1+ '''
2+ This algorithm calculates the Nth Fibonacci Nummber in log(N) time complexity
3+ The key idea behind this algorimth is to use matrix exponentiation technique (matrix multiplication + binary exponentiation)
4+ This algorithm uses the fact that the Fibonacci Numbers can be converted to a 2D matrix of 2x2 size, [[1, 1][1, 0]], and the Fibonacci
5+ Number which is to be calculated with the help of matrix multiplication will be nothing but cell, (0,0) of (Matrix)^n ,i.e, our resultant
6+ matrix.
7+ '''
8+ def fib (n ):
9+ 10+ F = [[1 , 1 ],
11+ [1 , 0 ]]
12+ if (n == 0 ):
13+ return 0
14+ power (F , n - 1 )
15+ 16+ return F [0 ][0 ]
17+ 18+ '''
19+ Matrix multiplication is calculated using Strassen Algorithm which calculates the product of Two Matrices in o(N^3) time complexity.
20+ Since, the size of matrix here is fixed, i.e, 2x2, so, the Time Complexity will be O(1).
21+ '''
22+ def multiply (F , M ):
23+ 24+ x = (F [0 ][0 ] * M [0 ][0 ] +
25+ F [0 ][1 ] * M [1 ][0 ])
26+ y = (F [0 ][0 ] * M [0 ][1 ] +
27+ F [0 ][1 ] * M [1 ][1 ])
28+ z = (F [1 ][0 ] * M [0 ][0 ] +
29+ F [1 ][1 ] * M [1 ][0 ])
30+ w = (F [1 ][0 ] * M [0 ][1 ] +
31+ F [1 ][1 ] * M [1 ][1 ])
32+ 33+ F [0 ][0 ] = x
34+ F [0 ][1 ] = y
35+ F [1 ][0 ] = z
36+ F [1 ][1 ] = w
37+ 38+ '''
39+ Power function uses Binary Exponentiation Technique to calculate the value of x^n in O(logN) time instead of O(N) time.
40+ '''
41+ def power (F , n ):
42+ 43+ if (n == 0 or n == 1 ):
44+ return
45+ M = [[1 , 1 ],
46+ [1 , 0 ]]
47+ 48+ power (F , n // 2 )
49+ multiply (F , F )
50+ 51+ if (n % 2 != 0 ):
52+ multiply (F , M )
53+ 54+ 55+ # Driver Code
56+ if __name__ == "__main__" :
57+ n = 5
58+ print (fib (n ))
59+ 60+ '''
61+ I hope you have a fun time reading this algorithm.
62+ Problems for practice:
63+ https://practice.geeksforgeeks.org/problems/generalised-fibonacci-numbers1820/1
64+ https://practice.geeksforgeeks.org/problems/nth-fibonacci-number1335/1
65+ https://practice.geeksforgeeks.org/problems/37f36b318a7d99c81f17f0a54fc46b5ce06bbe8c/0
66+ '''
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