Artificial Neural Networks
Notes based on Nilsson and Mitchell’s
Machine learning
Outline
Perceptrons (LTU)
Gradient descent
Multi

layer networks
Backpropagation
Biological Neural Systems
Neuron switching time : > 10

3
secs
Number of neurons in the human brain:
~10
10
Connections (synapses) per neuron : ~10
4
–
10
5
Face recognition : 0.1 secs
High degree of parallel computation
Distributed representations
Properties of Artificial Neural Nets (ANNs)
Many simple neuron

like threshold switching
units
Many weighted interconnections among units
Highly parallel, distributed processing
Learning by adaptation of the connection
weights
Appropriate Problem Domains for Neural Network Learning
Input is high

dimensional discrete or real

valued (e.g. raw sensor input)
Output is discrete or real valued
Output is a vector of values
Form of target function is unknown
Humans do not need to interpret the results
(black box model)
General Idea
•
A network of neurons. Each neuron is characterized
by:
•
number of input/output wires
•
weights on each wire
•
threshold value
•
These values are not explicitly programmed, but they
evolve through a training process.
•
During training phase, labeled samples are presented.
If the network classifies correctly, no weight changes.
Otherwise, the weights are adjusted.
•
backpropagation algorithm used to adjust weights.
ALVINN (Carnegie Mellon Univ)
Automated driving at 70 mph on a public highway
Camera
image
30x32 pixels
as inputs
30 outputs
for steering
30x32 weights
into one out of
four hidden
unit
4 hidden
units
Another Example
NETtalk
–
Program that learns to pronounce English text.
(Sejnowski and Rosenberg 1987).

A difficult task using conventional programming models.

Rule

based approaches are too complex since
pronunciations are very irregular.

NETtalk takes as input a sentence and produces a
sequence of phonemes and an associated stress for each
letter.
NETtalk
A phoneme is a basic unit of sound in a language.
Stress
–
relative loudness of that sound.
Because the pronunciation of a single letter depends upon its
context and the letters around it,
NETtalk
is given a seven
character window.
Each position is encoded by one of 29 symbols, (26 letters
and 3 punctuations.)
Letter in each position activates the corresponding unit.
NETtalk
The output units encode phonemes using 21 different
features of human articulation.
Remaining five units encode stress and syllable boundaries.
NETtalk also has a middle layer (hidden layer) that has 80
hidden units and nearly 18000 connections (edges).
NETtalk is trained by giving it a 7 character window so that
it learns the pronounce the middle character.
It learns by comparing the computed pronunciation to the
correct pronunciation.
Handwritten character recognition
This is another area in which neural networks have been
successful.
In fact, all the successful programs have a neural
network component.
Threshold Logic Unit (TLU)
x
1
x
2
x
n
.
.
.
w
1
w
2
w
n
a=
i=1
n
w
i
x
i
1 if
a
q
y
=
0 if
a
<
q
y
{
inputs
weights
activation
output
q
Activation Functions
a
y
a
y
a
y
a
y
threshold
linear
piece

wise linear
sigmoid
Decision Surface of a TLU
x
1
x
2
Decision line
w
1
x
1
+ w
2
x
2
=
q
w
1
1
1
0
0
0
0
0
1
Scalar Products & Projections
w • v > 0
v
w
w • v = 0
v
w
w • v < 0
v
w
j
j
j
w • v = wv cos
j
v
w
j
Geometric Interpretation
x
1
x
2
Decision line
w
x
w•x=
q
y=1
y=0
x
w
=
q
/w
The relation
w•x
=
q
defines the decision line
x
w
Geometric Interpretation
In n dimensions the relation
w • x
=
q
defines a n

1 dimensional hyper

plane, which is
perpendicular to the weight vector
w
.
On one side of the hyper

plane (
w • x
>
q
) all
patterns are classified by the TLU as “1”, while
those that get classified as “0” lie on the other
side of the hyper

plane.
If patterns can be not separated by a hyper

plane
then they cannot be correctly classified with a
TLU.
Linear Separability
x
1
x
2
1
0
0
0
Logical AND
x
1
x
2
a
y
0
0
0
0
0
1
1
0
1
0
1
0
1
1
2
1
w
1
=1
w
2
=1
q
=1.5
x
1
1
0
0
w
1
=?
w
2
=?
q
= ?
1
Logical XOR
x
1
x
2
y
0
0
0
0
1
1
1
0
1
1
1
0
Threshold as Weight
x
1
x
2
x
n
.
.
.
w
1
w
2
w
n
w
n+1
x
n+1
=

1
a=
i=1
n+1
w
i
x
i
y
1 if
a
0
y =
0 if
a
<
0
{
q
=w
n+1
Geometric Interpretation
x
1
x
2
Decision line
w
x
w•x=
0
y=1
y=0
The relation
w • x
=
0
defines the decision line
Training ANNs
Training set S of examples
{x, t}
x
is an input vector and
t
the desired target vector
Example: Logical And
S = {(0,0),0}, {(0,1),0}, {(1,0),0}, {(1,1),1}
Iterative process
Present a training example x , compute network output y,
compare output y with target t, adjust weights and thresholds
Learning rule
Specifies how to change the weights w and thresholds
q
of the
network as a function of the inputs x, output y and target t.
Adjusting the Weight Vector
Target t=1
Output y=0
x
w
j
>90
w
x
w’ = w +
a
x
a
x
Target t=0
Output y=1
w
j
<90
x
w
x
Move w in the direction of x

a
x
w’ = w

a
x
Move w away from the direction of x
Perceptron Learning Rule
w’
=
w
+
a
(t

y)
x
Or in components
w’
i
= w
i
+
D
w
i
= w
i
+
a
(t

y) x
i
(i=1..n+1)
With w
n+1
=
q
and x
n+1
=
–
1
The parameter
a
is called the
learning rate
. It
determines the magnitude of weight updates
D
w
i
.
If the output is correct (t = y) the weights are not
changed (
D
w
i
=0).
If the output is incorrect (t
y) the weights w
i
are
changed such that the output of the TLU for the
new weights w’
i
is
closer/further
to the input x
i
.
Perceptron Training Algorithm
repeat
for
each training vector pair (
x
,t)
evaluate the output y when
x
is the input
if
y
t t
hen
form a new weight vector
w’
according
to
w’
=
w
+
a
(t

y)
x
else
do nothing
end if
end for
until
y=t for all training vector pairs
Perceptron Convergence Theorem
The algorithm converges to the correct classification
if the training data is linearly separable
and
a
is sufficiently small
If two classes of vectors X
1
and X
2
are linearly separable,
the application of the perceptron training algorithm will
eventually result in a weight vector
w
0
, such that
w
0
defines a TLU whose decision hyper

plane separates X
1
and X
2
(Rosenblatt 1962).
Solution
w
0
is not unique, since if
w
0
x
=0 defines a
hyper

plane, so does
w’
0
= k
w
0
.
Example
x1 x2 output
1
1 1
9.4 6.4

1
2.5 2.1 1
8.0 7.7

1
0.5 2.2 1
7.9 8.4

1
7.0 7.0

1
2.8 0.8 1
1.2 3.0 1
7.8 6.1

1
Initial weights: (0.75,

0.5,

0.6)
Linear Unit
x
1
x
2
x
n
.
.
.
w
1
w
2
w
n
a=
i=1
n
w
i
x
i
y
y= a =
i=1
n
w
i
x
i
inputs
weights
activation
output
Gradient Descent Learning Rule
Consider linear unit without threshold and
continuous output o (not just
–
1,1)
0 =w
0
+ w
1
x
1
+ … + w
n
x
n
Train the w
i
’s such that they minimize the squared
error
e
=
a
D
(f
a

d
a
)
2
where D is the set of training examples
Here
f
a
is the actual output,
d
a
is the desired
output.
Gradient Descent rule:
We want to choose the weights w
i
so that
e
is minimized. Recall that
e
=
a
D
(f
a
–
d
a
)
2
Since our goal is to work this error function for one input at a time,
let us consider a fixed input x in D, and define
e =
(f
x
–
d
x
)
2
We will drop the subscript and just write this as:
e =
(f
–
d)
2
Our goal is to find the weights that will minimize this expression.
e
/
W
=
[
e
/
w
1
,…
e
/
w
n+1
]
Since s, the threshold function, is given by s =
X
.
W,
we have:
e
/
W =
e
/
s
*
s/
W.
However,
s/
W = X.
Thus,
e
/
W =
e
/
s *
X
Recall from the previous slide that
e =
(f
–
d)
2
So, we have:
e
/
s = 2(f
–
d)*
f /
s (note: d is constant)
This gives the expression:
e
/
W =
2(f
–
d)*
f /
s * X
A problem arises when dealing with TLU, namely f is not a
continuous function of s.
For a fixed input x, suppose the
desired
output is d, and the actual
output is f, then the above expression becomes:
D
w
=

2(d
–
f) x
This is what is known as the
Widrow

Hoff
procedure, with 2
replaced by c:
The key idea is to move the weight vector along the gradient.
When will this converge to the correct weights?
•
We are assuming that the data is linearly separable.
•
We are also assuming that the
desired
output from the linear
threshold gate is available for the training set.
•
Under these conditions, perceptron convergence theorem shows
that the above procedure will converge to the correct weights after
a finite number of iterations.
Neuron with Sigmoid

Function
x
1
x
2
x
n
.
.
.
w
1
w
2
w
n
a=
i=1
n
w
i
x
i
y=
s
(a) =1/(1+e

a
)
y
inputs
weights
activation
output
Sigmoid Unit
x
1
x
2
x
n
.
.
.
w
1
w
2
w
n
w
0
x
0=

1
a=
i=0
n
w
i
x
i
y
y=
s
(a)=1/(1+e

a
)
s
(x) is the sigmoid function: 1/(1+e

x
)
d
s
(x)/dx=
s
(x) (1

s
(x))
Derive gradient descent rules to train:
Sigmoid function
f
s
Gradient Descent Rule for Sigmoid Output Function
a
s
sigmoid
E
p
/
w
i
=
/
w
i
(t
p

y
p
)
2
=
/
w
i
(t
p

s
(
i
w
i
x
i
p
))
2
= (t
p

y
p
)
s
’(
i
w
i
x
i
p
) (

x
i
p
)
for y=
s
(a) = 1/(1+e

a
)
s
’(a)= e

a
/(1+e

a
)
2
=
s
(a) (1

s
(a))
E
p
[w
1
,…,w
n
] =
(t
p

y
p
)
2
w’
i
= w
i
+
D
w
i
= w
i
+
a
y(1

y)(t
p

y
p
) x
i
p
a
s
’
Presentation of Training Examples
Presenting all training examples once to the
ANN is called an
epoch
.
In incremental stochastic gradient descent
training examples can be presented in
Fixed order (1,2,3…,M)
Randomly permutated order (5,2,7,…,3)
Completely random (4,1,7,1,5,4,……) (repetitions
allowed arbitrarily)
Capabilities of Threshold Neurons
The threshold neuron can realize any
linearly
separable
function
R
n
{0, 1}.
Although we only looked at two

dimensional
input, our findings apply to
any dimensionality
n
.
For example, for n = 3, our neuron can realize
any function that divides the three

dimensional
input space along a two

dimension plane.
Capabilities of Threshold Neurons
What do we do if we need a more complex function?
We can
combine
multiple artificial neurons to form
networks with increased capabilities.
For example, we can build a two

layer network with any
number of neurons in the first layer giving input to a single
neuron in the second layer.
The neuron in the second layer could, for example,
implement an AND function.
Capabilities of Threshold Neurons
What kind of function can such a network realize?
o
1
o
2
o
1
o
2
o
1
o
2
.
.
.
o
i
Capabilities of Threshold Neurons
Assume that the dotted lines in the diagram represent the input

dividing lines implemented by the neurons in the first layer:
1
st
comp.
2
nd
comp.
Then, for example, the second

layer neuron could output 1 if the
input is within a
polygon
, and 0 otherwise.
Capabilities of Threshold Neurons
However, we still may want to implement functions that
are more complex than that.
An obvious idea is to extend our network even further.
Let us build a network that has
three layers
, with
arbitrary numbers of neurons in the first and second layers
and one neuron in the third layer.
The first and second layers are
completely connected
,
that is, each neuron in the first layer sends its output to
every neuron in the second layer.
Capabilities of Threshold Neurons
What type of function can a three

layer network realize?
o
1
o
2
o
1
o
2
o
1
o
2
.
.
.
o
i
.
.
.
Capabilities of Threshold Neurons
Assume that the polygons in the diagram indicate the
input regions for which each of the second

layer neurons
yields output 1:
1
st
comp.
2
nd
comp.
Then, for example, the third

layer neuron could output 1 if the
input is within
any of the polygons
, and 0 otherwise.
Capabilities of Threshold Neurons
The more neurons there are in the first layer, the more
vertices can the polygons have.
With a sufficient number of first

layer neurons, the
polygons can approximate
any
given shape.
The more neurons there are in the second layer, the more
of these polygons can be combined to form the output
function of the network.
With a sufficient number of neurons and appropriate
weight vectors
w
i
, a three

layer network of threshold neurons
can realize
any
function
R
n
{0, 1}.
Terminology
Usually, we draw neural networks in such a way that
the input enters at the bottom and the output is
generated at the top.
Arrows indicate the direction of data flow.
The first layer, termed
input layer
, just contains the
input vector and does not perform any computations.
The second layer, termed
hidden layer
, receives input
from the input layer and sends its output to the
output
layer
.
After applying their activation function, the neurons in
the output layer contain the output vector.
Terminology
Example:
Network function f:
R
3
{0, 1}
2
output layer
hidden layer
input layer
input vector
output vector
Multi

Layer Networks
input layer
hidden layer
output layer
Training

Rule for Weights to the Output Layer
y
j
x
i
w
ji
E
p
[w
ij
] = ½
j
(t
j
p

y
j
p
)
2
E
p
/
w
ij
=
/
w
ij
½
j
(t
j
p

y
j
p
)
2
= …
=

y
j
p
(1

y
p
j
)(t
p
j

y
p
j
) x
i
p
D
w
ij
=
a
y
j
p
(1

y
j
p
) (t
p
j

y
j
p
) x
i
p
=
a
d
j
p
x
i
p
with
d
j
p
:= y
j
p
(1

y
j
p
) (t
p
j

y
j
p
)
Training

Rule for Weights to the Hidden Layer
x
k
x
i
w
ki
Credit assignment problem:
No target values t for hidden
layer units.
Error for hidden units?
w
jk
d
j
d
k
y
j
d
k
=
j
w
jk
d
j
y
j
(1

y
j
)
D
w
ki
=
a
x
k
p
(1

x
k
p
)
d
k
p
x
i
p
Training

Rule for Weights to the Hidden Layer
x
k
E
p
[w
ki
] = ½
j
(t
j
p

y
j
p
)
2
E
p
/
w
ki
=
/
w
ki
½
j
(t
j
p

y
j
p
)
2
=
/
w
ki
½
j
(t
j
p

s(
k
w
jk
x
k
p
))
2
=
/
w
ki
½
j
(t
j
p

s(
k
w
jk
s(
i
w
ki
x
i
p
)))
2
=

j
(t
j
p

y
j
p
)
s
’
j
(
a) w
jk
s
’
k
(a) x
i
p
=

j
d
j
w
jk
s
’
k
(a) x
i
p
=

j
d
j
w
jk
x
k
(1

x
k
) x
i
p
d
j
x
i
w
k
i
w
j
k
d
k
y
j
D
w
ki
=
a
d
k
x
i
p
with
d
k
=
j
d
j
w
jk
x
k
(1

x
k
)
Backpropagation
x
k
x
i
w
ki
w
jk
d
j
d
k
y
j
Backward step:
propagate errors from
output to hidden layer
Forward step:
Propagate activation
from input to output layer
Backpropagation Algorithm
Initialize weights w
ij
with a small random value
repeat
for each training pair {(x
1
,…x
n
)
p
,(t
1
,...,t
m
)
p
} Do
Present (x
1
,…,x
n
)
p
to the network and compute the
outputs y
j
(forward step)
Compute the errors
d
j
in the output layer and
propagate them to the hidden layer (backward step)
Update the weights in both layers according to
D
w
ki
=
a
d
k
x
i
end for loop
until overall error E becomes acceptably low
Backpropagation Algorithm
Initialize each w
i
to some small random value
Until the termination condition is met, Do
For each training example <(x
1
,…x
n
),t> Do
Input the instance (x
1
,…,x
n
) to the network and compute the
network outputs y
k
For each output unit k
d
k
=y
k
(1

y
k
)(t
k

y
k
)
For each hidden unit h
d
h
=y
h
(1

y
h
)
k
w
h,k
d
k
For each network weight w
i,j
Do
w
i,j
=w
i,j
+
D
w
i,j
where
D
w
i,j
=
d
j
x
i,j
Backpropagation
Gradient descent over entire
network
weight vector
Easily generalized to arbitrary directed graphs
Will find a local, not necessarily global error minimum

in practice often works well (can be invoked multiple times
with different initial weights)
Often include weight
momentum
term
D
w
i,j
(n)=
d
j
x
i,j
+
a
D
w
i,j
(n

1)
Minimizes error training examples
Will it generalize well to unseen instances (over

fitting)?
Training can be slow typical 1000

10000 iterations
(Using network after training is fast)
Backpropagation
Easily generalized to arbitrary directed graphs without clear
layers.
BP finds a local, not necessarily global error minimum

in practice often works well (can be invoked multiple times
with different initial weights)
Minimizes error over training examples
How does it generalize to unseen instances ?
Training can be slow typical 1000

10000 iterations
(use more efficient optimization methods than gradient descent)
Using network after training is fast
Convergence of Backprop
Gradient descent to some local minimum perhaps not global
minimum
Add momentum term:
D
w
ki
(n)
D
w
ki
(n) =
a
d
k
(n) x
i
(n) +
l D
w
ki
(n

1)
with
l
=
嬰ⰱ[
=
Stochastic gradient descent
Train multiple nets with different initial weights
Nature of convergence
Initialize weights near zero
Therefore, initial networks near

linear
Increasingly non

linear functions possible as training progresses
Expressive Capabilities of ANN
Boolean functions
Every boolean function can be represented by network with
single hidden layer
But might require exponential (in number of inputs) hidden
units
Continuous functions
Every bounded continuous function can be approximated
with arbitrarily small error, by network with one hidden
layer [Cybenko 1989, Hornik 1989]
Any function can be approximated to arbitrary accuracy by a
network with two hidden layers [Cybenko 1988]
Comments 0
Log in to post a comment