Problem 259: Reachable Numbers

(see projecteuler.net/problem=259)

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

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:

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.

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:

(please click 'Go !')

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)

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>
#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_)
{}
 
// add
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)
{
// add
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());
 
// add all
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.

Benchmark

The correct solution to the original Project Euler problem was found in 1.5 seconds on an 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=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 4, 2017 submitted solution
July 4, 2017 added comments

Difficulty

70% Project Euler ranks this problem at 70% (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.

more about me can be found on my homepage, especially in my coding blog.
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