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Re: [xsl] RE: Using divide-and-conquer recursion to create a cumulative sequence

2009-12-23 13:46:11
Dimitre provided a solution that used XSLT 2.0 and Hermann provided a 
solution that used 1.0 plus EXSLT extensions.


I did so, because I believe nobody (almost) USES xslt 1.0 nowadays.

Exactly the same solution ( using <xsl:choose> and not XPath 2.0) is
part of the FXSL 1.x (for XSLT 1.0).


--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
Never fight an inanimate object
-------------------------------------
You've achieved success in your field when you don't know whether what
you're doing is work or play
-------------------------------------
I enjoy the massacre of ads. This sentence will slaughter ads without
a messy bloodbath.

On Wed, Dec 23, 2009 at 10:32 AM, Costello, Roger L. 
<costello(_at_)mitre(_dot_)org> wrote:

Hi Folks,

On 11 December 2009 I inquired on this list about how to create a recursive, 
divide-and-conquer XSLT transform that creates a cumulative sequence (see 
below for my original post).

Dimitre Novatchev and Hermann Stamm-Wilbrandt responded with excellent 
solutions. Dimitre provided a solution that used XSLT 2.0 and Hermann 
provided a solution that used 1.0 plus EXSLT extensions.

I was interested in an XSLT 1.0 solution without using extensions. After 
reflecting on Dimitre's and Hermann's solutions, I came up with the below 
solution. I believe that it has a time complexity of O(n log2n). If I err in 
my complexity calculation, please let me know.

Here is the XSLT:

   <xsl:template name="cumulative">
       <xsl:param name="numberList" />
       <xsl:param name="sum" select="0" />

       <xsl:choose>
           <xsl:when test="not($numberList)" />
           <xsl:when test="count($numberList) = 1">
               <xsl:value-of select="$numberList + $sum" />
               <xsl:text> </xsl:text>
           </xsl:when>
           <xsl:otherwise>
               <xsl:variable name="mid" select="floor(count($numberList) div 
2)"/>
               <xsl:call-template name="cumulative">
                   <xsl:with-param name="numberList" 
select="$numberList[position() &lt;= $mid]"/>
                   <xsl:with-param name="sum" select="$sum"/>
               </xsl:call-template>
               <xsl:call-template name="cumulative">
                   <xsl:with-param name="numberList" 
select="$numberList[position() &gt; $mid]"/>
                   <xsl:with-param name="sum" select="$sum + 
sum($numberList[position() &lt;= $mid])"/>
               </xsl:call-template>
           </xsl:otherwise>
       </xsl:choose>
   </xsl:template>

/Roger

-----Original Message-----
From: Costello, Roger L.
Sent: Friday, December 11, 2009 4:39 PM
To: 'xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com'
Subject: Using divide-and-conquer recursion to create a cumulative sequence

Hi Folks,

I wish to convert a sequence of N numbers:

  (23, 41, 70, 103, 99, 6)

Into a cumulative sequence, in which each number is the sum of the previous 
numbers:

  (23, 64, 134, 237, 336, 342)


One approach to solving this is to iterate through the N numbers and sum the 
preceding numbers:

  for i=1 to N
      sum(for j=1 to i return numbers[j])


However, that approach has a time complexity of:

  1 + 2 + 3 + ... + N = N**2/2

For large N, that will be very expensive.

An alternative approach is to create a recursive function that does a single 
pass through the sequence, carrying along (and adding) the accumulated total 
on each recursive call. This has a time complexity of N. Nice.

*********************************************************************
The above (paraphrases) from Michael Kay's book, XSLT 2.0 and XPath 2.0, p. 
993.
The below is from me.
*********************************************************************

However, that sequential recursive approach will entail N recursive calls, 
which will result in running out of memory for large N (let's assume that the 
XSLT processor does not do tail recursive optimization).

I would like a way of solving the problem using divide-and-conquer recursion. 
Can you provide a solution?

/Roger
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--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
Never fight an inanimate object
-------------------------------------
You've achieved success in your field when you don't know whether what
you're doing is work or play
-------------------------------------
I enjoy the massacre of ads. This sentence will slaughter ads without
a messy bloodbath.

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