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Coding Interview with Fibonacci Numbers: Down the Rabbit Hole

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27 Dec 2023CPOL24 min read 8.5K   43   5   2
Imaginary coding interview using Fibonacci numbers as single showcase for a surprising variety of programmers' skills.
This article presents the imaginary coding interview using Fibonacci numbers as a single showcase for the set of tasks. Despite the simplicity of this numeric sequence, we will see how it could be used to demonstrate a variety of software engineering skills: from basic coding through improving the code quality up to asymptotic complexity analysis and, for the dessert, to some algorithmic trick which appeared to be quite useful in some real-world applications.


What could software developers say about Fibonacci numbers and coding interviews? While most of us would likely agree that interview is aimed to expose the best of our knowledge and experience, Fibonacci numbers probably have a reputation of coding exercise for beginners. Is it possible, however, to blend these two notions into something useful?

Thinking a bit more about this subject, I came to quite an unexpected conclusion: despite the  seeming simplicity of Fibonacci numbers, we can use them to develop several code snippets using a plethora of programming techniques. We can also perform some decent asymptotic complexity analysis without diving into the math too deeply.

Putting it all together, I could imagine a proof-of-concept scenario of such interview session. This scenario is definitely artificial: the interviewer always asks the question expected by the candidate, who is ready to answer. In order to make things more close to reality of junior-grade interview, I used a programming language which is not completely familiar to me and performed all the math without looking at the textbooks (except for one beautiful formula) trading, therefore, rigor for simplicity. I should also notice that code and math used in this article are in no way something new so I would never claim for authorship. Finally, you will surely notice that I've written much more text than code. Mea culpa. I'm a fond of these words written by E. W. Dijkstra about 50 years ago [1, p. 39]:

... it is hard to claim that you know what you are doing unless you can present your act as a deliberate choice out of a possible set of things you could have done as well.

Nevertheless, I hope someone could find my writing at least amusing or even learn something useful.

Background and Conventions

The intended audience of this article are beginning software developers and/or students learning the basics of the computer science. Seasoned developers are welcome but they will unlikely see anything new for them.

Code snippets are written using Python (v3.10 from Anaconda 2023.03) so intermediate knowledge of this language is needed.

The text below uses some typesetting conventions to separate different parts of the dialogue:

  • Plain text (like this) is used for the candidate speech which comprises most of the text below.
  • Interviewer questions look like quotations:
    Could you tell us more about this topic, please?
  • Comments which are expected to be useful to the reader but inappropriate to be said aloud during the real interview are like this:

    Note the use of assignment statement.

Level 0: A Variety of Implementations

Ok, let's begin. How would you like to implement a program calculating a Fibonacci numeric sequence?

Our imaginary interviewer prefers the ``open'' questions, leaving much of the initiative to the candidate.

I would start from the definition of this sequence. Let's specify it with following recurrent rule:

$ F_0 = 0,\; F_1 = 1, \; F_i = F_{i-1} + F_{i-2} \; \mbox{for $i \ge 2$}. $

This ``user requirement'' seems to be too obvious to clarify it explicitly, but software developers should always make sure they understand their client properly. Despite the Fibonacci sequence looking too simple to make a mistake, some textbooks define it differently, omitting \(F_0\). I've also seen a programming exercise where Fibonacci sequence had been extended towards \(-\infty\), that is, \(F_{-1}=1\), \(F_{-2}=-1\), \(F_{-3}=2\) and so on.

Assuming that we could present our solution with printing the first \(n\) Fibonacci numbers, this piece of code seems be sufficient to begin a discussion:

n = int(input("Enter series length: "))

assert(n >= 1)

past_item, current_item = 1, 0

for i in range(n):
    past_item, current_item = current_item, past_item + current_item 

Note the use of so called ``parallel assignment'' statement which often allows avoiding temporary variables.

Also, this code takes me back to my high school years, except that I would use Pascal or even BASIC that time.

Looks good for beginning. Now, do you see any way to improve your solution?

User requirements are the top priority for improvement, but since we currently have no new ones, we can devote our attention to making our code better according to software engineering best practices.

Our current implementation doesn't meet two important criteria: the code is neither reusable nor maintainable. Indeed, we can see two interspersed activities: calculating Fibonacci numbers and performing input/output operations. We can't, therefore, easily use these calculations for some other purpose or modify them without understanding the output part and vice versa.

Talking about ``problems'' with reusability and maintainability of this simple peace of code is, surely, exaggerated. However, it could be very important to show that you are aware of these concepts and value them.

Therefore, we should separate Fibonacci numbers calculation from output.

Agree. Given that our time is limited, let's omit the I/O details and discuss the computation of our numbers only.

Current implementation uses global variables to keep the current and the past produced numbers (or the computation state). This, for example, makes it impossible to produce multiple number series at once. In order to improve that, we should store our state into some appropriate data structure. In our case, any data structure capable of storing two integer numbers is sufficient (e.g., pair, two-element array or list, whatever) but the most convenient choice would be an object in a sense of object-oriented programming paradigm because this makes it easier to integrate our solution with other code in our language of choice which encourages using this approach.

Despite of elements of functional programming becoming more and more popular, Python, in my opinion, remains more object-oriented than functional.

Next design decision we must take right now is even more important than the previous one: how should we produce our numbers? Different problems may require different ``access patterns'': sometimes, it is convenient to produce numbers in sequence, one by one or in blocks of some size, but sometimes we need only a subset of those numbers in random order.

Why do you think this decision is so important?

Because our choice directly determines the lifetime of our code. Most of the programs are not carved in stone, so we must modify them as time goes. Properly chosen abstraction can withstand these changes longer. However, we must also understand that given new and new requirements to our software, we will eventually overcome the reserve of flexibility of the original design choices and should be ready to take a step forward and review them, sometimes completely.

Frederick Brooks said ``plan to throw one away'' meaning the prototype version of some software system. I feel, this is also true regarding any other code as well.

So far, let's begin from one possible solution written in pretty old-school object-oriented style which could also suit well for other languages than Python:

class OldSchool:
    def __init__(self):
        self._past_item = 1
        self._current_item = 0

    def next_value(self):
        self._past_item, self._current_item = self._current_item, self._past_item + self._current_item

        return self._past_item 

Putting this code to separate source file, say,, we create a module which could be used in multiple programs.

In order to produce a sequence of Fibonacci numbers, we should now create an instance of OldSchool class and call next_value as many times as needed to retrieve one more item of this sequence:

import fib

fo = fib.OldSchool()

print([fo.next_value(), fo.next_value(), fo.next_value()])
print([fo.next_value(), fo.next_value()]) 

The output is:

[0, 1, 1]
[2, 3] 

Naming conventions in the code for this article came from my habit of using fully qualified names of imported entities (possibly introducing some short alias for long module name) whenever possible. Your preferences may vary.

Even more, we can create multiple objects and produce many sequences simultaneously:

fo1 = fib.OldSchool()
fo2 = fib.OldSchool()
print([[fo1.next_value(), fo1.next_value()], [fo2.next_value()], [fo1.next_value()], [fo2.next_value()]])
print([[fo1.next_value()], [fo2.next_value(), fo2.next_value()], [fo1.next_value()], [fo2.next_value()]]) 

The output is:

[[0, 1], [0], [1], [1]]
[[2], [1, 2], [3], [3]]

Given the code snippets above, we can conclude that our current implementation can be reused in various ways (either in different programs or producing multiple sequences within one program). Also, since the implementation details are hidden and can be changed without a need to rework the ``client'' code, we have also achieved some better degree of maintainability.

Automated unit tests could be even more important for maintainability since they provide early diagnostics of errors inevitable when code is changed. Making code reusable is, in turn, absolutely necessary for unit testsing.

Ok. I'm agree about the importance of unit tests but let's postpone discussing them for a time being. You have mentioned above that an alternative implementation is possible which generates blocks of numbers. Do you have any arguments to prefer producing the same sequence one item by one?

Yes, we could possibly change our implementation to something like this:

class Blocks:
    def __init__(self):
        self._past_item = 1
        self._current_item = 0

    def next_values(self, n=1):
        assert (type(n) is int and n >= 0)

        if n < 0:
            raise ValueError("Block length must be non-negative")

        result = []

        for i in range(n):
            self._past_item, self._current_item = self._current_item, self._past_item + self._current_item

        return result 

This could greatly simplify our current task:

fb = fib.Blocks()


The output is:

[0, 1, 1]
[2, 3, 5, 8] 

However, thinking out of the box, we can foresee an unwanted consequence of block-oriented approach: we generate all numbers in advance even if we consume them later item by item. Given that block size could be large, the unnecessary memory consumption and performance loss due to adding items to the list could be noticeable. So, it seems to be preferable to produce our sequence item by item which we can always pack to the list when needed using, for example, list comprehensions:

fo = fib.OldSchool()
print([fo.next_value() for i in range(5)])

[0, 1, 1, 2, 3] 

While problem with producing too large data blocks looks a bit exaggerated regarding Fibonacci numbers, this could be actual for other sequential data in the real life. Don't miss your chance to show that you are aware of it.

You have mentioned that your approach is somewhat old school and applicable to other languages. Can you show us some more improvements of your code which are specific to Python?

One possible way would be using special method named __call__ to make our code more concise. This kind of syntactic sugar doesn't change neither memory consumption nor execution speed but improves, so to say, our visual experience.

class Callable(OldSchool):
    def __call__(self):
        return self.next_value()

Note we reused our existing code one more time via inheritance.

fc = fib.Callable()
print([fc(), fc(), fc()])
print([fc() for i in range(4)])

The output is:

[0, 1, 1]
[2, 3, 5, 8]

For the purposes of the interview, using special methods also allows you to show that you are aware of some less known language features.

Another improvement could be making our class iterable:

class Iterable:
    def __iter__(self):
        class FibIter:
            def __init__(self):
                self._seq = OldSchool()

            def __iter__(self):
                return self

            def __next__(self):
                return self._seq.next_value()

        return FibIter()

This class looks completely different but it also reuses OldSchool, this time via aggregation.

While our former improvement with making object ``callable'' has been only cosmetic, the latter one has much greater impact: this way, we integrate our code into a generic language infrastructure using so called ``iterator protocol'':

import itertools

fi = iter(fib.Iterable())

print(list(itertools.islice(fi, 2)))
print(list(itertools.islice(fi, 4)))

The output is:

[1, 1]
[2, 3, 5, 8]

Note that itertools module allows us doing much more than simply collecting data from iterable object:

print(sum(itertools.takewhile(lambda x: x < 30, fib.Iterable())))

The output is:


This expression takes first items from the Fibonacci sequence until their value exceeds 30 and calculates their sum, all in one line of code.

Although this style of programming could look quite confusing for beginners, I would strongly advice to learn it.

Indeed, this programming style is very concise and expressive. Can you tell me, please, a bit more about its strong and weak points?

Putting aside expressive power, I could also say this approach does quite efficiently: the code implementing iterable simply produces one item by one as long as they are consumed by caller but no more.

Regarding its weakness, it is naturally applicable to one-pass algorithms consuming sequence items more or less in order of their appearance (one by one or by limited-size chunks), other item access patterns could require other abstractions. Also, as we could see, implementing iterator protocol requires some boilerplate code and could be non-intuitive.

Any ideas how to improve it?

The boilerplate code could be produced automatically if we rewrite our iterable as generator function:

def seq():
    past_item, current_item = 1, 0

    while True:
        yield current_item
        past_item, current_item = current_item, past_item + current_item

This special function (note the use of yield) being called effectively returns an iterable similar to our fib.Iterable but with all the needed machinery created automatically under the hood. It can be used exactly as above:

print(sum(itertools.takewhile(lambda x: x < 30, fib.seq())))

The output is:

So, can generators be considered as ultimate solution for making sequences?

It depends. In most cases, they are indeed more than enough but sometimes we could need an additional control over iteration. For example, some algorithms (e.g., parsers) could require an ability to ``look ahead'' or inspect the incoming item without consuming it. Similarly, some problems could require saving the current state of the iterator and restoring it back. In all these cases, custom iterable object could provide the needed methods.

Level 1: Fighting with Complexity

Well, let's now consider another usecase: we need to produce a subset of Fibonacci sequence referring to its items by index, in random order. Could we probably reuse some code we have recently discussed?

No, unfortunately, we can't. This time, we have got exactly that problem I mentioned before: every decision has its limits. We could try to reuse our existing code like this:

def nth_seq(n):
    assert (type(n) is int and n >= 0)

    if n < 0:
        raise ValueError("Item index must be non-negative")

    return next(itertools.islice(seq(), n, None))

At first glance, this works as expected and looks nice:

print([fib.nth_seq(i) for i in [3, 5, 6, 4]])

The output is:

[2, 5, 8, 3]

However, the performance of this solution is unfeasible. Indeed, producing number given index \(n\) we re-generate the entire subsequence \(F_0\), \(\ldots\), \(F_n\) under the hood. Using big \(O\) notation, the complexity of nth_seq(n) is \(O(n)\).

This complexity may look acceptable when we produce a single number but if we need, say, \(m\) numbers, these \(O(n)\)s are accumulated.

Can you provide some better estimate, please?

Better estimate requires some additional knowledge about \(n\) which could be different for every \(m\). For example, if we try to generate a sequence of randomly shuffled first \(m\) Fibonacci numbers from a set of pre-shuffled indices, we could call nth_seq \(m\) times, once for each value of \(0 \le i < m\), at random order. In this case, we produce \(i+1\) Fibonacci numbers for each \(i\) (remember, we index our sequence from zero) and the total number of produced items is

$ \sum_{i=0}^{m-1}(i+1) = \sum_{i=1}^m i = (1 + m) \frac{m}{2} $

which, in turn, corresponds to complexity \(O(m^2)\).

In real life, I would propose to produce a sequence of Fibonacci numbers first and then shuffle them, if possible, but restricting our problem setting with pre-shuffled indices allows us to showcase this estimate.

What about using recursive formula mentioned at the beginning of our conversation?

Putting this formula directly to the code, we could get quite an elegant function:

def nth_rec(n):
    assert (type(n) is int and n >= 0)

    if n < 0:
        raise ValueError("Item index must be non-negative")
    elif n == 0:
        return 0
    elif n <= 2:
        return 1
        return nth_rec(n - 2) + nth_rec(n - 1)

Unfortunately, this function has significantly worse performance estimation. Indeed, if we expand our formula:

$ F_i = F_{i-2} + F_{i-1} = F_{i-2} + (F_{i-2} + F_{i-3}), $

we can see that \(F_{i-2}\) is calculated twice. Therefore, the amount of work needed to calculate the \(i\mbox{-th}\) number is at least twice more than for its pre-predecessor. Since this doubling, in turn, occurs for the \((i-2)\mbox{-th}\) item as well and so on, we can conclude that the lower boundary of our complexity estimate would be

$ \Omega(2^{\lfloor \frac{n}{2}\rfloor}). $

Note this boundary is overly optimistic: we don't take into account the amount of work needed to calculate \(F_{i-3}\).

Looking at this function, we can conclude that it is stepwise, non-decreasing and grows at even values of \(n\):

$ 2^{\lfloor \frac{n}{2}\rfloor} = 2^{\frac{n}{2}} = (\sqrt{2})^{n} \approx 1.414214^{n} \quad \mbox{for even $n$}. $

This quick check leads us to unfortunate conclusion: our nth_rec has exponential complexity which is even worse than our initial attempt.

This estimation in fact sweeps too much under the carpet. Those who would like to see more detailed and precise reasoning could probably find it here.

Any ideas for further improvement?

Yes, we can get rid of the combinatory explosion occurring in this function with technique known as ``memoization'':

class _NthMem:
    def __init__(self):
        self._nums = {}

    def __call__(self, n):
        assert (type(n) is int and n >= 0)

        if n < 0:
            raise ValueError("Item index must be non-negative")
        elif n == 0:
            return 0
        elif n <= 2:
            return 1
            f_n = self._nums.get(n, None)

            if f_n is None:
                f_n = self(n - 2) + self(n - 1)
                self._nums[n] = f_n

            return f_n

nth_mem = _NthMem()

The basic idea of this approach: each Fibonacci number being calculated for the first time is saved to the map using number index as a key. When it is needed later, we get it from the map and reuse instead of recalculating from scratch.

This improvement, looking so simple, has drastic effect:




Recursive implementation will unlikely be able to produce this result in a reasonable amount of time.

Yes, this optimization does real magic. What is the time complexity of memoized implementation?

Let's calculate the cost of \(m\) calls of nth_mem where \(n_{max}\) is the largest Fibonacci number index used within those \(m\) calls. Given that we have a decent dictionary implementation providing us a constant amortized cost of individual add/lookup operations, we can say that the cost of filling the table of \(n_{max}\) Fibonacci numbers is \(O(n_{max})\). Indeed, calculating each next number from known predecessors would require a constant amount of work (getting two numbers from dictionary and storing their sum). After all those \(n_{max}\) numbers are calculated, all subsequent calls of fib_mem also require constant time (getting the precalculated number from dictionary).

Therefore, we have two cases: when \(n_{max}\) dominates over \(m\), the overall time complexity is \(O(n_{max})\), otherwise we have \(O(m)\). Putting it all together, we can write the resulting estimate as \(O(\max(n_{max}, m))\) or \(O(n_{max} + m)\).

By the way, this estimate particularly means that if we use nth_mem to produce a sequence of Fibonacci numbers calling it with index growing from \(0\) up to some \(n\) the entire time complexity of this process would also be \(O(n)\), the same as we could get using some of our sequential implementations, say, seq.

Looks great! From initial quadratic through exponential, we could finally get linear time! I should notice, however, that fib_mem contains code related to our application domain (calculating Fibonacci numbers) together with implementation details of memoization making it less clear. Can we do better?

Following our initial goal of making our code reusable, we can separate the implementation of table lookup from some specific calculations using Python feature known as decorator:

def memoize(f):
    class _MemF:
        def __init__(self, f):
            self._vals = {}
            self._f = f

        def __call__(self, x):
            y = self._vals.get(x, None)

            if y is None:
                y = self._f(x)
                self._vals[x] = y

            return y

    return _MemF(f)

def nth_dec(n):
    assert (type(n) is int and n >= 0)

    if n < 0:
        raise ValueError("Item index must be non-negative")
    elif n == 0:
        return 0
    elif n <= 2:
        return 1
        return nth_dec(n - 2) + nth_dec(n - 1)

As we can see, function nth_dec keeps the definition of Fibonacci number free from implementation details, while its performance is comparable to nth_mem. Meanwhile, the memoize decorator can also be used to add memoization to other functions with no efforts.

Level 2: Curiouser and Curiouser

Coming back to that you said earlier, can we use some of memoized implementations, nth_mem or nth_dec, as a replacement of our sequence-oriented code given they both have an \(O(n)\) time complexity when it comes to produce the first \(n\) Fibonacci numbers?

I wouldn't recommend making such a decision because this estimation is asymptotic and includes a hidden coefficient which could make a substantial difference in realistic scenarios. Putting it simpler, the sequential generator simply calculates the next number updating a pair of state variables while our memoized recursive function ends up with maintaining a dictionary requiring therefore more efforts. Besides that, we should also take into account memory consumption.

By the way, can we possibly estimate memory complexity of memoization as well?

Unlike time complexity, where we can usually analyze the performance of some data structure putting aside the nature of stored objects, memory complexity inherently depends from the nature of data, so we can't make any statements in general.

Fibonacci numbers, however, allows us to do something better. In order to make things simpler, we won't take into account the cost of bookkeeping, that is, the amount of memory needed to store a dictionary with \(n\) entries (this amount greatly depends on the implementation of this dictionary) and try to estimate the memory needed to store these numbers themselves.

Let's begin from simple fact that memory needed to store some number \(x\) can be estimated by its logarithm \(\log x\) (for asymptotic analysis, the logarithm base can be arbitrary).

Can we estimate a logarithm of some Fibonacci number? Fortunately, we can do that using the so called Binet's formula:

$ F_i = \frac{\varphi^i - \psi^i}{\sqrt 5}, $


$ \varphi = \frac{1 + \sqrt 5}{2} \approx 1.618034, \quad \psi = 1 - \varphi = \frac{1 - \sqrt 5}{2} \approx -0.618034. $

Let's try to estimate the lower boundary for this expression. For beginning, we can notice that \(\psi^i < 0\) for odd values of \(i\). Therefore, we can state that

$ F_i > \frac{\varphi^i}{\sqrt 5} \qquad \mbox{for odd $i$}. $

For even values of \(i\), we can use the fact that \(F_{i+1} > F_{i}\), therefore

$ F_i > F_{i-1} > \frac{\varphi^{i-1}}{\sqrt 5} \qquad \mbox{for even $i$}. $

The latter boundary for even indices is also applicable to odd indices as well, although it would be less accurate. However, it allows us to get a common estimate:

$ F_i > \frac{\varphi^{i-1}}{\sqrt 5}, $


$ F_i = \Omega(\varphi^i). $

The same approach is applicable to logarithm of \(F_i\):

$ \log F_i > \log \frac{\varphi^{i-1}}{\sqrt 5} = (i - 1) \log\varphi - \log\sqrt{5}, $


$ \log F_i = \Omega(i). $
Looks very impressive! I've never had an idea of this relationship between Fibonacci number and its index. Can we possibly get some evidence these estimations are correct?

We can try to calculate this lower boundary for some part of the sequence and see how it looks like. By the way, this would be a good opportunity to see how our sequence generator works together with some tools from standard library:

phi = (1 + math.sqrt(5)) / 2

logF_lb = map(lambda i, F_i: ((i - 1) * math.log(phi) - math.log(math.sqrt(5)), math.log(F_i)), 
              itertools.islice(fib.seq(), 1, None))

print(list(itertools.islice(logF_lb, 10)))

This code snippet is written in functional style. We process two infinite lists using built-in map function: the first is just a stream of indices and the second is a sequence of Fibonacci numbers. Starting zero item of the sequence is omitted to avoid ValueError raised by math.log. The lambda expression passed to map calculates both the lower boundary estimate and effective logarithm of Fibonacci number returning them as a pair.

Note that the real computations occur only when resulting sequence is effectively consumed, that is, when we extract the first 10 items from logF_lb converting them to list and printing.

The produced output is (split by lines for better readability):

[(-0.8047189562170503, 0.0), (-0.3235071311574468, 0.0), 
 (0.1577046939021567, 0.6931471805599453), (0.6389165189617603, 1.0986122886681098), 
 (1.1201283440213636, 1.6094379124341003), (1.601340169080967, 2.0794415416798357), 
 (2.082551994140571, 2.5649493574615367), (2.563763819200174, 3.044522437723423), 
 (3.0449756442597775, 3.5263605246161616), (3.5261874693193813, 4.007333185232471)]

As we can see, each lower boundary estimate (the first item) is indeed less than the corresponding logarithm value (the second item).

So, how can we use these estimates to get memory complexity?

As I've said earlier, amount of memory needed to store some number is proportional to its logarithm. Let's assume that we have produced some Fibonacci numbers whose indices are less or equal to \(n\). In this case, the amount of memory needed to store memoized values can be estimated as

$ \sum_{i=1}^{n} \log F_i. $

Note that we ignore \(F_0=0\) since its logarithm is undefined. Dropping fixed amount of items at the beginning of sequence is allowed when we do asymptotic analysis.

The lower boundary of this sum:

$ \sum_{i=1}^{n} \log F_i > \sum_{i=1}^{n} \left( (i - 1) \log\varphi - \log\sqrt{5} \right). $

The right part of this inequality can in turn be transformed as

$ \begin{align*} \sum_{i=1}^{n} \left( (i - 1) \log\varphi - \log\sqrt{5} \right) & = \log\varphi \sum_{i=1}^{n} (i - 1) - n \log\sqrt{5} \\ & = \log\varphi\, (n+1) \frac{n}{2} - n \left(\log\varphi+\log\sqrt{5}\right)\\ & = \frac{\log\varphi}{2} n^2 - \frac{\log\left(5\varphi\right)}{2} n. \end{align*} $

Putting it all together, we have

$ \sum_{i=1}^{n} \log F_i = \Omega(n^2). $

So, looks like we are doomed either to spend time calculating each Fibonacci number from the beginning or need a plenty of memory to use memoization. Both alternatives look painful when it comes to working with reasonably large numbers.

Is there any way to escape from this whack-a-mole situation? Could we possibly use the Binet's formula to calculate the needed Fibonacci number directly?

I wouldn't recommend doing this because using Binet's formula assumes working with irrational numbers which can't be represented at full precision. Also, the built-in support for real numbers usually implements floating-point arithmetics following the ubiquitous IEEE 754 Standard. Assuming we use double-precision data format, we can't expect that integers greater than \(2^{53}\) can always be represented without a loss of accuracy. Even worse, we can get inaccurate results even well below this upper boundary due to various subtle roundoffs and other pitfalls of floating-point computations.

Fortunately, Fibonacci numbers have some mathematical properties which allow us to propose one very elegant solution. Let's recall our sequential implementation: it has been based on saving two last items of Fibonacci sequence and updating them on each step. This process can be expressed using vector notation:

$ \left\langle F_{i-1}, F_i \right\rangle = f(\left\langle F_{i-2}, F_{i-1}\right\rangle), $

where \(f\) --- some transformation. Fibonacci sequence belongs to the class of so called linear recurrencies and this transformation can be represented as matrix multiplication:

$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \cdot \begin{pmatrix} F_{i-2} \\ F_{i-1} \\ \end{pmatrix} = \begin{pmatrix} 0 \cdot F_{i-2} + 1 \cdot F_{i-1} \\ 1 \cdot F_{i-2} + 1 \cdot F_{i-1} \\ \end{pmatrix} = \begin{pmatrix} F_{i-1} \\ F_i \\ \end{pmatrix}. $

Note that we used this matrix product to transform current state of our sequential Fibonacci sequence generator into the next state, that is, to calculate next sequence item which is always kept at the second item of state vector. Doing the next step requires one more matrix product:

$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \cdot \begin{pmatrix} F_{i-1} \\ F_{i} \\ \end{pmatrix} = \begin{pmatrix} F_{i} \\ F_{i+1} \\ \end{pmatrix}. $

Substituting our previous matrix product, we can see the pattern:

$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \cdot \begin{pmatrix} F_{i-2} \\ F_{i-1} \\ \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix}^2 \cdot \begin{pmatrix} F_{i-2} \\ F_{i-1} \\ \end{pmatrix} = \begin{pmatrix} F_{i} \\ F_{i+1} \\ \end{pmatrix}. $

That is, if we like to advance our current generator state by \(n\) steps forward, we can do that immediately by multiplying this state by our matrix raised to the power of \(n\). And here comes the magic: we can calculate this power in logarithmic time! The magic potion used to do that is known as exponentiation by squaring.

So, we can replace the Binet's formula with this one:

$ \begin{pmatrix} F_{n-1} \\ F_n \\ \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix}^n \cdot \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}, $

getting \(F_n\) in \(O(\log n)\) time (with \(F_{n-1}\) coming as a free bonus).

The implementation of this algorithm is as follows:

def _mul_2x2(x, y):
    return [[x[0][0] * y[0][0] + x[0][1] * y[1][0], x[0][0] * y[0][1] + x[0][1] * y[1][1]],
            [x[1][0] * y[0][0] + x[1][1] * y[1][0], x[1][0] * y[0][1] + x[1][1] * y[1][1]]]

def _pow_2x2(x, n):
    res = [[1, 0], [0, 1]]
    sq_i = x

    while n > 0:
        if n & 1 != 0:
            res = _mul_2x2(res, sq_i)
        sq_i = _mul_2x2(sq_i, sq_i)
        n >>= 1

    return res

def _mul_2x2_v(x, y):
    return [x[0][0] * y[0] + x[0][1] * y[1], x[1][0] * y[0] + x[1][1] * y[1]]

def nth_mtx(n):
    assert (type(n) is int and n >= 0)

    if n < 0:
        raise ValueError("Item index must be non-negative")
        A_n = _pow_2x2([[0, 1], [1, 1]], n)
        st_n = _mul_2x2_v(A_n, [1, 0])
        return st_n[1]

Let's try it:

print([(i, len(str(fib.nth_mtx(i)))) for i in [200, 400, 600]])

As you can see, the number of decimal digits in produced Fibonacci numbers indeed grows linearly:

[(200, 42), (400, 84), (600, 126)]

And finally, this statement is executed instantly:

print(fib.nth_mtx(1000000) % 1000000)



I should also notice that if you have a pair of consecutive Fibonacci numbers, say \(\left\langle F_{i-1}, F_i \right\rangle\), you can use the same approach to calculate the \(n\)-th number relatively to \(i\), that is, \(\left\langle F_{i+n-1}, F_{i+n} \right\rangle\). Similarly, we can step back, this time using another matrix. In case of Fibonacci sequence this matrix can also be written down with minimum effort.

Looks great! However, is there any real-life application of this formula or it is nothing more than a salon trick only capable of entertaining me and all those curious people who had the patience to read our conversation till this point?

This approach can be used for any linear recurrence if you know the corresponding matrix. Particularly, many popular pseudo-random number generators are based on linear recurrence of some kind and this theory allows us to split them on a number of independent ``streams''. These streams, in turn, are very useful when you need to parallelize or distribute your calculations. (More about this interesting problem could be read here.)

That's enough for today. Thank you for the opportunity to look at the Fibonacci numbers from this unusual point of view!


Our imaginary interview came to its end. Our imaginary candidate got his imaginary ``Yes'' from no less imaginary interviewer. Coming back to the real world, what aspects of software engineering and Python skills could we expose? Let's list them in order of appearance:

  • A good habit of clarifying all the details of the task before coding it.
  • Basic programming skills, like using the I/O and loops.
  • Understanding the difference between simply solving the problem and creating reusable and maintainable code --- very important for software engineers, but sometimes missed by people who came to software development from academia.
  • Understanding the fact that software design is a path to a proper balance between various possible solutions and, more importantly, that such balance is not carved in stone and will eventually require making some breaking changes.
  • Using object-oriented approach to improve our initial solution.
  • Using list comprehensions.
  • Making our code more readable with Python special methods.
  • Integrating our solution with the standard libraries provided by Python via iterator protocol.
  • Reusing code either by inheritance or by aggregation.
  • Doing elements of ``functional-like'' programming with itertools.
  • Making our solution even more clear with Python generator.
  • Performing some basic complexity analysis with ``big-\(O\) notation''.
  • Using recursion.
  • Improving recursive solution speed with memoization.
  • Implementing reusable memoization with Python decorator.
  • Using math to solve a problem which looked hopeless from purely engineering point of view (a quite rare stroke of luck in the real life, but in my opinion, a software engineer should be aware of this possibility and look for it whenever possible).

Looks not so bad for a ``freshmen-grade exercise'', doesn't it? Is this set of skills really enough to pass the interview? Different roles require different blend of design, math and mere coding, so there is no single answer. From my personal point of view, being aware about the discussed topics should give you at least some additional points and shouldn't harm.

Regarding my own takeaway experience gained from this article, I would consider it positive: while I knew some things in advance before starting it, quite a plenty of others were explored just in progress. Hope you learned something useful to you as well.


  • [1] Edsger W. Dijkstra. Notes on structured programming. In Structured Programming, pages 1-82. Academic Press Ltd., London and New York, 1972. (Available online.)


  • 27th December, 2023: Initial version


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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Comments and Discussions

GeneralMy vote of 5 Pin
Ștefan-Mihai MOGA29-Dec-23 18:44
professionalȘtefan-Mihai MOGA29-Dec-23 18:44 
GeneralRe: My vote of 5 Pin
Member 420181330-Dec-23 4:31
Member 420181330-Dec-23 4:31 
Thank you very much for support!

Happy New Year!

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