Tools: A Guide to Fibonacci Series and Recursion in Go Language

What is recursion? ## Problem Statement ## Algorithm ## Visual example ## Final Thougths The Fibonacci sequence is one of the most common problems you'll solve throughout your software career. Its implementation can be as simple or complex as you want. One of the most frequently used solutions is recursion—a core concept in computer science. Whether you're learning computer science, preparing for your next interview, or simply reinforcing old concepts, join me in writing a Fibonacci sequence using recursion with Go. Recursion means breaking a problem into smaller subproblems. Sometimes you'll add or remove something f(n), or you'll need to adjust the solution f(n - 1). In some cases, you might solve the problem for half of the dataset. In the Fibonacci sequence, the function calls itself with smaller inputs. Each recursive call works toward the base case. Given n calculate the nth Fibonacci number. The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, starting with 0 and 1. The sequence begins as follows: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... Input: an integer where the series will stop. Output: the sequence from 0 to n. Now that we've identified the input, output, and approach, it's time to implement it. First, since the function calls itself, I need to prevent infinite loops. To do this, I define a base case—a condition that tells the function when to stop. Every recursive function consists of two parts: the base case (when to stop) and the recursive case (when to call itself). I've already defined the base case. Now it's time to declare the recursive case. For Fibonacci, I need to call the function twice with smaller inputs—specifically n - 1 and n - 2. You might be asking: "Why does it call itself twice?" Good question. Each number is the sum of the two preceding ones. The Fibonacci function is mathematically defined as: F(n) = F(n-1) + F(n-2) for n > 1. 1 So to compute F(4), you need: F(3) (which needs F(2) and F(1)) F(2) (which needs F(1) and F(0)) Easy, right? And there you have it—you've solved the Fibonacci sequence using recursion. But I'm afraid to say it's the worst solution. Why learn something that's a bad solution? Well, because it's the core concept for complex solutions like dynamic programming or memoization. If you're a React programmer, you've used memoization plenty of times with the useMemo or useCallback hooks. So that's why you need to learn recursion first. An example will make this clearer. Let's find the Fibonacci sequence for the number 4. Let's trace through the recursion step by step: F(4) calls F(3) and F(2) F(3) calls F(2) and F(1) Notice how F(2) is calculated twice and F(1) is calculated three times. This redundancy is why the naive recursive solution is inefficient—it has exponential time complexity of O(2ⁿ). For larger values of n, the same calculations are repeated thousands or even millions of times. This is exactly why we need optimization techniques like memoization or dynamic programming, which store previously calculated values to avoid redundant work. Recursion is best used when it makes the solution clearer. When you call a function from another function, the calling function pauses in a partially completed state. Imagine what this does to memory. Despite its drawbacks, recursion powers many important algorithms, so it's worth understanding how it works. If you want to dive deeper or practice, try other exercises like factorial or the Tower of Hanoi. Happy coding, and I'll see you in the next one! Templates let you quickly answer FAQs or store snippets for re-use. Are you sure you want to hide this comment? It will become hidden in your post, but will still be visible via the comment's permalink. Hide child comments as well For further actions, you may consider blocking this person and/or reporting abuse CODE_BLOCK: if n <= 1 { return n } Enter fullscreen mode Exit fullscreen mode CODE_BLOCK: if n <= 1 { return n } CODE_BLOCK: if n <= 1 { return n } CODE_BLOCK: return fibonacci(n-1) + fibonacci(n-2) Enter fullscreen mode Exit fullscreen mode CODE_BLOCK: return fibonacci(n-1) + fibonacci(n-2) CODE_BLOCK: return fibonacci(n-1) + fibonacci(n-2) CODE_BLOCK: * `F(2)` calls `F(1)` and `F(0)` * `F(1)` returns 1 (base case) * `F(0)` returns 0 (base case) * So `F(2) = 1 + 0 = 1` * `F(1)` returns 1 (base case) * So `F(3) = 1 + 1 = 2` Enter fullscreen mode Exit fullscreen mode CODE_BLOCK: * `F(2)` calls `F(1)` and `F(0)` * `F(1)` returns 1 (base case) * `F(0)` returns 0 (base case) * So `F(2) = 1 + 0 = 1` * `F(1)` returns 1 (base case) * So `F(3) = 1 + 1 = 2` CODE_BLOCK: * `F(2)` calls `F(1)` and `F(0)` * `F(1)` returns 1 (base case) * `F(0)` returns 0 (base case) * So `F(2) = 1 + 0 = 1` * `F(1)` returns 1 (base case) * So `F(3) = 1 + 1 = 2` CODE_BLOCK: * `F(1)` returns 1 (base case) * `F(0)` returns 0 (base case) * So `F(2) = 1 + 0 = 1` Enter fullscreen mode Exit fullscreen mode CODE_BLOCK: * `F(1)` returns 1 (base case) * `F(0)` returns 0 (base case) * So `F(2) = 1 + 0 = 1` CODE_BLOCK: * `F(1)` returns 1 (base case) * `F(0)` returns 0 (base case) * So `F(2) = 1 + 0 = 1` - Input: an integer where the series will stop. - Output: the sequence from 0 to n. - F(3) (which needs F(2) and F(1)) - F(2) (which needs F(1) and F(0)) - F(4) calls F(3) and F(2) - F(3) calls F(2) and F(1) - F(2) calls F(1) and F(0) - Final result: F(4) = 2 + 1 = 3