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------=_NextPart_000_001F_01C3AA38.9E5B9B40 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable What is the command (if any) that returns (in a list) the disjoint = cyclic factors of a permutation? For instance, (1,2,3)(4,5,6) returns = [(1,2,3), (4,5,6)]. =20
Building a routine is easy enough, but it's better to use a built-in =
one, if it's there.
Thanks.
Bruce ------=_NextPart_000_001F_01C3AA38.9E5B9B40 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=3DContent-Type content=3D"text/html; = charset=3Diso-8859-1"> <META content=3D"MSHTML 6.00.2800.1264" name=3DGENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=3D#ffffff> <DIV><FONT face=3DArial size=3D2>What is the command (if any) that = returns (in a=20 list) the disjoint cyclic factors of a permutation? For instance,=20 (1,2,3)(4,5,6) returns [(1,2,3), (4,5,6)]. </FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Building a routine is easy enough, but = it's better=20 to use a built-in one, if it's there.</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Thanks.</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Bruce</FONT></DIV></BODY></HTML> ------=_NextPart_000_001F_01C3AA38.9E5B9B40--