I attempted to make better Fibonacci sequence calculation algorithm in C++. This was the best I could:

constexpr unsigned fibonacci(unsigned n) {
    unsigned result[2] = {1, 0};
    for (unsigned i = 1; i < n; ++i)
        result[i%2] = result[0] + result[1];
    return result[1 - n%2];

...which runs at O(n) time complexity and O(1) space complexity.
Is there any better?

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    \$\begingroup\$ Yes. Check out mathworld.wolfram.com/FibonacciQ-Matrix.html. Combined with exponentiation by squaring it yields \$O(\log{n})\$ time complexity, while still \$O(1)\$ space. \$\endgroup\$ – vnp May 12 '17 at 1:59
  • \$\begingroup\$ @vnp That involves multiplication. Doesn't multiplication have worse complexity than addition? \$\endgroup\$ – Dannyu NDos May 12 '17 at 2:09
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    \$\begingroup\$ Everything else equal, an individual multiplication is (usually) slower than an individual addition. However the suggested approach performs so much less operations that individual complexity doesn't count. There is a rigorous proof of this intuition. \$\endgroup\$ – vnp May 12 '17 at 2:16
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    \$\begingroup\$ O(1) time and space if you're willing to go via floating-point for the closed-form expression - there's probably a crossover value of n for which that's better. \$\endgroup\$ – Toby Speight Jan 15 '18 at 8:36
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    \$\begingroup\$ @greybeard, I don't consider solving a popular programming exercise to be reinventing the wheel. If there existed std::fibonacci (or similar in Boost etc), I think you'd be right. \$\endgroup\$ – Toby Speight Jan 15 '18 at 8:37

(Uh-oh: better - better define a quality measure!)

  • Your code doesn't tell what it's about.
  • constexpr looks good - "my" environment complains with C++11.
  • "The array&index manipulation" where "simultaneous assignment" is wanted is hard to read (could be worse: the elements of result[] could be non-interchangeable).
    Alas, what I found for "modern C++" is ghastly compared to python's a, b = b, a + b. I appreciate the attempt to avoid avoidable assignments; I'm mildly curious if it makes any difference in the code generated by an optimising compiler.

  • Is there any better? Well, with output size limited by a constant, there's a tighter limit:
    the runtime of your code is in O(1), just as any other.
    In a comment, you express concern about the complexity of multiplication. If you accept ("bit-wise") "shift" as a (very) cheap operation, you can take three steps in the Fibonacci sequence at once without an increase in "logic gate complexity":

#include <cstdlib>
#include <tuple>
#include <iostream>

/// Iterates an a,b = b,a+b sequence in steps of three.
static unsigned long tri(int previous, int current, const unsigned int n) {
    if (n < 2)
        return n ? current : previous;
    std::div_t split = std::div(n-2, 3);
    while (0 <= --split.rem)
        std::tie(previous, current)
            = std::make_tuple(current, current+previous);
    unsigned long
        a = current - previous,
        b = current + previous;      
    while (0 <= --split.quot)
        std::tie(a, b) = std::make_tuple(b, (b<<2)+a);

    return b;
/// Iterates the Fibonacci sequence in steps of three.
unsigned long fibonacci(const unsigned int n) {
    return tri(0, 1, n);
/// Iterates the Lucas numbers in steps of three.
unsigned long lucas(const unsigned int n) {
    return tri(2, 1, n);

(For variants using arrays of precomputed elements in stead of "the setup-loop" (and a main()), consult the edit history.)
(b*Phi³ coincidentally can be computed with just two summands (and no other power up to 2³² can).)

  • \$\begingroup\$ (Disclaimer: I haven't used C++ in earnest for over a decade, I never was up to speed with "modern C++", even C++03. Feel invited to improve (especially) the code above.) \$\endgroup\$ – greybeard Jan 15 '18 at 1:15
  • \$\begingroup\$ I think I would move fib and luc into the scope of triFib() and triLuc() respectively (as static constants) - no need for them to be globally visible. \$\endgroup\$ – Toby Speight Jan 15 '18 at 8:40
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    \$\begingroup\$ Woah... Never expected this old question to be answered! Thank you very much. \$\endgroup\$ – Dannyu NDos Jan 16 '18 at 0:23
  • \$\begingroup\$ (For the curious: multipliers and step sizes for three summands: 3/2, (7/4((x<<3)-x, shift&mask instead of divmod)), 18/6; four summands: (7/4, shift&mask), 11/5, (47/8((x<<5)+(x<<4)-x, s&m)), 76/9, 322/12, 521/13) \$\endgroup\$ – greybeard Jan 17 '18 at 8:04

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