3
$\begingroup$

Given M binary variables and R samples, what is the maximum number of leaves in a decision tree?

My first assumption was that the worst case would be a leaf for each sample, thus R leaves maximum. Am I wrong and there should be a kind of connection with the number of variables M? I know that the maximum depth of a decision tree is M as a variable can appear once in a branch, but I don't see the relation with the number of leaves.

Thanks in advance!

$\endgroup$

1 Answer 1

2
$\begingroup$

The maximum possible combinations with M binary variable is

$$ 2^M $$ so essentially if all these values have different classes, then the number of leaves should be equal to

$$ if \ R<2^M => R \\ else \ 2^M $$

A tree which has all possible leaves for M binary variables could at max contain 2^M combinations, think each leaf with written value of a possible combination eg: 0000,0001,0002,...,1110,1111, and these can come only once, because only 1 label can be associated with each leaf

In case a row has multiple labels, for same set of input, the max number of leaves would be equal to unique input combinations in R

    A   B   label
0   1   0   0
1   0   0   1
2   1   1   2
3   1   1   1

The number of leaves in this case would be 3 and not 4 (number of inputs)

enter image description here

$\endgroup$
2
  • $\begingroup$ Thank you very much! Assuming I am using the ID3 algorithm, a greedy algorithm, which may assign a variable once in a branch but multiple times in a tree- won't it have to affect the 2^M maximum? $\endgroup$
    – yaminlee
    Jun 27, 2021 at 17:00
  • 1
    $\begingroup$ an example row of input eg: 1110 with 4 variables would be found in the tree only once, irrespective of however the tree is split, extrapolating this information to all input rows, any leaf would only appear at max once in a tree $\endgroup$ Jun 27, 2021 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.