Royal Institution Masterclass in Mathematics – Problem sheet one: Finding
Eulerian circuits
Standard questions (30 minutes)
Q1 For the following graph, G, determine whether the graph is Eulerian. If the graph is Eulerian trace
a circuit (same start and end point) that starts and finishes at node x1, and which uses each edge
exactly once.
x7
x6
x2
x5
x4
x1
x3
Solution (indicative)
A quick check of the graph shows that all nodes have even degree, so a Eulerian circuit does exist.
There are a number of edge orderings that are possible, so one indicative solution is shown below
(although your ordering may be different depending on which edge is chosen at each point):
x1→x3→x4→x6→x3→x5→x4→x2→x5→x6→x7→x2→x1→x7→x5→x1
Q2 Can you put together a logical argument as to why the number of nodes with odd-degree has to
be zero or two when determining whether a graph is Eulerian (hint: think about entrances and exits)
Solution
The answer is fairly simple, but requires a little thought – if we are looking to complete a circuit we
must enter and exit the intermediate vertices the same number of times, otherwise I get stranded at
that point. At either end point, I need to either enter the vertex one more time than I leave it, or vice
versa, hence I can have at exactly two vertices where this takes place, or all vertices have even
degree (as for the example above) otherwise I will not end up at the same end point (circuit).
Q3 For the graph, G, below find a Eulerian circuit starting and ending at vertex x3 – trace this path
and show your working out.
x4
x3
x1
x6
x5
x7
x2
Solution (indicative)
A quick check of the graph shows that all nodes have even degree, so a Eulerian circuit does exist.
There are a number of edge orderings that are possible, so one indicative solution is shown below
(although your ordering may be different depending on which edge is chosen at each point):
x3→x1→x2→x7→x6→x3→x2→x6→x5→x4→x3
Q4 For the graph, G below, determine whether there is a Eulerian circuit – if one does not exist,
show how the graph might be expanded (that is, additional edges added) to allow a Eulerian circuit
to be found.
x6
x5
x7
x3
x4
x2
x1
Solution
It can be seen from the graph above that two vertices have odd degree – this means that the graph is
Eulerian, but we cannot find a Eulerian circuit. In this case we could find a path starting at x5 and
ending at x4. We can simple add in the dotted edge x5->x4 to make the Eulerian circuit:
x5→x6→x7→x3→x6→x4→x5→x1→x2→x4→x5
Q5 If we now add directions to each of the edges, as shown in the graph below, find a Eulerian
circuit – In a directed graph, there is a Eulerian circuit if the number of edges entering a node is the
same as the number of edges leaving – trace out the circuit that you find – you are free to choose
your start/end point, but remember that you are looking for a circuit.
x2
x1
x3
x5
x4
x6
x7
Solution (indicative)
I have chosen to start and end at vertex x1 – students may choose another start/end vertex, but as
long as the circuit starts and ends at this vertex, and each edge is followed in the correct direction,
then this is a valid solution:
x1→x2→x3→x7→x6→x5→x1→x4→x6→x1
Additional questions (if you don’t finish these in class then you can do them at home)
Q6 For the directed graph below, trace out a Eulerian circuit that begins and ends at x6
x2
x1
x5
x3
x4
x6
Solution (indicative)
The starting point is vertex x5 – students may choose different edge selections at each point, but as
the circuit starts and ends at this vertex, and each edge is followed in the correct direction, then this
is a valid solution:
x5→x2→x1→x6→x3→x5→x4→x3→x1→x5
Q7 Can you turn the method that you have been using in the questions above into an algorithm (set
of written steps) that someone can follow in order to find a Eulerian circuit? (You can gain help from
the adults in the room). Try and write out your method and then use it on one of the questions
above to test the correctness.
Solution (outline)
There is no single ‘correct’ method. However, the initial test should assess the degress of the nodes,
checking whether there are zero, or two, nodes with odd degree. If the answer to this is no, then the
graph cannot be Eulerian. If the graph passes this test then it is a case of traversing the edges of the
graph, at each point choosing an edge, checking whether it has already been visited (in which case
choose another). At each point we change the edge to being visited, and add it to the set of edges
that have been visited, and choose the edge point of the current edge to be the start point for the
next edge. We check before adding an edge whether the set of edges visited is exactly the same as
the number of edges in the graph, if so our job should be done, and we should have found our
Eulerian circuit.
This is a challenging question – and help (and a Raspberry Pi) will be on hand to help students who
reach this question to develop their algorithm.