In the classic arcade game Pac-Man, the player moves the title character through a maze. The aim of the game is to eat all of the pac-dots that are spread throughout the maze while avoiding the ghosts that prowl it.
While playing Pac-Man recently, my concentration drifted from the pac-dots and I began to think about the best route I could take to complete the level.
Seven Bridges of Königsberg
In the 1700s, Swiss mathematician Leonhard Euler studied a related problem. The city of Königsberg had seven bridges, which the residents would try to cross while walking around the town. However, they were unable to find a route crossing every bridge without repeating one of them.
In fact, the city dwellers could not find such a route because it is impossible to do so, as Euler proved in 1735. He first simplified the map of the city, by making the islands into vertices (or nodes) and the bridges into edges.
This type of diagram has (slightly confusingly) become known as a graph, the study of which is called graph theory. Euler represented Königsberg in this way as he realised that the shape of the islands is irrelevant to the problem: representing the problem as a graph gets rid of this useless information while keeping the important details of how the islands are connected.
Euler next noticed that if a route crossing all the bridges exactly once was possible then whenever the walker took a bridge onto an island, they must take another bridge off the island. In this way, the ends of the bridges at each island can be paired off. The only bridge ends that do not need a pair are those at the start and end of the circuit.
This means that all of the vertices of the graph except two (the first and last in the route) must have an even number of edges connected to them; otherwise there is no route around the graph travelling along each edge exactly once. In Königsberg, each island is connected to an odd number of bridges. Therefore the route that the residents were looking for did not exist (a route now exists due to two of the bridges being destroyed during World War II).
This same idea can be applied to Pac-Man. By ignoring the parts of the maze without pac-dots the pac-graph can be created, with the paths and the junctions forming the edges and vertices respectively. Once this is done there will be twenty-four vertices, twenty of which will be connected to an odd number of edges, and so it is impossible to eat all of the pac-dots without repeating some edges or travelling along parts of the maze with no pac-dots.
This is a start, but it does not give us the shortest route we can take to eat all of the pac-dots: in order to do this, we are going to have to look at the odd vertices in more detail.
The Chinese Postman Problem
The task of finding the shortest route covering all the edges of a graph has become known as the Chinese postman problem as it is faced by postmen—they need to walk along each street to post letters and want to minimise the time spent walking along roads twice—and it was first studied byChinese mathematician Kwan Mei-Ko.
As the seven bridges of Königsberg problem demonstrated, when trying to find a route, Pac-Man will get stuck at the odd vertices. To prevent this from happening, all the vertices can be made into even vertices by adding edges to the graph. Adding an edge to the graph corresponds to choosing an edge, or sequence of edges, for Pac-man to repeat or including a part of the maze without pac-dots. In order to complete the level with the shortest distance travelled, Pac-man wants to add the shortest total length of edges to the graph. Therefore, in order to find the best route, Pac-Man must look at different ways to pair off the odd vertices and choose the pairing which will add the least total distance to the graph.
The Chinese postman problem and the Pac-Man problem are slightly different: it is usually assumed that the postman wants to finish where he started so he can return home. Pac-Man however can finish the level wherever he likes but his starting point is fixed. Pac-Man may therefore leave one odd node unpaired but must add an edge to make the starting node odd.
One way to find the required route is to look at all possible ways to pair up the odd vertices. With a low number of odd vertices this method works fine, but as the number of odd vertices increases, the method quickly becomes slower.
With four odd vertices, there are three possible pairings. For the Pac-Man problem there will be over 13 billion (1.37 × 1010) pairings to check. These pairings can be checked by a laptop running overnight, but for not too many more vertices this method quickly becomes unfeasible.
With 46 odd nodes there will be more than one pairing per atom in the human body (2.53 × 1028). By 110 odd vertices there will be more pairings (3.47 × 1088) than there are estimated to be atoms in the universe. Even the greatest supercomputer will be unable to work its way through all these combinations.
Better algorithms are known for this problem that reduce the amount of work on larger graphs. The number of pairings to check in the method above increases like the factorial of the number of vertices. Algorithms are known for which the amount of work to be done increases like a polynomial in the number of vertices. These algorithms will become unfeasible at a much slower rate but will still be unable to deal with very large graphs.
Solution of the Pac-Man Problem
For the Pac-Man problem, the shortest pairing of the odd vertices requires the edges marked in red to be repeated. Any route which repeats these edges will be optimal. For example, the route in green will be optimal.
One important element of the Pac-Man gameplay that I have neglected are the ghosts (Blinky, Pinky, Inky and Clyde), which Pac-Man must avoid. There is a high chance that the ghosts will at some point block the route shown above and ruin Pac-Man’s optimality. However, any route repeating the red edges will be optimal: at many junctions Pac-man will have a choice of edges he could continue along. It may be possible for a quick thinking player to utilise this freedom to avoid the ghosts and complete an optimal game.
Additionally, the skilled player may choose when to take the edges that include the power pellets, which allow Pac-Man to reverse the roles and eat the ghosts. Again cleverly timing these may allow the player to complete an optimal route.
Unfortunately, as soon as the optimal route is completed, Pac-Man moves to the next level and the player has to do it all over again ad infinitum.
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Does it have any effect on optimal solutions if you take into account the “optional” edges in the Pac-Man graph (including the wrap-around path)? I believe that would turn it into a “Rural Postman Problem” instead of a traditional Chinese Postman Problem. Can you suggest any good, accessible reading material on Rural Postman Problems?
In this case, taking the optional edges into account did not affect the end solution. I agree that this turns it into a Rural Postman Problem, but I’ve never found anything accessible (or much at all) about these problems.
I have done something a bit similar…
Graph theory used to solve some levels of Manic Miner optimally:
This is pixel-perfect optimal and takes enemy positions into account, so it makes very “risky” jumps 🙂
Is this for the unweighted PacMan graph? It looks like, if you consider the distances in the maze, you can do better (with a little shuffling at the bottom, though I’m not sure exactly how you’re treating the starting point).
This is for the weighted graph (lengths of the paths are the weights).
I don’t think you can do better than this (unless there was a bug in the code I wrote to check it)…
I’m guessing it has to do with the starting position (or maybe I’m just missing something). Replace the two lower-central red paths by the four-pip paths in the very bottom-center and just left of Pac-Man’s starting position.
Similar to a comment on your blog post, I also wonder how much this changes if you measure by-pixel and can turn around immediately after picking up a pip…
(I tried to comment last week, but it still hasn’t appeared. Guess I’ll try again?)
I think the difference is probably in how to treat the starting position (or perhaps I’m missing something obvious). Remove the two bottom-center red paths and replace them by the four-pac-dot segments in the bottom-center and just left of the starting position?