<< problem 250 - 250250 Stone Game - problem 260 >>

# Problem 259: Reachable Numbers

A positive integer will be called reachable if it can result from an arithmetic expression obeying the following rules:

- Uses the digits 1 through 9, in that order and exactly once each.
- Any successive digits can be concatenated (for example, using the digits 2, 3 and 4 we obtain the number 234).
- Only the four usual binary arithmetic operations (addition, subtraction, multiplication and division) are allowed.
- Each operation can be used any number of times, or not at all.
- Unary minus is not allowed.
- Any number of (possibly nested) parentheses may be used to define the order of operations.

For example, 42 is reachable, since (1/23) * ((4*5)-6) * (78-9) = 42.

What is the sum of all positive reachable integers?

# My Algorithm

I wrote a very basic struct Fraction to represent a rational number. It supports addition, multiplication, division and comparison.
It doesn't care about signs, division-by-zero and so on.

My search function returns all fractions that can be generated by splitting its std::string parameter in any possible way
and applying any allowed operation:
- it splits its input into two parts
- then calls itself recursively to get all fractions generated by the left part and by the right part
- two nested loops add, subtract, multiply and divide all fractions from the left part with all fractions from the right part
- invalid things, like division-by-zero, are rejected and subtraction is simulated by adding with a negated numerator
- there might be several combinations that produce the same result, therefore the output is sorted and std::unique ensures only unique fractions are left
- (I don't reduce fractions and so it's possible to have the same fraction multiple times if their numerators and denominators differ by a factor)

main has to check whether these fractions are positive integers.
And finally all duplicates are removed and the sum of all unique integers is displayed.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <string>
#include <vector>
#include <algorithm>

// a rational number
// fractions are neither reduced nor is the sign kept consistent
// (both numerator and denominator might be negative)
struct Fraction
{
// create a new number
Fraction(unsigned int numerator_, unsigned int denominator_ = 1)
: numerator(numerator_), denominator(denominator_)
{}

Fraction operator+(const Fraction& other) const
{
return Fraction(numerator   * other.denominator + other.numerator * denominator,
denominator * other.denominator);
}
// multiply
Fraction operator*(const Fraction& other) const
{
return Fraction(numerator   * other.numerator,
denominator * other.denominator);
}
// divide
Fraction operator/(const Fraction& other) const
{
return Fraction(numerator * other.denominator,
denominator * other.numerator);
// note: I don't attempt to reduce the fraction
}

// for std::sort
bool operator<(const Fraction& other) const
{
return numerator * other.denominator <  denominator * other.numerator;
}
// for std::unique
bool operator==(const Fraction& other) const
{
return numerator * other.denominator == denominator * other.numerator;
}

// both might have a negative sign
int numerator;
int denominator;
};

// return all rational numbers that can be produced by the sequence of digits
std::vector<Fraction> search(const std::string& digits)
{
// if no operations are applied, then all digits might be a single number
std::vector<Fraction> result = { Fraction(std::stod(digits)) };

// split digits into two parts and apply an operation on them
for (size_t split = 1; split < digits.size(); split++)
{
auto left  = digits.substr(0, split);
auto right = digits.substr(split);

// recursively find all fractions that can be created with these parts
auto leftFractions  = search(left);
auto rightFractions = search(right);

// merge both with + - * /
for (auto x : leftFractions)
for (auto y : rightFractions)
{
result.push_back(x + y);
// subtract: not really implemented, just negate second number's numerator and then add
result.push_back(x + Fraction(-y.numerator, y.denominator));
// multiply
result.push_back(x * y);
// divide: disallow division by zero
if (y.numerator != 0)
result.push_back(x / y);
}
}

// prune redundant values (makes the code about 10x faster !)
if (result.size() > 1)
{
std::sort(result.begin(), result.end());
auto last = std::unique(result.begin(), result.end());
result.erase(last, result.end());
}

return result;
}

int main()
{
unsigned int lastDigit = 9;
std::cin >> lastDigit;

// create a string with all digits in ascending order (from 1 to lastDigit)
std::string digits = "123456789";
digits = digits.substr(0, lastDigit);

// find all possible fractions
auto fractions = search(digits);

// all found values
std::vector<int> found;

// extract all integers from these fractions
for (auto current: fractions)
{
// remove negative sign from denominator
if (current.denominator < 0)
{
current.numerator   *= -1;
current.denominator *= -1;
}
// fraction must be positive
if (current.numerator <= 0)
continue;

// numerator must be a multiple of its denominator (=> fraction must be an integer)
if (current.numerator % current.denominator == 0)
found.push_back(current.numerator / current.denominator);
}

// remove duplicates
std::sort(found.begin(), found.end());
auto last = std::unique(found.begin(), found.end());
found.erase(last, found.end());

unsigned long long sum = 0;
for (auto x : found)
sum += x;

// show result
std::cout << sum << std::endl;
return 0;
}


This solution contains 20 empty lines, 31 comments and 4 preprocessor commands.

# 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 the highest allowed digit.

This is equivalent to
echo 4 | ./259

Output:

Note: the original problem's input 9 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)

# Benchmark

The correct solution to the original Project Euler problem was found in 1.5 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 139 MByte.

(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++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 4, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 70% (out of 100%).

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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The 235 solved problems (level 9) had an average difficulty of 29.1% at Project Euler and
I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

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