Solving Linear Recurrence for Programming Contest

One of the topics that often arises in programming contests nowadays is about solving linear recurrence. It may come as a classic “find the n-th term of Fibonacci sequence” to more complex and creative forms of problems. Usually, after being simplified, the problem is merely asking you the n-th term of a linear recurrence. It should be able to solve with dynamic programming, however, the problem is that usually n can very large.

Because this topic starts to emerge in contests, I would like to share how I usually solve problems of this topic.

Definition

First, let’s start with a definition. A linear recurrence relation is a function or a sequence such that each term is a linear combination of previous terms. Each term can be described as a function of the previous terms. A famous example is the Fibonacci sequence: f(i) = f(i-1) + f(i-2). Linear means that the previous terms in the definition are only multiplied by a constant (possibly zero) and nothing else. So, this sequence: f(i) = f(i-1) * f(i-2) is not a linear recurrence.

Problem

Now, let’s state the problem formally.

Given f, a function defined as a linear recurrence relation. Compute f(N). N may be very large.

How to Solve

Here is the method I usually use in programming contests. I break down the method into four steps. Fibonacci sequence will be used as an example.

It is assumed that you have learned matrix and matrix operations, especially multiplication of two matrices.

Step 1. Determine K, the number of terms on which f(i) depends.

More precisely, K is the minimum integer such that f(i) doesn’t depend on f(i-M), for all M > K. For Fibonacci sequence, because the relation is:

,

therefore, K = 2. In this way, be careful for missing terms though, for example, this sequence:

has K = 4, because it can be rewritten explicitly as:

.

Step 2. Determine vector F₁, the initial values.

If each term of a recurrence relation depends on K previous terms, then it must have the first K terms defined, otherwise the whole sequence is undefined. For Fibonacci sequence (K = 2), the well-known initial values are:

• 
• 

Although some people may define other initial values, like f(1) = 0 and f(2) = 1, we will use the above definition in this post.

From now on, we will work extensively in matrices. We define a column vector F_i as a K x 1 matrix whose first row is f(i), second row is f(i+1), and so on, until K-th row is f(i+K-1). The initial values of f are given in column vector F₁ that has values f(1) through f(K):

Step 3. Determine T, the transformation matrix.

This is the most important step in solving recurrence relation. The heart of this method is to construct a K x K matrix T, called transformation matrix, such that

.

Here is how to construct it. Suppose that . You can solve this equation with any method, and obtain the result:

.

More precisely, T is a K x K matrix whose last row is a vector� , and for the other rows, the i-th row is a zero vector except that its (i+1)th element is 1.

Let’s apply it to our example. In Fibonacci sequence,

.

Exercise: find out how matrix T can transforms F_i into F_i+1, and memorize the pattern.

Step 4. Determine F_N.

After we construct the transformation matrix, the rest is very simple. We can obtain F_i for any i, by repeatedly multiplying T with F₁. For example, to obtain F₂,

.

To obtain F₃,

.

And so on. In general,

.

Therefore, the original problem is now (almost) solved: compute F_N as above, and then we can obtain f(N): it is exactly the first row of F_N. In case of our Fibonacci sequence, the N-th term in Fibonacci sequence is the first row of:

.

What remains is how to compute T^N-1 efficiently. The most popular method is to use exponentiation by squaring method that works in O(log N) time, with this recurrence:

• 
• 
• 

Multiplying two matrices takes O(K³) time using standard method, so the overall time complexity to solve a linear recurrence is O(K³ log N).

Sample code

Here is a sample C++ code to compute the N-th term of Fibonacci sequence, modulo 1,000,000,007.

#include <vector>
#define REP(i,n) for (int i = 1; i <= n; i++)
using namespace std;

typedef long long ll;
typedef vector<vector<ll> > matrix;
const ll MOD = 1000000007;
const int K = 2;

// computes A * B
matrix mul(matrix A, matrix B)
{
    matrix C(K+1, vector<ll>(K+1));
    REP(i, K) REP(j, K) REP(k, K)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
}

// computes A ^ p
matrix pow(matrix A, int p)
{
    if (p == 1)
        return A;
    if (p % 2)
        return mul(A, pow(A, p-1));
    matrix X = pow(A, p/2);
    return mul(X, X);
}

// returns the N-th term of Fibonacci sequence
int fib(int N)
{
    // create vector F1
    vector<ll> F1(K+1);
    F1[1] = 1;
    F1[2] = 1;

    // create matrix T
    matrix T(K+1, vector<ll>(K+1));
    T[1][1] = 0, T[1][2] = 1;
    T[2][1] = 1, T[2][2] = 1;

    // raise T to the (N-1)th power
    if (N == 1)
        return 1;
    T = pow(T, N-1);

    // the answer is the first row of T . F1
    ll res = 0;
    REP(i, K)
        res = (res + T[1][i] * F1[i]) % MOD;
    return res;
}

Variants

Those are the basics of solving linear recurrence. Here we will discuss the variants that often occur.

* Constants in the relation

The recurrence relation may include a constant, i.e., the function is of the form . In this variant, the vector F is enhanced to “remember” the value of d. It is of size (K+1) x 1 now:

.

We now want to construct a matrix T, of size (K+1) x (K+1), such that:

.

After learning the basic method, can you figure out the matrix T for this variant? Here it is:

.

For example, let . Then, the matrix equation for this function is:

.

* Odd/even conditional function

The function f may behave differently according to the parity of the argument. For example, if i is even, then , otherwise (if i is odd) .

How to solve this variant? It is possible to split the equation into even and odd parts. We construct two matrices T_even and T_odd such that:

• 
• 

We also construct matrix . Now, we can compute F_N with a single formula:

• 
• 

Note that T_even and T_odd must be of the same size, so in this example, convert into to make it look like the odd part, and then apply the standard method.

* Your variants

Have you ever encountered another variant of linear recurrence problem? Please share it in the comment section