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Consider a binary classification problem using 1-nearest neighbors with the Euclidean dis- and corresponding labels M are in ascend- 6. tance metric. We have N 1-dimensional training points x", . ya, a.. with xl ER and y" e 0. 1. Assume the points x. . ing order by vahue. If there are ties during the 1-NN algorithm, we break ties by choosing the label corresponding to the x" with lower value. Is it possible to build a decision tree that behaves exactly the same as the 1-nearest neighbor classifier? Assume that the decision at each node takes the form of" x st or x >1," wheret eR. O Yes O No (a) If your answer is yes, please explain how you will construct the decision tree. If your answer is no, explain why it's not possible. Your answer: Decision Trees, k-NN, Regression () where x" = k, ) eR and the decision at each node takes the form of" xy stor x) > t," where t eR and / E (1. 2. Give an example with at most 3 training points for which it isn't possible to build a decision tree that behaves exactly the same as a l-nearest neighbor classifier. Let's add a dimension! Now assume the training points are 2-dimensional Your answer:

Consider a binary classification problem using 1-nearest neighbors with the Euclidean dis- and corresponding labels M are in ascend- 6. tance metric. We have N 1-dimensional training points x", . ya, a.. with xl ER and y" e 0. 1. Assume the points x. . ing order by vahue. If there are ties during the 1-NN algorithm, we break ties by choosing the label corresponding to the x" with lower value. Is it possible to build a decision tree that behaves exactly the same as the 1-nearest neighbor classifier? Assume that the decision at each node takes the form of" x st or x >1," wheret eR. O Yes O No (a) If your answer is yes, please explain how you will construct the decision tree. If your answer is no, explain why it's not possible. Your answer: Decision Trees, k-NN, Regression () where x" = k, ) eR and the decision at each node takes the form of" xy stor x) > t," where t eR and / E (1. 2. Give an example with at most 3 training points for which it isn't possible to build a decision tree that behaves exactly the same as a l-nearest neighbor classifier. Let's add a dimension! Now assume the training points are 2-dimensional Your answer:

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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 6. Consider a binary classification problem using 1-nearest neighbors with the Euclidean distance metric. We have N 1-dimensional training points x(1), x(2), . . . x(N ) and corresponding labels
y(1), y(2), . . . y(N ) with x(i ) ∈ R and y(i ) ∈ {0, 1}. Assume the points x(1), x(2), . . . x(N ) are in ascending order by value. If there are ties during the 1-NN algorithm, we break ties by choosing the label
corresponding to the x(i ) with lower value. 

 

6. Consider a binary classification problem using l-nearest neighbors with the Euclidean dis- tance metric. We have N 1-dimensional training points x", x".... and corresponding labels are in ascend- ing order by vahue. If there are ties during the 1-NN algorithm, we break ties by choosing the label y". P..M with x ER and y" e (0. 1. Assume the points x. x. coresponding to the x" with lower value. Is it possible to build a decision tree that behaves exactly the same as the 1-nearest neighbor classifier? Assume that the decision at each node takes the fom of" x st or x >1," wheret ER. (a) Yes O No If your answer is yes, please explain how you will construct the decision tree. If your answer is no, explain why it's not possible. Your answer. Decision Trees, k-NN, Regression Now assume the training points are 2-dimensional where x" = . ) ER? and the decision at each node takes the form of “ xy st or x, > t," where t eR andj e (1. 2. Give an example with at most 3 training points for which it isn't possible to build a decision tree that behaves exactly the same as a l-nearest neighbor classifier. Let's add a dimension! Your answer: = . ) ER* and (c) the decision at each node takes the form of "x, st or x, > 1," wheret ER and j e 1, 2, under what conditions is it possible to buzild a decision tree that behaves exactly the same as a l-nearest neighbor classifier? Explain why it is possible to build the decision tree as stated in part (a) but, in general, not possible in part (6). Assuming we have 2-dimensional training points
Transcribed Image Text:6. Consider a binary classification problem using l-nearest neighbors with the Euclidean dis- tance metric. We have N 1-dimensional training points x", x".... and corresponding labels are in ascend- ing order by vahue. If there are ties during the 1-NN algorithm, we break ties by choosing the label y". P..M with x ER and y" e (0. 1. Assume the points x. x. coresponding to the x" with lower value. Is it possible to build a decision tree that behaves exactly the same as the 1-nearest neighbor classifier? Assume that the decision at each node takes the fom of" x st or x >1," wheret ER. (a) Yes O No If your answer is yes, please explain how you will construct the decision tree. If your answer is no, explain why it's not possible. Your answer. Decision Trees, k-NN, Regression Now assume the training points are 2-dimensional where x" = . ) ER? and the decision at each node takes the form of “ xy st or x, > t," where t eR andj e (1. 2. Give an example with at most 3 training points for which it isn't possible to build a decision tree that behaves exactly the same as a l-nearest neighbor classifier. Let's add a dimension! Your answer: = . ) ER* and (c) the decision at each node takes the form of "x, st or x, > 1," wheret ER and j e 1, 2, under what conditions is it possible to buzild a decision tree that behaves exactly the same as a l-nearest neighbor classifier? Explain why it is possible to build the decision tree as stated in part (a) but, in general, not possible in part (6). Assuming we have 2-dimensional training points
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