nips nips2002 nips2002-119 knowledge-graph by maker-knowledge-mining

119 nips-2002-Kernel Dependency Estimation

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Author: Jason Weston, Olivier Chapelle, Vladimir Vapnik, André Elisseeff, Bernhard Schölkopf

Abstract: We consider the learning problem of finding a dependency between a general class of objects and another, possibly different, general class of objects. The objects can be for example: vectors, images, strings, trees or graphs. Such a task is made possible by employing similarity measures in both input and output spaces using kernel functions, thus embedding the objects into vector spaces. We experimentally validate our approach on several tasks: mapping strings to strings, pattern recognition, and reconstruction from partial images. 1

Reference: text

Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 Such a task is made possible by employing similarity measures in both input and output spaces using kernel functions, thus embedding the objects into vector spaces. [sent-3, score-0.604]

2 We experimentally validate our approach on several tasks: mapping strings to strings, pattern recognition, and reconstruction from partial images. [sent-4, score-0.446]

3 1 Introduction In this article we consider the rather general learning problem of finding a dependency between inputs x E X and outputs y E Y given a training set (Xl,yl), . [sent-5, score-0.46]

4 g mapping of a certain class of strings to a certain class of graphs (as in text parsing) or the mapping of text descriptions to images. [sent-11, score-0.555]

5 , ex), ex E A} which provides the minimum value of the risk function R(ex) = r ix xY L(y , j(x,ex))dP(x, y) (1) where P is the (unknown) joint distribution ofx and y and L(y, 1]) is a loss function, a measure of distance between the estimate 1] and the true output y at a point x. [sent-13, score-0.406]

6 Hence in this setting one is given a priori knowledge of the similarity measure used in the space Y in the form of a loss function. [sent-14, score-0.251]

7 In pattern recognition this is often the zero-one loss, in regression often squared loss is chosen. [sent-15, score-0.305]

8 Most algorithms have focussed on the pattern recognition and regression problems and cannot deal with more general outputs. [sent-20, score-0.168]

9 Conversely, specialist algorithms have been made for structured outputs, for example the ones of text classification which calculate parse trees for natural language sentences, however these algorithms are specialized for their tasks. [sent-21, score-0.342]

10 Recently, kernel methods [12, 11] have been extended to deal with inputs that are structured objects such as strings or trees by linearly embedding the objects using the so-called kernel trick [5, 7]. [sent-22, score-1.349]

11 These objects are then used in pattern recognition or regression domains. [sent-23, score-0.224]

12 The algorithm ends up in an formulation which has a kernel function for the inputs and a kernel function (which will correspond to choosing a particular loss function) for the outputs. [sent-25, score-0.841]

13 This also enables us (in principle) to encode specific prior information about the outputs (such as special cost functions and/or invariances) in an elegant way, although this is not experimentally validated in this work. [sent-26, score-0.228]

14 In Section 2 it is shown how to use kernel functions to measure similarity between outputs as well as inputs. [sent-28, score-0.541]

15 For outputs one is usually given a loss function for measuring similarity (this can be, but is not always , inherent to the problem domain). [sent-36, score-0.438]

16 For inputs, one way of measuring similarity is by using a kernel function. [sent-37, score-0.377]

17 A kernel k is a symmetric function which is an inner product in some Hilbert space F, i. [sent-38, score-0.446]

18 We can think of the patterns as k(X) , k(X / ), and carry out geometric algorithms in the inner product space ("feature space") F. [sent-42, score-0.143]

19 Typical kernel functions are polynomials k(x, Xl) = (x . [sent-45, score-0.303]

20 Note that , like distances between examples in input space, it is also possible to think of the loss function as a distance measure in output space, we will denote this space 1:. [sent-47, score-0.377]

21 We can measure inner products in this space using a kernel function. [sent-48, score-0.41]

22 This map makes it possible to consider a large class of nonlinear loss functions. [sent-51, score-0.25]

23 l As in the traditional kernel trick for the inputs, the nonlinearity is only taken into account when computing the kernel matrix. [sent-52, score-0.666]

24 It also makes it possible to consider structured objects as outputs such as the ones described in [5]: strings, trees, graphs and so forth. [sent-56, score-0.364]

25 One embeds the output objects in the space I: using a kernel. [sent-57, score-0.235]

26 Let us define some kernel functions for output spaces. [sent-58, score-0.413]

27 IFor instance, assuming the outputs live in lI~n, usin~ an RBF kernel, one obtains a loss function II e(y) - e(Y/) 112 = 2 - 2 exp (-Ily - y'll /2(7 2 ). [sent-59, score-0.364]

28 This is a nonlinear loss function which takes the value 0 if Y and y' coincide, and 2 if they are maximally different . [sent-60, score-0.171]

29 To construct a corresponding inner product it is necessary to embed this distance into a Euclidean space, which can be done using the following kernel: £pat(y,y') = ~[y = y'], (2) as L(y, y')2 = Illf>f(Y) - If>f(y')11 2 = £(y, y) + £(y', y') - 2£(y, y') = 1 - [y = y']. [sent-68, score-0.136]

30 It corresponds to embedding into aM-dimensional Euclidean space via the map If>f( Y) = (0,0, . [sent-69, score-0.16]

31 It is also possible to describe multi-label classification (where anyone example belongs to an arbitrary subset of the M classes) in a similar way. [sent-73, score-0.171]

32 For regression estimation, one can use the usual inner product (3) £reg(y, y') = (y . [sent-77, score-0.211]

33 For outputs such as strings and other structured objects we require the corresponding string kernels and kernels for structured objects [5, 7]. [sent-79, score-1.035]

34 We give one example here, the string subsequence kernel employed in [7] for text categorization. [sent-80, score-0.591]

35 This kernel is an inner product in a feature space consisting of all ordered subsequences of length r, denoted ~r. [sent-81, score-0.489]

36 A fast way to compute this kernel is described in [7]. [sent-86, score-0.303]

37 Sometimes, one would also like apply the loss given by an (arbitrary) distance matrix D of the loss between training examples, i. [sent-87, score-0.411]

38 In general it is not always obvious to find an embedding of such data in an Euclidian space (in order to apply kernels) . [sent-89, score-0.143]

39 We wish to minimize the risk function (1) using the feature space F induced by the kernel k and the loss function measured in the space £ induced by the kernel £. [sent-98, score-0.925]

40 Our solution is the following: decompose If>e(Y) into p orthogonal directions using kernel principal components analysis (KPCA) (see, e. [sent-100, score-0.339]

41 One can then learn the mapping from If>k(X) to each direction independently using a standard kernel regression method, e. [sent-102, score-0.517]

42 Finally, to output an estimate Y given a test example x one must solve a pre-image problem as the solution of the algorithm is initially a solution in the space £. [sent-104, score-0.145]

43 1) Decomposition of outputs Let us construct the kernel matrix L on the training data such that Lij = f(Yi,Yj), and perform kernel principal components analysis on L. [sent-106, score-0.876]

44 One can learn the map by estimating each output independently. [sent-118, score-0.196]

45 In our experiments we use kernel ridge regression [9] , note that this requires only a single matrix inversion to learn all p directions. [sent-119, score-0.614]

46 , fp(x))11 For the kernel (3) it is possible to compute the solution explicit ely. [sent-132, score-0.303]

47 For the simple case of vectorial outputs with linear kernel (3), if the output is only one dimension the method of KDE boils down to the same solution as using ridge regression since the matrix L is rank 1 in this case. [sent-140, score-0.868]

48 However, when there are d outputs, the rank of L is d and the method trains ridge regression d times, but the kernel PCA step first decorrelates the outputs. [sent-141, score-0.565]

49 Thus, in the special case of multiple outputs regression with a linear kernel , the method is also related to the work of [2] (see [4, page 73] for an overview of other multiple output regression methods. [sent-142, score-0.84]

50 In the experiments we chose the parameters of KDE to be from the following : u* = {l0-3 , 10- 2,10-\ 10°,10\ 102, 103} where u = and the ridge parameter 3, 10- 2,10-\ 100 , 1O 1 }. [sent-145, score-0.244]

51 We chose them by five fold cross valida, = {l0-4, 10tion. [sent-146, score-0.312]

52 Three classes of strings consist ofletters from the same alphabet of 4 letters (a,b,c,d), and strings from all classes are generated with a random length between 10 to 15. [sent-149, score-0.506]

53 Strings from the first class are generated by a model where transitions from any letter to any other letter are equally likely. [sent-150, score-0.346]

54 The output is the string abad, corrupted with the following noise. [sent-151, score-0.311]

55 In the second class, transitions from one letter to itself (so the next letter is the same as the last) have probability 0. [sent-158, score-0.266]

56 The output is the string dbbd, but corrupted with the same noise as for class one. [sent-161, score-0.353]

57 In the third class only the letters c and d are used; transitions from one letter to itself have probability 0. [sent-162, score-0.197]

58 The output is the string aabc, but corrupted with the same noise as for class one. [sent-164, score-0.353]

59 For classes one and two any starting letter is equally likely, for the third class only c and d are (equally probable) starting letters. [sent-165, score-0.191]

60 input string ccdddddddd dccccdddcd adddccccccccc bbcdcdadbad cdaaccadcbccdd --+ --+ --+ --+ --+ output string aabc abc bb aebad abad Figure 1: Five examples from our artificial task (mapping strings to strings). [sent-166, score-0.818]

61 The task is to predict the output string given the input string. [sent-167, score-0.267]

62 Note that this is almost like a classification problem with three classes, apart from the noise on the outputs. [sent-168, score-0.171]

63 This construction was employed so we can also calculate classification error as a sanity check. [sent-169, score-0.199]

64 We use the string subsequence kernel (4) from [7] for both inputs and outputs, normalized such that k(x,x') = k(x,x' )/(Jk(x,x)Jk(x',x')). [sent-170, score-0.58]

65 In the space induced by the input kernel k we then chose a further nonlinear map using an RBF kernel: exp( - (k(x, x) + k(x',x') - 2k(x,x'))/2(J2). [sent-173, score-0.499]

66 We generated 200 such strings and measured the success by calculating the mean and standard error of the loss (computed via the output kernel) over 4 fold cross validation. [sent-174, score-0.766]

67 We chose (J (the width of the RBF kernel) and'Y (the ridge parameter) on each trial via a further level of 5 fold cross validation. [sent-175, score-0.506]

68 We compare our method to an adaptation of k-nearest neighbors for general outputs: if k = 1 it returns the output of the nearest neighbor , otherwise it returns the linear combination (in the space of outputs) of the k nearest neighbors (in input space) . [sent-176, score-0.145]

69 In the case of k > 1, as well as for KDE, we find a pre-image by finding the closest training example output to the given solution. [sent-177, score-0.298]

70 We choose k again via a further level of 5 fold cross validation. [sent-178, score-0.232]

71 026 Table 1: Performance of KDE and k-NN on the string to string mapping problem. [sent-188, score-0.376]

72 2 Multi-class classification problem We next tried a multi-class classification problem, a simple special case of the general dependency estimation problem. [sent-190, score-0.446]

73 We performed 5-fold cross validation on 1000 digits (the first 100 examples of each digit) of the USPS handwritten 16x16 pixel digit database, training with a single fold (200 examples) and testing on the remainder. [sent-191, score-0.651]

74 We used an RBF kernel for the inputs and the zero-one multi-class classification loss for the outputs using kernel (2). [sent-192, score-1.205]

75 We found k for k-NN and a and "( for the other methods (we employed a ridge also to the SVM method, reulting in a squared error penalization term) by another level of 5-fold cross validation. [sent-196, score-0.283]

76 0075 Table 2: Performance of KDE, 1-vs-rest SVMs and k-NN on a classification problem of handwritten digits. [sent-206, score-0.227]

77 3 Image reconstruction We then considered a problem of image reconstruction: given the top half (the first 8 pixel lines) of a USPS postal digit, it is required to estimate what the bottom half will be (we thus ignored the original labels of the data). [sent-208, score-0.387]

78 2 The loss function we choose for the outputs is induced by an RBF kernel. [sent-209, score-0.403]

79 Hence, the difficulty in this task is both choosing a good loss function (to reflect the end user's objectives) as well as an accurate estimator. [sent-211, score-0.171]

80 We chose the width a' of the output RBF kernel which maximized the kernel alignment [1] with a target kernel generated via k-means clustering. [sent-212, score-1.202]

81 We chose k=30 clusters and the target kernel is K ij = 1 if Xi and Xj are in the same cluster, and 0 otherwise. [sent-213, score-0.388]

82 We again performed 5-fold cross validation on the first 1000 digits of the USPS handwritten 16x16 pixel digit database, training with a single fold (200 examples) and testing on the remainder, comparing KDE to k-NN and a Hopfield net. [sent-217, score-0.618]

83 We then chose as output the pre-image from the training data that is closest to this solution (thus the possible outputs are the 2 A similar problem, of higher dimensionality, would be to learn the mapping from top half to complete digit . [sent-220, score-0.855]

84 3Note that training a naive regressor on each pixel output independently would not take into account that the combination of pixel outputs should resemble a digit . [sent-221, score-0.671]

85 Figure 2: Errors in the digit database image reconstruction problem. [sent-222, score-0.3]

86 Images have to be estimated using only the top half (first 8 rows of pixels) of the original image (top row) by KDE (middle row) and k-NN (bottom row). [sent-223, score-0.152]

87 We show all the test examples on the first fold of cross validation where k-NN makes an error in estimating the correct digit whilst KDE does not (73 mistakes) and vice-versa (23 mistakes). [sent-224, score-0.438]

88 0072 Table 3: Performance of KDE, k-NN and a Hopfield network on an image reconstruction problem of handwritten digits. [sent-241, score-0.156]

89 Note that we cannot easily compare classification rates on this problem using the pre-images selected since KDE outputs are not correlated well with the labels. [sent-243, score-0.364]

90 For example it will use the bottom stalk of a digit "7" or a digit "9" equally if they are identical, whereas k-NN will not: in the region of the input space which is the top half of "9"s it will only output the bottom half of "9"s. [sent-244, score-0.793]

91 This explains why measuring the class of the pre-images compared to the true class as a classification problem yields a lower loss for k-NN, 0. [sent-245, score-0.455]

92 Note that if we performed classification as in Section 4. [sent-252, score-0.171]

93 Finally, we note that nothing was stopping us from incorporating known invariances into our loss function in KDE via the kernel. [sent-260, score-0.259]

94 For example we could have used a kernel which takes into account local patches of pixels rendering spatial information or jittered kernels which take into account chosen transformations (translations, rotations, and so forth). [sent-261, score-0.404]

95 It may also be useful to add virtual examples to the output matrix 1:- before the decomposition step. [sent-262, score-0.143]

96 5 Discussion We have introduced a kernel method of learning general dependencies. [sent-264, score-0.303]

97 There are many applications of KDE to explore: problems with complex outputs (natural language parsing, image interpretation/manipulation, . [sent-266, score-0.237]

98 To make the approach work on more complex problems (where a test output is not so trivially close to a training output) improved pre-image approaches should be applied. [sent-273, score-0.151]

99 Although one can apply techniques such as [10] for vector based pre-images, efficiently finding pre-images for structured objects such as strings is an open problem. [sent-274, score-0.483]

100 Finally, the algorithm should be extended to deal with non-Euclidean loss functions directly, e. [sent-275, score-0.205]

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[8] uniﬁes the boosting process for two popular loss functions, the exponential-loss (denoted henceforth as ExpLoss) and logarithmic-loss (denoted as LogLoss) that bound the empir- ˜ ˜ Input: Labelled and unlabelled sets of examples: S = {(xi , yi )}m ; S = {˜i }m x i=1 i=1 Initialize: K ← 0 (all zeros matrix) For t = 1, 2, . . . , T : • Calculate distribution over pairs 1 ≤ i, j ≤ m: Dt (i, j) = exp(−yi yj K(xi , xj )) 1/(1 + exp(−yi yj K(xi , xj ))) ExpLoss LogLoss ˜ • Call base-kernel-learner with (Dt , S, S) and receive Kt • Calculate: + − St = {(i, j) | yi yj Kt (xi , xj ) > 0} ; St = {(i, j) | yi yj Kt (xi , xj ) < 0} + Wt = (i,j)∈S + Dt (i, j)|Kt (xi , xj )| ; Wt− = (i,j)∈S − Dt (i, j)|Kt (xi , xj )| t t 1 2 + Wt − Wt • Set: αt = ln ; K ← K + α t Kt . Return: kernel operator K : X × X → Figure 1: The skeleton of the boosting algorithm for kernels. ical classiﬁcation error. Given the prediction of a classiﬁer f on an instance x and a label y ∈ {−1, +1} the ExpLoss and the LogLoss are deﬁned as, ExpLoss(f (x), y) = exp(−yf (x)) LogLoss(f (x), y) = log(1 + exp(−yf (x))) . Collins et al. described a single algorithm for the two losses above that can be used within the boosting framework to construct a strong-hypothesis which is a classiﬁer f (x). This classiﬁer is a weighted combination of (possibly very simple) base classiﬁers. (In the boosting framework, the base classiﬁers are referred to as weak-hypotheses.) The strongT hypothesis is of the form f (x) = t=1 αt ht (x). Collins et al. discussed a few ways to select the weak-hypotheses ht and to ﬁnd a good of weights αt . Our starting point in this paper is the ﬁrst sequential algorithm from [8] that enables the construction or creation of weak-hypotheses on-the-ﬂy. We would like to note however that it is possible to use other variants of boosting to design kernels. In order to use boosting to design kernels we extend the algorithm to operate over pairs of instances. Building on the notion of alignment from [5, 6], we say that the inner-product of x1 and x2 is aligned with the labels y1 and y2 if sign(K(x1 , x2 )) = y1 y2 . Furthermore, we would like to make the magnitude of K(x, x ) to be as large as possible. We therefore use one of the following two alignment losses for a pair of examples (x 1 , y1 ) and (x2 , y2 ), ExpLoss(K(x1 , x2 ), y1 y2 ) = exp(−y1 y2 K(x1 , x2 )) LogLoss(K(x1 , x2 ), y1 y2 ) = log(1 + exp(−y1 y2 K(x1 , x2 ))) . Put another way, we view a pair of instances as a single example and cast the pairs of instances that attain the same label as positively labelled examples while pairs of opposite labels are cast as negatively labelled examples. Clearly, this approach can be applied to both losses. In the boosting process we therefore maintain a distribution over pairs of instances. The weight of each pair reﬂects how difﬁcult it is to predict whether the labels of the two instances are the same or different. The core boosting algorithm follows similar lines to boosting algorithms for classiﬁcation algorithm. The pseudo code of the booster is given in Fig. 1. The pseudo-code is an adaptation the to problem of kernel design of the sequentialupdate algorithm from [8]. As with other boosting algorithm, the base-learner, which in our case is charge of returning a good kernel with respect to the current distribution, is left unspeciﬁed. We therefore turn our attention to the algorithmic implementation of the base-learning algorithm for kernels. 3 Learning Base Kernels The base kernel learner is provided with a training set S and a distribution D t over a pairs ˜ of instances from the training set. It is also provided with a set of unlabelled examples S. Without any knowledge of the topology of the space of instances a learning algorithm is likely to fail. Therefore, we assume the existence of an initial inner-product over the input space. We assume for now that this initial inner-product is the standard scalar products over vectors in n . We later discuss a way to relax the assumption on the form of the inner-product. Equipped with an inner-product, we deﬁne the family of base kernels to be the possible outer-products Kw = wwT between a vector w ∈ n and itself. Using this deﬁnition we get, Kw (xi , xj ) = (xi ·w)(xj ·w) . Input: A distribution Dt . Labelled and unlabelled sets: ˜ ˜ Therefore, the similarity beS = {(xi , yi )}m ; S = {˜i }m . x i=1 i=1 tween two instances xi and Compute : xj is high iff both xi and xj • Calculate: ˜ are similar (w.r.t the standard A ∈ m×m , Ai,r = xi · xr ˜ inner-product) to a third vecm×m B∈ , Bi,j = Dt (i, j)yi yj tor w. Analogously, if both ˜ ˜ K ∈ m×m , Kr,s = xr · xs ˜ ˜ xi and xj seem to be dissim• Find the generalized eigenvector v ∈ m for ilar to the vector w then they the problem AT BAv = λKv which attains are similar to each other. Dethe largest eigenvalue λ spite the restrictive form of • Set: w = ( r vr xr )/ ˜ ˜ r vr xr . the inner-products, this famt ily is still too rich for our setReturn: Kernel operator Kw = ww . ting and we further impose two restrictions on the inner Figure 2: The base kernel learning algorithm. products. First, we assume ˜ that w is restricted to a linear combination of vectors from S. Second, since scaling of the base kernels is performed by the boosted, we constrain the norm of w to be 1. The m ˜ resulting class of kernels is therefore, C = {Kw = wwT | w = r=1 βr xr , w = 1} . ˜ In the boosting process we need to choose a speciﬁc base-kernel K w from C. We therefore need to devise a notion of how good a candidate for base kernel is given a labelled set S and a distribution function Dt . In this work we use the simplest version suggested by Collins et al. This version can been viewed as a linear approximation on the loss function. We deﬁne the score of a kernel Kw w.r.t to the current distribution Dt to be, Score(Kw ) = Dt (i, j)yi yj Kw (xi , xj ) . (1) i,j The higher the value of the score is, the better Kw ﬁts the training data. Note that if Dt (i, j) = 1/m2 (as is D0 ) then Score(Kw ) is proportional to the alignment since w = 1. Under mild assumptions the score can also provide a lower bound of the loss function. To see that let c be the derivative of the loss function at margin zero, c = Loss (0) . If all the √ training examples xi ∈ S lies in a ball of radius c, we get that Loss(Kw (xi , xj ), yi yj ) ≥ 1 − cKw (xi , xj )yi yj ≥ 0, and therefore, i,j Dt (i, j)Loss(Kw (xi , xj ), yi yj ) ≥ 1 − c Dt (i, j)Kw (xi , xj )yi yj . i,j Using the explicit form of Kw in the Score function (Eq. (1)) we get, Score(Kw ) = i,j D(i, j)yi yj (w·xi )(w·xj ) . Further developing the above equation using the constraint that w = m ˜ r=1 βr xr we get, ˜ Score(Kw ) = βs βr r,s i,j D(i, j)yi yj (xi · xr ) (xj · xs ) . ˜ ˜ To compute efﬁciently the base kernel score without an explicit enumeration we exploit the fact that if the initial distribution D0 is symmetric (D0 (i, j) = D0 (j, i)) then all the distributions generated along the run of the boosting process, D t , are also symmetric. We ˜ now deﬁne a matrix A ∈ m×m where Ai,r = xi · xr and a symmetric matrix B ∈ m×m ˜ with Bi,j = Dt (i, j)yi yj . Simple algebraic manipulations yield that the score function can be written as the following quadratic form, Score(β) = β T (AT BA)β , where β is m dimensional column vector. Note that since B is symmetric so is A T BA. Finding a ˜ good base kernel is equivalent to ﬁnding a vector β which maximizes this quadratic form 2 m ˜ under the norm equality constraint w = ˜ 2 = β T Kβ = 1 where Kr,s = r=1 βr xr xr · xs . Finding the maximum of Score(β) subject to the norm constraint is a well known ˜ ˜ maximization problem known as the generalized eigen vector problem (cf. [10]). Applying simple algebraic manipulations it is easy to show that the matrix AT BA is positive semideﬁnite. Assuming that the matrix K is invertible, the the vector β which maximizes the quadratic form is proportional the eigenvector of K −1 AT BA which is associated with the m ˜ generalized largest eigenvalue. Denoting this vector by v we get that w ∝ ˜ r=1 vr xr . m ˜ m ˜ Adding the norm constraint we get that w = ( r=1 vr xr )/ ˜ vr xr . The skeleton ˜ r=1 of the algorithm for ﬁnding a base kernels is given in Fig. 3. To conclude the description of the kernel learning algorithm we describe how to the extend the algorithm to be employed with general kernel functions. Kernelizing the Kernel: As described above, we assumed that the standard scalarproduct constitutes the template for the class of base-kernels C. However, since the proce˜ dure for choosing a base kernel depends on S and S only through the inner-products matrix A, we can replace the scalar-product itself with a general kernel operator κ : X × X → , where κ(xi , xj ) = φ(xi ) · φ(xj ). Using a general kernel function κ we can not compute however the vector w explicitly. We therefore need to show that the norm of w, and evaluation Kw on any two examples can still be performed efﬁciently. First note that given the vector v we can compute the norm of w as follows, T w 2 = vr xr ˜ vs xr ˜ r s = vr vs κ(˜r , xs ) . x ˜ r,s Next, given two vectors xi and xj the value of their inner-product is, Kw (xi , xj ) = vr vs κ(xi , xr )κ(xj , xs ) . ˜ ˜ r,s Therefore, although we cannot compute the vector w explicitly we can still compute its norm and evaluate any of the kernels from the class C. 4 Experiments Synthetic data: We generated binary-labelled data using as input space the vectors in 100 . The labels, in {−1, +1}, were picked uniformly at random. Let y designate the label of a particular example. Then, the ﬁrst two components of each instance were drawn from a two-dimensional normal distribution, N (µ, ∆ ∆−1 ) with the following parameters, µ=y 0.03 0.03 1 ∆= √ 2 1 −1 1 1 = 0.1 0 0 0.01 . That is, the label of each examples determined the mean of the distribution from which the ﬁrst two components were generated. The rest of the components in the vector (98 8 0.2 6 50 50 100 100 150 150 200 200 4 2 0 0 −2 −4 −6 250 250 −0.2 −8 −0.2 0 0.2 −8 −6 −4 −2 0 2 4 6 8 300 20 40 60 80 100 120 140 160 180 200 300 20 40 60 80 100 120 140 160 180 Figure 3: Results on a toy data set prior to learning a kernel (ﬁrst and third from left) and after learning (second and fourth). For each of the two settings we show the ﬁrst two components of the training data (left) and the matrix of inner products between the train and the test data (right). altogether) were generated independently using the normal distribution with a zero mean and a standard deviation of 0.05. We generated 100 training and test sets of size 300 and 200 respectively. We used the standard dot-product as the initial kernel operator. On each experiment we ﬁrst learned a linear classier that separates the classes using the Perceptron [11] algorithm. We ran the algorithm for 10 epochs on the training set. After each epoch we evaluated the performance of the current classiﬁer on the test set. We then used the boosting algorithm for kernels with the LogLoss for 30 rounds to build a kernel for each random training set. After learning the kernel we re-trained a classiﬁer with the Perceptron algorithm and recorded the results. A summary of the online performance is given in Fig. 4. The plot on the left-hand-side of the ﬁgure shows the instantaneous error (achieved during the run of the algorithm). Clearly, the Perceptron algorithm with the learned kernel converges much faster than the original kernel. The middle plot shows the test error after each epoch. The plot on the right shows the test error on a noisy test set in which we added a Gaussian noise of zero mean and a standard deviation of 0.03 to the ﬁrst two features. In all plots, each bar indicates a 95% conﬁdence level. It is clear from the ﬁgure that the original kernel is much slower to converge than the learned kernel. Furthermore, though the kernel learning algorithm was not expoed to the test set noise, the learned kernel reﬂects better the structure of the feature space which makes the learned kernel more robust to noise. Fig. 3 further illustrates the beneﬁts of using a boutique kernel. The ﬁrst and third plots from the left correspond to results obtained using the original kernel and the second and fourth plots show results using the learned kernel. The left plots show the empirical distribution of the two informative components on the test data. For the learned kernel we took each input vector and projected it onto the two eigenvectors of the learned kernel operator matrix that correspond to the two largest eigenvalues. Note that the distribution after the projection is bimodal and well separated along the ﬁrst eigen direction (x-axis) and shows rather little deviation along the second eigen direction (y-axis). This indicates that the kernel learning algorithm indeed found the most informative projection for separating the labelled data with large margin. It is worth noting that, in this particular setting, any algorithm which chooses a single feature at a time is prone to failure since both the ﬁrst and second features are mandatory for correctly classifying the data. The two plots on the right hand side of Fig. 3 use a gray level color-map to designate the value of the inner-product between each pairs instances, one from training set (y-axis) and the other from the test set. The examples were ordered such that the ﬁrst group consists of the positively labelled instances while the second group consists of the negatively labelled instances. Since most of the features are non-relevant the original inner-products are noisy and do not exhibit any structure. In contrast, the inner-products using the learned kernel yields in a 2 × 2 block matrix indicating that the inner-products between instances sharing the same label obtain large positive values. Similarly, for instances of opposite 200 1 12 Regular Kernel Learned Kernel 0.8 17 0.7 16 0.5 0.4 0.3 Test Error % 8 0.6 Regular Kernel Learned Kernel 18 10 Test Error % Averaged Cumulative Error % 19 Regular Kernel Learned Kernel 0.9 6 4 15 14 13 12 0.2 11 2 0.1 10 0 0 10 1 10 2 10 Round 3 10 4 10 0 2 4 6 Epochs 8 10 9 2 4 6 Epochs 8 10 Figure 4: The online training error (left), test error (middle) on clean synthetic data using a standard kernel and a learned kernel. Right: the online test error for the two kernels on a noisy test set. labels the inner products are large and negative. The form of the inner-products matrix of the learned kernel indicates that the learning problem itself becomes much easier. Indeed, the Perceptron algorithm with the standard kernel required around 94 training examples on the average before converging to a hyperplane which perfectly separates the training data while using the Perceptron algorithm with learned kernel required a single example to reach a perfect separation on all 100 random training sets. USPS dataset: The USPS (US Postal Service) dataset is known as a challenging classiﬁcation problem in which the training set and the test set were collected in a different manner. The USPS contains 7, 291 training examples and 2, 007 test examples. Each example is represented as a 16 × 16 matrix where each entry in the matrix is a pixel that can take values in {0, . . . , 255}. Each example is associated with a label in {0, . . . , 9} which is the digit content of the image. Since the kernel learning algorithm is designed for binary problems, we broke the 10-class problem into 45 binary problems by comparing all pairs of classes. The interesting question of how to learn kernels for multiclass problems is beyond the scopre of this short paper. We thus constraint on the binary error results for the 45 binary problem described above. For the original kernel we chose a RBF kernel with σ = 1 which is the value employed in the experiments reported in [12]. We used the kernelized version of the kernel design algorithm to learn a different kernel operator for each of the binary problems. We then used a variant of the Perceptron [11] and with the original RBF kernel and with the learned kernels. One of the motivations for using the Perceptron is its simplicity which can underscore differences in the kernels. We ran the kernel learning al˜ gorithm with LogLoss and ExpLoss, using bith the training set and the test test as S. Thus, we obtained four different sets of kernels where each set consists of 45 kernels. By examining the training loss, we set the number of rounds of boosting to be 30 for the LogLoss and 50 for the ExpLoss, when using the trainin set. When using the test set, the number of rounds of boosting was set to 100 for both losses. Since the algorithm exhibits slower rate of convergence with the test data, we choose a a higher value without attempting to optimize the actual value. The left plot of Fig. 5 is a scatter plot comparing the test error of each of the binary classiﬁers when trained with the original RBF a kernel versus the performance achieved on the same binary problem with a learned kernel. The kernels were built ˜ using boosting with the LogLoss and S was the training data. In almost all of the 45 binary classiﬁcation problems, the learned kernels yielded lower error rates when combined with the Perceptron algorithm. The right plot of Fig. 5 compares two learned kernels: the ﬁrst ˜ was build using the training instances as the templates constituing S while the second used the test instances. Although the differenece between the two versions is not as signiﬁcant as the difference on the left plot, we still achieve an overall improvement in about 25% of the binary problems by using the test instances. 6 4.5 4 5 Learned Kernel (Test) Learned Kernel (Train) 3.5 4 3 2 3 2.5 2 1.5 1 1 0.5 0 0 1 2 3 Base Kernel 4 5 6 0 0 1 2 3 Learned Kernel (Train) 4 5 Figure 5: Left: a scatter plot comparing the error rate of 45 binary classiﬁers trained using an RBF kernel (x-axis) and a learned kernel with training instances. Right: a similar scatter plot for a learned kernel only constructed from training instances (x-axis) and test instances. 5 Discussion In this paper we showed how to use the boosting framework to design kernels. Our approach is especially appealing in transductive learning tasks where the test data distribution is different than the the distribution of the training data. For example, in speech recognition tasks the training data is often clean and well recorded while the test data often passes through a noisy channel that distorts the signal. An interesting and challanging question that stem from this research is how to extend the framework to accommodate more complex decision tasks such as multiclass and regression problems. Finally, we would like to note alternative approaches to the kernel design problem has been devised in parallel and independently. See [13, 14] for further details. Acknowledgements: Special thanks to Cyril Goutte and to John Show-Taylor for pointing the connection to the generalized eigen vector problem. Thanks also to the anonymous reviewers for constructive comments. References [1] V. N. Vapnik. Statistical Learning Theory. Wiley, 1998. [2] N. Cristianini and J. Shawe-Taylor. An Introduction to Support Vector Machines. Cambridge University Press, 2000. [3] Huma Lodhi, John Shawe-Taylor, Nello Cristianini, and Christopher J. C. H. Watkins. Text classiﬁcation using string kernels. Journal of Machine Learning Research, 2:419–444, 2002. [4] C. Leslie, E. Eskin, and W. Stafford Noble. The spectrum kernel: A string kernel for svm protein classiﬁcation. In Proceedings of the Paciﬁc Symposium on Biocomputing, 2002. [5] Nello Cristianini, Andre Elisseeff, John Shawe-Taylor, and Jaz Kandla. On kernel target alignment. In Advances in Neural Information Processing Systems 14, 2001. [6] G. Lanckriet, N. Cristianini, P. Bartlett, L. El Ghaoui, and M. Jordan. Learning the kernel matrix with semi-deﬁnite programming. In Proc. of the 19th Intl. Conf. on Machine Learning, 2002. [7] Jerome Friedman, Trevor Hastie, and Robert Tibshirani. Additive logistic regression: a statistical view of boosting. Annals of Statistics, 28(2):337–374, April 2000. [8] Michael Collins, Robert E. Schapire, and Yoram Singer. Logistic regression, adaboost and bregman distances. Machine Learning, 47(2/3):253–285, 2002. [9] Llew Mason, Jonathan Baxter, Peter Bartlett, and Marcus Frean. Functional gradient techniques for combining hypotheses. In Advances in Large Margin Classiﬁers. MIT Press, 1999. [10] Roger A. Horn and Charles R. Johnson. Matrix Analysis. Cambridge University Press, 1985. [11] F. Rosenblatt. The perceptron: A probabilistic model for information storage and organization in the brain. Psychological Review, 65:386–407, 1958. [12] B. Sch¨ lkopf, S. Mika, C.J.C. Burges, P. Knirsch, K. M¨ ller, G. R¨ tsch, and A.J. Smola. Input o u a space vs. feature space in kernel-based methods. IEEE Trans. on NN, 10(5):1000–1017, 1999. [13] O. Bosquet and D.J.L. Herrmann. On the complexity of learning the kernel matrix. NIPS, 2002. [14] C.S. Ong, A.J. Smola, and R.C. Williamson. Superkenels. NIPS, 2002.

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