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Re: Re: Keeping a running total?

Subject: Re: Re: Keeping a running total?
From: Andrew Franz <afranz0@xxxxxxxxxxxxxxxx>
Date: Sat, 15 Jul 2006 15:33:16 +1000
Re:  Re: Keeping a running total?
Dimitre Novatchev wrote:

> What about having a bigger than 2 (variable) number of quotas? How
> would your solution change to tackle this?
I suppose a general solution would have a list of quotas (and a
corresponding list of quota names), e.g. :
       <quota id="Widget" left="{$Widget_quota}" />
       <quota id="Gadget" left="{$Gadget_quota}" />

Updating this list on every template call leads to O(Qn)*O(Fn) time complexity, where Fn is the number of factories and Qn the number of quotas.

In the general case this is quite similar to quadratic time complexity.

In the past I've 'cheated' by using the stack as a list, e.g. as follows:

<xsl:template match="xml">
  <xsl:apply-templates select="factory">
      <xsl:with-param name="left0" select="$Widget_quota" />
      <xsl:with-param name="left1" select="$Gadget_quota" />
      <xsl:with-param name="name0">Widget</xsl:with-param>
      <xsl:with-param name="name1">Gadget</xsl:with-param>

then in the factory template, each factory would consume $left0 until it
was depleted and then shift by assigning $left1 to $left0, $left2 to
$left1 and so on through parameters.
       <xsl:with-param name="left0" select="$left1" />   <!-- shift to
next quota -->
       <xsl:with-param name="left1" select="$left2" />
       <xsl:with-param name="left2" select="$left3" />
       <xsl:with-param name="left3" select="$left4" />

Quota-recursion would end when $left0 was empty then capacity would
accumulate in $excess until factories were depleted.

Having to perform such shifts on template calls again leads to quadratic-like time complexity.

This is not correct.

A template call will occur in 2 cases:
1. quota < capacity, in which case quota is consumed and quotas are shifted
2. capacity < quota, in which case you shift to the next factory

Order(Qn) + Order(Fn)

Like I said, "It is O(N) because it monotonically advances through both lists (factories and quotas) without backtracking"

Better time complexity can be achieved if the running totals for *both* factories and quotas are calculated with an O(N) (linear) algorithm.

The implementation I provided could have better than quadratic time
complexity if it used a more efficient way to find overlapping
intervals -- a binary search with an f:binSearch() - like function
would be O(Log2(Qn)). In this case the time complexity would be:


For simplicity, in the implementation I provided the identification of
overlapping intervals traverses all quotas intervals for every

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