Kernels and the Kernel Trick. Machine Learning Fall 2017

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1 Kernels and the Kernel Trick Machine Learning Fall

2 Support vector machines Training by maximizing margin The SVM objective Solving the SVM optimization problem Support vectors, duals and kernels 2

3 Support vector machines Training by maximizing margin The SVM objective Solving the SVM optimization problem Support vectors, duals and kernels 3

4 This lecture 1. Support vectors 2. Kernels 3. The kernel trick 4. Properties of kernels 5. Another example of the kernel trick 4

5 This lecture 1. Support vectors 2. Kernels 3. The kernel trick 4. Properties of kernels 5. Another example of the kernel trick 5

6 So far we have seen Support vector machines Hinge loss and optimizing the regularized loss More broadly, different algorithms for learning linear classifiers 6

7 So far we have seen Support vector machines Hinge loss and optimizing the regularized loss More broadly, different algorithms for learning linear classifiers What about non-linear models? 7

8 One way to learn non-linear models Explicitly introduce non-linearity into the feature space If the true separator is quadratic 8

9 One way to learn non-linear models Explicitly introduce non-linearity into the feature space If the true separator is quadratic Transform all input points as 9

10 One way to learn non-linear models Explicitly introduce non-linearity into the feature space If the true separator is quadratic Transform all input points as Now, we can try to find a weight vector in this higher dimensional space That is, predict using w T Á(x 1, x 2 ) b 10

11 SVM: Primals and duals The SVM objective This is called the primal form of the objective This can be converted to its dual form, which will let us prove a very useful property 11

12 SVM: Primals and duals The SVM objective This is called the primal form of the objective This can be converted to its dual form, which will let us prove a very useful property Another optimization problem Has the property that max Dual = min Primal 12

13 Support vector machines Let w be the minimizer of the SVM problem for some dataset with m examples: {(x i, y i )} Then, for i = 1 m, there exist i 0 such that the optimum w can be written as 13

14 Support vector machines Let w be the minimizer of the SVM problem for some dataset with m examples: {(x i, y i )} Then, for i = 1 m, there exist i 0 such that the optimum w can be written as Furthermore, All points outside the margin 14

15 Support vector machines Let w be the minimizer of the SVM problem for some dataset with m examples: {(x i, y i )} Then, for i = 1 m, there exist i 0 such that the optimum w can be written as Furthermore, All points on the wrong side of the margin 15

16 Support vector machines Let w be the minimizer of the SVM problem for some dataset with m examples: {(x i, y i )} Then, for i = 1 m, there exist i 0 such that the optimum w can be written as Furthermore, All points on the margin 16

17 Support vectors The weight vector is completely defined by training examples whose i s are not zero These examples are called the support vectors 17

18 This lecture ü Support vectors 2. Kernels 3. The kernel trick 4. Properties of kernels 5. Another example of the kernel trick 18

19 Predicting with linear classifiers Prediction = and That is, we just showed that We only need to compute dot products between training examples and the new example x 19

20 Predicting with linear classifiers Prediction = and That is, we just showed that We only need to compute dot products between training examples and the new example x 20

21 Predicting with linear classifiers Prediction = and That is, we just showed that We only need to compute dot products between training examples and the new example x 21

22 Predicting with linear classifiers Prediction = and That is, we just showed that We only need to compute dot products between training examples and the new example x This is true even if we map examples to a high dimensional space 22

23 Predicting with linear classifiers Prediction = and That is, we just showed that That is we only need to compute dot products between training examples and the new example x This is true even if we map examples to a high dimensional space 23

24 Dot products in high dimensional spaces Let us define a dot product in the high dimensional space So prediction with this high dimensional lifting map is because 24

25 Dot products in high dimensional spaces Let us define a dot product in the high dimensional space So prediction with this high dimensional lifting map is because 25

26 Dot products in high dimensional spaces Let us define a dot product in the high dimensional space So prediction with this high dimensional lifting map is because 26

27 Kernel based methods Predict using What does this new formulation give us? If we have to compute Á every time anyway, we gain nothing If we can compute the value of K without explicitly writing the blown up representation, then we will have a computational advantage 27

28 Kernel based methods Predict using What does this new formulation give us? If we have to compute Á every time anyway, we gain nothing If we can compute the value of K without explicitly writing the blown up representation, then we will have a computational advantage. 28

29 This lecture ü Support vectors ü Kernels 3. The kernel trick 4. Properties of kernels 5. Another example of the kernel trick 29

30 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] 30

31 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] All degree zero terms 31

32 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] All degree zero terms All degree one terms 32

33 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] All degree zero terms All degree one terms All degree two terms 33

34 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] All degree zero terms All degree one terms All degree two terms and compute the dot product A = Á(x) T Á (z) [takes time ] 34

35 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] and compute the dot product A = Á(x) T Á (z) [takes time ] Instead, in the original space, compute Theorem: A = B (Coefficients do not really matter) 35

36 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] and compute the dot product A = Á(x) T Á (z) [takes time ] Instead, in the original space, compute Theorem: A = B (Coefficients do not really matter) 36

37 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] and compute the dot product A = Á(x) T Á (z) [takes time ] Instead, in the original space, compute Theorem: A = B (Coefficients do not really matter) 37

38 Example: Polynomial Kernel Given two examples x and z we want to map them to a high dimensional space [for example, quadratic] and compute the dot product A = Á(x) T Á (z) [takes time ] Instead, in the original space, compute Claim: A = B (Coefficients do not really matter) 38

39 Example: Two dimensions, quadratic kernel A = Á(x) T Á (z) 39

40 The Kernel Trick Suppose we wish to compute K(x, z) = Á(x) T Á (z) Here Á maps x and z to a high dimensional space The Kernel Trick: Save time/space by computing the value of K(x, z) by performing operations in the original space (without a feature transformation!) 40

41 Computing dot products efficiently Kernel Trick: You want to work with degree 2 polynomial features, Á(x). Then, your dot product will be operate using vectors in a space of dimensionality n(n+1)/2. The kernel trick allows you to save time/space and compute dot products in an n dimensional space. (Not just for degree 2 polynomials) Can we use any function K(.,.)? No! A function K(x,z) is a valid kernel if it corresponds to an inner product in some (perhaps infinite dimensional) feature space. General condition: construct the Gram matrix {K(x i,z j )}; check that it s positive semi definite 41

42 This lecture ü Support vectors ü Kernels ü The kernel trick 4. Properties of kernels 5. Another example of the kernel trick 42

43 Which functions are kernels? Kernel Trick: You want to work with degree 2 polynomial features, Á(x). Then, your dot product will be operate using vectors in a space of dimensionality n(n+1)/2. The kernel trick allows you to save time/space and compute dot products in an n dimensional space. (Not just for degree 2 polynomials) Can we use any function K(.,.)? No! A function K(x,z) is a valid kernel if it corresponds to an inner product in some (perhaps infinite dimensional) feature space. General condition: construct the Gram matrix {K(x i,z j )}; check that it s positive semi definite 43

44 Which functions are kernels? Kernel Trick: You want to work with degree 2 polynomial features, Á(x). Then, your dot product will be operate using vectors in a space of dimensionality n(n+1)/2. The kernel trick allows you to save time/space and compute dot products in an n dimensional space. (Not just for degree 2 polynomials) Can we use any function K(.,.)? No! A function K(x,z) is a valid kernel if it corresponds to an inner product in some (perhaps infinite dimensional) feature space. General condition: construct the Gram matrix {K(x i,z j )}; check that it s positive semi definite 44

45 Which functions are kernels? Kernel Trick: You want to work with degree 2 polynomial features, Á(x). Then, your dot product will be operate using vectors in a space of dimensionality n(n+1)/2. The kernel trick allows you to save time/space and compute dot products in an n dimensional space. (Not just for degree 2 polynomials) Can we use any function K(.,.)? No! A function K(x,z) is a valid kernel if it corresponds to an inner product in some (perhaps infinite dimensional) feature space. General condition: construct the Gram matrix {K(x i,z j )}; check that it s positive semi definite 45

46 Reminder: Positive semi-definite matrices A symmetric matrix M is positive semi-definite if it is For any vector non-zero z, we have z T Mz 0 (A useful property characterizing many interesting mathematical objects) 46

47 The Kernel Matrix The Gram matrix of a set of n vectors S = {x 1 x n } is the n n matrix G with G ij = x it x j The kernel matrix is the Gram matrix of {φ(x 1 ),,φ(x n )} (size depends on the # of examples, not dimensionality) Showing that a function K is a valid kernel Direct approach: If you have the φ(x i ), you have the Gram matrix (and it s easy to see that it will be positive semi-definite). Why? Indirect: If you have the Kernel, write down the Kernel matrix K ij, and show that it is a legitimate kernel, without an explicit construction of φ(x i ) 47

48 The Kernel Matrix The Gram matrix of a set of n vectors S = {x 1 x n } is the n n matrix G with G ij = x it x j The kernel matrix is the Gram matrix of {φ(x 1 ),,φ(x n )} (size depends on the # of examples, not dimensionality) Showing that a function K is a valid kernel Direct approach: If you have the φ(x i ), you have the Gram matrix (and it s easy to see that it will be positive semi-definite). Why? Indirect: If you have the Kernel, write down the Kernel matrix K ij, and show that it is a legitimate kernel, without an explicit construction of φ(x i ) 48

49 Mercer s condition Let K(x, z) be a function that maps two n dimensional vectors to a real number K is a valid kernel if for every finite set {x 1, x 2,! }, for any choice of real valued c 1, c 2,!, we have 49

50 Polynomial kernels Linear kernel: k(x, z) = x T z Polynomial kernel of degree d: k(x, z) = (x T z) d only dth-order interactions Polynomial kernel up to degree d: k(x, z) = (x T z + c) d (c>0) all interactions of order d or lower 50

51 Gaussian Kernel (or the radial basis function kernel) (x z) 2 : squared Euclidean distance between x and z c = σ 2 : a free parameter very small c: K identity matrix (every item is different) very large c: K unit matrix (all items are the same) k(x, z) 1 when x, z close k(x, z) 0 when x, z dissimilar 51

52 Gaussian Kernel (or the radial basis function kernel) (x z) 2 : squared Euclidean distance between x and z c = σ 2 : a free parameter very small c: K identity matrix (every item is different) very large c: K unit matrix (all items are the same) k(x, z) 1 when x, z close k(x, z) 0 when x, z dissimilar Exercises: 1. Prove that this is a kernel. 2. What is the blown up feature space for this kernel? 52

53 Constructing New Kernels You can construct new kernels k (x, x ) from existing ones: Multiplying k(x, x ) by a constant c ck(x, x ) Multiplying k(x, x ) by a function f applied to x and x f(x)k(x, x )f(x ) Applying a polynomial (with non-negative coefficients) to k(x, x ) P( k(x, x ) ) with P(z) = i a i z i and a i 0 Exponentiating k(x, x ) exp(k(x, x )) 53

54 Constructing New Kernels (2) You can construct k (x, x ) from k 1 (x, x ), k 2 (x, x ) by: Adding k 1 (x, x ) and k 2 (x, x ): k 1 (x, x ) + k 2 (x, x ) Multiplying k 1 (x, x ) and k 2 (x, x ): k 1 (x, x )k 2 (x, x ) Also: If φ(x)2 R m and k m (z, z ) a valid kernel in R m, k(x, x ) = k m (φ(x), φ(x )) is also a valid kernel If A is a symmetric positive semi-definite matrix, k(x, x ) = xax is also a valid kernel 54

55 Constructing New Kernels (2) You can construct k (x, x ) from k 1 (x, x ), k 2 (x, x ) by: Adding k 1 (x, x ) and k 2 (x, x ): k 1 (x, x ) + k 2 (x, x ) Multiplying k 1 (x, x ) and k 2 (x, x ): k 1 (x, x )k 2 (x, x ) Also: If φ(x)2 R m and k m (z, z ) a valid kernel in R m, k(x, x ) = k m (φ(x), φ(x )) is also a valid kernel If A is a symmetric positive semi-definite matrix, k(x, x ) = xax is also a valid kernel 55

56 Kernel Trick: An example Let the blown up feature space represent the space of all 3 n conjunctions. Then, where same(x,z) is the number of features that have the same value for both x and z Example: Take n=3; x=(001), z=(011), we have conjunctions of size 0,1,2,3 Proof: let m=same(x, z); construct surviving conjunctions by 1. choosing to include one of these k literals with the right polarity in the conjunctions, or 2. choosing to not include it at all. Conjunctions with literals outside this set disappear. 56

57 This lecture ü Support vectors ü Kernels ü The kernel trick ü Properties of kernels 5. Another example of the kernel trick 57

58 Kernel Trick: An example Let the blown up feature space represent the space of all 3 n conjunctions. Then, where same(x,z) is the number of features that have the same value for both x and z Example: Take n=3; x=(001), z=(011), we have conjunctions of size 0,1,2,3 Proof: let m=same(x, z); construct surviving conjunctions by 1. choosing to include one of these k literals with the right polarity in the conjunctions, or 2. choosing to not include it at all. Conjunctions with literals outside this set disappear. 58

59 Kernel Trick: An example Let the blown up feature space represent the space of all 3 n conjunctions. Then, where same(x,z) is the number of features that have the same value for both x and z Example: Take n=3; x=(001), z=(011), we have conjunctions of size 0,1,2,3 Proof: let m=same(x, z); construct surviving conjunctions by 1. choosing to include one of these k literals with the right polarity in the conjunctions, or 2. choosing to not include it at all. Conjunctions with literals outside this set disappear. 59

60 Kernel Trick: An example Let the blown up feature space represent the space of all 3 n conjunctions. Then, where same(x,z) is the number of features that have the same value for both x and z Example: Take n=3; x=(001), z=(011), we have conjunctions of size 0,1,2,3 Proof: let m=same(x, z); construct surviving conjunctions by 1. choosing to include one of these k literals with the right polarity in the conjunctions, or 2. choosing to not include it at all. Conjunctions with literals outside this set disappear. 60

61 Kernel Trick: An example Let the blown up feature space represent the space of all 3 n conjunctions. Then, where same(x,z) is the number of features that have the same value for both x and z Example: Take n=3; x=(001), z=(011), we have conjunctions of size 0,1,2,3 Proof: let m=same(x, z); construct surviving conjunctions by 1. choosing to include one of these k literals with the right polarity in the conjunctions, or 2. choosing to not include it at all. Conjunctions with literals outside this set disappear. 61

62 Kernel Trick: An example Let the blown up feature space represent the space of all 3 n conjunctions. Then, where same(x,z) is the number of features that have the same value for both x and z Example: Take n=3; x=(001), z=(011), we have conjunctions of size 0,1,2,3 Proof: let m=same(x, z); construct surviving conjunctions by 1. choosing to include one of these k literals with the right polarity in the conjunctions, or 2. choosing to not include it at all. Conjunctions with literals outside this set disappear. 62

63 Exercises 1. Show that this argument works for a specific example Take X={x 1, x 2, x 3, x 4 } Á(x) = The space of all 3 n conjunctions ; Á(x) = 81 Consider x=(1100), z=(1101) Write Á(x), Á(z), the representation of x, z in the Á space Compute Á(x) T Á(z) Show that K(x,z) =Á(x) T Á(z)= å i Á i (z) Á i (x) = 2 same(x,z) = 8 2. Try to develop another kernel, e.g., where the space of all conjunctions of size 3 (exactly) 63

64 Summary: Kernel trick To make the final prediction, we are computing dot products The kernel trick is a computational trick to compute dot products in higher dimensional spaces This is applicable not just to SVMs. The same idea can be extended to Perceptron too: the Kernel Perceptron Important: All the bounds we have seen (eg: Perceptron bound, etc) depend on the underlying dimensionality By moving to a higher dimensional space, we are incurring a penalty on sample complexity 64

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