# Matrix Exponentiation
# Matrix Exponentiation to Solve Example Problems
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.
Prerequisites:
- Given two matrices, know how to find their product. Further, given the product matrix of two matrices, and one of them, know how to find the other matrix.
- Given a matrix of size d X d, know how to find its nth power in O(d3log(n)).
Patterns:
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:
Matrix A Matrix B
| f(n) | | f(n+1) |
| f(n-1) | | f(n) |
Now for f(n+1) = a X f(n) + b X f(n-1), we need [a, b] in the first row of objective matrix M. And for the 2nd item in B, i.e. f(n) we already have that in matrix A, so we just take that, which leads, the 2nd row of the matrix M to [1 0]. This time we get:
| a b | X | f(n) | = | f(n+1) |
| 1 0 | | f(n-1) | | f(n) |
Pretty simple, eh?
Type 3:
If you've survived through to this stage, you've grown much older, now let's face a bit complex relation: find f(n) = a X f(n-1) + c X f(n-3)?
Ooops! A few minutes ago, all we saw were contiguous states, but here, the state f(n-2) is missing. Now?
Actually this is not a problem anymore, we can convert the relation as follows: f(n) = a X f(n-1) + 0 X f(n-2) + c X f(n-3), deducing f(n+1) = a X f(n) + 0 X f(n-1) + c X f(n-2). Now, we see that, this is actually a form described in Type 2. So here the objective matrix M will be 3 X 3, and the elements are:
| a 0 c | | f(n) | | f(n+1) |
| 1 0 0 | X | f(n-1) | = | f(n) |
| 0 1 0 | | f(n-2) | | f(n-1) |
These are calculated in the same way as type 2, if you find it difficult, try it on pen and paper.
Type 4:
Life is getting complex as hell, and Mr, Problem now asks you to find f(n) = f(n-1) + f(n-2) + c where c is any constant.
Now this is a new one and all we have seen in past, after the multiplication, each state in A transforms to its next state in B.
f(n) = f(n-1) + f(n-2) + c
f(n+1) = f(n) + f(n-1) + c
f(n+2) = f(n+1) + f(n) + c
.................... so on
So , normally we can't get it through previous fashion, but how about we add c as a state:
| f(n) | | f(n+1) |
M X | f(n-1) | = | f(n) |
| c | | c |
Now, its not much hard to design M. Here's how its done, but don't forget to verify:
| 1 1 1 | | f(n) | | f(n+1) |
| 1 0 0 | X | f(n-1) | = | f(n) |
| 0 0 1 | | c | | c |
Type 5:
Let's put it altogether: find f(n) = a X f(n-1) + c X f(n-3) + d X f(n-4) + e.
Let's leave it as an exercise for you. First try to find out the states and matrix M. And check if it matches with your solution. Also find matrix A and B.
| a 0 c d 1 |
| 1 0 0 0 0 |
| 0 1 0 0 0 |
| 0 0 1 0 0 |
| 0 0 0 0 1 |
Type 6:
Sometimes the recurrence is given like this:
f(n) = f(n-1) -> if n is odd
f(n) = f(n-2) -> if n is even
In short:
f(n) = (n&1) X f(n-1) + (!(n&1)) X f(n-2)
Here, we can split the functions in the basis of odd even and keep 2 different matrix for both of them and calculate them separately.
Type 7:
Feeling little too confident? Good for you. Sometimes we may need to maintain more than one recurrence, where they are interested. For example, let a recurrence re;atopm be:
g(n) = 2g(n-1) + 2g(n-2) + f(n)
Here, recurrence g(n) is dependent upon f(n) and this can be calculated in the same matrix but of increased dimensions. From these let's at first design the matrices A and B.
Matrix A Matrix B
| g(n) | | g(n+1) |
| g(n-1) | | g(n) |
| f(n+1) | | f(n+2) |
| f(n) | | f(n+1) |
Here, g(n+1) = 2g(n-1) + f(n+1) and f(n+2) = 2f(n+1) + 2f(n).
Now, using the processes stated above, we can find the objective matrix M to be:
| 2 2 1 0 |
| 1 0 0 0 |
| 0 0 2 2 |
| 0 0 1 0 |
So, these are the basic categories of recurrence relations which are used to solveby this simple technique.