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Re: When Wolfram's technical support cannot help

  • To: mathgroup at
  • Subject: [mg105290] Re: When Wolfram's technical support cannot help
  • From: "Nasser M. Abbasi" <nma at>
  • Date: Wed, 25 Nov 2009 23:04:53 -0500 (EST)
  • References: <hej3pj$d77$>

On Nov 25, 5:18 am, "Ant King" <mathstutor... at> wrote:
> 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

May be using the algorithm for finding longest repeated subsequence
might help with your second case data2?


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