Packing Degree Sequences
3.5 Potential Multigraph Packing
A sequence is multigraphic if it is the degree sequence of a multigraph.
James Sellers [57] posed the following question. Suppose you have two multigraphic degree sequences α and β. Do there exist multigraph realizations A and B of α and β on the same set of vertices such that no two vertices have both A-edges and B-edges going between them? While the simple graph version of degree sequence packing is NP-complete [20], there is some evidence to suggest the multigraph case might be easier.
For example multigraph realizability is easier than simple graph realizability in that you need only satisfy the first Erd˝os-Gallai inequality, not all n of them [33]. However, in this section we will see that muligraph degree sequence packing is also NP-complete, and the reduction is more straightforward than in the simple graph case.
The multigraphs we consider have no loops. Let π1= (x1, . . . , xn), π2 = (y1, . . . , yn) be sequences of numbers with even sum. A sequence is multigraphic if there exists a multigraph G having the sequence as its degree sequence. Hakimi[33] characterized when a sequence is multigraphic:
Theorem 22. (Hakimi) π = (z1≥z2 ≥. . . ≥zn)is multigraphic if and only if it has even sum and z1 ≤∑ni=2zi.
We say that π1and π2 multipack if there exist multigraph realizations G1of π1and G2
of π2on the same vertex set V such that
1. v ∈V gets G1degree xi and G2degree yi for some i, and
2. For every u, v∈ V, there is not simultaneously an edge between u and v in both G1 and G2.
Let MULTIPACK be the language of all pairs of degree sequences that pack. Let SUBSET SUM be the decision problem to determine if a set of positive integers S has a subset with sum k. SUBSET SUM is known to be NP-complete (see KNAPSACK from Karp’s original paper [39]).
Proposition 23. MULTIPACK is reducible to SUBSET SUM.
Proof. Let S be a set, and let k be the target value for the sum of a subset. Let~S be an ordered list of elements of S. Let N be the sum of all the elements in S. We will assume that N+k is even, though a minor change in the below argument handles the case where N+k is odd. Let
π1 = (2(N+1) +k, N+1, N+1, N+1, N+1,~S), π2 = (2(N+1) + (N−k), N+1, N+1, N+1, N+1,~S).
Notice the degree sum of π1and π2 is even under the assumption N+k is even.
We will show that π1and π2multipack if and only if S has a subset that sums to k.
(⇐) First assume that S has a subset that sums to k. Let v be the vertex to receive degree 2(N+1) +k in G1 and 2(N+1) + (N−k) in G2. We will think of G1 being colored red, and G2 being colored blue. In the red graph G1, attach v to the first two N+1 degree vertices, and to the subset of~S that sums to k. In the blue graph G2, attach v to the second two vertices of N+1 degree, and to the complement of the subset that sums to k (this subset sums to N−k). This will exhaust the degree of the first vertex in both graphs.
The nonzero red degrees that are leftover at this point are on a disjoint vertex set from the leftover blue degrees that are nonzero. Therefore, π1 and π2will multipack as long as these two leftover degree sequences are multigraphic. They will be: the largest degree and the second largest degree of both are N+1, and thus satisfy Hakimi’s criterion.
(⇒) Assume now that π1 and π2 can multipack. Consider the vertex v that has red degree 2(N+1) +k and blue degree 2(N+1) + (N−k). For every other vertex u, either u and v have red edges between them, or u and v have blue edges between them. Notice for u 6= v, the red degree and blue degree of u are the same. Therefore, the most total degree the other vertices can absorb from v is∑u6=vdegred(u) =4(N+1) +N, which is the total degree of v. Therefore, if you look at a neighbor u of v that receives red edges, v must absorb all of u’s red degree. In order to hit the right amount, v has red edges to exactly two of the vertices of degree N+1. Thus v has red adjacencies to a subset of S that sums to exactly k.
We could also ask this same question about degree sequence packing about bipartite multigraphs. However, it is NP-complete in this case as well.
Let π1= (x1, . . . xn; w1, . . . wm) and π2= (y1, . . . , yn; z1, . . . , zm)be sequences, where
Under this assumption, π1and π2represent the degree sequences of bipartite multigraphs.
We say that π1 and π2 bimultipack if they have realizations G1and G2 such that
1. the ith vertex on the left gets degree xi in G1 and yiin G2, and the jth vertex on the right side gets degree wj in G1 and zjin G2.
2. For every u, v∈ V, there is not simultaneously an edge between u and v in both G1 and G2.
The resulting decision problem BIMULTIPACK is also reducible to SUBSET SUM.
Proposition 24. BIMULTIPACK is reducible to SUBSET SUM.
Proof. Let S be a set, and let k be the target value for the sum of a subset. Again let~S be a vector of the elements of S, and let N be the sum of the elements of S. Let
π1 = (k, N−k;~S), π2 = (N−k, k;~S).
We want to show that π1 and π2 bimultipack if and only if S has a subset with sum k.
(⇐) Assume S has a subset S0 with sum k. Let v1 be the vertex of red degree k, blue degree N−k, and let v2 be the vertex of red degree N−k, and blue degree k. Thus v1 and v2 are the sole vertices on the left hand side. Attach v1 to the S0 with red edges and to the complement of S0 with blue edges. Similarly, connect v2to S0 with blue edges and the complement of S0 with red edges. In this way v1and v2exactly absorb all the degrees of the vertices on the right hand side.
(⇒) Assume that π1 and π2 bimultipack. Since S consists of positive values, every vertex on the right hand side has positive red and blue degree. Therefore, each edge must have red edges going to v1 or v2, and blue edges going to the other one. If you look at the vertices on the right hand side with red edges going to v1, the red degree on these vertices must sum to k. This gives a subset of S that sums to k.