RE: [xsl] XSLT 2.0 function - fastest node comparison

Subject: RE: [xsl] XSLT 2.0 function - fastest node comparison
From: "Kirkham, Pete (UK)" <pete.kirkham@xxxxxxxxxxxxxx>
Date: Thu, 10 Mar 2005 14:39:06 +0000
Assuming the ranges don't overlap, you could transform the list of ranges into
a sorted binary tree:

<range from="988" to="8197">
  <range to="1036">
    <range to="1009">
      <range from="988" to="989"/>
      <range from="1008" to="1009"/>
    <range from="1014">
      <range from="1014" to="1014"/>
      <range from="1025" to="1036"/>
  <range from="1038">
    <range to="1116">
      <range from="1038" to="1103"/>
      <range from="1105" to="1116"/>
    <range from="1118">
      <range to="4150">
        <range from="1118" to="1119"/>
        <range from="4150" to="4150"/>
      <range from="8194" to="8197"/>

which gives O(log N) range tests rather than O(N) range tests. Then you have
to find out whether a recursive function to navigate the tree is slower by a
sufficient factor to kill any advantage in the number of tests being lower for
large N.


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