Re: how to suppress carefully a recursive tree
Date: Wed, 23 Jan 2008 04:40:00 -0800 (PST)
On 22 jan, 17:10, fj <francois.j..._at_irsn.fr> wrote:
> On 22 jan, 15:52, Jan Hidders <hidd..._at_gmail.com> wrote:
> > On 22 jan, 12:04, fj <francois.j..._at_irsn.fr> wrote:
> > > I know how to suppress a normal tree but I meet the following kind of
> > > situation :
> > I'm guessing that when you say "suppress" you mean "represent in a
> > database". Correct?
> No : I want to destroy, remove, kill ...
Try "delete", that's a better translation of "supprimer". To suppress is more like "reprimer", "bannir" or "inhiber".
> a part of the data (a
> complete tree or just a branch), but without destroying data shared by
> other trees or branches.
> > > r1
> > > -> b1
> > > -> b2 -> ...
> > > -> b3 -> ...
> > > -> b2
> > > -> b1 -> ...
> > > -> b1 -> ...
> > > -> b3
> > > -> b4
> > > r2
> > > -> b3 -> ...
> > So, a directed graph with multiple roots. Correct? So basically you
> > want to store arbitrary directed graphs.
> Yes and no. The two "roots" r1 r2 could perhaps belong to a bigger
> > > A same node can be referenced at several places. Each node is
> > > associated to a storage count :
> > > r1(0) b1(3) b2(2) b3(3) b4(1) r2(0)
> > Which would correspond to the number of incoming edges, yes?
> > > In a normal tree without recursion (in the example above, recursion
> > > occurs because b1 contains b2 and vice versa), a node is destroyed
> > > when its count storage is equal to zero else its count storage is
> > > simply decremented.
> > > What algorithm should be applied ? I want for instance to cleanup r1
> > > but, of course, r2 must remain valid (=> b3 and b4 are not destroyed
> > > during the process and their storage count must be b3(1) b4(1)).
> > > Notice that the deletion of of a tree must be possible even if the
> > > count storage of the root is not equal to zero :
> > > r1 -> b1 -> b2 -> r1 -> ...
> > It all depends a bit on how large your typical graphs are, how long on
> > average the simple paths, what type of operations and queries you want
> > to do on it and how often. My first guess for the representation
> > would a simple straightforward adjacency list representation, (a
> > binary relation that contains all the edges) and if it's not too big
> > and your paths are often long it might be interesting to maintain an
> > extra table with the transitive closure of the graph.
> The graph may be quite large. Number of vertices (nodes) : usually
> 100000, sometimes much more (a very big computation may lead to about
> 100 millions). This corresponds to a 3D meshing, each mesh (a
> particular node) containing information about chemical composition (a
> sub-node), temperature, fluid characteristics (another sub-node) ...
> Let us precise that simple paths are always short when one excludes
> recursive points (a maximum of 10 nodes).
> > If you go for the adjacency list approach, make sure that you do as
> > much as possible in one SQL statement when you start following the
> > edges. So look up all nodes that are reachable in one step in one
> > statement, update those, and store the ones that have to be deleted in
> > a temporary table. Then again with one statement look up those that
> > can be reached from in one step from the nodes in the temporary table.
> > Et cetera.
> I don't use SQL but it does not matter :
It might. Outside the context of an SQL database my advice might actually be counterproductive.
> I can build up easily the
> list of nodes related to a particular starting point (the node "env"
> in my example). I am also able to compute, for each node, its
> "external count" (0 for all the nodes except b3 which has the value
> After that I can start to destroy really : I only go down the nodes
> which have an external count equal to zero (this will protect b4 in
> the example). The problem is that the algorithm is not very efficient
> when the list of nodes is high, simply because I need, for each node
> having a storage count greater than 0, to look for that node in the
> list of all the nodes and to check the external count (CPU cost
> proportional to O(n2) if n is the number of concerned nodes).
You might consider defining an index over the set of edges that allows a quicker look-up on the basis of the begin node. Of course there is a price to pay for that because updates get more expensive.
- Jan Hidders