Answers.Body: <p>I'd say that the leaves <em>per se</em> count, too, but only if they're real leaves, like e.g., checkmate positions in chess.</p>
  
  <p>Such a node has really no children and no further calculation is needed. Unlike with nodes which weren't expanded yet.</p>
  
  <p><em>Note that always counting the leaves <a href="https://math.stackexchange.com/a/454646/14435">provably</a> leads to <code>(n-1)/n</code> for every <code>n</code>-node thee!</em></p>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-04-10T03:37:51.487Z
Answers.LastActivityDate: 2018-04-10T03:37:51.487Z
Answers.UserDisplayName: maaartinus
Body: <p>In the search tree below, there are 11 nodes, 5 of which are leaves. There are 10 branches.</p>
  <p><a href="https://i.sstatic.net/MCRn3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MCRn3.png" alt="Search tree" /></a></p>
  <p>Is the average branching factor given by 10/6, or 10/11?</p>
  <p>Are leaves included in the calculation? Intuitively, I would think not, since we are interested in nodes with branches. However, a definition given to me by my professor was &quot;The average number of branches of all nodes in the tree&quot;, which would imply leaves are included.</p>
  
Tags: <search><branching-factors>
Title: Are leaf nodes included in the calculation of average branching factor for search trees?
UserDisplayName: Data
ThreadText: # Are leaf nodes included in the calculation of average branching factor for search trees?
  
  (Asked by: username_0 on 2018-04-04)
  
  In the search tree below, there are 11 nodes, 5 of which are leaves. There are 10 branches.
  
  [![Search tree](https://i.sstatic.net/MCRn3.png)](https://i.sstatic.net/MCRn3.png)
  
  Is the average branching factor given by 10/6, or 10/11?
  
  Are leaves included in the calculation? Intuitively, I would think not, since we are interested in nodes with branches. However, a definition given to me by my professor was "The average number of branches of all nodes in the tree", which would imply leaves are included.
  
  **Tags:** search, branching-factors
  
  
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  ## 1 answers
  
  ### Answer by username_1 on 2018-04-10. Score: 1
  
  I'd say that the leaves *per se* count, too, but only if they're real leaves, like e.g., checkmate positions in chess.
  
  Such a node has really no children and no further calculation is needed. Unlike with nodes which weren't expanded yet.
  
  *Note that always counting the leaves [provably](https://math.stackexchange.com/a/454646/14435) leads to `(n-1)/n` for every `n`-node thee!*
  
  
  ---
  
  