Problem 297: Zeckendorf Representation

(see projecteuler.net/problem=297)

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:

(please click 'Go !')

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.

#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 https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem" target="_blank" alt="Zeckendorf's theorem" title="Zeckendorf's theorem">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
 
// start with Fibonacci numbers F(2) and F(3)
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
July 25, 2017 added comments

Difficulty

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

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
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