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Artificial Neural Networks Notes based on Nilsson and Mitchell’s Machine learningPowerPoint Presentation

Artificial Neural Networks Notes based on Nilsson and Mitchell’s Machine learning

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Artificial Neural Networks Notes based on Nilsson and Mitchell’s Machine learning

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Artificial Neural Networks Notes based on Nilsson and Mitchell’s Machine learning

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Artificial Neural Networks

Notes based on Nilsson and Mitchell’s

Machine learning

- Perceptrons (LTU)
- Gradient descent
- Multi-layer networks
- Backpropagation

- Neuron switching time : > 10-3 secs
- Number of neurons in the human brain: ~1010
- Connections (synapses) per neuron : ~104–105
- Face recognition : 0.1 secs
- High degree of parallel computation
- Distributed representations

- Many simple neuron-like threshold switching units
- Many weighted interconnections among units
- Highly parallel, distributed processing
- Learning by adaptation of the connection weights

- 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.

Automated driving at 70 mph on a public highway

Camera

image

30 outputs

for steering

30x32 weights

into one out of

four hidden

unit

4 hidden

units

30x32 pixels

as inputs

- 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.

- This is another area in which neural networks have been successful.
- In fact, all the successful programs have a neural network component.

x1

x2

xn

inputs

weights

w1

output

activation

w2

y

q

.

.

.

a=i=1n wi xi

wn

1 if a q

y=

0 if a< q

{

threshold

linear

y

y

a

a

sigmoid

piece-wise linear

y

y

a

a

1

1

Decision line

w1 x1 + w2 x2 = q

x2

w

1

0

0

0

x1

1

0

0

v

v

v

w

w

w

j

j

j

w • v > 0

w • v = 0

w • v < 0

v

w

j

w • v = |w||v| cos j

The relation w•x=q defines the decision line

x2

Decision line

w

w•x=q

y=1

|xw|=q/|w|

xw

x1

x

y=0

- 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.

x2

w1=?

w2=?

q= ?

w1=1

w2=1

q=1.5

0

1

0

1

x1

x1

1

0

0

0

Logical XOR

Logical AND

x1

x2

xn

q=wn+1

1 if a 0

y =

0 if a<0

xn+1=-1

w1

wn+1

w2

y

.

.

.

a= i=1n+1 wi xi

wn

{

The relation w • x=0 defines the decision line

x2

Decision line

w

w•x=0

y=1

x1

y=0

x

- 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.

x

x

w’ = w + ax

j>90

ax

w

w

Target t=1

Output y=0

Move w in the direction of x

x

x

w

-ax

j<90

w

w’ = w - ax

Target t=0

Output y=1

Move w away from the direction of x

- w’= w + a (t-y) x
Or in components

- w’i = wi + Dwi = wi + a (t-y) xi (i=1..n+1)
With wn+1 = q and xn+1= –1

- The parameter a is called the learning rate. It determines the magnitude of weight updates Dwi .
- If the output is correct (t = y) the weights are not changed (Dwi =0).
- If the output is incorrect (t y) the weights wi are changed such that the output of the TLU for the new weights w’i is closer/further to the input xi.

repeat

for each training vector pair (x,t)

evaluate the output y when x is the input

if yt then

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

- The algorithm converges to the correct classification
- if the training data is linearly separable
- and a is sufficiently small

- If two classes of vectors X1 and X2 are linearly separable, the application of the perceptron training algorithm will eventually result in a weight vector w0, such that w0 defines a TLU whose decision hyper-plane separates X1 and X2 (Rosenblatt 1962).
- Solution w0 is not unique, since if w0 x =0 defines a hyper-plane, so does w’0= k w0.

- x1 x2 output
- 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)

x1

x2

xn

inputs

weights

w1

output

activation

w2

y

.

.

.

y= a = i=1n wi xi

a=i=1n wi xi

wn

- Consider linear unit without threshold and continuous output o (not just –1,1)
- 0 =w0 + w1 x1 + … + wn xn

- Train the wi’s such that they minimize the squared error
- e = aD (fa-da)2
where D is the set of training examples

Here fa is the actual output, da is the desired output.

- e = aD (fa-da)2

- Gradient Descent rule:
- We want to choose the weights wi so that e is minimized. Recall that
e = aD (fa – da)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 = (fx – dx)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/w1,… e/wn+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:
- 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.

x1

x2

xn

inputs

weights

w1

output

activation

w2

y

.

.

.

a=i=1n wi xi

wn

y=s(a) =1/(1+e-a)

x1

x2

xn

x0=-1

w1

a=i=0n wi xi

w0

y=(a)=1/(1+e-a)

w2

y

.

.

.

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

wn

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

Derive gradient descent rules to train:

Sigmoid function

f

s

s

sigmoid

Ep[w1,…,wn] = (tp-yp)2

Ep/wi = /wi (tp-yp)2

= /wi(tp- s(Si wi xip))2

= (tp-yp) s’(Si wi xip) (-xip)

for y=s(a) = 1/(1+e-a)

s’(a)= e-a/(1+e-a)2=s(a) (1-s(a))

a

s’

a

w’i= wi + wi = wi + a y(1-y)(tp-yp) xip

- 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)

- The threshold neuron can realize any linearly separable function Rn {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.

- What do we do if we need a more complex function?
- We can combinemultiple 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.

- o1

- o1

- o1

- o2

- o2

- o2

- oi

- .
- .
- .

- What kind of function can such a network realize?

- 2nd comp.

- 1st comp.

- Assume that the dotted lines in the diagram represent the input-dividing lines implemented by the neurons in the first layer:

- Then, for example, the second-layer neuron could output 1 if the input is within a polygon, and 0 otherwise.

- 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.

- o1

- o1

- o1

- o2

- o2

- o2

- oi

- .
- .
- .

- .
- .
- .

- What type of function can a three-layer network realize?

- 2nd comp.

- 1st comp.

- Assume that the polygons in the diagram indicate the input regions for which each of the second-layer neurons yields output 1:

- Then, for example, the third-layer neuron could output 1 if the input is within any of the polygons, and 0 otherwise.

- 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 wi, a three-layer network of threshold neurons can realize any function Rn {0, 1}.

- 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.

- output vector

- output layer

- Example:Network function f: R3 {0, 1}2

- hidden layer

- input layer

- input vector

output layer

hidden layer

input layer

Ep[wij] = ½ j (tjp-yjp)2

yj

Ep/wij = /wij ½ Sj (tjp-yjp)2

= …

= - yjp(1-ypj)(tpj-ypj) xip

wji

xi

wij = a yjp(1-yjp) (tpj-yjp) xip

= adjp xip

with djp := yjp(1-yjp) (tpj-yjp)

yj

Credit assignment problem:

No target values t for hidden layer units.

dj

wjk

xk

Error for hidden units?

dk

dk = Sj wjkdj yj (1-yj)

wki

wki = a xkp(1-xkp) dkp xip

xi

yj

Ep[wki] = ½ j (tjp-yjp)2

dj

wjk

Ep/wki = /wki ½ Sj (tjp-yjp)2

=/wki ½Sj (tjp-s(Skwjk xkp))2

=/wki ½Sj (tjp-s(Skwjks(Siwki xip)))2

= -j (tjp-yjp) s’j(a) wjks’k(a) xip

= -jdj wjks’k(a) xip

= -jdj wjk xk (1-xk) xip

xk

dk

wki

xi

wki = adk xip

withdk = jdj wjkxk(1-xk)

Backward step:

propagate errors from output to hidden layer

yj

dj

wjk

xk

dk

wki

Forward step:

Propagate activation

from input to output layer

xi

- Initialize weights wij with a small random value
- repeat
for each training pair {(x1,…xn)p,(t1,...,tm)p} Do

- Present (x1,…,xn)p to the network and compute the outputs yj (forward step)
- Compute the errors dj in the output layer and propagate them to the hidden layer (backward step)
- Update the weights in both layers according to
wki = adk xi

end for loop

until overall error E becomes acceptably low

- Initialize each wi to some small random value
- Until the termination condition is met, Do
- For each training example <(x1,…xn),t> Do
- Input the instance (x1,…,xn) to the network and compute the network outputs yk
- For each output unit k
- k=yk(1-yk)(tk-yk)

- For each hidden unit h
- h=yh(1-yh) k wh,k k

- For each network weight wi,j Do
- wi,j=wi,j+wi,j where
wi,j= j xi,j

- For each training example <(x1,…xn),t> Do

- 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
wi,j(n)= j xi,j + wi,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)

- 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

Gradient descent to some local minimum perhaps not global minimum

- Add momentum term: wki(n)
- wki(n) = adk(n) xi (n) + l Dwki(n-1)
with l [0,1]

- wki(n) = adk(n) xi (n) + l Dwki(n-1)
- 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

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]