If I've understood the problem correctly...
You can use a key to map a location (@area,@cite) to a sequence of
element(tree). Iterating over the tree elements, do not copy a loc
unless it is first in the sequence returned by that key: This
guarantees that all locs are covered. It may happen that this
eliminates all locs from a tree: if this must not happen, copy an
arbitrary loc.
(If the pass over all tree elements is written as a recursive
function, you may pass down a set of loc's that has been used, and
select one from this set if it can be used to avoid the empty set
within a tree.)
This may not produce the smalles set, but it should at least deal with
all necessities.
-W
On 09/01/2013, Graydon <graydon(_at_)marost(_dot_)ca> wrote:
So I have a lot (~600) tree elements that look like:
<tree parent="attribution">
<count count="1"/>
<locations>
<loc area="loiprop" cite="2010-07-prop10"/>
</locations>
</tree>
<tree parent="block">
<count count="10"/>
<locations>
<loc area="loiprop" cite="2004-09-bud2004"/>
<loc area="loiprop" cite="2010-07-prop10"/>
<loc area="loiprop" cite="2010-08-allprop"/>
</locations>
</tree>
<tree parent="block">
<key>section</key>
<count count="6689"/>
<locations>
<loc area="CRCc945-en" cite="2606"/>
<loc area="CRCc945-en" cite="402"/>
<loc area="CRCc945-en" cite="6804"/>
<loc area="CRCc945-en" cite="8301"/>
<loc area="CRCc945-en" cite="8308.1"/>
....
</locations>
<child>section</child>
</tree>
The @area,@cite pairs represent documents with the locations of particular
patterns (via the @parent and the child elements of the tree elements) of
documentation content, one tree element per pattern. (So there are lots of
tree/@parent="block" elements, etc.)
What's wanted is the minimum number of documents that contain _all_ the
patterns, for output testing purposes.
So I need to produce the, or at least _a_, shortest list of <loc/> elements
so that every (all ~600) tree element contains at least one loc element on
the list.
This has turned out to be much less straightforward than I thought. I can't
shake the feeling that there's a grouping solution, but I'm not seeing it if
there is. So I'd appreciate any algorithm hints anybody has got;
unfortunately, efficiency is a concern.
Demonstration that this is really an NP-complete problem also grateful
accepted!
Thanks!
Graydon
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