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How to find modulus patterns using Mathematica

开发者 https://www.devze.com 2023-03-16 17:20 出处:网络
Is there any way to find the lowest modulus of a list of integers? I\'m not sure how to say it correctly, so I\'m going to clarify with an example.

Is there any way to find the lowest modulus of a list of integers? I'm not sure how to say it correctly, so I'm going to clarify with an example.

I'd like to input a list (mod x) and output the "same" list, modulus y (< x). For example, the list {0, 4, 6, 10, 12, 16, 18, 22} (mod 24) is essential开发者_开发百科ly the same as {0, 4} (mod 6).

Thank you for all your help.


You are looking for a set of arithmetic sequences. We'll consider your example

ee = {0, 4, 6, 10, 12, 16, 18, 22};

which has two such sequences, and an example with four of them.

ff = {0, 3, 7, 11, 17, 20, 24, 28, 34, 37, 41, 45};

In this second one we start with {0,3,7,11} and then increase by 17. So what is the general way to get from the nth term to the n+1th? If the set has k sequences (k=2 for ee and 4 for ff) then add the modulus to the n-k+1th term. What is the modulus? It is the difference between the nth and n-kth terms.

Putting this together and assuming we know k (we don't in general, but we'll get to that) we have a recurrence of the form f(n+1)=f(n-k+1) + (f(n)-f(n-k)). So we need to find a recurrence (if one exists), check that it is of the correct form, and post-process if so.

Here is code to do all this. Note that it in effect solves for k.

findArithmeticSequences[ll : {_Integer ..}] := With[
  {rec = FindLinearRecurrence[ll]},
  {Take[ll, Length[rec] - 1], ll[[Length[rec]]]} /;
   ListQ[rec] &&
    (rec === {1, 1, -1} || MatchQ[rec, {1, 0 .., 1, -1}])
  ]

(Afficionados of pure functions might prefer the variant below. Failure cases are handled a bit differently, for no compelling reason.)

findArithmeticSequences2[ll : {_Integer ..}] :=
 If[ListQ[#] &&
     (# === {1, 1, -1} || MatchQ[#, {1, 0 .., 1, -1}]), {Take[ll, 
      Length[#] - 1], ll[[Length[#]]]}, $Failed] &[
  FindLinearRecurrence[ll]]

Tests:

In[115]:= findArithmeticSequences[ee]

Out[115]= {{0, 4}, 6}

In[116]:= findArithmeticSequences[ff]

Out[116]= {{0, 3, 7, 11}, 17}

Note that one can "almost" do such problems by polynomial factorization (if the input has no partial sequences at the end). For example, the polynomial

In[117]:= poly = Plus @@ (x^ee)

Out[117]= 1 + x^4 + x^6 + x^10 + x^12 + x^16 + x^18 + x^22

factors into

(1+x^4)*(1+x^6+x^12+x^18)

which contains the needed information in a way that is easy to see. Unfortunately for this particular purpose, Factor will factor beyond this point, and obscure the information in so doing.

I keep wondering if there might be a signal processing way to go about this sort of thing, e.g. via DFTs. But I've not come up with anything.

Daniel Lichtblau


Wow, thank you Daniel for this! It works nearly the way I want it to. Your method is just a bit "too restrictive". It doesn't return anything useful if 'FindLinearRecurrence' doesn't find any recurrence. I've modified your method a bit, so it suits my needs better. I hope you don't mind. Here's my code.

findArithmeticSequences[ll_List] := Module[{rec = FindLinearRecurrence[ll]}, If[! MatchQ[rec, {1, 0 ..., 1, -1}], Return[ll], Return[{ll[[Length[rec]]], Take[ll, Length[rec] - 1]}]; ]; ];

I had a feeling it'd have to involve recurrence, I just don't have enough experience with Mathematica to implement it. Thank you again for your time!


Mod is listable, and you can remove duplicate elements by DeleteDuplicates. So

DeleteDuplicates[Mod[{0, 4, 6, 10, 12, 16, 18, 22}, 6]]
(*
-> {0,4}
*)
0

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