Forced Edges and Graph Structure

Abstract

For a degree sequence, we define the set of edges that appear in every labeled realization of that sequence as forced, while the edges that appear in none are define as forbidden. We examine the structure of graphs in which the degree sequences contain either forced or forbidden edges. The results include the determination of the structure of the forced or forbidden edge sets, the relationship between the sizes of forced and forbidden sets for a sequence, and the structural consequences to their realizations. This includes showing that the diameter of every realization of a degree sequence containing forced or forbidden edges is no greater than 3, and that these graphs are maximally edge-connected.

1. Introduction

A degree sequence is a sequence of non-negative integers where there exists a simple and undirected graph in which the node degrees correspond with the values in the sequence. Any graph for which the node degrees correspond to a integer sequence is said to a realization of that sequence. A degree sequence may have a number of labeled realizations. If an edge appears between two labeled nodes for every realization, we denote that edge as forced for the degree sequence. Conversely, if an edge never occurs in any labeled realization of the sequence, we denote that edge as forbidden for the sequence.

The simplest example of a forced edge is when there is a dominating value in a degree sequence, i.e., a value equal to the number of other entries. Then for every realization of the degree sequence, there must be forced edges from the dominating vertex to every other vertex in the graph. Likewise, an empty value in a degree sequence causes forbidden edges between the vertex and all the other vertices. A non-trivial example of a forced edge for a degree sequence is shown in Fig. 1.

Fig. 1.

Every labeled realization of the degree sequence π = (4,4,3,3,3,1) is isomorphic to one of these two graphs. Since there is always an edge between the two degree 4 nodes, then we say that the edge {1,2} as forced for π.

Consideration of the forced and forbidden edges for a degree sequence has both algorithmic and theoretical applications. For instance, the creation of random graphs from a given degree sequence is useful for tasks from counting graphs with a given degree sequence to creating models of networks. There are two principal approaches to creating a random realization. The most common method involves a Monte Carlo Markov chain (MCMC) approach, where one starts with an initial non-random realization for a degree sequence, and then performs a random walk to a new realization through a series of edge switches [1].

Obviously, if an edge is forced, then it is in every realization, and it can never be swapped out during the random walk. For some realizations, it is computationally expensive to find edges to switch. We can optimize this edge selection process by excluding forced edges.

A more striking example is in the creation of a sequential importance sampling (SIS) method for creating random realizations. In the SIS approach, edges are randomly selected until a realization is built.

The difficulty with this approach is it is possible to become stuck while selecting edge. In other words, we can create a partial graph that is impossible to complete into a realization for the given degree sequence. The first algorithm to overcome this difficulty was proposed by Blitzstein and Diaconis [2]. Their breakthrough idea was to show that by carefully selecting the edges that are not forbidden, a realization can always be created.

A recent application area where forced edges provide a fundamental limitation is in degree sequence packing [3, 4]. The degree sequence packing problem entails examining two degree sequences and determining whether labeled realizations exist for both sequences that are edge-disjoint. Obviously, if the two sequences contain the same forced edge then the two sequences cannot pack.

2. Basic Definitions and Results

We begin with some needed formal definitions and results. All notation that is not explicitly defined is taken from Diestel [5]. A sequence of non-negative integers n that corresponds to the vertex degrees of some simple graph is called a degree sequence or is said to be graphic. In this article, we assume that for the degree sequence n, the values are ordered n − 1 ≥ π₁ ≥ π₂ ≥ … ≥ π_n ≥ 0. Thus, the largest value I π is π₁ and the smallest value is π_n. The complementary sequence for π is the sequence $\overline{\pi }$ where ${\overline{\pi }}_i=n-{\pi }_{n+1-i}-1$. To represent the degree sequence of a given graph G, we will use the notation deg(G).

We say that a realization G of a degree sequence π is labeled, when the degree of the vertex v_i in G corresponds to the ith value in its sequence π. The neighborhood, or set of adjacent vertices, of v_i is represented as N(v_i), i.e., ∣N(v_i)∣ = π_i. For a subset of vertices S ⊆ V, the induced subgraph on this subset is represented as G[S]. An edge between two vertices, v_i and v_j, is designated as {i, j}. For a graph G, we denote the diameter of the graph as diam(G).

For a degree sequence π, we define its forbidden set $\mathcal{B}\left(\pi \right)$ as the set of all edges between labeled nodes that do not appear in any realization of n, while its forced set $\mathcal{F}\left(\pi \right)$ is the set of all edges between labeled nodes that appear in every realization of π. We will also define the set $\mathcal{P}\left(\pi \right)$ to contain all the vertices that are in some forced edge in π, i.e., $\mathcal{P}\left(\pi \right)=\left\{v_i\mid \exists v_j:\left\{i,j\right\}\in \mathcal{F}\left(\pi \right)\right\}$.

In order to compare integer sequences, we will use the following partial ordering of majorization. A sequence S majorizes (or dominates) the integer sequence T, denoted by S ≽ T, if for all k from 1 to n

$$\sum _{i=1}^k{S}_i\ge \sum _{i=1}^k{T}_i,$$ (1)

and if the sums of the two sequences are equal. A convenient fact that we will use is that the majorization order is preserved by the complementary sequences.

Claim 2.1.

If S ≽ T, then $\overline{S}\succcurlyeq \overline{T}$.

A degree sequence that has precisely one labeled realization is called a threshold sequence and the resulting realization is called a threshold graph [6]. In the context of our discussion about forced edges, threshold graphs can be seen as graphs where every edge is forced, and where every non-edge is forbidden.

For convenience, we introduce a notation for showing increments or decrements to specific indices in a sequence. For the sequence S, the sequences ⊖i₁,…,i_kS and ⊕i₁,…,i_kS are defined by

$${\left({\ominus }_{i_1}{}_{,\dots ,i_k}S\right)}_i=\{\begin{array}{l}{S}_i-1\text{for}i\in \left\{i_1,\dots ,i_k\right\}\\ S_i\text{otherwise},\end{array}$$ (2)

$${\left({\ominus }_{i_1}{}_{,\dots ,i_k}S\right)}_i=\{\begin{array}{l}{S}_i+1\text{for}i\in \left\{i_1,\dots ,i_k\right\}\\ S_i\text{otherwise},\end{array}$$ (3)

There is a straightforward but nontrivial relationship between majorization and the decrementing and incrementing operations.

Theorem 2.1 (Fulkerson and Ryser [7], Lemma 3.1).

If S ≽ T and ω₁ = {i₁,…, i_k} and ω₂ = {j₁,…, j_k}, where i₁ ≥ j₁,…, i_k ≥ j_k then ⊖_ω1S ≽ ⊖_ω2T. and ⊕_ω2S ≽ ⊕_ω1T

For our purposes, the usefulness of comparing degree sequences using majorization stems from the following result.

Theorem 2.2 (Ruch and Gutman [8], Theorem 1).

If the integer sequence S is graphic and S ≽ T, then T is graphic.

Finally, we will use another classic result.

Theorem 2.3 (Kleitman and Wang [9], Theorem 2.1).

For an integer sequence S and an index i, let S′ be the sequence created by subtracting 1 from the first S_i values in S not including index i and then setting S_i = 0. Then, the sequence S is graphic if and only if the sequence S′ is graphic.

3. Forced and Forbidden Edges

A simple observation about forced and forbidden edge sets is that they have a dual relationship through their complementary degree sequences.

Claim 3.1.

$$\mathcal{F}\left(\pi \right)=\mathcal{B}\left(\overline{\pi }\right).$$

A method for determining whether an edge is either forced or forbidden for a degree sequence is given by the next theorem.

Theorem 3.1 (Blitzstein and Diaconis [2], Proposition 6.2).

Let π be a degree sequence and i, j ∈ {1,…, n} with i ≠ j. The edge $\left\{i,j\right\}\in \mathcal{F}\left(\pi \right)$ if and only if ⊕_i,jπ is not graphic, while the edge $\left\{i,j\right\}\in \mathcal{B}\left(\pi \right)$ if and only if ⊖ _i,jπ.

A useful fact for examining the structure of graphs containing forced edges is that a forced edge remains forced across any induced subgraphs containing that edge.

Claim 3.2.

Let π be a degree sequence and i, j ∈ {1,…,n} with i ≠ j. If the edge $\left\{i,j\right\}\in \mathcal{F}\left(\pi \right)$, then for any realization G of π and for every vertex set S, where {i, j} ⊆ S ⊆ V, $\left\{i,j\right\}\in \mathcal{F}\left(\text{deg}\left(G\left[S\right]\right)\right)$.

There is a simple extension of Theorem 3.1 for sets of forbidden edges through the complementary sequence. This includes a test for determining if an edge {i, j} is forbidden by testing whether or not ⊖_i,jπ is graphic. There is a simple edge-inclusion result for forced and forbidden sets of a degree sequence.

Theorem 3.2 (Barrus [10], Theorem 3.1).

For the degree sequence π, if $\left\{i,j\right\}\in \mathcal{F}\left(\pi \right)$, then for all indices p, q where 1 ≤ p ≤ i, 1 ≤ q ≤ j, and p ≠ q, $\left\{p,q\right\}\in \mathcal{F}\left(\pi \right)$.

An immediate consequence of this proposition is that if there exist any forced edges for a degree sequence, then the edge {1,2} must be one of them. This provides linear-time methods to determine if a sequence has any forced or forbidden edges by testing whether ⊕₁,₂π or ⊖_n−1,_nπ are graphic respectively. Extending this observation establishes conditions for degree sequences that cannot have forced edges.

Theorem 3.3.

For the degree sequence π where π_n > 0, if

$$n\ge \text{min}\left\{\frac{{\left({\pi }_1+{\pi }_n+2\right)}^2}{4{\pi }_n},\frac{{\left({\pi }_1+{\pi }_n\right)}^2}{2{\pi }_n}\right\},$$ (4)

then $\mathcal{F}\left(\pi \right)=0$.

Proof. The first term in this bound comes by substituting π₁ + 1 for the maximum degree into the graphic bound given by Zverovich and Zverovich (Theorem 6, [11]); it follows that if the above bound holds then ⊕_1,2π is graphic. The second term uses a degree sequence packing result. Using the observation that two degree sequences cannot pack where both have the same forced edge {1,2}, we apply Theorem 2.2 of Busch et al. [3] to pack a sequence π with the sequence (1, 1,0,…,0) and after some algebraic manipulation establish the second bound.

Theorem 3.2 is also enough to establish the structure of the sets of forced and forbidden edges. While the induced subsets $S\subseteq \mathcal{P}\left(\pi \right)$ do not necessarily need to be threshold graphs, the sets of forced edges for a degree sequence always do form a threshold graph.

Theorem 3.4 (Barrus [10], Theorem 3.2).

For a degree sequence π, the graph $G=\left(\mathcal{P}\left(\pi \right),\mathcal{F}\left(\pi \right)\right)$ is a threshold graph.

Over the set of partitions for some positive integer p, majorization forms a lattice [12]. In these partition lattices, at the top of the graphic sequences there are the threshold sequences in which every edge is forced. In contrast, Theorem 3.2 can be extended to show that the near-regular sequences (the sequences where π₁ ≤ π_n + 1), which occupy the bottom of the lattice, cannot have any forced or forbidden edges (other than trivially with the complete or empty sequences). We formalize this observation by showing a strict ordering of forced and forbidden sets by subset down chains in this lattice.

Theorem 3.5 (Barrus [10], Theorem 4.1).

For the degree sequences π and π′, if π ≽ π′ then $\mathcal{F}\left(\pi \right)\supseteq \mathcal{F}\left(\pi \prime \right)$ and $\mathcal{B}\left(\pi \right)\supseteq \mathcal{B}\left(\pi \prime \right)$.

4. Structure of the Realizations

A useful result with structural implications is that forced (and forbidden) edges for a degree sequence imply independent sets (or cliques) in the realizations of the degree sequence.

Theorem 4.1.

Let π be a degree sequence.

1. If edge $\left\{i,j\right\}\in \mathcal{F}\left(\pi \right)$, then for any realization G of π, the set of vertices V − (N(vi) ⋃ N (vj)) forms an independent set,

2. If n − 1 > d₁ and 0 < d_n and $\left\{i,j\right\}\in \mathcal{B}\left(\pi \right)$, then for any realization G of π, the set of vertices N(v_i) ⋃ N(v_j) forms a clique.

Proof. For the first statement, suppose there are vertices {v_p, v_q} ⊆ V − (N(v_i) ⋃ N(v_j)) that have an edge between them. We can replace the edges {{p, q}, {i, j}} with {{p, i}, {q, j}} forming a realization of π without the edge {i, j}, in turn causing a contraction.

For the second statement, suppose that {i, j} is forbidden. If v_l ∈ N(v_i) and v_p ∈ N(v_j) and there is no edge {l, p} a realization of π, then {{i, l}, {j, p}} can be swapped out with {{i, j}, {l, p}} contradicting the assumption that {i, j} is forbidden. Thus for N(v_i) ⋃ N(v_j) not to be a clique requires that there exist two vertices v_l and v_k where {v_l, v_k} ⊆ N(v_i) but {v_l, v_k} ⊈ N(v_j) and {k, l} is not in a realization.

Consider that if the edge {i, p} is also not in the realization, then we have the induced graph shown in Fig. 2a. As the caption shows, the edge {i, j} would not be forbidden in this case. Thus if the vertex v_p is connected to {i, j, k, l} and since π_p < n − 1, then there must exist a vertex v_q in the realization that is not adjacent to v_p.

Fig. 2.

This figure shows the cases used in the proof of Theorem 4.1. For each of the above graphs, the edge {i, j} is not forbidden as shown by the edge replacements in each caption.

Again, since π_q > 0, then v_q must be adjacent to another vertex, but we have already established that it cannot be adjacent to v_i. Thus either it is adjacent to v_j, v_k (or v_l), or a completely separate node v_r. For each of those cases, Figs. 2b, 2c, and 2d show that these arrangements invalidate the assumption that edge {i, j} is forbidden. Thus the edge {k, l} must exist, and the vertices N(v_i) ⋃ N(v_j) form a clique.

Extending these results, we now relate the size of the sets of forced edges to forbidden edges for a degree sequence, by showing that forbidden edges in a degree sequence imply cliques of forced edges.

Theorem 4.2.

Let π be a degree sequence where μ_n > 0. If $\left\{i,j\right\}\in \mathcal{B}\left(\pi \right)$ then there exists a clique of μ_i nodes in $\mathcal{F}\left(\pi \right)$.

Proof. ¿From Theorem 4.1, for any realization of π, the vertices N(v_i) ⋃ N(v_j) form a clique. Using Theorem 2.3, we construct a realization H of π where the first π_i vertices are connected to v_i, and so these first π_i vertices form a $K_{{\pi }_i}$ -clique. For the degree sequence b created by removing the vertex v_i and its adjacent edges from H, this clique of the first π_i vertices must exist in every realization, i.e., $K_{{\pi }_i}\subseteq \mathcal{F}\left(\pi \prime \right)$.

Now take an arbitrary realization G of π. If we remove v_i along with its adjacent edges from G, then for the resulting graph G′ with its degree sequence deg(G′), it is straightforward to see that b ≼ deg(G′). ¿From Theorem 3.5, $\mathcal{F}\left(\pi \prime \right)\subseteq \mathcal{F}\left(\text{deg}\left(G\prime \right)\right)$ and so any realization of deg(G′) must contain all the edges in $\mathcal{F}\left(\pi \prime \right)$, specifically $K_{{\pi }_i}$. By adding back the vertex v_i, we see that every realization G will also contain those edges.

We can extend this result to show forced cliques based on the minimum degree value.

Corollary 4.3.

For the degree sequence π where π₁ < n − 2 and $\mathcal{F}\left(\pi \right)\ne \cancel{0}$, then $\mathcal{F}\left(\pi \right)$ contains a clique of size π_n.

Proof. Applying Theorem 4.2 to the complementary sequence $\overline{\pi }$, there must exist a forbidden set of size n − 1 – π₂ in π. Since π₁ < n − 2, then $\left|\mathcal{B}\left(\pi \right)\right|\ge 2$. Thus π_n must be in a forbidden edge with π_n−1. Then applying Theorem 4.2 again, we conclude that $\mathcal{F}\left(\pi \right)$ must contain a clique of size π_n.

We now show that having forced or forbidden edges for a degree sequence limits the diameter of its realizations.

Theorem 4.4.

For the degree sequence π where π_n ≥ 1, if $\mathcal{F}\left(\pi \right)\ne \cancel{0}$, or $\mathcal{B}\left(\pi \right)\ne \cancel{0}$, then for any realization G of π,

$$\text{diam}\left(G\right)\le 3.$$ (5)

Proof. If π₁ = n − 1 then trivially diam(G) ≤ 2, thus we will assume that π₁ < n − 1. We begin with a consideration of the case when $\mathcal{F}\left(\pi \right)\ne \cancel{0}$, by partitioning the set of vertices of G into three sets, $V=\mathcal{P}\left(\pi \right)\cup Q\cup R$. We define the set Q as all the vertices in G that are adjacent to a vertex in $\mathcal{P}\left(\pi \right)$, but are not themselves in $\mathcal{P}\left(\pi \right)$. We next define the set R as all the remaining vertices, $R=V-\mathcal{P}\left(\pi \right)-Q$. By performing a case analysis, we show that for any two vertices v_i and v_j in G, there is a path between them of length no greater than 3.

Case $\left\{v_i,v_j\right\}\subseteq \mathcal{P}\left(\pi \right)$: Theorem 3.4 says that the forced edges between the vertices in $\mathcal{P}\left(\pi \right)$ form a connected threshold graph, implying that the minimum path length between any two vertices in $\mathcal{P}\left(\pi \right)$ is no more than 2.

Case $v_i\in Q,v_j\in \mathcal{P}\left(\pi \right)$: ¿From the definition of Q and Theorem 3.4, the path length between v_i and v_j is no more than 3.

Case {v_i, v_j} ⊆ Q: Let v_k ∈ N(v_i) and v_l ∈ N(v_j) where $\left\{v_k,v_l\right\}\subseteq \mathcal{P}\left(\pi \right)$. If v_k = v_l or the edge {k, l} ∈ E, then we have found a path of length no more than 3 between v_i and v_j. Else, from Theorem 3.4 we can find a path with length 2 composed of forced edges from v_k to v_l. Let us assume that the path goes through v_m. If the edge {i, m} ∉ E then we could replace the edges {{i, k}, {l, m}} with {{i, m}, {k, l}} violating the assumption that $\left\{l,m\right\}\subseteq \mathcal{F}\left(\pi \right)$. A similar argument establishes that {j, m} must also be in G, giving a path of length 2 from v_i to v_j through v_m.

Case v_i ∈ R, $v_j\in \mathcal{P}\left(\pi \right)$: ¿From Theorem 4.1, since N(v_i) ⊆ Q, then any vertex v_k ∈ N(vi) that we choose will be in Q. Now select two vertices $\left\{v_m,v_n\right\}\subseteq \mathcal{P}\left(\pi \right)$ such that v_m ∈ N(v_k) and $\left\{m,n\right\}\in \mathcal{F}\left(\pi \right)$. We first note that G also must contain the edge {k, n}, because if {k, n} did not exist then we could replace the edges {{i, k}, {m, n}} in G with the set {{i, m}, {k, n}} violating the assumption that $\left\{m,n\right\}\in \mathcal{F}\left(\pi \right)$. Now because all the forced edges are connected, we can inductively extend this argument to show that every forced edge must be in a triangle with v_k. Thus v_i can reach any vertex $v_j\in \mathcal{P}\left(\pi \right)$ with a path of length 2,

Case v_i ∈ R, v_j ∈ Q: The argument for the preceding case shows that v_i can reach any vertex in Q with a path of no more than length 3 by going through some vertex in $\mathcal{P}\left(\pi \right)$.

Case {v_i, v_j} ⊆ R: Choose two vertices v_k ∈ N(v_i) and v_l ∈ N(v_j) and an edge $\left\{m,n\right\}\in \mathcal{F}\left(\pi \right)$. If v_k = v_l then we found a path of length 2. If not then the edge {k, l} must be in E, or else we could replace the edges in {{i, k}, {j, l}, {m, n}} with {{k, l}, {i, m}, {j, n}} violating the assumption that $\left\{m,n\right\}\in \mathcal{F}\left(\pi \right)$. Thus there is a path of no more than length 3 between v_i and v_j.

For the second part of the statement when π_n > 1 and $\mathcal{B}\left(\pi \right)\ne 0$, we note that Theorem 4.2 coupled with the proof of the first part of Theorem 4.4 is almost enough to prove the second part; it only fails when the forbidden edges are strictly between vertices of degree 1. To show the complete statement, assume that $\left\{m,n\right\}\in \mathcal{B}\left(\pi \right)$. Since v_m and v_n are not isolated, then we choose the vertices v_p ∈ N(v_m) and v_q ∈ N(v_n) where v_p and v_q are not necessarily distinct. Theorem 4.1 says that for every realization G of π, the vertices in N(m) ⋃ N(n) form a clique. If the diameter of the graph is greater than 3, then there would have to exist a minimal path of length 4 in G between two vertices v_i and v_j. Without a loss of generality, we can assume that neither the vertex v_i nor its neighbor in that path v_k is in N(m) ⋃ N(n), or else we could find a path of length 3 from v_i to v_j. However, we could replace the edges in {{i, k}, {m, p}, {n, q}} with {{m, n}, {i, p}, {k, q}} violating the assumption that $\left\{m,n\right\}\in \mathcal{B}\left(\pi \right)$.

We now examine the edge connectivity of a graph in which the degree sequence contains either a forced or forbidden edge. The edge connectivity λ(G) is the minimum cardinality of an edge-cut over all edge-cuts of G. There is a trivial upper bound for λ(G) ≤ π_n where π = deg(G). When a graph G has this edge connectivity of λ(G) = π_n, then it is said to be maximally edge-connected. Any realization of a degree sequence with either forced or forbidden edges is maximally edge-connected.

Theorem 4.5.

For the degree sequence π where π_n ≥1, if $\mathcal{B}\left(\pi \right)\ne \cancel{0}$ or $\mathcal{F}\left(\pi \right)\ne \cancel{0}$, then for any realization G of π,

$$\lambda \left(G\right)={\pi }_n.$$ (6)

Proof. We begin with some simple observations about what is required for a graph to be maximally edge-connected. If π_n = 1, then for the connected graph G, λ(G) = π_n is trivially true; thus we assume that π_n ≥ 2. In addition, a result by Plesnrk [13] establishes that if diam(G) ≤ 2, then is maximally edge-connected. Thus, from Theorem 4.4, if G is not maximally edge-connected, then diam(G) = 3.

For a contradiction, we assume that there is a realization of G where λ(G) < π_n. We denote the edge set S as an arbitrary minimum edge-cut of G, and the two components of G with S removed as P and Q. For each set P and Q, we partition each into two sets, P = P_s ⋃ P_n (and Q = Q_s ⋃ Q_n), where P_s (Q_s) is the set of vertices in P (Q) with an adjacent edge in S, and P_n (Q_n) are the remaining vertices.

Using an argument first presented by Hellwig and Volkmann [14], we show that ∣P_n∣ ≥ 2. From the assumption that λ(G) ≤ π_n − 1, then

$${\pi }_n\left|P\right|\le \sum _{p\in P}\text{deg}\left(p\right)\le \left|P\right|\left(\left|P\right|-1\right)+{\pi }_n-1,$$ (7)

which implies that ∣P∣ ≥ π_n + 1. Along with the assumption that ∣P_s∣ ≤ λ(G) ≤ π_n − 1, it follows that ∣P_n∣ = ∣P∣ - ∣P_s∣ > 2. There is a similar argument to show that ∣Q_n∣ ≥ 2 also. One implication from this result is that since G is not maximally edge-connected, then π₁ ≤ n − 3.

We now show that if $\left\{i,j\right\}\in \mathcal{F}\left(\pi \right)$, then v_i and v_j must be in separate components. Suppose that {v_i, v_j} ⊆ Q; then from the proceding argument, there must be at least one edge {k, l} strictly in P. This edge {k, l} would allow us to replace {{i, j}, {k, l}} with {{i, k}, {j, l}} violating the assumption that {i, j} is forced; thus, each forced edge must be in S. Extension of this observation shows that if $K_3\subseteq \mathcal{F}\left(\pi \right)$, then we would have a contradiction with G not being maximally edge-connected.

Let us consider the case where the forbidden edge set for n is not empty, $\mathcal{B}\left(\pi \right)\ne \cancel{0}$. ¿From Theorem 4.2, since π_n ≥ 2, then G has a clique of π_n in $\mathcal{F}\left(\pi \right)$, and so if π_n > 3, then $K_3\in \mathcal{F}\left(\pi \right)$ proving that G is not maximally edge-connected. Thus the only possible case for π not covered by this result is when π_n = 2 and the resulting forced edge {i, j} makes up the set S. Assuming that {v_k, v_l} ⊆ P_n and v_p ∈ Q_n, then if there would exist an edge between v_k and v_l, then the induced subgraph G[{v_i, v_j, v_k, v_l, v_p}] would be one of the three cases in Fig. 3. Since in all three cases the edge {i, j} is not forced, then the edge {k, l} cannot exist. This means that in general any vertex in P_n (or Q_n) must be connected to members of P_s (or Q_s) only, and specifically, in this case, deg(v_k) = deg(v_l) = 1. This is a contradiction to π_n ≥ 2, and thus it follows that if $\mathcal{B}\left(\pi \right)\ne \cancel{0}$, then λ(G) = π_n.

Fig. 3.

The three possible cases when {k, l} ∈ E and {k, l} ⊆ P_n. In all three cases, the edge {i, j} is not forced causing a contradiction.

When the forced edge set is not empty, we again use Theorem 4.2, this time on the complementary sequence $\overline{\pi }$, to show that forbidden edge set $\mathcal{B}\left(\pi \right)$ has a clique of size n − 1 – π₁. Since π₁ ≤ n − 3, then the forbidden edge set is not empty, and thus to avoid a contradiction, then G must be maximally edge-connected.

Biography

About the author: Brian Cloteaux is a Computer Scientist in the Applied and Computation Mathematics Division of the Information Technology Laboratory of NIST The National Institute of Standards and Technology is an agency of the U.S. Department of Commerce.