Related articles 

[4 earlier articles] 
Re: finding all dominator trees s1l3ntR3p@gmail.com (s1l3ntr3p) (20061128) 
Re: finding all dominator trees amichail@gmail.com (Amir Michail) (20061129) 
Re: finding all dominator trees bmosesnospam@cits1.stanford.edu (Brooks Moses) (20061129) 
Re: finding all dominator trees s1l3ntR3p@gmail.com (s1l3ntr3p) (20061129) 
Re: finding all dominator trees DrDiettrich1@aol.com (HansPeter Diettrich) (20061129) 
Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (20061130) 
Re: finding all dominator trees amichail@gmail.com (Amir Michail) (20061203) 
Re: finding all dominator trees diablovision@yahoo.com (20061204) 
Re: finding all dominator trees amichail@gmail.com (Amir Michail) (20061205) 
Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (20061205) 
Re: finding all dominator trees martin@gkc.org.uk (Martin Ward) (20061205) 
Re: finding all dominator trees mailbox@dmitrykazakov.de (Dmitry A. Kazakov) (20061206) 
Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (20061206) 
From:  "Amir Michail" <amichail@gmail.com> 
Newsgroups:  comp.compilers 
Date:  3 Dec 2006 21:32:18 0500 
Organization:  Compilers Central 
References:  06110960611117 0611125 0611131 
Keywords:  analysis, question 
Chris F Clark wrote:
> ...
>
> Yes, there are more efficient algorithms for computing the sets of
> dominators in a graph considering each vertex in the graph as a root.
> Both reachable vertexes and stronglyconnectedcomponents (cycles)* are
> part of the algorithm.
>
> First, if there is one unique vertex that can reach all other vertexes
> (verticies if you prefer) in the graph, consider that vertex the root.
> The dominator algorithm given that root will calculate the correct
> dominators (the dominator tree) for every vertex (v1) in the graph
> assuming that each other vertex (v2) is the root. That is, for any
> target vertex v1 and root vertex v2, some vertex in the dominator tree
> of v1 will be the dominator of v1 given the root v2.
>
I don't understand this solution. How would it work in this example?
A > B <> C
The dominator tree for A:
A

B

C
The dominator tree for B:
B

C
The dominator tree for C:
C

B
Amir
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