# Graph labyrinth solving sequence

Starting from a vertex of an unknown, finite, strongly connected directed graph, we want to 'get out' (reach the vertex of the labyrinth called 'end'). Each vertex has two exits (edge which goes from vertex in question to an other one), one exit is labeled 'a', the other exit is labeled 'b'.

We have limitless 'memory' but, we don't recognize when we arrive at the same vertex again, so at each step we can only pick if we go exit a or exit b, or we recognize when we have entered the exit vertex. Show that there is an algorithm to get out of any maze!

Write the algorithm. If n is its input, then its output is a sequence 'a', 'b' that exits any maze with at most n vertices.

I got this assignment (math student) in a course to do with algorithms. I don't believe the actual code outputting the 'a', 'b' sequence is particularly difficult once the structure of the function is found mathematically.

I've had multiple ideas, it is clear, that were we to find a sequence that would guarantee a visit to all vertices, we would be done, as one of them must be the "end" vertex. This would be easy, if the edges weren't directed, because we would try all possible 1-long sequences by doing a 1-long sequence and tracing our steps back and doing the next, if we didnt reach the end. Then do the same with 2-long sequences, and so on. I think the longest sequnces we would have to try would be $$2^n$$ in length and we would get an incredibly long sequence but it could be shown to work. But this solution relies heavily on the fact that we could always trace our way back to the vertex at which we start and thereby doing something like a BFS. That doesn't work here, as we can't go back on edges we came from (at least not necessarily).

I also stumbled upon De Bruijn sequences. A De Bruijn sequence for a given alphabet (in this case, 'a' and 'b') and a given length 'n' is a cyclic sequence in which every possible subsequence of length 'n' appears exactly once. I don't see how this would work, but maybe if we collate this sequence 'n+1' times, we would visit all vertices, because we would be guaranteed to have tried all 'walks' from all vertices, but I don't see this rigorously at all.

I found similar posts, but their answers are not conclusive at all, and the fact that we do do not recognize a vertex we have already been at significantly complicates this version of the problem. That is why I come to you, I need some serious guidance, as I have no clue about this at all

Edit:

from collections import deque

def generate_graphs(n):

Generate all strongly connected directed graphs with n vertices and out-degree 2.

graphs = []
# Generate all possible graphs with the given constraints
return graphs

def find_exit_path(graph, start, target):
"""
Find the path from the start vertex to the target vertex using BFS.
"""
queue = deque([(start, [])])
visited = set()

while queue:
vertex, path = queue.popleft()
if vertex == target:
return path

for neighbor in graph[vertex]:
if neighbor not in visited:
queue.append((neighbor, path + ['a'] if neighbor == graph[vertex][0] else path + ['b']))

return []  # No path found

def solve_labyrinth(graphs):
"""
Find the solution path that works for any labyrinth in the given set of graphs.
"""
solution_path = []
current_vertex = None

def solve_labyrinth(graphs):
"""
Find the solution path that works for any labyrinth in the given set of graphs.
"""
solution_path = []
current_vertex = None

for graph, start, target in graphs:
if not solution_path:
current_vertex = start
else:
# Walk along the existing solution path from the start vertex
current_vertex = start
for step in solution_path:
neighbor_a, neighbor_b = graph[current_vertex]
current_vertex = neighbor_a if step == 'a' else neighbor_b

exit_path = find_exit_path(graph, current_vertex, target)
solution_path.extend(exit_path)
current_vertex = graph[current_vertex][0] if exit_path[-1] == 'a' else graph[current_vertex][1]

return solution_path

# Generate all graphs with n vertices and out-degree 2
n = 5  # Example number of vertices
graphs = generate_graphs(n)

# Find the solution path
solution_path = solve_labyrinth(graphs)
print("Solution path:", "".join(solution_path))
$$$$
`

Let $$G_1, \dots, G_m$$ be an enumeration of all strongly connected directed graphs on at most $$n$$ vertices in which every vertex has out-degree 2 (the corresponding edges labelled $$a,b$$).
The algorithm proceeds in $$n^2m$$ rounds. In round $$(i-1) n^2 + (j-1) n + k$$ (where $$1 \leq i \leq m$$ and $$1 \leq j,k \leq n$$), the algorithm guesses that the graph in question is $$G_i$$, that at the very beginning we were at vertex $$\min(j,|V(G_i)|)$$, and that the end vertex is $$\min(k,|V(G_i)|)$$. If this is the case, then given what happened in the preceding rounds, the algorithm knows that we are at some vertex $$v$$. Since $$G_i$$ is strongly connected, there is a path from $$v$$ to $$k$$. The algorithm follows this path, and checks whether it has reached the (real) end vertex. If so, it terminates, and otherwise, it proceeds to the following round.
The sequence output by such an algorithm is known as a universal traversal sequence or a universal exploration sequence, and there are much more efficient constructions in the literature. Of particular note is the undirected $$d$$-regular case, in which there is a universal traversal sequence of polynomial length.
• We imagine all strongly connect directed graphs with all $n$ vertices having out-degree 2 and proceed to number them somehow. So, in round $i$ we suppose that we are in $G_i$ and from our previous steps and our supposed $s_i$ starting vertex (which can be chosen arbitrarily, as the graph is known to be strongly connected) we calculate where we think we are, namely $v_i$. Now, I don't understand why a path to one arbitrarily chosen $t_i$ would be good enough, as we might miss vertices in $G_i$, wouldn't we instead have to do a path on all vertices and then move on to $G_{i+1}$ ? Mar 23 at 20:00
• Your solution also works. In my solution, you go over all possible triples $(G,s,t)$ where $G$ is a graph and $s,t$ are vertices. Mar 24 at 9:18
• Oh wait, so we iterate for every $G_i$, seperately for every $s_i$ and every $t_i$? Mar 24 at 13:29
• You can look at it this way if you wish. In that case you will be iterating over all triplets $(G_i,s_j,t_k)$. But it's just a different way to enumerate the triplets. Mar 24 at 19:55