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I spent the better part of a long car ride today thinking about the following problem:

Consider an alphabet of $n$ letters. How many $k$-letter words of this alphabet are there such that two certain letters (say A and B) are not next to each other? A word may contain the same letter multiple times.

My thoughts: There are $n^k$ words with length $k$ over the alphabet. We need to find and subtract away the number of words where A and B are next to each other. I modeled any such word as one with the string AB or BA at an index $i$:

$$x_1x_2x_3\cdots x_{i-1}ABx_{i+2}\cdots x_k$$

The $x_j$ here are all letters of the alphabet. We can think of the above word as an $(i-1)$-letter word, followed by AB, followed by a $(k-i-2)$-letter word. Thus, there are $n^i \cdot n^{k-i-2} = n^{k-2}$ words where AB or BA is at an arbitrary index $i$. Summing over all $i$, we get that there are $(k-2)n^{k-2}$ words containing AB and $2(k-2)n^{k-2}$ words containing AB or BA. Our final expression for the number of $k$-letter words such that A and B are not adjacent is

$$n^k - 2(k-2)n^{k-2}$$

There’s an issue though. I’m over-counting words that contain AB or BA by a lot. I ran a simulation, and at $n=k=10$, for example, the correct value is $8441614754$, and my answer is $8400000000$. By this method, any word that contains AB or BA more than once is counted more than once. Is there any way I can modify this approach to yield the correct result? If not, how should I proceed?

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3 Answers 3

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You can handle this with coupled recurrences. Let $A(k)$ be the number of words without $AB$ or $BA$ that end in a character other than $A$ or $B$. Let $B(k)$ be the number of words without $AB$ or $BA$ that end in $A$ or $B$. You can add any character to an $A(k)$ word, but only $n-1$ characters to a $B(k)$ word, one of which leaves it as a $B(k+1)$ word. The recurrences are $$A(k+1)=(n-2)A(k)+(n-2)B(k)\\ B(k+1)=2A(k)+B(k)\\ A(0)=1\\ B(0)=0$$ For small $k$ you can make a spreadsheet to do these. For large $k$ you can do the eigenvalue/eigenvector thing on the matrix $$\begin {pmatrix} n-2&n-2\\2&1 \end {pmatrix}$$ The leading eigenvalue is $\frac 12\left(\sqrt{n^2+2n-7}+n-1\right)\approx n$ when both $k$ and $n$ are large. You want $A(k)+B(k)$ I made a spreadsheet for $n=10$, shown below. In the line for $n=2$ the $98$ under $A+B$ shows there are $98$ two character strings that are not $AB$ or $BA$. As there are $10^2=100$ unrestricted strings and we rule out $2$, this is correct. The $18\ B$ strings have one of $9$ characters (not a $B$) then an $A$, or one of $9$ characters (not an $A$) then a $B$. enter image description here

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  • $\begingroup$ I'm a little confused as to how you came up with the recurrences. What do you mean when you say that you can add any character to an $A(k)$ word, but only $n-1$ characters to a $B(k)$ word? $\endgroup$
    – Crosby
    Dec 28, 2020 at 4:14
  • $\begingroup$ Because a $B(k)$ word ends in $A$ or $B$ there is one character you can't add without the word becoming unacceptable. Because the $A(k)$ words do not end with $A$ or $B$, no addition will result in an $AB$ or $BA$ $\endgroup$ Dec 28, 2020 at 4:19
  • $\begingroup$ That makes sense. Thanks for your answer! $\endgroup$
    – Crosby
    Dec 28, 2020 at 4:33
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Consider the following automaton that accepts the strings not containing AB or BA, enter image description here

From the DFA, using Chomsky-Schutzemberger method, we get, $$q_0 = 1+xq_1+xq_2+(n-2)xq_0$$ $$q_1 = 1+xq_1+(n-2)xq_0$$ $$q_2 = 1+xq_2+(n-2)xq_0$$ (We omit $q_3$ and $q_4$ because it is a dead state)

On solving the above equations (i.e. substituting $q_1$ and $q_2$) we get the following expression for $q_0$, $$q_0 = \frac{1+x}{1-(n-1)x-(n-2)x^{2}}$$ This is our generating function. The coefficient of $x^{k}$ gives the number of strings of length k not having AB or BA. For example, for $n = k = 10$, Wolfram Alpha gives the following Taylor Series,

enter image description here

And as you can see, the coefficient of $x^{10}$ is equal to the answer you have provided.

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This answer is based upon the Goulden-Jackson Cluster Method. We consider words of length $k\geq 0$ built from an alphabet $\mathcal{V}$ with $|\mathcal{V}|=n$ and the set $B=\{AB,BA\}$ of bad words, which are not allowed to be part of the words we are looking for.

We derive a generating function $A_n(z)$ with the coefficient of $z^k$ being the number of wanted words of length $k$. According to the paper (p.7) the generating function $A_n(z)$ is \begin{align*} A_n(z)=\frac{1}{1-dz-\text{weight}(\mathcal{C})} \end{align*} with $d=|\mathcal{V}|=n$, the size of the alphabet and $\mathcal{C}$ the weight-numerator with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[AB])+\text{weight}(\mathcal{C}[BA]) \end{align*}

We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[AB])&=-z^2-\text{weight}(\mathcal{C}[BA])z\\ \text{weight}(\mathcal{C}[BA])&=-z^2-\text{weight}(\mathcal{C}[AB])z\\ \end{align*} and get

\begin{align*} \text{weight}(\mathcal{C}[AB])=\text{weight}(\mathcal{C}[BA])=-\frac{z^2}{1+z} \end{align*}

It follows \begin{align*} \color{blue}{A_n(z)}&=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\\ &=\frac{1}{1-nz+2\frac{z^2}{1+z}}\\ &\,\,\color{blue}{=\frac{1+z}{1-(n-1)z-(n-2)z^2}}\tag{1} \end{align*}

Denoting with $[z^k]$ the coefficient of $z^k$ of a series, we obtain from (1) the number of valid words of length $k$ as \begin{align*} \color{blue}{[z^k]}\color{blue}{A_n(z)}&=[z^k]\frac{1+z}{1-(n-1)z-(n-2)z^2}\\ &=[z^k]\sum_{j=0}^\infty z^j\left((n-1)+(n-2)z\right)^j(1+z)\\ &=\sum_{j=0}^k[z^{k-j}]\left((n-1)+(n-2)z\right)^{j}(1+z)\\ &=\sum_{j=0}^k[z^j]\left((n-1)+(n-2)z\right)^{k-j}(1+z)\\ &\,\,\color{blue}{=\sum_{j=0}^k\binom{k-j}{j}(n-2)^j(n-1)^{k-2j}}\\ &\qquad\qquad\color{blue}{+\sum_{j=1}^k\binom{k-j}{j-1}(n-2)^{j-1}(n-1)^{k-2j+1}}\tag{2} \end{align*}

Plausibility check: Evaluating (2) at $n=k=10$ we obtain \begin{align*} [z^{10}]A_{10}(z)=8\ 441\ 614\ 754 \end{align*} in accordance with OPs calculation.

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