Recursion function in Python

Question

Consider this basic recursion in Python:

def fibonacci(number):
    if number == 0: return 0
    elif number == 1:
        return 1
    else:
        return          


        

Answer 1

Short Answer

Each time Python "sees" fibonacci() it makes another function call and doesn't progress further until it has finished that function call.

Example

So let's say it's evaluating fibonacci(4).

Once it gets to the line return fibonacci(number-1) + fibonacci(number-2), it "sees" the call fibonacci(number-1).

So now it runs fibonacci(3) - it hasn't seen fibonacci(number-2) at all yet. To run fibonacci(3), it must figure out fibonacci(2)+fibonacci(1). Again, it runs the first function it sees, which this time is fibonacci(2).

Now it finally hits a base case when fibonacci(2) is run. It evaluates fibonacci(1), which returns 1, then, for the first time, it can continue to the + fibonacci(number-2) part of a fibonacci() call. fibonacci(0) returns 0, which then lets fibonacci(2) return 1.

Now that fibonacci(3) has gotten the value returned from fibonacci(2), it can progress to evaluating fibonacci(number-2) (fibonacci(1)).

This process continues until everything has been evaluated and fibonacci(4) can return 3.

To see how the whole evaluation goes, follow the arrows in this diagram:

Answer 2

You can use the rcviz module to visualize recursions by simply adding a decorator to your recursive function.

Here's the visualization for your code above:

The edges are numbered by the order in which they were traversed by the execution. The edges fade from black to grey to indicate order of traversal: black edges first, grey edges last.

(I wrote the rcviz module on GitHub.)

Answer 3

You can figure this out yourself, by putting a print function in the function, and adding a depth so we can print it out prettier:

def fibonacci(number, depth = 0):
    print " " * depth, number
    if number == 0:
        return 0
    elif number == 1:
        return 1
    else:
        return fibonacci(number-1, depth + 1) + fibonacci(number-2, depth + 1)

Calling fibonacci(5) gives us:

5
 4
  3
   2
    1
    0
   1
  2
   1
   0
 3
  2
   1
   0
  1

We can see that 5 calls 4, which goes to completion, and then it calls 3, which then goes to completion.

Answer 4

I would really recommend that you put your code in the Python tutor.

You will the be able to get it on the fly. See the stackframe, references, etc.

Answer 5

  def fib(x):
    if x == 0 or x == 1:
    return 1
  else:
    return fib(x-1) + fib(x-2)

print(fib(4))

Answer 6

does the 'finobacci(number-1)' completes all the recursion until it reaches '1' and then it does the same with 'fibonacci(number-2)' and add them?

Yes, that's exactly right. In other words, the following

return fibonacci(number-1) + fibonacci(number-2)

is equivalent to

f1 = fibonacci(number-1)
f2 = fibonacci(number-2)
return f1 + f2