Hi,<br><br>I'm trying to implement a fast kd-tree in Haskell. <a href="http://en.wikipedia.org/wiki/Kd-tree">http://en.wikipedia.org/wiki/Kd-tree</a> It's a binary tree representing space partitions.<br><br>One of the fastest ways to build kd-trees is outlined in this paper:
<a href="http://www.cgg.cvut.cz/~havran/ARTICLES/ingo06rtKdtree.pdf">http://www.cgg.cvut.cz/~havran/ARTICLES/ingo06rtKdtree.pdf</a> (I'm trying to implement Algorithm 5 on page 5 and the steps outlined in section 4.3.1
.)<br><br>I'll try to outline a simplified analogous algorithm to demonstrate my problem. It's all about classifying which children should go in a left or right branch using heuristics.<br><br>Say I want to put the words 'foo', 'bar' and 'baz' into a binary tree. The heuristic requires I split the words into letters and sort them:
<br>'aabbfoorz'. The heuristic then may decide, based on the sorted letters, that 'bar' and 'foo' should go in the left child and 'baz' goes in the right. Typically we'd then simply recurse and for example, the left child's words would be re-sorted into 'abfoor' and the heuristic is reapplied.
<br><br>If we assume that sorting is relatively expensive, we can avoid the re-sort for the children by unmerging the parent's sorted list of letters. Two sublists of a sorted list should already be sorted. If we know which word each letter belongs to it would be more efficient to tag the letters with 'left' or 'right' as the words are classified. Then we can iterate down the sorted letter list and produce new sorted sublists rather simply.
<br><br>So it's not actually that complicated, and I can imagine exactly how it could be done in C but I really don't know how to approach this in Haskell. The problem I'm having is how to keep a map between the words and its letters (which in the real problem is a map between a list of vectors and 6 tuples containing Doubles and enumerations) while keeping in mind you can't just map any letter to a word, but specific letters.
e.g., the 'b's in the example must remember specifically whether they belong to 'bar' or 'baz'.<br><br>Thanks for any insights,<br>Toby.<br><br>