A Polynomial-Time Algorithm for Outerplanar Diameter Improvement

Abstract

The Outerplanar Diameter Improvement problem asks, given a graph G and an integer D, whether it is possible to add edges to G in a way that the resulting graph is outerplanar and has diameter at most D. We provide a dynamic programming algorithm that solves this problem in polynomial time. Outerplanar Diameter Improvement demonstrates several structural analogues to the celebrated and challenging Planar Diameter Improvement problem, where the resulting graph should, instead, be planar. The complexity status of this latter problem is open.

Research supported by the Languedoc-Roussillon Project “Chercheur d’avenir” KERNEL and the French project EGOS (ANR-12-JS02-002-01). The sixth author was co-financed by the European Union (European Social Fund ESF) and Greek national funds through the Operational Program “Education and Lifelong Learning” of the National Strategic Reference Framework (NSRF) - Research Funding Program: ARISTEIA II.

Appendix

Proof

(of Lemma 1 ). Let $G'$ be an outerplanar completion of G such that ${\text {ecc}}(v,G') = {\text {ecc}}^*(v,G)$. Contracting the edges of $G'$ that have at least one endpoint out of V(H) one obtains an outerplanar completion $H'$ of H (as outerplanar graphs are minor-closed). As contracting an edge does not elongate any shortest path, we have that ${\text {dist}}_{H'}(v,u) \le {\text {dist}}_{G'}(v,u)$ for any vertex $u\in V(H)$, and in particular the diameter of $H'$ is at most the diameter of $G'$, so ${\text {ecc}}^*(v,H) \le {\text {ecc}}(v,H') \le {\text {ecc}}(v,G') = {\text {ecc}}^*(v,G)$.    $\square$

Proof

(of Lemma 2 ). “$\Leftarrow$”: If ${\text {ecc}}^*(v,B_{1\ldots 6}) + {\text {ecc}}^*(v,B_7) \le D$, gluing on v the outerplanar completions of $G[B_{1\ldots 6}], G[B_7],\ldots ,G[B_t]$, respectively achieving ${\text {ecc}}^*(v,B_{1\ldots 6}), {\text {ecc}}^*(v,B_7), \ldots , {\text {ecc}}^*(v,B_t)$, one obtains a diameter-D outerplanar completion $G'$ of G. Indeed,

• The graph obtained is outerplanar and contains G.

• Two vertices x, y of $G[B_{1\ldots 6}]$ (resp. of $G[B_i]$ for $7\le i\le t$) are at distance at most D from each other, as ${\text {ecc}}^*(v,B_{1\ldots 6}) < \infty$ (resp. as ${\text {ecc}}^*(v,B_i) < \infty$).

• Any vertex x of $G[B_{1\ldots 6}]$ and y of $G[B_i]$, with $7\le i\le t$, are respectively at distance at most ${\text {ecc}}^*(v,B_{1\ldots 6})$ and ${\text {ecc}}^*(v,B_i)\le {\text {ecc}}^*(v,B_7)$ from v. They are thus at distance at most ${\text {ecc}}^*(v,B_{1\ldots 6}) + {\text {ecc}}^*(v,B_7) \le D$ from each other.

• Any vertex x of $G[B_i]$ and y of $G[B_j]$, with $7\le i<j\le t$, are respectively at distance at most ${\text {ecc}}^*(v,B_i)\le {\text {ecc}}^*(v,B_1) \le {\text {ecc}}^*(v,B_{1\ldots 6})$ (By Lemma 1) and ${\text {ecc}}^*(v,B_j)\le {\text {ecc}}^*(v,B_7)$ from v. They are thus at distance at most D from each other.

“$\Rightarrow$”: In the following, we consider towards a contradiction an outerplanar graph G admitting a diameter-D outerplanar completion, but such that

$$\begin{aligned} {\text {ecc}}^*(v,B_{1\ldots 6}) + {\text {ecc}}^*(v,B_7) > D. \end{aligned}$$

(1)

Among the triangulated diameter-D outerplanar completions of G, let $G'$ be one that maximizes the number of branches at v. Let $t'>0$ be the number of branches at v in $G'$, and denote these branches $B'_1,\ldots ,B'_{t'}$, in such a way that ${\text {ecc}}^*(v,G')={\text {ecc}}^*(v,B'_1) \ge {\text {ecc}}^*(v,B'_2) \ge \ldots \ge {\text {ecc}}^*(v,B'_{t'})$. Let $S_{i'}:=\{i\mid B_i\subseteq B'_{i'}\}$ for all $1\le i'\le t'$ (note that $\{S_1,\ldots ,S_{t'}\}$ is a partition of $\{1,\ldots ,t\}$). Furthermore, among all $B'_{i'}$ maximizing ${\text {ecc}}^*(v,B'_{i'})$, we choose $B'_1$ such that $\min S_1$ is minimal. Then, since $G'$ has diameter at most D and shortest paths among distinct branches of $G'$ contain v, it is clear that

$$\begin{aligned} \underset{1\le i'<j'\le t'}{\forall }{\text {ecc}}^*(v,B'_{i'}) + {\text {ecc}}^*(v,B'_{j'}) \le D. \end{aligned}$$

(2)

The branches $B'_{i'}$ with $|S_{i'}|=1$ are called atomic.

Claim 1

Let $B'_{i'}$ be a non-atomic branch and let $S'\subsetneq S_{i'}$. Then, ${\text {ecc}}^*(v,\bigcup _{i\in S'}B_i) + {\text {ecc}}^*(v, \bigcup _{i\in S_{i'} \setminus S'}B_i ) > D$.

Proof

Let $\mathcal {B}:=\bigcup _{i\in S'} B_i$ and $\bar{\mathcal {B}}:=B'_{i'}\setminus \mathcal {B}$. If the claim is false, then ${\text {ecc}}^*(v,\mathcal {B})+{\text {ecc}}^*(v,\bar{\mathcal {B}})\le D$. Furthermore, for all $j'\ne i'$,

$$\begin{aligned} {\text {ecc}}^*(v,\mathcal {B}) + {\text {ecc}}^*(v,B'_{j'}) \mathop {\le }\limits ^{\text {Lemma 1}} {\text {ecc}}^*(v,B'_{i'}) + {\text {ecc}}^*(v,B'_{j'}) \mathop {\le }\limits ^{(2)} D\\ \end{aligned}$$

and, likewise, ${\text {ecc}}^*(v,\bar{\mathcal {B}})+{\text {ecc}}^*(v,B'_{j'})\le D$. Thus, the result of replacing $G'[B'_{i'}]$ with the disjoint union of an outerplanar completion achieving ${\text {ecc}}^*(v,\mathcal {B})$ and an outerplanar completion achieving ${\text {ecc}}^*(v,\bar{\mathcal {B}})$ yields a diameter-D outerplanar completion containing more branches than $G'$, contradicting our choice of $G'$.

In the following, we abbreviate $|S_1|=:s$.

Claim 2

$S_1=\{j \mid 1\le j\le s\}$.

Proof

Towards a contradiction, assume that there is some $i\notin S_1$ with $i+1\in S_1$. Let $i'>1$ be such that $B_i\subseteq B'_{i'}$. Note that $B'_1$ is not atomic, as otherwise ${\text {ecc}}^*(v,B'_1) ={\text {ecc}}^*(v,B_{i+1}) \le {\text {ecc}}^*(v,B_{i}) \le {\text {ecc}}^*(v,B'_{i'})$, contradicting the numbering of the $B'_j$’s. Then,

$$\begin{aligned} {\text {ecc}}^*(v,B'_1\setminus (B_{i+1}\setminus v))) + {\text {ecc}}^*(v, B_{i+1})&\mathop {\le }\limits ^{\text {Lemma 1}} {\text {ecc}}^*(v,B'_1) + {\text {ecc}}^*(v, B_{i+1})\\&\le \,\,\,\,\,\,\, {\text {ecc}}^*(v,B'_1) + {\text {ecc}}^*(v, B_i)\\&\le \,\,\,\,\,\,\, {\text {ecc}}^*(v,B'_1) + {\text {ecc}}^*(v, B'_{i'}) \mathop {\le }\limits ^{(2)} D, \end{aligned}$$

contradicting Claim 1.

Claim 3

For all i, we have ${\text {ecc}}^*(v,B_{1\ldots i}) + {\text {ecc}}^*(v,B_{i+1})> D$ if and only if $i< s$.

Proof

“$\Leftarrow$”: Towards a contradiction, assume there is some $i< s$ such that ${\text {ecc}}^*(v,B_{1\ldots i}) + {\text {ecc}}^*(v,B_{i+1}) \le D$. Then the graph obtained from the diameter-D outerplanar completions of $B_{1\ldots i}$ and $B_j$ for all $i<j\le t$, respectively achieving ${\text {ecc}}^*(v,B_{1\ldots i})$ and ${\text {ecc}}^*(v,B_j)$, would be a diameter-D outerplanar completion of G with more branches than $G'$, a contradiction.

“$\Rightarrow$”: Assume towards a contradiction that there is some $i\ge s$ such that ${\text {ecc}}^*(v,B_{1\ldots i}) + {\text {ecc}}^*(v,B_{i+1}) > D$. By (2) and Lemma 1, we have $D\ge {\text {ecc}}^*(v,B_{1\ldots s}) + {\text {ecc}}^*(v,B_{i+1})$ and, hence ${\text {ecc}}^*(v,B_{1\ldots i}) > {\text {ecc}}^*(v,B_{1\ldots s})$. But this contradicts Lemma 1, as ${\text {ecc}}^*(v,B_{1\ldots s})= {\text {ecc}}(v,G') \ge {\text {ecc}}^*(v,G)$.

By (1), Claim 3 implies that $s\ge 7$.

Claim 4

Let $S'\subseteq \{1,\ldots ,t\}$ and let $\mathcal {B}:=\bigcup _{i\in S'}B_i$. Then, there is a vertex in $\mathcal {B}$ that is, in $G'$, at distance at least ${\text {ecc}}^*(v,\mathcal {B})$ to every vertex of $V(G)\setminus (\mathcal {B}\setminus v)$.

Proof

Towards a contradiction, assume that for every vertex $u \in \mathcal {B}$ there exists a vertex $w \in V(G)\setminus (\mathcal {B}\setminus v))$ such that ${\text {dist}}_{G'}(u,w) < {\text {ecc}}^*(v,\mathcal {B})$. From $G'$, contracting all vertices of $V(G)\setminus \mathcal {B}$ onto v yields a graph H with a path between u and v of length strictly smaller than ${\text {ecc}}^*(v,\mathcal {B})$. As this argument holds for every vertex $u \in \mathcal {B}$, it implies that ${\text {ecc}}(v,H)<{\text {ecc}}^*(v,\mathcal {B})$. Since H is an outerplanar completion of $G[\mathcal {B}]$, this contradicts the definition of ${\text {ecc}}^*$.

Two sub-branches $B_i$ and $B_j$ of $B'_1$ are linked if $G'$ contains an edge from a vertex of $B_i\setminus \{v\}$ to a vertex of $B_j\setminus \{v\}$.

Claim 5

Let $1\le i< j \le s$ and let ${\text {ecc}}^*(v,B_{1\ldots i}) + {\text {ecc}}^*(v,B_j) > D$. Then ${\text {ecc}}^*(v,B_{1\ldots i}) + {\text {ecc}}^*(v,B_j) = D+1$, and $B_{j}$ is linked to one of $B_1,\ldots ,B_i$.

Proof

By Claim 4, there is a vertex $x\in B_j$ that is, in $G'$, at distance at least ${\text {ecc}}^*(v,B_j)$ to every vertex in $B_{1\ldots i}\subseteq V(G)\setminus (\mathcal {B}\setminus v)$. Likewise, there is a vertex $y\in B_{1\ldots i}$ that is, in $G'$, at distance at least ${\text {ecc}}^*(v,B_{1\ldots i})$ to every vertex in $B_j$. Let P be any shortest path of $G'$ between x and y (hence P has length at most D). By construction, the maximal subpath of P in $B_{j}\setminus v$ containing x has length at least ${\text {ecc}}^*(v,B_j)-1$ and the maximal subpath of P in $B_{1\ldots i}\setminus v$ containing y has length at least ${\text {ecc}}^*(v,B_{1\ldots i})-1$. Since these subpaths are vertex disjoint the remaining part of P has length $d_P\ge 1$. Hence $D\ge {\text {ecc}}^*(v,B_j) + {\text {ecc}}^*(v,B_{1\ldots i}) +d_P-2$. As ${\text {ecc}}^*(v,B_{1\ldots i}) + {\text {ecc}}^*(v,B_{j}) > D$, we have that $d_P=1$, and thus there is a single edge in P linking $B_{j}$ and $B_{1\ldots i}$. This also yields to ${\text {ecc}}^*(v,B_j) + {\text {ecc}}^*(v,B_{1\ldots i}) = D+1$.

In the following, consider the graph L on the vertex set $\{1,\ldots ,t\}$ such that ij is an edge of L if and only if $B_i$ is linked to $B_j$ in $G'$. For all $1\le k\le t$, let $L_k$ be the subgraph of L that is induced by $\{1,\ldots ,k\}$. A consequence of the next claim will be that $B_{i+1}$ is linked to exactly one of these branches.

Claim 6

For each $1\le k\le s$, the graph $L_k$ is a path.

Proof

Let $1\le k\le s$. Then,

• 1. $L_k$ is connected. Indeed Claims 3 and 5 clearly imply that for any $1\le i< s$, $B_{i+1}$ is linked to one of $B_1,\ldots ,B_i$, i.e. that there is a path from any k to 1 in $L_k$.

• 2. $L_k$ has maximum degree 2: towards a contradiction, assume that some branch $B_i$ is linked to three branches $B_{j_1}$, $B_{j_2}$, and $B_{j_3}$. As each of $B_i\setminus v$, $B_{j_1}\setminus v$, $B_{j_2}\setminus v$, and $B_{j_3}\setminus v$ induces a connected graph in $G'$, these four sets together with v induce a $K_{2,3}$-minor in $G'$, contradicting its outerplanarity.

• 3. $L_k$ is not a cycle since otherwise, as each $B_i\setminus v$ induces a connected graph in $G'$, these sets together with v would induce a $K_4$-minor in $G'$ (since $s\ge 3$), contradicting its outerplanarity.

Hence, for any $1\le i\le s$, the graph $G'[B_{1\ldots i}\setminus v]$ is connected.

Claim 7

For any $3\le i < s$, ${\text {ecc}}^*(v,B_{1\ldots i}) > {\text {ecc}}^*(v,B_{1\ldots i-2})$.

Proof

The monotonicity property given by Lemma 1 implies that ${\text {ecc}}^*(v,B_{1\ldots i})\ge {\text {ecc}}^*(v,B_{1\ldots i-1}) \ge {\text {ecc}}^*(v,B_{1\ldots i-2})$. Towards a contradiction suppose that ${\text {ecc}}^*(v,B_{1\ldots i}) = {\text {ecc}}^*(v,B_{1\ldots i-1}) = {\text {ecc}}^*(v,B_{1\ldots i-2}) =: c$. Then, Claim 3 implies that $c + {\text {ecc}}^*(v,B_{j}) > D$ for all $j\in \{i-1,i,i+1\}$. Thus, by Claim 5, each of $B_{i-1}$, $B_{i}$, and $B_{i+1}$ is linked to one of $B_1,\ldots ,B_{i-2}$. As each of $B_{1\ldots i-2}\setminus v$, $B_{i-1}\setminus v$, $B_{i}\setminus v$, and $B_{i+1}\setminus v$ induces a connected graph in $G'$, these sets together with vertex v induce a $K_{2,3}$-minor, contradicting the outerplanarity of $G'$.

In the following let q be any integer such that $3\le q\le s$ and $B_q$ is not linked to $B_1$. Let $p<q$ be such that $B_p$ and $B_q$ are linked. Note that p is unique since otherwise, $L_q$ would not be a path, contradicting Claim 6.

Consider a shortest cycle containing v, a vertex $u\in B_p$ and some vertex of $B_q$. Since $G'$ is triangulated, this cycle is a triangle. Thus, u is a neighbor of v (in $G'$) with $u\in B_p$ and u is adjacent to some vertex in $B_q\setminus v$ (see Fig. 3 for an illustration).

Fig. 3.

Structure of $G'[B_{1\ldots q}]$.

Since, by Claim 6, all paths in $G'$ between a vertex in $B_1$ and a vertex in $B_q$ contain u or v, it is clear that $\{v,u\}$ separates $B_1\setminus v$ and $B_q\setminus v$. Let (X, Y) be a separation of $G'$ (that is, two sets $X,Y \subseteq V(G')$ such that $X \cup Y = V(G')$ and such that there are no edges between $X \setminus Y$ and $Y \setminus X$) such that $X\cap Y =\{v,u\}$, $B_{1\ldots q-1}\setminus B_p\subsetneq X$ and $B_q\subseteq Y$ (such a separation exists by Claim 6).

Claim 8

${\text {ecc}}^*(v,B_{1\ldots q}) = {\text {ecc}}^*(v,B_{1\ldots q-1})$.

Proof

By Lemma 1, it suffices to show ${\text {ecc}}^*(v,B_{1\ldots q}) \le {\text {ecc}}^*(v,B_{1\ldots q-1})$. To this end, let H be the outerplanar completion of $B_{1\ldots q}$ obtained from $G'$ by contracting every branch $B_i$, with $i>q$, onto v. Since H is a minor of $G'$, H is a diameter-D outerplanar completion of $B_{1\ldots q}$. We show ${\text {ecc}}(v,H)\le {\text {ecc}}^*(v,B_{1\ldots q-1})$.

Consider any vertex $x\in X$, and let $y\in B_q \subseteq Y$ be a vertex that is at distance at least ${\text {ecc}}^*(v,B_q)$ to both v and u (such a vertex y exists by Claim 4). As all shortest paths between x and y (of length at most D) contain v or u, the vertex x is at distance at most $D -{\text {ecc}}^*(v,B_q)$ to v or u. As v and u are adjacent, the vertex x is at distance at most $D +1 -{\text {ecc}}^*(v,B_q)$ ($= {\text {ecc}}^*(v,B_{1\ldots q-1})$ by Claim 5, which is applicable since, by Claim 3, ${\text {ecc}}^*(v,B_{1\ldots q-1}) + {\text {ecc}}^*(v,B_q) > D$) to v. Since x is chosen arbitrarily in X, every vertex in X is at distance at most ${\text {ecc}}^*(v,B_{1\ldots q-1})$ to v in H.

Consider now any vertex $y\in Y\cap V(H)$, and let $x\in B_1 \subsetneq X$ be a vertex that is at distance at least ${\text {ecc}}^*(v,B_1)$ to both v and u (such a vertex x exists by Claim 4). As a shortest path between x and y (of length at most D) goes through v or u, the vertex y is thus at distance at most $D -{\text {ecc}}^*(v,B_1)$ to v or u. As v and u are adjacent, the vertex y is at distance at most $D +1 -{\text {ecc}}^*(v,B_1)$ ($= {\text {ecc}}^*(v,B_2)$ by Claim 5, which is applicable since, by Claim 3, ${\text {ecc}}^*(v,B_1)+{\text {ecc}}^*(v,B_2)>D$) to v. As ${\text {ecc}}^*(v,B_2) \le {\text {ecc}}^*(v,B_{1\ldots q-1})$ by Lemma 1, every vertex $y\in Y\cap V(H)$ is at distance at most ${\text {ecc}}^*(v,B_{1\ldots q-1})$ to v in H.    $\square$

We now claim that there exist two consecutive such values q between 3 and 6. Indeed, by Claim 6 $B_1$ is linked to at most two other branches, and by Claims 3 and 5 $B_2$ is linked to $B_1$, so it follows that $B_1$ is linked to at most one branch $B_j$ with $j \ge 3$. Therefore, for $3 \le q \le 6$, there are at least two consecutive values of q such that $B_q$ is not linked to $B_1$. Once we have these two consecutive values, say $i-1$ and i, we have by Claim 8 that ${\text {ecc}}^*(v,B_{1\ldots i-2}) = {\text {ecc}}^*(v,B_{1\ldots i})$, for some $i\le 6$, contradicting Claim 7. This concludes the proof of the lemma.   $\square$

Proof

(of Corollary 1 ). Recall that the property of having an outerplanar completion with bounded diameter is minor closed. Thus $G_i$ being a minor of G, we have that if G admits a diameter-D outerplanar completion, then so does $G_i$.

On the other hand, if $G_i$ admits a diameter-D outerplanar completion, by Lemma 2 applied to $G_i$ we have that ${\text {ecc}}^*(v,B_{1\ldots 6}) + {\text {ecc}}^*(v,B_7) \le D$. Thus gluing on v the outerplanar completions of $G[B_{1\ldots 6}], G[B_7],\ldots ,G[B_t]$, respectively achieving ${\text {ecc}}^*(v,B_{1\ldots 6}), {\text {ecc}}^*(v,B_7), \ldots , {\text {ecc}}^*(v,B_t)$, one obtains a diameter-D outerplanar completion of G.   $\square$

Proof

(of Lemma 3 ). It is clear from the fact that a shortest path from $X_u \setminus \{u\}$ to u does not go through $X_v \setminus \{w\}$ (as it should go through $w\in N(u)$), from the fact that a shortest path from $X_u$ to v goes through $\{u,w\}\subseteq N(v)$, and from the fact that any subpath of a shortest path is a shortest path (for some pair of vertices).   $\square$

Proof

(of Lemma 5 ). In a diameter-D outerplanar completion of $G\setminus C$ there is a vertex v with eccentricity at most $\lfloor D/2\rfloor +1$, by Lemma 4. In this completion, adding the completion of $C+v$ achieving $r^+({C}) < D/2$, yields a diameter-D outerplanar completion of G.   $\square$

Proof

(of Observaton 2 ). Let the result of contracting all vertices in $G' \setminus (C\cup C')$ onto vertices in C and contracting $C'$ onto a single vertex u be $G''$. Then, $G''$ is a subgraph of an outerplanar completion of the result of adding u as isolated vertex to $G'[C]$. By definition, ${\text {ecc}}(u,G'')\ge r^+({C})$, implying that there is a vertex $v\in C$ at distance at least $r^+({C})$ to u in $G''$. Thus, v is at distance at least $r^+({C})$ to each vertex of $C'$ in $G'$.   $\square$

Proof

(of Lemma 6 ). The first statement comes from the above comments. The proof of the second statement is similar to the one of Lemma 5. For some component C of G, let v be such that $ecc(v,C)= r^*({C}) \le r^+({C})\le D/2$, and complete C in order to achieve this value. Then for the other components $C'$ consider their escalated completion with respect to v. As $r^+({C'})\le D/2$ this graph has diameter at most D.   $\square$

Proof

(of Lemma 7 ). Same proof as Lemma 6   $\square$

Proof

(of Lemma 8 ). Indeed, by Lemma 5 it is sufficient to consider the component $\mathcal {C}_{\text {max}}$ alone. As $r^+({\mathcal {C}_{\text {max}}}) < \infty$, $\mathcal {C}_{\text {max}}$ has a diameter-D outerplanar completion, and so does G.   $\square$

Proof

(of Lemma 9 ). By Corollary 3, in a diameter-D outerplanar completion $G'$ of G the p components are pairwise adjacent, and any of the q components is adjacent to every of the p ones. For $p=2$, as $q\ge 3$, this would induce a $K_{2,3}$-minor in $G'$, a contradition. For the other cases, this would induce a $K_{4}$-minor.   $\square$

Proof

(of Lemma 10 ). Let us denote $C_1,C_2,C_3,$ and $C_4$ the connected components such that $r^+({C_i}) \ge D/2$, distinct from $\mathcal {C}_{\text {max}}$. Assume for contradiction that G admits a diameter-D outerplanar completion, denoted $G'$.

Claim 9

For each $C_i,C_j$, either $C_i$ and $C_j$ are adjacent in $G'$, or $C_i$ and $C_j$ have a common neighbor in $G'$.

Proof

Assume for contradiction that $C_i$ and $C_j$ are not adjacent and do not have a common neighbor in $G'$. Let us now construct the graph $G''$ from $G'$ as follows. For any component C of $G'\setminus (C_i\cup C_j)$ that is not adjacent to both $C_i$ and $C_j$, contract C onto vertices of $C_i$ or $C_j$ (According to the one C is neighboring). As $G''$ is obtained from $G'$ by contracting edges, $G''$ also is a diameter-D outerplanar completion (for some graph containing $C_i$ and $C_j$). Let $N_i:=N_{G''}(C_i)$, let $N_j:=N_{G''}(C_j)$, and note that $C_i\cap N_j=\emptyset$, $N_i\cap C_j=\emptyset$, and $N_i\cap N_j=\emptyset$. Then, by Observation 2 (as $G''\setminus C_i$ and $G''\setminus C_j$ are connected), there are vertices $v_i\in C_i$ and $v_j\in C_j$ at distance at least D / 2 to each vertex in $N_i$ and $N_j$, respectively, in $G''$. Since $N_i$ and $N_j$ are at distance at least one, $v_i$ and $v_j$ are at distance at least $D+1$, contradicting $G''$ having diameter D.

Claim 10

There is a vertex $u\in \mathcal {C}_{\text {max}}$ that is adjacent in $G'$ to 3 of the components $C_1,C_2,C_3,$ and $C_4$.

Proof

First, note that there is a vertex u and 3 components, say $C_1,C_2,C_3$, with $u\in N_{G'}[C_i]$ for all $1\le i\le 3$, since otherwise, there would be internally vertex-disjoint paths between each two of the four components $C_i$, implying the existence of a $K_4$-minor in $G'$.

If u is neither in $\mathcal {C}_{\text {max}}$ nor in $C_i$, for $1\le i\le 3$, then, since all the $C_i$ are adjacent to $\mathcal {C}_{\text {max}}$ (by Corollary 3), $G'$ would have a $K_{2,3}$-minor on the vertex sets $\{u,\mathcal {C}_{\text {max}}\}$ and $\{C_1,C_2,C_3\}$.

Hence, in the following, we assume that $u\in C_1$. Let z be a neighbor of $C_1$ in $\mathcal {C}_{\text {max}}$ and, for $i\in \{2,3\}$ let $w_i$ denote a neighbor of $C_4$ in $N[C_i]$. We note that $w_2\ne z$ and $w_3\ne z$, since otherwise, the claim follows and we are done. Furthermore, $w_2\ne u$ and $w_3\ne u$, since otherwise there is a $K_{2,3}$-minor on the vertex sets $\{u,\mathcal {C}_{\text {max}}\}$ and $\{C_2,C_3,C_4\}$. Let $X:=(C_4\cup \{w_2,w_3\})\setminus (C_2\cup C_3)$ and note that X is adjacent to $C_2$ and $C_3$, respectively. Let Y be the connected component of $\mathcal {C}_{\text {max}}\setminus \{w_2,w_3\}$ containing z, and note that Y is adjacent to $C_1$ and X. Finally, since X, Y, $C_1$, $C_2$, and $C_3$ are pairwise disjoint, $G'$ has a $K_{2,3}$-minor on the vertex sets $\{X,C_1\}$ and $\{C_2,C_3,Y\}$.

Let v denote a vertex of $\mathcal {C}_{\text {max}}$ that is at distance at least $D/2+1$ to u in $G'$ and consider the result $G' \setminus \{u\}$ of removing u from $G'$. Let C denote the connected component of $G' \setminus \{u\}$ that contains v. Towards a contradiction, assume there is a connected component $C_i$ that is adjacent to u but not to C in $G'$, then all paths between v and any vertex in $C_i$ contain u. Since $G'$ has diameter D, all vertices in $C_i$ are at distance at most $D/2-1$ to u in $G'$, contradicting $r^+({C_i})\ge D/2$. Thus there is a $K_{2,3}$-minor in $G'$ on the vertex sets $\{C_1,C_2,C_3\}$ and $\{u,X\}$ where X is the connected component of $G'\setminus (C_1\cup C_2\cup C_3\cup \{u\})$ containing v. This concludes the proof of the lemma.   $\square$

Proof

(of Theorem 2 ). Indeed, the algorithm runs in time $O(n^7)$ for odd D (at most $O(n^4)$ at line 16, times $O(n^3)$ for the call to OPDI-Connected in line 18). The algorithm runs in $O(n^{2p+2q+1})$ time for even D ($O(n^{2p-2})$ in line 16, times $O(n^{2q})$ in line 22, times $O(n^3)$ for the call to OPDI-Connected in line 23), where p and q respectively denote the number of connected components C such that $r^+({C}) > D/2$ and $r^+({C}) = D/2$. As $p+q \le 4$, we are done.