Representing a Maze in an Embedded System: Some Thoughts (posted 19 Nov 2010)
I’ve started thinking about maze mapping and generation algorithms for micromouse, and I thought what I came up with was worth sharing.
So, first of all: A representation of a maze, in a computer, is usually going to come down to some form of a graph–that is, the kind with nodes (the squares one may stand in) connected by edges (areas without walls). Of course, for a typical 2d square maze there are going to be some assumptions we can make:

The graph is bidirectional. That is, if you can go from box 1 to box 2, you can also go from box 2 back to box 1.

The graph is what are called “simple.” This means that boxes don’t really have edges connecting to themselves (for example, there’s no warp portal that will take you from box 1 to, umm, box 1), nor are there multiple, equivalent edges (There aren’t two edges that go from box 1 to box 2).
So, for example’s sake, suppose we have the following graph:
[1]a[2]
 \ 
b c d
 \ 
[3]e[4]
Hopefully, that makes sense. I don’t feel like drawing pictures—I’ve done enough of that lately.
Based on what I know about graph data structures, there are a few ways to represent them:

A variation of linked lists: In this case, you have all these variables that represent your nodes, and each node has links to other nodes you can jump to. If the links are bidirectional, one needs another link to take them back. Maybe in JSON, we could encode it like this:
{ ‘n1’: [‘n2’, ‘n3’, ‘n4’],
‘n2’: [‘n1’, ‘n4’],
‘n3’: [‘n1’, ‘n4’],
‘n4’: [‘n1’, ‘n2’, ‘n3’] }
Keep in mind that, in C(++) land, these links are usually more pointeresque.
This model makes sense if all you want to do is traverse the tree–You just start at The Beginning and follow the links all the way down. On the other hand, it’s harder to ask questions like, “what does link 7 look like?” because, typically, you have to traverse your way down the list to GET to link 7. So: Pretty intuitive and awesome in some ways, but not right for what we’re doing.

Incidence matrix: Here, you have a matrix that looks something like:
(1) (2) (3) (4) (a) 1 1 0 0 (b) 1 0 1 0 (c) 1 0 0 1 (d) 0 1 0 1 (e) 0 0 1 1
Basically, your rows represent edges, and your columns represent nodes. To read the incidence matrix, you read across a row and see which nodes your edge connects. For example, row “a” has ones at columns (1) and (2), meaning edge “a” connects nodes (1) and (2). This can be extended to encode simple loops as well, though we don’t really need to concern ourselves with that for this case since simple loops shouldn’t happen in a maze.
This approach works great if you care as much about your edges as you do your nodes. However, I think their identities only matter incidentally in terms of the maze, as it’s just getting to other nodes that we care about. Also, its size is equal to the number of edges times the number of nodes–for an 8x8 maze, that ends up being (8 x 8) x (2 x 8 x 7) = 7168. That’s pretty big (Numerical wizards: Shush.)!

Adjacency Matrix: This matrix looks more like this:
(1) (2) (3) (4) (1) 0 1 1 1 (2) 1 0 0 1 (3) 1 0 0 1 (4) 1 1 1 0
This matrix can be read as, “Does node {row} connect to node {column}?” For example, if we want to know if there is an edge connecting nodes 2 and 3, we may check A(2, 3) and see that, in fact, there is no edge. In other words, there is a wall.
This method, I think, is the right one for our particular problem. First, it allows us to quickly look up whether a wall exists or not. Second, it can actually be encoded in a fairly compact manner, even without talking about sparseness of a matrix, given our initial assumptions:
 The lack of simple loops means that the diagonal is always all zeros.
 The bidirectionality of the graph means our matrix is diagonal.
This means that we only actually need to hold the part of the matrix above the diagonal in memory. That is, there are only (n^2  n)/2 unique elements we have to be concerned about, a’priori. For a 8x8 matrix, that’s 2016 elements. It’s an improvement over the incidence matrix for sure, but can we do better?
Of course we can.
I began to write down the incidence matrix for a 3x3 “maze” with no walls:
(1)(2)(3)
  
(4)(5)(6)
  
(7)(8)(9)
and it looked like this:
[R1] [R2] [R3]
_ (1) (2) (3)  (4) (5) (6)  (7) (8) (9)
(1) 0 1 0  1 0 0  0 0 0
[R1] (2) 1 0 1  0 1 0  0 0 0
(3) 0 1 0  0 0 1  0 0 0

(4) (look up &  0 1 0  1 0 0
[R2] (5) to the  1 0 1  0 1 0
(6) left!)  0 1 0  0 0 1

(7) (look in  (look up &  0 1 0
[R3] (8) the upper  to the  1 0 1
(9) right!)  left!)  0 1 0
We can see that the semidiagonal blocks are identity matrices. In retrospect, maybe this isn’t surprising, since blocks in a row connect to the ones directly above them. Of course, the other blocks not on the diagonal are zero matrices, since, nonadjacent rows do not connect to each other directly. Meanwhile, the ondiagonal blocks have an interesting form that in only 3d is hard to see.
So, let’s write down the adjacency matrix for a row of four elements:
[1][2][3][4]
(1) (2) (3) (4)
(1) 0 1 0 0
(2) 1 0 1 0
(3) 0 1 0 1
(4) 0 0 1 0
So, in an interestingly selfsimilar manner, the block matrices on the diagonal have ones on their offdiagonals, and zeros everywhere else! Fascinating.
Knowing that we have this definite pattern should be able to help us more efficiently store our graph in memory (keep in mind, memory is at a premium on a microcontroller). There are two parts to this: First, there’s the question of how many slots do we need for an nxn maze?
First, we know that for an nxn maze, we have nxn block matrices, and that we only need to encode (n^2 + n)/2 of those, due to symmetry. For example, for our 3x3 maze we only have 6 independent block matrices to concern ourselves with. Second, we know that n of those matrices are on the diagonal, (n1) of those are just off the diagonal, and that the rest are zero anyway. So, that means we only actually have to keep track of 2n1 block matrices in total (5 in the 3x3 case). Moreover, for the diagonal matrices, we have (n1) unique values to track (due again to symmetry), and for the offdiagonal matrices, we have n unique values. This brings our total number of unique values to n(n1) + (n1)n, or 2(n^2  n). In the case of an 8x8 matrix, this comes out to 112.
Not bad.
(Incidentally, this is a roundabout way of counting the total number of walls.)
Second, there is figuring out if there is a nice way to translate a (row, column) pair in the adjacency matrix into, say, a simple 1D list of all the values that could potentially be 1 from left to right. I’ll ignore that for now–I already feel like I’m prematurely optimizing. Plus, typical sparse matrix notation (i, j, value) isn’t too bad.