nips nips2002 nips2002-120 knowledge-graph by maker-knowledge-mining

120 nips-2002-Kernel Design Using Boosting

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Author: Koby Crammer, Joseph Keshet, Yoram Singer

Abstract: The focus of the paper is the problem of learning kernel operators from empirical data. We cast the kernel design problem as the construction of an accurate kernel from simple (and less accurate) base kernels. We use the boosting paradigm to perform the kernel construction process. To do so, we modify the booster so as to accommodate kernel operators. We also devise an efﬁcient weak-learner for simple kernels that is based on generalized eigen vector decomposition. We demonstrate the effectiveness of our approach on synthetic data and on the USPS dataset. On the USPS dataset, the performance of the Perceptron algorithm with learned kernels is systematically better than a ﬁxed RBF kernel. 1 Introduction and problem Setting The last decade brought voluminous amount of work on the design, analysis and experimentation of kernel machines. Algorithm based on kernels can be used for various machine learning tasks such as classiﬁcation, regression, ranking, and principle component analysis. The most prominent learning algorithm that employs kernels is the Support Vector Machines (SVM) [1, 2] designed for classiﬁcation and regression. A key component in a kernel machine is a kernel operator which computes for any pair of instances their inner-product in some abstract vector space. Intuitively and informally, a kernel operator is a means for measuring similarity between instances. Almost all of the work that employed kernel operators concentrated on various machine learning problems that involved a predeﬁned kernel. A typical approach when using kernels is to choose a kernel before learning starts. Examples to popular predeﬁned kernels are the Radial Basis Functions and the polynomial kernels (see for instance [1]). Despite the simplicity required in modifying a learning algorithm to a “kernelized” version, the success of such algorithms is not well understood yet. More recently, special efforts have been devoted to crafting kernels for speciﬁc tasks such as text categorization [3] and protein classiﬁcation problems [4]. Our work attempts to give a computational alternative to predeﬁned kernels by learning kernel operators from data. We start with a few deﬁnitions. Let X be an instance space. A kernel is an inner-product operator K : X × X → . An explicit way to describe K is via a mapping φ : X → H from X to an inner-products space H such that K(x, x ) = φ(x)·φ(x ). Given a kernel operator and a ﬁnite set of instances S = {xi , yi }m , the kernel i=1 matrix (a.k.a the Gram matrix) is the matrix of all possible inner-products of pairs from S, Ki,j = K(xi , xj ). We therefore refer to the general form of K as the kernel operator and to the application of the kernel operator to a set of pairs of instances as the kernel matrix. The speciﬁc setting of kernel design we consider assumes that we have access to a base kernel learner and we are given a target kernel K manifested as a kernel matrix on a set of examples. Upon calling the base kernel learner it returns a kernel operator denote Kj . The goal thereafter is to ﬁnd a weighted combination of kernels ˆ K(x, x ) = j αj Kj (x, x ) that is similar, in a sense that will be deﬁned shortly, to ˆ the target kernel, K ∼ K . Cristianini et al. [5] in their pioneering work on kernel target alignment employed as the notion of similarity the inner-product between the kernel matrices < K, K >F = m K(xi , xj )K (xi , xj ). Given this deﬁnition, they deﬁned the i,j=1 kernel-similarity, or alignment, to be the above inner-product normalized by the norm of ˆ ˆ ˆ ˆ ˆ each kernel, A(S, K, K ) = < K, K >F / < K, K >F < K , K >F , where S is, as above, a ﬁnite sample of m instances. Put another way, the kernel alignment Cristianini et al. employed is the cosine of the angle between the kernel matrices where each matrix is “ﬂattened” into a vector of dimension m2 . Therefore, this deﬁnition implies that the alignment is bounded above by 1 and can attain this value iff the two kernel matrices are identical. Given a (column) vector of m labels y where yi ∈ {−1, +1} is the label of the instance xi , Cristianini et al. used the outer-product of y as the the target kernel, ˆ K = yy T . Therefore, an optimal alignment is achieved if K(xi , xj ) = yi yj . Clearly, if such a kernel is used for classifying instances from X , then the kernel itself sufﬁces to construct an excellent classiﬁer f : X → {−1, +1} by setting, f (x) = sign(y i K(xi , x)) where (xi , yi ) is any instance-label pair. Cristianini et al. then devised a procedure that works with both labelled and unlabelled examples to ﬁnd a Gram matrix which attains a good alignment with K on the labelled part of the matrix. While this approach can clearly construct powerful kernels, a few problems arise from the notion of kernel alignment they employed. For instance, a kernel operator such that the sign(K(x i , xj )) is equal to yi yj but its magnitude, |K(xi , xj )|, is not necessarily 1, might achieve a poor alignment score while it can constitute a classiﬁer whose empirical loss is zero. Furthermore, the task of ﬁnding a good kernel when it is not always possible to ﬁnd a kernel whose sign on each pair of instances is equal to the products of the labels (termed the soft-margin case in [5, 6]) becomes rather tricky. We thus propose a different approach which attempts to overcome some of the difﬁculties above. Like Cristianini et al. we assume that we are given a set of labelled instances S = {(xi , yi ) | xi ∈ X , yi ∈ {−1, +1}, i = 1, . . . , m} . We are also given a set of unlabelled m ˜ ˜ examples S = {˜i }i=1 . If such a set is not provided we can simply use the labelled inx ˜ ˜ stances (without the labels themselves) as the set S. The set S is used for constructing the ˆ primitive kernels that are combined to constitute the learned kernel K. The labelled set is used to form the target kernel matrix and its instances are used for evaluating the learned ˆ kernel K. This approach, known as transductive learning, was suggested in [5, 6] for kernel alignment tasks when the distribution of the instances in the test data is different from that of the training data. This setting becomes in particular handy in datasets where the test data was collected in a different scheme than the training data. We next discuss the notion of kernel goodness employed in this paper. This notion builds on the objective function that several variants of boosting algorithms maintain [7, 8]. We therefore ﬁrst discuss in brief the form of boosting algorithms for kernels. 2 Using Boosting to Combine Kernels Numerous interpretations of AdaBoost and its variants cast the boosting process as a procedure that attempts to minimize, or make small, a continuous bound on the classiﬁcation error (see for instance [9, 7] and the references therein). A recent work by Collins et al. [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.

Reference: text

Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 il Abstract The focus of the paper is the problem of learning kernel operators from empirical data. [sent-4, score-0.512]

2 We cast the kernel design problem as the construction of an accurate kernel from simple (and less accurate) base kernels. [sent-5, score-1.184]

3 We use the boosting paradigm to perform the kernel construction process. [sent-6, score-0.701]

4 To do so, we modify the booster so as to accommodate kernel operators. [sent-7, score-0.541]

5 We also devise an efﬁcient weak-learner for simple kernels that is based on generalized eigen vector decomposition. [sent-8, score-0.304]

6 On the USPS dataset, the performance of the Perceptron algorithm with learned kernels is systematically better than a ﬁxed RBF kernel. [sent-10, score-0.295]

7 1 Introduction and problem Setting The last decade brought voluminous amount of work on the design, analysis and experimentation of kernel machines. [sent-11, score-0.478]

8 A key component in a kernel machine is a kernel operator which computes for any pair of instances their inner-product in some abstract vector space. [sent-14, score-1.25]

9 Intuitively and informally, a kernel operator is a means for measuring similarity between instances. [sent-15, score-0.588]

10 Almost all of the work that employed kernel operators concentrated on various machine learning problems that involved a predeﬁned kernel. [sent-16, score-0.554]

11 A typical approach when using kernels is to choose a kernel before learning starts. [sent-17, score-0.632]

12 Examples to popular predeﬁned kernels are the Radial Basis Functions and the polynomial kernels (see for instance [1]). [sent-18, score-0.34]

13 Our work attempts to give a computational alternative to predeﬁned kernels by learning kernel operators from data. [sent-21, score-0.666]

14 A kernel is an inner-product operator K : X × X → . [sent-24, score-0.588]

15 Given a kernel operator and a ﬁnite set of instances S = {xi , yi }m , the kernel i=1 matrix (a. [sent-26, score-1.426]

16 a the Gram matrix) is the matrix of all possible inner-products of pairs from S, Ki,j = K(xi , xj ). [sent-28, score-0.319]

17 We therefore refer to the general form of K as the kernel operator and to the application of the kernel operator to a set of pairs of instances as the kernel matrix. [sent-29, score-1.888]

18 The speciﬁc setting of kernel design we consider assumes that we have access to a base kernel learner and we are given a target kernel K manifested as a kernel matrix on a set of examples. [sent-30, score-2.201]

19 Upon calling the base kernel learner it returns a kernel operator denote Kj . [sent-31, score-1.224]

20 [5] in their pioneering work on kernel target alignment employed as the notion of similarity the inner-product between the kernel matrices < K, K >F = m K(xi , xj )K (xi , xj ). [sent-34, score-1.625]

21 Put another way, the kernel alignment Cristianini et al. [sent-36, score-0.626]

22 employed is the cosine of the angle between the kernel matrices where each matrix is “ﬂattened” into a vector of dimension m2 . [sent-37, score-0.589]

23 Therefore, this deﬁnition implies that the alignment is bounded above by 1 and can attain this value iff the two kernel matrices are identical. [sent-38, score-0.616]

24 Given a (column) vector of m labels y where yi ∈ {−1, +1} is the label of the instance xi , Cristianini et al. [sent-39, score-0.506]

25 Therefore, an optimal alignment is achieved if K(xi , xj ) = yi yj . [sent-41, score-0.683]

26 Clearly, if such a kernel is used for classifying instances from X , then the kernel itself sufﬁces to construct an excellent classiﬁer f : X → {−1, +1} by setting, f (x) = sign(y i K(xi , x)) where (xi , yi ) is any instance-label pair. [sent-42, score-1.271]

27 then devised a procedure that works with both labelled and unlabelled examples to ﬁnd a Gram matrix which attains a good alignment with K on the labelled part of the matrix. [sent-44, score-0.6]

28 While this approach can clearly construct powerful kernels, a few problems arise from the notion of kernel alignment they employed. [sent-45, score-0.619]

29 For instance, a kernel operator such that the sign(K(x i , xj )) is equal to yi yj but its magnitude, |K(xi , xj )|, is not necessarily 1, might achieve a poor alignment score while it can constitute a classiﬁer whose empirical loss is zero. [sent-46, score-1.636]

30 Furthermore, the task of ﬁnding a good kernel when it is not always possible to ﬁnd a kernel whose sign on each pair of instances is equal to the products of the labels (termed the soft-margin case in [5, 6]) becomes rather tricky. [sent-47, score-1.225]

31 we assume that we are given a set of labelled instances S = {(xi , yi ) | xi ∈ X , yi ∈ {−1, +1}, i = 1, . [sent-50, score-0.781]

32 The set S is used for constructing the ˆ primitive kernels that are combined to constitute the learned kernel K. [sent-56, score-0.74]

33 The labelled set is used to form the target kernel matrix and its instances are used for evaluating the learned ˆ kernel K. [sent-57, score-1.435]

34 This approach, known as transductive learning, was suggested in [5, 6] for kernel alignment tasks when the distribution of the instances in the test data is different from that of the training data. [sent-58, score-0.938]

35 We next discuss the notion of kernel goodness employed in this paper. [sent-60, score-0.553]

36 This notion builds on the objective function that several variants of boosting algorithms maintain [7, 8]. [sent-61, score-0.286]

37 2 Using Boosting to Combine Kernels Numerous interpretations of AdaBoost and its variants cast the boosting process as a procedure that attempts to minimize, or make small, a continuous bound on the classiﬁcation error (see for instance [9, 7] and the references therein). [sent-63, score-0.358]

38 Return: kernel operator K : X × X → Figure 1: The skeleton of the boosting algorithm for kernels. [sent-69, score-0.884]

39 (In the boosting framework, the base classiﬁers are referred to as weak-hypotheses. [sent-75, score-0.349]

40 We would like to note however that it is possible to use other variants of boosting to design kernels. [sent-80, score-0.309]

41 In order to use boosting to design kernels we extend the algorithm to operate over pairs of instances. [sent-81, score-0.512]

42 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. [sent-85, score-0.978]

43 In the boosting process we therefore maintain a distribution over pairs of instances. [sent-87, score-0.297]

44 The core boosting algorithm follows similar lines to boosting algorithms for classiﬁcation algorithm. [sent-89, score-0.479]

45 The pseudo-code is an adaptation the to problem of kernel design of the sequentialupdate algorithm from [8]. [sent-92, score-0.567]

46 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. [sent-93, score-0.723]

47 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. [sent-95, score-0.95]

48 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. [sent-101, score-0.304]

49 x i=1 i=1 tween two instances xi and Compute : xj is high iff both xi and xj • Calculate: ˜ are similar (w. [sent-105, score-0.966]

50 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. [sent-107, score-0.606]

51 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. [sent-108, score-0.783]

52 Dethe largest eigenvalue λ spite the restrictive form of • Set: w = ( r vr xr )/ ˜ ˜ r vr xr . [sent-109, score-0.684]

53 ting and we further impose two restrictions on the inner Figure 2: The base kernel learning algorithm. [sent-111, score-0.627]

54 Second, since scaling of the base kernels is performed by the boosted, we constrain the norm of w to be 1. [sent-114, score-0.337]

55 The m ˜ resulting class of kernels is therefore, C = {Kw = wwT | w = r=1 βr xr , w = 1} . [sent-115, score-0.393]

56 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 . [sent-117, score-0.831]

57 t to the current distribution Dt to be, Score(Kw ) = Dt (i, j)yi yj Kw (xi , xj ) . [sent-122, score-0.42]

58 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 . [sent-127, score-2.285]

59 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 ) . [sent-130, score-0.757]

60 ˜ ˜ 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. [sent-131, score-0.92]

61 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 . [sent-132, score-0.696]

62 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 . [sent-135, score-1.25]

63 Denoting this vector by v we get that w ∝ ˜ r=1 vr xr . [sent-140, score-0.392]

64 m ˜ m ˜ Adding the norm constraint we get that w = ( r=1 vr xr )/ ˜ vr xr . [sent-141, score-0.79]

65 The skeleton ˜ r=1 of the algorithm for ﬁnding a base kernels is given in Fig. [sent-142, score-0.353]

66 To conclude the description of the kernel learning algorithm we describe how to the extend the algorithm to be employed with general kernel functions. [sent-144, score-1.064]

67 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 ). [sent-146, score-1.465]

68 Using a general kernel function κ we can not compute however the vector w explicitly. [sent-147, score-0.502]

69 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 ) . [sent-149, score-0.891]

70 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 ) . [sent-150, score-1.068]

71 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). [sent-165, score-0.478]

72 We used the standard dot-product as the initial kernel operator. [sent-170, score-0.478]

73 We then used the boosting algorithm for kernels with the LogLoss for 30 rounds to build a kernel for each random training set. [sent-174, score-0.98]

74 After learning the kernel we re-trained a classiﬁer with the Perceptron algorithm and recorded the results. [sent-175, score-0.511]

75 Clearly, the Perceptron algorithm with the learned kernel converges much faster than the original kernel. [sent-179, score-0.619]

76 It is clear from the ﬁgure that the original kernel is much slower to converge than the learned kernel. [sent-184, score-0.608]

77 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. [sent-185, score-1.755]

78 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. [sent-188, score-0.684]

79 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. [sent-190, score-1.351]

80 This indicates that the kernel learning algorithm indeed found the most informative projection for separating the labelled data with large margin. [sent-192, score-0.647]

81 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. [sent-196, score-0.524]

82 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. [sent-198, score-0.83]

83 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. [sent-208, score-0.788]

84 Right: the online test error for the two kernels on a noisy test set. [sent-209, score-0.348]

85 The form of the inner-products matrix of the learned kernel indicates that the learning problem itself becomes much easier. [sent-211, score-0.631]

86 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. [sent-223, score-0.639]

87 For the original kernel we chose a RBF kernel with σ = 1 which is the value employed in the experiments reported in [12]. [sent-226, score-0.998]

88 We used the kernelized version of the kernel design algorithm to learn a different kernel operator for each of the binary problems. [sent-227, score-1.222]

89 We then used a variant of the Perceptron [11] and with the original RBF kernel and with the learned kernels. [sent-228, score-0.586]

90 We ran the kernel learning al˜ gorithm with LogLoss and ExpLoss, using bith the training set and the test test as S. [sent-230, score-0.676]

91 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. [sent-232, score-0.315]

92 When using the test set, the number of rounds of boosting was set to 100 for both losses. [sent-233, score-0.333]

93 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. [sent-236, score-0.857]

94 The kernels were built ˜ using boosting with the LogLoss and S was the training data. [sent-237, score-0.431]

95 In almost all of the 45 binary classiﬁcation problems, the learned kernels yielded lower error rates when combined with the Perceptron algorithm. [sent-238, score-0.33]

96 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. [sent-240, score-0.394]

97 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. [sent-247, score-1.276]

98 Right: a similar scatter plot for a learned kernel only constructed from training instances (x-axis) and test instances. [sent-248, score-0.962]

99 Finally, we would like to note alternative approaches to the kernel design problem has been devised in parallel and independently. [sent-253, score-0.561]

100 The spectrum kernel: A string kernel for svm protein classiﬁcation. [sent-277, score-0.478]

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