Re: how to suppress carefully a recursive tree

From: David BL <>
Date: Wed, 23 Jan 2008 05:32:42 -0800 (PST)
Message-ID: <>

On Jan 23, 4:10 pm, fj <> wrote:
> On 23 jan, 01:12, David BL <> wrote:
> > On Jan 23, 1:10 am, fj <> wrote:
> > > On 22 jan, 15:52, Jan Hidders <> wrote:
> > > > On 22 jan, 12:04, fj <> 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 ... a part of the data (a
> > > complete tree or just a branch), but without destroying data shared by
> > > other trees or branches.
> > Would the mark and sweep algorithm suit you purposes?
> No
> the mark and sweep algorithm needs to know all the tree roots in order
> to mark all the used objects.
> In my example I indicates that the node b3 belongs to the root r2
> just for information to explain the storage count. But the deletion
> routine I want to write just receives "r1" as argument, nothing else.
> It does not know that r2 exists too ... It is just able to detect the
> existence of other objects via the storage count of b3 which is a
> little bit too high for being just referenced by r1 !

For a directed edge from node n1 towards node n2, let n2 be referred to as the "destination" node.

   n1 ---->---- n2
  source destination

Maybe the following approach would work:

It is assumed that node r1 is the root of the "tree" to be deleted.

Step 1: Perform a trace (in the manner of a mark phase) treating r1 as the one and only root and decrementing the storage counter on the destination node for each edge that is traversed. ie find the set X of nodes that are reachable from r1. Nodes that have already been visited are flagged to ensure each reachable node is visited exactly once. For each node during the recurse, retrieve the set of outgoing edges yielding a set of destination nodes. Decrement each destination node's storage counter (this must be done for all the destination nodes - ie including the nodes that have already been visited). Then recurse into the destination nodes that haven't been visited before.

Step 2: Find the subset R of X corresponding to nodes whose storage counter hasn't fallen to zero.

Step 3: Run a mark and sweep using roots R and sweeping over X. For each edge traversed in the mark phase, increment the storage counter on the destination node.

In your example I think step 1 will find X = {r1,b1,b2,b3,b4} and only b3 will end up with a nonzero storage counter, so step 2 yields R = {b3}. Then in step 3, a trace from roots {b3} will bump the storage counter of b4 back up to 1 and only {r1,b1,b2} will be deleted.

Am I making sense? Received on Wed Jan 23 2008 - 14:32:42 CET

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