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Graphs with loops
Let us first specify the object of our investigation. Definition 1. By a graph-with-loops we mean a graph, without multiple edges, in which there is at most one loop at each vertex.
Our main interest in this paper is to make the connection between graph-with-loops and bipartite graphs. Recall that for every simple graph G, there is a natural associated bipartite simple graphĜ called the bipartite double-cover of G. For convenience in the following discussion, assume that G is connected. From a practical point of view,Ĝ can be constructed as follows: remove from G the minimum possible number of edges so as to make the resulting graph bipartite. Take two copies G 1 , G 2 of this bipartite graph, and two copies of the removed edges, and reattach the edges to their original vertices but in such a way that each attached edge joins G 1 with G 2 . The resulting bipartite graphĜ is independent of the choices made in its construction. From a theoretical point of view,Ĝ can be conceived in (at least) two natural ways: (1)Ĝ is the tensor product G × K 2 of G with the connected graph with 2 vertices; the vertex set of G × K 2 is the Cartesian product of the vertices of G and K 2 , there are edges in G × K 2 between (a, 0) and (b, 1) and between (a, 1) and (b, 0) if and only if there is an edge in G between a and b. (2)Ĝ has a topological definition as a covering space of G. Give G the obvious topology, and let H denote the subgroup of the fundamental group π 1 (G) composed of loops of even length, where length is defined by taking each edge to have length one. Clearly H has index two in π 1 (G), so it is a normal subgroup. Thus, by standard covering space theory, there is a two-fold normal covering spaceĜ of G, and by construction, G is bipartite; see [7,Chapter 1.3]. Now consider the situation where G is a graph-with-loops. Here the two constructions described above do not produce the same graph. The distinction is that in the tensor product, each loop in G produces just one edge in G × K 2 , while in the covering space, each loop lifts to two edges. Figure 1 shows the constructions for the complete graph-with-loops G on three vertices. In the tensor product construction, G × K 2 is not a covering of G in the topological sense, and in particular, the vertices in G × K 2 do not have the same degree as Figure 1. The complete graph-with-loops on three vertices, and its two "bipartite covers" those in G. In the topological construction, the covering space is not a simple graph, but a multigraph, however its vertex degrees do have the same degree as those in G. We will argue below that the tensor product construction is the appropriate concept for graphs-with-loops. In the following we will need to refer to the Erdős-Gallai Theorem, which we recall for convenience.
Erdős-Gallai Theorem. A sequence d = (d 1 , . . . , d n ) of nonnegative integers in decreasing order is graphic if and only if its sum is even and, for each integer k with 1 ≤ k ≤ n, Note that graphs-with-loops are a special family of multigraphs [5] and that for multigraphs, the degree of a vertex is usually taken to be the number of edges incident to the vertex, with loops counted twice. We have the following straightforward generalisation of the Erdős-Gallai Theorem. We postpone the proof of this theorem to the final section. Theorem 1. Let d = (d 1 , . . . , d n ) be a sequence of nonnegative integers in decreasing order. Then d is the sequence of vertex degrees of a graph-with-loops if and only if its sum is even and, for each integer k with 1 ≤ k ≤ n, Despite the above result, we claim that in the context of graphs-with-loops, a different definition of degree is more appropriate. We introduce the following definition.
Definition 2. For a graph-with-loops, the reduced degree of a vertex is taken to be the number of edges incident to the vertex, with loops counted once.
So, for example, in the complete graph-with-loops G on three vertices, shown on the left in Figure 1, the vertices each have reduced degree three. Notice the vertices in the tensor product G × K 2 also have the same degrees as the reduced degrees of the vertices of G. In our view, the tensor product construction is the appropriate concept of bipartite double cover for graphs-with-loops; as we discussed above, it does not produce a covering space, in the topological sense, but it does produce a bipartite graph, and the vertex degrees are preserved provided we use the notion of reduced degree. We have the following Erdős-Gallai type result; the proof is given in the final section.
Theorem 2. Let d = (d 1 , . . . , d n ) be a sequence of nonnegative integers in decreasing order. Then d is the sequence of reduced degrees of the vertices of a graph-with-loops if and only if for each integer k with 1 ≤ k ≤ n, Let us say that a finite sequence d of nonnegative integers is bipartite graphic if the pair (d, d ) can be realized as the degree sequences of the parts of a bipartite simple graph; such sequences have been considered in [1]. The utility of the notion of reduced degree is apparent in the following result. Proof. If d is the sequence of reduced degrees of the vertices of a graph-with-loops G, then forming the tensor product G×K 2 we obtain a bipartite graph having degree sequence (d, d ). Conversely, if d is bipartite graphic, then by the Gale-Ryser Theorem [6,10], for each k with and so by Theorem 2, d is the sequence of reduced degrees of the vertices of a graph-withloops.

Some Remarks
Remark 1. Corollary 1 has two consequences. First, from Theorem 2 and Corollary 1, condition (2) gives an Erdős-Gallai type condition for a sequence to be bipartite graphic, which is analogous to the Gale-Ryser condition. Secondly, consider a bipartite graphic sequence d. So (d, d) can be realised as the degree sequences of the parts of a bipartite grapĥ G. Corollary 1 shows that this can be done in a symmetric manner, in thatĜ is a tensor productĜ = G × K 2 , and in particular there is an involutive graph automorphism that interchanges the two parts ofĜ.
Remark 3. There are several results in the literature of the following kind: if d is graphic, and if d ′ is obtained from d using a particular construction, then d ′ is also graphic. The Kleitman-Wang Theorem is of this kind [8]. Another useful result is implicit in Choudum's proof [4] of the Erdős-Gallai Theorem: If a decreasing sequence d = (d 1 , . . . , d n ) of positive integers is graphic, then so is the sequence d ′ obtained by reducing both d 1 and d n by one. Analogously, our proofs of Theorems 1 and 2, which are modelled on Choudum's proof, also establish the following result: If a decreasing sequence d = (d 1 , . . . , d n ) of positive integers is bipartite graphic, then so is the sequence d ′ obtained by reducing both d 1 and d n by one.
Remark 4. There are other facts for graphs that generalise easily to graphs-with-loops. For example, if d = (d 1 , . . . , d n ) has a realization as a graph G, then one can consider G as a subgraph of the complete graph K n . Considering the complement of G in K n , one obtains the well known result: If (d 1 , . . . , d n ) is graphic, then so too is (n − 1 − d n , n − 1 − d n−1 , . . . , n − 1 − d 1 ). Analogously, by replacing the complete graph on n vertices by the complete graph-with-loops on n vertices, one immediately obtains the following equivalent result: If a sequence d = (d 1 , . . . , d n ) of nonnegative integers is bipartite graphic, then so is the sequence d ′ = (n − d n , n − d n−1 , . . . , n − d 1 ).

Proofs of Theorems 1 and 2
The following two proofs are modelled on Choudum's proof of the Erdős-Gallai Theorem [4].
Proof of Theorem 1. For the proof of necessity, first note that for every graph-with-loops with vertex degree sequence d, the sum n i=1 d i is twice the number of edges, so it is even. Now consider the set S comprised of the first k vertices. The left hand side of (1) is the number of half-edges incident to S, with each loop counting as two. On the right hand side, k(k + 1) is the number of half-edges in the complete graph-with-loops on S, again with each loop counting as two, while n i=k+1 min{k, d i } is the maximum number of edges that could join vertices in S to vertices outside S. So (1) is obvious.
The proof of sufficiency is by induction on n i=1 d i . It is obvious for n i=1 d i = 2. Suppose that we have a decreasing sequence d = (d 1 , . . . , d n ) of positive integers which has even sum and satisfies (1). As in Choudum's proof of the Erdős-Gallai Theorem, consider the sequence d ′ obtained by reducing both d 1 and d n by 1. Let d ′′ denote the sequence obtained by reordering d ′ so as to be decreasing.
Suppose that d ′′ satisfies (1) and hence by the inductive hypothesis, there is a graphwith-loops G ′ that realizes d ′′ . We will show how d can be realized. Let the vertices of G ′ be labelled v 1 , . . . , v n . [Note that v n may be an isolated vertex]. If there is no edge in G ′ connecting v 1 to v n , then add one; this gives a graph-with-loops G that realizes d. So it remains to treat the case where there is an edge in G ′ connecting v 1 to v n . If there is no loop at either v 1 or v n , remove the edge between v 1 and v n , and add loops at both v 1 and v n . Now, for the moment, let us assume there is a loop in G ′ at v 1 . Applying the hypothesis to d, using k = 1 gives  for which there is no edge from v 1 to v i . Note that d ′ i > d ′ n . If there is a loop in G ′ at v n , or if there is no loop at v i nor at v n , then there is a vertex v j such that there is an edge in G ′ from v i to v j , but there is no edge from v j to v n . Now remove the edge v i v j , and put in edges from v 1 to v i , and from v j to v n , as in Figure 2. This gives a graph-with-loops G that realizes d. If there is no loop in G ′ at v n , but there is a loop at v i , we consider the two cases according to whether or not there is an edge between v i and v n . If there is an edge between v i and v n , then remove this edge, add an edge v 1 v i and add a loop at v n , as in Figure 3. If there is no edge between v i and v n , add edges v 1 v i and v i v n and remove the loop at v i , as in Figure 4. In either case, we again obtain a graph-with-loops G that realizes d.
Finally, assume there is no loop in G ′ at v 1 , but there is a loop in G ′ at v n . So, apart from the loop, there are a further d n − 3 edges incident to v n . Since d 1 ≥ d n , we have d 1 − 1 > d n − 3, and so there is a vertex v i such that there is an edge in G ′ from v 1 to v i , but there is no edge from v i to v n . Remove the edge v 1 v i , put in an edge v i v n and add a loop at v 1 , as in Figure 5. The resulting graph-with-loops G realizes d.
It remains to show that d ′′ satisfies (1 d ′′ : So it remains to deal with the case where k < m and Notice that as d i = d ′′ i except for i = m, n, we have min{k, d ′′ i } = min{k, d i } except possibly for i = m, n. In fact, as k < m, we have d m = d k > k + 1 and so min{k, We have d k+1 = d m > k + 1 and so min{k, d k+1 } = k. Note that n i=k+2 min{k, d i } = 0 as k + 2 ≤ n, since k < m ≤ n − 1. So contradicting (1). Hence d ′′ satisfies (3), as claimed. This completes the proof. Figure 6.
Proof of Theorem 2. The proof mimics the above proof of Theorem 1. For the proof of necessity, consider the set S comprised of the first k vertices. The left hand side of (1) is the number of half-edges incident to S, with each loop counting as one. On the right hand side, k 2 is the number of half-edges in the complete graph-with-loops on S, again with each loop counting as one, while n i=k+1 min{k, d i } is the maximum number of edges that could join vertices in S to vertices outside S. So (1) is obvious.
Conversely, suppose that d = (d 1 , . . . , d n ) verifies (2) and consider the sequence d ′ obtained by reducing both d 1 and d n by 1. Let d ′′ denote the sequence obtained by reordering d ′ so as to be decreasing. Suppose that d ′′ satisfies (2) and hence by the inductive hypothesis, there is a graph-with-loops G ′ that realizes d ′ . We will show how d can be realized. Let the vertices of G ′ be labelled v 1 , . . . , v n . If there is no edge in G ′ connecting v 1 to v n , then add one; this gives a graph-with-loops G that realizes d. Similarly, if there is no loop at either v 1 or v n , just add loops at both v 1 and v n . So it remains to treat the case where there is an edge in G ′ connecting v 1 to v n , and at least one of the vertices v 1 , v n has a loop. Now, for the moment, let us assume there is a loop in G ′ at v 1 . Applying the hypothesis to d, using k = 1 gives and so d 1 − 2 < n − 1. Now in G ′ , the degree of v 1 is d 1 − 1 and so apart from the loop at v 1 , there are a further If there is a loop in G ′ at v n , or if there is no loop at v i nor at v n , then there is a vertex v j such that there is an edge in G ′ from v i to v j , but there is no edge from v j to v n . Now remove the edge v i v j , and put in edges from v 1 to v i , and from v j to v n , as in Figure 6. This gives a graph-with-loops G that realizes d. If there is no loop in G ′ at v n , but there is a loop at v i , remove the loop at v i , add the edge v 1 v i and add a loop at v n , as in Figure 7.
Finally, assume there is no loop in G ′ at v 1 , but there is a loop in G ′ at v n . So, apart from the loop, there are a further d n − 2 edges incident to v n . Since d 1 ≥ d n , we have d 1 − 1 > d n − 2, and so there is a vertex v i such that there is an edge in G ′ from v 1 to v i , but there is no edge from v i to v n . Note that d ′ i > d ′ n , so as there is a loop in G ′ at v n , there is a vertex v j such that there is an edge in G ′ from v i to v j , but there is no edge from v j to v n . Now remove the loop at v n and the edge v i v j , and put edges v j v n and v i v n and add a loop at v 1 , as in Figure 8. This gives a graph-with-loops G that realizes d.
It remains to show that d ′′ satisfies (2). Define m as follows: if the d i are all equal, put m = n − 1, otherwise, define m by the condition that d 1 = · · · = d m and d m > d m+1 . We have d ′′ i = d i for all i = m, n, while d ′′ m = d m − 1 and d ′′ n = d n − 1. Consider condition (2) for d ′′ : For m ≤ k < n, we have k i=1 d ′′ i = k i=1 d i −1, while n i=k+1 min{k, d ′′ i } ≥ n i=k+1 min{k, d i }− 1, and so (4) holds. For k = n, k i=1 d ′′ i = k i=1 d i − 2 < k 2 , and so (4) again holds. For k < m, first note that if d k ≤ k, then k i=1 d ′′ i = k i=1 d i ≤ k 2 ≤ k 2 + n i=k+1 min{k, d ′′ i }. So it remains to deal with the case where k < m and d k > k. We have Notice that as d i = d ′′ i except for i = m, n, we have min{k, d ′′ i } = min{k, d i } except possibly for i = m, n. In fact, as k < m, we have d m = d k > k and d ′′ m = d m − 1 ≥ k and so min{k, d m } = k = min{k, d ′′ m }. Hence n i=k+1 min{k, d ′′ i } ≥ n i=k+1 min{k, d i } − 1. Thus, in order to establish (4), it suffices to show that k i=1 d i < k 2 + n i=k+1 min{k, d i }. Suppose instead that k i=1 d i = k 2 + n i=k+1 min{k, d i }. We have We have d k+1 = d m > k and so min{k, d k+1 } = k. Note that n i=k+2 min{k, d i } = 0 as k + 2 ≤ n, since k < m ≤ n − 1. So contradicting (2). Hence d ′′ satisfies (4), as claimed. This completes the proof.