In the basic decision tree algorithm (assuming we always create binary splits), we choose the fea- ture/value pair with the maximum information gain as the test to use at each internal node of the decision tree. Suppose we modified the algorithm to choose at random from among those feature/value combinations that had non-zero mutual information, and we kept all other parts of the algorithm unchanged. (a) What is the maximum number of leaf nodes that such a decision tree could contain if it were

Database System Concepts
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ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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3. In the basic decision tree algorithm (assuming we always create binary splits), we choose the fea-
ture/value pair with the maximum information gain as the test to use at each internal node of the
decision tree. Suppose we modified the algorithm to choose at random from among those feature/value
combinations that had non-zero mutual information, and we kept all other parts of the algorithm
unchanged.
(a) What is the maximum number of leaf nodes that such a decision tree could contain if it were
trained on m training examples?
(b) What is the maximum number of leaf nodes that a decision tree could contain if it were trained on
m training examples using the original maximum mutual information version of the algorithm?
Is it bigger, smaller, or the same as your answer to (b)?
(c) How do you think this change (using random splits vs. maximum information mutual information
splits) would affect the accuracy of the decision trees produced on average? Why?
Transcribed Image Text:3. In the basic decision tree algorithm (assuming we always create binary splits), we choose the fea- ture/value pair with the maximum information gain as the test to use at each internal node of the decision tree. Suppose we modified the algorithm to choose at random from among those feature/value combinations that had non-zero mutual information, and we kept all other parts of the algorithm unchanged. (a) What is the maximum number of leaf nodes that such a decision tree could contain if it were trained on m training examples? (b) What is the maximum number of leaf nodes that a decision tree could contain if it were trained on m training examples using the original maximum mutual information version of the algorithm? Is it bigger, smaller, or the same as your answer to (b)? (c) How do you think this change (using random splits vs. maximum information mutual information splits) would affect the accuracy of the decision trees produced on average? Why?
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