## Computing Eulerian paths is harder than you think

Everyone who has studied any graph theory at all knows the celebrated story of the Seven Bridges of Königsberg, and how Euler gave birth to modern graph theory while solving the problem.

Euler’s proof is clever, incisive, not hard to understand, and a great introduction to the kind of abstract reasoning we can do about graphs. There’s little wonder that it is often used as one of the first nontrivial graph theory results students are introduced to, e.g. in a discrete mathematics course. (Indeed, I will be teaching discrete mathematics in the spring and certainly plan to talk about Eulerian paths!)

Euler’s 1735 solution was not constructive, and in fact he really only established one direction of the “if and only if”:

If a graph has an Eulerian path, then it has exactly zero or two vertices with odd degree.

This can be used to rule out the existence of Eulerian paths in graphs without the right vertex degrees, which was Euler’s specific motivation. However, one suspects that Euler knew it was an if and only if, and didn’t write about the other direction (if a graph has exactly zero or two vertices with odd degree, then it has an Eulerian path) because he thought it was trivial.1

The first person to publish a full proof of both directions, including an actual algorithm for finding an Eulerian path, seems to be Carl Hierholzer, whose friend published a posthumous paper in Hierholzer’s name after his untimely death in 1871, a few weeks before his 31st birthday.2 (Notice that this was almost 150 years after Euler’s original paper!) If the vertex degrees cooperate, finding an Eulerian path is almost embarrassingly easy according to Hierholzer’s algorithm: starting at one of the odd-degree vertices (or anywhere you like if there are none), just start walking through the graph—any which way you please, it doesn’t matter!—visiting each edge at most once, until you get stuck. Then pick another part of the graph you haven’t visited, walk through it randomly, and splice that path into your original path. Repeat until you’ve explored the whole graph. And generalizing all of this to directed graphs isn’t much more complicated.

So, in summary, this is a well-studied problem, solved hundreds of years ago, that we present to students as a first example of a nontrivial yet still simple-to-understand graph proof and algorithm. So it should be pretty easy to code, right?

## So what’s the problem?

Recently I came across the eulerianpath problem on Open Kattis, and I realized that although I have understood this algorithm on a theoretical level for almost two decades (I almost certainly learned it as a young undergraduate), I have never actually implemented it! So I set out to solve it.

Right away the difficulty rating of 5.7 tells us that something strange is going on. “Easy” problems—the kind of problems you can give to an undergraduate at the point in their education when they might first be presented with the problem of finding Eulerian paths—typically have a difficulty rating below 3. As I dove into trying to implement it, I quickly realized two things. First of all, given an arbitrary graph, there’s a lot of somewhat finicky work that has to be done to check whether the graph even has an Eulerian path, before running the algorithm proper:

1. Calculate the degree of all graph vertices (e.g. by iterating through all the edges and incrementing appropriate counters for the endpoints of each edge).
2. Check if the degrees satisfy Euler’s criteria for the existence of a solution, by iterating through all vertices and making sure their degrees are all even, but also counting the number of vertices with an odd degree to make sure it is either zero or two. At the same time, if we see an odd-degree vertex, remember it so we can be sure to start the path there.
3. If all vertices have even degree, pick an arbitrary node as the start vertex.
4. Ensure the graph is connected (e.g. by doing a depth-first search)—Euler kind of took this for granted, but this technically has to be part of a correct statement of the theorem. If we have a disconnected graph, each component could have an Eulerian path or cycle without the entire graph having one.

And if the graph is directed—as it is in the eulerianpath problem on Kattis—then the above steps get even more finicky. In step 1, we have to count the in- and outdegree of each vertex separately; in step 2, we have to check that the in- and outdegrees of all vertices are equal, except for possibly two vertices where one of them has exactly one more outgoing than incoming edge (which must be the start vertex), and vice versa for the other vertex; in step 4, we have to make sure to start the DFS from the chosen start vertex, because the graph need not be strongly connected, it’s enough for the entire graph to be reachable from the start vertex.

The second thing I realized is that Hierholzer’s algorithm proper—walk around until getting stuck, then repeatedly explore unexplored parts of the graph and splice them into the path being built—is still rather vague, and it’s nontrivial to figure out how to do it, and what data structures to use, so that everything runs in time linear in the number of edges. For example, we don’t want to iterate over the whole graph—or even just the whole path built so far—to find the next unexplored part of the graph every time we get stuck. We also need to be able to do the path splicing in constant time; so, for example, we can’t just store the path in a list or array, since then splicing in a new path segment would require copying the entire path after that point to make space. I finally found a clever solution that pushes the nodes being explored on a stack; when we get stuck, we start popping nodes, placing them into an array which will hold the final path (starting from the end), and keep popping until we find a node with an unexplored outgoing edge, then switch back into exploration mode, pushing things on the stack until we get stuck again, and so on. But this is also nontrivial to code correctly since there are many lurking off-by-one errors and so on. And I haven’t even talked about how we keep track of which edges have been explored and quickly find the next unexplored edge from a vertex.

I think it’s worth writing another blog post or two with more details of how the implementation works, both in an imperative language and in a pure functional language, and I may very well do just that. But in any case, what is it about this problem that results in such a large gap between the ease of understanding its solution theoretically, and the difficulty of actually implementing it?

1. Actually, the way I have stated the other direction of the if and only if is technically false!—can you spot the reason why?

2. Though apparently someone named Listing published the basic idea of the proof, with some details omitted, some decades earlier. I’ve gotten all this from Herbert Fleischner, Eulerian Graphs and Related Topics, Annals of Discrete Mathematics 45, Elsevier 1990. Fleischner reproduces Euler’s original paper as well as Hierholzer’s, together with English translations.

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 5 Responses to Computing Eulerian paths is harder than you think

1. Derek Elkins says:

My snap reaction to your last question is that the non-determinism causes the problem. The stack is typical backtracking search code. However, it’s not *just* the non-determinism but the admixture of state that you’ll likely want to keep track of things like which nodes have been visited. Using a *linear* logic programming language, like Linear Meld or Lolli(Mon), would likely let you push (almost?) all the issues onto the language implementation. In general, linear logic programming languages handle graph algorithms rather well

As I imagine it, for the undirected case, the “non-determinism” would mostly be the non-determinism in splitting the linear context. State would be handled by facts in the linear context, e.g. a linear unvisited_edge/2 fact for each edge. I think you can handle path splicing by maintaining linear facts that correspond to a linked list of edges and just consuming and reproducing the appropriate linear facts. I also think that (at the cost of doing the traversal twice) the non-determinism with respect to which odd degree node to start at will just naturally do the right thing. With what I’m imagining, the typical implementation approaches for predicates which provide constant time lookup (assuming the arguments are primitive types) would lead to a linear time algorithm. I suspect you could make a very elegant and performant implementation in such a language. I have, however, just thought about this and not produced any code so take it with a huge grain of salt.

• Brent says:

Fascinating, thanks for your comment. I’m not too familiar with linear logic programming languages, I’ll have to look into these more. I do like the thought that there exists a language in which the implementation of this algorithm really is as easy as it sounds like it should be, and all the other complexity is really just an incidental consequence of a mismatch between problem and programming model.

• Derek Elkins says:

I wrote up some LolliMon code, see (https://gist.github.com/derekelkins/d32af329b6e44ed814de6d6589534fd6). It could be made a little bit better if LolliMon supported a simple form of negation. It probably does NOT run in linear time even for cases where any start node works. (It definitely doesn’t run in linear time when there is an in-degree /= out-degree node as I just try every node rather than filtering to the appropriate start node, but that would be easy enough to change.) I think being able to specify a secondary index on `untraversed_edge` is the main thing you’d need to do to get a better expectation of linear asymptotic complexity (plus doing something to ensure `consume_connections` only runs when it will succeed).

The core of the code is the `euler_step` and `euler_prestep` predicates. `euler_prestep` just adds all the untraversed outgoing edges of a node the first time it is visited to a (dynamic) collection of edges that are incident to the path so far. `euler_step` has three main cases: either there’s an untraversed edge adjacent to the current node, in which case we traverse it consuming the linear assumption, or there is no untraversed edge adjacent to the current node so we arbitrarily pick one of the incident edges and start a walk from there which we’ll splice into the path, or there are no untraversed edges at all in which case we go to `consume_connections` which collects all the linear `connection` assumptions into a list clearing out the linear context. `connection` is essentially a linked list of edges represented as a collection of linear assumptions. To “mutate” this list, we simply consume an assumption and introduce new ones. This is how splicing is handled.

`incident_edge` is also a bit more subtle than I described it above. When we first visit a node, we are adding a hypothetical rule to the `incident_edge` predicate. We don’t actually consume any `untraversed_edge` assumptions at that point as my earlier description misleadingly suggests. What’s particularly powerful here is that the rule has free variables. A different way of implementing/thinking about it would be to use a materialized view approach. Here we would actually copy the relevant part of `untraversed_edge` into `incident_edge` and then removing an entry from either one (by consuming the linear assumption) would remove the corresponding entry in the other.

Another minor but crucial detail is the affine as opposed to linear assumption of `unvisited_node` for the starting node. This could be handled in a different way, but essentially the issue is that we arguably “visit” the starting node but since we don’t yet have a previous edge we can’t populate `incident_edge`. So I leave the start node marked as unvisited so that when it gets visited again it will go through the first time visit stuff. The problem is that it may never get visited again in which case the linear context won’t be empty at the end and the `euler_path` predicate will fail. In particular, the second case of the `consume_connections` predicate fails because `one` only succeeds if the linear context is empty. Using the affine context allows this extra assumption to be ignored.

2. Daniel Beecham says:

The link to Kattis has moved to https://open.kattis.com/problems/eulerianpath

• Brent says:

Thanks, fixed!

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