Artificial Neural Networks

Outline Biological Motivation Perceptron Gradient Descent Least Mean Square Error Multi-layer networks Sigmoid node Backpropagation

Biological Neural Systems 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

Artificial Neural Networks Many simple neuron-like threshold units Many weighted interconnections Multiple outputs Highly parallel and distributed processing Learning by tuning the connection weights

Perceptron: Linear threshold unit x0=1 x1 w1 w0 w2 x2 S o . i=0n wi xi wn xn 1 if i=0n wi xi >0 o(xi)= -1 otherwise {

Decision Surface of a Perceptron + - x1 x2 Xor x2 + + + - - x1 + - - Linearly Separable Theorem: VC-dim = n+1

Perceptron Learning Rule S sample xi input vector t=c(x) is the target value o is the perceptron output  learning rate (a small constant ), assume =1 wi = wi + wi wi =  (t - o) xi

Perceptron Algo. Correct Output (t=o) False Negative (t=-1 and o=1) Weights are unchanged Incorrect Output (to) Change weights ! False Positive (t=1 and o=-1) Add x to w False Negative (t=-1 and o=1) Subtract x from w

Perceptron Learning Rule (x,t)=([2,1],-1) o=sgn(0.45-0.6+0.3) =1 w=[0.25 –0.1 0.5] x2 = 0.2 x1 – 0.5 o=-1 w=[0.2 –0.2 –0.2] (x,t)=([-1,-1],1) o=sgn(0.25+0.1-0.5) =-1 (x,t)=([1,1],1) o=sgn(0.25-0.7+0.1) =-1 w=[0.2 0.2 0.2] w=[-0.2 –0.4 –0.2]

Perceptron Algorithm: Analysis Theorem: The number of errors of the Perceptron Algorithm is bounded Proof: Make all examples positive change <xi,bi> to <bixi, +1> Margin of hyperplan w

Perceptron Algorithm: Analysis II Let mi be the number of errors of xi M=  mi From the algorithm: w=  mixi Let w* be a separating hyperplane

Perceptron Algorithm: Analysis III Change in weights: Since w errs on xi , we have wxi <0 Total weight:

Perceptron Algorithm: Analysis IV Consider the angle between w and w* Putting it all together

Gradient Descent Learning Rule Consider linear unit without threshold and continuous output o (not just –1,1) o=w0 + w1 x1 + … + wn xn Train the wi’s such that they minimize the squared error E[w1,…,wn] = ½ dS (td-od)2 where S is the set of training examples

Gradient Descent S={<(1,1),1>,<(-1,-1),1>, <(1,-1),-1>,<(-1,1),-1>} (w1,w2) Gradient: E[w]=[E/w0,… E/wn] w=- E[w] (w1+w1,w2 +w2) wi=- E/wi =/wi 1/2d(td-od)2 = /wi 1/2d(td-i wi xi)2 = d(td- od)(-xi)

Gradient Descent Gradient-Descent(S:training_examples, ) Until TERMINATION Do Initialize each wi to zero For each <x,t> in S Do Compute o=<x,w> For each weight wi Do wi= wi +  (t-o) xi For each weight wi Do1 wi=wi+wi

Incremental Stochastic Gradient Descent Batch mode : Gradient Descent w=w -  ES[w] over the entire data S ES[w]=1/2d(td-od)2 Incremental mode: gradient descent w=w -  Ed[w] over individual training examples d Ed[w]=1/2 (td-od)2 Incremental Gradient Descent can approximate Batch Gradient Descent arbitrarily closely if  is small enough

Comparison Perceptron and Gradient Descent Rule Perceptron learning rule guaranteed to succeed if Training examples are linearly separable No guarantee otherwise Linear unit using Gradient Descent Converges to hypothesis with minimum squared error. Given sufficiently small learning rate  Even when training data contains noise Even when training data not linearly separable

Multi-Layer Networks output layer hidden layer(s) input layer

Sigmoid Unit S . x0=1 x1 w1 w0 w2 x2 o wn xn z=i=0n wi xi o=(z)=1/(1+e-z) w2 x2 S o . wn (z) =1/(1+e-z) sigmoid function. xn

Sigmoid Function (z) =1/(1+e-z) d(z)/dz= (z) (1- (z)) Gradient Decent Rule: one sigmoid function E/wi = -d(td-od) od (1-od) xi Multilayer networks of sigmoid units: backpropagation

Backpropagation: overview Make threshold units differentiable Use sigmoid functions Given a sample compute: The error The Gradient Use the chain rule to compute the Gradient

Backpropagation Motivation Consider the square error ES[w]=1/2d  S k  output (td,k-od,k)2 Gradient: ES[w] Update: w=w -  ES[w] How do we compute the Gradient?

Backpropagation: Algorithm Forward phase: Given input x, compute the output of each unit Backward phase: For each output k compute

Backpropagation: Algorithm Backward phase For each hidden unit h compute: Update weights: wi,j=wi,j+wi,j where wi,j=  j xi

Backpropagation: output node

Backpropagation: output node

Backpropagation: inner node

Backpropagation: inner node

Backpropagation: Summary Gradient descent over entire network weight vector Easily generalized to arbitrary directed graphs Finds a local, not necessarily global error minimum in practice often works well requires multiple invocations with different initial weights A variation is to include momentum term wi,j(n)=  j xi +  wi,j (n-1) Minimizes error training examples Training is fairly slow, yet prediction is fast

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]

VC-dim of ANN A more general bound. Concept class F(C,G): G : Directed acyclic graph C: concept class, d=VC-dim(C) n: input nodes s : inner nodes (of degree r) Theorem: VC-dim(F(C,G)) < 2ds log (es)

Proof: Bound |F(C,G)(m)| Find smallest d s.t. |F(C,G)(m)| <2m Let S={x1, … , xm} For each fixed G we define a matrix U U[i,j]= ci(xj), where ci is a specific i-th concept U describes the computations of S in G TF(C,G) = number of different matrices.

Proof (continue) Clearly |F(C,G)(m)|  TF(C,G) Let G’ be G without the root. |F(C,G)(m)|  TF(C,G)  TF(C,G’) |C(m)| Inductively, |F(C,G)(m)|  |C(m)|s Recall VC Bound: |C(m)|  (em/d)d Combined bound |F(C,G)(m)| (em/d)ds

Proof (cont.) Solve for: (em/d)ds2m Holds for m  2ds log(es) QED Back to ANN: VC-dim(C)=n+1 VC(ANN)  2(n+1) log (es)