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

 Subject: Re: [xsl] Re: Keeping a running total? From: "Dimitre Novatchev" Date: Sat, 15 Jul 2006 09:00:11 -0700
> So, how many times will parameter shift be required?

```Parameter shift will occur whenever a factory's capacity is used up (Fn
times)
```
```So we have Fn shifts of parameters. Every shift of parameters takes
O(Qn) -because every parameter takes the velue of the next parameter.```

```Therefore, the time only the shift of parameters will require will be
Fn * O(Qn)```

which, of course can be written as:

O(Fn) * O(Qn).

This was my initial estimation and you now confirm it.

```A better time complexity can be achieved if the "shift parameters"
operation can be replaced with something else.```

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

On 7/15/06, Andrew Franz <afranz0@xxxxxxxxxxxxxxxx> wrote:
Dimitre Novatchev wrote:

```>> > Having to perform such shifts on template calls again leads to
>>
>>
>> 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?```

```Parameter shift will occur whenever a factory's capacity is used up (Fn
times)```

```1. At any instant, you're comparing the head of the factory list with
the head of the quota list.
2. For each comparison, either the quota list will advance (shift
parameters) or the factory list will advance.(successor axis)```

Number of operations = O(Qn)+O(Fn)

```Is it usual to talk of O(n) with respect to an implementation, or should
the term apply strictly to an algorithm?```

```>
>
> 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}" />
>> >>        ...
>> >>    </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: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
>>
>>
>> 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.
```