![]() If your goal is to create a list of Fibonacci numbers, then this method is not recommended. Please note that this method is efficient in predicting the nth term of the Fibonacci sequence. Here, the program uses assignments and swapping of values in just a single line. The main part of the code is at line no.4. You can see how simple and beautiful the code is written in this method. So, if you want to find the nth term in this series, you can do this in a few lines of code as follows. Iterative Solution to find the n th term in Fibonacci Sequence You will get an output like the one that is given below. You can just copy and paste this code and run it on your python IDLE. ![]() It is so easy to code when you compare it with any other programming language. Python language has the built-in capability to do this to reduce your coding efforts. Note that the program uses assignment and swapping of values in a single line. We decrement the value of n and print the Fibonacci series till n-2 is greater than 0. ![]() We use a while loop to find the sum of the first two terms and proceed with the series by interchanging the variables. The user must enter the number of terms to be printed in the Fibonacci sequence. N=int(input("Enter the number of terms in the sequence: ")) The source code of the Python Program to find the Fibonacci series without using recursion is given below. In Python, we can solve the Fibonacci sequence in both recursive as well as iterative ways, but the iterative way is the best and easiest way to do it. Iterative Solution to find Fibonacci Sequence Let’s look at how can we write the fastest solution to the Fibonacci sequence. In python, you can write the solution to this problem very easily, to avoid all these complexities. Instead, we compute each number from scratch. The major problem of this approach is that with each Fibonacci number we calculate in our list, we don’t use the previous numbers we have knowledge of to make the computation faster. This gets worse and worse the higher the number you want to compute. The primitive recursive solution takes a huge amount of time because for each number calculated, it needs to calculate all the previous numbers more than once. One = sp.There are many ways to solve this problem. It may look something like this: > def fib(n, computed = -1/2 + (1/4 + 3*sqrt(5)/20)*(2+sqrt(5))**n + (1/4 - 3*sqrt(5)/20)*(2-sqrt(5))**nĭespite the sqrt, this is integral for integral n, so this can be conveniently computed using the handy functions from my previous answer, or using a package such as sympy to handle the roots exactly. Since your problem is that you re-compute values you have already computed, you can instead choose to save the values you have already computed in a dict, and try to get them from that before you recompute them. Memoization (relatively advanced technique)Īnother alternative to make it faster exists, but it is a little more complicated as well. Or you can get the 17th fibonacci number from a list of the first 40 by doing > fib_to(40)Ģ. Then you can get the first 20 fibonacci numbers by doing > fib_to(20) This approach would look something like this: > def fib_to(n): If you have a list of the fibonacci numbers, you can use the last two numbers in that list to create the next number. If you do that, you build "from the bottom up" or so to speak, and you can reuse previous numbers to create the next one. The easiest way is to just create a list of fibonacci numbers up to the number you want. There are a few options to make this faster: What makes it even worse is that with each fibonacci number you calculate in your list, you don't use the previous numbers you have knowledge of to speed up the computation – you compute each number "from scratch." That just gets worse and worse the higher the number you want to compute. As you can see, it computes the value of Fibonacci(2) three times, and the value of Fibonacci(1) five times. It represents calculating Fibonacci(5) with your function. ![]() The reason for this is that for each number calculated, it needs to calculate all the previous numbers more than once. The primitive recursive solution takes a lot of time.
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