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When Wolfram's technical support cannot help

• To: mathgroup at smc.vnet.net
• Subject: [mg105255] When Wolfram's technical support cannot help
• From: "Ant King" <mathstutoring at ntlworld.com>
• Date: Wed, 25 Nov 2009 06:14:07 -0500 (EST)

Hi

I sent this email to technical support (as I hold a premier licence)

I am looking for a single function that will extract the cyclic part of (a)
a completely cyclic sequence and (b) an eventually cyclic sequence. So if

data1={1,2,3,1,2,3,1,2,3,1,2,3} then cyclicpart[data1] should return {1,2,3}

And if

data2={5,9,11,8,1,2,3,1,2,3,1,2,3,1,2,3} then cyclicpart[data2] should also
return {1,2,3}

And this was the reply that I got

Here is one way to extract out the cyclic part:

lis = Flatten[Join[Table[{1, 2, 3}, {8}]]]

p = Position[Table[BitXor[lis, RotateLeft[lis, i]], {i, 1, 10}],
    ConstantArray[0, Length[lis]]][[1]] /. {a_} -> a
lis[[1 ;; p]]

The above however will only work for case where the list always contains
the repeated pattern.

There is no built-in function as such that will extract out the pattern
automatically. I have filed a suggestion with our developers and you will
be notified when this suggestion gets implemented. Again, my apologies for
the delay and my thanks for your patience.

Now I don't believe that. I think that it should be quite achievable. Anyone
got any ideas

Thanks a lot

Ant

• Follow-Ups:
  • Re: When Wolfram's technical support cannot help
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
  • Re: When Wolfram's technical support cannot help
    • From: Daniel Lichtblau <danl@wolfram.com>
  • Re: When Wolfram's technical support cannot help
    • From: DrMajorBob <btreat1@austin.rr.com>