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Fibonacci without recursiveness in Python - a better way

joaomcteixeira profile image João M.C. Teixeira Updated on ・3 min read

¡Short post! (not so short after the edit)

Some time ago we had a great discussion in a post (see also comments) about From 100% to 0% CPU with memoization applied to Fibonacci series.

From that post comes the idea that explaining recursive functions using the Fibonacci example might fail the very purpose of Python in the first place.

What about solving the Fibonacci problem this way?

EDIT 20 Feb 2021

The original for loop was for i in range(2, i -1):, I corrected it to the following bellow. Thanks to @paddy3118 for spotting that repetition of i.

# first definitions
i = 50  # number of elements in the final Fibonacci series
fib = [0, 1, 1]  # fixed seed
fiba = fib.append  # short cut to the append method

# Calculate the Fibonacci series
for _ in range(3, i): fiba(fib[-2] + fib[-1])
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@paddy3118 in his/her comment also suggested the use of numpy to store the values and just fill the results in the indexes. Obviously, that requires using numpy and that might not be an option, however, if it is, let's investigate how much time it costs to use that approach:

This is how I understood @paddy3118 comment, please Paddy correct me I am wrong.

i = 50
fib = np.zeros(i, dtype=np.int)
fib[0:3] = [0, 1, 1]
for k in range(3, i):
    fib[k] = fib[k - 2] + fib[k - 1]
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In my machine with Jupyter %%timeit I receive:

The slowest run took 5.29 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 5: 23 µs per loop
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From which: 100000 loops, best of 5: 2.42 µs per loop refer to the array creation and 10000 loops, best of 5: 21.5 µs per loop to the for loop.

While if I %%timeit the approach with lists, we have for the whole process:

100000 loops, best of 5: 4.66 µs per loop
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Actually, it is much faster to perform this operation with lists. Maybe contrarily to expectations?

Is that what you were referring to @paddy3118 ? Thanks for your comment, I enjoy it a lot!

Second EDIT

On a later discussion with a friend, we saw that if the list grows considerably, the append version becomes much slower than the version where the list is defined beforehand (see example below). Hence, if the number i is known, and that number is large, the version presented next is much faster.

This latter version avoids the list being (internally) copied over to larger allocated memory blocks as new values are appended. So, depending on the size of the target lists you expect to produce and on the performance requirements, you could choose one or the other. I believe this last version is more in line with what @paddy3118 was referring to before.

i = 50_000
fib = [0 for _ in range(i)]
fib[1:3] = [1, 1]

for k in range(3, i):
    fib[k] = fib[-2] + fib[-1]
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Runs in:

100 loops, best of 5: 4.1 ms per loop
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While the first version with i = 50_000 runs in:

10 loops, best of 5: 68.8 ms per loop
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I started this post to discuss that the Fib series might not be a good example to explain recursion. We ended up finding a very nice situation where Fib can (maybe) serve as a more adequate example. I will look forward to writing a small post on how lists work internally in Python and explain the behavior here observed.

Cheers,

Discussion (5)

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Paddy3118 • Edited

Hi again, I was researching a similar sequence, the padovan sequence which is like the fibonacci, but P[n] = P[n-2] + P[n-3]. In that code I keep only the last four terms of a list turned into a deque with the following code:

def padovan_r() -> Generator[int, None, None]:
    last = deque([1, 1, 1], 4)
    while True:
        last.append(last[-2] + last[-3])
        yield last.popleft()
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Similar could be done to save memory in Fibonacci.

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João M.C. Teixeira Author

Hi,
That is also a very nice implementation if you don't want to keep a list of the whole sequence. Going forward in the discussion, we can actually avoid using deque and increase the speed almost by two.

def padovan_j():
    last = 1
    prev1 = 1
    prev2 = 1
    prev3 = 1
    while True:
        yield prev3
        last = prev2 + prev3
        prev1, prev2, prev3 = last, prev1, prev2
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Paddy3118
  • You reuse name i in the for loop.
  • Since you know the total number of terms, you could create the whole array at the start, with the initial values followed by anything; then loop over an index to create next values without needing to append, which takes time.
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João M.C. Teixeira Author

Hi, thanks a lot for your comment. I updated the post in response. Let me know your thoughts. Cheers!

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João M.C. Teixeira Author

added more to the cause