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In n-space the problem is almost as easy as in 2-space if you can only go in forward in each dimension. If you can go in both directions and you want to find the number of non self intersecting paths, the problem is hard even in 2-space (as far as I can tell). If you allow paths to be self-intersecting, then the count is obviously infinite. Note that the actual interview question asks you to list all paths, not count them.

To count paths in 2-space from (0,0) to (5,6), you have the operations "X++" and "Y++" which go right and up, respectively. Each increasing path from (0,0) to (5,6) has to be some permutation of 5 times X++ and 6 times Y++. So the count is (5+6)!/(5!6!) where the exclamation mark denotes factorial. This extends to higher dimensions by adding an additional operation "Z++". Then the count to go from (0,0,0) to (5,6,7) is (5+6+7)!/(5!6!7!).

"I vaguely recall counting lattice paths in 3-space being a wickedly hard problem with no known polynomial time solution."

I think you can easily write the recursion:

  p(x,y,z) = p(x-1,y,z) + p(x,y-1,z) + p(x,y,z-1)

Leave out the term with x-1, y-1, or z-1 for the edge cases for x, y, and/or z equal to zero.

With that in hand, it is easy to compute all values bottom up, starting with those where x+y+z = 0, 1, 2, etc.

Definitely fewer than (x+y+z)^3 values to compute, all of them in O(1) (disregarding cases where the numbers become bignums)

Generalization to any number of dimensions seems easy, too.

A closed form solution, that might be harder.

You're assuming that all movement must be in one of three directions, but in general, you can go in six directions -- forward or backward along the x, y, and z axes. Your cubic-time DP algorithm doesn't solve the self-avoiding walk problem, which I think is what jsolson vagely recalls.

http://en.wikipedia.org/wiki/Self-avoiding_walk

So, the professor in question was Dana Randall. I believe she was talking about this topic: http://en.scientificcommons.org/42983059

Basically, yes, self-avoiding paths.

This was basically a brief aside in her honors undergrad algorithms course, so the topic was a bit beyond what I was prepared for at the time :)

I wasn't assuming, I was following what the OP wrote: "by going only up or right one point with each move (that is, each move is either (0,1) or (1,0)))" for 3D.

Counting self-avoiding walks is hard in 2D, too (http://oeis.org/A007764)

Glad I'm not the only one who thought this problem statement must be missing constraints if they expect a code solution on a whiteboard.

I assumed it was intentionally ambiguous (like a lot of problem statements in practice), to see how the interviewees respond. There are a few interesting interpretations, some of which are obviously exponential time, infinite time, or impossible (ie uncountably many solutions).