Can I solve this recurrence using the Fast Fibonacci transform?

Question

Write a program to calculate the total number of strings that are made of exactly N characters. None of the strings can have "13" as a substring. The strings may contain any integer from "0-9", repeated any number of times.

From above question I derived a recursive equation which gives total "13" s as follow:

$$F_{n} = 10F_{n-1} + 10^{n-2} - F_{n-2}$$

I am trying to solve the problem using Fast Fibonacci Transform with O(logn) time complexity as described in this link.

Taking reference to this post I tried to convert the obtained recursive equation into matrix recursive form:

I need to find A such that:

$$\begin{bmatrix} F_n \\\ F_{n-1} \end{bmatrix} = A \begin{bmatrix} F_{n-1} \\\ F_{n-2} \end{bmatrix}$$

But because of the presence of 10^n-2 I am not getting the constant.

My $A$ looks like the following:

$$ A = \begin{bmatrix} 10 & -10^{n-2} \\\ 1 & 0 \end{bmatrix}$$

Thus matrix $A$ is not constant.

What should I do in this case? Please shed some light

Answer 1

One solution is to add a dimension, and write $$ \pmatrix{F_n\\F_{n-1}\\1} = \pmatrix {10 & -1 & 10^{n-2} \\ 1 & 0 & 0 \\ 0 & 0 & 1 }\pmatrix{F_{n-1}\\F_{n-2}\\1} $$ Now at least you have a matrix multiplication. But as you'll surely note, that $10^{n-2}$ term isn't a constant. But you can fix that with $$ \pmatrix{F_n\\F_{n-1}\\10^{n-1}} = \pmatrix {10 & -1 & 10 \\ 1 & 0 & 0 \\ 0 & 0 & 10 }\pmatrix{F_{n-1}\\F_{n-2}\\10^{n-2}} $$

I can't say whether this'll help you or not, but it's at least a way to express the recurrence as a matrix multiplication.

Answer 2

The removal of the exponential term may be done by finding a particular solution to the recurrence. Letting $F_n=C\times10^n$, one has

$$C\times10^n=(C+0.01-0.01C)10^n$$

$$C=1.01C+0.01$$

$$C=-1$$

and thus we may consider $F_n=G_n-10^n$ to get

$$G_n=10G_{n-1}-G_{n-2}$$

$$\begin{bmatrix}G_n\\G_{n-1}\end{bmatrix}=\begin{bmatrix}10&-1\\1&0\end{bmatrix}\begin{bmatrix}G_{n-1}\\G_{n-2}\end{bmatrix}$$

$$\begin{bmatrix}G_{n+1}\\G_n\end{bmatrix}=\begin{bmatrix}10&-1\\1&0\end{bmatrix}^n\begin{bmatrix}G_1\\G_0\end{bmatrix}$$

and in terms of the original sequence,

$$\begin{bmatrix}F_{n+1}\\F_n\end{bmatrix}=\begin{bmatrix}10&-1\\1&0\end{bmatrix}^n\begin{bmatrix}F_1+10\\F_0+1\end{bmatrix}-10^n\begin{bmatrix}10\\1\end{bmatrix}$$