Added matrix exponentiation approach for finding fibonacci number. (T…

…heAlgorithms#1042)

* Added matrix exponentiation approach for finding fibonacci number.

* Implemented the way of finding nth fibonacci.
* Complexity is about O(log(n)*8)

* Updated the matrix exponentiation approach of finding nth fibonacci.

- Removed some extra spaces    
- Added the complexity of bruteforce algorithm  
- Removed unused function called zerro()  
- Added some docktest based on request

* Updated the matrix exponentiation approach of finding nth fibonacci.

- Removed some extra spaces
- Added the complexity of bruteforce algorithm
- Removed unused function called zerro()
- Added some docktest based on request

* Tighten up main() and add comments on performance

matrix/nth_fibonacci_using_matrix_exponentiation.py

@@ -0,0 +1,88 @@
+"""
+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.
+"""
+from __future__ import print_function
+
+
+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():
+    fmt = "{} fibonacci number using matrix exponentiation is {} and using bruteforce is {}\n"
+    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(fmt.format(ordinal, nth_fibonacci(n), nth_fibonacci_test(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__":
+    main()