27 Nov 2006 17:44:34 -0500

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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