<< problem 501 - Eight Divisors Tangent Circles - problem 510 >>

# Problem 504: Square on the Inside

Let ABCD be a quadrilateral whose vertices are lattice points lying on the coordinate axes as follows:

A(a, 0), B(0, b), C(-c, 0), D(0, -d), where 1 <= a, b, c, d <= m and a, b, c, d, m are integers.

It can be shown that for m = 4 there are exactly 256 valid ways to construct ABCD.
Of these 256 quadrilaterals, 42 of them strictly contain a square number of lattice points.

How many quadrilaterals ABCD strictly contain a square number of lattice points for m = 100?

# My Algorithm

Pick's theorem states A = i + b/2 - 1 (see en.wikipedia.org/wiki/Pick's_theorem):
if a simple polygon has i interior integer points and b boundary integer points, then A must be a perfect square to match the conditions of the problem.
So all I have to do is finding i and b - that's what my function countLatticePoints is for.

The bounding box that enclosed the quadrilateral and is parallel to the x- and y-axis has an area of
ab + bc + cd + ad = a(b + d) + c(b + d) = (a + c)(b + d)
The four triangles cover only half the area of the bounding box. My variable inside equals i of the formula:
i = dfrac{1}{2} (a + c)(b + d)

The boundary integer points can be subdivided into two groups:

• 4 corners A(a, 0), B(0, b), C(-c, 0), D(0, -d)
• the edges between A and B, B and C, C and D, D and A
An edge between between two points X(x, 0) and Y(0, y) covers the same number of integer lattice points as an edge between X_2(0,0) and Y_(x,y).
And there are gcd(x,y) + 1 such integer lattice points: the origin plus all multiples of k * ( dfrac{x}{gcd(x,y)}, dfrac{y}{gcd(x,y)} ).
However, each corner appears twice: A in AB and DA, B in AB and BC, ...
Using just gcd(x,y) (without minus one) solves this problem.

Unfortunately, my results of countLatticePoints were off by 4:
the corners A, B, C and D are not part of the original bounding-box and must be subtracted, too.

## Alternative Approaches

There are many symmetries. For example I could assume a <= b and reduce the search space accordingly.

## Note

I set up a few caches to speed up the computations considerably:

• the same GCDs have to by computed quite many times, therefore I store them in pointsOnEdge
• sqrt is pretty slow and a simple container isSquare stores a true flag for each square up to the largest area of an quadrilateral (100,100,100,100) → about 20000.

# 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 4 | ./504

Output:

Note: the original problem's input 100 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 <vector>

// cache gcd(a,b)
std::vector<std::vector<unsigned int>> pointsOnEdge;

// greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}

// count lattice points with Pick's theorem
// A(a, 0), B(b, 0), C(-c, 0), D(-d, 0)
unsigned int countLatticePoints(unsigned int a, unsigned int b, unsigned int c, unsigned int d)
{
// Pick's theorem: A = i + b/2 - 1
// where i is the area inside the quadrilateral
// and b is the number of lattice points on the boundary
// https://en.wikipedia.org/wiki/Pick%27s_theorem

// total area of the bounding box
auto rectangle = (a + c) * (b + d);
// only half of the area belongs to the quadrilateral
auto inside = rectangle / 2;

// points on the edges of the quadrilateral
auto boundary = pointsOnEdge[a][b] + pointsOnEdge[b][c] + pointsOnEdge[c][d] + pointsOnEdge[a][d];
// the four corners are not inside the bounding box (they are on its edges)
boundary -= 4;

// Pick's formula
return inside - (boundary / 2) - 1;
}

int main()
{
unsigned int limit = 100;
std::cin >> limit;

// precompute number of lattice points lying exactly on the sides/edges
pointsOnEdge.resize(limit + 1);
for (auto& x : pointsOnEdge)
x.resize(limit + 1);
for (unsigned int a = 1; a <= limit; a++)
for (unsigned int b = 1; b <= limit; b++)
// note: only one endpoint is included for each edge
//       e.g. a=4 b=2 gcd(4,2)=2
//       => has actually 3 lattice points:
//          both endpoints plus point (3,1)
//       but each endpoint is part of two edges therefore it would be counted twice
//       that means that keeping only one endpoint is okay and much simpler
pointsOnEdge[a][b] = pointsOnEdge[b][a] = gcd(a, b);

// precompute square numbers
// note: std::vector<bool> would be sufficient but std::vector<char> is about 20% faster
std::vector<unsigned char> isSquare(countLatticePoints(limit, limit, limit, limit) + 1, false);
for (size_t i = 1; i*i < isSquare.size(); i++)
isSquare[i*i] = true;

// plain enumeration of all combinations, 100^4 = 10^8 iterations
auto count = 0;
for (unsigned int a = 1; a <= limit; a++)
for (unsigned int b = 1; b <= limit; b++)
for (unsigned int c = 1; c <= limit; c++)
for (unsigned int d = 1; d <= limit; d++)
// count only perfect squares
if (isSquare[countLatticePoints(a, b, c, d)])
count++;

// display result
std::cout << count << std::endl;
return 0;
}


This solution contains 11 empty lines, 25 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.22 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

August 11, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 15% (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.

 << problem 501 - Eight Divisors Tangent Circles - problem 510 >>
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