决策树学习算法
我想先说这是一项家庭作业。
我得到一组 Q 二进制输入变量，用于对 Y 的输出进行分类，Y 也是二进制的。
问题的第一部分是：我最多需要多少个例子来枚举Q的所有可能组合？我目前认为，由于它最多要求我需要 Q ，因为 Q-1 之前的所有值可能都是相同的，例如 1 ，而 Q 处的项目是 0 。
问题的第二部分是：树最多可以有多少个叶子节点给出Z个例子？
我目前的答案是，树最多有 2 个叶节点，一个代表 true，一个代表 false，因为它处理的是二进制输入和二进制输出。
这是检查这个问题的正确方法还是我太深入地概括了我的答案？
编辑
在查看卡梅伦的回复后，我现在将我的第一个答案变成 2^Q 并以他的 Q = 3 的示例为基础，我将得到 2^3 或 8 (2*2*2)。如果有不正确的想法请指正。
编辑 #2
问题的第二部分看起来应该是 (2^Q) * Z 或提供一个示例：(2^3) * 3) 或 8*3 = 24 个叶节点。回顾一下，如果我有 3 个二进制输入，我最初会取 2^3，现在得到 8 个，现在我想看 3 个例子。因此我应该得到 8*3 或 24。
编辑#3
事后看来，无论我使用多少个例子，叶节点的数量都不应该增加，因为它是基于每棵树的。
I want to preface this by saying that this is a homework assignment.
I am given a set of Q binary input variables that will be used to classify output of Y which is also binary.
The first part of the question is: at most how many examples do I need to enumarate all possibile combinations of Q? I am currently think that since it asks for at most I will need Q as it is possible that all values up to Q-1 are the same for instance 1 and the item at Q is 0 .
The second part of the question is: at most how many leaf nodes can the tree have given Z examples?
My current answer is that at most the tree would have 2 leaf nodes, one representing true and one representing false since it is dealing with binary inputs and binary outputs.
Is this the correct way of examining this problem or am I generalizing my answers too deeply?
Edit
After looking at Cameron's response, I would now turn my first answer into 2^Q and to build on his example of Q = 3, I would get 2^3 or 8 (2*2*2). Please correct if that is incorrect thinking.
Edit #2
The second part of the question it appears as though it should be (2^Q) * Z or to provide an example: (2^3) * 3) or 8*3 = 24 leaf nodes. To recap if I have 3 inputs that are binary I would initially take 2^3 and get 8 now I want to go over 3 examples. Therefore I should get 8*3 or 24.
Edit #3
In hindsight it seems that no matter how many examples I use the number of leaf nodes should never increase, as it is a per tree basis.
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评论（1）
我建议您通过手动计算小示例来解决该问题。
对于第一部分，为
Q选择一个较小的值，例如 3，并写下Q的所有可能组合。然后你就可以算出你需要多少个例子。增加Q并再做一次。对于问题的第二部分，选择一个小的
Z并手动运行决策树算法。看看你得到了多少片叶子。然后选择另一个Z并查看它是否/如何变化。尝试生成不同的示例（使用相同的Z），看看是否可以更改叶子的数量。I'd suggest you approach the problem by working out small example cases by hand.
For the first part, choose a small value for
Q, say 3, and write down all possible combinations ofQ. Then you can figure out how many examples you need. IncreaseQand do it again.For the second part of your question, pick a small
Zand run the decision tree algorithm by hand. See how many leaves you get. Then pick anotherZand see if/how it changes. Try generating different examples (with the sameZ) and see if you can change the number of leaves.