Net
 Continuous Hopfield Net

3. Discrete Hopfield Net
 The net is a fully interconnected neural net, in the sense that each
unit is connected to every other unit.
 The net has symmetric weights with no self-connections, i.e.,
and

4.  Hopfield net differ from iterative auto associative net
in 2 things.
1. Only one unit updates its activation at a
time (based on the signal it receives from each other
unit)
2. Each unit continues to receive an external
signal in addition to the signal from the other units in
the net.

5.  The asynchronous updating of the units allows a function, known
as an energy function, to be found for the net.
 The existence of such a function enables us to prove that the net
will converge to a stable set of activations, rather than oscillating.
 The original formulation of the discrete Hopfield net showed the
usefulness of the net as content-addressable memory.

6. Architecture

7. Algorithm
There are several versions of the discrete Hopfield net.
 Binary Input Vectors
To store a set of binary patterns s ( p ) ,
p = 1 , . . . , P, where
))().....().....(()( 1 pspspsps ni

8.  The weight matrix W = is given by
and
}{ ijw
]12][12[ )()( pj
p
piij ssw for ji
.0iiw

9.  Bipolar Inputs
To store a set of binary patterns s ( p ) ,
p = 1 , . . . , P, where
The weight matrix W = is given by,
for
and
))().....().....(()( 1 pspspsps ni
}{ ijw
)()( pj
p
piij ssw ji
0iiw

10.  Application algorithm is stated for binary patterns.
 The activation functions can be modified easily to accommodate
bipolar patterns.

11. Application Algorithm for the Discrete Hopfield Net
Step 0. Initialize weights to store patterns.
(Use Hebb rule.)
While activations of the net are not converged, do Steps 1-7.
Step 1. For each input vector x, do Steps 2-6.
Step 2. Set initial activations of net equal to the
external input vector x:
, ( i=1,2…n)
Step 3. Do Steps 4-6 for each unit
(Units should be updated in random order.)
Step 4. Compute net input:
ii xy
iY
j
jijii wyxiny _

12. Step 5. Determine activation (output signal):
Step 6. Broadcast the value of to all other
units.
(This updates the activation vector.)
Step 7. Test for convergence.
 The threshold, is usually taken to be zero.
 The order of update of the units is random, but each unit must be
updated at the same average rate.
iy
i

13. Applications
 A binary Hopfield net can be used to determine whether an input
vector is a "known” or an "unknown" vector.
 The net recognizes a "known" vector by producing a pattern of
activation on the units of the net that is the same as the vector
stored in the net.
 If the input vector is an "unknown" vector, the activation vectors
produced as the net iterates will converge to an activation vector
that is not one of the stored patterns.

14. Example
 Consider an Example in which the vector (1, 1, 1,0) (or its bipolar
equivalent (1, 1, 1, - 1)) was stored in a net. The binary input vector
corresponding to the input vector used (with mistakes in the first
and second components) is (0, 0, 1, 0). Although the Hopfield net
uses binary vectors, the weight matrix is bipolar. The units update
their activations in a random order. For this example the update
order is 2341 yyyy

15.  Step 0. Initialize weights to store patterns:
 Step 1. The input vector is x = (0, 0, 1, 0). For this vector,
Step 2. y = (0, 0, 1, 0).
Step 3. Choose unit to update its activation:
step 4.
step 5.
step 6. y=(1,0,1,0).
iY
j
jj wyxiny 10_ 111
10_ 11 yiny

16. Step 3. Choose unit to update its activation:
step 4.
step 5.
step 6. y=(1,0,1,0).
step 3. Choose unit to update its activation:
step 4.
step 5.
Step 6. y=(1,0,1,0).
4y
j
jjwyxiny )2(0444
00 44 yiny
.11333
j
jjwyxiny
10 33 yiny
3y

17. step 3. Choose unit to update its activation:
step 4.
step 5.
step 6. y=(1,1,1,0)
Step 7. Test for convergence.
2y
20_ 222
j
jj wyxiny
10_ 22 yiny

18. Analysis
Energy Function.
 An energy function is a function that is bounded below and is a
non increasing function of the state of the system.
 For a neural net, the state of the system is the vector of activations
of the units.
 Thus, if an energy function can be found for an iterative neural
net, the net will converge to a stable set of activations.

19.  Energy function for the discrete Hopfield net is given by,
 If the activation of the net changes by an amount , the
energy changes by an amount,
iy

20.  consider the two cases in which a change will occur in
the activation of neuron
 If is positive, it will change to zero if,
This gives a negative change for In this case,
 If is zero, it will change to positive if,
This gives a negative change for In this case,
iy
iy
iy
j
ijiji wyx
iy .0E
iy
j
ijiji wyx
iy .0E

21. Storage Capacity.
 Hopfield found experimentally that the number of binary patterns
that can be stored and recalled in a net with reasonable accuracy,
is given approximately by,
n= The number of neurons in the net.

22. Continuous Hopfield Net
 A modification of the discrete Hopfield net with continuous-
valued output functions, can be used either for associative
memory problems or constrained optimization problems such as
the travelling salesman problem.
 Here, denote the internal activity of a neuron.
 Output signal is
iu
).( ii ugv

23.  If we define an energy function,
 Then the net will converge to a stable configuration that is a
minimum of the energy function as long as,
 For this form of the energy function, the net will converge if the
activity of each neuron changes with time according to the
differential equation

24.  In the original presentation of the continuous Hopfield net the
energy function is,
where „T‟ is the time constant.
 If the activity of each neuron changes with time according to the
differential equation
the net will converge.

25.  In the Hopfield-Tank solution of the travelling salesman problem,
each unit has two indices.
 The first index--x, y, etc. denotes the city, the second-i,j, etc.-the
position in the tour.
 The Hopfield-Tank energy function for the travelling salesman
problem is,

26.  The differential equation for the activity of unit UX,I is,
 The output signal is given by applying the sigmoid function (with
range between 0 and 1), which Hopfield and Tank expressed as

27. Architecture
 The units used to solve the 10-city travelling salesman
problem are arranged as,

28.  There is a contribution to the energy if two units in the same row
are "on."More explicitly, the weights between units Uxi and Uyj
are,
 Where is Dirac delta, which is 1 if i = j and 0 otherwise. In
addition, each unit receives an external input signal.
The parameter N is usually taken to be somewhat larger than the
number of cities, n.

29. Algorithm

30. Application
 Hopfield and Tank used the following parameter values in their solution
of the problem:
A = B = 500, C = 200, D = 500, N = 15, = 50.
 Hopfield and Tank claimed a high rate of success in finding valid tours; they
found 16 from 20 starting configurations. Approximately half of the trials
produced one of the two shortest paths. The best tour found was
A D E F G H I J B C
with length 2.71

31. Best tour for travelling salesman problem found by Hopfield and Tank