The problem is that your return y is within the loop of your function. So after the first iteration, it will already stop and return the first value: 1. Except when n is 0, in which case the function is made to return 0 itself, and in case n is 1, when the for loop … Read more
The complexity of a naive recursive fibonacci is indeed 2ⁿ. T(n) = T(n-1) + T(n-2) = T(n-2) + T(n-3) + T(n-3) + T(n-4) = = T(n-3) + T(n-4) + T(n-4) + T(n-5) + T(n-4) + T(n-5) + T(n-5) + T(n-6) = … In each step you call T twice, thus will provide eventual asymptotic barrier … Read more
Here’s a different and simpler function that calculates the n’th Fibonacci number: fib :: Integer -> Integer fib 0 = 0 fib 1 = 1 fib n = fib (n-1) + fib (n-2) The implementation you are referring to relays on some observations about how values in Fibonacci relate to each other (and how Haskell … Read more
How to write a generator class? You’re almost there, writing an Iterator class (I show a Generator at the end of the answer), but __next__ gets called every time you call the object with next, returning a generator object. Instead, to make your code work with the least changes, and the fewest lines of code, … Read more
You have never declared fib to be an array. Use var fib = []; to solve this. Also, you’re never modifying the y variable, neither using it. The code below makes more sense, plus, it doesn’t create unused variables: var i; var fib = [0, 1]; // Initialize array! for (i = 2; i <= … Read more
I would use this method: Python 2 a = int(raw_input(‘Give amount: ‘)) def fib(n): a, b = 0, 1 for _ in xrange(n): yield a a, b = b, a + b print list(fib(a)) Python 3 a = int(input(‘Give amount: ‘)) def fib(n): a, b = 0, 1 for _ in range(n): yield a a, … Read more
You can use the matrix exponentiation method (linear recurrence method). You can find detailed explanation and procedure in this or this blog. Run time is O(log n). I don’t think there is a better way of doing this.
In fibonacci sequence each item is the sum of the previous two. So, you wrote a recursive algorithm. So, fibonacci(5) = fibonacci(4) + fibonacci(3) fibonacci(3) = fibonacci(2) + fibonacci(1) fibonacci(4) = fibonacci(3) + fibonacci(2) fibonacci(2) = fibonacci(1) + fibonacci(0) Now you already know fibonacci(1)==1 and fibonacci(0) == 0. So, you can subsequently calculate the other … Read more
Yes. The primitive recursive solution takes a lot of time. The reason for this is that for each number calculated, it needs to calculate all the previous numbers more than once. Take a look at the following image. It represents calculating Fibonacci(5) with your function. As you can see, it computes the value of Fibonacci(2) … Read more