Graphs with prescribed degree sequences. Given a list of n positive integers d1; d2,..., d...

Graphs with prescribed degree sequences. Given a list of n positive integers d1; d2,..., dn, we want to efficiently determine whether there exists an undirected graph G = (V, E) whose nodes have degrees precisely d1; d2,...,dn. That is, if V = {v1;..., vn}, then the degree of vi should be exactly di. We call (d1;..., dn) the degree sequence of G. This graph G should not contain self-loops (edges with both endpoints equal to the same node) or multiple edges between the same pair of nodes.

(a) Give an example of d1; d2, d3, d4 where all the di ≤ 3 and d1 + d2 + d3 + d4 is even, but for which no graph with degree sequence (d1; d2, d3, d4) exists.

(b) Suppose that d1 ≥ d2 ≥...≥dn and that there exists a graph G = (V, E) with degree sequence (d1,..., dn). We want to show that there must exist a graph that has this degree sequence and where in addition the neighbors of v1 are  The idea is to gradually transform G into a graph with the desired additional property.

i. Suppose the neighbors of v1 in G are not  Show that there exists i < j ≤ n and u ∈ V such that {v1; vi}, {u, vj} ∉ E and {v1, vj}, {u, vi} ∈ E.

ii. Specify the changes you would make to G to obtain a new graph G' = (V, E') with the same degree sequence as G and where (v1; vi) ∈ E ’.

iii. Now show that there must be a graph with the given degree sequence but in which v1 has neighbors

(c) Using the result from part (b), describe an algorithm that on input d1,...,dn (not necessarily sorted) decides whether there exists a graph with this degree sequence. Your algorithm should run in time polynomial in n.