MATH 829: Introduction to Data Mining and Analysis Support vector machines

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1 1/10 MATH 829: Introduction to Data Mining and Analysis Support vector machines Dominique Guillot Departments of Mathematical Sciences University of Delaware March 11, 2016

2 Hyperplanes 2/10 Recall: A hyperplane H in V = R n is a subspace of V of dimension n 1 (i.e., a subspace of codimension 1). Each hyperplane is determined by a nonzero vector β R n via H = {x R n : β T x = 0} = span(β).

3 Hyperplanes 2/10 Recall: A hyperplane H in V = R n is a subspace of V of dimension n 1 (i.e., a subspace of codimension 1). Each hyperplane is determined by a nonzero vector β R n via H = {x R n : β T x = 0} = span(β). An ane hyperplane H in R n is a subset of the form where β 0 R, β R n. H = {x R n : β 0 + β T x = 0}

4 Hyperplanes Recall: A hyperplane H in V = R n is a subspace of V of dimension n 1 (i.e., a subspace of codimension 1). Each hyperplane is determined by a nonzero vector β R n via H = {x R n : β T x = 0} = span(β). An ane hyperplane H in R n is a subset of the form where β 0 R, β R n. H = {x R n : β 0 + β T x = 0} We often use the term hyperplane for ane hyperplane. 2/10

5 Hyperplanes (cont.) 3/10 Let H = {x R n : β 0 + β T x = 0}.

6 Hyperplanes (cont.) Let H = {x R n : β 0 + β T x = 0}. Note that for x 0, x 1 H, β T (x 0 x 1 ) = 0. Thus β is perpendicular to H. It follows that for x R n, d(x, H) = βt β (x x 0) = β 0 + β T x. β 3/10

7 4/10 Separating hyperplane Suppose we have binary data with labels {+1, 1}. We want to separate data using an (ane) hyperplane. ESL, Figure (Orange = least-squares)

8 4/10 Separating hyperplane Suppose we have binary data with labels {+1, 1}. We want to separate data using an (ane) hyperplane. ESL, Figure (Orange = least-squares) Classify using G(x) = sgn(x T β + β 0 ).

9 4/10 Separating hyperplane Suppose we have binary data with labels {+1, 1}. We want to separate data using an (ane) hyperplane. ESL, Figure (Orange = least-squares) Classify using G(x) = sgn(x T β + β 0 ). Separating hyperplane may not be unique. Separating hyperplane may not exist (i.e., data may not be separable).

10 Margins 5/10 Uniqueness problem: when the data is separable, choose the hyperplane to maximize the margin (the no man's land).

11 Margins 5/10 Uniqueness problem: when the data is separable, choose the hyperplane to maximize the margin (the no man's land). Data: (y i, x i ) {+1, 1} R p (i = 1,..., n). Suppose β 0 + β T x is a separating hyperplane with β = 1.

12 Margins 5/10 Uniqueness problem: when the data is separable, choose the hyperplane to maximize the margin (the no man's land). Data: (y i, x i ) {+1, 1} R p (i = 1,..., n). Suppose β 0 + β T x is a separating hyperplane with β = 1. Note that: y i (x T i β + β 0 ) > 0 Correct classication y i (x T i β + β 0 ) < 0 Incorrect classication

13 Margins Uniqueness problem: when the data is separable, choose the hyperplane to maximize the margin (the no man's land). Data: (y i, x i ) {+1, 1} R p (i = 1,..., n). Suppose β 0 + β T x is a separating hyperplane with β = 1. Note that: y i (x T i β + β 0 ) > 0 Correct classication y i (x T i β + β 0 ) < 0 Incorrect classication Also, y i (x T i β + β 0) = distance between x and hyperplane (since β = 1). 5/10

14 Margins (cont.) 6/10 Thus, if the data is separable, we can solve max M β 0,β R p, β =1 subject to y i (x T i β + β 0 ) M (i = 1,..., n).

15 Margins (cont.) 6/10 Thus, if the data is separable, we can solve max M β 0,β R p, β =1 subject to y i (x T i β + β 0 ) M (i = 1,..., n). We will transform the problem into a usual form used in convex optimization.

16 6/10 Margins (cont.) Thus, if the data is separable, we can solve max M β 0,β R p, β =1 subject to y i (x T i β + β 0 ) M (i = 1,..., n). We will transform the problem into a usual form used in convex optimization. We can remove β = 1 by replacing the constraint by 1 β y i(x T i β+β 0 ) M, or equivalently, y i (x T i β+β 0 ) M β.

17 Margins (cont.) 6/10 Thus, if the data is separable, we can solve max M β 0,β R p, β =1 subject to y i (x T i β + β 0 ) M (i = 1,..., n). We will transform the problem into a usual form used in convex optimization. We can remove β = 1 by replacing the constraint by 1 β y i(x T i β+β 0 ) M, or equivalently, y i (x T i β+β 0 ) M β. We can always rescale (β, β 0 ) so that β = 1/M. Our problem is therefore equivalent to 1 min β 0,β R p 2 β 2 subject to y i (x T i β + β 0 ) 1 (i = 1,..., n).

18 Margins (cont.) Thus, if the data is separable, we can solve max M β 0,β R p, β =1 subject to y i (x T i β + β 0 ) M (i = 1,..., n). We will transform the problem into a usual form used in convex optimization. We can remove β = 1 by replacing the constraint by 1 β y i(x T i β+β 0 ) M, or equivalently, y i (x T i β+β 0 ) M β. We can always rescale (β, β 0 ) so that β = 1/M. Our problem is therefore equivalent to 1 min β 0,β R p 2 β 2 subject to y i (x T i β + β 0 ) 1 (i = 1,..., n). We now recognize the problem as a convex optimization problem with a quadratic objective, and linear inequality constraints. 6/10

19 Support vector machines 7/10 The previous problem works well when the data is separable. What happens if there is no way to nd a margin?

20 Support vector machines 7/10 The previous problem works well when the data is separable. What happens if there is no way to nd a margin? We allow some points to be on the wrong side of the margin, but keep control on the error.

21 Support vector machines 7/10 The previous problem works well when the data is separable. What happens if there is no way to nd a margin? We allow some points to be on the wrong side of the margin, but keep control on the error. We replace y i (x T i β + β 0) M by y i (x T i β + β 0 ) M(1 ξ i ), ξ i 0, and add the constraint n ξ i C for some xed constant C > 0. i=1

22 Support vector machines 7/10 The previous problem works well when the data is separable. What happens if there is no way to nd a margin? We allow some points to be on the wrong side of the margin, but keep control on the error. We replace y i (x T i β + β 0) M by y i (x T i β + β 0 ) M(1 ξ i ), ξ i 0, and add the constraint n ξ i C for some xed constant C > 0. i=1 The problem becomes: max M β 0,β R p, β =1 subject to y i (x T i β + β 0 ) M(1 ξ i ) n ξ i 0, ξ i C. i=1

23 Support vector machines (cont.) 8/10 As before, we can transform the problem into its normal form: 1 min β 0,β 2 β 2 subject to y i (x T i β + β 0 ) 1 ξ i n ξ i 0, ξ i C. i=1 Problem can be solved using standard optimization packages.

24 Multiple classes of data 9/10 The SVM is a binary classier. How can we classify data with K > 2 classes?

25 Multiple classes of data 9/10 The SVM is a binary classier. How can we classify data with K > 2 classes? One versus all:(or one versus the rest) Fit the model to separate each class against the remaining classes. Label a new point x according to the model for which x T β + β 0 is the largest. Need to t the model K times.

26 Multiple classes of data (cont.) 10/10 One versus one: 1 Train a classier for each possible pair of classes. Note: There are ( K 2 ) = K(K 1)/2 such pairs. 2 Classify a new points according to a majority vote: count the number of times the new point is assign to a given class, and pick the class with the largest number.

27 Multiple classes of data (cont.) One versus one: 1 Train a classier for each possible pair of classes. Note: There are ( K 2 ) = K(K 1)/2 such pairs. 2 Classify a new points according to a majority vote: count the number of times the new point is assign to a given class, and pick the class with the largest number. Need to t the model ( K 2 ) times (computationally intensive). 10/10

Support Vector Machines

Support Vector Machines Wien, June, 2010 Paul Hofmarcher, Stefan Theussl, WU Wien Hofmarcher/Theussl SVM 1/21 Linear Separable Separating Hyperplanes Non-Linear Separable Soft-Margin Hyperplanes Hofmarcher/Theussl SVM 2/21 (SVM)