<< problem 336 - Maximix Arrangements Fractional Sequences - problem 343 >>

# Problem 341: Golomb's self-describing sequence

The Golomb's self-describing sequence G(n) is the only nondecreasing sequence of natural numbers such that n appears exactly G(n) times in the sequence.
The values of G(n) for the first few n are

n123456789101112131415...
G(n)122334445556666...

You are given that G(10^3) = 86, G(10^6) = 6137.
You are also given that sum{G(n^3)} = 153506976 for 1 <= n < 10^3.

Find sum{G(n^3)} for 1 <= n < 10^6.

# My Algorithm

I brute forced the first values up to 100 (far more can be found here: oeis.org/A001462/b001462.txt) and observed a certain pattern:

• G(1) = 1 and there is exactly one G(x) = 1
• G(2) = 2 and there are two x such that G(x) = 2 (it's 2 and 3)
• G(4) = 3 and there are two x such that G(x) = 3 (it's 4 and 5)
• G(6) = 4 and there are three x such that G(x) = 4 (it's 6, 7 and 8)
• G(9) = 5 and there are three x such that G(x) = 5 (it's 9, 10 and 11)
• G(12) = 6 and there are four x such that G(x) = 6 (it's 12, 13, 14 and 15)
• ...
→ the number of identical G(x) increments by one after sum{x * G(x)} steps:
1 * G(1) = 1 * 1 = 1 → all G(x > 1) appear at least twice
1 * G(1) + 2 * G(2) = 1 + 4 = 5 → all G(x > 5) appear at least three times
1 * G(1) + 2 * G(2) + 3 * G(3) = 1 + 4 + 6 = 11 → all G(x > 11) appear at least four times
I store these products in products[] (or compute them on-the-fly in the LOW_MEMORY code path).

It turns out that sum{x * G(x)} > {10^6}^3 for x approx 10 million.
Brute-forcing the first few million values of the Golomb sequence is no problem using the "recursive" Wikipedia formula:
a(1) = 1
a(n+1) = 1 + a (n + 1 - a( a(n) ) )
which can be simplified for a(n)
a(n) = 1 + a (n - a( a(n - 1) ) )

Once I know the upper limit of the number of elements I make use of a second observation:
I already demonstrated that G(1..3) is sufficient to know that any number 6 <= x <= 11 maps to a value to appears three times in the Golomb sequence.
But I still don't know which number ! (it can be 4 or 5)
The sums G(1) + G(2) = 3 and G(1) + G(2) + G(3) = 5 have a nice property:
I know there 6 values between those sums (it's 6 to 11). If I find the relative position ratio between the relevant products, then the sum at the same relative position contains the correct value.

Let's compute G(x) for x = 10:
from = products = 5
to = products = 11

ratio = dfrac{10 - from}{to - from} = dfrac{10 - 5}{11 - 5} = dfrac{5}{6} approx 0.83333

low = sums = G(1) + G(2) = 3
high = sums = G(1) + G(2) + G(3) = 5

G(10) = low + roundUp((high - low) * ratio) = 3 + roundUp((5 - 3) * 0.83333) = 5

(that's good old en.wikipedia.org/wiki/Linear_interpolation)

My code became a mess when I replaced the temporary containers sums and produces by variables.
However, it cut the memory usage by 50%: from about 270 MByte (slightly above the inofficial limit of 256 MByte) down to 135 MByte.
#define LOW_MEMORY is also about twice as fast.

## Note

The Wikipedia page describes an O(1) formula which is unfortunately only asymptotic correct:
the result is more or less pretty close to the actual value but can't be used to solve this problem.
Nevertheless, the fast function helped me a lot in my debugging session when some intermediate values were "strange".

Let's be honest: I discovered the connections between G(x) the sums and the products more or less by chance when I spent time on the commuter train home.
It was trial'n'error all the way - and therefore it's definitely not amongst my favorite problems.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 1000 | ./341

Output:

Note: the original problem's input 1000000 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.

       #include <iostream>
#include <vector>
#include <cmath>

// find the asymptotic value, quite often off by a few numbers (first four digits tend to be okay)
// note: unused code, but helped me a lot during debugging
double fast(unsigned long long n)
{
// golden ratio
const auto phi = (1 + sqrt(5.0)) / 2;
const auto constant = pow(phi, 2 - phi);
// see https://en.wikipedia.org/wiki/Golomb_sequence
return constant * pow(n, phi - 1);
}

int main()
{
unsigned int limit = 1000000;
std::cin >> limit;

// (10^6)^3 = 10^18
auto cubicLimit = (unsigned long long) limit * limit * limit;

// index 0 is not used, golomb(1) = 1
std::vector<unsigned long long> golomb = { 0, 1 };

#define LOW_MEMORY
#ifdef LOW_MEMORY
// precompute golomb[i] = G(i)
// stop when 1*G(1) + 2*G(2) + 3*G(3) + ... + i*G(i) >= 10^18
unsigned long long products = 1;
for (unsigned long long i = 2; products < cubicLimit; i++)
{
// https://en.wikipedia.org/wiki/Golomb_sequence
auto current = 1 + golomb[i - golomb[golomb[i - 1]]];
golomb.push_back(current);
products += current * i;
}

#else

// precompute golomb[i] = G(i)
// and sums[i]     =   G(1) +   G(2) +   G(3) + ... +   G(i)
// and products[i] = 1*G(1) + 2*G(2) + 3*G(3) + ... + i*G(i)
std::vector<unsigned long long> sums     = { 0, 1 };
std::vector<unsigned long long> products = { 0, 1 };
// stop when products[i] >= 10^18
for (unsigned long long i = 2; products.back() < cubicLimit; i++)
{
auto current = 1 + golomb[i - golomb[golomb[i - 1]]];

golomb  .push_back(current);
sums    .push_back(current     + sums.back());
products.push_back(current * i + products.back());
}
#endif

// will contain the result
unsigned long long sum = 0;

#ifdef LOW_MEMORY
unsigned long long lastSums     = 0;
unsigned long long sums         = 1;

unsigned long long lastProducts = 0;
products     = 1;
#endif

// find products[index - 1] < i <= products[index]
auto index = 1;
for (unsigned long long i = 1; i < limit; i++)
{
// n = i^3
auto n = i * i * i;

// find products[index - 1] < i <= products[index]
#ifdef LOW_MEMORY
while (products < n)
{
index++;
lastSums     = sums;
sums        += golomb[index];
lastProducts = products;
products    += golomb[index] * index;
}
#else
while (products[index] < n)
index++;
#endif
// note: n will be in ascending order, therefore I re-use index from previous iterations
//       in most cases it will be already the correct value

// find linear interpolation between products[index - 1] and products[index]
#ifdef LOW_MEMORY
auto from = lastProducts;
auto to   = products;
#else
auto from = products[index - 1];
auto to   = products[index];
#endif
auto ratio = (n - from) / double(to - from);

// and apply it to sums[index - 1] and sums[index]
#ifdef LOW_MEMORY
auto low  = lastSums;
auto high = sums;
#else
auto low  = sums[index - 1];
auto high = sums[index];
#endif

// round up
auto offset = ceil((high - low) * ratio);
// note: convert to integer as soon as possible to avoid losing digits due to double's limited precision
auto result = (unsigned long long)offset + low;

// finished another number ...
sum += result;
}

// solved another problem !
std::cout << sum << std::endl;
return 0;
}


This solution contains 18 empty lines, 25 comments and 18 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.11 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 133 MByte.

(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 15, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 45% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

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I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
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.

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