[SOLVED] CS代写 Introduction

Introduction
The Perceptron Algorithm: A Geometrical Proof
Marc de Kamps
Institute for Artificial and Biological Intelligence University of Leeds

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Marc de Algorithm

Introduction
The Perceptron What is the definition?
We introduced the perceptron:
h = w1x1 +w2x2 +w3x3 −θ
if h > 0 then o = 1 if h ≤ 0 then o = 0
h is sometimes called the local field
Perceptron Algorithm
Marc de Kamps

Introduction
The Perceptron What is the definition?
Data set: linearly separable
Perceptron Algorithm: will always converge on weights, bias
Weights and bias will classify the data set correctly
Figure: Data set
Perceptron Algorithm
Marc de Kamps

Introduction
The Perceptron What is the definition?
Data set: linearly separable
Perceptron Algorithm: will always converge on weights, bias
Weights and bias will classify the data set correctly
Figure: linearly separable
Perceptron Algorithm
Marc de Kamps

Introduction
Perceptron Algorithm Proof
We introduced the perceptron:
h = w1x1 +w2x2 +w3x3 −θ
if h > 0 then o = 1 if h ≤ 0 then o = 0
h is sometimes called the local field
Perceptron Algorithm
Marc de Kamps

Introduction
Perceptron Algorithm Get Rid of Bias
w1 Consider a 2D classification w2
Find two weights and one bias But note:
w1x1 + w2x2 − θ is equivalent to:
w1x1 + w2x2 + w3x3 ifx3 ≡1andw3 ≡−θ
Perceptron Algorithm
Marc de Kamps

Introduction
Perceptron Algorithm Get Rid of Bias
In the data set this amounts to adding an extra column, e.g.:
classification
classification
Perceptron Algorithm
Marc de Kamps

Introduction
Perceptron Algorithm Get Rid of Bias
Effectively: new data set 3D not 2D
data points hover at z = 1
plane through origin (hard to see, but easy to verify algebraically:)
3x + y − 3z = 0 z = 1: y = −3x + 1
Perceptron Algorithm
Marc de Kamps

Introduction
The Perceptron Algorithm
Linearly Classifiable Dataset
Assume that there are two classes of input patterns in an N-dimensional space: C1 and C2. Also assume that a set of weights w⃗ ∗ exists, such that for all patterns ⃗x ∈ C1:
w⃗ ∗ · ⃗x < 0 and for all patterns ⃗y ∈ C2:w⃗ ∗ · ⃗y ≥ 0 Marc de Algorithm Introduction The Perceptron AlgorithmThe Perceptron Algorithm1 Start with a random set of weights w⃗0.2 Pick an arbitrary pattern ⃗x ∈ C1 ∪ C2.3 If ⃗x ∈ C1 and ⃗x · w⃗ < 0 goto 2. Else w⃗ → w⃗ − λ⃗x. Goto 24 If ⃗x ∈ C2 and ⃗x · w⃗ ≥ 0 goto 2. Else w⃗ → w⃗ + λ⃗x. Goto 2The Perceptron Theorem states that this algorithm will hit the ’Else’ clauses only a finite number of times if and only if the data set is linearly classifiable.0 < λ ≤ 1 is the learning rate.Marc de AlgorithmIntroduction Perceptron Algorithm Leave Only One ClassConstruct a single class C as follows:Take a pattern ⃗x from C1 Replace it by −⃗x and addRemove the originalpattern from C1Repeat until C1 is emptyPerceptron AlgorithmMarc de Kamps IntroductionPerceptron Algorithm Leave Only One Class Simplified ProblemGiven that we have a set of input vectors, and given the existance of a plane through the origin such that all input vectors of a class C are on one side of the plane, find such a plane. That plane is also the solution of our orginal classification problem of separating the two classes C1 and C2. Marc de Algorithm Introduction The Perceptron AlgorithmThe Simplified Perceptron Algorithm1 Start with a random set of weights w⃗0.2 Pick an arbitrary pattern ⃗x ∈ C.3 If⃗x∈Cand⃗x·w⃗ ≥0goto2.Elsew⃗ →w⃗ +λ⃗x.Goto2The Perceptron Theorem states that this algorithm will hit the ’Else’ clauses only a finite number of times if and only if the data set is linearly classifiable.0 < λ ≤ 1 is the learning rate.Marc de AlgorithmIntroduction Perceptron Algorithm Leave Only One Class Normalizing:Normaling does not alter classificationAmounts to multiplying all weights by the same positive constantPut patterns on the unit sphereBut still on the same side of the planePerceptron AlgorithmMarc de Kamps IntroductionPerceptron TheoremWe now have a single class C of patterns, x⃗i with | x⃗i |= 1. We assume the existence of a set of weights w⃗∗ such that for each ⃗x ∈ C ,⃗x · w⃗ ≥ δ > 0 .
w⃗∗ is a solution to our classifcation problem, that we do not know, but that we know exists. (Without loss of generality:
| w⃗∗ |= 1)
We will look at what happens to
w⃗ ∗ · w⃗ n | w⃗ n |
during the execution of the Perceptron Algorithm.
Marc de Algorithm

Introduction
Perceptron Theorem
Strategy: assume we have had n misclassifications so far. The current set of weights is w⃗n, reflecting that the weights have been updated n times. What happens to:
w⃗ ∗ · w⃗ n | w⃗ n |
We will show that when the weights change, this increases by a finite amount
Marc de Algorithm

Introduction
Perceptron Theorem
Observe that:
w⃗ ∗ · w⃗ n | w⃗ n |
is nothing but cos(w⃗∗, w⃗n), so < 1 Observe numerator under change: Denominator:w⃗ ∗ · w⃗ → w⃗ ∗ · w ⃗w⃗ · w⃗ → w ⃗ · w ⃗n n n+1 n+1Marc de Algorithm Introduction For an input pattern ⃗x:w⃗∗·w⃗ =w⃗∗·w⃗ +w⃗∗·⃗xWe know that:w⃗ ∗ · ⃗x ≥ δ > 0
So the numerator increases with every step by at least a fixed amount δ:
w⃗∗·w⃗ ≥w⃗∗·w⃗+δ n+1 n
Marc de Algorithm

Introduction
Denominator
For an input pattern ⃗x:
w⃗ ·w⃗ =|w⃗ |2+2⃗x·w⃗+|⃗x·⃗x|
since w⃗n · ⃗x < 0 (Why?)w ⃗ · w ⃗ < | w⃗ | 2 + 1 , n+1 n+1 nMarc de Algorithm IntroductionConclusionConclusionSince after the n-th application of the weight adding, w ∗ · w⃗ n ≥ n δit follows that:i.e. increases by a fixed amount. This means, there can only be a finite number of steps or cos(w⃗∗,w⃗n) > 1, which is a contradiction.
| w⃗ n | 2 < n , w⃗ ∗ · w⃗ n n δ| w⃗ n | > √ n ,
Marc de Algorithm

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