<< problem 347 - Largest integer divisible by two primes Langton's ant - problem 349 >>

# Problem 348: Sum of a square and a cube

Many numbers can be expressed as the sum of a square and a cube. Some of them in more than one way.

Consider the palindromic numbers that can be expressed as the sum of a square and a cube, both greater than 1, in exactly 4 different ways.
For example, 5229225 is a palindromic number and it can be expressed in exactly 4 different ways:

2285^2 + 20^3
2223^2 + 66^3
1810^2 + 125^3
1197^2 + 156^3

Find the sum of the five smallest such palindromic numbers.

# My Algorithm

My SquareCube objects are simple structs that store the value of a square and a cube.
It contains two comparison functions: operator== returns true when two objects are equal (obviously ...).
However, operator< returns the "wrong" result: I flipped true, if the current object is bigger than the parameter.
The reason behind my weird logic is that I use the std::priority_queue is returns always the largest object.
By flipping operator< I can modifiy it to returns the smallest object.

I create all numbers which can be expressed as the sum of square and a cube in ascending order.
Initially an object SquareCube which represents 2^2 + 3^2 is stored in a std::priority_queue.
Whenever I take an object from the top of the queue, I add two objects (square+1)^2 + cube^3 and square^2 + (cube+1)^3.
And whenever I can pick exactly four objects in a row with the same value, then I check whether they are palindromes.

## Alternative Approaches

A much simpler alternative is to create a huge array (about 10^9 entries) which contains only zeros.
Then generate all pairs (square,cube) and increment myarray[square*square + cube*cube*cube].
Look for the 5 smallest indices where myerray[] = 5.

Remember that this approach needs about 1 GByte of memory. My solution needs less than 5 MByte.

## Note

Many numbers will occur twice in the queue, e.g. when I pick 35 = 3^2 + 3^3 from the queue then it was inserted after
picking 16 = 3^2 + 2^3 and after picking 31 = 2^2 + 3^3. I can easily spot those duplicates with my overloaded operator==.

# 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 1 | ./348

Output:

Note: the original problem's input 5 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>
#include <queue>
#include <string>

// represent a number where "value = square^2 + cube^3"
struct SquareCube
{
// create a new object made of a square and a cube
SquareCube(unsigned int square_, unsigned int cube_)
: square(square_), cube(cube_),
value(cube_*(unsigned long long)cube_*cube_ + square_*(unsigned long long)square_)
{ }

// compare two objects (for std::priority_queue)
bool operator<(const SquareCube& other) const
{
if (value != other.value)
return value > other.value; // deliberately switched the sign => min-heap

return cube > other.cube;
}
bool operator==(const SquareCube& other) const
{
return value == other.value && cube == other.cube;
}

unsigned int square;
unsigned int cube;
// equal to square^2 + cube^3
unsigned int value;
};

int main()
{
// find the first 5 numbers
unsigned int maxFound = 5;
std::cin >> maxFound;

// all cubes and squares must be greater than 1
std::priority_queue<SquareCube> todo;
todo.push(SquareCube(2, 2));

// find first 5
unsigned int numFound = 0;
// and their sum
unsigned int sum = 0;

while (numFound < maxFound)
{
// start a new value
auto current = todo.top();
todo.emplace(current.square + 1, current.cube);
todo.emplace(current.square, current.cube + 1);

while (todo.top() == current)
todo.pop();

// count all combinations of squares and cube with the same value
unsigned int numSame = 1;
while (todo.top().value == current.value)
{
numSame++;
auto same = todo.top();

// same value, remove it and add its successors
while (todo.top() == same)
todo.pop();
todo.emplace(same.square + 1, same.cube);
todo.emplace(same.square, same.cube + 1);
}

// exactly four combinations ?
if (numSame == 4)
{
// palindrome ?
unsigned int reverse = 0;
auto reduce = current.value;
while (reduce > 0)
{
reverse *= 10;
reverse += reduce % 10;
reduce  /= 10;
}

// yes, a match
if (current.value == reverse)
{
numFound++;
sum += current.value;
}
}
}

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


This solution contains 14 empty lines, 15 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 5.6 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 23, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 25% (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 347 - Largest integer divisible by two primes Langton's ant - problem 349 >>
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