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# Problem 247: Squares under a hyperbola

Consider the region constrained by 1 <= x and 0 <= y <= 1/x.

Let S_1 be the largest square that can fit under the curve.
Let S_2 be the largest square that fits in the remaining area, and so on.
Let the index of Sn be the pair (left, below) indicating the number of squares to the left of S_n and the number of squares below S_n.

The diagram shows some such squares labelled by number.
S_2 has one square to its left and none below, so the index of S_2 is (1,0).
It can be seen that the index of S_32 is (1,1) as is the index of S_50.
50 is the largest n for which the index of S_n is (1,1).

What is the largest n for which the index of S_n is (3,3)?

# My Algorithm

Each square has a lower-left corner (x_0,y_0) and an upper-right corner (x_1,y_1) such that:
(1) x_1 - x_0 = y_1 - y_0 = side (length of each side of that square)

The hyperbola y = 1/x must go through the upper-right corner:
(2) y_1 = dfrac{1}{x_1}

Substituting (2) in (1):
(3) x_1 - x_0 = dfrac{1}{x_1} - y_0

And solve for x_1:
(4) x_1 - dfrac{1}{x_1} = x_0 - y_0
(5) dfrac{x^2_1 - 1}{x_1} = x_0 - y_0
(6) x^2_1 - 1 = x_1 (x_0 - y_0)
(7) 0 = x^2_1 - x_1 (x_0 - y_0) - 1
(8) 0 = x^2_1 + x_1 (y_0 - x_0) - 1

That's a quadratic equation with parameters:
(9) p = y_0 - x_0, q = -1
(10) x = -frac{p}{2} \pm sqrt{(frac{p}{2})^2 - q}

Omitting the second solution:
(11) x_1 = -frac{y_0 - x_0}{2} + sqrt{(frac{y_0 - x_0}{2})^2 + 1}
(12) x_1 = frac{x_0 - y_0}{2} + frac{1}{2} sqrt{(x_0 - y_0)^2 + 4}
(13) x_1 = 0.5 * (x_0 - y_0 + sqrt{(x_0 - y_0)^2 + 4})

Therefore the length of a side in (1) becomes:
(14) side = x_1 - x_0
(15) side = 0.5 * (x_0 - y_0 + sqrt{(x_0 - y_0)^2 + 4}) - x_0

And a bit simplified:
(16) side = 0.5 * (-x_0 - y_0 + sqrt{(x_0 - y_0)^2 + 4})
(17) side = 0.5 * (sqrt{(x_0 - y_0)^2 + 4} - x_0 - y_0)

That formula can be found in Square::setSideLength.

My program starts with a single square whose lower-left corner (x, y) is at (1, 0).
All squares are stored in an std::set named todo which is sorted descendingly by the square's size
(the overloaded Square::operator< returns the "wrong" result on purpose because an std::set is sorted ascendingly by default).

I always pick the largest (→ the first) square from todo and replace it by its upper and right neighbor.
Each square's position (x, y) and index (left, below) are tracked.

In order to have a square at index (3, 3) I have to have a square at (2, 3) or (3, 2).
Generalized, only if some squares are in todo where left <= 3 and below <= 3 and left + below < 3+3 then it's
still possible to generate a square at index (3, 3).
candidates counts how many squares fulfil that condition. If candidates becomes zero then no more squares can be at (3,3).
The most recent will be printed.

# 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 1" | ./247

Output:

Note: the original problem's input 3 3 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 <set>
#include <cmath>

// a square is described by its lower left corner (x,y) and its index along x and y axis
struct Square
{
Square(double x_, double y_, unsigned int left_, unsigned int below_)
: x(x_), y(y_), left(left_), below(below_)
{
setSideLength();
}

// lower-left corner
double x;
double y;
// number of squares on the left side
unsigned int left;
// number of squares below
unsigned int below;

// length of a side
double side;

// note: sort in reverse, therefore return true if it's the larger (!) value
bool operator<(const Square& other) const
{
return side > other.side;
}

private:
// find length of edges of the largest square whose lower-left corner is at x,y
void setSideLength()
{
// see my explanations above
side = 0.5 * (sqrt((x - y) * (x - y) + 4) - x - y);
}
};

int main()
{
unsigned int indexLeft  = 3;
unsigned int indexBelow = 3;
std::cin >> indexLeft >> indexBelow;

// square's number
unsigned int result     = 0;

// create first square with lower-left corner at (1,0) and no other squares on its left or bottom side
std::set<Square> todo = { Square(1, 0, 0, 0) };
unsigned int candidates = 1; // => the first square is left of and below all other squares
while (candidates > 0)
{
result++;

// pick first (= largest) square
auto current = *(todo.begin());
// and remove it
todo.erase(todo.begin());

// create a new square on top of the current square
Square top  (current.x,    current.y + current.side,
current.left, current.below + 1);
todo.insert(top);
// and a square on the right side
Square right(current.x + current.side, current.y,
current.left + 1, current.below);
todo.insert(right);

// count how many squares could be on the left or bottom side of a square with index (3, 3)
if (top    .left <= indexLeft && top    .below <= indexBelow)
candidates++;
if (right  .left <= indexLeft && right  .below <= indexBelow)
candidates++;
if (current.left <= indexLeft && current.below <= indexBelow)
candidates--;
}

// show ID of last square with index (3, 3)
std::cout << result << std::endl;
return 0;
}


This solution contains 12 empty lines, 16 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.19 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 25 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

August 14, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 65% (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 the flashing problem is the one I solved most recently
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The 299 solved problems (that's level 11) had an average difficulty of 32.4% 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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