Find subsequence A in sequence B A007908 - OEIS

Question

I have sequence A as A007908 - OEIS and pattern B which is a sub-sequence of that sequence A. I need to find index of first occurrence of the given pattern. For example pattern ..1010...

I've looked at this. Could anyone explain how to solve the problem?

Finding a formula for a pattern

Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! — Michael E2, Oct 02 '15 at 19:06
Your question could be a question about the software system Mathematica, but it seems from the link like it might belong on math.SE instead. Could you say more? (See http://mathematica.stackexchange.com/help/on-topic) — Michael E2, Oct 02 '15 at 19:07
Can you construct a simple example that you can include in the question instead of making folks go to some external site? — george2079, Oct 02 '15 at 19:08
possible duplicate: http://mathematica.stackexchange.com/questions/941/finding-a-subsequence-in-a-list ( Unless the specific input sequence makes a difference ) — george2079, Oct 02 '15 at 19:17
In fact, that sequence has such a simple description, you ought to include its definition, maybe even instead of the link. — Michael E2, Oct 02 '15 at 19:19
@george2079 I guess one difference between the questions is that here, the list is indefinitely long. While the pattern n = B would certainly occur by the n-th term, there may be efficient ways of finding the pattern earlier. (Or not. I don't know anything about this problem.) — Michael E2, Oct 02 '15 at 19:27
Do you mean the mathematical definition of sub-sequence? e.g., 2,4,6 is a sub-sequence of 1,2,3,4,5,6,7, or do you mean some consecutive sequence of members of a larger sequence that matches a pattern? These are very different, the latter trivial to do for your case. — ciao, Oct 03 '15 at 07:37
@ciao, funny nobody pointed out this issue on that other question (and maybe should). Having looked up the mathematical definition of subsequence that seems even more trivial as you just search ahead digit by digit. — george2079, Oct 03 '15 at 11:21

score 5 · Answer 1 · edited Apr 13 '17 at 12:55

Using the findSubsequence function from the linked answer: https://mathematica.stackexchange.com/a/942/2079

 findSubsequence[list_, {ss__}] := 
      ReplaceList[list, {pre___, ss, ___} :> Length[{pre}] + 1]

suppose we have b:

 b = {8, 4, 9, 1, 2}

In principle you search the full sequence:

 findSubsequence[Flatten[IntegerDigits[Range@#] & /@ Range[FromDigits[b]] ,b] 
      (*huge, dont do it*)

Find the first subsequence long enough to contain b:

 i = 1; While[ Length@Flatten[IntegerDigits[Range@i] ] < Length@b, 
    i++]; i

5

check if b occurs early..

 findSubsequence[Flatten[IntegerDigits[Range@#] & /@ Range[i]] , b]

{}

beyond here we need only look in adjacent pairs:

 While[
   findSubsequence[ 
     Flatten[Join[IntegerDigits[Range[i]], 
        IntegerDigits[Range[i + 1]]]], b] == {}, ++i]; i

49

now a manageable* full search up to that i:

 findSubsequence[Flatten[IntegerDigits[Range@#] & /@ Range[i + 1]] , b]

{2043}

*manageable because I picked a sequence I knew occurred pretty early..You can readily eliminate that last search by keeping a cumulative tally of all the sub-sequence lengths.

Edit: a little cleaner version tracking the lengths...

 b = {8, 4, 9, 1, 2}
 i = 1; While[ Length@Flatten[IntegerDigits[Range@i] ] < Length@b, 
  i++]; i
 early = Flatten[IntegerDigits[Range@#] & /@ Range[i]];
 findSubsequence[early , b]
 res = NestWhile[  {
     z = Flatten@IntegerDigits[Range[#[[2]] + 1]],
     #[[2]] + 1, #[[3]] + Length@z } & , 
     {Flatten@IntegerDigits[Range[i]], i , Length@early} ,
      findSubsequence[ Flatten[Join[#1[[1]], #2[[1]]]], b] == {} &, 
           2 ];
 lastseq = Flatten@IntegerDigits[Range[res[[2]] - 1]]~Join~
                Flatten@IntegerDigits[Range[res[[2]]]];
 res[[3]] - Length@lastseq + First@findSubsequence[ lastseq, b]

2043

Pending clarification from the OP, I don't think this answers the question, unless they're abusing terminology in subsequence... — ciao, Oct 03 '15 at 07:36

Find subsequence A in sequence B A007908 - OEIS

1 Answers1