<< problem 229 - Four Representations using Squares The prime factorisation of binomial coefficients - problem 231 >>

# Problem 230: Fibonacci Words

For any two strings of digits, A and B, we define F_{A,B} to be the sequence (A,B,AB,BAB,ABBAB,...)
in which each term is the concatenation of the previous two.

Further, we define D_{A,B}(n) to be the nth digit in the first term of F_{A,B} that contains at least n digits.

Example:

Let A=1415926535, B=8979323846. We wish to find D_{A,B}(35), say.

The first few terms of F_{A,B} are:
1415926535
8979323846
14159265358979323846
897932384614159265358979323846
14159265358979323846897932384614159265358979323846

Then D_{A,B}(35) is the 35th digit in the fifth term, which is 9.

Now we use for A the first 100 digits of pi behind the decimal point:

14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679

and for B the next hundred digits:

82148086513282306647093844609550582231725359408128
48111745028410270193852110555964462294895493038196 .

Find sum_{n = 0,1,...,17} 10^n * D_{A,B}((127+19n) * 7n) .

# My Algorithm

I build a small lookup table of all Fibonacci numbers up to 2^64.

A simple loop from 17 to 0 processes one digit at a time.
Note: I go from 17 to 0 instead of the more common 0 to 17 because then all digits are printed
in the correct order and I can avoid multiplying them by 10^n.

The index formula needs to be adjusted by minus one because my indexing scheme starts at zero instead of one.

Now comes the first step of the core algorithm:
a simple loop looks for the first Fibonacci number that represents enough digits for index.
The value fibo will be incremented steadily, therefore I call it "step forward".

The second step subdivides the current Fibonacci number in a left and right part in order to find out which
building block (A or B) contains the desired digit.
The value fibo will be decremented steadily, therefore I call it "step backwards".

## Note

I don't store the true Fibonacci numbers in fibonacci → I pre-multiply them by BlockSize.
It saves me a few multiplications when stepping forward and backwards and - more important to me - keeps the code cleaner.

# 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):
Note: Enter two strings A and B of the same size. The may contain letters, too.

This is equivalent to
echo "1415926535 8979323846" | ./230

Output:

Note: the original problem's input 1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679 8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196 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 <string>
#include <vector>

int main()
{
// digits of pi
std::string A = "1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679";
std::string B = "8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196";

std::cin >> A >> B;
// live test only: detect invalid input
if (A.size() != B.size() || A.empty())
return 1;

// 100 digits are one "block"
const auto BlockSize = A.size(); // == B.size()

// first two elements of the Fibonacci series (and dummy Fibonacci number 0 is 0)
std::vector<unsigned long long> fibonacci = { 0, BlockSize };

// find all Fibonacci numbers below 2^64
while (fibonacci.back() < 1ULL << 63)
fibonacci.push_back(fibonacci[fibonacci.size() - 2] + fibonacci[fibonacci.size() - 1]);

// it's easier to reverse the loop and print all digits in the correct order ...
for (auto n = 17; n >= 0; n--)
{
// compute index according to problem description
unsigned long long index = 127 + 19*n;
// times 7^n
for (auto seven = 1; seven <= n; seven++)
index *= 7;

// my indexing scheme is zero-based but Project Euler's index starts at 1
index--;

// "step forward":
// find the first Fibonacci number with enough digits
unsigned int current = 1;
while (index >= fibonacci[current])
current++;

// "step backwards":
// deduce which Fibonacci number generated the part where the index is located
while (true)
{
// reduced to a single block A or B ?
if (current == 1)
{
std::cout << A[index];
break;
}
if (current == 2)
{
std::cout << B[index];
break;
}

// the mid-point divides left part (fibonacci minus 2) and right part (fibonacci minus 1)
auto mid = fibonacci[current - 2];
if (index < mid)
{
// left part
current -= 2;
}
else
{
// right part
index   -= mid;
current -= 1;
}
}
}

std::cout << std::endl;
return 0;
}


This solution contains 11 empty lines, 17 comments and 3 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 12, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 50% (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.

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