On 1/17/06, David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk> wrote:
Dimitre asked a similar question a while back:
http://www.xslt.com/html/xsl-list/2005-06/msg00844.html
Yes. I owe an apology to David and to everyone for not posting the
responses to the XSLT puzzle.
I planned to do this before New Year, but just then someone I loved
died and I was overseas for three weeks. I will summarize David's and
mine solution most probably this weekend.
Dimitre
at least, the first part of his question involved calculating fibonacci
numbers efficiently.
Yes, this algorithm has a time complexity of O(log2 N).
I'm not sure he ever published the results that
people sent him.
For fibonacci, I suggested:
<xsl:function name="x:fib" as="xs:integer">
<xsl:param name="n" as="xs:integer"/>
<xsl:choose>
<xsl:when test="$n=0">
<xsl:sequence select="1"/>
</xsl:when>
<xsl:when test="$n =1">
<xsl:sequence select="1"/>
</xsl:when>
<xsl:when test="$n mod 2 = 0">
<xsl:variable name="n2" select="$n idiv 2"/>
<xsl:variable name="fn" select="x:fib($n2)"/>
<xsl:variable name="fn-1" select="x:fib($n2 -1)"/>
<xsl:sequence select="$fn * $fn + $fn-1 * $fn-1"/>
</xsl:when>
<xsl:when test="$n mod 2 = 1">
<xsl:variable name="n2" select="($n - 1) idiv 2"/>
<xsl:variable name="fn" select="x:fib($n2)"/>
<xsl:variable name="fn-1" select="x:fib($n2 -1)"/>
<xsl:sequence select="(2 * $fn-1 + $fn) * $fn"/>
</xsl:when>
</xsl:choose>
</xsl:function>
which should I think me considerably quicker than the function that you
suggested for large numbers.
David
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Cheers,
Dimitre Novatchev
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