"""
Implementation of finding nth fibonacci number using matrix exponentiation.
Time Complexity is about O(log(n)*8), where 8 is the complexity of matrix
multiplication of size 2 by 2.
And on the other hand complexity of bruteforce solution is O(n).
As we know
f[n] = f[n-1] + f[n-1]
Converting to matrix,
[f(n),f(n-1)] = [[1,1],[1,0]] * [f(n-1),f(n-2)]
-> [f(n),f(n-1)] = [[1,1],[1,0]]^2 * [f(n-2),f(n-3)]
...
...
-> [f(n),f(n-1)] = [[1,1],[1,0]]^(n-1) * [f(1),f(0)]
So we just need the n times multiplication of the matrix [1,1],[1,0]].
We can decrease the n times multiplication by following the divide and conquer approach.
"""
def multiply(matrix_a, matrix_b):
matrix_c = []
n = len(matrix_a)
for i in range(n):
list_1 = []
for j in range(n):
val = 0
for k in range(n):
val = val + matrix_a[i][k] * matrix_b[k][j]
list_1.append(val)
matrix_c.append(list_1)
return matrix_c
def identity(n):
return [[int(row == column) for column in range(n)] for row in range(n)]
def nth_fibonacci_matrix(n):
"""
>>> nth_fibonacci_matrix(100)
354224848179261915075
>>> nth_fibonacci_matrix(-100)
-100
"""
if n <= 1:
return n
res_matrix = identity(2)
fibonacci_matrix = [[1, 1], [1, 0]]
n = n - 1
while n > 0:
if n % 2 == 1:
res_matrix = multiply(res_matrix, fibonacci_matrix)
fibonacci_matrix = multiply(fibonacci_matrix, fibonacci_matrix)
n = int(n / 2)
return res_matrix[0][0]
def nth_fibonacci_bruteforce(n):
"""
>>> nth_fibonacci_bruteforce(100)
354224848179261915075
>>> nth_fibonacci_bruteforce(-100)
-100
"""
if n <= 1:
return n
fib0 = 0
fib1 = 1
for i in range(2, n + 1):
fib0, fib1 = fib1, fib0 + fib1
return fib1
def main():
for ordinal in "0th 1st 2nd 3rd 10th 100th 1000th".split():
n = int("".join(c for c in ordinal if c in "0123456789"))
print(
f"{ordinal} fibonacci number using matrix exponentiation is "
f"{nth_fibonacci_matrix(n)} and using bruteforce is "
f"{nth_fibonacci_bruteforce(n)}\n"
)
if __name__ == "__main__":
main()