Tribonacci series with different initial values

Question

How to find nth tribonacci number with matrix multiplication method if the initial values are some arbitrary numbers say 1, 2 3 i.e T(1) = 1, T(2) =2 and T(3) = 3.

If T(n) = T(n-1) + T(n-2) + T(n-3) then how to find T(n) if n is very very large, I would appreciate if anyone can explain with matrix multiplication method. How to construct initial matrix.

Answer 1

The matrix multiplication method involves using the matrix recurrence relation.

For the Fibonacci series, we can define a vector of length 2 to represent adjacent Fibonacci numbers. Using this vector, we can define a recurrence relation with a matrix multiplication:

Similarly, the Tribonacci series recurrence relation can be written in this way:

The only difference is that the vector and matrix sizes are different.

Now, to calculate a large Tribonacci number, we just apply the matrix multiplication n times, and we get:

The matrix to the power of n (Mⁿ) can be efficiently calculated, because we can use an exponentiation algorithm.

Many efficient exponentiation algorithms for scalars are described by Wikipedia in Exponentiation by Squaring. We can use the same idea for matrix exponentiation.

I will describe a simple way to do this. First we write n as a binary number, eg:

n = 37 = 100101

Then, calculate M to each power of 2 by squaring the previous power of 2: M¹, M² = M¹M¹, M⁴ = M²M², M⁸ = M⁴M⁴, M¹⁶ = M⁸M⁸, M³² = M¹⁶M¹⁶, ...

And finally, multiply the powers of M corresponding to the binary digits of n. In this case, Mⁿ = M¹M⁴M³².

After calculating that, we can multiply the matrix with the Tribonacci vector for the first 3 values, ie.

Because the matrices have fixed size, each matrix multiplication takes constant time. We must do O(log n) matrix multiplications. Thus, we can calculate the n^th Tribonacci number in O(log n) time.

Compare this to the normal dynamic programming method, where it takes O(n) time, by calculating each Tribonacci number up to the n^th Tribonacci number (ie. for (i = 3 to n) {T[i] = T[i-1]+T[i-2]+T[i-3];} return T[n];).

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I will assume that you know how to code up matrix multiplication in the language of your choice.