Related articles |
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finding all dominator trees amichail@gmail.com (Amir Michail) (2006-11-22) |
Re: finding all dominator trees diablovision@yahoo.com (2006-11-26) |
Re: finding all dominator trees martin@gkc.org.uk (Martin Ward) (2006-11-27) |
Re: finding all dominator trees amichail@gmail.com (Amir Michail) (2006-11-27) |
Re: finding all dominator trees s1l3ntR3p@gmail.com (s1l3ntr3p) (2006-11-28) |
Re: finding all dominator trees amichail@gmail.com (Amir Michail) (2006-11-29) |
Re: finding all dominator trees bmoses-nospam@cits1.stanford.edu (Brooks Moses) (2006-11-29) |
Re: finding all dominator trees s1l3ntR3p@gmail.com (s1l3ntr3p) (2006-11-29) |
Re: finding all dominator trees DrDiettrich1@aol.com (Hans-Peter Diettrich) (2006-11-29) |
Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (2006-11-30) |
[7 later articles] |
From: | "Amir Michail" <amichail@gmail.com> |
Newsgroups: | comp.compilers |
Date: | 27 Nov 2006 17:44:34 -0500 |
Organization: | Compilers Central |
References: | 06-11-09606-11-106 |
Keywords: | analysis |
Posted-Date: | 27 Nov 2006 17:44:34 EST |
diablovision@yahoo.com wrote:
> > Is there an efficient algorithm for finding all dominator trees of a
> > graph? That is, for every node, find its dominator tree. I'm looking
> > for something better than simply running a dominator tree algorithm
> > for each node in the graph.
>
> Unless I am missing some subtlety of your situation, you should just be
> able to run the dominator tree algorithm for the root node of the
> graph. Dominator trees have the property that each subtree for a node
> corresponds to the dominator tree for that node.
>
There is no root. This is an arbitrary graph.
Amir
> See "Efficiently Computing Static Single Assignment Form and the
> Control Dependence Graph" by Cytron et al.
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