Introduction to SVMs.

Introduction to SVMs

SVMs Geometric Kernel Methods Capacity Maximizing Margin
Making nonlinear decision boundaries linear Efficiently! Capacity Structural Risk Minimization

SVM History SVM is a classifier derived from statistical learning theory by Vapnik and Chervonenkis SVM was first introduced by Boser, Guyon and Vapnik in COLT-92 SVM became famous when, using pixel maps as input, it gave accuracy comparable to NNs with hand-designed features in a handwriting recognition task SVM is closely related to: Kernel machines (a generalization of SVMs), large margin classifiers, reproducing kernel Hilbert space, Gaussian process, Boosting

Linear Classifiers f a yest x f(x,w,b) = sign(w. x - b) denotes +1
How would you classify this data?

Linear Classifiers f a yest x f(x,w,b) = sign(w. x - b) denotes +1
Any of these would be fine.. ..but which is best?

Classifier Margin f a yest x
f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 Define the margin of a linear classifier as the width that the boundary could be increased by before hitting a datapoint.

Maximum Margin f a yest x
f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 The maximum margin linear classifier is the linear classifier with the maximum margin. This is the simplest kind of SVM (Called an LSVM) Linear SVM

Maximum Margin f a yest x
f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 The maximum margin linear classifier is the linear classifier with the, um, maximum margin. This is the simplest kind of SVM (Called an LSVM) Support Vectors are those datapoints that the margin pushes up against Linear SVM

Why Maximum Margin? Intuitively this feels safest. If we’ve made a small error in the location of the boundary (it’s been jolted in its perpendicular direction) this gives us least chance of causing a misclassification. There’s some theory (using VC dimension) that is related to (but not the same as) the proposition that this is a good thing. Empirically it works very very well. f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 The maximum margin linear classifier is the linear classifier with the, um, maximum margin. This is the simplest kind of SVM (Called an LSVM) Support Vectors are those datapoints that the margin pushes up against

A “Good” Separator O O X X O X X O O X O X O X O X

Noise in the Observations
X X O X X O O X O X O X O X

Ruling Out Some Separators
X X O X X O O X O X O X O X

Lots of Noise O O X X O X X O O X O X O X O X

Maximizing the Margin O O X X O X X O O X O X O X O X

Specifying a line and margin
Plus-Plane “Predict Class = +1” zone Classifier Boundary Minus-Plane “Predict Class = -1” zone How do we represent this mathematically? …in m input dimensions?

Specifying a line and margin
Plus-Plane “Predict Class = +1” zone Classifier Boundary Minus-Plane “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Plus-plane = { x : w . x + b = +1 } Minus-plane = { x : w . x + b = -1 } Classify as.. +1 if w . x + b >= 1 -1 if w . x + b <= -1 Universe explodes if -1 < w . x + b < 1

Computing the margin width
M = Margin Width “Predict Class = +1” zone How do we compute M in terms of w and b? “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Plus-plane = { x : w . x + b = +1 } Minus-plane = { x : w . x + b = -1 } Claim: The vector w is perpendicular to the Plus Plane. Why?

M = Margin Width “Predict Class = +1” zone How do we compute M in terms of w and b? “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Plus-plane = { x : w . x + b = +1 } Minus-plane = { x : w . x + b = -1 } Claim: The vector w is perpendicular to the Plus Plane. Why? Let u and v be two vectors on the Plus Plane. What is w . ( u – v ) ? And so of course the vector w is also perpendicular to the Minus Plane

x+ M = Margin Width “Predict Class = +1” zone How do we compute M in terms of w and b? x- “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Plus-plane = { x : w . x + b = +1 } Minus-plane = { x : w . x + b = -1 } The vector w is perpendicular to the Plus Plane Let x- be any point on the minus plane Let x+ be the closest plus-plane-point to x-. Any location in m: not necessarily a datapoint Any location in Rm: not necessarily a datapoint

x+ M = Margin Width “Predict Class = +1” zone How do we compute M in terms of w and b? x- “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Plus-plane = { x : w . x + b = +1 } Minus-plane = { x : w . x + b = -1 } The vector w is perpendicular to the Plus Plane Let x- be any point on the minus plane Let x+ be the closest plus-plane-point to x-. Claim: x+ = x- + l w for some value of l. Why?

x+ M = Margin Width “Predict Class = +1” zone The line from x- to x+ is perpendicular to the planes. So to get from x- to x+ travel some distance in direction w. How do we compute M in terms of w and b? x- “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Plus-plane = { x : w . x + b = +1 } Minus-plane = { x : w . x + b = -1 } The vector w is perpendicular to the Plus Plane Let x- be any point on the minus plane Let x+ be the closest plus-plane-point to x-. Claim: x+ = x- + l w for some value of l. Why?

x+ M = Margin Width “Predict Class = +1” zone x- “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 What we know: w . x+ + b = +1 w . x- + b = -1 x+ = x- + l w |x+ - x- | = M It’s now easy to get M in terms of w and b

x+ M = Margin Width “Predict Class = +1” zone x- “Predict Class = -1” zone wx+b=1 w . (x - + l w) + b = 1 => w . x - + b + l w .w = 1 -1 + l w .w = 1 wx+b=0 wx+b=-1 What we know: w . x+ + b = +1 w . x- + b = -1 x+ = x- + l w |x+ - x- | = M It’s now easy to get M in terms of w and b

x+ M = Margin Width = “Predict Class = +1” zone x- “Predict Class = -1” zone wx+b=1 M = |x+ - x- | =| l w |= wx+b=0 wx+b=-1 What we know: w . x+ + b = +1 w . x- + b = -1 x+ = x- + l w |x+ - x- | = M

Learning the Maximum Margin Classifier
M = Margin Width = “Predict Class = +1” zone x- “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 Given a guess of w and b we can Compute whether all data points in the correct half-planes Compute the width of the margin So now we just need to write a program to search the space of w’s and b’s to find the widest margin that matches all the datapoints. How? Gradient descent? Simulated Annealing? Matrix Inversion? EM? Newton’s Method?

Linear Programming find w argmax cw subject to
Don’t worry… it’s good for you… Linear Programming find w argmax cw subject to wai  bi, for i = 1, …, m wj  0 for j = 1, …, n There are fast algorithms for solving linear programs including the simplex algorithm and Karmarkar’s algorithm

Learning via Quadratic Programming
QP is a well-studied class of optimization algorithms to maximize a quadratic function of some real-valued variables subject to linear constraints.

Quadratic Programming
Quadratic criterion Find Subject to n additional linear inequality constraints And subject to e additional linear equality constraints

Quadratic Programming
Quadratic criterion Find There exist algorithms for finding such constrained quadratic optima much more efficiently and reliably than gradient ascent. (But they are very fiddly…you probably don’t want to write one yourself) Subject to n additional linear inequality constraints And subject to e additional linear equality constraints

Learning the Maximum Margin Classifier
Given guess of w , b we can Compute whether all data points are in the correct half-planes Compute the margin width Assume R datapoints, each (xk,yk) where yk = +/- 1 “Predict Class = +1” zone “Predict Class = -1” zone wx+b=1 wx+b=0 wx+b=-1 What should our quadratic optimization criterion be? How many constraints will we have? What should they be? R w . xk + b >= 1 if yk = 1 w . xk + b <= -1 if yk = -1 Minimize w.w

Uh-oh! This is going to be a problem! What should we do? Idea 1:
Find minimum w.w, while minimizing number of training set errors. Problem: Two things to minimize makes for an ill-defined optimization Uh-oh! denotes +1 denotes -1

Uh-oh! This is going to be a problem! What should we do? Idea 1.1:
Minimize w.w + C (#train errors) There’s a serious practical problem that’s about to make us reject this approach. Can you guess what it is? Uh-oh! denotes +1 denotes -1 Tradeoff parameter

Minimize w.w + C (#train errors) There’s a serious practical problem that’s about to make us reject this approach. Can you guess what it is? Uh-oh! denotes +1 denotes -1 Tradeoff parameter Can’t be expressed as a Quadratic Programming problem. Solving it may be too slow. (Also, doesn’t distinguish between disastrous errors and near misses) So… any other ideas?

Minimize w.w + C (distance of error points to their correct place) Uh-oh! denotes +1 denotes -1

Learning Maximum Margin with Noise
Given guess of w , b we can Compute sum of distances of points to their correct zones Compute the margin width Assume R datapoints, each (xk,yk) where yk = +/- 1 wx+b=1 wx+b=0 wx+b=-1 What should our quadratic optimization criterion be? How many constraints will we have? What should they be?

Large-margin Decision Boundary
The decision boundary should be as far away from the data of both classes as possible We should maximize the margin, m Distance between the origin and the line wtx=k is k/||w|| Class 2 m Class 1 7/31/2018

Finding the Decision Boundary
Let {x1, ..., xn} be our data set and let yi Î {1,-1} be the class label of xi The decision boundary should classify all points correctly Þ The decision boundary can be found by solving the following constrained optimization problem This is a constrained optimization problem. Solving it requires some new tools Feel free to ignore the following several slides; what is important is the constrained optimization problem above 7/31/2018

Back to the Original Problem
The Lagrangian is Note that ||w||2 = wTw Setting the gradient of w.r.t. w and b to zero, we have 7/31/2018

The Karush-Kuhn-Tucker conditions,
7/31/2018

The Dual Problem If we substitute to , we have Note that
This is a function of ai only 7/31/2018

The Dual Problem The new objective function is in terms of ai only
It is known as the dual problem: if we know w, we know all ai; if we know all ai, we know w The original problem is known as the primal problem The objective function of the dual problem needs to be maximized! The dual problem is therefore: Properties of ai when we introduce the Lagrange multipliers The result when we differentiate the original Lagrangian w.r.t. b

The Dual Problem This is a quadratic programming (QP) problem
A global maximum of ai can always be found w can be recovered by Let x(1) and x(-1) be two S.V. Then b = -1/2( w^T x(1) + w^T x(-1) ) 7/31/2018

Characteristics of the Solution
Many of the ai are zero w is a linear combination of a small number of data points This “sparse” representation can be viewed as data compression as in the construction of knn classifier xi with non-zero ai are called support vectors (SV) The decision boundary is determined only by the SV Let tj (j=1, ..., s) be the indices of the s support vectors. We can write For testing with a new data z Compute and classify z as class 1 if the sum is positive, and class 2 otherwise Note: w need not be formed explicitly 7/31/2018

A Geometrical Interpretation
Class 2 a10=0 a8=0.6 a7=0 a2=0 a5=0 a1=0.8 a4=0 a6=1.4 So, if change internal points, no effect on the decision boundary a9=0 a3=0 Class 1 7/31/2018

Non-linearly Separable Problems
We allow “error” xi in classification; it is based on the output of the discriminant function wTx+b xi approximates the number of misclassified samples Class 1 Class 2 7/31/2018

Given guess of w , b we can Compute sum of distances of points to their correct zones Compute the margin width Assume R datapoints, each (xk,yk) where yk = +/- 1 e2 wx+b=1 e7 wx+b=0 wx+b=-1 What should our quadratic optimization criterion be? Minimize How many constraints will we have? R What should they be? w . xk + b >= 1-ek if yk = 1 w . xk + b <= -1+ek if yk = -1

m = # input dimensions M = Given guess of w , b we can Compute sum of distances of points to their correct zones Compute the margin width Assume R datapoints, each (xk,yk) where yk = +/- 1 e11 e2 Our original (noiseless data) QP had m+1 variables: w1, w2, … wm, and b. Our new (noisy data) QP has m+1+R variables: w1, w2, … wm, b, ek , e1 ,… eR wx+b=1 e7 wx+b=0 wx+b=-1 What should our quadratic optimization criterion be? Minimize How many constraints will we have? R What should they be? w . xk + b >= 1-ek if yk = 1 w . xk + b <= -1+ek if yk = -1 R= # records

Given guess of w , b we can Compute sum of distances of points to their correct zones Compute the margin width Assume R datapoints, each (xk,yk) where yk = +/- 1 e11 e2 wx+b=1 e7 wx+b=0 wx+b=-1 What should our quadratic optimization criterion be? Minimize How many constraints will we have? R What should they be? w . xk + b >= 1-ek if yk = 1 w . xk + b <= -1+ek if yk = -1 There’s a bug in this QP. Can you spot it?

Given guess of w , b we can Compute sum of distances of points to their correct zones Compute the margin width Assume R datapoints, each (xk,yk) where yk = +/- 1 e11 e2 wx+b=1 e7 wx+b=0 wx+b=-1 How many constraints will we have? 2R What should they be? w . xk + b >= 1-ek if yk = 1 w . xk + b <= -1+ek if yk = -1 ek >= 0 for all k What should our quadratic optimization criterion be? Minimize

An Equivalent Dual QP Minimize w . xk + b >= 1-ek if yk = 1
ek >= 0， for all k Maximize where Subject to these constraints:

An Equivalent Dual QP Maximize where Subject to these constraints:
Then define: Then classify with: f(x,w,b) = sign(w. x - b)

Example XOR problem revisited:
Let the nonlinear mapping be : f(x) = (1,x12, 21/2 x1x2, x22, 21/2 x1 , 21/2 x2)T And: f(xi)=(1,xi12, 21/2 xi1xi2, xi22, 21/2 xi1 , 21/2 xi2)T Therefore the feature space is in 6D with input data in 2D x1 = (-1,-1), d1= - 1 x2 = (-1, 1), d2= 1 x3 = ( 1,-1), d3= 1 x4 = (-1,-1), d4= -1

Q(a)= S ai – ½ S S ai aj di dj f(xi) Tf(xj)
=a1 +a2 +a3 +a4 – ½(9 a1 a1 - 2a1 a2 -2 a1 a3 +2a1 a4 +9a2 a2 + 2a2 a3 -2a2 a4 +9a3 a3 -2a3 a a4 a4 ) To minimize Q, we only need to calculate (due to optimality conditions) which gives 1 = 9 a1 - a2 - a3 + a4 1 = -a a2 + a3 - a4 1 = -a1 + a2 + 9 a3 - a4 1 = a1 - a2 - a a4

The solution of which gives the optimal values:
a0,1 =a0,2 =a0,3 =a0,4 =1/8 w0 = S a0,i di f(xi) = 1/8[f(x1)- f(x2)- f(x3)+ f(x4)] Where the first element of w0 gives the bias b

From earlier we have that the optimal hyperplane is defined by:
w0T f(x) = 0 That is: w0T f(x) which is the optimal decision boundary for the XOR problem. Furthermore we note that the solution is unique since the optimal decision boundary is unique

Output for polynomial RBF

Harder 1-dimensional dataset
Remember how permitting non-linear basis functions made linear regression so much nicer? Let’s permit them here too x=0

For a non-linearly separable problem we have to first map data onto
feature space so that they are linear separable xi f(xi) Given the training data sample {(xi,yi), i=1, …,N}, find the optimum values of the weight vector w and bias b w = S a0,i yi f(xi) where a0,i are the optimal Lagrange multipliers determined by maximizing the following objective function subject to the constraints S ai yi =0 ; ai >0

SVM building procedure:
Pick a nonlinear mapping f Solve for the optimal weight vector However: how do we pick the function f? In practical applications, if it is not totally impossible to find f, it is very hard In the previous example, the function f is quite complex: How would we find it? Answer: the Kernel Trick

Notice that in the dual problem the image of input vectors only involved as an inner product meaning that the optimization can be performed in the (lower dimensional) input space and that the inner product can be replaced by an inner-product kernel How do we relate the output of the SVM to the kernel K? Look at the equation of the boundary in the feature space and use the optimality conditions derived from the Lagrangian formulations

In the XOR problem, we chose to use the kernel function:
K(x, xi) = (x T xi+1)2 = 1+ x12 xi x1x2 xi1xi2 + x22 xi22 + 2x1xi1 ,+ 2x2xi2 Which implied the form of our nonlinear functions: f(x) = (1,x12, 21/2 x1x2, x22, 21/2 x1 , 21/2 x2)T And: f(xi)=(1,xi12, 21/2 xi1xi2, xi22, 21/2 xi1 , 21/2 xi2)T However, we did not need to calculate f at all and could simply have used the kernel to calculate: Q(a) = S ai – ½ S S ai aj di dj K(xi, xj) Maximized and solved for ai and derived the hyperplane via:

We therefore only need a suitable choice of kernel function cf:
Mercer’s Theorem: Let K(x,y) be a continuous symmetric kernel that defined in the closed interval [a,b]. The kernel K can be expanded in the form K (x,y) = f(x) T f(y) provided it is positive definite. Some of the usual choices for K are: Polynomial SVM (x T xi+1)p p specified by user RBF SVM exp(-1/(2s2) || x – xi||2) s specified by user MLP SVM tanh(s0 x T xi + s1)

An Equivalent Dual QP Maximize where Subject to these constraints:
Then define: Datapoints with ak > 0 will be the support vectors Then classify with: f(x,w,b) = sign(w. x - b) ..so this sum only needs to be over the support vectors.

Quadratic Basis Functions
Constant Term Quadratic Basis Functions Linear Terms Pure Quadratic Terms Number of terms (assuming m input dimensions) = (m+2)-choose-2 = (m+2)(m+1)/2 = (as near as makes no difference) m2/2 You may be wondering what those ’s are doing. You should be happy that they do no harm You’ll find out why they’re there soon. Quadratic Cross-Terms

QP with basis functions
Maximize where Subject to these constraints: Then define: Then classify with: f(x,w,b) = sign(w. f(x) - b)

QP with basis functions
Maximize where We must do R2/2 dot products to get this matrix ready. Each dot product requires m2/2 additions and multiplications The whole thing costs R2 m2 /4. Yeeks! …or does it? Subject to these constraints: Then define: Then classify with: f(x,w,b) = sign(w. f(x) - b)

Quadratic Dot Products
+ + Quadratic Dot Products +

Just out of casual, innocent, interest, let’s look at another function of a and b: Quadratic Dot Products

Just out of casual, innocent, interest, let’s look at another function of a and b: Quadratic Dot Products They’re the same! And this is only O(m) to compute!

QP with Quadratic basis functions
Maximize where We must do R2/2 dot products to get this matrix ready. Each dot product now only requires m additions and multiplications Subject to these constraints: Then define: Then classify with: f(x,w,b) = sign(w. f(x) - b)

Higher Order Polynomials
f(x) Cost to build Qkl matrix traditionally Cost if 100 inputs f(a).f(b) Cost to build Qkl matrix efficiently Quadratic All m2/2 terms up to degree 2 m2 R2 /4 2,500 R2 (a.b+1)2 m R2 / 2 50 R2 Cubic All m3/6 terms up to degree 3 m3 R2 /12 83,000 R2 (a.b+1)3 Quartic All m4/24 terms up to degree 4 m4 R2 /48 1,960,000 R2 (a.b+1)4

QP with Quintic basis functions
Maximize where Subject to these constraints: Then define: Then classify with: f(x,w,b) = sign(w. f(x) - b)

We must do R2/2 dot products to get this matrix ready. In 100-d, each dot product now needs 103 operations instead of 75 million But there are still worrying things lurking away. What are they? Maximize where Subject to these constraints: The fear of overfitting with this enormous number of terms The evaluation phase (doing a set of predictions on a test set) will be very expensive (why?) Then define: Then classify with: f(x,w,b) = sign(w. f(x) - b)

We must do R2/2 dot products to get this matrix ready. In 100-d, each dot product now needs 103 operations instead of 75 million But there are still worrying things lurking away. What are they? Maximize where The use of Maximum Margin magically makes this not a problem Subject to these constraints: The fear of overfitting with this enormous number of terms The evaluation phase (doing a set of predictions on a test set) will be very expensive (why?) Then define: Because each w. f(x) (see below) needs 75 million operations. What can be done? Then classify with: f(x,w,b) = sign(w. f(x) - b)

We must do R2/2 dot products to get this matrix ready. In 100-d, each dot product now needs 103 operations instead of 75 million But there are still worrying things lurking away. What are they? Maximize where The use of Maximum Margin magically makes this not a problem Subject to these constraints: The fear of overfitting with this enormous number of terms The evaluation phase (doing a set of predictions on a test set) will be very expensive (why?) Then define: Because each w. f(x) (see below) needs 75 million operations. What can be done? Only Sm operations (S=#support vectors) Then classify with: f(x,w,b) = sign(w. f(x) - b)

Maximize where Why SVMs don’t overfit as much as you’d think: No matter what the basis function, there are really only up to R parameters: a1, a2 .. aR, and usually most are set to zero by the Maximum Margin. Asking for small w.w is like “weight decay” in Neural Nets and like Ridge Regression parameters in Linear regression and like the use of Priors in Bayesian Regression---all designed to smooth the function and reduce overfitting. Subject to these constraints: Then define: Only Sm operations (S=#support vectors) Then classify with: f(x,w,b) = sign(w. f(x) - b)

SVM Kernel Functions K(a,b)=(a . b +1)d is an example of an SVM Kernel Function Beyond polynomials there are other very high dimensional basis functions that can be made practical by finding the right Kernel Function Radial-Basis-style Kernel Function: Neural-net-style Kernel Function: s, k and d are magic parameters that must be chosen by a model selection method such as CV or VCSRM

SVM Implementations Sequential Minimal Optimization, SMO, efficient implementation of SVMs, Platt in Weka SVMlight

References Tutorial on VC-dimension and Support Vector Machines:
C. Burges. A tutorial on support vector machines for pattern recognition. Data Mining and Knowledge Discovery, 2(2): , The VC/SRM/SVM Bible: Statistical Learning Theory by Vladimir Vapnik, Wiley-Interscience; 1998

Introduction to SVMs.

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