Note that each step takes O(1) meaning constant time,since it does only one comparison to check value of n in if block.Recursion tree would look like n Recursive algorithm's time complexity can be better estimated by drawing recursion tree, In this case the recurrence relation for drawing recursion tree would be T(n)=T(n-1)+T(n-2)+O(1) Now, in terms of complexity: O( F(6) ) = O(2^6) Which means, total times F() gets called when n=6 is 2x32=64=2^6. If we mentally move all the *'s from F(6) to F(2) lines into F(1) line, we see that F(1) and F(0) lines are now equal in length. Now, we want to know how many times F(x) gets called at all, and we can see the number of times F(0) is called is only a part of that. We see F(0) gets called 32 times, which is 2^5, which for this sample case is 2^(n-1). Now, the question is, how fast is the base of this pyramid enlarging as n grows? So, when F() is called for a number n, the number of times F() is called for a given number between 0 and n-1 grows as we approach 0.Īs a first impression, it seems to me that if we put it in a visual way, drawing a unit per time F() is called for a given number, wet get a sort of pyramid shape (that is, if we center units horizontally). If it gets called once per number in the sequence 0 to n, then we have O(n), if it gets called n times for each number, then we get O(n*n), or O(n^2), and so on. I came to the same conclusion by a rather simplistic but I believe still valid reasoning.įirst, it's all about figuring out how many times recursive fibonacci function ( F() from now on ) gets called when calculating the Nth fibonacci number. I agree with pgaur and rickerbh, recursive-fibonacci's complexity is O(2^n). You can find out this tight bound by using generating functions as I'd mentioned above. Consequently, the tight bound for this function is the Fibonacci sequence itself (~ θ(1.6 n )). Since each leaf will take O(1) to compute, T(n) is equal to Fib(n) x O(1). The value of Fib(n) is sum of all values returned by the leaves in the recursion tree which is equal to the count of leaves. The leaves of the recursion tree will always return 1. An interesting fact about this function is that the T(n) is asymptotically the same as the value of Fib(n) since both are defined as However, as noted in a comment, this is not the tight bound. You can then prove your conjecture by induction. You solve this recurrence relation (using generating functions, for instance) and you'll end up with the answer.Īlternatively, you can draw the recursion tree, which will have depth n and intuitively figure out that this function is asymptotically O(2 n ). This is assuming that repeated evaluations of the same Fib(n) take the same time - i.e. Output: Enter the Number of terms 6 Method 4: Fibonacci Sequence using lambda and map function n = int(input('Enter the Number of terms'))def fibonacci(count): fib_list = any(map(lambda _: fib_list.You model the time function to calculate Fib(n) as sum of time to calculate Fib(n-1) plus the time to calculate Fib(n-2) plus the time to add them together ( O(1)). Output: Enter the number of terms 6 Method 3: Fibonacci Sequence Using Lambda and Reduce from functools import reducen = int(input('Enter the Number of terms'))def fib(n): return reduce(lambda x, _: x++x], range(n-2), )print(fib(n)) Output: Enter the number of terms 6 0 1 1 2 3 5 Method 2: Fibonacci Sequence Using For Loop n = int(input('Enter the number of terms'))def Fibonacci(n): f0, f1 = 0, 1 for _ in range(n): yield f0 f0, f1 = f1, f0+f1fibs = list(Fibonacci(n))print(fibs) Method 1: Fibonacci Sequence Using Recursion n = input('Enter the number of terms')def fibo(n): if n <= 1: return n else: return(fibo(n-1) + fibo(n-2))for i in range(int(n)): print(fibo(i), end=' ') The beauty of Python is that there is always more than one way to tackle the same problem in this article we will go over some of the best methods to generate Fibonacci series in Python. Example Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21 Solution
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