Re: Ordered union of sequences
Am 08.04.2010 um 16:09 schrieb Michael Kay: > It seems to me that you first want to create (or imagine) a graph: in your > example there are arcs k->o, o->p, p->c, a->b, b->c, etc. > > Then you want to look for cycles in this graph. If any cycles exist, there > is no solution to your problem. Thanks Michael, I guess I have to finally understand the graph stuff. I will turn to the example in the 4th edition p.251 ff. for a starter. Am 08.04.2010 um 16:47 schrieb Imsieke, Gerrit, le-tex: > [... cool, ready-made example ...] > Does that make sense? > > If I include <o/> at the other position, i.e., > <seq><k/><f/><z/><o/></seq>, > I receive "Too many nested function calls. May be due to infinite recursion." as expected. Gerrit, I am at early stages to understand the algorithm. The function counts the maximum number of preceding siblings on any available path and uses this as a sort key. This is some sort of creating a graph, backtracing to the beginning... which is what Michael suggested, isn't it. Thanks a lot for the most valuable input! I need a result and the inconsistency report, so the user can fix the input. I will be looking into stopping the infinite recursion somehow. - Michael >>>> There is an arbitrary number of sequences, sometimes >>> containing items >>>> with the same name: >>>> >>>> (k, o, p, c, f) >>>> (d, e, f, g) >>>> (k, f, z, o) >>>> (a, b, c, d) >>>> >>>> I want to create a master sequence which contains every item once, >>>> preserving the original order. >>> -- _______________________________________________________________ Michael M|ller-Hillebrand: Dokumentations-Technologie Adobe Certified Expert, FrameMaker Lvsungen und Training, FrameScript, XML/XSL, Unicode Blog: http://cap-studio.de/ - Tel. +49 (9131) 28747
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