|
| 1 | +""" |
| 2 | +Implementation of finding nth fibonacci number using matrix exponentiation. |
| 3 | +Time Complexity is about O(log(n)*8) |
| 4 | +As we know |
| 5 | + f[n] = f[n-1] + f[n-1] |
| 6 | +Converting to matrix, |
| 7 | + [f(n),f(n-1)] = [[1,1],[1,0]] * [f(n-1),f(n-2)] |
| 8 | +-> [f(n),f(n-1)] = [[1,1],[1,0]]^2 * [f(n-2),f(n-3)] |
| 9 | + ... |
| 10 | + ... |
| 11 | +-> [f(n),f(n-1)] = [[1,1],[1,0]]^(n-1) * [f(1),f(0)] |
| 12 | +
|
| 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 | +""" |
| 17 | +from __future__ import print_function |
| 18 | + |
| 19 | + |
| 20 | +def multiply(matrix_a, matrix_b): |
| 21 | + matrix_c = [] |
| 22 | + n = len(matrix_a) |
| 23 | + for i in range(n): |
| 24 | + list_1 = [] |
| 25 | + for j in range(n): |
| 26 | + val = 0 |
| 27 | + for k in range(n): |
| 28 | + val = val + matrix_a[i][k] * matrix_b[k][j] |
| 29 | + list_1.append(val) |
| 30 | + matrix_c.append(list_1) |
| 31 | + return matrix_c |
| 32 | + |
| 33 | + |
| 34 | +def identity(n): |
| 35 | + return [[int(row == column) for column in range(n)] for row in range(n)] |
| 36 | + |
| 37 | + |
| 38 | +def zerro(n): |
| 39 | + return [[int(row == column) for column in range(n)] for row in range(n)] |
| 40 | + |
| 41 | + |
| 42 | +def nth_fibonacci(n): |
| 43 | + if n <= 1: |
| 44 | + return n |
| 45 | + res_matrix = identity(2) |
| 46 | + fibonacci_matrix = [[1, 1], [1, 0]] |
| 47 | + n = n - 1 |
| 48 | + while n > 0: |
| 49 | + if n % 2 == 1: |
| 50 | + res_matrix = multiply(res_matrix, fibonacci_matrix) |
| 51 | + fibonacci_matrix = multiply(fibonacci_matrix, fibonacci_matrix) |
| 52 | + n = int(n / 2) |
| 53 | + return res_matrix[0][0] |
| 54 | + |
| 55 | + |
| 56 | +def nth_fibonnaci_test(n): |
| 57 | + if n <= 1: |
| 58 | + return n |
| 59 | + fib0 = 0 |
| 60 | + fib1 = 1 |
| 61 | + for i in range(2, n + 1): |
| 62 | + tem = fib1 + fib0 |
| 63 | + fib0 = fib1 |
| 64 | + fib1 = tem |
| 65 | + return fib1 |
| 66 | + |
| 67 | + |
| 68 | +def main(): |
| 69 | + print( |
| 70 | + "0th fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 71 | + % (nth_fibonacci(0), nth_fibonnaci_test(0)) |
| 72 | + ) |
| 73 | + print( |
| 74 | + "1st fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 75 | + % (nth_fibonacci(1), nth_fibonnaci_test(1)) |
| 76 | + ) |
| 77 | + print( |
| 78 | + "2nd fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 79 | + % (nth_fibonacci(2), nth_fibonnaci_test(2)) |
| 80 | + ) |
| 81 | + print( |
| 82 | + "3rd fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 83 | + % (nth_fibonacci(3), nth_fibonnaci_test(3)) |
| 84 | + ) |
| 85 | + print( |
| 86 | + "10th fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 87 | + % (nth_fibonacci(10), nth_fibonnaci_test(10)) |
| 88 | + ) |
| 89 | + print( |
| 90 | + "100th fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 91 | + % (nth_fibonacci(100), nth_fibonnaci_test(100)) |
| 92 | + ) |
| 93 | + print( |
| 94 | + "1000th fibonnacsi number using matrix exponentiation is %s and using bruteforce is %s \n" |
| 95 | + % (nth_fibonacci(1000), nth_fibonnaci_test(1000)) |
| 96 | + ) |
| 97 | + |
| 98 | + |
| 99 | +if __name__ == "__main__": |
| 100 | + main() |
0 commit comments