Find f(n): nth Fibonacci number. The problem is quite easy when n is relatively small. We can use simple recursion, f(n) = f(n-1) + f(n-2), or we can use dynamic programming approach to avoid the calculation of same function over and over again. But what will you do if the problem says, Given 0 < n < 10⁹, find f(n) mod 999983? Dynamic programming will fail, so how do we tackle this problem?

First let’s see how matrix exponentiation can help to represent recursive relation.



At first we need a recursive relation and we want to find a matrix M which can lead us to the desired state from a set of already known states. Let’s assume that, we know the k states of a given recurrence relation and we want to find the (k+1)th state. Let M be a k X k matrix, and we build a matrix A:[k X 1] from the known states of the recurrence relation, now we want to get a matrix B:[k X 1] which will represent the set of next states, i. e. M X A = B as shown below:

|  f(n)  |     | f(n+1) |
| f(n-1) |     |  f(n)  |
M X | f(n-2) |  =  | f(n-1) |
| ...... |     | ...... |
| f(n-k) |     |f(n-k+1)|

So, if we can design M accordingly, our job will be done! The matrix will then be used to represent the recurrence relation.

Type 1: Let’s start with the simplest one, f(n) = f(n-1) + f(n-2) We get, f(n+1) = f(n) + f(n-1). Let’s assume, we know f(n) and f(n-1); We want to find out f(n+1). From the situation stated above, matrix A and matrix B can be formed as shown below:

Matrix A          Matrix B

|  f(n)  |        | f(n+1) |
| f(n-1) |        |  f(n)  |

[Note: Matrix A will be always designed in such a way that, every state on which f(n+1) depends, will be present] Now, we need to design a 2X2 matrix M such that, it satisfies M X A = B as stated above. The first element of B is f(n+1) which is actually f(n) + f(n-1). To get this, from matrix A, we need, 1 X f(n) and 1 X f(n-1). So the first row of M will be [1 1].

| 1   1 |  X  |  f(n)  |  =  | f(n+1) |
| ----- |     | f(n-1) |     | ------ |

[Note: —– means we are not concerned about this value.] Similarly, 2nd item of B is f(n) which can be got by simply taking 1 X f(n) from A, so the 2nd row of M is [1 0].

| ----- |  X  |  f(n)  |  =  | ------ |
| 1   0 |     | f(n-1) |     |  f(n)  |

Then we get our desired 2 X 2 matrix M.

| 1   1 |  X  |  f(n)  |  =  | f(n+1) |
| 1   0 |     | f(n-1) |     |  f(n)  |

These matrices are simply derived using matrix multiplication.

Type 2:

Let’s make it a little complex: find f(n) = a X f(n-1) + b X f(n-2), where a and b are constants. This tells us, f(n+1) = a X f(n) + b X f(n-1). By this far, this should be clear that the dimension of the matrices will be equal to the number of dependencies, i.e. in this particular example, again 2. So for A and B, we can build two matrices of size 2 X 1: