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**An Introduction of Support Vector Machine**

Courtesy of Jinwei Gu

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**Today: Support Vector Machine (SVM)**

A classifier derived from statistical learning theory by Vapnik, et al. in 1992 SVM became famous when, using images as input, it gave accuracy comparable to neural-network with hand-designed features in a handwriting recognition task Currently, SVM is widely used in object detection & recognition, content-based image retrieval, text recognition, biometrics, speech recognition, etc.

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**Discriminant Function**

It can be arbitrary functions of x, such as: Decision Tree Linear Functions Nonlinear Functions

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**Linear Discriminant Function**

g(x) is a linear function: x2 wT x + b > 0 g(x) is a hyper-plane in the n-dimensional feature space, w and x vectors w1x1+w2x2+b=0 wT x + b = 0 n (Unit-length) normal vector of the hyper-plane: wT x + b < 0 x1

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**Linear Discriminant Function**

denotes +1 denotes -1 How would you classify these points using a linear discriminant function in order to minimize the error rate? x2 Infinite number of answers! x1

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**Linear Discriminant Function**

denotes +1 denotes -1 How would you classify these points using a linear discriminant function in order to minimize the error rate? x2 Infinite number of answers! x1

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**Linear Discriminant Function**

denotes +1 denotes -1 How would you classify these points using a linear discriminant function in order to minimize the error rate? x2 Infinite number of answers! x1

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**Linear Discriminant Function**

denotes +1 denotes -1 How would you classify these points using a linear discriminant function in order to minimize the error rate? x2 Infinite number of answers! Which one is the best? x1

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**Large Margin Linear Classifier**

denotes +1 denotes -1 The linear discriminant function (classifier) with the maximum margin is the best x2 Margin “safe zone” Margin is defined as the width that the boundary could be increased by before hitting a data point Why it is the best? Robust to outliners and thus strong generalization ability x1

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**Large Margin Linear Classifier**

denotes +1 denotes -1 Given a set of data points: x2 , where Note: class is +1,-1 NOT 1,0) With a scale transformation on both w and b, the above is equivalent to x1

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**Large Margin Linear Classifier**

denotes +1 denotes -1 We know that x2 Margin x+ x- wT x + b = 1 Support Vectors wT x + b = 0 wT x + b = -1 The margin width is: n x1

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**Large Margin Linear Classifier**

denotes +1 denotes -1 Formulation: x2 Margin x+ x- wT x + b = 1 such that wT x + b = 0 wT x + b = -1 n x1 Remember: our objective is finding w, b such that margins are maximized

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**Large Margin Linear Classifier**

denotes +1 denotes -1 Equivalent formulation: x2 Margin x+ x- wT x + b = 1 such that wT x + b = 0 wT x + b = -1 n x1

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**Large Margin Linear Classifier**

denotes +1 denotes -1 Equivalent formulation: x2 Margin x+ x- wT x + b = 1 such that wT x + b = 0 wT x + b = -1 n x1

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**Solving the Optimization Problem**

s.t. Quadratic programming with linear constraints s.t. Lagrangian Function

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Lagrangian Given a function f(x) and a set of constraints c1..cn, a Lagrangian is a function L(f,c1,..cn,1,.. n) that “incorporates” constraints in the optimization problem The optimum is at a point where (Karush-Kuhn-Tucker (KKT) conditions): The third condition is known as complementarity condition

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**Solving the Optimization Problem**

s.t. To minimize L we need to maximize the red box

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**Dual Problem Formulation (2)**

s.t. Since and Because of complementary condition 3 Lagrangian Dual Problem s.t. , and Each non-zero αi indicates that corresponding xi is a support vector SV.

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Why only SV? The complementarity condition (the third one of KKT, see previous definition of Lagrangians) implies that one of the 2 multiplicands is zero. However, for non-SV xi. Therefore only data points on the margin (=SV) will have a non-zero α (because they are the only ones for wich yi (wTxi + b ) = 1 )

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**Final Formulation To summaryze:**

Again, remember that the constraint (2) can be verified with a non-equality only for the SV!! Find α1…αn such that Q(α) =Σαi - ½ΣΣαiαjyiyjxiTxj is maximized and (1) Σαiyi = 0 (2) αi ≥ 0 for all αi

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**The Optimization Problem Solution**

Given a solution α1…αn to the dual problem Q(α) , solution to the primal is: Each non-zero αi indicates that corresponding xi is a support vector. Then the classifying function for a new point x is (note that we don’t need w explicitly): Notice that it relies on an inner product xiTx between the test point x and the support vectors xi . Only SV are useful for classification of new instances!!. Also keep in mind that solving the optimization problem involved computing the inner products xiTxj between all training points. w =Σαiyixi b = yk - Σαiyixi Txk for any αk > 0 g(x) = ΣαiyixiTx + b (xi are SV)

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Working example A 3-point training set: <(1,1)+><(2,0)+><(2,3),-> The maximum margin weight vector will be parallel to the shortest line connecting points of the two classes, that is, the line between (1,1) and (2,3) (the 2 Support Vectors) W

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Margin computation wTx1+b=+1 wTx2+b=-1

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**Soft Margin Classification**

What if the training set is not linearly separable? Slack variables ξi can be added to allow misclassification of difficult or noisy examples, resulting margins are called soft. ξi ξi xi are the misclassified examples

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**Large Margin Linear Classifier**

Formulation: such that Parameter C can be viewed as a way to control over-fitting.

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**Large Margin Linear Classifier**

Formulation: (Lagrangian Dual Problem) such that

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Non-linear SVMs Datasets that are linearly separable with noise work out great: x x But what are we going to do if the dataset is just too hard? How about… mapping data to a higher-dimensional space: x x2 This slide is courtesy of

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**Non-linear SVMs: Feature Space**

General idea: the original input space can be mapped to some higher-dimensional feature space where the training set is separable: Φ: x → φ(x) This slide is courtesy of

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**The original objects (left side of the schematic) are mapped, i. e**

The original objects (left side of the schematic) are mapped, i.e., rearranged, using a set of mathematical functions, known as kernels. φ(x)

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**Nonlinear SVMs: The Kernel Trick**

With this mapping, our discriminant function is now: Original formulation was g(x) = ΣαiyixiTx + b No need to know this mapping explicitly, because we only use the dot product of feature vectors in both the training and test. A kernel function is defined as a function that corresponds to a dot product of two feature vectors in some expanded feature space:

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**Nonlinear SVMs: The Kernel Trick**

An example: 2-dimensional vectors x=[x1 x2]; let K(xi,xj)=(1 + xiTxj)2, Need to show that K(xi,xj) = φ(xi) Tφ(xj): K(xi,xj)=(1 + xiTxj)2, = 1+ xi12xj xi1xj1 xi2xj2+ xi22xj22 + 2xi1xj1 + 2xi2xj2 = [1 xi12 √2 xi1xi2 xi22 √2xi1 √2xi2]T [1 xj12 √2 xj1xj2 xj22 √2xj1 √2xj2] = φ(xi) Tφ(xj), where φ(x) = [1 x12 √2 x1x2 x22 √2x1 √2x2] This slide is courtesy of

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**Nonlinear SVMs: The Kernel Trick**

Examples of commonly-used kernel functions: Linear kernel: Polynomial kernel: Gaussian (Radial-Basis Function (RBF) ) kernel: Sigmoid: In general, functions that satisfy Mercer’s condition can be kernel functions.

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**Nonlinear SVM: Optimization**

Formulation: (Lagrangian Dual Problem) such that The solution of the discriminant function is The optimization technique is the same.

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**Support Vector Machine: Algorithm**

1. Choose a kernel function 2. Choose a value for C 3. Solve the quadratic programming problem (many software packages available) 4. Construct the discriminant function from the support vectors

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**Some Issues Choice of kernel Choice of kernel parameters**

- Gaussian or polynomial kernel is default - if ineffective, more elaborate kernels are needed - domain experts can give assistance in formulating appropriate similarity measures Choice of kernel parameters - e.g. σ in Gaussian kernel - σ is the distance between closest points with different classifications - In the absence of reliable criteria, applications rely on the use of a validation set or cross-validation to set such parameters. Optimization criterion – Hard margin v.s. Soft margin - a lengthy series of experiments in which various parameters are tested This slide is courtesy of

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**Summary: Support Vector Machine**

1. Large Margin Classifier Better generalization ability & less over-fitting 2. The Kernel Trick Map data points to higher dimensional space in order to make them linearly separable. Since only dot product is used, we do not need to represent the mapping explicitly.

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Additional Resource

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**LibSVM (best implementation for SVM)**

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