Javascript Fibonacci method

Question

So I created the fibonacci series, but the way I found on w3resources is a bit confusing to me as a beginner. What exactly happens when n = 2? s becomes fibonnacci_series(1), but what does this even imply? In the first place why do we have var fibonacci_series = function(n) , why not just have a function called function fibonacci_series(n)?

var fibonacci_series = function (n)   
{  
  if (n===1)   
  {  
    return [0, 1];  
  }   
  else   
  {  
    var s = fibonacci_series(n - 1);  
    s.push(s[s.length - 1] + s[s.length - 2]);  
    return s;  
  }  
};  

 console.log(fibonacci_series(8));

score 2 · Answer 1 · edited May 23 '17 at 11:52

2

These two lines do almost the same thing (see here for more details but as said in the comments, this is probably not relevant for you right now):

var fibonacci_series = function (n)   

function fibonacci_series(n)

Both define a function called fibonacci_series that take n as parameter, it's just a variant.

The fibonacci series function is a recursive function, it means that it calls itself in its body (e.g there is fibonnacci_series(n-1) inside the function).

When n = 2, s equals the value of fibonnacci_series(1), which is what is returned by the if statement.

edited May 23 '17 at 11:52

Community

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1

answered Jun 09 '16 at 19:28

Graham Slick

6,692
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There are actually some subtle differences, but they are probably irrelevant for the OPs question. See [for example](http://stackoverflow.com/a/5857498/1250301) – Matt Burland Jun 09 '16 at 19:30

score 2 · Answer 2 · answered Jun 09 '16 at 19:32

2

When n = 2, s becomes the result of the recursive call of n - 1 (which is [0,1]). Then, s.push(s[s.length - 1] + s[s.length - 2]) is pushing the addition of the 2nd last element with the last element, which is how the fibonacci sequence is obtained.

This is a reccursive implementation of the fibonacci sequence. It can also be done through iteration.

answered Jun 09 '16 at 19:32

Chris Wissmach

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When s = 2, s = fibonacci_series(1). What does this mean? I dont understand what fibonacci_series(1) exactly means. Does this just mean s = [0,1]? – Jason Genova Jun 09 '16 at 19:37
Basically, yes. The statement `s = fibonacci_series(n - 1)` is a recursive call. Since `n = 2` in this case, the function will be called again with `n = 1`. If you look at the return value for the case when `n = 1`, you'll see that it returns `[0,1]`. Thus, `s = [0,1]`. – Chris Wissmach Jun 09 '16 at 19:44

score 1 · Accepted Answer · edited May 23 '17 at 12:08

What exactly happens when n = 2? s becomes fibonacci_series(1), but what does this even imply?

This means you're going to get the same result as if you had called fibonacci_series(1) by itself. So s will be [0, 1]. Afterwards, you add the sum of those two numbers to the end of the array, resulting in [0, 1, 1]. This relationship (Nm = N(m-1) + N(m - 2)) is what defines the Fibonacci sequence.

Writing it in the this way means that this same process will happen recursively. Following this logic, try writing down what would happen if n = 3. What would happen during the first call to fibonacci_series? How would the next call to fibonacci_series change what the function returns?

In the first place why do we have var fibonacci_series = function(n), why not just have a function called function fibonacci_series(n)?

For your purposes, there's really no difference. The technical answer is that the first is a function expression and the second is a function declaration. If you're interested in the details of how they're different, check out this Stack Overflow question.

wow. This thing suddenly makes all sense. It was really creative how they thought of using .length. Honestly would have never thought of that. I was just doing it iteratively before. Thanks so much — Jason Genova, Jun 09 '16 at 19:40
@Munt No worries! Not going to lie, I remember this thing blowing my mind when I first learned about it. — Mike Cluck, Jun 09 '16 at 19:41

Javascript Fibonacci method

3 Answers3