Chess board position numbers in 6-rooted-binary tree?

Posted by HH on Stack Overflow See other posts from Stack Overflow or by HH
Published on 2010-05-25T18:56:26Z Indexed on 2010/05/25 19:01 UTC
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The maximum number of adjacent vertices is 6 that corresponds to the number of roots. By the term root, I mean the number of children for each node. If adjacent square is empty, fill it with Z-node. So every square will have 6 nodes.

  1. How can you formulate it with binary tree?
  2. Is the structure just 6-rooted-binary tree?
  3. What is the structure called if nodes change their positions?
  4. Suppose partially ordered list where its units store a large randomly expanding board. I want a self-adjusting data structure, where it is easy to calculate distances between nodes. What is its name?

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