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() <= $mid]"/>
<xsl:with-param name="sum" select="$sum"/>
</xsl:call-template>
<xsl:call-template name="cumulative">
<xsl:with-param name="numberList"
select="$numberList[position() > $mid]"/>
<xsl:with-param name="sum" select="$sum +
sum($numberList[position() <= $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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