|
| 1 | +""" |
| 2 | +Implementation of finding nth fibonacci number using matrix exponentiation. |
| 3 | +Time Complexity is about O(log(n)*8), where 8 is the complexity of matrix multiplication of size 2 by 2. |
| 4 | +And on the other hand complexity of bruteforce solution is O(n). |
| 5 | +As we know |
| 6 | + f[n] = f[n-1] + f[n-1] |
| 7 | +Converting to matrix, |
| 8 | + [f(n),f(n-1)] = [[1,1],[1,0]] * [f(n-1),f(n-2)] |
| 9 | +-> [f(n),f(n-1)] = [[1,1],[1,0]]^2 * [f(n-2),f(n-3)] |
| 10 | + ... |
| 11 | + ... |
| 12 | +-> [f(n),f(n-1)] = [[1,1],[1,0]]^(n-1) * [f(1),f(0)] |
| 13 | +So we just need the n times multiplication of the matrix [1,1],[1,0]]. |
| 14 | +We can decrease the n times multiplication by following the divide and conquer approach. |
| 15 | +""" |
| 16 | +from __future__ import print_function |
| 17 | + |
| 18 | + |
| 19 | +def multiply(matrix_a, matrix_b): |
| 20 | + matrix_c = [] |
| 21 | + n = len(matrix_a) |
| 22 | + for i in range(n): |
| 23 | + list_1 = [] |
| 24 | + for j in range(n): |
| 25 | + val = 0 |
| 26 | + for k in range(n): |
| 27 | + val = val + matrix_a[i][k] * matrix_b[k][j] |
| 28 | + list_1.append(val) |
| 29 | + matrix_c.append(list_1) |
| 30 | + return matrix_c |
| 31 | + |
| 32 | + |
| 33 | +def identity(n): |
| 34 | + return [[int(row == column) for column in range(n)] for row in range(n)] |
| 35 | + |
| 36 | + |
| 37 | +def nth_fibonacci_matrix(n): |
| 38 | + """ |
| 39 | + >>> nth_fibonacci_matrix(100) |
| 40 | + 354224848179261915075 |
| 41 | + >>> nth_fibonacci_matrix(-100) |
| 42 | + -100 |
| 43 | + """ |
| 44 | + if n <= 1: |
| 45 | + return n |
| 46 | + res_matrix = identity(2) |
| 47 | + fibonacci_matrix = [[1, 1], [1, 0]] |
| 48 | + n = n - 1 |
| 49 | + while n > 0: |
| 50 | + if n % 2 == 1: |
| 51 | + res_matrix = multiply(res_matrix, fibonacci_matrix) |
| 52 | + fibonacci_matrix = multiply(fibonacci_matrix, fibonacci_matrix) |
| 53 | + n = int(n / 2) |
| 54 | + return res_matrix[0][0] |
| 55 | + |
| 56 | + |
| 57 | +def nth_fibonacci_bruteforce(n): |
| 58 | + """ |
| 59 | + >>> nth_fibonacci_bruteforce(100) |
| 60 | + 354224848179261915075 |
| 61 | + >>> nth_fibonacci_bruteforce(-100) |
| 62 | + -100 |
| 63 | + """ |
| 64 | + if n <= 1: |
| 65 | + return n |
| 66 | + fib0 = 0 |
| 67 | + fib1 = 1 |
| 68 | + for i in range(2, n + 1): |
| 69 | + fib0, fib1 = fib1, fib0 + fib1 |
| 70 | + return fib1 |
| 71 | + |
| 72 | + |
| 73 | +def main(): |
| 74 | + fmt = "{} fibonacci number using matrix exponentiation is {} and using bruteforce is {}\n" |
| 75 | + for ordinal in "0th 1st 2nd 3rd 10th 100th 1000th".split(): |
| 76 | + n = int("".join(c for c in ordinal if c in "0123456789")) # 1000th --> 1000 |
| 77 | + print(fmt.format(ordinal, nth_fibonacci(n), nth_fibonacci_test(n))) |
| 78 | + # from timeit import timeit |
| 79 | + # print(timeit("nth_fibonacci_matrix(1000000)", |
| 80 | + # "from main import nth_fibonacci_matrix", number=5)) |
| 81 | + # print(timeit("nth_fibonacci_bruteforce(1000000)", |
| 82 | + # "from main import nth_fibonacci_bruteforce", number=5)) |
| 83 | + # 2.3342058970001744 |
| 84 | + # 57.256506615000035 |
| 85 | + |
| 86 | + |
| 87 | +if __name__ == "__main__": |
| 88 | + main() |
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