Support Vector Machines
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1 Support Vector Machines Tobias Pohlen Selected Topics in Human Language Technology and Pattern Recognition February 10, 2014 Human Language Technology and Pattern Recognition Lehrstuhl für Informatik 6 Computer Science Department RWTH Aachen University, Germany Tobias Pohlen: Support Vector Machines 1 / 48 February 10, 2014
2 Outline 1. Introduction 2. Recap: Linear classifiers 3. Feature space mappings and kernel functions 4. Support Vector Machines (a) Motivation (b) Learning (c) Limitations 5. Conclusion Tobias Pohlen: Support Vector Machines 2 / 48 February 10, 2014
3 Literature N. Christiani, J. Shawe-Taylor An introduction to support vector machines, Cambridge University Press, Indepth introduction to SVMs (theoretical and practical concepts) V. N. Vapnik The nature of statistical learning theory, Springer, 1995 Theoretical background of SVMs C. J. C. Burges A Tutorial on Support Vector Machines for Pattern Recognition. Data Mining and Knowledge Discovery 2, 1998, pages Good and short introduction to SVMs (Only 40 pages) Tobias Pohlen: Support Vector Machines 3 / 48 February 10, 2014
4 Classification Problem We want to solve the binary classification problem Training set X = {(x 1, y 1 ),..., (x N, y N )} R m { 1, 1} Input space R m training examples x i and class labels y i Goal: Assign the vector x to one of the classes 1 or 1 We will consider the multiclass problem later. Tobias Pohlen: Support Vector Machines 4 / 48 February 10, 2014
5 Classification Problem - Visualization x 2 x x 1 Tobias Pohlen: Support Vector Machines 5 / 48 February 10, 2014
6 Linear discriminant functions and linear classifiers Definition 1. Linear discriminant function defined as f : R m R, x w T x + b = m w i x i + b (1) w R m is called weight vector b R is called bias Definition 2. Linear classifier defined as h f : R m { 1, 1}, x sign(f(x)) = { 1 if f(x) 0 1 otherwise (2) Tobias Pohlen: Support Vector Machines 6 / 48 February 10, 2014
7 Decision surface and decision regions Definition 3. Decision surface of a linear classifier h f is defined by: f(x) = 0 x 2 Class 1 Class 1 f(x) < 0 f(x) > 0 w x 1 f(x) = 0 Decision surface of a linear classifier (red) is an (m 1)-dimensional hyperplane. Tobias Pohlen: Support Vector Machines 7 / 48 February 10, 2014
8 Linear separability Definition 4. X = {(x 1, y 1 ),..., (x N, y N )} is called linearly separable if (i.e. there exists a classifier h f that classifies all points correctly). h f : h f (x i ) = y i, i = 1,..., N (3) Tobias Pohlen: Support Vector Machines 8 / 48 February 10, 2014
9 Linear separability - Visualization x 2 Linearly separable training set Linearly inseparable training set x x 1 x 1 1 Tobias Pohlen: Support Vector Machines 9 / 48 February 10, 2014
10 Linear classifiers and inseparable training sets How can we use linear classifiers for linearly inseparable training sets? Idea: Map the data into another space in which it is linearly separable x 2 φ 2 (x) x 5 x 6 φ(x 6 ) φ(x 3 ) φ(x 5 ) x 1 x 2 x 3 x 4 x 7 φ(x 1 ) φ(x 7 ) x 1 φ(x 4 ) φ(x 2 ) φ 1 (x) Tobias Pohlen: Support Vector Machines 10 / 48 February 10, 2014
11 Feature Space Mappings Definition 5. Let H be a D-dimensional Hilbert space, D N { }. A feature space mapping is defined as φ i are called basis functions H is called feature space φ : R m H, x (φ i (x)) D (4) Linear discriminant function with feature space mapping φ, is the inner product of H f(x) = w, φ(x) + b, w H (5) A Hilbert space is a vector space with inner product,. Tobias Pohlen: Support Vector Machines 11 / 48 February 10, 2014
12 Separate the XOR Example Let H R 2 with x, y x T y = x y (dot product) ( ) φ : R 2 x1 H, x x 1 x 2 (6) f(x) = (0 1) T φ(x) + 0 = x 1 x 2 (7) Decision surface in the input space x 2 Decision surface in the feature space x 1 x x x Tobias Pohlen: Support Vector Machines 12 / 48 February 10, 2014
13 Feature Space Mappings - Complexity Evaluation of f : R m R without feature space mapping Complexity: O(m) f(x) = w T x + b, w R m (8) Evaluation of f with feature space mapping Complexity: O(dim(H)) Typically dim(h) m. f(x) = w, φ(x) + b, w H (9) Tobias Pohlen: Support Vector Machines 13 / 48 February 10, 2014
14 Reducing complexity Idea: The inner product φ(x), φ(y) can often be computed very efficiently. Simple example: φ(x) = x 2 1 2x1 x 2 x 2 2 (10) Let x, y R 2 arbitrary. Then φ(x), φ(y) =φ(x) T φ(y) (11) =x 2 1 y x 1x 2 y 1 y 2 + x 2 2 y2 2 (12) =(x T y) 2 (13) Naive computation: 11 multiplications, 2 additions Optimized computation: 3 multiplications, 2 additions Tobias Pohlen: Support Vector Machines 14 / 48 February 10, 2014
15 Kernel Functions Definition 6. Let φ : R m H be a feature space mapping. is called kernel function or kernel. k(x, y) = φ(x), φ(y) (14) Interpretation: A kernel measures similarity. Tobias Pohlen: Support Vector Machines 15 / 48 February 10, 2014
16 Dual form In order to use kernels, we need to formulate our algorithms in the so called dual form. φ(x) occurs only as argument of an inner product, Definition 7. Let X = {(x 1, y 1 ),..., (x N, y N )} be a training set. Linear discriminant function f in dual form: f(x) = θ i k(x i, x) + b, θ i R, i = 1,..., N (15) (16) We will see later how to determine θ i. Tobias Pohlen: Support Vector Machines 16 / 48 February 10, 2014
17 Kernel Functions - Complexity Evaluation of f : R m R without feature space mapping f(x) = w T x + b, w R m (17) Complexity: O(m) Evaluation of f with feature space mapping f(x) = w, φ(x) + b, w H (18) Complexity: O(dim(H)) Evaluation of f in the dual form f(x) = θ i k(x, x i ) + b, θ i R (19) Complexity: O(N) (Assuming k(, ) can be evaluated efficiently) Tobias Pohlen: Support Vector Machines 17 / 48 February 10, 2014
18 Popular kernel functions There are many known kernels. Popular examples: Polynomial kernel k(x, y) = (x T y + c) d, c R, d N 0 (20) Gaussian kernel ( x y 2) k(x, y) = exp 2, σ R + (21) σ 2 Both kernels are widely used in practice. Tobias Pohlen: Support Vector Machines 18 / 48 February 10, 2014
19 SVMs - Motivation Support Vector Machines (SVMs) arose from the theoretical question: Given a training set. What is the optimal linear classifier? x 2 x 1 Tobias Pohlen: Support Vector Machines 19 / 48 February 10, 2014
20 Concept of margins Definition 8. h f linear classifier X training set (geometric) margin γ of h f on X is the smallest distance from a point to the decision surface x 2 γ x 1 Tobias Pohlen: Support Vector Machines 20 / 48 February 10, 2014
21 Expected generalization error Vapnik answered the question using statistical learning theory Idea: The optimal classifier has the lowest expected error on unknown data Data (points and labels) is generated i.i.d. according to a fixed but unknown distribution D Upper bound on the generalization error i { 1,1} R m yh f(x) p(x, y)dx ɛ(h f, X) (22) Tobias Pohlen: Support Vector Machines 21 / 48 February 10, 2014
22 Maximum margin classifiers Given a linearly separable training set. What is the optimal linear classifier in terms of the upper bound ɛ? Result: The linear classifier with the largest margin γ on the training set. Such classifiers are called maximum margin classifiers See [Vapnik 95] for an introduction to statistical learning theory [Boser & Guyon + 92] and [Vapnik 82] for more theoretical background on SVMs [Christiani & Shawe-Taylor 00] for a good overview [Shawe-Taylor & Bartlett + 98] and [Bartlett & Shawe Taylor 99] for Data Dependent Structural Risk Minimization Tobias Pohlen: Support Vector Machines 22 / 48 February 10, 2014
23 Maximum margin visualized Decision surface of a maximum margin classifier: x 2 γ γ x 1 Tobias Pohlen: Support Vector Machines 23 / 48 February 10, 2014
24 Computing margins Lemma 1. Margin of h f on X can be computed by γ = min,...,n y i f(x i ) w 2 (23) y if(x i ) w 2 is the euclidean distance from the point x i to the decision surface Tobias Pohlen: Support Vector Machines 24 / 48 February 10, 2014
25 SVM Learning Assume the training set X = {(x 1, y 1 ),..., (x N, y N )} is linearly separable. How can we determine w, b such that the margin is maximal? Observation: max w,b γ = max w,b min,...,n If we scale w and b, the decision surface does not move: Idea: y i f(x i ) w 2 (24) λf(x) = 0 f(x) = 0, for λ > 0 (25) (w, b) and (λw, λb) induce the same decision surface Scale w, b such that y i f(x i ) = 1 (26) for the training examples x i that have the smallest distance to the decision surface Tobias Pohlen: Support Vector Machines 25 / 48 February 10, 2014
26 Optimization problem (primal form) We can maximize γ by minimizing w 2 : max w,b γ = max w,b min,...,n y i f(x i ) w 2 1 = max w,b w 2 (27) 1 = min w,b w 2 (28) Resulting optimization problem in the linearly separable case (primal form): 1 min w R m,b R 2 w 2 2 (29) subject to y i f(x i ) 1, i = 1,..., N (30) Tobias Pohlen: Support Vector Machines 26 / 48 February 10, 2014
27 Optimization theory: Lagrange multipliers Introduce Langrange function and Lagrange multipliers. Optimization problem: { ci (x) = 0, i E min f(x) subject to x Rm c i (x) 0, i I (31) Lagrange function: L(x, λ) = f(x) i E I λ i c i (x) (32) λ = (λ 1,..., λ l ) are called Lagrange mutlipliers Minimizing f subject to the constraints is equivalent to Minimizing L w.r.t. x Maximizing L w.r.t. λ Tobias Pohlen: Support Vector Machines 27 / 48 February 10, 2014
28 Optimization theory: KKT conditions First order necessary conditions If x is local minimum of f (respecting the constraints), then there exists λ such that x L(x, λ ) = 0 (33) c i (x ) = 0, i E (34) c i (x ) 0, i I (35) λ i 0, i I (36) λ i c i(x ) = 0, i E I (37) All these conditions are called Karush-Kuhn-Tucker conditions (KKT conditions) The last conditions are called complementary conditions See [Nocedal & Wrigt 06] for an in-depth introduction. Tobias Pohlen: Support Vector Machines 28 / 48 February 10, 2014
29 Lagrangian and KKT conditions SVM training problem: X = {(x 1, y 1 ),..., (x N, y N )} is linearly separable Lagrange function: KKT conditions: 1 min w R m,b R 2 w 2 2 (38) subject to y i ( w, φ(x i ) + b) 1 0, i = 1,..., N (39) L(w, b, α) = 1 N 2 w 2 2 α i (y i ( w, φ(x i ) + b) 1) (40) L(w, b, α) = w y i α i φ(x i ) =! 0 w = y i α i φ(x i ) (41) w! L(w, b, α) = y i α i = 0 (42) b Tobias Pohlen: Support Vector Machines 29 / 48 February 10, 2014
30 Finding the dual (part 1) KKT condition (part 1) L(w, b, α) = w w y i α i φ(x i ) =! 0 w = y i α i φ(x i ) (43) Substituting this condition back into the Langrange function: L(w, b, α) = 1 N 2 w 2 2 α i (y i ( w, φ(x i ) + b) 1) (44) = 1 N y i α i φ(x i ), 2 = 1 2 = y i α i φ(x i ) y i y j α i α j φ(x i ), φ(x j ) j=1 α i 1 2 α i N y i i=j y j α j φ(x j ), φ(x i ) y i y j α i α j φ(x i ), φ(x j ) b j=1 y i y j α i α j φ(x i ), φ(x j ) b j=1 + b 1 (45) α i y i + (46) α i y i (47) α i Tobias Pohlen: Support Vector Machines 30 / 48 February 10, 2014
31 Finding the dual (part 2) KKT condition (part 2) L(w, b, α) = b y i α i! = 0 (48) Substituting this condition back into the Langrange function: L(w, b, α) = α i 1 2 y i y j α i α j φ(x i ), φ(x j ) b α i y i (49) = α i 1 2 j=1 j=1 y i y j α i α j φ(x i ), φ(x j ) }{{} =k(x i,x j ) } {{ } =0 (50) Tobias Pohlen: Support Vector Machines 31 / 48 February 10, 2014
32 KKT-conditions for inequality constraints: Optimization problem (dual form) λ i 0, i I (51) Optimization problem in the dual form: X = {(x 1, y 1 ),..., (x N, y N )} is linearly separable max α R N subject to α i 1 y i y j α i α j k(x i, x j ) (52) 2 j=1 y i α i = 0 (53) α i 0, i = 1,..., N (54) This is a quadratic programming problem Quadratic objective function, Linear constraints Convex problem Unique solution Can be solved using standard solvers Complexity O(N 3 ), N = # training examples Tobias Pohlen: Support Vector Machines 32 / 48 February 10, 2014
33 How do we classify new data points? SVM classification Need to determine θ i for the dual form f(x) = N θ ik(x i, x) KKT conditions yield: L(w, b, α) = w y i α i φ(x i ) =! 0 (55) w w = y i α i φ(x i ) (56) Insert into linear discriminant function (primal form): f(x) = w, φ(x) + b (57) N = y i α i φ(x i ), φ(x) + b (58) = y }{{} i α i k(x i, x) + b (59) θ i Tobias Pohlen: Support Vector Machines 33 / 48 February 10, 2014
34 Sparsity Why is it called a sparse kernel machine? Recall: Evaluating f(x) in the dual form costs O(N) f(x) = y i α i k(x, x i ) + b, α i R (60) Optimization problem (primal form): Complementary conditions: Meaning Either α i = 0 Or y i f(x i ) = 1 1 min w R m,b R 2 w 2 2 (61) subject to y i f(x i ) 1 0, i = 1,..., N (62) α i (y i f(x i ) 1) = 0, i = 1,..., N (63) Tobias Pohlen: Support Vector Machines 34 / 48 February 10, 2014
35 Support vectors α i (y i f(x i ) 1) = 0, i = 1,..., N (64) α i = 0 y i f(x i ) = 1 (65) only the Lagrange multipliers of the points closest to the hyperplane are non-zero. We call those points support vectors Set of support vector indices is denoted by SV Evaluating f in the dual form is linear in the number of support vectors f(x) = y i α i k(x, x i ) + b = i SV y i α i k(x, x i ) + b (66) Typically: N SV (67) Tobias Pohlen: Support Vector Machines 35 / 48 February 10, 2014
36 Support vectors visualized x 2 x 1 Support vectors are circled Dashed lines are called margin boundaries Tobias Pohlen: Support Vector Machines 36 / 48 February 10, 2014
37 Support vectors - Implications Only the support vectors influence the decision surface The support vectors are the points hardest to classify Give insight into the classification problem After learning, only the support vectors and their respective weights have to be saved Modelsize is small compared to the size of the training set Tobias Pohlen: Support Vector Machines 37 / 48 February 10, 2014
38 Remaining questions What do we do if X is not linearly separable? What do we do if we have more than two classes? Tobias Pohlen: Support Vector Machines 38 / 48 February 10, 2014
39 Learning inseparable case X is not linearly separable. Introduce a penalty for points that violate the maximum margin constraint Penalty increases linearly in the distance from the respective boundary Introduce so called slack variables ξ i y = 1 y = 0 ξ > 1 y = 1 ξ < 1 ξ = 0 ξ = 0 Optimization problem (primal form): 1 min w R m,b R,ξ R N 2 w C ξ 1 (68) subject to y i ( w, φ(x i ) + b) 1 + ξ i 0, i = 1,..., N (69) ξ i 0, i = 1,..., N (70) Imagesource: C. M. Bishop Tobias Pohlen: Support Vector Machines 39 / 48 February 10, 2014
40 Learning inseparable case - Dual form The dual form can be found in the same manner as in the linearly separable case. Result: max α R N subject to α i 1 y i y j α i α j k(x i, x j ) (71) 2 j=1 y i α i = 0 (72) 0 α i C, i = 1,..., N (73) C is a tradeoff parameter that determines how strongly a point is punished b can be calculated as before Points for which α i 0 are still support vectors Tobias Pohlen: Support Vector Machines 40 / 48 February 10, 2014
41 SVMs for multiclass problems If we have K classes 1,..., K. Train K SVMs (one-against-all) Get K linear discriminant functions f 1,..., f K Assign a point to the class whose hyperplane is furthest from it Resulting classifier h f1,...,f K (x) = argmax,...,k f i (x) (74) Tobias Pohlen: Support Vector Machines 41 / 48 February 10, 2014
42 SVMs in practice In MATLAB Implement on your own (5 lines of code) Use quadprog to solve the quadratic programming problem Or use the built in library (better optimizer) svmtrain to train the SVM svmclassify to classify new data points Documentation: doc svmtrain or doc quadprog In C++ Very good and easy to use libraries are available such as LIBSVM SVMLight Highly optimized quadratic programming solvers Significantly faster than MATLAB Tobias Pohlen: Support Vector Machines 42 / 48 February 10, 2014
43 USPS Data Set USPS Data set ( images) Used feature vectors: R 256 (75) Tobias Pohlen: Support Vector Machines 43 / 48 February 10, 2014
44 Performance comparison SVM performance benchmark as summarized by Vapnik in [Vapnik 95] Classifier Best parameter choice SV Raw error in % Human performance % Decision tree, C % Best two-layer neural network % Five-layer network (LeNet 1) % SVM with polynomial kernel d = % SVM with Gaussian kernel σ 2 = % Tobias Pohlen: Support Vector Machines 44 / 48 February 10, 2014
45 Limitations of SVMs It s not clear how an appropiate kernel should be chosen for a given problem Computationally expensive: Training in the dual form (N training examples): O(N 3 ) Infeasible for large-scale applications Training in the primal form (m-dimensional input space): O(m 3 ) Evaluation of f in the dual form more expensive than in the primal form Evaluation practically infeasible if number of support vectors is very large Tobias Pohlen: Support Vector Machines 45 / 48 February 10, 2014
46 Conclusion SVMs use a simple linear model Feature space mappings enlarge the range of linearly separable training sets They can efficiently be used by enabling kernel functions Good generalization performance Margin concept Convex optimization problem In practice SVMs are good blackbox classifiers They give reasonable good results without much effort When dealing with new classification problems, it s often a good choice to try SVMs using different kernels Tobias Pohlen: Support Vector Machines 46 / 48 February 10, 2014
47 Support Vector Machines - References See [Nocedal & Wrigt 06] for an indepth introduction to numerical optimization (theory and practice) [Bishop 06] and [Christiani & Shawe-Taylor 00] for the derivations of the dual forms [Burges 98] for some thoughts about limitations [Dalal & Triggs 05] for a practical application of SVMs in computer vision [Vapnik 95] For a performance comparison (Kernel SVM vs. Neural Network) [Shawe-Taylor & Cristianini 06] for more background on kernels [Mercer 09] for a proof of Mercer s theorem Tobias Pohlen: Support Vector Machines 47 / 48 February 10, 2014
48 Thank you for your attention Tobias Pohlen Tobias Pohlen: Support Vector Machines 48 / 48 February 10, 2014
49 GoBack Tobias Pohlen: Support Vector Machines 49 / 48 February 10, 2014
50 References [Bartlett & Shawe Taylor 99] P. Bartlett, J. Shawe Taylor: Generalization Performance of Support Vector Machines and Other Pattern Classifiers. In B. Schölkopf, C.J.C. Burges, A.J. Smola, editors, Advances in Kernel Methods Support Vector Learning, pp , Cambridge, MA, MIT Press. 22 [Bishop 06] C.M. Bishop: Pattern Recognition and Machine Learning. Springer, New York, [Boser & Guyon + 92] B.E. Boser, I.M. Guyon, V.N. Vapnik: A Training Algorithm for Optimal Margin Classifiers. In Proceedings of the Fifth Annual Workshop on Computational Learning Theory, COLT 92, pp , New York, NY, USA, ACM. 22 [Burges 98] C.J.C. Burges: A Tutorial on Support Vector Machines for Pattern Recognition. Data Min. Knowl. Discov., Vol. 2, No. 2, pp , June [Christiani & Shawe-Taylor 00] N. Christiani, J. Shawe-Taylor: An introduction to support vector machines. Cambridge University Press, , 47 [Dalal & Triggs 05] N. Dalal, B. Triggs: Histograms of Oriented Gradients for Human Detection. In C. Schmid, S. Soatto, C. Tomasi, editors, International Conference on Computer Vision & Pattern Recognition, Vol. 2, pp , INRIA Rhône-Alpes, ZIRST-655, av. de l Europe, Montbonnot-38334, June Tobias Pohlen: Support Vector Machines 50 / 48 February 10, 2014
51 [Mercer 09] J. Mercer: Functions of positive and negative type, and their connection with the theory of integral equations. Philosophical Transactions of the Royal Society, London, Vol. 209, pp , [Nocedal & Wrigt 06] J. Nocedal, S.J. Wrigt: Numerical optimization. Springer, second edition edition, , 47 [Shawe-Taylor & Bartlett + 98] J. Shawe-Taylor, P.L. Bartlett, R.C. Williamson, M. Anthony: Structural Risk Minimization Over Data-Dependent Hierarchies. IEEE Transactions on Information Theory, Vol. 44, No. 5, pp , [Shawe-Taylor & Cristianini 06] J. Shawe-Taylor, N. Cristianini: Kernel Methods for Pattern Analysis. Cambridge University Press, [Vapnik 82] V. Vapnik: Estimation of Dependences Based on Empirical Data: Springer Series in Statistics (Springer Series in Statistics). Springer-Verlag New York, Inc., Secaucus, NJ, USA, [Vapnik 95] V.N. Vapnik: The Nature of Statistical Learning Theory. Springer, , 44, 47 Tobias Pohlen: Support Vector Machines 51 / 48 February 10, 2014
52 The Blackslide GoBack
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