Determining graph biconnection from degree sequence The title is self-explanatory: having the degree sequence of a graph, how can I find whether it is biconnected? The fact that I can't manage to draw one with such property does not mean that it does not exist...
Also, if I manage to draw a biconnected graph from the degree sequence, does this imply that all connected graphs with that degree sequence are biconnected?
Thanks in advance.
 A: Two graphs with degree sequence $(2,2,2,2,2,2)$ are the (biconnected) $6$-cycle, and the (disconnected) graph consisting of two disjoint $3$-cycles. So the answer to your question is that the degree sequence can't even tell you about connectedness in general, let alone biconnectedness.
There are also degree sequences that are realizable by both biconnected graphs and connected graphs that are not biconnected. For example, for both graphs in the above example, add a single edge between two non-adjacent vertices. You get two graphs with degree sequence $(2,2,2,2,3,3)$: a $6$-cycle with a chord, and two disjoint triangles joined by a single edge. Of course one is biconnected, the other is connected but not biconnected.
However, see this question for a sufficient condition on the degree sequence that will ensure $k$-connectedness.
A: There are indeed counterexamples; graphs with the same degree sequence but different biconnectivity. As a visual summary of the examples given, see this:

The vertices are coloured by degree. There is a simple procedure, given a graph with a particular degree sequence, to produce another graph with the same degree sequence (isodegree?). As shown in the upper example (from this answer) two edges can be swapped between four vertices to make another graph.
