# Number of binary strings without adjacent char more than $n$

$$N$$ is a natural number, let $$0\leq\alpha\leq 1$$ a fixed real number, I want to estimate the number of length-N binary strings which statisfy: the same character (i.e $$0$$ or $$1$$) could not occur repeatedly and subsequently more than $$\alpha N$$ times.

For example: substrings like $$000....000$$ (more than $$\alpha N$$ many $$0$$) could not exist.

I only need to know how the number $$I(N)$$ of these strings grows when $$N\to \infty$$.

• What is $I(N)$? – orlp Jan 11 at 23:10
• A closed form formula can be found here; maybe you can find out asymptotics starting from there. – BillyJoe Jan 12 at 10:47
• @orlp It is the number of the strings satisfying the condition above – Tibeku Jan 12 at 13:05

The answer is close to $$2^n$$. Here's an easy computation which suggests something like this: for a random binary string, the expected number of times $$k$$ consecutive zeroes or ones appear is $$\frac{2(n-k+1)}{2^k}$$. Already from this simple calculation we expect that a "typical" binary string has runs of length about $$\log_2 n$$. It turns out to be unlikely for a binary string to have runs much longer than this, which can be proven as follows.

The number $$I(n, k)$$ of length-$$n$$ binary strings not containing $$k$$ consecutive $$0$$s or $$1$$s can be computed as follows. We can encode such a string by recording first its initial digit and second a word of length $$n-1$$ using letters $$D$$ and $$S$$ (for "different" and "same") which record whether each successive digit is the same or different than the previous one. This is readily seen to be a bijection, and moreover the original string does not contain $$k$$ consecutive $$0$$s or $$1$$s iff the new string does not contain $$k-1$$ consecutive $$S$$'es. The first digit is irrelevant. It follows that

$$I(n, k) = 2 J(n-1, k-1)$$

where $$J(n-1, k-1)$$ is the number of length-$$(n-1)$$ strings of $$D$$s and $$S$$'es not containing $$k-1$$ consecutive $$S$$'es. The behavior of this sequence is thoroughly analyzed in Example V.4 of Flajolet and Sedgewick's Analytic Combinatorics on page 308, which contains in particular the following estimate: if $$L$$ denotes the random variable given by the longest run of $$S$$'es in a random string of length length $$n$$, then

$$\mathbb{P}(L < \lfloor \log_2 n \rfloor + h) = \exp(-f(n) 2^{-h-1}) + O \left( \frac{\log n}{\sqrt{n}} \right)$$

where $$f(n) = 2^{ \{ \log_2 n \} }$$ and $$\{ x \}$$ denotes the fractional part; in particular $$1 \le f(n) \le 2$$. Replacing $$n$$ with $$n-1$$ and setting $$h = k - \lfloor \log_2 n \rfloor$$ gives

$$\frac{I(n, k)}{2^n} = \frac{J(n-1, k-1)}{2^{n-1}} = \exp \left( -f(n-1) 2^{-k - \lfloor \log_2 n \rfloor - 1} \right) + O \left( \frac{\log n}{\sqrt{n}} \right)$$

from which we conclude that for $$\alpha \in (0, 1]$$, $$I(n, \alpha n)$$ is close to $$2^n$$ as $$n \to \infty$$; their ratio is asymptotically $$1 + O \left( \frac{\log n}{\sqrt{n}} \right)$$ (for $$k$$ this large the exponential term is $$1$$ plus a term so small that the error term dominates), which in particular approaches $$1$$.

The weaker conclusion that $$\lim_{n \to \infty} \frac{I(n, \alpha n)}{2^n} = 1$$ can probably be proven by estimating the variance of the number of times $$k$$ consecutive $$S$$'es appear and then applying Chebyshev's inequality. Flajolet and Sedgewick give an asymptotic expansion for the expected length $$\mathbb{E}(L)$$ of the longest run (which is $$\log_2 n + O(1)$$ but they give a more precise asymptotic than this) and mention that the variance is $$O(1)$$.

Partial answer: You can make (at least) somewhere around $$2^{N-2/\alpha}$$ different valid bitstrings (there is some rounding going on, but that's roughly the result you get). You do this by inserting $$01$$ every $$\alpha N$$ bits to break any potential consecutive substrings, and then choosing the remaining bits freely.

I don't know whether I am missing a significant amount of valid strings this way, though. It depends on what "significant" means.