No. One can still visit 2, while the other can still visit 1. The visited set changes future legality. Equal set and current vertex would be sufficient under the fixed edge rules here.

For vertices 0..n-1, bit v is set when v has been visited. mask & (1 << v) checks membership; mask | (1 << v) includes the vertex. Count vertex sequences that start at 0 and visit all vertices exactly once; the endpoint is unrestricted. Assume n>=1, valid labels, and no duplicate neighbors.

from functools import cache

def complete_routes(graph):
    n = len(graph)
    all_seen = (1 << n) - 1
    @cache
    def ways(node, mask):
        if mask == all_seen:
            return 1
        return sum(ways(v, mask | (1 << v))
                   for v in graph[node] if not mask & (1 << v))
    return ways(0, 1)

assert complete_routes([[1, 2], [0, 2], [0, 1]]) == 2
assert complete_routes([[1], [], []]) == 0
assert complete_routes([[]]) == 1

The two triangle routes are 0→1→2 and 0→2→1. This does not require returning to the start. Adding that requirement changes the terminal condition.

There are at most n·2**n states, each inspecting up to n neighbors: O(n²·2**n) time and O(n·2**n) cached space, plus O(n) stack depth. This can improve on enumerating every ordering, but remains exponential. A bitmask stored in one integer does not make the state space one small linear dimension.

Analogy limit: “compress” here means encode a set compactly. It does not justify forgetting its members or turning an exponential problem into a polynomial one.

No. For a fixed graph, ordinary reachability needs only the graph vertex as the search state and a global discovered set to avoid repeated exploration. The exactly-once route constraint created the exponential state requirement.