mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-30 16:31:08 +00:00
46842e8c5b
* Update count_islands_in_matrix.py * Update matrix_class.py * Update matrix_operation.py * Update nth_fibonacci_using_matrix_exponentiation.py * Update searching_in_sorted_matrix.py * Update count_islands_in_matrix.py * Update matrix_class.py * Update matrix_operation.py * Update rotate_matrix.py * Update sherman_morrison.py * Update spiral_print.py * Update count_islands_in_matrix.py * formatting * formatting * Update matrix_class.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
95 lines
2.7 KiB
Python
95 lines
2.7 KiB
Python
"""
|
|
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: list[list[int]], matrix_b: list[list[int]]) -> list[list[int]]:
|
|
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: int) -> list[list[int]]:
|
|
return [[int(row == column) for column in range(n)] for row in range(n)]
|
|
|
|
|
|
def nth_fibonacci_matrix(n: int) -> int:
|
|
"""
|
|
>>> 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: int) -> int:
|
|
"""
|
|
>>> 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() -> None:
|
|
for ordinal in "0th 1st 2nd 3rd 10th 100th 1000th".split():
|
|
n = int("".join(c for c in ordinal if c in "0123456789")) # 1000th --> 1000
|
|
print(
|
|
f"{ordinal} fibonacci number using matrix exponentiation is "
|
|
f"{nth_fibonacci_matrix(n)} and using bruteforce is "
|
|
f"{nth_fibonacci_bruteforce(n)}\n"
|
|
)
|
|
# from timeit import timeit
|
|
# print(timeit("nth_fibonacci_matrix(1000000)",
|
|
# "from main import nth_fibonacci_matrix", number=5))
|
|
# print(timeit("nth_fibonacci_bruteforce(1000000)",
|
|
# "from main import nth_fibonacci_bruteforce", number=5))
|
|
# 2.3342058970001744
|
|
# 57.256506615000035
|
|
|
|
|
|
if __name__ == "__main__":
|
|
import doctest
|
|
|
|
doctest.testmod()
|
|
main()
|