Python: Compute a Huge Fibonacci Number Modulo m - python-3.x

Question

# Uses python3
# Given two integers n and m, output Fn mod m (that is, the remainder of Fn when divided by m
def Huge_Fib(n,m):
if n == 0 : return 0
elif n == 1: return 1
else:
a,b = 0,1
for i in range(1,n):
a, b = b, (a+b) % m
print(b);
n,m = map(int, input().split());
Huge_Fib(n,m);
The code works very well. However, when I run a case as n = 99999999999999999, m = 2, it takes me much time. Do you have any better solutions?

Answer 1

Here is my solution, you don't have to go through 99999999999999999 iterations if you find the pisano period.
I also recommend that you watch this video: https://www.youtube.com/watch?v=Nu-lW-Ifyec
# Uses python3
import sys
def get_fibonacci_huge(n, m):
if n <= 1:
return n
arr = [0, 1]
previousMod = 0
currentMod = 1
for i in range(n - 1):
tempMod = previousMod
previousMod = currentMod % m
currentMod = (tempMod + currentMod) % m
arr.append(currentMod)
if currentMod == 1 and previousMod == 0:
index = (n % (i + 1))
return arr[index]
return currentMod
if __name__ == '__main__':
input = sys.stdin.read();
n, m = map(int, input.split())
print(get_fibonacci_huge(n,m))

Answer 2

# Uses python3
# Given two integers n and m, output Fn mod m (that is, the remainder of Fn when divided by m
def Huge_Fib(n,m):
# Initialize a matrix [[1,1],[1,0]]
v1, v2, v3 = 1, 1, 0
# Perform fast exponentiation of the matrix (quickly raise it to the nth power)
for rec in bin(n)[3:]:
calc = (v2*v2) % m
v1, v2, v3 = (v1*v1+calc) % m, ((v1+v3)*v2) % m, (calc+v3*v3) % m
if rec == '1': v1, v2, v3 = (v1+v2) % m, v1, v2
print(v2);
n,m = map(int, input().split());
Huge_Fib(n,m);
This is a superfast solution refer to https://stackoverflow.com/a/23462371/3700852

Answer 3

I solved it in Python 3. This the fastest algorithm to compute a huge Fibonacci number modulo m.For example for n =2816213588, m = 239, it took Max time used: 0.01/5.00, max memory used: 9424896/536870912.)
def pisanoPeriod(m):
previous, current = 0, 1
for i in range(0, m * m):
previous, current = current, (previous + current) % m
# A Pisano Period starts with 01
if (previous == 0 and current == 1):
return i + 1
def calc_fib(n,m):
p = pisanoPeriod(m)
n = n % p
if (n <= 1):
return n
else:
previous,current = 0,1
for i in range(2,n+1):
previous,current = current,(previous+current)
return current%m
n,m =map(int,input().split(" "))
print(calc_fib(n,m))

Answer 4

In the below code we are using two concepts of Fibonacci series:
Pisano periods follows a Fibonacci sequence and hence each repetition(pattern) begins with 0 and 1 appearing consecutively one after the other.
fib(n) divides fib(m) only when n divides m which means if fib(4)%3==0,then fib(4+4)%3==0,fib(4+4+4)%3==0 and so on.This helps us in finding the Pisano period.
To know about Pisano periods,I recommend that you watch this video: https://www.youtube.com/watch?v=Nu-lW-Ifyec
#python3
def pisano_length(m):
i=2
while(fib(i)%m!=0):
i+=1
if(fib(i+1)%m!=1):
while(fib(i+1)%m!=1):
i+=i
print("The pisano length for mod {} is: {}".format(m,i))
return(i)
def fib(n):
a,b=0,1
if(n==0 or n==1):
return n
else:
for i in range(2,n+1):
b,a=a+b,b
return(b)
#we want to calculate fib(n)%m for big numbers
n,m=map(int,input().split())
remainder=n%pisano_length(m)
print(fib(remainder)%m)

Answer 5

You should look up Pisano periods.
https://en.wikipedia.org/wiki/Pisano_period and
http://webspace.ship.edu/msrenault/fibonacci/fibfactory.htm should give you a good understanding of what they are.
edit: Just googling "fibonacci modulo" gives you those two as the top two results.

Answer 6

For any integer m>=2, the sequence fn modulo m is periodic - Pisano Period.
So no need to store and find fn. Instead, find a repeating pattern for given m.

Answer 7

This is how i have done by calculating the pisano period.(Java)
public static long get_pisano_period(long m) {
long a = 0, b = 1;
long c;
for (int i = 0; i < m * m; i++) {
c = (a + b) % m;
a = b;
b = c;
if (a == 0 && b == 1)
return i + 1;
}
return 0;
}
public static BigInteger get_fibonacci_huge(long n,long m) {
long remainder = n % get_pisano_period(m);
BigInteger first = BigInteger.valueOf(0);
BigInteger second=BigInteger.valueOf(1);
BigInteger m1=BigInteger.valueOf(m);
BigInteger res = BigInteger.valueOf(remainder);
for (long i = 1; i < remainder; i++) {
res = (first.add(second)).mod(m1);
first = second;
second = res;
}
return res.mod(m1);
}