4.1 DCMs and linear centralities

A (graph) centrality [21] is a function associating with each graph a map such that for any two graphs , if is an isomorphism then

for all .

Centralities that depend essentially only on the rows of the DCM are called geometric [7,22]; within this class, particularly relevant are the so-called linear geometric centralities.

Let : we view a as an infinite vector of real numbers. If is an matrix, we write as a shortcut for

where stands for the vector of the first n entries of a.

Definition 4.2 (Linear (geometric) centrality). Given (the coefficient vector), the centrality is defined by

A centrality is strictly linear (geometric) if it is for some . A centrality is linear (geometric) if it is equivalent to (i.e., it produces the same node ranking as) a strictly linear (geometric) centrality.

One special instance of (strictly) linear centralities are the so-called exponential-decay centralities [23]:

Definition 4.3 (Exponential-decay centrality). Given , the exponential-decay centrality of a node is defined by

where a is obtained for every k > 0 as , and a₀ = 0.

Observe that , seen as a function of , is a polynomial whose coefficients are those in the i-th row of (lowest-degree first). We use to denote such a polynomial. More explicitly,

The degree of is at most

4.2 Graphical sequences for undirected graphs

A good reference about distances in graphs is [24]: this book includes material on DCMs under different terminology for the undirected case. The distance degree sequence (or dds) of node v of an undirected graph (see [24, Section 9.2]) is the vector

where is the number of nodes w such that d(v,w) = i, and e(v) is the eccentricity of v, which is the maximum value of d(v,w) for . Thus, dds(v) is row v of the DCM, truncated to remove the final zeros.

The distance degree sequence (or dds) of the graph is the list of all dds’s of its nodes, listed including multiplicity (see [24], Sect 9.2). This is essentially the DCM with zero entries removed.

Also of interest is the degree sequence of an undirected graph (see [24, Section 9.2]), which is the list of degrees of nodes arranged in nonincreasing order. Up to permutation of the nodes, this is column 1 of the DCM of .

Not all sequences of positive integers can be a degree sequence. A graphical degree sequence is a degree sequence that can be realised by an undirected graph. A complete characterization of graphical degree sequences for undirected graphs is given by the Erdős-Gallai theorem:

Theorem 4.4 ([15]). Consider a sequence of p positive integers. This sequence is a graphical degree sequence if and only if its sum is even and for every

(1)

The Havel-Hakimi algorithm [16,17] provides a constructive way to check in polynomial time if a sequence is a graphical degree sequence. The algorithm is essentially described by the following [24, Theorem 9.1]:

Theorem 4.5. The sequence with for positive integers is a graphical degree sequence if and only if

when re-ordered in nonincreasing order, is a graphical degree sequence.

The Havel-Hakimi algorithm applies Theorem 4.5 inductively: if the process stops at the empty sequence, the original sequence is a graphical degree sequence; if any entry becomes negative before this happens, it is not a graphical degree sequence.

As explained above, column 1 of the DCM of an undirected graph is always a graphical degree sequence. This fact is more general: all columns of a DCM are graphical degree sequences; more precisely, column k of the DCM of an undirected graph is the degree sequence of the graph having the same vertices as and with an edge iff .

4.3 Graphical in-degree sequences

Mutatis mutandis, we can define the in-degree sequence of a directed graph as the list of in-degrees of nodes arranged in non-increasing order. Up to permutation of the nodes, this is column 1 of the DCM of . We can ask whether the algorithm described in Theorem 4.5 can be adapted to in-degree sequences. In fact, the result for directed graphs is much simpler:

Theorem 4.6. The sequence with for natural numbers is always the in-degree sequence of a directed graph.

Proof. Just use and for every , add edges, whose head is node i and whose tails are distinct but arbitrarily chosen elements of . □

A more sophisticated question is whether a given sequence of pairs is the sequence of in- and out-degrees of a graph: the Kleitman–Wang Algorithm [18] extends the Havel-Hakimi Algorithm for this case.

4.4 Some basic properties of (C)DCMs

In this brief section we collect some observations about (C)DCMs. Let us start with some trivial ones.

Proposition 4.7. In any graph with n nodes, for any node i of , and for any :

1. (a) .
2. (b) .
3. (c) and where is the in-degree (number of predecessors) of i.
4. (d) .
5. (e) .
6. (f) .
7. (g) .
8. (h) If M is a DCM (resp. CDCM), so is any matrix obtained by permuting the rows of M.
9. (i) if and only if d(j,i) is finite for all nodes j; in particular, this is always true if is strongly connected.

Observe that the (C)DCM of a graph is uniquely defined only up to row permutation, but by (h) we can use lexicographic order to define a unique (C)DCM of a graph, that we shall call the canonical (C)DCM of .

We can extend the function to subsets: for , define the p-neighbourhood of to be

Clearly

(2)

Proposition 4.8. For any graph and node i:

1. (a) If then for .
2. (b) If is strongly connected and then and for all .

Proof. (a) If then case (d) of Proposition 4.7 implies that . We use induction on r. The result holds for r = 1. For the induction step:

(b) This follows from (a), since for any i by connectedness of .

Corollary 4.9. For any graph and node i the sequence for (which is the i-th row of the CDCM) is monotonic strictly increasing until it reaches a certain value k, at which point all subsequent entries equal k. The value k is exactly the number of nodes j such that d(j,i) is finite. If is strongly connected, k = n.

Definition 4.10 (Good sequence). For a given n, a sequence of positive integers is good if a₀ = 1 and there exists j and such that and for all . If further k = n, we say that the sequence is very good.

Corollary 4.9 states that every row of a CDCM is good. If furthermore the graph is strongly connected, its rows are very good. We observe:

Theorem 4.11. Every good sequence occurs in the CDCM of some graph (in particular, some undirected graph).

Proof. Let the good sequence be , and define for all i > 0, and b₀ = 1. Define to be a tree with root node 0. Node 0 is connected to nodes . One of those nodes connects to nodes , and so on inductively. Then the 0-th row of is , so the 0-th row of is . See Fig 3. □

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Fig 3. Construction for a graph with a given good row.

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We end this section with some necessary conditions for a matrix to be a CDCM, in terms of inequalities. We already know:

Theorem 4.12. For the CDCM of a strongly connected graph:

1. (a) The column 0 entry in any row is 1.
2. (b) The column 1 entry in row i is . That is, the CDCM (hence also DCM) tells us the in-degree of node i.
3. (c) Each row is very good.

We now prove two further inequalities.

Theorem 4.13. Let be the CDCM of a graph . Then for any , with , there exist distinct such that:

(3)

and

(4)

for all .

Proof. Let be the predecessors of i in . To prove (3), observe that since , we have . Therefore . This holds for all k such that , and this proves (3). To prove (4), observe that by (2),

Therefore

which is (4). □

That is, the vector of entries of row i after the first is bounded below, termwise, by the maximum of other distinct rows shifted one place to the right. It is bounded above, termwise, by the sum of the same distinct rows, again shifted one place to the right. Of course, we do not know which rows to choose, but one such set of rows must exist. If no such set exists, the matrix under consideration cannot be a CDCM.

5.1 The three-partition problem

Informally, the three-partition problem (shortened hereafter as TPP) consists in deciding if it is possible to partition a list of natural numbers into groups of exactly three integers each, all having the same sum. This problem is very hard: it is strongly NP-complete (i.e., NP-complete even when the input sequence is provided in unary), and it remains so in many restricted cases.

Definition 5.1 (TPP instance). A TPP instance is a sequence of natural numbers such that and for all , where .

Formally, the three-partition problem is

Problem 5.2 [Three-partition problem (TPP)]. Given a TPP instance , decide whether there exist subsets such that:

1. – the ’s are pairwise disjoint;
2. – for all ;
3. – for all ,

where . If this partition exists, the instance is positive, otherwise it is negative.

We consider a very mild variant, where we assume that all integers are distinct and separated from one another by a certain threshold. This variant is still strongly NP-complete, as shown by the following result.

Theorem 5.3. Problem 5.2 (TPP) is strongly NP-complete (i.e., it is NP-complete even if the input sequence is assumed to be encoded in unary); it remains so even under the following further assumptions:

1. – all ’s are distinct,
2. – ,
3. – for a fixed positive K, and for all .

Proof. The problem is strongly NP-complete as shown in [25]. It remains so even if the input integers are all distinct, as proved in [26]; the fact that the latter result holds also in the presence of the bounds is easy (by adding 2t to each entry in the sequence).

If , then all ’s must be zero or one (because ), so the instance is negative because it contains only one or two elements (recall that the elements in the sequence are all distinct).

Hence, the problem remains strongly NP-complete for distinct integers with . Note that in this case , so all are strictly positive. Multiplying all the entries by K we can assume that and for all (multiplying all elements by a constant does not change the problem, because the sequence obtained after the multiplication is a positive instance of the problem if and only if the original sequence was itself a positive instance). □

As a consequence, we can assume that the input sequence of a TPP instance is a TPP instance in the sense of the following definition, which includes all the assumptions of Theorem 5.3 for K = 3.

Definition 5.4 (Special TPP instance). A special TPP instance is a TPP instance with the following additional properties: , (for all ) and .

5.2 NP-Completeness of (C)DCM

Given a special TPP instance , let

We define a matrix as follows:

1. – the first 3m rows of have the form for ,
2. – the following s rows of have the form ,
3. – the last m rows of have the form .

Intuitively, if a is a positive instance of TPP, it is possible to show that is the DCM of an undirected graph . This fact will be part of the proof of Theorem 5.5, but we want to provide immediately a visual clue to this fact for . To keep the graph small, we use a sequence that fails to satisfy the more restrictive conditions of Theorem 5.3. The graph (see Fig 4) contains m (in this case, 2) connected components, one for each . All components contain three vertices (representing the three integers of the input sequence belonging to ) connected to t vertices (this is possible because the sum of the three integers is always t), which are further connected to a single vertex (of degree t).

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Fig 4. The graph of the construction of Theorem 5.5, for the sequence ; recall that this is a positive instance of TPP (without the further assumptions of Theorem 5.3), because 1 + 5 + 9 = 2 + 6 + 7 = 15.

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Theorem 5.5. Given a special TPP instance , the following two statements are equivalent:

1. a is a positive instance of Problem 5.2;
2. is the DCM of an undirected graph.

Proof. (1) (2). Let be a partition of showing that the sequence is a positive instance. Build an undirected graph with n vertices as follows:

1. – there is one vertex for every integer in the sequence a; we will call these vertices ;
2. – there are t vertices for every class ; we will call them for each ;
3. – there is one vertex for every class , ;
4. – for every , suppose : we add edges
   • for every ;
   • for every ;
   • for every ;
5. – for every and , we add an edge .

Each vertex has degree , and each vertex has degree 2 (because it is connected to for exactly one , and to ). Each has degree t. All vertices related to j (i.e., for , for , and ) form one connected component with t + 4 vertices. There are m such components.

If , has only vertex at distance two, and vertices at distance three (all the vertices of the form except those that are neighbors of ). Finally, has two vertices at distance four (exactly the other two vertices of the form for , and : there are two of them, because ). Hence the row related to is of the form .

The vertices at distance two from are all the other (there are of them). The vertices at distance three are 2 (the vertices of the form for , except the only one that is a direct neighbor of ). Hence, the row related to is of the form .

Finally, the vertices at distance two from are all the vertices of the form for (there are 3 of them). Thus, the row related to is of the form .

It is immediate to check that the DCM of is , meaning that is a DCM.

(2) (1). Suppose that is the DCM of some undirected graph . The graph has n vertices, divided into three groups:

1. – Group I: the first 3m vertices all have degree larger than 2 (because a is a TPP instance such that all of its elements are ≥3); group I corresponds to rows of the form ;
2. – Group II: the next s vertices all have degree 2; group II corresponds to rows of the form ;
3. – Group III: the last m vertices all have degree t (which is larger than a₁, hence of all ’s, because ); group III corresponds to rows of the form ;

Observe that every row of has sum equal to t + 4. Hence the graph has m connected components, with t + 4 vertices each. It is also worth taking note of the form of the rows of the CDCM of :

1. – Group I: ,
2. – Group II: ,
3. – Group III: .

Consider any component X of , and suppose that this component contains A vertices of group I, B vertices of group II and C vertices of group III. The degree sequence of this component is (after reordering):

for some choice of . Note that A + B + C = t + 4 (the size of each component).

Start by considering a vertex of group II: it is connected to two vertices j₁ and j₂, which may each belong, in principle, to one of the three groups. We shall apply Theorem 4.13 to the row of vertex i in the CDCM of : Table 1 describes the possible intervals of integers for the third value in the row of i depending on how we choose j₁ and j₂.

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Table 1. The third value in the row of node i of group II depending on the group its two in-neighbors j₁ and j₂ belong to. Here, we are assuming without loss of generality that .

https://doi.org/10.1371/journal.pone.0352427.t001

The third value in the row of the CDCM for group II is t + 2, and the only intervals containing t + 2 are those in the last column (since , t is certainly larger than 6). This means that every vertex of group II is connected to at least one vertex of group III. As a consequence, B > 0 implies C > 0.

On the other hand, applying again Theorem 4.13, no vertex of group I can be connected to a node of group III, because . Also, no two vertices of group I can be connected to each other: suppose, by contradiction that a vertex of degree is connected to a vertex of degree with . Then we should have (by Theorem 4.13) , that is , which is impossible because . We conclude that vertices of group I are connected only to vertices of group II. As a consequence, A > 0 implies B > 0.

For the reasons above, every component of must contain at least one vertex of group III. Since there are m components and exactly m vertices of group III, we conclude that each component contains exactly one vertex of group III, that is, C = 1.

The A vertices of group I in component X are all connected to vertices of group II (this is, as we discussed above, the only possibility for them), and those vertices of group II must be distinct (they have degree 2, and each of them is connected to at least one vertex of group III); so there must be at least vertices of group II in the component, say for some . Assume for the moment that .

The only vertex of group III is itself connected to t vertices of group II, so we conclude that and B = t. Since A + B + C = t + 4, we have A + t + 1 = t + 4, hence A = 3.

These conditions, together with the fact that this property holds for all components of , allow us to conclude that a is a positive instance of TPP.

Finally, note that must be zero, because if even one component contained some extra vertex of group II, then altogether there would be more than s vertices of group II in the graph, which is impossible. □

Corollary 5.6. Recognizing whether a given matrix is the (C)DCM of an undirected graph is strongly NP-complete.

Proof. Theorem 5.5 reduces TPP to recognizing DCMs, hence the result follows by Theorem 5.3. The fact that TPP is strongly NP-complete is needed for the construction to be admissible (the size of the matrix is pseudopolynomial in the input sequence a). □

Observe that our construction does not directly tell us if the problem of recognizing whether a matrix is the (C)DCM of an arbitrary directed graph is also NP-complete. Incidentally, it is easy to see that is a negative instance of TPP satisfying the conditions in the statement of Theorem 5.3: still, we were able to find a directed graph whose DCM is . In other words, the construction we exhibited is not enough to conclude that recognizing general (C)DCMs is itself an NP-complete problem, although we believe so.

6.1 Disjoint union

The disjoint union of graphs is possibly the simplest binary graph operation: it consists of taking the actual union of nodes and arcs, without any extra arcs added between the two graphs. Since the two node sets may partially overlap, it is formally necessary to label each node so that we know where it comes from:

Definition 6.2 (Disjoint union). Given graphs and , their disjoint union has node set and contains an arc ((x,0),(y,0)) for every arc , and an arc ((x,1),(y,1)) for every arc .

It is nonetheless convenient (and in fact equivalent, up to node renaming) to assume that and are disjoint sets, and to simply let

Now paths (hence, shortest paths) in are simply paths in the two original graphs, which implies that (Fig 5):

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Fig 5. The graph , where is an undirected 8-cycle and is an undirected 3-path.

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Proposition 6.3 Let and be two graphs, and denote their disjoint union. Then, for all we have:

Hence:

Theorem 6.4 Let and be two graphs, and denote their disjoint union. Then for every and natural number k,

Corollary 6.5 Disjoint union ⊕ of graphs is DCM-stable.

6.2 Standard graph products

Graph products [13], in general, are binary graph operations between a graph and another graph that yield a graph whose node set is the cartesian product . The arcs in the product can be defined in different ways, yielding different forms of graph product. It is often useful to think of the arcs in and as possible “moves in a game”, and the product as a way to compose those moves. In particular, [13, Chapter 5] considers four so-called standard products: strong, cartesian, lexicographic, and Kronecker product. Note that some results about distance degree sequences in these products are already known in the literature [27–30]; here we will extend those results to the full distance-count matrix.

6.2.1 Strong product.

Let us start with the strong product (see Fig 6). Here, moves in the product correspond to moves in at least one graph (possibly both). The following is a formal definition:

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Fig 6. The strong product , where is an undirected 8-cycle and is an undirected 3-path (see Fig 5).

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Definition 6.6 (Strong product). Given graphs and , their strong product has node set and contains an arc if and only if one of the following is true:

• and ;
• and ;
• and .

A path in is a sequence of moves in either graph, or in both of them. More precisely, any path can be projected onto (, respectively) by taking (, resp.) and removing repeated consecutive occurrences of the same node. Hence, every path of length k from (x,y) to in yields a path of length at most k from x to in and a path of length at most k from y to in . Hence [13],

On the other hand, it is easy to see that this is an equality, by taking a path that follows the shortest path from x to in and the shortest path from y to in in parallel until this is possible, and then remaining still in either graph and continuing in the other one for the remaining steps. As a consequence:

Proposition 6.7 (Proposition 5.4 of [13]). Let and be two graphs, and denote their strong product. Then for every we have

Note that this proposition (like other similar ones that we will be using from [13]) holds also in the directed case. A consequence of Proposition 6.7 is the following:

Theorem 6.8. Let and be two graphs, and let denote their strong product. Then for every and natural number k,

Proof. The set A of nodes such that consists of those for which

Consider all the pairs of natural numbers such that : we can characterize A as the union of all pairs such that and . In other words, for we must have either h₁ = k or h₂ = k (or both). So we can count the cardinality of A as follows:

where the last summand is needed to remove the pairs that are both at distance k. □

Corollary 6.9. Strong product of graphs is DCM-stable.

6.2.2 Cartesian product.

The cartesian product (see Fig 7) is a different form of product where moves in the product correspond to moves in one of the graphs, but not both. The following is a formal definition:

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Fig 7. The cartesian product , where is an undirected 8-cycle and is an undirected 3-path (see Fig 5).

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Definition 6.10 (Cartesian product). Given graphs and , their cartesian product has node set and contains an arc if and only if one of the following is true:

• and ;
• and .

Following the same line of reasoning as for the strong product, we have:

Proposition 6.11 (Proposition 5.1 of [13]). Let and be two graphs, and denote their cartesian product. Then for every we have

Interestingly, the converse of this fact is also true

Proposition 6.12 Let and be two graphs, and be a graph with . Suppose that for every we have

Then .

Proof. We have that if and only if , which happens if and only if . The latter is equivalent to and , or and , so .

An immediate consequence of Proposition 6.11 is the following

Theorem 6.13. Let and be two graphs, and denote their cartesian product. Then for every and natural number k,

Recalling the definition of exponential-decay centrality, we have that

hence

for every . More is true:

Theorem 6.14. Let and be two graphs, and be a graph with node set . Assume moreover that for any and for at least distinct values we have

(5)

Then (5) holds for any ; moreover, for every and natural number k,

Proof. Remember that and similarly for and , so we have

for at least values . Since on both sides of the equation we have polynomials of degree at most , we conclude that those polynomials in fact coincide; that is,

i.e., for all t

□

Corollary 6.15. Cartesian product of graphs is DCM-stable.

6.2.3 Lexicographic product.

The third common type of graph product [13] is the following (see Fig 8):

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Fig 8. The lexicographic product , where is an undirected 8-cycle and is an undirected 3-path (see Fig 5).

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Definition 6.16 (Lexicographic product). Given graphs and , their lexicographic product has node set and contains an arc if and only if one of the following is true:

• ;
• and .

Distances in the lexicographic product can be characterized as follows for the undirected case:

Proposition 6.17 (Proposition 5.12 of [13]). Let and be two undirected graphs, and . Then:

The directed version of the same property is trickier: the 2 appearing in the lemma would become the length of the shortest cycle of involving x, information that cannot be deduced from the DCM of itself.

Theorem 6.18 Let and be two undirected graphs, and denote their lexicographic product. Then for every and natural number k,

1. •. if then

1. •. else

Proof. Case 1. Let us consider the case first. By Proposition 6.17, in this case we have

that is, more explicitly,

Now, for any given (x,y) and natural number k, the pairs that are at distance k from (x,y) are:

• just (x,y), if k = 0;
• if k = 1, we have all the pairs with , for every possible , plus all the pairs with ;
• if k = 2, we have all the pairs with , for every possible , plus all the pairs with (all possible , except when or );
• if k > 2, we have all the pairs with , for every possible .

Case 2. In the case , then for all k > 0. Hence

Now, for any given (x,y) and natural number k, the pairs that are at distance k > 0 from (x,y) are those with (there are none), plus all those of the form with .□

Corollary 6.19. Lexicographic product ∘ of graphs is DCM-stable for undirected graphs.

This result is anyway not true for directed graphs. To prove that the lexicographic product is not DCM-stable, we must provide four digraphs , , , such that

whereas

This fact implies, clearly, that cannot be computed directly from and .

We choose the graphs , and as in Fig 9 (and let ). It is immediate to observe that

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Fig 9. The graphs and have the same DCM, but has a different DCM than .

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To show that

we need the following:

Definition 6.20 (Girth). For a graph and a node , the local girth of at x, , is the length (number of arcs) of the shortest non-empty path from x to itself in (∞ if such a path does not exist). Define

Then we have:

Theorem 6.21. Given two graphs and we have

Proof. If are two nodes realizing , then for any , the distance from (x,y) to in is exactly . On the other hand, for any and we can build in a path from (x,y) to of length (by leaving the first coordinate fixed and moving the second coordinate along the path from y to in ), and another path of length (by moving along the non-empty path from x to x in , while jumping from y to in at any point along the path). □

As a consequence:

Corollary 6.22. The lexicographic product is not DCM-stable on directed graphs.

Proof. We must prove that , for the graphs in Fig 9. We observe that , , , . Hence:

Two graphs with different finite diameters must have different DCMs. □

6.2.4 Kronecker product.

Not all natural graph operations (in particular: not all graph products) are DCM-stable. One example [13] is the following (see Fig 10):

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Fig 10. The Kronecker product , where is an undirected 8-cycle and is an undirected 3-path (see Fig 5).

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Definition 6.23 (Kronecker (or tensor) product). Given graphs and , their Kronecker product has node set and contains an arc if and only if and .

Note that a path in can always be obtained by travelling in parallel along two paths of the same length (one in and one in ). Even if either can be non-simple, the result might still be simple and in fact even a shortest path.

As we did for the lexicographic product, we provide an undirected counterexample: consider , and defined in Fig 11. We have that, up to node permutation,

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Fig 11. The graphs , and used to prove that the Kronecker product is not DCM-stable.

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Nonetheless, from Theorem 1 of [31] we know that the Kronecker product of two connected graphs is connected if and only if one of them contains an odd cycle. In our example, only contains an odd cycle. This implies that the graph is connected, while is not, which means that .

Corollary 6.24. Kronecker product is not DCM-stable. It is not even DCM-stable for undirected graphs.

6.3 Corona product

The so-called corona product was originally defined by Harary and Frucht [32] for undirected graphs; here, we are going to borrow the three extensions of this product for the directed case proposed in [33], called forward, backward, and symmetric.

Definition 6.25 (Corona product). Given graphs and , their forward corona product , backward corona product , and symmetric corona product are the graphs having node set and such that:

• all the arcs of also belong to all the corona products;
• for every and any , there is an arc from (x,y) to in all the corona products;
• for every and
  1. – there is an arc from x to (x,y) in ;
  2. – there is an arc from (x,y) to x in ;
  3. – there are two symmetric arcs connecting x with (x,y) in .

Alternatively, we can say that contains one copy of (the so-called “center graph”) and one copy of for each node of , and the three products differ only in the direction of the arcs between each node and the nodes of the copy of associated with it. Note that yields the same result as the original definition of the product if and are undirected. In such a case, we can refer to this operation simply as (refer to Fig 12 for an example).

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Fig 12. The corona product , where is an undirected 8-cycle and is an undirected 3-path (see Fig 5).

In the directed versions, the edges connecting to the copies of are directed outwards in , inwards in .

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Proposition 6.26. Let and be two graphs, and let , , and . Then, for every and , the distances between nodes in , and are shown in Table 2.

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Table 2. The distance between two nodes in the different versions of corona product.

https://doi.org/10.1371/journal.pone.0352427.t002

As a consequence:

Theorem 6.27. Let and be two graphs, and let , , and . Then, for every and :

• and ;
• and ;
• and

Proof. Let us start with . Since it belongs to the center graph, regardless of the type of corona product we are considering, at a given distance k it has at least the same number of nodes it has at distance k in . Additionally, for each at distance in , in and node x has at distance k the nodes belonging to the copy of associated with (and, thus, connected to) . Note that when k = 0, , thus , and when k = 1 the nodes at distance 1 to x in and are those belonging to the copy of associated to x itself. Hence, in both cases, everything holds.

Moving to node (x,y), it is easy to see that in and it has at least the same number of nodes that y has at distance k in . Additionally, in in k steps (x,y) can also be reached by those nodes that can reach x in steps in . In , instead, there are several cases to consider, based on the value of k (where k = 0 trivially holds). If k = 1, the nodes whose distance to (x,y) is 1 are the in-neighbors of y in , plus x. When k = 2, instead, the nodes that can reach (x,y) in 2 steps are the in-neighbors of x in , plus the nodes of the copy of associated with x, from which we have to remove the in-neighbors of y and y itself. Finally, when k > 2, the nodes whose distance to (x,y) is k are those at distance to x in , plus those belonging to copies of associated with the nodes such that . □

Corollary 6.28. The forward , backward , and symmetric corona products of graphs are DCM-stable.