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minimum disjoint path covering of a graph - algorithm needed

hello
i have a problem - i need an algorithm which computes the minimum
number of disjoint path in a labirynth.

but i think that using a graph to do it is a good idea, but all i've
found is that F.T. Boesch had written "Covering the points of a graph
with point-disjoint paths..." in 1974. or something like that.

i couldn't find that article in Internet (at least a free one)

so if someone knows where should i look for that algorithm (link would
be usefull) , please tell me.

thank you

SubNeo
zg*****@tenbit.pl

Oct 26 '05 #1
2 2351
SubNeo wrote:
hello
i have a problem - i need an algorithm which computes the minimum
number of disjoint path in a labirynth.
This is off topic in this group. Please see the FAQ for what which topics to
take here and which topics you should take some other place.
but i think that using a graph to do it is a good idea, but all i've
found is that F.T. Boesch had written "Covering the points of a graph
with point-disjoint paths..." in 1974. or something like that.

i couldn't find that article in Internet (at least a free one)


What about a library? They specialize in archiving and retrieving paper.
Probably, they can help you finding stuff that is not online.
Best

Kai-Uwe Bux

Oct 26 '05 #2

"SubNeo" <zg*****@tenbit.pl> wrote in message
news:11**********************@g44g2000cwa.googlegr oups.com...
hello
i have a problem - i need an algorithm which computes the minimum
number of disjoint path in a labirynth.

but i think that using a graph to do it is a good idea, but all i've
found is that F.T. Boesch had written "Covering the points of a graph
with point-disjoint paths..." in 1974. or something like that.

i couldn't find that article in Internet (at least a free one)

so if someone knows where should i look for that algorithm (link would
be usefull) , please tell me.


How about:

http://www.boost.org/libs/graph/doc/..._contents.html

Jeff Flinn
Oct 26 '05 #3

This thread has been closed and replies have been disabled. Please start a new discussion.

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