Upper bound of of fib(n+2)

I have a homework problem that is perplexing me because the math is beyond what I have done, although we were told that it was unnecessary to solve this mathematically. Just provide a close upper bound and justify it.

Let $$f(n) = |\{w ∈ \{a, b\}^n : aa \notin w \}|.$$ Provide an asymptotic upper bound on $f$ as $n\to\infty$.

So far: $$\begin{array}{c|c|c} n & \text{strings} & \text{compared to } 2^n \\\hline 1 & 2 & 2^n - 0 \\ 2 & 3 & 2^n - 1 \\ 3 & 5 & 2^n - 3 \\ 4 & 8 & 2^n - 8 \\ 5 & 13 & 2^n - 18 \\ 6 & 21 & 2^n - 43 \\ \end{array}$$

The math that will give me an exact bound is beyond me. Obviously $O(2^n)$ is an upper bound, though it is not particularly tight.

Any suggestions on what I should try?

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• Many people would consider fib (n+1) as a perfectly fine expression for an upper bound. Even better as it is exact :-) – gnasher729 Nov 18 '16 at 11:37

So I'm not completely sure, but I think you're asking to count the number of strings of size $n$ (over the alphabet $\{a, b\}$) where the factor/substring $aa$ does not appear right?

In this case, there are a few combinatorial approaches that you can take. Both Yuval and ADG have given simpler and more intuitive arguments, so I definitely suggest checking their answers out! Here's one of my favorites, it's a little strange, but it's a very general (and kind of a fun) approach.

Let's start with a simpler language, that of words which starts and ends with $b$ (also with no substrings of $aa$). We can look at an admissible string (e.g. $bbb\textbf{a}b\textbf{a}bbbb$) as a list of sequences of $b$s separated by singular $a$s. This gives the construction: $$w = (b^+a)^*b^+$$ Now, how do we count sentences that belongs to this language?

Let's imagine that we're expanding these expressions out. What does $e^*$ denote? Well, it's simply $$e^* = \epsilon \mid e \mid ee \mid eee \mid eeee \mid \dots$$ Now, this will make very little sense, but let's imagine that $e$ is a variable over some numerical field. In particular, we'll treat $\epsilon \to 1$, $a \mid b \to a + b$, and $abc \to a \times b \times c$. This then says that $$e^* \to 1 + e + ee + eee + \dots$$ Let's try to see the motivation behind this strange interpretation. This is almost a bijective transformation. In particular, we want to preserve the count of each $e^n$ word, which, as you can readily see, we do. However, there's one crucial difference between the string expressions and numerical expressions: the multiplication (concatenation in strings, $\times$ in numerical expressions) is now commutative! Intuitively, commutativity lets us treat all permutations of the same word as the same; that is, we don't disambiguate between the expression $bbbab$ and $bbabb$; they both represent a string with 4 $b$s and one $a$. Therefore, this transformation allows us to preserve the count of each word of a certain number of $a$s and $b$s, but it now allows us to turn a blind eye on the superfluous details that we don't care about.

If you go back to precalculus, you might recognize this series as $\frac{1}{1 - e}$. I know that it doesn't make any sense to rewrite this regular expression as a numerically-valued function, but just bare with me for a moment.

Similarly, $e^+ = ee^* \to \frac{e}{1 - e}$. Which means that we can translate $w$ into $$w \to \frac{1}{1 - (\frac{b}{1 - b} \times a)} \times \frac{b}{1 - b}$$

In turn, we can simplify this down to $$w(a, b) = b \times \frac{1}{1 - (b + ba)}$$

This tells us that the language $w$ is isomorphic to the language $b(b \mid ab)^*$ (whose direct translation is already $\frac{b}{1 - b - ba}$) without ever having to resort to any language-theoretic tools! This is one of the power of treating these series as closed-form functions: we can perform simplifications on them that are nearly impossible to perform otherwise, hence reducing it down to a simpler problem.

Now, if you still recall any of your calculus course, you'll recall that certain types of functions (being well-behaving enough) admits these series representations known as Taylor expansions. Don't worry, we won't really have to worry about those pesky Calc 1 problem-sets; I'm just merely pointing out that these functions can be represented as the sum \begin{align*} w(a, b) &= \sum_{i,j} w_{ij} a^i b^j \end{align*} so that $w_{ij}$ gives the number of words that satisfies $w$ such that it has exactly $i$ occurrences of $a$ and $j$ occurrences of $b$. However, we don't particularly care about whether something is an $a$ or a $b$. Rather, we just care about the total number of characters in the string. To turn a "blind eye" between $a$ and $b$, we can just (literally) treat them the same, e.g. let $z = a = b$ and get $$w(z) = w(z, z) = \frac{z}{1 - z - z^2} = \sum_k w_k z^k$$

where $w_k$ counts the number of satisfiable words of length $k$.

Now, all that's left to do is to find $w_k$. The usual combinatorial approach here would be to decompose this rational function into its partial-fraction: that is, given the denominator $1 - z - z^2 = (z - \phi)(z - \psi)$, we can rewrite $\frac{z}{(z - \phi)(z - \psi)} = \frac{A}{z - \phi} + \frac{B}{z - \psi}$ (There's a bit of algebra involved here, but this is a universal property of rational (one polynomial dividing another) functions). To solve this, you can refactor $$\frac{A}{z - \phi} + \frac{B}{z - \psi} = \frac{z}{(z - \phi)(z - \psi)}$$ which generates the constraints $A + B = 1, A\psi + B\phi = 0$. Irregardless of what $A$ and $B$ are, recall that $\frac{1}{1 - x} = 1 + x + x^2 + \dots$, well, we can rearrange \begin{align*} w(z) &= \frac{-A}{\phi - z} + \frac{-B}{\psi - z} \\ &= (-A\phi)\frac{1}{1 - \frac{z}{\phi}} + (-B\psi) \frac{1}{1 - \frac{z}{\psi}} \\ &= (-A\phi)\left(1 + \phi^{-1} z + \phi^{-2} z^2 + \dots\right) + (-B\psi)\left(1 + \psi^{-1} z + \psi^{-2} z^2 + \dots\right) \end{align*} therefore $$w_k = (-A\phi)\phi^{-k} + (-B\psi)\psi^{-k}$$ Here, $\phi$ is the golden ratio $\frac{1 + \sqrt{5}}{2}$ and $\psi = -\phi^{-1}$ is its conjugate. We then have an easy description of the asymptotic behavior of the $w$ language: it runs in $\Theta(\phi^n)$. In fact, if you expand everything out, you'll find that $$w_k = \frac{\phi^{k} - \psi^{k}}{\sqrt{5}} = \left\lceil \frac{\phi^{k}}{\sqrt{5}} \right\rceil$$ There's also an intricate connection to another common combinatorial class. This is just the Fibonacci numbers!

Now, suppose you have $w_k$, which counts the number of strings of size $k$ that starts and ends with $k$ (and also contains no $aa$ substrings), how can we build a string that can start or end with an $a$? Well, it's simple: an admissible string is either in $w$ (starts and ends with $b$), or it is $a w$ (starts with $a$), or it is $w a$ (ends with $a$), or it is $a w a$ (starts and ends with $a$). Therefore: $$f(n) = w_n + w_{n-2} + 2 * w_{n - 1}$$ Recall that $w_n$ is the fibonacci sequence, so $w_{n-1} + w_{n-2} = w_n$, which means that \begin{align*} f(n) &= (w_n + w_{n-1}) + (w_{n-2} + w_{n-1}) \\ &= w_{n+1} + w_{n} \\ &= w_{n+2} \end{align*} Therefore, $f(n) = \text{fib}(n + 2) = \left\lceil \frac{\phi^{n+2}}{\sqrt{5}} \right\rceil$

Now you probably don't have to do this analysis, but just by having the insight that this sequence is a shifted Fibonacci sequence ought to give you an idea of some other combinatorial interpretations that you can try.

Lee Gao's answer is excellent. Here is a different account. Consider the following automaton:

This is an unambiguous finite automaton (UFA) without $\epsilon$ transitions: an NFA such that each word has exactly one accepting path. The number of words of length $n$ is thus the number of paths of length $n$ from the starting state to an accepting state (since there are no $\epsilon$ transitions).

We can count the number of paths in a graph using linear algebra. Let $M$ be the transition matrix of the automaton: $M(q_i,q_j)$ is the number of arrows from $q_j$ to $q_i$ (each arrow is associated with a single symbol). Then $$M^2(q_i,q_j) = \sum_k M(q_i,q_k) M(q_k,q_j),$$ which is exactly the number of paths of length 2 from $q_j$ to $q_i$. Similarly, $M^n(q_i,q_j)$ is the number of paths of length $n$ from $q_j$ to $q_i$. In our case, we want to count the number of paths of length $n$ from $q_0$ to $\{q_0,q_1\}$, and so $$f(n) = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} \, .$$ The standard way to calculate such an expression is by diagonalizing $M$ (or, more generally, by computing the Jordan form of $M$). The eigenvalues of $M$ are easily calculated to be $\frac{1 \pm \sqrt{5}}{2}$, and so for some matrix $P$, \begin{align*} f(n) &= \begin{pmatrix} 1 & 1 \end{pmatrix} P \begin{pmatrix} \left(\frac{1+\sqrt{5}}{2}\right)^n & 0 \\ 0 & \left(\frac{1-\sqrt{5}}{2}\right)^n \end{pmatrix} P^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ &= A \left(\frac{1+\sqrt{5}}{2}\right)^n + B \left(\frac{1-\sqrt{5}}{2}\right)^n, \end{align*} for some coefficients $A,B$. To determine $A,B$ we can either diagnoalize $M$ explicitly, or just set up a linear system using known values of $f$. Since $f(0) = 1$ and $f(1) = 2$, the latter approach shows that $$A + B = 1 \\ \frac{1+\sqrt{5}}{2} A + \frac{1-\sqrt{5}}{2} B = 2$$ Solving this system (e.g. using Gaussian elimination), we find out that $A = \frac{5+3\sqrt{5}}{10}$ and $B = \frac{5-3\sqrt{5}}{10}$. Therefore \begin{align*} f(n) &= \frac{5+3\sqrt{5}}{10} \left(\frac{1+\sqrt{5}}{2}\right)^n + \frac{5-3\sqrt{5}}{10} \left(\frac{1-\sqrt{5}}{2}\right)^n \\ &= \Theta(\lambda_{\max}^n), \text{ where } \lambda_{\max} = \frac{1+\sqrt{5}}{2}. \end{align*} The same approach works for every regular language.

• This is a fun approach! I always love seeing linear algebraic reductions for path related problems :) – Lee Nov 16 '16 at 17:42

@Lee Gao's is too complex (I haven't even read the whole thing), here is a simplistic approach:

Let f(n) be all desired strings out of which let a(n) be strings that end at a and b(n) be strings that end at b.

Now for every string that ends at b we can directly add a to get ba in ending and a valid string: $$a(n)=b(n-1)\tag{1}$$ Note than we cannot add a to end of strings that end at a otherwise we will have aa at the end.

We can add b to any string: $$b(n)=a(n-1)+b(n-1)\tag{2}$$

Now $n\mapsto n-1$ in $(1)$ and substitue in $(2)$: $$b(n)=b(n-2)+b(n-1)$$ So b(n) is fib(n) and since a(n) is b(n-1) hence a(n) is fib(n-1). Now f(n) is: $$f(n)=a(n)+b(n)={\rm fib}(n)+{\rm fib}(n-1)={\rm fib}(n+1)$$ As fib(n) is $(\varphi^n-\varphi^{-n})/\sqrt{5}$, hence f(n) is ${\rm O}(\varphi^n)$, $\varphi=\frac{1+\sqrt{5}}{2} \approx 1.618$. (Taking $\varphi/\sqrt5$ as constant and neglecting $\varphi^{-n}$ for large $n$ to get an asymptote)

Note: fib(0)=0, fib(1)=1.

• Your upper bound isn't very good, however: every language over $\{a,b\}$ has at most $2^n$ words of length $n$! – Yuval Filmus Nov 16 '16 at 7:32
• @YuvalFilmus what's the problem, you want a closer bound then use ${\rm fib}(n)=(\phi^n-\phi^{-n})/\sqrt{5}$ – RE60K Nov 16 '16 at 7:50
• This is great! The pairs lock-step inductive constructions of $a$ and $b$ was what I was thinking of as well. – Lee Nov 16 '16 at 17:37