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:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 10" | ./137

Output:

(please click 'Go !')

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.

The code contains #ifdefs 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

50% 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.

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
  the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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