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

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

> > 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"


So, how many times will parameter shift be required?


Cheers,
Dimitre Novatchev



On 7/14/06, Andrew Franz <afranz0@xxxxxxxxxxxxxxxx> wrote:
> 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.


--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.