highest tower of cubes (with numbers on sides)

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Problem:
There are N cubes. There are M numbers. Each side of cube has number from 1 to M. You can stack one cube on another if their touching sides have same number (top side of bottom cube and bottom side of top cube has same number). Find the highest tower of cubes.

Input: number N of cubes and number M.
Example:

INPUT: N=5, M=6. Now we generate 5 random cubes with 6 sides = <1,M>.
[2, 4, 3, 1, 4, 1]
[5, 1, 6, 6, 2, 5]
[2, 5, 3, 1, 1, 6]
[3, 5, 6, 1, 3, 4]
[2, 4, 4, 5, 5, 5]

how you interpret single array of 6 numbers is up to you. Opposite sides in cube might be index, 5-index (for first cube opposite side of 4 would be 4). Opposite sides in cube might also be index and index+1 or index-1 if index%2==0 or 1 respectively. I used the second one. Now let's say first cube is our current tower. Depending on the rotation top color might be one of 1, 2, 3, 4. If the 1 is color on top we can stack on top of it second, third or fourth cube. All of them has color 1 on their sides. Third cube even has two sides with color 1 so we can stack it in two different ways.
I won't analyse it till the end because this post would be too long. Final answer for these (max height of the tower) is 5.

My current solution (you can SKIP this part):
Now I'm just building the tower recursively. Each function has this subproblem to solve: find highest tower given the top color of current tower and current unused cubes (or current used cubes). This way I can memoize and store results for tuple(top color of tower, array of used cubes). Despite memoization I think that in the worst case (for small M) this solution has to store M*(2^N) values (and this many cases to solve).

What I'm looking for:
I'm looking for something that would help me solve this efficiently for small M. I know that there is tile stacking problem (which uses Dynamic Programming) and tower of cubes (which uses DAG longest path) but I don't see the applicability of these solutions to my problem.

dynamic-programming algorithm
1 Answers

You won't find a polynomial time solution- if you did, we'd be able to solve the decision variant of the longest path problem (which is NP-Complete) in polynomial time. The reduction is as follows: for every edge in an undirected graph G, create a cube with opposing faces (u, v), where u and v are unique identifiers for the vertices of the edge. For the remaining 4 faces, assign globally unique identifiers. Solve for the tallest cube tower, this tower's height will be the length of the longest path of G, return if path length equals the queried value (yes/no).

However, you could still solve it in something like O(M^3*(N/2)!*log(N)) time (I think that bound is a bit loose, but its close). Use divide and conquer with memoization. Find all longest paths using cubes [0, N) beginning with a value B in range [0, M) and ending with a value E in range [0, M), for all possible B and E. To compute this, recurse, partitioning the cubes evenly in every possible way. Keep recursing until you hit the bottom (just one cube). Then begin merging them (by combining cube stacks that end in X with those beginning with X, for all X in [0, M). Once that's all done, at the topmost level just take the max of all the tower heights.

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