Time/Space cost of Taxicab algorithm?

Question

The following is an algorithm for generating "Taxicab numbers" using a priority queue (pq). Vector is an arbitrary data type that allows for storage of two number and their cubed sum. For those unaware (although you really don't need to know), a taxicab number is an integer that can be expressed as the sum of two cubes of integers in two different ways: $a^3+b^3 = c^3+d^3$. An example would be $1729 = 12^3 + 1^3 = 10^3 + 9^3$.

for i = 1..n
 pq.insert( Vector(i^3+i^3,i,i) )

prev = Vector(0, 0, 0)

while not pq.empty()
 curr = pq.deleteMin()
 if prev[0] == curr[0]
      print curr[0] is a Taxicab number that can be expressed as
      prev[1]^3 + prev[2]^3 and curr[1]^3 + curr[2]^3
 prev = curr
 if curr[2] < N
     j = curr[2] + 1
     pq.insert( Vector(curr[1]^3 + j^3, curr[1], j) )

I know inserting an item into the priority queue is $O(\log n)$ but I am not sure how this relates to space usage and runtime. Can someone help?

Answer 1

If you think about this program, you will realize pretty quickly that the vectors you're pushing onto the stack iterate over the values [a^3 + b^3, a, b] for all $1\leq a, b\leq N$ This means you're pushing $N^2$ items onto the stack over the course of the program.

Since each removal (and push, depending on the implementation of the PQ) takes $\Theta (\lg N)$ time, your algorithm would run in $\Theta (N^2\lg N)$ time. There's a straightforward $\Theta (N^2)$ solution, so you're not optimal yet.