How to find all taxicab numbers less than N?

Question

A taxicab number is an integer that can be expressed as the sum of two cubes of integers in two different ways: a^3+b^3 = c^3+d^3. Design an algorithm to find all t

Answer 1

But this takes N^2 space. Can we do better?

There exists an O(N) space solution based on a priority queue. Time complexity is O(N^2 logN). To sketch out the idea of the algorithm, here is the matrix M such that M[i][j] = i^3 + j^3 (of course, the matrix is never created in memory):

0 1 8 27 64 125 1 2 9 28 65 126 8 9 16 35 72 133 27 28 35 54 91 152 64 65 72 91 128 189 125 126 133 152 189 250
Observe that every line and every row is sorted in ascending order. Let PQ be the priority queue. First we put the biggest element in the priority queue. Then perform the following, as long as the PQ is not empty:

1. Pop the biggest element from PQ
2. add adajcent element above if the PQ doesn't have any element from that row
3. add adajcent element on the left if the PQ doesn't have any element from that column, and if it is not under the diagonal of the matrix (to avoid redundant elements)

Note that

1. You don' t need to create the matrix in memory to implement the algorithm
2. The elements will be popped from the PQ in descending order, from the biggest element of the matrix to its smallest one (avoiding elements from the redundant half part of the matrix).

Everytime the PQ issues the same value twice then we have found a taxicab number.

As an illustration, here is an implementation in C++. The time complexity is O(N^2 logN) and space complexity O(N).

#include 
#include 
#include 

using namespace std;

typedef unsigned int value_type;

struct Square
{
   value_type i;
   value_type j;
   value_type sum_of_cubes;
   Square(value_type i, value_type j) : i(i), j(j), sum_of_cubes(i*i*i+j*j*j) {}
   friend class SquareCompare;
   bool taxicab(const Square& sq) const
   {
     return sum_of_cubes == sq.sum_of_cubes && i != sq.i && i != sq.j;
   }
   friend ostream& operator<<(ostream& os, const Square& sq);
};

class SquareCompare
{
public:
  bool operator()(const Square& a, const Square& b)
  {
    return a.sum_of_cubes < b.sum_of_cubes;
  }
};

ostream& operator<<(ostream& os, const Square& sq)
{
  return os << sq.i << "^3 + " << sq.j << "^3 = " << sq.sum_of_cubes;
}

int main()
{
  const value_type N=2001;
  value_type count = 0;
  bool in_i [N];
  bool in_j [N];

  for (value_type i=0; i, SquareCompare> p_queue;

  p_queue.push(Square(N-1, N-1));
  in_i[N-1] = true;
  in_j[N-1] = true;

  while(!p_queue.empty()) {
    Square sq = p_queue.top();

    p_queue.pop();
    in_i[sq.i] = false;
    in_j[sq.j] = false;

    // cout << "pop " << sq.i << " " << sq.j << endl;

    if (sq.i > 0 && !in_i[sq.i - 1] && sq.i-1 >= sq.j) {
      p_queue.push(Square(sq.i-1, sq.j));
      in_i[sq.i-1] = true;
      in_j[sq.j] = true;
      // cout << "push " << sq.i-1 << " " << sq.j << endl;
    }
    if (sq.j > 0 && !in_j[sq.j-1] && sq.i >= sq.j - 1) {
      p_queue.push(Square(sq.i, sq.j-1));
      in_i[sq.i] = true;
      in_j[sq.j - 1] = true;
      // cout << "push " << sq.i << " " << sq.j-1 << endl;
    }

    if (sq.taxicab(p_queue.top())) {
      /* taxicab number */
      cout << sq << " " << p_queue.top() << endl;
      count++;
    }
  }
  cout << endl;

  cout << "there are " << count << " taxicab numbers with a, b, c, d < " << N << endl;

  return 0;
}