POTW #14-17 Five Room Puzzle
Finding A Eulerian Path That Visits Each Of The Five Rooms
John Snyder, FSA
January 10, 2014
23:55 EST
Puzzle
One of the oldest topological puzzles consists of drawing a continuous line across the closed network
shown in the figure below so that the line crosses each of the 16 segments of the network only once.
The curved line shown here does not solve the puzzle because it leaves one segment uncrossed. No
”trick”solutions are allowed, such as passing the line through a vertex or along one of the segments,
folding the paper, and so on.
Can the puzzle be solved on a planar surface? What about a sphere? Or a torus (donut)? Prove if each
of the cases can or cannot be solved.
Solution
This puzzle is sometimes called the 5 room problem where each rectangular area represents a room
that must be entered or left through a door in each of the line segments.
Here is a graph of the 5 rectangles in the plane. This is a connected graph because there is a path from
any one vertex to any other in the graph. To solve the puzzle we would need to find a Eulerian path that
visits each of the edges of the graph exactly once. This is not possible because Euler proved that such
a path exists if and only if the graph has at most two vertices of odd degree. The graph of the five rooms
shown below has 4 vertices having odd degree (the degree of a vertex is the number of edges connected to the vertex). Therefore, the puzzle cannot be solved in the plane.
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FiveRooms1417.nb
Another way to see that the puzzle cannot be solved in the plane is to consider the fact that a continuous line that enters and leaves one of the rectangular spaces must cross two line segments. Since the
two top and central bottom spaces in the original diagram are each surrounded by an odd number of
segments, then an end of a line must be inside each if all the segments are crossed. But a continuous
line has only two ends, so the puzzle is insoluble on a plane surface. This same reasoning applies if the
rectangles are drawn on a sphere, so the puzzle cannot be solved on a sphere.
As we see in the diagram below, the puzzle can be solved on a torus by having the path pass through
the torus’hole, traveling underneath, and then up the side to the top of the torus where the path enters
the final rectangle. This solves the puzzle on the torus.