<< problem 293 - Pseudo-Fortunate Numbers Protein folding - problem 300 >>

# Problem 297: Zeckendorf Representation

Each new term in the Fibonacci sequence is generated by adding the previous two terms.
Starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.

Every positive integer can be uniquely written as a sum of nonconsecutive terms of the Fibonacci sequence.
For example, 100 = 3 + 8 + 89.
Such a sum is called the Zeckendorf representation of the number.

For any integer n>0, let z(n) be the number of terms in the Zeckendorf representation of n.
Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc.
Also, for 0 < n < 10^6, sum{z(n)} = 7894453.

Find sum{z(n)} for 0 < n < 10^17.

# My Algorithm

Often I write a simple brute-force program which produces the correct result for small inputs.
The function zeckendorf returns the number of terms in the Zeckendorf representation of a single number.
According to the Wikipedia page on the Zeckendorf theorem (en.wikipedia.org/wiki/Zeckendorf's_theorem) a greedy repeated search
for the largest Fibonacci numbers <= the current number is sufficient.

Then I printed the first 100 numbers (and their sum) and discovered the pattern:
the number of terms repeats after each new Fibonacci number, abeit increased by one. For example:

numberzeckendorf(number)sumfiboSum
1111
2122
3133
425
5166
628
7210
811111
9213
10215
11217
12320
1312121
14223
15225
16227
17330
18232
19335
20338
2113939

The Zeckendorf representation is always 1 for a Fibonacci number (which are bold in the table above).
My array fiboSum contains the running total (the sum) for each n-th Fibonacci number:
fiboSum[n] = fiboSum[n - 1] + fiboSum[n - 2] + fibonacci[n - 2] - 1

Using that relationship my function search recursively subtracts the largest possible Fibonacci number
while adding the relevant value fiboSum.

## Note

The zeckendorf function isn't needed anymore but I didn't remove it because my gut tells me I could use it in the future ...

# 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 1000000 | ./297

Output:

Note: the original problem's input 100000000000000000 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>

// all Fibonacci numbers below 10^17
std::vector<unsigned long long> fibonacci;
// and the sum according to my algorithm description
std::vector<unsigned long long> fiboSum; // fibonacci.size() == fiboSum.size()

// see en.wikipedia.org/wiki/Zeckendorf's_theorem
// return the length of the Zeckendorf representation of x
unsigned int zeckendorf(unsigned long long x)
{
unsigned int result = 0;

// greedy search:
// in each step subtract the largest possible Fibonacci number
unsigned int pos = fibonacci.size() - 1;
while (x > 0)
{
while (fibonacci[pos] > x)
pos--;

x -= fibonacci[pos];
result++;
}

return result;
}

// compute sum of the length of all Zeckendorf representations from 1 to x
unsigned long long search(unsigned long long x)
{
// find largest Fibonacci number <= x
auto pos = 0;
while (fibonacci[pos + 1] <= x)
pos++;

// strip off that Fibonacci number
auto reduced = x - fibonacci[pos];

// done ?
if (reduced == 0)
return fiboSum[pos];

// still more 1s in the binary Zeckendorf representation left ...
return fiboSum[pos] + reduced + search(reduced);
}

int main()
{
unsigned long long limit = 100000000000000000ULL;
std::cin >> limit;

// note: it took me a while to figure out that because of F(1) = F(2) = 1
//       I am not allowed to represent 3 as F(1) + F(3) (which is 3 = 1 + 2)
//       even though F(1) and F(3) are not consecutive
//       if I remove F(1) (or F(2)) then that ambiguity disappears and everything's fine

fibonacci = { 1, 2 };

// F(2) has length 1, F(3) as length 1 as well plus 1 from F(2)
fiboSum   = { 1, 1+1 };

// find all Fibonacci number below 10^17
while (fibonacci.back() < limit)
{
auto size = fibonacci.size();
auto nextFibo = fibonacci[size - 1] + fibonacci[size - 2];
fibonacci.push_back(nextFibo);

// "special" sum of Fibonacci numbers
auto nextSum  = fiboSum  [size - 1] + fiboSum  [size - 2] + fibonacci[size - 2] - 1;
fiboSum.push_back(nextSum);
}

// NOT including 10^17
limit--;
// display result
std::cout << search(limit) << std::endl;

// my old test code for the first 100 numbers
//auto sum = 0;
//for (auto i = 1; i < 100; i++)
//{
//  auto current = zeckendorf(i);
//  sum += current;
//  std::cout << i << "=" << current << " " << sum << " / " << search(i) << std::endl;
//}

return 0;
}


This solution contains 18 empty lines, 29 comments and 2 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

July 25, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 35% (out of 100%).

# 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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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.

 << problem 293 - Pseudo-Fortunate Numbers Protein folding - problem 300 >>
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