A regular bipartite graph has the same number of vertices in the two partions. So we need to add vertices also. I'm not sure that it is always possible to add edges to get a $\Delta$-regular bipartite graph, even if we have the same number of vertices. See the figure below. B and E both have degree two, but we cannot make them degree 3 Am I right ?
Hint: If a graph is bipartite, it means that you can color the vertices such that every black vertex is connected to a white vertex and vice versa. Hint: Consider parity of the sum of coordinates.
Of course, the definition of "bipartite" is easily generalised to graphs that are not simple, and we might want to do this in some cases: for instance if we are studying graph colourability, we might want to use "bipartite" as synonymous with " $2$ -colourable". However, other times, it does make sense to restrict ourselves to simple graphs.
Can someone explain to me with an example how to create the adjacency matrix of a bipartite graph? And why the diagonal elements of it are not zero? Thanks.
If the matrix is now in the canonical form of a bipartite adjacency matrix (where the upper-left and lower-right blocks are all zero), the graph is bipartite; quit and return BIPARTITE. Otherwise, the graph isn't bipartite — quit and return NOT BIPARTITE.
How can we prove that a graph is bipartite if and only if all of its cycles have even order? Also, does this theorem have a common name? I found it in a maths Olympiad toolbox.
A bipartite graph has a unique bipartition (except for interchanging the two partite sets) if and only if it is connected. Let G be a bipartite graph. If u and v are vertices in distinct components, then there is a bipartition in whic h u and v are in the same partite set and another in whic h they are in opposite partite sets. If G is connected, then from a fixed vertex u we can walk to all ...
Connected bipartite graph which is neither path or (even) cycle must have a vertex of degree greater than $2$ Adjacency Matrix of Connected Bipartite Graph graph-theory matrix-equations bipartite-graphs
So by definition a bipartite graph has some edges that are not used (i.e. the edges between vertices of the same set). That would then mean that there are unused edges and so the graph cannot be Eulerian.
