Algorithm for Simple Squared Squares - algorithm

Question

I want to split a square in unequal squares.
After some search on the web found this Link.
This is an output i need :
Does anyone have idea for this?

Answer 1

As Yves Daoust said the algorithm to solve this is going to be slow. The first challenge is to determine what squares COULD be combined to fit into your big square. Then figure out if they WILL fit in there.
I would first filter by area.
To answer the first part you need to look for a combination of squares that will fit into your big one. There are likely multiple combinations as a 5x5 square takes up the same area as a 3x3 with a 4x4 square. This is a O(2^n) problem in itself.
Then attempt to arrange them.
I would make a matrix that is the size of your big square. Then starting at the topmost then right most index add in a square by marking the matrix positions as occupied by that square. Then move to the next unoccupied space, based on the previous rules adding an unused square. If no square fits then remove the previous square and continue to the next. This is a method begging for recursion.
As I said at the beginning this is a SLOW way to do it but it will give you a solution if one exists.

Answer 2

I used a dynamic programming approach for solving this. but it works until n ~ 50. I stored a solution as a bitset for efficiency:
You can compile the code yourself with:
$ g++ -O3 -std=c++11 squares.cpp -o squares
#include <bitset>
#include <iostream>
#include <list>
#include <vector>
using namespace std;
constexpr auto N = 116;
class FastSquareList {
public:
FastSquareList() = default;
FastSquareList(int i) { mask_.set(i); }
FastSquareList operator+(int i) const {
FastSquareList result = *this;
result.mask_.set(i);
return result;
}
bool has(int i) const { return mask_.test(i); }
void display() const {
for (auto i = 1; i <= N; ++i) {
if (has(i)) {
cout << i * i << " ";
}
}
cout << endl;
}
private:
bitset<N + 1> mask_;
};
int main() {
int n;
cin >> n;
vector<list<FastSquareList> > v(n * n + 1);
for (int i = 1; i <= n; ++i) {
v[i * i].push_back(i);
for (int a = i * i + 1; a <= n * n; ++a) {
int p = a - i * i;
for (const auto& l : v[p]) {
if (l.has(i)) {
continue;
}
v[a].emplace_back(l + i);
}
}
}
for (const auto& l : v[n * n]) {
l.display();
}
cout << "solutions count = " << v[n*n].size() << endl;
return 0;
}
an example:
$ ./Squares
15
9 16 36 64 100
25 36 64 100
1 4 9 16 25 49 121
4 36 64 121
4 100 121
4 16 25 36 144
1 16 64 144
81 144
4 16 36 169
4 9 16 196
4 25 196
225
solutions count = 12