Hierholzers Algorithm to find the Euler tour

Question

Does anyone know how to use Hierholzers Algorithm to find a Euler tour of a graph with Java?
Or anyone know of any sites that contain a program that already does this?
Thanx

Answer 1

Does anyone know how to use Hierholzers Algorithm to find a Euler tour of a graph with Java?The construction of an Euler tour is not that difficult. The core of the
proof whether or not such a tour exists is the following lemma --
An Euler tour exists if and only if every vertex has an even number of
edges connected to it.
Hierholzer came up with the following construction; for the sake of the
example, we'll use this graph; A is the start/end vertex --     B-------D
   /|\     /|\
  / | \   / | \
/  |  \ /  |  \
A   |   /   |   F
\  |  / \  |  /
  \ | /   \ | /
   \|/     \|/
    C-------E Note that all vertexes have an even degree, so an Euler tour can be
constructed. Step /zero one in Hierholzer's construction is --
0) let 'level' == 1; 'start' == A;
1) Start at vertex 'start' and generate a random tour;
This is easy, simply wander around over any unused edge until you reach
vertex A again. Suppose we created the tour A -> B -> C -> A.
2) label all the used edges using 'level' and remove all used edges;
This leaves us with the following graph --     B-------D
     \     /|\
      \   / | \
       \ /  |  \
A       /   |   F
       / \  |  /
      /   \ | /
     /     \|/
    C-------E Note that every vertex still has an even number of edges connected to it.
3) For any vertex on the tour so far, find a vertex that still has unused
edges. We randomly pick vertex B; if we can't find such vertex, goto step 5;
4) level= level + 1; start= B; goto step 1;
If we've reached this step for a second time we could have generated a tour
B -> D -> E -> B. All the edges are removed, after labeling them with a
level equal to 2. The resulting graph looks like this then --     B       D
A       /       F
    C-------E It's not difficult to see that, after we get at step 4, we have to complete
the last tour, say, D -> F -> E -> C -> D and we remove the corresponding
edges after labeling them with a level equal to 3.
Finally we arrive at this step --
5) All edges are removed and we've collected three tours --
level 1: A -> B -> C -> A
level 2: B -> D -> E -> B
level 3: D -> F -> E -> C -> D
6) Walk the first level tour until you reach a vertex that connects to a higher
level tour (at vertex B); leave the level 1 tour and follow the next level
tour until a similar situation occurs. Apply this step continuously until
you've reached your original starting vertex (A) again.
You've found an Euler tour --
This: A -> B -> C -> A expands to A -> B -> D -> E -> B -> C -> A.
Which in turn expands to A -> B -> D -> F -> E -> C -> D -> E -> B -> C -> A.
Et voila!
kind regards,
Jos