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1 Support Vector Machines CS 6243 Machine Learning Modified from the slides by Dr. Raymond J. Mooney And by Dr. Andrew W. Moore

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2 Perceptron Revisited: Linear Separators Binary classification can be viewed as the task of separating classes in feature space: w T x + b = 0 w T x + b < 0 w T x + b > 0 Classification function: f(x) = sign(w T x + b)

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3 Linear Separators Which of the linear separators is optimal? Need to find a decision boundary: ax + by − c = 0 Lots of possible solutions for a, b, c. Some methods find a separating hyperplane, but not the optimal one E.g., perceptron Support Vector Machine (SVM) finds an optimal solution.

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4 Maximum Margin Classification SVMs maximize the margin around the separating hyperplane. A.k.a. large margin classifiers Efficient algorithms to find optimal solution Elegantly extensions to handle non-linearly-separable data Currently widely seen as one of the best classification methods. w T x + b = 0 w T x + b < -ρ/2 w T x + b > ρ/2 Classification function: f(x) = sign(w T x + b) ρ/2

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5 Why Maximum Margin? 1.Intuitively this feels safest. 2.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. 3.LOOCV is easy since the model is immune to removal of any non- support-vector datapoints. 4.There’s some theory (using VC dimension) that is related to (but not the same as) the proposition that this is a good thing. 5.Empirically it works very very well.

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6 Classification Margin Distance from example x i to the separator is Examples closest to the hyperplane are support vectors. Margin ρ of the separator is the distance between support vectors. r ρ Derivation of finding r: Dotted line x’ − x is perpendicular to decision boundary so parallel to w. Unit vector is w/|w|, so line is rw/|w|. x’ = x – yrw/|w|. x’ satisfies w T x’+b = 0. So w T (x –yrw/|w|) + b = 0 Recall that |w| = sqrt(w T w). So, solving for r gives: r = y(w T x + b)/|w| x x′x′

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7 Linear SVM Mathematically Let training set {(x i, y i )} i=1..n, x i R d, y i {-1, 1} be separated by a hyperplane with margin ρ. Then for each training example (x i, y i ): For every support vector x s the above inequality is an equality. After rescaling w and b by ρ/2 in the equality, we obtain that distance between each x s and the hyperplane is Then the margin can be expressed through (rescaled) w and b as: w T x i + b ≤ - ρ/2 if y i = -1 w T x i + b ≥ ρ/2 if y i = 1 y i (w T x i + b) ≥ ρ/2

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8 Linear SVMs Mathematically (cont.) Then we can formulate the quadratic optimization problem: Which can be reformulated as: Find w and b such that is maximized and for all (x i, y i ), i=1..n : y i (w T x i + b) ≥ 1 Find w and b such that Φ(w) = ||w|| 2 =w T w is minimized and for all (x i, y i ), i=1..n : y i (w T x i + b) ≥ 1

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9 Solving the Optimization Problem Need to optimize a quadratic function subject to linear constraints. Quadratic optimization problems are a well-known class of mathematical programming problems for which several (non-trivial) algorithms exist. The solution involves constructing a dual problem where a Lagrange multiplier α i is associated with every inequality constraint in the primal (original) problem: Find w and b such that Φ(w) =w T w is minimized and for all (x i, y i ), i=1..n : y i (w T x i + b) ≥ 1 Find α 1 …α n such that Q( α ) = Σ α i - ½ ΣΣ α i α j y i y j x i T x j is maximized and (1) Σ α i y i = 0 (2) α i ≥ 0 for all α i

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10 The Optimization Problem Solution Given a solution α 1 …α n to the dual problem, solution to the primal is: Each non-zero α i indicates that corresponding x i is a support vector. Then the classifying function is (note that we don’t need w explicitly): Notice that it relies on an inner product between the test point x and the support vectors x i – we will return to this later. Also keep in mind that solving the optimization problem involved computing the inner products x i T x j between all training points. w = Σ α i y i x i b = y k - Σ α i y i x i T x k for any α k > 0 f(x) = Σ α i y i x i T x + b

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11 Soft Margin Classification What if the training set is not linearly separable? Slack variables ξ i can be added to allow misclassification of difficult or noisy examples, resulting margin called soft. ξjξj ξiξi

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12 Soft Margin Classification Mathematically The old formulation: Modified formulation incorporates slack variables: Parameter C can be viewed as a way to control overfitting: it “trades off” the relative importance of maximizing the margin and fitting the training data. Find w and b such that Φ(w) =w T w is minimized and for all (x i,y i ), i=1..n : y i (w T x i + b) ≥ 1 Find w and b such that Φ(w) =w T w + C Σ ξ i is minimized and for all (x i,y i ), i=1..n : y i (w T x i + b) ≥ 1 – ξ i,, ξ i ≥ 0

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13 Soft Margin Classification – Solution Dual problem is identical to separable case (would not be identical if the 2- norm penalty for slack variables CΣξ i 2 was used in primal objective, we would need additional Lagrange multipliers for slack variables): Again, x i with non-zero α i will be support vectors. Solution to the dual problem is: Find α 1 …α N such that Q( α ) = Σ α i - ½ ΣΣ α i α j y i y j x i T x j is maximized and (1) Σ α i y i = 0 (2) 0 ≤ α i ≤ C for all α i w = Σ α i y i x i b= y k (1- ξ k ) - Σ α i y i x i T x k for any k s.t. α k >0 f(x) = Σ α i y i x i T x + b Again, we don’t need to compute w explicitly for classification:

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14 *Theoretical Justification for Maximum Margins Vapnik has proved the following: The class of optimal linear separators has VC dimension h bounded from above as where ρ is the margin, D is the diameter of the smallest sphere that can enclose all of the training examples, and m 0 is the dimensionality. Intuitively, this implies that regardless of dimensionality m 0 we can minimize the VC dimension by maximizing the margin ρ. Thus, complexity of the classifier is kept small regardless of dimensionality.

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15 Linear SVMs: Overview The classifier is a separating hyperplane. Most “important” training points are support vectors; they define the hyperplane. Quadratic optimization algorithms can identify which training points x i are support vectors with non-zero Lagrangian multipliers α i. Both in the dual formulation of the problem and in the solution training points appear only inside inner products: Find α 1 …α N such that Q( α ) = Σ α i - ½ ΣΣ α i α j y i y j x i T x j is maximized and (1) Σ α i y i = 0 (2) 0 ≤ α i ≤ C for all α i f(x) = Σ α i y i x i T x + b

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16 Non-linear SVMs Datasets that are linearly separable with some noise work out great: But what are we going to do if the dataset is just too hard? How about… mapping data to a higher-dimensional space: x2x2 x x x

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17 Non-linear SVMs: Feature spaces General idea: the original feature space can be mapped to some higher- dimensional feature space where the training set is separable: Φ: x → φ(x) x1x1 x2x2 x12x12 x22x22

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18 Quadratic Basis Functions Constant Term Linear Terms Pure Quadratic Terms Quadratic Cross-Terms Number of terms (assuming d input dimensions) = (d+2)-choose-2 = (d+2)(d+1)/2 = (as near as makes no difference) d 2 /2 Potential problems with high- dimensional feature space: Overfitting Separation in high dimensional space is trivial High computational cost

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19 Quadratic Dot Products O(d 2 ) to compute

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20 Quadratic Dot Products

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21 Quadratic Dot Products They’re the same! And this is only O(d) to compute!

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22 Higher Order Polynomials Poly-nomial (x)(x) Cost to build Q kl matrix traditionally Cost if 100 inputs (a). (b) Cost to build Q kl matrix sneakily Cost if 100 inputs QuadraticAll d 2 /2 terms up to degree 2 m 2 R 2 /42,500 R 2 (a.b+1) 2 m R 2 / 250 R 2 CubicAll d 3 /6 terms up to degree 3 m 3 R 2 /1283,000 R 2 (a.b+1) 3 m R 2 / 250 R 2 QuarticAll d 4 /24 terms up to degree 4 m 4 R 2 /481,960,000 R 2 (a.b+1) 4 m R 2 / 250 R 2 R: number of instances m: number of attributes

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23 The “Kernel Trick” The linear classifier relies on inner product between vectors K(x i,x j )=x i T x j If every datapoint is mapped into high-dimensional space via some transformation Φ: x → φ(x), the inner product becomes: K(x i,x j )= φ(x i ) T φ(x j ) A kernel function is a function that is eqiuvalent to an inner product in some feature space. Example: 2-dimensional vectors x=[x 1 x 2 ]; let K(x i,x j )=(1 + x i T x j ) 2, Need to show that K(x i,x j )= φ(x i ) T φ(x j ): K(x i,x j )=(1 + x i T x j ) 2, = 1+ x i1 2 x j x i1 x j1 x i2 x j2 + x i2 2 x j x i1 x j1 + 2x i2 x j2 = = [1 x i1 2 √2 x i1 x i2 x i2 2 √2x i1 √2x i2 ] T [1 x j1 2 √2 x j1 x j2 x j2 2 √2x j1 √2x j2 ] = = φ(x i ) T φ(x j ), where φ(x) = [1 x 1 2 √2 x 1 x 2 x 2 2 √2x 1 √2x 2 ] Thus, a kernel function implicitly maps data to a high-dimensional space (without the need to compute each φ(x) explicitly).

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24 Examples of Kernel Functions Linear: K(x i,x j )= x i T x j –Mapping Φ: x → φ(x), where φ(x) is x itself Polynomial of power p: K(x i,x j )= (1+ x i T x j ) p –Mapping Φ: x → φ(x), where φ(x) has dimensions Gaussian (radial-basis function): K(x i,x j ) = –Mapping Φ: x → φ(x), where φ(x) is infinite-dimensional: every point is mapped to a function (a Gaussian); combination of functions for support vectors is the separator. Higher-dimensional space still has intrinsic dimensionality d (the mapping is not onto), but linear separators in it correspond to non-linear separators in original space.

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25 Non-linear SVMs Mathematically Dual problem formulation: The solution is: Optimization techniques for finding α i ’s remain the same! Find α 1 …α n such that Q(α) = Σ α i - ½ ΣΣ α i α j y i y j K(x i, x j ) is maximized and (1) Σ α i y i = 0 (2) α i ≥ 0 for all α i f(x) = Σ α i y i K(x i, x j )+ b

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26 SVM applications SVMs were originally proposed by Boser, Guyon and Vapnik in 1992 and gained increasing popularity in late 1990s. SVMs are currently among the best performers for a number of classification tasks ranging from text to genomic data. SVMs can be applied to complex data types beyond feature vectors (e.g. graphs, sequences, relational data) by designing kernel functions for such data. SVM techniques have been extended to a number of tasks such as regression [Vapnik et al. ’97], principal component analysis [Schölkopf et al. ’99], etc. Most popular optimization algorithms for SVMs use decomposition to hill- climb over a subset of α i ’s at a time, e.g. SMO [Platt ’99] and [Joachims ’99] Tuning SVMs remains a black art: selecting a specific kernel and parameters is usually done in a try-and-see manner.

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27 SVM software and resources SVM software –LibSVM (C++) –SVMLight (C) As well as complete machine learning toolboxes that include SVMs: – Torch (C++) – Spider (Matlab) – Weka (Java) Useful websites –www.kernel-machines.org –svms.org

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