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Possible Duplicate:
Counting subsets containing three consecutive elements (previously Summation over large values of nCr)

Given an integer $N$, we have to count the number of possible binary strings(string is madeup of only $0$'s and $1$'s) of length $N$, and matches the regular expression pattern $[111]$. For example if $N$ is $4$, then $1110$ , $0111$ , $1111$ are the possibilities.

I have worked on it, and have got the following recurrence:

$\text{count}(a,N) = 4\text{count}(0,N-3) + 2\text{count}(1,N-3)+ \text{count}(2,N-3)$ ; if$(a == 0)$

$\text{count}(a,N) = 2\text{count}(0,N-2)+ \text{count}(1,N-1)$ ; if$(a == 1)$

$\text{count}(a,N) = \text{count}(0,N-1)$ ; if$(a == 2)$

Answer is $[(2^N) - \text{count}(0,N)]$. By using memorization, we can compute the the $\text{count}(0,N)$ in $O(N)$
but is there any better algorithm other than $O(N)$

Community
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Bhanu Kishore
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    Probably not: see this answer and the OEIS references cited there. – Brian M. Scott Sep 07 '12 at 20:51
  • @BrianM.Scott: do you agree that is a duplicate? – Ross Millikan Sep 07 '12 at 21:17
  • @Ross: I’m undecided: it actually does ask a slightly different (and more specific) question, but I doubt that we can come up with a better answer than we had before. – Brian M. Scott Sep 07 '12 at 21:20
  • "memorization"? – Gerry Myerson Sep 07 '12 at 23:12
  • @Gerry: Are you asking what it means, or do you just want it to be spelled without the "r"? – hmakholm left over Monica Sep 08 '12 at 01:32
  • @Henning, a bit of both, actually. It's only recently I've come across the word, "memoization", and at first I thought it was a typo for "memorization". I still don't know quite what it means (so I should just go look it up somewhere, I know), and I think it's funny that here we may have "memorization" as a typo for "memoization". – Gerry Myerson Sep 08 '12 at 03:10
  • See this and this to solve the problem in $O(\lg n)$ time. – Saurabh Sep 08 '12 at 05:02
  • This question is equivalent to the newer question http://math.stackexchange.com/questions/192693/ (now closed as duplicate of the older http://math.stackexchange.com/questions/191702) and a special case of http://math.stackexchange.com/questions/192658. I find it very suspect that essentially the same question is asked, in different guises, four times in a couple of days. Voting to close. – Marc van Leeuwen Sep 09 '12 at 12:24

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Let $f(N)$ be the number of binary strings of length $N$ that do not contain "111" as a substring. (So you are interested in $2^N - f(N)$.) Every such binary string ends in exactly one of the strings "0", "01", or "011". This gives rise to the recurrence $$ f(N) = f(N-1) + f(N-2) + f(N-3), $$ the "tribonacci recurrence". The inital values are $f(1)=1$, $f(2)=2$, and $f(3)=4$. Therefore a closed form can be given that is similar to the one for the tribonacci numbers, but with different constants in front of the exponential terms.

In fact, now that I look at the tribonacci numbers, I see that this exact problem has already been linked to them (see the comment by Emeric Deutsch, Apr 27 2006).

Greg Martin
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