The story goes like this: There were seven bridges
crossing the river Pregel in the town of Konigsberg (18th century
Prussia). See map and picture below.
Two of the remaining bridges of Konisgsberg (now
Kaliningrad in Russia). Photo is from
The town people had long pondered the question of
whether it is possible to walk through the town using every bridge
just once, and returning home at the end.
Euler solved the
long-standing question by mapping the problem to a graph problem,
which we examine next.
Make the observation that for a circuit that traverses every edge
exactly once to exist, every node must be entered and exited an
equal number of times, and since no edge can be used twice, the node
degree must be even.
This suggest the following conjecture.
Conjecture: It is impossible to find a circuit
that traverses all edges in a graph exactly once unless all nodes
have an even degree.
Proof: By contradiction.
Since nodes A, B, C, and D have odd degrees, it is obvious that
this cannot be done in the above graph -- putting the question
of the town of Konigsberg to rest.
For example, to carry the story of the town of Konigsberg
further, upon discovery of the above theorem (that an even degree
for all nodes is a necessary condition for Eulerian circuits to
exist), the town demolished the bridge from A to D and built two new
bridges from C to d and from B to D, as shown below.
Now, as it turns out, it is possible indeed to cross every bridge
exactly once and end up back where we started. Example: A g C e D d
C f A a B c D B b A.
Trying this on a number of small graphs, one will realize that
this is always possible. But is it true for all possible graphs
(satisfying the above condition)? For this we need to prove the
Conjecture: There exists a circuit that
traverses every edge in a connected graph whose nodes are all of
Proof: By induction.
Base: Show that this must be the case for the graph with the
smallest number of nodes -- namely three nodes in a cycle.
Step: Assume that the conjecture holds for all graphs (connected
with even-degree nodes) of size less than N nodes, and consider a
graph of size N.
Start in any node and follow an edge to another node. Since nodes
have even degrees, I am guaranteed to exit any node I enter until I
enter one of the nodes I have already visited -- call such a node X.
When this happens, one of two things will happen:
There are edges out of X that I
am yet to use. If that is the case, then keep going.
There are no more edges out of X
that I am yet to use. Call the circuit we obtained "C".
Under case b, we have one of two possibilities -- either:
All edges of the graph are in C,
in which case we are done since we proved the conjecture, or
There are still some edges that
are not in C. If that is the case, then consider the graph
defined by the edges not in C. This graph must be of size
smaller than N (since the node we stopped at cannot be part of
it). Also, all nodes in this smaller graph must be of even
degree. By the assumption of the induction step, we are
guaranteed to find a circuit C' that traverses all edges of that
smaller graph. The circuit C' can be joined to the circuit C
since there must be common nodes in C and C' (otherwise the
original graph would have been disconnected). Joining C and C'
gives us a circuit that covers all edges of the original graph,
proving the conjecture.