Question

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.

Answer 1

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 of Q. Then you can figure out how many examples you need. Increase Q and do it again.

For the second part of your question, pick a small Z and run the decision tree algorithm by hand. See how many leaves you get. Then pick another Z and see if/how it changes. Try generating different examples (with the same Z) and see if you can change the number of leaves.