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61f3119467
* f-string update rsa_cipher.py * f-string update rsa_key_generator.py * f-string update burrows_wheeler.py * f-string update non_recursive_segment_tree.py * f-string update red_black_tree.py * f-string update deque_doubly.py * f-string update climbing_stairs.py * f-string update iterating_through_submasks.py * f-string update knn_sklearn.py * f-string update 3n_plus_1.py * f-string update quadratic_equations_complex_numbers.py * f-string update nth_fibonacci_using_matrix_exponentiation.py * f-string update sherman_morrison.py * f-string update levenshtein_distance.py * fix lines that were too long
92 lines
2.6 KiB
Python
92 lines
2.6 KiB
Python
"""
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Implementation of finding nth fibonacci number using matrix exponentiation.
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Time Complexity is about O(log(n)*8), where 8 is the complexity of matrix
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multiplication of size 2 by 2.
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And on the other hand complexity of bruteforce solution is O(n).
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As we know
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f[n] = f[n-1] + f[n-1]
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Converting to matrix,
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[f(n),f(n-1)] = [[1,1],[1,0]] * [f(n-1),f(n-2)]
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-> [f(n),f(n-1)] = [[1,1],[1,0]]^2 * [f(n-2),f(n-3)]
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...
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...
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-> [f(n),f(n-1)] = [[1,1],[1,0]]^(n-1) * [f(1),f(0)]
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So we just need the n times multiplication of the matrix [1,1],[1,0]].
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We can decrease the n times multiplication by following the divide and conquer approach.
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"""
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def multiply(matrix_a, matrix_b):
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matrix_c = []
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n = len(matrix_a)
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for i in range(n):
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list_1 = []
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for j in range(n):
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val = 0
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for k in range(n):
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val = val + matrix_a[i][k] * matrix_b[k][j]
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list_1.append(val)
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matrix_c.append(list_1)
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return matrix_c
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def identity(n):
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return [[int(row == column) for column in range(n)] for row in range(n)]
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def nth_fibonacci_matrix(n):
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"""
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>>> nth_fibonacci_matrix(100)
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354224848179261915075
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>>> nth_fibonacci_matrix(-100)
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-100
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"""
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if n <= 1:
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return n
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res_matrix = identity(2)
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fibonacci_matrix = [[1, 1], [1, 0]]
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n = n - 1
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while n > 0:
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if n % 2 == 1:
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res_matrix = multiply(res_matrix, fibonacci_matrix)
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fibonacci_matrix = multiply(fibonacci_matrix, fibonacci_matrix)
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n = int(n / 2)
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return res_matrix[0][0]
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def nth_fibonacci_bruteforce(n):
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"""
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>>> nth_fibonacci_bruteforce(100)
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354224848179261915075
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>>> nth_fibonacci_bruteforce(-100)
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-100
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"""
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if n <= 1:
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return n
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fib0 = 0
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fib1 = 1
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for i in range(2, n + 1):
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fib0, fib1 = fib1, fib0 + fib1
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return fib1
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def main():
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for ordinal in "0th 1st 2nd 3rd 10th 100th 1000th".split():
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n = int("".join(c for c in ordinal if c in "0123456789")) # 1000th --> 1000
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print(
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f"{ordinal} fibonacci number using matrix exponentiation is "
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f"{nth_fibonacci_matrix(n)} and using bruteforce is "
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f"{nth_fibonacci_bruteforce(n)}\n"
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)
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# from timeit import timeit
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# print(timeit("nth_fibonacci_matrix(1000000)",
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# "from main import nth_fibonacci_matrix", number=5))
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# print(timeit("nth_fibonacci_bruteforce(1000000)",
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# "from main import nth_fibonacci_bruteforce", number=5))
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# 2.3342058970001744
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# 57.256506615000035
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if __name__ == "__main__":
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main()
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