Find taxicab numbers in $O(n)$ time

Question

This is a final exam question in my algorithms class:

$k$ is a taxicab number if $k = a^3+b^3=c^3+d^3$, $a,b,c,d$ are distinct positive integers. Find all taxicab number $k$ such that $a,b,c,d < n$ in $O(n)$ time.

I don't know if the problem had a typo or not, because $O(n^3)$ seems more reasonable. The best I can come up with is $O(n^2 \log n)$, and that's the best anyone I know can come up with.

The $O(n^2 \log n)$ algorithm:

1. Try all possible $a^3+b^3=k$ pairs, for each $k$, store $(k,1)$ into a binary tree(indexed by $k$) if $(k,i)$ doesn't exist, if $(k,i)$ exists, replace $(k,i)$ with $(k,i+1)$

2. Transverse the binary tree, output all $(k,i)$ where $i\geq 2$

Are there any faster methods? This should be the best possible method without using any number theoretical result because the program might output $O(n^2)$ taxicab numbers.

Is $O(n)$ even possible? One have to prove there are only $O(n)$ taxicab numbers lesser than $2n^3$ in order to prove there exist a $O(n)$ algorithm.

Edit: The professor admit it was a typo, it should have been $O(n^3)$. I'm happy he made the typo, since the answer Tomer Vromen suggested is amazing.

Answer 1

I don't know about $O(n)$, but I can do it in $O(n^2)$. The main idea is to use initialization of an array in $O(1)$ (this is the best reference that I've found, which is weird since this seems like a very important concept). Then you iterate through all the possible $(a,\ b)$ pairs and do the same as step 1 in your proposed algorithm. Since $a^3+b^3 \leq 2n^3$, the array needs to be of size $2n^3$, but it's still initialized in $O(1)$. Accessing an array element is $O(1)$ like in a regular array.

Answer 2

Apparently this is a solved problem and every rational solution to $x^3 + y^3 = z^3 + w^3$ is proportional to

$x = 1 − (p − 3q)(p^2 + 3q^2)$

$y = −1 + (p + 3q)(p^2 + 3q^2)$

$z = (p + 3q) − (p^2 + 3q^2)^2$

$w = −(p − 3q) + (p^2 + 3q^2)^2$

See: http://129.81.170.14/~erowland/papers/koyama.pdf, Page 2.

Also see: http://sites.google.com/site/tpiezas/010 (search for J. Binet).

So it seems like an O(n^2) algorithm might be possible. Perhaps using this more cleverly can give us an O(n) time algorithm.

Hope that helps. Please do update us with the solution given by your professor.

Answer 3

For $O(n^2)$ (randomized) time, you can also use a hashtable of size $\Theta(n^2)$. Looking up will be constant time because the number of taxicab numbers is $O(n^2)$.