CS 590D Lecture Notes
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1 CS 590D Lecture Notes David Wemhoener December Introduction Figure 1: Various Separators We previously looked at the perceptron learning algorithm, which gives us a separator for linearly separable data. One issue, however, is that we are not guaranteed to find the separator which maximizes the margin. We will discuss how to maximize the margin as well as what to do when our data is not linearly separable. Margin Maximization The margin of a solution w to (w T a i )l i > 0, 1 i n where a i = 1 is δ = min i (w T a i)l i w. Note that (wt a i)l i w δ 1 for all a i w w δ Let v = be the modified weight vector, as dividing the normalized weight by δ normalizes the margin to one. vector w w 1
2 Maximizing δ is equivalent to minimizing v so the optimization problem becomes: minimize v subject to (v T a i )l i > 0, 1 i n Although v is a convex function of the coordinates of v, a better convex function to minimize is v since v is differentiable. Thus we have: minimize v subject to (v T a i )l i 0, 1 i n An optimal solution v to this problem has the following property: Let V be the space spanned by the examples a i for which (v T a i )l i = 1. Then v lies in V. If V is full dimensional, then there are d independent examples for which (v T a i )l i = 1 holds. These d equations then have v as a unique solution. These examples are called support vectors, and uniquely determine the maximum margin separator. 3 Kernels If the data is not linearly separable we can separate it by: Applying a transformation to the data Replacing the dot product with another function: a kernel kernel examples: linear e.x. k(u, v) = u T v + 1 quadratic e.x. k(u, v) = (u T v + 1) other polynomials radial basis 4 Neural Networks We can use multiple perceptrons connected together to solve classification problems for which no linear separator exists. For example, a XOR gate has a value of 1 when both inputs have different values, and 0 when they are both 0 or both 1. We can use a network of three perceptrons to model this gate. The first two perceptrons are in the input layer, and each can be thought of as creating a line in the space of the data. The third perceptron combines the outputs from the first two perceptrons, and determines what part of the space has the positive classification.
3 Figure : XOR Graph If we graph the XOR function, we can use the equations of the lines of the individual separators to provide us with the weights we need for our perceptrons. Figure 3: XOR Network We can extend this to many layers of pereptrons, with the first layer being the input layer, the final layer being the output layer, and the zero or more middle layers being hidden layers. The process of assigning a classification to an input example is called forward propagation. 5 Boosting Suppose we have n labeled examples a1,..., an. A strong learner is an algorithm that takes n labeled examples and produces a classifier that correctly labels all of them. A weak learner is a learner that only gets a fraction 1 + γ correct. Assume we have a nonnegative real weight wi oneachexampleai, then the classifier 3
4 correctly labels a subset of exmaples with total weight at least ( 1 + γ) n i=1 w i. A strong learner can be built from a weak learner using boosting. 5.1 The boosting algorithm: Make the first call to the weak learner with all w i set equal to 1 At time t + 1 multiply the weight of each example that was misclassified the previous time by 1 + ɛ. Make a call to the weak learner. After T steps, stop and output the classifier FC FC: Label each of the examples by the label given to it by a majority of calls to the weak learner. (Assume T is off, so there is no tie) 5. Proof Let m be the number of examples FC classifies incorrectly. Each of these m examples we misclassified at least T/ times and thus each has a weight of at least (1 + ɛ) T/ and the total weight is at least m(1 + ɛ) T/. At time t+1, only the weight of the example misclassified at the previous step time t were increased. By the property of weak learning, the total weight of misclassified examples is at most f = 1 + γ of the total weight at time t. Let weight(t) be the total weight at time t. Then weight(t + 1) ((1 + ɛ)f + 1 f) weight(t) = (1 + ɛf) weight(t) (1 + ɛ γɛ) weight(t) Thus, since weight(0) = n, weight(t + 1) ((1 + ɛ)f + 1 f) weight(t) = (1 + ɛf) weight(t) (1 + ɛ γɛ) weight(t) Taking logarithms: ln m + T ln (1 + ɛ) ln n + T ln (1 + ɛ γɛ) 4
5 To a first order approximation, ln(1 + δ) δ for small δ. If we make ɛ a small constant, then ln m ln m T γɛ. Let T = (1 + lnn)/γɛ. Then ln m 1 and m 1 ɛ, which means the number of misclassified items is less than one and thus must be zero. 5
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