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.
>>>

--
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