![]() ![]() Simple test and bench mark for all four examples with Fibonacci 40 is giving me: 102334155 - bench simple took - 35742.329ms.Īs you can see the pure recursion is really slow and inefficient in comparison to the other methods. Java 8 fast non recursive Fibonacci Benchmark of the methods Using class and method is also fast and efficient way.: class Fibon: The sequence is easily calculated by just adding the last two items together repeatedly, while some definition might lead you into a much more complex solution from the other direction, involving a fully grown tree of recursions. In some cases could be slightly better than memoization :Ĭalc 6 įibonacci numbers with class and method OOP The Fibonacci sequence is one of the main examples why not to use recursion and think before implementing. It's similar approach to memoization without using recursion. Similar approach to this one is bottom-up approach which is not using recursion but for loop. We are using hint for the values which are calculated already and we are saving calculations - for 3 we have only 1. ![]() Memoization will allow us to calculate each number in the sequence only once: def fibMemo(i): In order to improve the result we can use memoization which is part of dynamic programing. For example we are calculating 3 - 3 times which is a performance issue for bigger numbers. Result: calc 3 calc 4 calc 3 calc 5 calc 3 calc 4 calc 6Īs you can see we are calculating some numbers several times. The simplest is the closest to the definition for producing numbers of Fibonacci is: def fib(i): Fibonacci sequence with bottom-up and not recursion.Fibonacci numbers with memoization and recursion.But memos are in fact a way not just to greatly speed some kinds of recursive algorithms, but also to apply them to some problems that "recurse too deep" without a memo constructed to limit the max call depth. I thought I'd mention that because hardly anyone ever does when answering a "memo" question. This ensures that recursion never has to go more than 100 levels deep to find an argument already memorized in the cache. For example, insert this after the if block: for i in range(100, n, 100): You can worm around that too, by exploiting the cache to call smaller arguments first, working your way up to the actual argument. That's not due to using a cache, it's just inherent in very deep recursion. Then the cache will persist across top-level calls too.īut now there's another problem -) If the argument is large enough (say, 30000), you're likely to get a RecursionError (too many levels of recursive calls). Simplest is to define the cache outside the function, and simply return at once if the argument is in the cache: fib_cache = įib_cache = result = fib(n-1) + fib(n-2) So you're not actually saving the work of calling. The point of "a memo" is to save calls, but you're making recursive calls regardless of whether the result for an argument has already been memorized. You'll find that this is very fast for even very high values: > fibonacci(300)Ģ22232244629420445529739893461909967206666939096499764990979600īut if you define the exact same function without the it gets very slow because it's not benefitting from the cache. If you're allowed to not reinvent the wheel, you could also just use functools.lru_cache, which adds memoization to any function through the magic of decorators: from functools import fibonacci(n): The way to fix this implementation would be something like: def fibonacci(n): The point of memoization is to be careful to avoid doing the computation (in this case a lengthy recursive call) in cases where you already know the answer. This is because setdefault doesn't do any magic that would prevent the arguments from being evaluated before they're passed into the function - you make the recursive call before the dict has checked to see whether it contains the value. Your function isn't memoized (at least not effectively) because you call fibonacci_helper regardless of whether you already have a memoized value. ![]()
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