Spanning eulerian subdigraphs avoiding k prescribed arcs in tournaments

TL;DR

This paper investigates the existence of spanning eulerian subdigraphs in tournaments that avoid a set number of prescribed arcs, providing new bounds on the arc-strongness needed for their existence.

Contribution

The authors establish a new upper bound on the arc-strongness function f(k), improving previous bounds and supporting the conjecture that f(k)=k+1 for all k.

Findings

Proved that f(k) ≤ ⌈(6k+1)/5⌉ for all k.

Improved the upper bound on the minimum arc-strongness for spanning eulerian subdigraphs.

Supported the conjecture that f(k)=k+1 for all k.

Abstract

A digraph is {\bf eulerian} if it is connected and every vertex has its in-degree equal to its out-degree. Having a spanning eulerian subdigraph is thus a weakening of having a hamiltonian cycle. A digraph is {\bf semicomplete} if it has no pair of non-adjacent vertices. A {\bf tournament} is a semicomplete digraph without directed cycles of length 2. Fraise and Thomassen \cite{fraisseGC3} proved that every $(k+1)$-strong tournament has a hamiltonian cycle which avoids any prescribed set of $k$ arcs. In \cite{bangsupereuler} the authors demonstrated that a number of results concerning vertex-connectivity and hamiltonian cycles in tournaments and have analogues when we replace vertex connectivity by arc-connectivity and hamiltonian cycles by spanning eulerian subdigraphs. They showed the existence of a smallest function $f(k)$ such that every $f(k)$-arc-strong semicomplete digraph has a spanning eulerian subdigraph which avoids any prescribed set of $k$ arcs. They proved that $f(k)\leq \frac{(k+1)^2}{4}+1$ and also proved that $f(k)=k+1$ when $k=2,3$. Based on this they conjectured that $f(k)=k+1$ for all $k\geq 0$. In this paper we prove that $f(k)\leq (\lceil\frac{6k+1}{5}\rceil)$.

Equations

$$|V(D^{\prime })|=0n_0+1n_1+2n_2=n_1+2n_2$$

$$2+2n_0+n_1\ge f(x,y,D)-f(x,y,D^{\prime })=\frac{24k+4}{5}-(2|V(D^{\prime })|-1)$$

$$2+2n_0+n_1\ge \frac{24k+4}{5}-(2(n_1+2n_2)-1)$$

$$4k=4(n_0+n_1+n_2)\ge 2n_0+3n_1+4n_2\ge \frac{24k-1}{5}$$

$$d_{C_p}^{-}(x_1)\le d_{C_p}^{-}(x_2)\le d_{C_p}^{-}(x_3)\le \cdots \le d_{C_p}^{-}(x_r).$$

$$f(v)\le (s_M-s_{M-1})+(s_{M-1}-s_{M-2})+\cdots +(s_m-s_{m-1})+1=s_M-s_m+1\le s_M+1-s^{\ast }$$

$$\begin{array}[]{rcl}\vspace{0.1cm}{}\sum_{v\in V(L)}f(v)&=&|I_{1}|+|W_{I_{1}}|(s_{1}+1)+|W_{I_{2}}|(s_{2}-s_{1})+\cdots+|W_{I_{\theta}}|(s_{\theta}-s_{\theta-1})\\ &=&|I_{1}|+|W_{I_{\theta}}|(s_{\theta}+1)+(|W_{I_{\theta-1}}|-|W_{I_{\theta}}|)(s_{\theta-1}+1)+\\ &&(|W_{I_{\theta-2}}|-|W_{I_{\theta}-1}|)(s_{\theta-2}+1)+\cdots+(|W_{I_{1}}|-|W_{I_{2}}|)(s_{1}+1)\hskip 17.07182pt(*)\\ \end{array}$$

$$\begin{array}[]{rcl}\sum_{v\in V(L)}f(v)&\geq&|I_{1}|+(t_{\theta}+|I_{\theta}|)(s_{\theta}+1)+(s_{\theta-1}+|I_{\theta-1}|-|I_{\theta}|)(s_{\theta-1}+1)\\ {}\vspace{0.1cm}\hfil&&\hskip 14.22636pt+(s_{\theta-2}+|I_{\theta-2}|-|I_{\theta-1}|)(s_{\theta-2}+1)+\cdots+(s_{1}+|I_{1}|-|I_{2}|)(s_{1}+1)\\ &\geq&|I_{1}|+(t_{\theta}+1)(s_{\theta}+1)+(s_{\theta-1}+1)^{2}+(s_{\theta-2}+1)^{2}+\cdots+(s_{1}+1)^{2}\\ \end{array}$$

$$\sum _{v\in V(L)}f(v)\ge (t_{\theta }+1{)}^2+1+\sum _{i=1}^{\theta -1}((s_i+1{)}^2+1)$$

$$\sum _{v\in V(L)}f(v)\ge 5t_{\theta }+\sum _{i=1}^{\theta -1}5s_i=5q$$

Full text

Keywords: Arc-connectivity, Eulerian subdigraph, Tournament, Semicomplete digraph, avoiding prescribed arcs

1 Introduction

The terminology is consistent with [4]. A classical result on digraphs is Camion’s Theorem (it was originally formulated only for tournaments but easily extends to semicomplete digraphs).

Theorem 1** (Camion [6]).**

Every strongly connected semicomplete digraph has a hamiltonian cycle.

A digraph is connected if its underlying undirected graph is connected. A digraph $D$ is eulerian if it contains a spanning eulerian trail $W$ such that $A(W)=A(D)$, or, equivalently by Euler’s theorem, if $D$ is connected and $d^{+}(v)=d^{-}(v)$ for all $v\in V(D)$. Finally we say that $D=(V,A)$ is supereulerian if it has a spanning eulerian subdigraph $D^{\prime}=(V,A^{\prime})$. By Camion’s theorem, every strong semicomplete digraph is supereulerian. Bang-Jensen and Thomassé made the following conjecture in 2011 (see e.g. [2]) which may be seen as a generalization of Camion’s theorem.

Conjecture 2**.**

Every digraph $D$ with $\lambda{}(D)\geq\alpha{}(D)$ is supereulerian.

This conjecture is open even for digraphs of independence number 2 but has been verified for a couple of classes of digraphs ([1, 2].

An eulerian factor of a digraph $D=(V,A)$ is a spanning subdigraph $H=(V,A^{\prime})$ so that $d_{H}^{+}(v)=d_{H}^{-}(v)>0$ for all $v\in V$. By a component of the eulerian factor $H$ we mean a connected component of the digraph $H$. Bang-Jensen and Maddaloni proved the following result in support of Conjecture 2.

Theorem 3**.**

[2*]**

Every digraph $D$ with $\lambda{}(D)\geq\alpha{}(D)$ has an eulerian factor. Furthermore, such a factor can be found in polynomial time.*

There are many results on hamiltonian cycles in tournaments and semicomplete digraphs (see e.g. [5, 8]). One of these is the following, due to Fraisse and Thomassen. A digraph $D$ is $\mathbf{k}$-strong for some natural number $k\geq 1$ if it has at least $k+1$ vertices and remains strongly connected whenever we delete at most $k-1$ vertices.

Theorem 4**.**

[7*]**

Let $T=(V,A)$ be a $(k+1)$-strong tournament and let $\hat{A}\subset A$ have size $k$. Then $T\setminus\hat{A}$ has a hamiltonian cycle.*

This is best possible in terms of the connectivity since there are infinitely many $(k+1)$-strong tournaments that have a vertex of out-degree $k+1$.

The authors of [3] showed that a number of results concerning hamiltonian paths and cycles in tournaments and semicomplete digraphs and vertex-connectivity have analogues in results on spanning eulerian subdigraphs in semicomplete digraphs where we can replace vertex-connectivity by arc-connectivity. A digraph $D$ is $\mathbf{k}$-arc-strong for some natural number $k\geq 1$ if it is strongly connected and remains so when we delete an arbitrary set of at most $k-1$ of its arcs.

It was shown in [3] that there exists a smallest function $f(k)$ such that every $f(k)$-arc-strong semicomplete digraph has a spanning eulerian subdigraph avoiding any set of $k$ prescribed arcs. They proved that $f(k)\leq\frac{(k+1)^{2}}{4}+1$ and that $f(2)=3,f(3)=4$. Note that, if true, Conjecture 2 would imply that $f(k)\leq k+2\sqrt{k}$ since deleting $k$ edges from a complete graph cannot generate an independent set of size more than $2\sqrt{k}$ (the exact bound is not important here).

Lemma 5**.**

[3*]**

Let $D=(V,A)$ be a $(k+1)$-arc-strong semicomplete digraph and let $\hat{A}$ be a set of $k$ arcs from $D$. Then $D\setminus\hat{A}$ has an eulerian factor.*

Based on these observations the authors of [3] posed the following conjecture that can be seen as such an arc-analogue of Theorem 4.

Conjecture 6**.**

Let $D=(V,A)$ be a $(k+1)$-arc-strong semicomplete digraph and let $A^{\prime}\subset A$ be any set of $k$ arcs of $D$. Then $D\setminus A^{\prime}$ has is supereulerian.

In this paper we prove the following theorem thus providing additional support for Conjecture 2.

Theorem 7**.**

Let $D=(V,A)$ be a semicomplete digraph with $\lambda{}(D)\geq\frac{6k+1}{5}$ and let $A\subset A$ be an arbitrary set of $k$ arcs of $D$. Then $D\setminus A^{\prime}$ is supereulerian.

We shall use the following theorem due to Meyniel.

Theorem 8**.**

[9*]**

Let $G$ be a strongly connected digraph on $n\geq 2$ vertices. If $d^{+}(x)+d^{-}(x)+d^{+}(y)+d^{-}(y)\geq 2|V(G)|-1$ for all pairs of non-adjacent vertices in $G$, then $G$ has a Hamilton cycle.*

2 Proof of Theorem 7

Proof.

Let $D$ and $A^{\prime}$ be defined as in the theorem and let $D^{\prime}=D-A$. Note that $D^{\prime}$ is $\frac{k+1}{5}$-arc-strong and let $q=\frac{k+1}{5}$. By Lemma 5 we note that $D^{\prime}$ contains an eulerian factor, $F$. Assume that $F=C_{1}\cup C_{2}\cup\cdots\cup C_{p}$, where $C_{i}$ denotes the components in $F$ and that $p$ is minimum possible. We may assume that $p\geq 2$ as otherwise we are done. Any vertex $u\in V(D^{\prime})$ which is adjacent to all vertices in $V(D^{\prime})\setminus\{u\}$ is called universal. We now prove the following claim.

Claim 1: There is a universal vertex, $u$, in $D^{\prime}$.

Proof of Claim 1: For the sake of contradiction, assume that there is no universal vertex in $D^{\prime}$. By Theorem 8 there exists two non-adjacent vertices $x$ and $y$ in $D^{\prime}$ such that $d_{D^{\prime}}^{+}(x)+d_{D^{\prime}}^{-}(x)+d_{D^{\prime}}^{+}(y)+d_{D^{\prime}}^{-}(y)<2|V(D^{\prime})|-1$, as otherwise there is a Hamilton cycle in $D^{\prime}$ and we are done.

Let $A^{\prime}=\{a_{1},a_{2},\ldots,a_{k}\}$. Start with $D$ and remove $a_{1}$, then remove $a_{2}$, then $a_{3}$ and continue this process until we obtain $D^{\prime}$. When removing $e_{i}$ above let $r_{i}$ denote how many more vertices become non-universal (that is, they previously had no non-neighbour, but after the removal of $e_{i}$ they get a non-neighbour) for $i\in[k]$. Note that $r_{i}\in\{0,1,2\}$ as the only vertices that can become non-universal are the endpoints of $e_{i}$. Let $n_{j}$ be the number of $r_{i}$’s of value j for $j\in\{0,1,2\}$.

As there is no universal vertex in $D^{\prime}$ the following is the number of non-universal vertices in $D^{\prime}$.

[TABLE]

Let $f(x,y,D)=d_{D}^{+}(x)+d_{D}^{-}(x)+d_{D}^{+}(y)+d_{D}^{-}(y)$ and $f(x,y,D^{\prime})=d_{D^{\prime}}^{+}(x)+d_{D^{\prime}}^{-}(x)+d_{D^{\prime}}^{+}(y)+d_{D^{\prime}}^{-}(y)$. Note that $f(x,y,D^{\prime})<2|V(D^{\prime})|-1$ and $f(x,y,D)\geq 4\frac{6k+1}{5}$ (as $D$ is $\frac{6k+1}{5}$-arc-strong, implying that every vertex has at least $\frac{6k+1}{5}$ arcs into it and out of it). Therefore, by removing the arcs of $A^{\prime}$ the $f$-value needs to drop by at least $\frac{24k+4}{5}-(2|V(D^{\prime})|-1)$.

Without loss of generality we may assume that $a_{1}$ is an arc between $x$ and $y$, and if there is two arcs between $x$ and $y$ then $a_{2}$ is the second such arc. Note that after removing the arcs between $x$ and $y$ all other arcs, $a_{i}$, touching $\{x,y\}$ will have $r_{i}\in\{0,1\}$.

If there are two arcs between $x$ and $y$ we note that the $r_{1}=0$ and $r_{2}=2$ and the value of $f(x,y,D)$ drops by at most $3+n_{0}+n_{1}\leq 2+2n_{0}+n_{1}$ (as after removing the first two arcs it has dropped by $4$ and we have one arc with $r$-value [math]).

If there is only one arc between $x$ and $y$ we note that $r_{1}=2$ and the value of $f(x,y,D)$ drops by at most $2+n_{0}+n_{1}$ (as after removing the first arc it has dropped by $2$ and we have one arc with $r$-value $2$). Therefore the following holds.

[TABLE]

We have previously shown that $|V(D^{\prime})|=n_{1}+2n_{2}$, which by the above implies the following.

[TABLE]

Simplifying the above gives us the following, as $n_{0}+n_{1}+n_{2}=k$.

[TABLE]

However the above is a contradiction, as $20k\geq 24k-1$ is impossible when $k\geq 1$.  ($\Box$)

By Claim 1, we will assume that $F$ is chosen such that $C_{1}$ contains the universal vertex, $u$, given by Claim 1 and that $|V(C_{p})|$ is as large as possible. That is, we want $|V(C_{p})|$ to be as large as possible, without containing all universal vertices. We call a vertex in $V\setminus V(C_{p})$ $C_{p}$-out-universal if it dominates all vertices in $C_{p}$ and analogously we call it $C_{p}$-in-universal if it is dominated by all vertices in $C_{p}$. We call a vertex $u\in V(L)$ $C_{p}$-almost-out-universal if it dominates all vertices in $C_{p}$ except at most one vertex. We now have the following observation.

Observation 2: Any vertex $w\in V(D^{\prime})\setminus V(C_{p})$ with no non-neighbour in $C_{p}$ (in $D^{\prime}$) is either $C_{p}$-out-universal or $C_{p}$-in-universal.

We may assume without loss of generality that $u$ (defined in Claim 1) is $C_{p}$-out-universal.

Proof of Observation 2: Let $w\in V(D^{\prime})\setminus V(C_{p})$ have no non-neighbour in $C_{p}$ (in $D^{\prime}$), but for the sake of contradiction assume that $w$ is not $C_{p}$-out-universal and not $C_{p}$-in-universal. As $w$ is not $C_{p}$-out-universal there exists a $x\in V(C_{p})$, such that $xw\in A(D^{\prime})$. Starting in $x$ and moving along an eulerian trail in $C_{p}$ we continue until we come to a vertex $v$ which does not dominate $w$ (which exists since $w$ is not $C_{p}$-in-universal). So there exists an arc $x^{\prime}v\in A(C_{p})$ such that $x^{\prime}w\in A(D^{\prime})$ but $vw\not\in A(D^{\prime})$. However as $w$ is universal we must then have that $wv\in A(D^{\prime})$. Now removing the arc $x^{\prime}v$ and adding the arcs $x^{\prime}w$ and $wv$ gives us an eulerian factor in $D^{\prime}$ with fewer components, a contradiction. This completes the first part of the proof of Observation 2.

Without loss of generality we may assume that $u$ is $C_{p}$-out-universal, as if it is is $C_{p}$-in-universal we can reverse all arcs.  ($\Box$)

Let $L_{i}$ be an inbranching in $C_{i}$ containing only $F$-arcs for all $i\in[p-1]$, such that the root of $L_{1}$ is $u$ (defined in Claim 1 and mentioned in Observation 2). Such an inbranching exists as $C_{i}$ is strongly connected. Let $L=L_{1}\cup L_{2}\cup\cdots\cup L_{p-1}$. Note that $V(L)$ and $V(C_{p})$ partition $V(D^{\prime})$. Let $I\subseteq V(L)$ be a set of vertices such that every vertex in $I$ is dominated by at least one vertex in $C_{p}$ (that is, $I\subseteq N^{+}(C_{p})\cap V(L)$). We now define an $I^{*}$-path as follows.

Definition ($I^{*}$-path): an $I^{*}$-path is a sequence of vertices $p_{0}p_{1}p_{2}\ldots p_{l}$, such that the following holds.

• •

$p_{i}\in V(L)$ for all $0\leq i\leq l$ and $p_{l}\in I$.

• •

$p_{i}p_{i+1}$ is an $(F-L)$-arc (i.e. $p_{i}p_{i+1}\in A(F)\setminus A(L)$) or $p_{i+1}p_{i}$ is a non-$F$-arc (i.e $p_{i+1}p_{i}\in A(D^{\prime})\setminus A(F)$) for all $i\in[l-1]$.

We will now prove the following claims.

Claim 3: If $I\subseteq V(L)\cap N^{+}(C_{p})$ and $P=p_{0}p_{1}p_{2}\ldots p_{l}$ is an $I^{*}$-path, then the following holds.

(i):

There is no $w\in V(C_{p})$ such that $p_{0}w,wp_{l}\in A(D^{\prime})$.

(ii):

There is no arc $ww^{+}\in A(C_{p})$ such that $p_{0}w^{+},wp_{l}\in A(D^{\prime})$.

(iii):

If $vp_{0}\in A(F)$, then there is no $w\in V(C_{p})$ such that $vw,wp_{l}\in A(D^{\prime})$.

(iv):

If $vp_{0}\in A(F)$, then there is no $ww^{+}\in A(C_{p})$ such that $vw^{+},wp_{l}\in A(D^{\prime})$.

Proof of Claim 3: Let $I\subseteq V(L)\cap N^{+}(C_{p})$ and let $P=p_{0}p_{1}p_{2}\ldots p_{l}$ be an $I^{*}$-path. For the sake of contradiction assume that one of the following holds.

(i):

There is a $w\in V(C_{p})$ such that $p_{0}w,wp_{l}\in A(D^{\prime})$. In this case let $C^{*}=p_{0}p_{1}\ldots p_{l}wp_{0}$.

(ii):

There is an arc $ww^{+}\in A(C_{p})$ such that $p_{0}w^{+},wp_{l}\in A(D^{\prime})$. In this case let $C^{*}=p_{0}p_{1}\ldots p_{l}ww^{+}p_{0}$.

(iii):

There is a $vp_{0}\in A(F)$ and a $w\in V(C_{p})$ such that $vw,wp_{l}\in A(D^{\prime})$. In this case let $C^{*}=p_{0}p_{1}\ldots p_{l}wvp_{0}$.

(iv):

There is a $vp_{0}\in A(F)$ and a $ww^{+}\in A(C_{p})$ such that $vw^{+},wp_{l}\in A(D^{\prime})$. In this case let $C^{*}=p_{0}p_{1}\ldots p_{l}ww^{+}vp_{0}$.

Denote $C*$ as we defined above, we note that $v_{i}=p_{i}$ when $i\leq l$ and $c_{l^{\prime}}=c_{0}$. Also note that by the construction either $c_{i}c_{i+1}\in A(F)$ or $c_{i+1}c_{i}\in A(D^{\prime})\setminus A(F)$. We now construct a new eulerian factor $F^{\prime}$ as follows. Initially let $F^{\prime}=F$ and for all $i\in\{0,1,2,\ldots,l^{\prime}-1\}$, if $c_{i+1}c_{i}\in A(D^{\prime})\setminus A(F)$ then add $c_{i+1}c_{i}$ to $F^{\prime}$ and if this is not the case then we must have $c_{i}c_{i+1}\in A(F)$ and in this case remove $c_{i}c_{i+1}$ from $F^{\prime}$. By construction we note that $d_{F^{\prime}}^{+}(x)=d_{F^{\prime}}^{-}(x)$ for all $x\in V(D^{\prime})$. Furthermore $F^{\prime}$ contains all arcs from $C_{p}$ except possibly one, so all vertices in $V(C_{p})$ belong to the same factor in $F^{\prime}$.

In case (i) and (ii) above all arcs in $A(L)$ also belong to $A(F^{\prime})$, so if two vertices belonged to the same factor in $F$ then they also belong to the same factor in $F^{\prime}$. As we furthermore have added the arc $wp_{l}$ to $F^{\prime}$ we have merged two factors of $F$, implying that there are fewer factors in $F^{\prime}$ than in $F$, contradicting the minimality of $p$. This completes the proof for case (i) and (ii).

Now assume case (iii) or (iv) holds above. Now all arcs of $L$, except the arc $vp_{0}$, belong to $A(F^{\prime})$. Assume that $v\in C_{j}$. When we remove the arc $vp_{0}$ from $L_{j}$ we obtain two connected components $R_{v}$ and $R_{p_{0}}$, where $v\in V(R_{v})$ and $p_{0}\in V(R_{p_{0}})$. Furthermore note that the root of $C_{j}$ belongs to $V(R_{p_{0}})$. As we merge the vertices of $R_{v}$ with $C_{p}$ (due to the arc $vw$ or $vw^{+}$) $F^{\prime}$ also contains at most $p$ factors. Moreover if $u$ is merged with $C_{p}$ then $j=1$ and $V(R_{p_{0}})$ has also been merged with $C_{p}$ contradicting the minimality of $p$. So, $u$ has not been merged with $C_{p}$, implying that the factor of $F^{\prime}$ containing the vertices of $V(C_{p})\cup V(R_{v})$ has more vertices than $C_{p}$ and does not contain all universal vertices, a contradiction to the maximality of $|V(C_{p})|$ in $F$.  ($\Box$)

Claim 4: There is no $C_{p}$-in-universal vertex in $V(L)$. This also implies that every vertex in $N^{+}(C_{p})\cap V(L)$ has a non-neighbour in $C_{p}$.

This is in fact true even if $D$ is only $(k+1)$-arc-strong (instead of $\frac{6k+1}{5}$-arc-strong).

Proof of Claim 4: Assume for the sake of contradiction that there is a $C_{p}$-in-universal vertex in $V(L)$. Let $I$ be all $C_{p}$-in-universal vertices in $V(L)$ and let $X\subseteq V(L)$ contain all vertices $p_{0}$ such that there exists an $I^{*}$-path starting at $p_{0}$.

For the sake of contradiction assume that there is an arc $xy\in A(D^{\prime})\setminus A(F)$ out of $X$ (that is, $x\in X$ and $y\not\in X$). If $y\in V(C_{p})$ then we get a contradiction to Claim 3 (i) (where $w=y$ and the end of the $I^{*}$-path is $C_{p}$-in-universal implying that $y$ dominates it). So $y\in A(L)\setminus X$. However in this case there is an $I^{*}$-path starting in $y$ (starting with $yx\ldots$), contradicting the fact that $y\not\in X$. Therefore there is no arc in $V(D^{\prime})\setminus A(F)$ leaving $X$.

This implies that all arcs of $D^{\prime}$ leaving $X$ belong to $F$. Assume there are $b$ such arcs. As $D^{\prime}$ is strong we note that $b\geq 1$. As $F$ is an eulerian factor there are also $b$ arcs in $A(F)$ entering $X$. Let the $F$-arcs entering $X$ be $y_{1}x_{1},y_{2}x_{2},\ldots,y_{b}x_{b}$. Note that $y_{1},y_{2},\ldots,y_{b}$ are distinct (as if $y_{i}=y_{j}$ then either $y_{i}x_{i}$ or $y_{j}x_{j}$ would not belong to $L$ and $y_{i}$ would belong to $X$, a contradiction). By the construction of $I$ we note that no $y_{i}$ is $C_{p}$-in-universal, implying that there is a vertex $w_{i}\in V(C_{p})$ which does not dominate $y_{i}$. By Claim 3 (iii) we note that $y_{i}$ and $w_{i}$ are non-adjacent in $D^{\prime}$, implying that there are $b$ pairs of non-neighbours between $\{y_{1},y_{2},\ldots,y_{b}\}$ and $V(C_{p})$, with $b\leq k$.

As there are only $b$ arcs in $D^{\prime}$ leaving $X$, we note that there must be $k+1-b$ pairs of non-neighbours between $X$ and $V(D^{\prime})\setminus X$ (as $D$ is $(k+1)$-arc-strong). However this implies that there are at least $k+1-b+b=k+1$ arcs in $A^{\prime}$, a contradiction. This proves the first part of the claim.

Let $w\in N^{+}(C_{p})\cap V(L)$ be arbitrary. If $w$ has no non-neighbour in $C_{p}$ then by Observation 2 it is either $C_{p}$-in-universal or $C_{p}$-out-universal. As $w\in N^{+}(C_{p})\cap V(L)$ it is not $C_{p}$-out-universal and by the above it is not $C_{p}$-in-universal. Therefore $w$ has a non-neighbour in $C_{p}$.  ($\Box$)

Recall that $q=\frac{k+1}{5}$.

Claim 5: Let $I\subseteq N^{+}(C_{p})\cap V(L)$ such that every vertex in $I$ is dominated by at least $i^{*}$ vertices of $C_{p}$. Furthermore assume that there are $q-a$ arcs from $C_{p}$ to $V(L)\setminus I$, where $a\geq 0$.

Then there exists a set $W_{I}\subseteq V(L)\setminus I$, such that $|W_{I}|\geq a+|I|$ and every vertex $w\in W_{I}$ has at least $i^{*}+1$ vertices in $V(C_{p})$ which are not dominated by $w$.

Proof of Claim 5: Let $I$, $i^{*}$ and $a$ be defined as in the statement of the Claim. Let $X$ be the set of vertices that are the starting point of an $I^{*}$-path and let $Y=V(L)\setminus X$. There is no non-$F$-arc (that is arc in $A(D^{\prime})\setminus A(F)$) that goes from $X$ to $Y$ by the definition of $X$ and $I^{*}$-paths. By Claim 4 we note that every vertex in $I$ has a non-neighbour in $C_{p}$ implying that there are at most $k-|I|$ non-arcs between $Y$ and $C_{p}$. Let $z$ denote the number of $F$-arcs that enter $Y$ and note that the total number of arcs entering $Y$ in $D$ is at most $z+(q-a)+(k-|I|)$. As this number has to be at least $k+q$ we get that $z\geq a+|I|$. Therefore (as there are $z$ $F$-arcs entering $Y$) we observe that there are $z$ $F$-arcs leaving $Y$, say $\{y_{1}x_{1},y_{2}x_{2},\ldots,y_{z}x_{z}\}$, where $z\geq a+|I|$. By the definition of $X$, these arcs are $L$-arcs, and as that the $L_{i}$ are in-branchings, we note that $y_{1},y_{2},\ldots,y_{z}$ are distinct vertices and that $W_{I}=\{y_{1},y_{2},\ldots,y_{z}\}$ therefore has size at least $a+|I|$.

Let $w\in W_{I}$ be arbitrary and let $p_{0}p_{1}\ldots p_{l}$ be an $I^{*}$-path where $wp_{0}=y_{i}x_{i}$ for some $i$. As $p_{l}\in I$ we note that $N^{-}(p_{l})\cap V(C_{p})$ has size at least $i^{*}$ and as $p_{l}$ is not $C_{p}$-in-universal (by Claim 4) there is an arc $vv^{\prime}\in A(C_{p})$ entering $N^{-}(p_{l})\cap V(C_{p})$. By Claim 3 (iii) and (iv) we note that there are no arcs from $w$ to $(N^{-}(p_{l})\cap V(C_{p}))\cup\{v\}$, completing the proof of Claim 5.  ($\Box$)

Recall that $q=\frac{k+1}{5}$.

Claim 6: There are at least $5q$ non-arcs between $V(L)$ and $V(C_{q})$.

This contradicts $|A^{\prime}|\leq k$ and thereby completes the proof of the theorem.

Proof of Claim 6: Let $\{x_{1},x_{2},\ldots,x_{r}\}=N^{+}(C_{p})\cap V(L)$, such that the following holds in $D^{\prime}$.

[TABLE]

As $D^{\prime}$ is $q$-arc-strong, we note that there are at least $q$ arcs into $V(L)$ in $D^{\prime}$. Therefore $\sum_{i=1}^{r}d_{C_{p}}^{-}(x_{i})\geq q$. Define $\theta$ such that it is as large as possible and $\sum_{i=1}^{\theta-1}d_{C_{p}}^{-}(x_{i})<q$. Note that $\theta\leq r$. Let $I_{j}=\{x_{j},x_{j+1},\ldots,x_{r}\}$ for $j\in[\theta]$. Let $s_{i}=d_{C_{p}}^{-}(x_{i})$ for all $i\in[\theta]$ and let $t_{\theta}=q-\sum_{i=1}^{\theta-1}s_{i}$. Note that $t_{\theta}+\sum_{i=1}^{\theta-1}s_{i}=q$ and $\sum_{i=1}^{\theta}s_{i}=q+(s_{\theta}-t_{\theta})$, where $t_{\theta}\leq s_{\theta}$.

We now define a function $f$ as follows. Initially let $f(v)=0$ for all vertices $v$ in $V(L)$. We now modify $f$ as follows.

• •

For all $v\in W_{I_{1}}$ ($I_{1}$ is defined above and $W_{I}$ is defined in Claim 5), we let $f(v)=s_{1}+1$ and for all $v\in I_{1}$ let $f(v)=1$.

• •

For all $v\in W_{I_{j}}$, for $j=2,3,\ldots,\theta$, increase $f(v)$ by $s_{j}-s_{j-1}$.

We will now show the following two subclaims, where Subclaim 6.2 will complete the proof of Claim 6.

Subclaim 6.1: Every $v\in V(L)$ has at least $f(v)$ non-neighbours in $C_{p}$.

Proof of Subclaim 6.1: Let $v\in V(L)$ be arbitrary. Let $W^{*}=\cup_{i=1}^{\theta}W_{I_{i}}$. First consider the case when $v\not\in W^{*}$. If $v\in I_{1}$, then $f(v)=1$ and by Claim 4 there is a non-neighbour of $v$ in $C_{p}$ and therefore Subclaim 6.1 holds in this case. So we may assume that $v\not\in I_{1}$, in which case $f(v)=0$ and again Subclaim 6.1 holds. We may therefore assume that $v\in W^{*}$.

Let $m$ be the minimum value such that $v\in W_{I_{m}}$ and let $M$ be the maximum value such that $v\in W_{I_{M}}$. First assume that there are no arcs from $C_{p}$ to $v$. In this case, by Claim 5, $v$ has at least $s_{M}+1$ non-neighbours in $C_{p}$. As $f(v)\leq(s_{M}-s_{M-1})+(s_{M-1}-s_{M-2})+\cdots(s_{2}-s_{1})+s_{1}+1=s_{M}+1$, we are done in this case. So we may assume that there are $s^{*}$ arcs from $C_{p}$ to $v$, where $s^{*}\geq 1$. As $v\in W_{I_{m}}$ we note that $v\not\in I_{m}$ which implies that $s^{*}\leq s_{m}$. Therefore the following holds.

[TABLE]

By Claim 5 we note that there are at least $s_{M}+1$ vertices in $C_{p}$ which are not dominated by $v$, which (by the definition of $s^{*}$) implies that $v$ has at least $s_{M}+1-s^{*}$ non-neighbours in $C_{p}$. This completes the proof of Subclaim 6.1.  ($\Box$)

Subclaim 6.2: There are at least $5q$ non-arcs between $V(L)$ and $V(C_{q})$

Proof of Subclaim 6.2: By the definition of $f$ we note that the following holds.

[TABLE]

By Claim 5 we note that $|W_{I_{j}}|\geq(q-\sum_{i=1}^{j-1}s_{i})+|I_{j}|$ for all $j\in[\theta]$. As $s_{1}\leq s_{2}\leq\cdots\leq s_{\theta}$ we note that (*) above is minimum when $|W_{I_{j}}|=(q-\sum_{i=1}^{j-1}s_{i})+|I_{j}|$ for all $j\in[\theta]$, which implies that the following holds (as $\sum_{i=1}^{\theta-1}s_{i}=q-t_{\theta}$).

[TABLE]

As $s_{\theta}\geq t_{\theta}$ and $|I_{1}|\geq\theta$ the above implies the following.

[TABLE]

Note that for every positive integer $x$ we have $0\leq(x-1)(x-2)=(x+1)^{2}+1-5x$, which implies that $5x\leq(x+1)^{2}+1$. Using the fact that $t_{\theta}+\sum_{i=1}^{\theta-1}s_{i}=q$ we now obtain the following.

[TABLE]

By Subclaim 6.1, we now note that Subclaim 6.2 holds.  ($\Box$)

∎