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In the general case this is quite similar to quadratic time complexity.
O(Fn)*O(Log2(Qn))
Re: [xsl] Re: Keeping a running total?
Subject: Re: [xsl] Re: Keeping a running total? From: "Dimitre Novatchev" <dnovatchev@xxxxxxxxx> Date: Thu, 13 Jul 2006 13:48:07 -0700 |
> 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. : <quotas> <quota id="Widget" left="{$Widget_quota}" /> <quota id="Gadget" left="{$Gadget_quota}" /> ... </quotas>
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> ... </xsl:apply-templates> </xsl:template>
then in the factory template, each factory would consume $left0 until it was depleted and then shift by assigning $left1 to $left0, $left2 to $left1 and so on through parameters. e.g. <xsl:with-param name="left0" select="$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.
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:
O(Fn)*O(Log2(Qn))
For simplicity, in the implementation I provided the identification of overlapping intervals traverses all quotas intervals for every factory.
-- Cheers, Dimitre Novatchev --------------------------------------- Truly great madness cannot be achieved without significant intelligence.
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