<< problem 136 - Singleton differences | Special isosceles triangles - problem 138 >> |
Problem 137: Fibonacci golden nuggets
(see projecteuler.net/problem=137)
Consider the infinite polynomial series A_F(x) = xF_1 + x^2 F_2 + x^3 F_3 + ..., where F_k is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ;
that is, F_k = F_{k-1} + F_{k-2}, F_1 = 1 and F_2 = 1.
For this problem we shall be interested in values of x for which A_F(x) is a positive integer.
Surprisingly A_F(1/2) = (1/2) * 1 + (1/2)^2 * 1 + (1/2)^3 * 2 + (1/2)^4 * 3 + (1/2)^5 * 5 + ...
= 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
= 2
The corresponding values of x for the first five natural numbers are shown below.
xA_F(x)
2-11
1/22
(sqrt{13}-2)/33
(sqrt{89}-5)/84
(sqrt{34}-3)/55
We shall call A_F(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.
Find the 15th golden nugget.
My Algorithm
Admittedly, I had no idea. And then I just searched the OEIS for 74049690 (the 10th nugget).
Fortunately, the first hit oeis.org/A081018 contained a list of numbers where 74049690 was at the 10th position. Jackpot ...
According to that website, the n-th golden nugget is the product of F(2n) * F(2n+1).
All I have to solve is F(2 * 15) * F(2 * 15 + 1) = F(30) * F(31) - I can do that instantly !
Modifications by HackerRank
This problem turned from "no idea" to "extremely easy" to "quite hard" when I saw Hackerrank's modifications:
I have to process up to 10^5 input values in 2 seconds where some input values can be as large as 10^18.
(And the result has to be printed modulo 10^9+7.)
Searching the web for fast Fibonacci algorithms I came across the "matrix form" (see en.wikipedia.org/wiki/Fibonacci_number):
\left( \begin{matrix} F_{n+2} \\ F_{n+1} \end{matrix} \right) = \left( \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right) \left( \begin{matrix} F_{n+1} \\ F_{n} \end{matrix} \right)
This looks to be even slower than the old-fashioned way of adding ... but there is a twist:
\left( \begin{matrix} F_{n+1} \\ F_{n} \end{matrix} \right) = \left( \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right)^n \left( \begin{matrix} F_1 \\ F_0 \end{matrix} \right)
Raising the matrix to the n-th power can be done pretty quickly: fast exponentiation needs about log_2(10^18) approx 60 iterations to calculate the final matrix for F(10^18).
The concept is the same as in powmod
but instead of multiplying single numbers I have to perform a matrix multiplication.
The most beautiful property of the matrix form is that I get the successor of the desired Fibonacci for free.
Using these tricks my program solves 100% at Hackerrank.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho "1 10" | ./137
Output:
Note: the original problem's input 15
cannot be entered
because just copying results is a soft skill reserved for idiots.
(this interactive test is still under development, computations will be aborted after one second)
My code
… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.
The code contains #ifdef
s to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL
to produce the result for the original problem (default setting for most problems).
#include <iostream>
// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;
// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;
// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);
// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}
// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;
// next bit
b >>= 1;
}
return result;
}
// Fibonacci matrix algorithm (taken from my solution of problem 304)
// return (Fibonacci(2n) * Fibonacci(2n+1)) % modulo
unsigned long long nugget(unsigned long long n, unsigned long long modulo)
{
// n-th nugget is based on Fibonacci(2n) and Fibonacci(2n+1)
n *= 2;
// fast exponentiation: same idea as powmod from my toolbox
// matrix values from https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
unsigned long long fibo [2][2]= { { 1, 1 },
{ 1, 0 } };
// initially identity matrix
unsigned long long result[2][2]= { { 1, 0 }, // { { F(n+1), F(n) },
{ 0, 1 } }; // { F(n), F(n-1) } }
while (n > 0)
{
// fast exponentation:
// odd exponent ? a^n = a*a^(n-1)
if (n & 1)
{
// compute new values, store them in temporaries
auto t00 = mulmod(result[0][0], fibo[0][0], modulo) + mulmod(result[0][1], fibo[1][0], modulo);
auto t01 = mulmod(result[0][0], fibo[0][1], modulo) + mulmod(result[0][1], fibo[1][1], modulo);
auto t10 = mulmod(result[1][0], fibo[0][0], modulo) + mulmod(result[1][1], fibo[1][0], modulo);
auto t11 = mulmod(result[1][0], fibo[0][1], modulo) + mulmod(result[1][1], fibo[1][1], modulo);
if (t00 >= modulo) t00 -= modulo;
if (t01 >= modulo) t01 -= modulo;
if (t10 >= modulo) t10 -= modulo;
if (t11 >= modulo) t11 -= modulo;
// copy back to matrix
result[0][0] = t00; result[0][1] = t01;
result[1][0] = t10; result[1][1] = t11;
}
// even exponent ? a^n = (a*a)^(n/2)
// compute new values, store them in temporaries
auto t00 = mulmod(fibo[0][0], fibo[0][0], modulo) + mulmod(fibo[0][1], fibo[1][0], modulo);
auto t01 = mulmod(fibo[0][0], fibo[0][1], modulo) + mulmod(fibo[0][1], fibo[1][1], modulo);
auto t10 = mulmod(fibo[1][0], fibo[0][0], modulo) + mulmod(fibo[1][1], fibo[1][0], modulo);
auto t11 = mulmod(fibo[1][0], fibo[0][1], modulo) + mulmod(fibo[1][1], fibo[1][1], modulo);
if (t00 >= modulo) t00 -= modulo;
if (t01 >= modulo) t01 -= modulo;
if (t10 >= modulo) t10 -= modulo;
if (t11 >= modulo) t11 -= modulo;
// copy back to matrix
fibo[0][0] = t00; fibo[0][1] = t01;
fibo[1][0] = t10; fibo[1][1] = t11;
n >>= 1;
}
// F(2n + 1) * F(2n)
return (result[0][0] * result[0][1]) % modulo;
}
int main()
{
#define ORIGINAL
#ifdef ORIGINAL
// a number large enough such that it basically disables modular arithmetic for the original problem
const unsigned long long Modulo = 10000000000000ULL;
#else
// 10^9 + 7
const unsigned long long Modulo = 1000000007;
#endif
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned long long n;
std::cin >> n;
std::cout << nugget(n, Modulo) << std::endl;
}
return 0;
}
This solution contains 21 empty lines, 26 comments and 5 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL
)
See here for a comparison of all solutions.
Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL
.
Changelog
August 1, 2017 submitted solution
August 1, 2017 added comments
Hackerrank
see https://www.hackerrank.com/contests/projecteuler/challenges/euler137
My code solves 4 out of 4 test cases (score: 100%)
Difficulty
Project Euler ranks this problem at 50% (out of 100%).
Hackerrank describes this problem as easy.
Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.
Links
projecteuler.net/thread=137 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-137-fibonacci-golden-nuggets/ (written by Kristian Edlund)
C# github.com/HaochenLiu/My-Project-Euler/blob/master/137.cs (written by Haochen Liu)
Python github.com/smacke/project-euler/blob/master/python/137.py (written by Stephen Macke)
C++ github.com/Meng-Gen/ProjectEuler/blob/master/137.cc (written by Meng-Gen Tsai)
C++ github.com/smacke/project-euler/blob/master/cpp/137.cpp (written by Stephen Macke)
Java github.com/dcrousso/ProjectEuler/blob/master/PE137.java (written by Devin Rousso)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem137.java (written by Magnus Solheim Thrap)
Go github.com/frrad/project-euler/blob/master/golang/Problem137.go (written by Frederick Robinson)
Mathematica github.com/steve98654/ProjectEuler/blob/master/137.nb
Perl github.com/shlomif/project-euler/blob/master/project-euler/137/euler-137.pl (written by Shlomi Fish)
Rust github.com/gifnksm/ProjectEulerRust/blob/master/src/bin/p137.rs
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
Heatmap
Please click on a problem's number to open my solution to that problem:
green | solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too | |
yellow | solutions score less than 100% at Hackerrank (but still solve the original problem easily) | |
gray | problems are already solved but I haven't published my solution yet | |
blue | solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much | |
orange | problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte | |
red | problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too | |
black | problems are solved but access to the solution is blocked for a few days until the next problem is published | |
[new] | the flashing problem is the one I solved most recently |
I stopped working on Project Euler problems around the time they released 617.
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I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
Look at my progress and performance pages to get more details.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
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