Better Fibonacci sequence calculation











up vote
7
down vote

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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?






c++ c++11 fibonacci-sequence







share|improve this question


















  • 4




    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.
    – vnp
    May 12 '17 at 1:59










  • @vnp That involves multiplication. Doesn't multiplication have worse complexity than addition?
    – Dannyu NDos
    May 12 '17 at 2:09






  • 5




    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.
    – vnp
    May 12 '17 at 2:16






  • 1




    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.
    – Toby Speight
    Jan 15 at 8:36






  • 1




    @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.
    – Toby Speight
    Jan 15 at 8:37















up vote
7
down vote

favorite
1












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?






c++ c++11 fibonacci-sequence







share|improve this question


















  • 4




    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.
    – vnp
    May 12 '17 at 1:59










  • @vnp That involves multiplication. Doesn't multiplication have worse complexity than addition?
    – Dannyu NDos
    May 12 '17 at 2:09






  • 5




    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.
    – vnp
    May 12 '17 at 2:16






  • 1




    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.
    – Toby Speight
    Jan 15 at 8:36






  • 1




    @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.
    – Toby Speight
    Jan 15 at 8:37













up vote
7
down vote

favorite
1









up vote
7
down vote

favorite
1






1





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?






c++ c++11 fibonacci-sequence







share|improve this question













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?






c++ c++11 fibonacci-sequence




c++ c++11 fibonacci-sequence






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked May 12 '17 at 1:31









Dannyu NDos

190113




190113








  • 4




    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.
    – vnp
    May 12 '17 at 1:59










  • @vnp That involves multiplication. Doesn't multiplication have worse complexity than addition?
    – Dannyu NDos
    May 12 '17 at 2:09






  • 5




    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.
    – vnp
    May 12 '17 at 2:16






  • 1




    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.
    – Toby Speight
    Jan 15 at 8:36






  • 1




    @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.
    – Toby Speight
    Jan 15 at 8:37














  • 4




    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.
    – vnp
    May 12 '17 at 1:59










  • @vnp That involves multiplication. Doesn't multiplication have worse complexity than addition?
    – Dannyu NDos
    May 12 '17 at 2:09






  • 5




    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.
    – vnp
    May 12 '17 at 2:16






  • 1




    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.
    – Toby Speight
    Jan 15 at 8:36






  • 1




    @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.
    – Toby Speight
    Jan 15 at 8:37








4




4




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.
– vnp
May 12 '17 at 1:59




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.
– vnp
May 12 '17 at 1:59












@vnp That involves multiplication. Doesn't multiplication have worse complexity than addition?
– Dannyu NDos
May 12 '17 at 2:09




@vnp That involves multiplication. Doesn't multiplication have worse complexity than addition?
– Dannyu NDos
May 12 '17 at 2:09




5




5




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.
– vnp
May 12 '17 at 2:16




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.
– vnp
May 12 '17 at 2:16




1




1




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.
– Toby Speight
Jan 15 at 8:36




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.
– Toby Speight
Jan 15 at 8:36




1




1




@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.
– Toby Speight
Jan 15 at 8:37




@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.
– Toby Speight
Jan 15 at 8:37










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










(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.

//constexpr

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).)






share|improve this answer























  • (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.)
    – greybeard
    Jan 15 at 1:15










  • 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.
    – Toby Speight
    Jan 15 at 8:40






  • 1




    Woah... Never expected this old question to be answered! Thank you very much.
    – Dannyu NDos
    Jan 16 at 0:23










  • (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)
    – greybeard
    Jan 17 at 8:04











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










(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.

//constexpr

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).)






share|improve this answer























  • (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.)
    – greybeard
    Jan 15 at 1:15










  • 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.
    – Toby Speight
    Jan 15 at 8:40






  • 1




    Woah... Never expected this old question to be answered! Thank you very much.
    – Dannyu NDos
    Jan 16 at 0:23










  • (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)
    – greybeard
    Jan 17 at 8:04















up vote
3
down vote



accepted










(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.

//constexpr

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).)






share|improve this answer























  • (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.)
    – greybeard
    Jan 15 at 1:15










  • 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.
    – Toby Speight
    Jan 15 at 8:40






  • 1




    Woah... Never expected this old question to be answered! Thank you very much.
    – Dannyu NDos
    Jan 16 at 0:23










  • (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)
    – greybeard
    Jan 17 at 8:04













up vote
3
down vote



accepted







up vote
3
down vote



accepted






(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.

//constexpr

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).)






share|improve this answer














(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.

//constexpr

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).)







share|improve this answer














share|improve this answer



share|improve this answer








edited 12 mins ago









albert

1071




1071










answered Jan 15 at 1:14









greybeard

1,3491521




1,3491521












  • (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.)
    – greybeard
    Jan 15 at 1:15










  • 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.
    – Toby Speight
    Jan 15 at 8:40






  • 1




    Woah... Never expected this old question to be answered! Thank you very much.
    – Dannyu NDos
    Jan 16 at 0:23










  • (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)
    – greybeard
    Jan 17 at 8:04


















  • (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.)
    – greybeard
    Jan 15 at 1:15










  • 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.
    – Toby Speight
    Jan 15 at 8:40






  • 1




    Woah... Never expected this old question to be answered! Thank you very much.
    – Dannyu NDos
    Jan 16 at 0:23










  • (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)
    – greybeard
    Jan 17 at 8:04
















(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.)
– greybeard
Jan 15 at 1:15




(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.)
– greybeard
Jan 15 at 1:15












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.
– Toby Speight
Jan 15 at 8:40




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.
– Toby Speight
Jan 15 at 8:40




1




1




Woah... Never expected this old question to be answered! Thank you very much.
– Dannyu NDos
Jan 16 at 0:23




Woah... Never expected this old question to be answered! Thank you very much.
– Dannyu NDos
Jan 16 at 0:23












(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)
– greybeard
Jan 17 at 8:04




(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)
– greybeard
Jan 17 at 8:04


















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