10 Aug 2002 02:32:49 -0400

Related articles |
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[3 earlier articles] |

Re: Finding the set of recursive calls Martin.Ward@durham.ac.uk (Martin Ward) (2002-07-24) |

Re: Finding the set of recursive calls vbdis@aol.com (VBDis) (2002-07-24) |

Re: Finding the set of recursive calls jeremy.wright@microfocus.com (Jeremy Wright) (2002-07-25) |

Re: Finding the set of recursive calls vbdis@aol.com (VBDis) (2002-07-31) |

Re: Finding the set of recursive calls haberg@matematik.su.se (Hans Aberg) (2002-08-04) |

Re: Finding the set of recursive calls vbdis@aol.com (VBDis) (2002-08-10) |

Re: Finding the set of recursive calls vbdis@aol.com (VBDis) (2002-08-10) |

Re: Finding the set of recursive calls haberg@matematik.su.se (Hans Aberg) (2002-08-14) |

From: | "VBDis" <vbdis@aol.com> |

Newsgroups: | comp.compilers |

Date: | 10 Aug 2002 02:32:49 -0400 |

Organization: | AOL Bertelsmann Online GmbH & Co. KG http://www.germany.aol.com |

References: | 02-08-007 |

Keywords: | analysis |

Posted-Date: | 10 Aug 2002 02:32:49 EDT |

"Hans Aberg" <haberg@matematik.su.se> schreibt:

*>In a directed graph, the equivalence relation used to define strongly*

*>connected components is that there are paths both ways between vertices.*

In a discussion with Prof. Eppstein it turned out, that a vertex can

form a strongly connected component, even if no explicit path exists

from the vertex back to itself. This is due to the reflexive property

of equivalence relations. Thus the set of recursive procedures is

only a subset of the strongly connected components of an graph.

So in the case of the set of recursive calls I'm not sure, which

algorithms really are applicable without modificiations. The set of

strongly connected components can be used as a base, from which all

single-vertex components must be removed, which have no edges to

themselves.

(I knew that there was a problem... ;-)

DoDi

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