# For a given value of $n$ and $m$, find $\text{fib}(n)$ $\text{mod } m$ where $n$ is very huge. (Pisano Period) [closed]

Input

Integers $$'n'$$ (up to $$10^{14}$$) and $$'m'$$(up to $$10^3$$)

Output

$$\text{Fib}(n)$$ $$\text{modulo}$$ $$m$$

My questions

For example : Why $$\text{fib}(n=2015)$$ $$\text{mod}$$ $$3$$ is equivalent to $$\text{fib}(7)$$ $$\text{mod } 3$$? (for $$ð = 3$$ the period is $$01120221$$ and has length $$8$$ and $$2015=251*8 + 7$$)

In general, after getting the remainder sequence, how (mathematical proof) it is used for computing $$\text{Fib}(n)$$ $$\text{mod } m$$?

The wikipedia article is quite enlightening on the Pisano period, but from the algorithmic viewpoint, it shows that you only need to compute the period for $$n$$ a prime power $$p^k,$$, and in that case it divides either $$p^{k-1}(p-1)$$ or $$p^{k-1}2(p+1).$$ For your range of $$n$$ brute force will tell you the period quickly, by computing the multiplicative order of $$\begin{pmatrix}0&1\\1&1\end{pmatrix}$$ modulo $$n.$$

• Thanks! @Igor, I had figured how to compute "period" but I have difficulties on how to use period length to get fib(n) mod m. Also, I am not able to understand the relation between fib(n) and n (as mentioned in the question I don't understand why fib(2015) mod 3 = fib(7) mod 3 ). More code-oriented question is on StackOverflow(stackoverflow.com/questions/61706899/…) May 10 '20 at 6:03
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From an algorithmic viewpoint, you can compute $$F_n\bmod m$$ efficiently in time $$\tilde O\bigl((\log n)(\log m))$$ [or $$O\bigl((\log n)(\log m)^2)$$ when employing a naive schoolbook multiplication algorithm] by computing

$$\begin{pmatrix}1&1\\1&0\end{pmatrix}^n\begin{pmatrix}1\\0\end{pmatrix}\bmod m$$

where the matrix power is evaluated by repeated squaring modulo $$m$$. Stated in a different way, this amounts to using the recurrences

\begin{align*} F_{2n-1}&=F_n^2 + F_{n-1}^2,\\ F_{2n}&=(2F_{n-1}+F_n)F_n \end{align*}

modulo $$m$$.

In contrast, I donât think there is any known method to compute the Pisano period faster than factorizing $$m$$ (which takes exponential time $$O\bigl(2^{(\log m)^\alpha}\bigr)$$ for some $$\alpha>0$$).

• Thanks, @Emil, I have computed period and length of period, as question says, how can I find fib(n) mod m using that period? and why fib(2015) mod 3 = fib(7) mod 3(for m=3 the period is 01120221 and has length 8 and 2015=251â8+7)? May 10 '20 at 7:00
• I can read myself, thank you. I am telling you it is more efficient to not compute the period. May 10 '20 at 7:02
• your answer doesn't explain why fib(2015) mod 3 = fib(7) mod 3(for m=3 the period is 01120221 and has length 8 and 2015=251â8+7) May 10 '20 at 7:04