Linear Recurrences

How often has it happened to you in a programming contest (or elsewhere) that you thought it was impossible to solve it faster than O(N) and yet the limits imposed suggest that it has to be done faster. Well, if not all, atleast a majority of them have a solution based on the idea of linear recurrences. In this blog post, I intend to help you out on this !!

In this post, we are going to do a – Solve and Learn strategy ; You will be given a question and I will show you how to apply  the concepts on them.

TYPE 1 :: The Simplest :

If a post mentions recurrences, then it has to mention Fibonacci (Gosh, if only I had a penny for every mention of Fibo in tutorials. )

The recurrence is of type : F(n) = F(n-1) + F(n-2).

I am pretty sure you know to code the linear version of it which runs in O(N) but can you do it in O(log N) ? If you throw google to good use, you will come up with a solution which says there is a Matrix M which when raised to power N, will give you the N-th fibonacci number. And since you can always exponentiate in logN time, you have your answer. But to those, who wondered if this Matrix is magical- read on!

Firstly the answer- No; Its not magical. How. Lets do a little Algebra (yumm… My favourite! )
F(n)=F(n-1)+F(n-2)\\ \\ F(n+1) =F(n)+F(n-1)\\ \\ F(n+2)=F(n+1)+F(n)

Obviously enough, the value of N-th term, depends on two previous terms (or states). This implies that all values depend on just the first two states in the sequence. As you can see here –

\begin{pmatrix}F(n+2)\\ F(n+1)\end{pmatrix}=\begin{pmatrix}1&1\\ 1&0\\ \end{pmatrix}\times\begin{pmatrix}F(n+1)\\ F(n)\end{pmatrix}\\ \\ and\\ \\ \begin{pmatrix}F(n+1)\\ F(n)\end{pmatrix}=\begin{pmatrix}1&1\\ 1&0\\ \end{pmatrix} \times \begin{pmatrix}F(n)\\ F(n-1)\end{pmatrix} \\ \\ Hence \\ \\ \begin{pmatrix}F(n+2)\\ F(n+1)\end{pmatrix}=\begin{pmatrix}1&1\\ 1&0\\ \end{pmatrix} ^2 \times \begin{pmatrix}F(n)\\ F(n-1)\end{pmatrix} \\ \\ \begin{pmatrix}F(n+2)\\ F(n+1)\end{pmatrix}=\begin{pmatrix}1&1\\ 1&0\\ \end{pmatrix}^3 \times \begin{pmatrix}F(n-1)\\ F(n-2)\end{pmatrix}

Hence in General, we may write ::
\begin{pmatrix}F(n)\\ F(n-1)\end{pmatrix}=\begin{pmatrix}1&1\\ 1&0\\ \end{pmatrix}^{n-1} \times \begin{pmatrix}1\\ 0\end{pmatrix}

I hope that has helped you in understanding how to frame such equations and solving it with a matrix.

TYPE 2 : Simplest ++

Now that we have a basic understanding. Try the following recurrence :

F(n) = F(n-1) + F(n-2) + F(n-3).

It is the same as the previous recurrence but with an additional state. I won’t go on explaining the hows (again!). I am going to share the solution.
\begin{pmatrix}F(n)\\ F(n-1)\\ F(n-2) \end{pmatrix}=\begin{pmatrix}1&1&1\\ 1&0&0\\ 0&1&0 \end{pmatrix}^{n-2} \times \begin{pmatrix}2\\ 1\\ 1\end{pmatrix}

TYPE 3: Simplest << 1

Consider the following scenario ::

G(n) = a . G(n-1) + b . G(n-2) + c . H(n)\\ \\ and \\ \\ H(n)= d . H(n-1) + e . H(n-2)

This one is a lot trickier. First thing to notice is that we will need 4 states in a matrix to fully define the next state. The reason for using 4 and not 3 is that H(n) depends on 2 states and thus we need 2 states (and not just 1) to represent it.

If you carefully write down the LHS matrix and the RHS matrix, then we can frame the solution as . . .

\begin{pmatrix}G(n)\\ G(n-1)\\ H(n+1)\\ H(n) \end{pmatrix}=\begin{pmatrix}a&b&c&0\\ 1&0&0&0\\ 0&0&d&e\\ 0&0&1&0 \end{pmatrix}^{n-1} \times \begin{pmatrix}G(1)\\ G(0)\\ H(2)\\ H(1)\end{pmatrix}

TYPE 4 : Ohhh !

The final hurdle can come in the name of a constant. If we add a constant C to the above recurrence we get –

G(n) = a . G(n-1) + b . G(n-2) + c . H(n) + C\\ \\ and \\ \\ H(n)= d . H(n-1) + e . H(n-2)

But to tell you the truth, its not that difficult if your concepts are clean. Now there is another additional state to hold the information about C. The solution will look like –

\begin{pmatrix}G(n)\\ G(n-1)\\ H(n+1)\\ H(n)\\ C \end{pmatrix}=\begin{pmatrix}a&b&c&0&1\\ 1&0&0&0&0\\ 0&0&d&e&0\\ 0&0&1&0&0\\ 0&0&0&0&1 \end{pmatrix}^{n-1} \times \begin{pmatrix}G(1)\\ G(0)\\ H(2)\\ H(1)\\ C\end{pmatrix}

I hope this post lived up to your expectations and I hope it was worth the wait :P. Please feel free to post comments/corrections/improvements to this post to make it really useful.


6 responses to “Linear Recurrences

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