Re: [xsl] Re: Keeping a running total?

[Andrew Franz <afranz0@xxxxxxxxxxxxxxxx>]

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. :
> >    <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.


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:
>
>  O(Fn)*O(Log2(Qn))
>
> For simplicity, in the implementation I provided the identification of
> overlapping intervals traverses all quotas intervals for every
> factory.