Highly connected infinite digraphs without edge-disjoint back and forth paths between a certain vertex pair

TL;DR

This paper constructs highly connected infinite directed graphs that lack edge-disjoint paths in both directions between a specific vertex pair, using self-similar graph techniques that may have broader applications.

Contribution

It introduces a method to build highly connected digraphs with no edge-disjoint s-t and t-s paths, employing self-similar graph constructions.

Findings

Constructed for all natural numbers k a k-edge-connected digraph

Demonstrated the absence of edge-disjoint s-t and t-s paths in these graphs

Utilized self-similar graphs as a key technique

Abstract

We construct for all $ k\in \mathbb{N} $ a $ k $-edge-connected digraph $ D $ with $ s,t\in V(D) $ such that there are no edge-disjoint $ s \rightarrow t $ and $t\rightarrow s $ paths. We use in our construction "self-similar" graphs which technique could be useful in other problems as well.

Equations

$$t,P_{s_{2k-2},t_{2k-2}},\dots ,P_{s_{2k-2j},t_{2k-2j}},\dots ,P_{s_0,t_0},P_{s_1,t_1},s$$

$$s,P_{t_1,s_1},P_{t_3,s_3},\dots ,P_{t_{2j+1},s_{2j+1}},\dots ,P_{t_{2k-1},s_{2k-1}}$$

$$P_{s_{2k-1},t_{2k-1}},P_{s_{2k-3},t_{2k-3}},\dots ,P_{s_{2k-1-2j},t_{2k-1-2j}},\dots ,P_{s_1,t_1},s$$

$$t,P_{s_{2k-2},t_{2k-2}},P_{s_{2k-4},t_{2k-4}},\dots ,P_{s_{2k-2-2j},t_{2k-2-2j}}\dots ,P_{s_0,t_0}$$

$$P_{t_0,s_0},P_{t_2,s_2},\dots ,P_{t_{2j},s_{2j}},\dots ,P_{t_{2k-2},s_{2k-2}},t$$

$$C\subseteq \bigcap _{n=0}^{\infty }\mathsf{span}(V_{i_0\dots i_n})=\mathsf{span}\left(\bigcap _{n=0}^{\infty }{V}_{i_0\dots i_n}\right)=\mathsf{span}(\varnothing )=\varnothing$$

Full text

This is the peer reviewed version of the following article: [3], which has been published in final form at http://dx.doi.org/10.1002/jgt.22046. This article may be used for non-commercial purposes in accordance with Wiley Terms and Conditions for Self-Archiving.

1 Introduction

1.1 Basic notions

In this paper by “path” we mean a finite, simple, directed path. Sometimes we define a path of a digraph $D=(V,A)$ by a finite sequence $v_{0},\dots,v_{n}$ of vertices of $D$. If there are more than one edges from $v_{i}$ to $v_{i+1}$ for some $i<n$, then it is not specified which edge is used by the path, so we use this kind of definition only if it does not matter. An $u\rightarrow v$ path is a path with initial vertex $u$ and terminal vertex $v$. Its length is the number of its edges. We call a digraph $D$ connected if for all $u,v\in V(D)$ there is a $u\rightarrow v$ path in $D$. For $U\subseteq V$ let $\mathsf{span}_{D}(U)$ be the set of those edges of $D$ whose heads and tails are contained in $U$ and let $D[U]=(U,\mathsf{span}_{D}(U))$. If it is clear what digraph we talk about, then we omit the subscripts.

1.2 Background and Motivation

R. Aharoni and C. Thomassen proved by a construction the following theorem that shows that several theorems about edge-connectivity properties of finite graphs and digraphs become “very” false in the infinite case.

Theorem 1** (R. Aharoni, C. Thomassen [1]).**

For all $k\in\mathbb{N}$ there is an infinite graph $G=(V,E)$ and $s,t\in V$ such that $E$ has a $k$-edge-connected orientation but for each path $P$ between $s$ and $t$ the graph $G=(V,E\setminus E(P))$ is not connected.

In this article we would like to introduce a similar result. If $D$ is a $k$-edge-connected finite digraph, then for all $s_{1},t_{1},\dots,s_{k},t_{k}\in V(D)$ there are pairwise edge-disjoint paths $P_{1},\dots,P_{k}$ such that $P_{i}$ is an $s_{i}\rightarrow t_{i}$ path. This fact is implied by the following Theorem of W. Mader as well as the (strong form of) Edmonds’ Branching theorem (see [2] p. $349$ Theorem $10.2.1$).

Theorem 2** (W. Mader [4]).**

Let $D=(V,A)$ be a $k+1$-edge-connected, finite digraph and $s,t\in V$. Then there is an $s\rightarrow t$ path $P$ such that $(V,A\setminus A(P))$ is $k$-edge-connected.

We will show that in the infinite case there is no $k\in\mathbb{N}$ such that $k$-edge-connectivity guarantees even the existence of edge-disjoint $s_{1}\rightarrow t_{1}$ and $s_{2}\rightarrow t_{2}$ paths for all $s_{1},t_{1},s_{2},t_{2}$ vertices. Not even in the special case where the two ordered vertex pair is the reverse of each other.

2 Main result

Theorem 3**.**

For all $k\in\mathbb{N}$ there exists a $k$-edge-connected digraph without back and forth edge-disjoint paths between a certain vertex pair.

Proof.

Let $k\geq 2$ be fixed, $I=\{0,\dots,2k-1\}$, $I_{e}=\{i\in I:i\text{ is even }\}$, $I_{o}=I\setminus I_{e}$. Denote by $I^{*}$ the set of finite sequences from $I$. Let the vertex set $V$ of the digraph is the union of the disjoint sets $\{s_{\mu}:\mu\in I^{*}\}$ ( we mean $s_{\mu}=s_{\nu}$ iff $\mu=\nu$) and $\{t_{\mu}:\mu\in I^{*}\}$ ( $t_{\mu}=t_{\nu}$ iff $\mu=\nu$). If $\mu$ is the empty sequence we write simply $s,t$ and we denote the concatenation of sequences by writing them successively. For $\nu\in I^{*}$ let denote the set $\{r_{\nu\mu}:r\in\{s,t\},\ \mu\in I^{*}\}\subseteq V$ by $V_{\nu}$. The edge-set $A$ of the digraph consists of the following edges. For all $\mu\in I^{*}$ there are $k$ edges in both directions between the two elements of the following pairs: $\{s_{\mu},t_{\mu 1}\},\ \{s_{\mu i},t_{\mu(i+2)}\}\ (i=0,\dots,2k-3),\ \{s_{\mu(2k-2)},t_{\mu}\}$. Simple directed edges are $(s_{\mu},t_{\mu 0}),(t_{\mu i},s_{\mu(i+1)})_{i\in I_{e}},(s_{\mu i},t_{\mu(i+1)})_{i\in I_{o}\setminus\{2k-1\}},$ (s_{\mu(2k-1)},t_{\mu})\ for all $\mu\in I^{*}$. Finally $D\stackrel{{\scriptstyle\text{def}}}{{=}}(V,A)$ (see figure 1).

Remark 4*.*

One can avoid using parallel edges (without losing the desired properties of the digraph) by dividing each of these edges with one-one new vertex and drawing between them $k(k-1)$-many new directed edges, one-one for each ordered pair. One can also achieve $k$-connectivity instead of $k$-edge-connectivity by using some similarly easy modification.

Proposition 5**.**

For $\nu\in I^{*}$ the function $f_{\nu}:V\rightarrow V_{\nu},\ f_{\nu}(r_{\mu})\stackrel{{\scriptstyle\text{def}}}{{=}}r_{\nu\mu}\ (r\in\{s,t\})$ is an isomorphism between $D$ and $D[V_{\nu}]$.

Proof:  It is a direct consequence of the definition of the edges since the number of edges from $r_{\mu}$ to $r^{\prime}_{\mu^{\prime}}$ are the same as from $r_{\nu\mu}$ to $r^{\prime}_{\nu\mu^{\prime}}$ for all $r,r^{\prime}\in\{s,t\},\ \nu,\mu,\mu^{\prime}\in I^{*}$. 🌑

Proposition 6**.**

Denote by $D_{v}$ the digraph that we obtain from $D$ by contracting for all $i\in I$ the set $V_{i}$ to a vertex $v_{i}$. Then $D_{v}$ is $k$-edge-connected.

Proof:  In the vertex-sequence $s,v_{1},v_{3},\dots,v_{2k-1}$ there are $k$ edges in both directions between the neighboring vertices such as in the sequence $v_{0},v_{2},\dots,v_{2k-2},t$. Finally there are in both directions at least $k$ edges between the vertex sets of the sequences above. 🌑

For $u\neq v$ we denote by $\lambda(u,v)$ the local edge-connectivity from $u$ to $v$ in $D$ (i.e. $\lambda(u,v)=\min\{\left|A^{\prime}\right|:A^{\prime}\subseteq A,\text{ there is no path from }u\text{ to }v\text{ in }(V,A\setminus A^{\prime})\}$) and let $\lambda\{u,v\}\stackrel{{\scriptstyle\text{def}}}{{=}}\min\{\lambda(u,v),\lambda(v,u)\}$.

Proposition 7**.**

$D$* is connected.*

Proof:  We will show that $\lambda\{s,r_{\mu}\}\geq 1$ for all $r\in\{s,t\},\ \mu\in I^{*}$. We will use induction on length of $\mu$ (which is denoted by $\left|\mu\right|$). Consider first the $\left|\mu\right|=0,1$ cases directly.

The path $s,t_{0},s_{1},t_{2},s_{3},\dots,t_{2k-2},s_{2k-1},t$ shows that $\lambda(s,t)\geq 1$. Using the isomorphism $f_{i}$ (see Proposition 5) we may fix an $s_{i}\rightarrow t_{i}$ path $P_{s_{i},t_{i}}$ in $D[V_{i}]$ for all $i\in I$. The path

[TABLE]

justifies that $\lambda(t,s)\geq 1$ (thus $\lambda\{s,t\}\geq 1$). Then we may fix a $t_{i}\rightarrow s_{i}$ path $P_{t_{i},s_{i}}$ in $D[V_{i}]\ (i\in I)$. The paths

[TABLE]

certify that $\lambda\{s,r_{i}\}\geq 1$ if $r\in\{s,t\},\ i\in I_{o}$. The paths

[TABLE]

certify that $\lambda\{t,r_{i}\}\geq 1$ if $r\in\{s,t\}\geq 1,\ i\in I_{e}$ and thus (by $\lambda\{s,t\}\geq 1$ and by transitivity) $\lambda\{s,r_{i}\}\geq 1$ if $r\in\{s,t\},\ i\in I_{e}$. Hence the cases $\mu\in I*$ with $|\mu|\leq 1$ are settled.

Let be $l\geq 1$ and suppose $\lambda\{s,r_{\mu}\}\geq 1$ if $r\in\{s,t\},\ \mu\in I^{*},\ \left|\mu\right|\leq l$. Let $\nu=\mu i$, where $i\in I$ and $\left|\mu\right|=l$. By the induction hypothesis we have $\lambda\{s,s_{\mu}\}\geq 1$. By the induction hypothesis for $l=1$ we have $\lambda\{s,r_{i}\}\geq 1$ and so $\lambda\{s_{\mu},r_{\mu i}\}\geq 1$ by the isomorphism $f_{\mu}$. Combining these, we get $\lambda\{s,r_{\mu i}\}\geq 1$. 🌑

Lemma 8**.**

$D$* is $k$-edge-connected.*

Proof:  Let $k>l\geq 1$.

Proposition 9**.**

Let $\mu\in I^{*}$ arbitrary. If we delete at most $l$ edges of the digraph $D[V_{\mu}]$ in such a way that its subgraphs $D[V_{\mu i}]\ (i\in I)$ remain connected after the deletion, then $D[V_{\mu}]$ also remains connected after the deletion.

Proof:  Because the isomorphism $f_{\mu}$ it is enough to deal with the case where $\mu$ is the empty sequence. Denote by $D^{\prime}$ the digraph that we have after the deletion. Let $D^{\prime}_{v}$ be the digraph that we get from $D^{\prime}$ by contracting the sets $V_{i}$ to a vertex $v_{i}$ for all $i\in I$. The digraphs $D^{\prime}[V_{i}]\ (i\in I)$ are connected by assumption, thus $D^{\prime}$ is connected iff $D^{\prime}_{v}$ is connected. The digraph $D^{\prime}_{v}$ arises by deleting at most $l<k$ edges of the $k$-edge-connected digraph $D_{v}$ (see Proposition 6) hence it is connected. 🌑

We will prove that if $D$ is $l$-edge-connected, then it is also $l+1$ edge-connected. This is enough since we have already proved $1$-connectivity of $D$ in Proposition 7. Assume that $D$ is $l$-edge-connected. Let $C\subseteq A,\ \left|C\right|=l$ arbitrary and $D^{\prime}\stackrel{{\scriptstyle\text{def}}}{{=}}(V,A\setminus C)$. By the definition of $l+1$-edge connectivity we need to show that $D^{\prime}$ is connected. Suppose for contradiction that it is not. Since the connectivity of the subgraphs $D^{\prime}[V_{i}]\ (i\in I)$ implies the connectivity of $D^{\prime}$ (by Proposition 9) there is an $i_{0}\in I$ such that $D^{\prime}[V_{i_{0}}]$ is not connected. Since the connectivity of the subgraphs $D^{\prime}[V_{i_{0}i}]\ (i\in I)$ implies the connectivity of $D^{\prime}[V_{i_{0}}]$ there is an $i_{1}\in I$ such that $D^{\prime}[V_{i_{0}i_{1}}]$ is not connected$\dots$ By recursion we obtain an infinite sequence $(i_{n})_{n\in\mathbb{N}}$ such that the digraphs $D^{\prime}[V_{i_{0}\dots i_{n}}]\ (n\in\mathbb{N})$ are all disconnected. Note that the digraphs $D[V_{i_{0}\dots i_{n}}]\ (n\in\mathbb{N})$ are $l$-connected because $D$ is $l$-connected by assumption and they are isomorphic to it, hence necessarily $C\subseteq\mathsf{span}(V_{i_{0}\dots i_{n}})$ for all $n\in\mathbb{N}$. But then

[TABLE]

which is a contradiction since $\left|C\right|=l\geq 1$.

Lemma 10**.**

There are no edge-disjoint back and forth paths between $s$ and $t$ in $D$.

Proof:  Suppose, seeking a contradiction, that there are. Let $P_{s,t}$ be an $s\rightarrow t$ path and $P_{t,s}$ be a $t\rightarrow s$ path such that they are edge-disjoint and have a minimal sum of lengths among these path pairs. For $u,v\in V$ call a set $U\subseteq V$ an $uv$-cut iff $u\in U$ and $v\notin U$. The set $\{t\}\cup\bigcup\{V_{i}:i\in I_{e}\}$ is a $ts$-cut and its outgoing edges are $\{(t_{i},s_{i+1})\}_{i\in I_{e}}$. Let $i_{0}\in I_{e}$ be the maximal index such that $P_{t,s}$ uses the edge $(t_{i_{0}}s_{i_{0}+1})$. Then an initial segment of $P_{t,s}$ is necessarily of the form $t,P_{s_{2k-2},t_{2k-2}},P_{s_{2k-4},t_{2k-4}},\dots,P_{s_{i_{0}},t_{i_{0}}},s_{i_{0}+1}$ where $P_{s_{i},t_{i}}$ is an $s_{i}\rightarrow t_{i}$ path in $D[V_{i}]$. The set $T\stackrel{{\scriptstyle\text{def}}}{{=}}\{t\}\cup\bigcup\{V_{i}:i_{0}\leq i\in I\}$ is also a $ts$-cut and all the tails of its outgoing edges are in $\{t_{i_{0}},t_{i_{0}+1}\}$. $P_{t,s}$ has already used the edge $(t_{i_{0}},s_{i_{0}+1})$ so it may not use another edge with tail $t_{i_{0}}$ hence $P_{t,s}$ leave $T$ using an edge with tail $t_{i_{0}+1}$. But then $P_{t,s}$ contains an $s_{i_{0}+1}\rightarrow t_{i_{0}+1}$ subpath $P_{s_{i_{0}+1},t_{i_{0}+1}}$ in $D[V_{i_{0}+1}]$.

$S\stackrel{{\scriptstyle\text{def}}}{{=}}\{s\}\cup\bigcup\{V_{i}:i_{0}+1\geq i\in I\}$ is an $st$-cut and all the tails of its outgoing edges are in $\{s_{i_{0}},s_{i_{0}+1}\}$. Therefore $P_{s,t}$ has an initial segment in $D[S]$ that terminates in this set. We know that $P_{s,t}$ does not use the edge $(t_{i_{0}},s_{i_{0}+1})$ because $P_{t,s}$ has already used it. Therefore there is an $m\in\{i_{0},i_{0}+1\}$ such that $P_{s,t}$ has a $t_{m}\rightarrow s_{m}$ subpath $P_{t_{m},s_{m}}$ in $D[V_{m}]$. But then the paths $P_{t_{m},s_{m}}$ and $P_{s_{m},t_{m}}$ are proper subpaths of $P_{s,t}$ and $P_{t,s}$ respectively. By Proposition 5 $f_{m}$ is an isomorphism between $D$ and $D[V_{m}]$ and thus the inverse-images of the paths $P_{t_{m},s_{m}}$ and $P_{s_{m},t_{m}}$ are edge-disjoint back and forth paths between $s$ and $t$ with strictly less sum of lengths than the added length of paths $P_{s,t}$ and $P_{t,s}$, which contradicts with the choice of $P_{s,t}$ and $P_{t,s}$.

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