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\title{Disjoint Cycles of Different Lengths \\ in Graphs and Digraphs\thanks{This paper is dedicated to the memory of Nicolas Lichiardopol.}}

\author{Julien Bensmail\thanks{Supported by ERC Advanced Grant GRACOL, project no. 320812.}\\
\small Universit\'e C\^ote d'Azur, CNRS, Inria, I3S, France\\
\small\tt julien.bensmail@unice.fr\\
\and
Ararat Harutyunyan\thanks{Supported by an FQRNT postdoctoral research grant and CIMI research fellowship. }\\
\small LAMSADE, CNRS \\ [-0.8ex] 
\small Universit\'e Paris-Dauphine, PSL Research University \\ [-0.8ex] 
\small 75016 Paris, France\\
\small\tt ararat.harutyunyan@lamsade.dauphine.fr\\
\and
Ngoc Khang Le\\
\small Laboratoire d'Informatique du Parall\'elisme \\ [-0.8ex] 
\small \'Ecole Normale Sup\'erieure de Lyon \\ [-0.8ex] 
\small 69364 Lyon Cedex 07, France \\
\small\tt ngoc-khang.le@ens-lyon.fr\\
\and
Binlong Li\thanks{Supported by NSFC (11601429, 11671320) and the Natural Science Foundation of Shaanxi Province (2016JQ1002, 2014JK1353).}\\
\small Department of Applied Mathematics \\ [-0.8ex] 
\small Northwestern Polytechnical University \\ [-0.8ex] 
\small Xi'an, Shaanxi 710072, P.R. China \\
\small\tt libinlong@mail.nwpu.edu.cn\\
\and
Nicolas Lichiardopol\\
\small Lyc\'{e}e A. de Craponne \\ [-0.8ex] 
\small Salon, France
}

\date{\dateline{XXX}{XXX}\\
\small Mathematics Subject Classifications: 05C07, 05C20, 05C38}


\begin{document}

\maketitle

\begin{abstract}
In this paper, we study the question of finding a set of $k$ vertex-disjoint cycles (resp. directed cycles) of distinct lengths
in a given graph (resp. digraph). In the context of undirected graphs, we prove that, for every $k \geq 1$, every graph with minimum degree at least $\frac{k^2+5k-2}{2}$
has~$k$ vertex-disjoint cycles of different lengths, where the degree bound is best possible.
We also consider other cases such as when the graph is triangle-free, or 
the $k$ cycles are required to have different lengths modulo some value $r$.
In the context of directed graphs, we consider a conjecture of Lichiardopol concerning the least minimum out-degree
required for a digraph to have $k$ vertex-disjoint directed cycles of different lengths.
We verify this conjecture for tournaments, and, by using the probabilistic method,
for some regular digraphs and digraphs of small order.

\bigskip\noindent \textbf{Keywords:} vertex-disjoint cycles; different lengths; minimum degree
\end{abstract}


\section{Introduction} \label{section:introduction}

In this paper, we study degree conditions guaranteeing the existence in a graph (resp. digraph)
of a certain number of vertex-disjoint cycles (resp. directed cycles) whose lengths satisfy particular properties.
More precisely, not only we want cycles (resp. directed cycles) being vertex-disjoint, but we also require their lengths to be different somehow.
Namely, we first ask for the lengths to be different only, but then also request additional properties on the lengths such as having 
the same remainder to some modulo.

After giving some preliminary results in Section~\ref{section:prelim},
we start in Section~\ref{section:undirected} by studying how the number of vertex-disjoint cycles of different lengths in an undirected 
graph behaves in relation to the minimum degree of that graph. More precisely, we consider, given some $k \geq 1$, the minimum degree 
required for a graph to have at least~$k$ vertex-disjoint cycles of different lengths (and sometimes additional length properties). 
We show that this value is precisely $\frac{k^2+5k-2}{2}$ for every $k$ (Theorem~\ref{ThMinimumDegree}). 
Several more constrained situations are then considered, \textit{e.g.} when the graph is triangle-free or the vertex-disjoint cycles are required
to be more than just of different lengths; in these situations as well, we exhibit bounds (most of which are tight) on the least minimum degree required
to guarantee the existence of the $k$ desired vertex-disjoint cycles.
We also conjecture that for every $D \geq 3$, every graph $G$ of large enough order
verifying $k+1 \leq \delta(G) \leq \Delta(G) \leq D$ has $k$ vertex-disjoint cycles
of different lengths (see Conjecture~\ref{ConjMaximumDegree}).
To support this conjecture, we prove it for $k=2$ (Theorem~\ref{ThConjMaximumDegree}).
This in particular yields that every cubic graph of order more than $14$ has two vertex-disjoint cycles
of different lengths, which is tight (see Theorem~\ref{ThCubicGraph}).

We then consider, in Section~\ref{section:directed}, the same question for digraphs: What minimum out-degree is required for a 
digraph to have at least~$k$ vertex-disjoint directed cycles of different lengths? 
The existence of such a minimum out-degree was conjectured by Lichiardopol in~\cite{Lic14}, who verified it for $k=2$.
We here give further support to Lichiardopol's Conjecture by showing it to hold for tournaments (see Corollary~\ref{corollary:tournaments}),
and, using the probabilistic method, for some regular digraphs (Theorem~\ref{theorem:regular}) and digraphs of small order (Theorem~\ref{theorem:order}).


\section{Preliminaries} \label{section:prelim}

Let $G$ be a graph and $X$ a subset of $V(G)$. We use $G[X]$ to
denote the subgraph of $G$ induced by $X$, and $G-X$ to denote the
subgraph of $G$ induced by $V(G)\backslash X$. For two disjoint
subsets $X,Y$ of $V(G)$, we denote by $(X,Y)_G$ the bipartite subgraph of
$G$ with all edges between $X$ and $Y$. For a subgraph $H$ of $G$,
we set $G-H=G-V(H)$.

\medskip

We now recall results, some of which are folklore,
that guarantee the existence of cycles under particular circumstances.
These results will be used to obtain our main results.
First, by considering a longest path of a graph, the following result is easily observed.

\begin{proposition}\label{ThCDL}
For every $k\geq 1$, every graph of minimum degree at least $k+1$ contains
$k$ cycles of different lengths.
\end{proposition}

Now, by considering a maximum cut of a graph, the following is also clear.

\begin{proposition}\label{ThBSMD}
Let $G$ be a graph of minimum degree at least $2k-1$, where $k\geq
1$. Then $V(G)$ can be partitioned into sets $S$ and $T$ such that
the bipartite subgraph $(S,T)_G$ has minimum degree at least $k$.
\end{proposition}

An immediate consequence is:

\begin{theorem}\label{ThECDL}
For every $k\geq 1$, every graph of minimum degree at least $2k+1$
contains $k$ even cycles of different lengths.
\end{theorem}

\begin{proof}
By Proposition \ref{ThBSMD}, we can partition $V(G)$ into sets $S$ and
$T$ such that the bipartite subgraph $G'=(S,T)_G$ has minimum degree at least $k+1$. By Proposition \ref{ThCDL}, graph $G'$
contains $k$ cycles of different lengths. Since $G'$ is bipartite, all these cycles have even length.
\end{proof}

Proposition \ref{ThBSMD} shows the existence, in any graph, 
of a cut such that every vertex has `many' neighbors in the different partite
set. In a different flavour, the following three theorems concern cuts of graphs
such that every vertex has `many' neighbors in the same partite set.

\begin{theorem}[Stiebitz \cite{Stiebitz}]\label{ThSt}
If $s$ and $t$ are non-negative integers, and $G$ is a graph with
minimum degree at least $s+t+1$, then the vertex set of $G$ can be
partitioned into two sets which induce subgraphs of minimum degree
at least $s$ and $t$, respectively.
\end{theorem}

\begin{theorem}[Kaneko \cite{Kaneko}]\label{ThKa}

If $s$ and $t$ are non-negative integers, and $G$ is a triangle-free graph with
minimum degree at least $s+t$, then the vertex set of $G$ can be
partitioned into two sets which induce subgraphs of minimum degree
at least $s$ and $t$, respectively.
\end{theorem}

\begin{theorem}[Diwan \cite{Diwan}]\label{ThDi}
If $s, t \geq 2$ are two integers, and $G$ is a graph with
minimum degree at least $s+t-1$ and girth at least~$5$, then the vertex set of $G$ can be
partitioned into two sets which induce subgraphs of minimum degree
at least $s$ and $t$, respectively.
\end{theorem}

Diwan proved in \cite{Diwan'} the following result:

\begin{theorem}[Diwan \cite{Diwan'}]\label{ThDi'}
For every $r\geq 2$, and for any natural number $m$, every graph of
minimum degree at least $2r-1$ contains a cycle of length $2m$
modulo $r$.
\end{theorem}

Theorem~\ref{ThDi'} yields that when $r$ is odd,
every graph of minimum degree at least $2r-1$ contains a cycle of
length $m$ modulo $r$,
for any natural number $m$.

\begin{corollary}\label{CoThDi'}
For every $k\geq 1$ and $r\geq 2$, every graph $G$ with
$$\delta(G)\geq\left\{\begin{array}{ll}
2(k+1)r-1,  & \mbox{if } k \mbox{ is even and } r \mbox{ is odd};\\
2kr-1,      & \mbox{otherwise},
\end{array}\right.$$
has $k$ cycles of different lengths all divisible by $r$.
\end{corollary}

\begin{proof}
If $r$ is even, then $\delta(G)\geq 2kr-1$. By Theorem \ref{ThDi'},
there are $k$ cycles $C_0,C_1,\ldots,C_{k-1}$ such that for each
$i$, cycle $C_i$ has length $ir$ (mod $kr$). Clearly the $k$ cycles have
different lengths all divisible by $r$.

If both $k$ and $r$ are odd, then $\delta(G)\geq 2kr-1$. By Theorem
\ref{ThDi'}, there exist $k$ cycles $C_0,C_1,\ldots,C_{k-1}$ such
that for each $i$, cycle $C_i$ has length $2ir$ (mod $kr$). Since $kr$ is
odd, the $k$ cycles have different lengths all divisible by $r$.

If $k$ is even and $r$ is odd, then $\delta(G)\geq 2(k+1)r-1$. Note
that $k+1$ is odd. By the analysis above, graph $G$ contains $k+1$ cycles
of different lengths all divisible by $r$.
\end{proof}

Let us also mention the following result due to Dirac,
about cycles in $2$-connected graphs with given minimum degree.

\begin{theorem}[Dirac~\cite{Dir52}] \label{theorem:dirac-degree}
Every $2$-connected graph with order~$n$ and minimum degree at least~$\delta$ contains a cycle of length at least $\min \{2\delta,n\}$.
\end{theorem}

When it comes to oriented graphs,
a result of Camion implies that every strongly-connected tournament is Hamiltonian.

\begin{theorem}[Camion~\cite{Cam59}] \label{theorem:camion}
Every strongly-connected tournament contains a Hamiltonian cycle.
\end{theorem}

Still in the oriented context,
the following result of Moon asserts that every strongly-connected tournament is pancyclic,
\textit{i.e.} has directed cycles of every possible length.

\begin{theorem}[Moon~\cite{Moon68}]\label{theorem:moon}
Every strongly-connected tournament with order~$n$ has a directed cycle of length~$i$,
for every $i=3,...,n$.
\end{theorem}


\section{Disjoint cycles of different lengths in undirected graphs} \label{section:undirected}

Throughout this section, we exhibit conditions, notably in terms of minimum degree,
implying the existence of a certain number of vertex-disjoint cycles of different lengths in a graph.


\subsection{Distinct lengths and relation to girth}

The central question of this work is the following:
What is the smallest minimum degree $f(k)$ required
so that a graph with minimum degree $f(k)$ has at least $k$ vertex-disjoint
cycles of different lengths?
We answer this question in the following result.

\begin{theorem}\label{ThMinimumDegree}
{\bf a)} For every $k \geq 1$, there exists a minimum integer
$f(k)$ such that every graph of minimum degree at least $f(k)$
contains $k$ vertex-disjoint cycles of different lengths. \newline
{\bf b)} We have $f(k)=\dfrac{k^2+5k-2}{2}$.
\end{theorem}

\begin{proof}
{\bf a)} We proceed by induction on $k$. Clearly $f(1)$ exists (and
we have $f(1)=2$). Suppose that the assertion is true for $k-1$ (where $k\geq 2$), 
and let us study it for $k$. Let $G$ be an arbitrary graph
of minimum degree at least $f(k-1)+k+2$. By Theorem~\ref{ThSt},
there exists a partition $(V_1,V_2)$ of $V(G)$ such that $G\lbrack
V_1\rbrack$ is of minimum degree at least $f(k-1)$ and $G\lbrack
V_2\rbrack$ is of minimum degree at least $k+1$. Then $G\lbrack
V_1\rbrack$ contains $k-1$ disjoint cycles $C_1,\ldots, C_{k-1}$ of
different lengths. By Proposition~\ref{ThCDL}, subgraph $G\lbrack V_2\rbrack$
contains $k$ cycles of different lengths. Then one of these cycles
has length distinct from those of the cycles $C_1,\ldots, C_{k-1}$.
We get then a collection of $k$ vertex-disjoint cycles of different
lengths. So the assertion is true for $k$, and the result is
proved. Furthermore, we have $f(k)\leq f(k-1)+k+2$.

\medskip

{\bf b)} We have $f(i)\leq f(i-1)+i+2$ for $2\leq i\leq k$. By
addition and simplification we get $f(k)\leq f(1)+4+\cdots + k+2$,
hence $f(k)\leq 2+\dfrac{(k+2)(k+3)}{2}-6$, which yields
\setcounter{equation}{0}
\begin{align}
f(k)\leq \dfrac{k^2+5k-2}{2}.
\end{align}
On the other hand since the complete graph on $f(k)+1$ vertices is
of minimum degree $f(k)$, it contains $k$ vertex-disjoint cycles of
different lengths. It follows that $f(k)+1\geq 3+\cdots+(k+2)$, hence
$f(k)+1\geq\dfrac{(k+2)(k+3)}{2}-3$, which yields
\begin{align}
f(k)\geq \dfrac{k^2+5k-2}{2}.
\end{align}
From Inequalities (1) and (2), we get $f(k)=\dfrac{k^2+5k-2}{2}$.
\end{proof}

One may naturally wonder how the girth parameter acts on the task of finding vertex-disjoint cycles of distinct lengths.
Towards this question, we show that, for triangle-free graphs, the function $f(k)$ from Theorem~\ref{ThMinimumDegree}
can be improved to the function $g(k)$ below, which is again best possible.

\begin{theorem}\label{ThTriangleFree}
{\bf a)} For every $k \geq 1$, there exists a minimum integer
$g(k)$ such that every triangle-free graph of minimum degree at
least $g(k)$ contains $k$ vertex-disjoint cycles of different
lengths.
\newline {\bf b)} We have $g(k)=\dfrac{k^2+3k}{2}$.
\end{theorem}

\begin{proof}
Using Theorem~\ref{ThKa} in place of Theorem~\ref{ThSt},
mimicking the proof of Theorem~\ref{ThMinimumDegree} results in a 
straightforward proof of the first part of the statement.
To see that the degree bound is indeed best possible,
consider the complete bipartite graph
$K_{g(k),g(k)}$, which is triangle-free and has minimum degree $g(k)$,
and has $k$ vertex-disjoint cycles of distinct lengths.
\end{proof}

Quite similarly as in the proofs of Theorems~\ref{ThMinimumDegree} and~\ref{ThTriangleFree},
using Theorem~\ref{ThDi} we can refine the function $f(k)$ for graphs with girth at least~$5$:

\begin{theorem}\label{ThGirth5}
{\bf a)} For every $k \geq 1$ and $g \geq 5$, there exists a minimum integer
$g_g(k)$ such that every graph of girth $g$ and minimum degree at
least $g_g(k)$ contains $k$ vertex-disjoint cycles of different
lengths.
\newline {\bf b)} We have $g_g(k) \leq \dfrac{k^2+k+2}{2}$.
\end{theorem}

\noindent We have no clue, however, whether the degree bound is tight here.


\subsection{Distinct lengths modulo $r$}

We now investigate the impact on the function $f(k)$ in Theorem~\ref{ThMinimumDegree}
if we additionally require the vertex-disjoint cycles to be of specific lengths modulo some integer $r$.
We start by determining the best degree bound $h(k)$ for even cycles (\textit{i.e.} $r=2$).

\begin{theorem}\label{ThEvenCycle}
{\bf a)} For every $k \geq 1$, there exists a minimum integer
$h(k)$ such that every graph of minimum degree at least $h(k)$
contains $k$ vertex-disjoint even cycles of different lengths.
\newline {\bf b)} We have $h(k)=k^2+3k-1$.
\end{theorem}

\begin{proof}
{\bf a)} We proceed by induction on $k$. Clearly $h(1)$ exists (and,
by Theorem \ref{ThECDL}, we have $h(1)=3$). Suppose that the
assertion is true for $k-1$, for some $k\geq 2$, and let us study it for $k$.
Let $G$ be an arbitrary graph of minimum degree at least
$h(k-1)+2k+2$. By Theorem \ref{ThSt}, there exists a partition
$(V_1,V_2)$ of $V(G)$ such that $G\lbrack V_1\rbrack$ is of minimum
degree at least $h(k-1)$ and $G\lbrack V_2\rbrack$ is of minimum
degree at least $2k+1$. Then $G\lbrack V_1\rbrack$ contains $k-1$
vertex-disjoint even cycles $C_1,\ldots, C_{k-1}$ being of different
lengths. According to Theorem~\ref{ThECDL}, subgraph $G\lbrack V_2\rbrack$ contains $k$
even cycles of different lengths. Then one of these cycles has
length distinct from those of the cycles $C_1,\ldots, C_{k-1}$. We
get then a collection of $k$ vertex-disjoint even cycles of
different lengths. So, the assertion is true for $k$, and the result
is proved. We also deduce $h(k)\leq h(k-1)+2k+2$.

\medskip

{\bf b)} We have $h(i)\leq h(i-1)+2i+2$ for $2 \leq i \leq k$. By addition
and simplification, we get $h(k)\leq h(1)+2(3+\cdots + k+1)$, hence
$h(k)\leq 3+2\left(\dfrac{(k+1)(k+2)}{2}-3\right)$, which yields
\begin{align}
h(k)\leq k^2+3k-1.
\end{align}
On the other hand since the complete graph on $h(k)+1$ vertices is
of minimum degree $h(k)$, it contains $k$ vertex-disjoint even
cycles of different lengths. It follows that $h(k)+1\geq 2\cdot
2+\cdots+2(k+1)$, hence $h(k)+1\geq
2\left(\dfrac{(k+1)(k+2)}{2}-1\right)$, which yields
\begin{align}
h(k)\geq k^2+3k-1.
\end{align}
We thus deduce $h(k)=k^2+3k-1$ from Inequalities~(3) and~(4).
\end{proof}

We extend Theorems \ref{ThMinimumDegree} and \ref{ThEvenCycle} to cycles having lengths multiple to a same value $r$.
Recall that the exact value of $f_2(k)$ is given in Theorem~\ref{ThEvenCycle}.

\begin{theorem}\label{ThLengthDivisible}
{\bf a)} Let $r\geq 3$ be an integer. For every $k \geq 1$, there
exists a minimum integer $f_r(k)$ such that every graph of minimum
degree at least $f_r(k)$ contains $k$ vertex-disjoint cycles of
different lengths all divisible by $r$. \newline {\bf b)} We have
$f_r(k)\leq\left\{\begin{array}{ll}
k(k+1)r-1,      & \mbox{if } r \mbox{ is even};\\
2(k^2+k-1)r-1,  & \mbox{if both } k \mbox{ and } r \mbox{ are odd};\\
2k^2r-1,      & \mbox{if } k \mbox{ is even and } r \mbox{ is
odd},
\end{array}\right.$ \\
${\rm ~ ~  ~  ~ ~~~~~~\textit{and}}~f_r(k)\geq\left\{\begin{array}{ll}
\dfrac{k(k+1)r}{2}-1,      & \mbox{if } r \mbox{ is even};\\
\dfrac{k(k+1)r}{2},      & \mbox{if } r \mbox{ is odd}.
\end{array}\right.$
\end{theorem}

\begin{proof}
{\bf a)} We proceed by induction on $k$. Clearly $f_r(1)$ exists
(and, by Theorem~\ref{ThDi'}, we have $f_r(1)\leq 2r-1$). Suppose
that the assertion is true for $k-1$, where $k\geq 2$, and let us study
it for $k$. Let $G$ be an arbitrary graph of minimum degree at least
$$f_r(k-1)+\left\{\begin{array}{ll}
  2(k+1)r,  & \mbox{if } k \mbox{ is even and } r \mbox{ is odd};\\
  2kr       & \mbox{otherwise}.
\end{array}\right.$$
According to Theorem~\ref{ThSt}, there exists a partition $(V_1,V_2)$ of
$V(G)$ such that $G\lbrack V_1\rbrack$ is of minimum degree at least
$f_r(k-1)$ and $G\lbrack V_2\rbrack$ is of minimum degree at least
$2(k+1)r-1$ for $k$ even and $r$ odd, and of minimum degree at least
$2kr-1$ otherwise. Then $G\lbrack V_1\rbrack$ contains $k-1$
disjoint cycles $C_1,\ldots, C_{k-1}$ of different lengths all
divisible by $r$. By Corollary~\ref{CoThDi'}, subgraph $G\lbrack V_2\rbrack$
contains $k$ cycles of different lengths all divisible by $r$. Then
one of these cycles has length distinct from those of the cycles
$C_1,\ldots, C_{k-1}$. We get then a collection of $k$ disjoint
cycles of different lengths all divisible by $r$. So the assertion
is true for $k$, and the existence is proved.

\medskip

{\bf b)} If $r$ is even, then we have
\begin{align*}
f_r(k)  & \leq f_r(k-1)+2kr\\
        & \leq f_r(k-2)+2(k-1)r+2kr\\
        & \leq f_r(1)+\sum_{i=2}^k2ir\\
        & =k(k+1)r-1.
\end{align*}

If both $k$ and $r$ are odd, then we have
\begin{align*}
f_r(k)  & \leq f_r(k-1)+2kr\\
        & \leq f_r(k-2)+2kr+2kr\\
        & \leq f_r(k-3)+2(k-2)r+2kr+2kr\\
        & \leq f_r(1)+\sum_{i=1}^{(k-1)/2}4(2i+1)r\\
        & =2(k^2+k-1)r-1.
\end{align*}

If $k$ is even and $r$ is odd, then
\begin{align*}
f_r(k)  &\leq f_r(k-1)+2(k+1)r\\
        &\leq 2((k-1)^2 + k -2)r - 1+2(k+1)r\\
        &=2k^2r-1.
\end{align*}
So the first assertion of \textbf{b)} is proved.

We now focus on the second assertion of \textbf{b)}. For even $r\geq 4$
and $k\geq 1$, the complete graph $K_{f_r(k)+1}$ contains $k$
vertex-disjoint cycles of different lengths, all divisible by $r$.
It follows that $f_r(k)+1\geq r+2r+\cdots +kr$, which yields $f_r(k)\geq
\dfrac{k(k+1)r}{2}-1$. For odd $r\geq 3$ and $k\geq 1$, the complete
bipartite graph $K_{f_r(k),f_r(k)}$ (of order $2f_r(k)$) contains
$k$ disjoint cycles of different lengths, all divisible by $r$. It
follows that $2f_r(k)\geq 2r+4r+\cdots +2kr$, which yields $f_r(k)\geq
\dfrac{k(k+1)r}{2}$. This concludes the proof.
\end{proof}

We conclude this section by considering the opposite direction,
namely cycles having distinct lengths modulo some value $r$.

\begin{theorem}
{\bf a)} Let $r\geq 3$ be an odd integer. There exists a minimum
integer $\phi(r)$ such that every graph of minimum degree at least
$\phi(r)$ contains $r$ vertex-disjoint cycles $C_0,\ldots, C_{r-1}$ with
$v(C_i)\equiv i$ (mod $r$) for $0\leq i\leq r-1$.
\newline {\bf b)} We have $\dfrac{r^2+5r-2}{2}\leq\phi(r)\leq 2r^2-1$.
 \end{theorem}

\begin{proof}
\textbf{a)} Let $G$ be a graph with minimum degree at least
$2r^2-1$. By repeated applications of Theorem~\ref{ThSt}, we get a
partition $(V_0,V_1,\ldots,V_{r-1})$ of $V(G)$ such that for every
$i$, subgraph $G[V_i]$ has minimum degree at least $2r-1$. By Theorem~\ref{ThDi'}, 
subgraph $G[V_i]$ has a cycle $C_i$ of length $i$ (mod $r$).
Thus the collection of $r$ cycles $C_0,C_1,\ldots,C_{r-1}$ is as
required. So $\phi(r)$ exists, and $\phi(r)\leq
2r^2-1$.

\medskip

\textbf{b)} The complete graph $K_{\phi(r)+1}$ contains $r$ disjoint
cycles with the required conditions. It follows that $\phi(r)+1\geq
3+\cdots + r+2$, hence $\phi(r)+1\geq \dfrac{(r+2)(r+3)}{2}-3$,
which implies $\dfrac{r^2+5r-2}{2}\leq\phi(r)$, as claimed.
\end{proof}


\subsection{Large graphs}

Recall that Theorem~\ref{ThMinimumDegree} implies that every graph of 
minimum degree at least $f(k)=\dfrac{k^2+5k-2}{2}$ has $k$
vertex-disjoint cycles of different lengths. Furthermore, the complete graph
$K_{f(k)+1}$ shows that the bound on the minimum degree is best
possible. However, if we allow finite exceptions, then this bound
can be improved.
That is, the degree bound can be reduced further for graphs with sufficiently large order.

\medskip

In the proof of the main result of this section, Theorem~\ref{ThLargeOrder},
we will make use of the following notion.
Consider an integer $k\geq 2$ and a graph $G$ of minimum degree
$\delta(G) \geq k+1$. By a \textit{$k$-path-vertex schema} of $G$ we mean a pair
$\mathcal{S}=(P,x)$ consisting of a path $P$ and a vertex $x$ not in
$P$ having exactly $k+1$ neighbors in $P$. Using a longest path argument, 
it can be observed that every graph contains a path-vertex schema. Observe
also that the subgraph induced by $V(P)\cup\{x\}$ contains $k$
cycles of different length (all containing $x$). For convenience, we
sometimes consider $\mathcal{S}$ as a subgraph of $G$. So
$V(\mathcal{S})=V(P)\cup\{x\}$, and
$G-\mathcal{S}=G-(V(P)\cup\{x\})$. The \textit{cardinality} of $\mathcal{S}$
is $\vert V(\mathcal{S}) \rvert$. It is obvious that this cardinality is at
least $k+2$. An \textit{optimal} $k$-path-vertex schema of $G$ is a schema of
of $G$ of minimum cardinality.
Observe that in this case the extremities of $P$ are neighbors of $x$.

We will make use of the following result concerning optimal path-vertex schemas.

\begin{lemma}\label{LeOptimalSchema}
Let $G$ be a graph with $\delta(G) \geq k+1$, and $\mathcal{S}=(P,x)$ be an optimal $k$-path-vertex schema of $G$. 
Then every vertex $y$ of $G-\mathcal{S}$ has at most $k+2$ neighbors in
$V(\mathcal{S})$. Moreover, if there exists a vertex $y$ in
$G-\mathcal{S}$ having exactly $k+2$ neighbors in $V(\mathcal{S})$,
then $x$ is a neighbor of $y$, all the vertices of $P$ are neighbors
of $x$, and $|V(\mathcal{S})|=k+2$.
\end{lemma}

\begin{proof}
Suppose $P=(u_1,u_2,\ldots,u_p)$. If $y$ has at least $k+2$
neighbors in $V(P)$, then there exists $i$ (with $2\leq i\leq p$) such
that $y$ has exactly $k+1$ neighbors in $P'=(u_i,\ldots,u_{p})$.
Thus $P'$ and $y$ form a $k$-path-vertex schema of cardinality less
than $\mathcal{S}$, a contradiction. This implies that $y$ has at
most $k+2$ neighbors in $V(\mathcal{S})$.

Suppose that $y$ has exactly $k+2$ neighbors in $V(\mathcal{S})$. By
the analysis above, we can see that $y$ has exactly $k+1$ neighbors
on $P$ and $xy\in E(G)$. Thus $(P,y)$ is an optimal $k$-path-vertex
schema, and then $u_1$ and $u_p$ are neighbors of $y$. Suppose for
the sake of a contradiction that there exists a vertex of $P$ which
is not a neighbor of $x$. Clearly $u_1$ and $u_p$ are neighbors of
$x$. Let $i$ be the maximum index such that $xu_i\notin E(G)$. Thus
$2\leq i\leq p-1$. If $i\leq p-2$, then let
$P'=(u_{i+2},\ldots,u_p,y,u_1,\ldots,u_{i-1})$; otherwise, if $i=p-1$, let
$P'=(y,u_1,\ldots,u_{p-2})$. Then $P'$ and $x$ form a
$k$-path-vertex schema of cardinality less than $\mathcal{S}$, a
contradiction. So, all vertices of $P$ are neighbors of $x$, and
clearly we have $|V(\mathcal{S})|=k+2$.
\end{proof}

We are now ready to prove a refinement of Theorem~\ref{ThMinimumDegree} for sufficiently large graphs.

\begin{theorem}\label{ThLargeOrder}
{\bf a)} For every $k\geq 2$, every graph $G$ of order $n\geq
7\cdot\left\lfloor\dfrac{k^2}{4}\right\rfloor$ and $\delta(G)\geq
\dfrac{k^2+3k}{2}$ has $k$ vertex-disjoint cycles of different
lengths.\\
{\bf b)} The bound on $\delta(G)$ is best possible.
\end{theorem}

\begin{proof}
{\bf a)} We prove the claim by induction on~$k$, starting from the row $k=2$.
In other words, we prove that if $G$ is a graph of order $n\geq 7$ and $\delta(G) \geq 5$,
then $G$ contains two vertex-disjoint cycles of different lengths.

If $G$ is triangle-free, then we are done by Theorem
\ref{ThTriangleFree}. Suppose now that $G$ contains a triangle
$C=(x_1,x_2,x_3,x_1)$. We claim that there exists a vertex $x_4$ of $G-C$ having
three neighbors in $C$. Indeed, if it was not the case, then $G-C$ would be
of minimum degree at least $3$ and then it would contain two cycles
of different lengths. One of these cycles and $C$ would  then form a pair
of vertex-disjoint cycles of different lengths, and we would be
done. Clearly the subgraph $H$ induced by $x_1$, $x_2$, $x_3$ and
$x_4$ is complete. Suppose that there exists at most one vertex of
$G-H$ having four neighbors in $H$. Then the graph $G-H$ has at most
one vertex of degree at most $1$. Then $G-H$ contains a cycle (since
any longest path in $G-H$ has one of its end-vertices having at least two
neighbors on the path) and since $H$ contains a triangle and a
4-cycle, we get two vertex-disjoint cycles of different lengths and we are done. 
Suppose thus that there exist two vertices $x_5$
and $x_6$ both having $x_1$, $x_2$, $x_3$ and $x_4$ as neighbors.

Let $H'$ be the subgraph induced by $x_1,\ldots,x_6$. If every
vertex in $G-H'$ has at most three neighbors in $H'$, then
$\delta(G-H')\geq 2$ and $G-H'$ contains a cycle. Since $H\subset
H'$ contains a triangle and a 4-cycle, we get two vertex-disjoint
cycles of different lengths and we are done. So assume now that
there exists a vertex $x_7$ having at least four neighbors in $\lbrace
x_1,\ldots,x_6\rbrace$. Then $x_7$ has at least two neighbors in
$\lbrace x_1,\ldots,x_4\rbrace$, say $x_7x_1,x_7x_2\in E(G)$. Then
$(x_1,x_2,x_7,x_1)$ and $(x_3,x_5,x_4,x_6,x_3)$ are two
vertex-disjoint cycles of different lengths. This achieves the
proof of the case $k=2$.

\medskip

Suppose now that the assertion is true up to row
$k-1$, where $k\geq 3$, and let us study it for $k$. So $G$ is a graph of
order $n\geq 7\cdot\left\lfloor\dfrac{k^2}{4}\right\rfloor$ and of
minimum degree
$\delta(G)\geq\dfrac{k(k+3)}{2}=\dfrac{(k+1)(k+2)}{2}-1$.
By applying Theorem \ref{ThSt} repeatedly, we get a partition
$(V_1,V_2,\ldots,V_t)$ of $V(G)$ where $t=\left\lfloor\dfrac{k+1}{2}\right\rfloor$,
such that each $V_i$ induces a graph of minimum degree at least
$k+1$. Hence each $G[V_i]$ has a $k$-path-vertex schema. This
implies that $G$ has a $k$-path-vertex schema of order at most
$\dfrac{n}{\lfloor(k+1)/2\rfloor}$.

Suppose first that there exists an optimal $k$-path-vertex schema
$\mathcal{S}=(P,x)$ such that every vertex $y$ of $G-\mathcal{S}$
has at most $k+1$ neighbors in $\mathcal{S}$. Recall that
$|V(\mathcal{S})|\leq\dfrac{n}{\lfloor(k+1)/2\rfloor}$. Let
$G'=G-\mathcal{S}$. Then the order of $G'$ is
$$n'=n-|V(\mathcal{S})|\geq\dfrac{\lfloor(k-1)/2\rfloor}{\lfloor(k+1)/2\rfloor}n
\geq\left\{\begin{array}{ll} 7k(k-2)/4, & k \mbox{ is even};\\
7(k-1)^2/4, & k \mbox{ is odd}, \end{array}\right.$$ \textit{i.e.} $n'\geq
7\cdot\left\lfloor\dfrac{(k-1)^2}{4}\right\rfloor$. Note that every
vertex $y$ of $G'$ has at least
$$\dfrac{k^2+3k}{2}-k-1=\dfrac{(k-1)^2+3(k-1)}{2}$$ neighbors in $G'$,
so $\delta(G')\geq\dfrac{(k-1)^2+3(k-1)}{2}$. By the induction
hypothesis $G'$ contains $k-1$ vertex-disjoint cycles $C_1,\dots,
C_{k-1}$ of different lengths. Plus, the induced subgraph $G\lbrack
V(\mathcal{S})\rbrack$ contains $k$ cycles of different lengths. Then
one of these cycles and the cycles $C_1,\dots, C_{k-1}$ form a
collection of $k$ vertex-disjoint cycles of different lengths. So in
this case the assertion is proved for $k$.

Suppose now that there does not exist an optimal $k$-path-vertex
schema $\mathcal{S}$ such that every vertex $y$ of $G-\mathcal{S}$
has at most $k+1$ neighbors in $V(\mathcal{S})$. We take an
arbitrary optimal $k$-path-vertex schema $\mathcal{S}=(P,x)$, and
let $y\in V(G-\mathcal{S})$ be a vertex having $k+2$
neighbors in $\mathcal{S}$. By Lemma \ref{LeOptimalSchema}, vertices $x$ and
$y$ are adjacent, all the vertices of $P$ are neighbors of $x$ (and
also of $y$), and $|V(\mathcal{S})|=k+2$. We put
$P=(u_1,u_2,\ldots,u_{k+1})$ and $\varOmega=\lbrace
u_1,u_2\ldots,u_{k-1},x,y\rbrace$. Clearly $G\lbrack
\varOmega\rbrack$ contains $k-1$ cycles of different lengths (ranging from
$3$ to $k+1$).

Let $G'=G-\varOmega$. Note that $G'$ has minimum degree
at least $\dfrac{(k-1)^2+3(k-1)}{2}$ and has order at least
$7\cdot\left\lfloor\dfrac{(k-1)^2}{4}\right\rfloor$. By the induction
hypothesis $G'$ contains $k-1$ disjoint cycles
$C_1,C_2,\ldots,C_{k-1}$ of different lengths (where we label the $C_i$'s so that their lengths are non-decreasing).
Since $G\lbrack \varOmega\rbrack$ contains $k-1$ cycles of all
lengths from $3$ to $k+1$, we have $\lvert
V(C_i)\rvert=i+2$ for $1\leq i\leq k-1$. In particular, cycle $C_{k-1}$ is
of order $k+1$ and $C_{k-2}$ is of order $k$.

Let
$G''=G-\left(\varOmega\cup\bigcup\limits_{i=1}^{k-1}V(C_i)\right)$,
and let $z$ be a vertex of $G''$. We claim that $z$ has at most one
neighbor in $\varOmega$. Indeed, we would get
that $z$ and the vertices of $\varOmega$ form a cycle of order $k+2$;
so, together with $C_1, ..., C_{k-2}$ we would then get $k$
disjoint cycles of different lengths. We also claim that $z$ has at most
$\dfrac{k+1}{2}$ neighbors in $C_{k-1}$. Indeed, otherwise the
vertices of $C_{k-1}$ and $z$ would form a cycle $C'_{k-1}$ of order
at least $k+2$; so $C_1,\ldots,C_{k-2},C'_{k-1}$ and an
appropriate cycle of $G\lbrack \varOmega\rbrack$ of order $k+1$
would form a collection of $k$ disjoint cycles of different lengths.
It follows that $z$ has at most $1+3+\cdots+k+\dfrac{k+1}{2}=\dfrac
{k^2+2k-3}{2}$ neighbors in
$\varOmega\cup\bigcup\limits_{i=1}^{k-1}V(C_i)$, and $z$ has at
least $\dfrac{k+3}{2}\geq 3$ neighbors in $G''$. Let
$P'=(v_1,v_2,\ldots,v_s)$ be a longest path in $G''$. Since
$\delta(G'')\geq 3$, we have $s\geq 4$.

Note that all the neighbors of $v_1$ in $G''$ are in $P'$. It
follows that $v_1$ has at most $k$ neighbors in $G''$ (for otherwise
we would have a cycle of length at least $k+2$ and, together with the
cycles $C_1, \ldots C_{k-1}$, we would have the desired cycles). We have seen that
$v_1$ has at most one neighbor in $\varOmega$, and that $v_1$ has at
most $\dfrac{k+1}{2}$ neighbors in $C_{k-1}$. We claim that $v_1$
has more than $\dfrac{k}{2}$ neighbors in $C_{k-2}$. Indeed, suppose
the opposite. Then $v_1$ has at most $1+3+\cdots
+k-1+\dfrac{k}{2}+\dfrac{k+1}{2}=\dfrac{k^2+k-3}{2}$ neighbors in
$\varOmega\cup\bigcup\limits_{i=1}^{k-1}V(C_i)$. This means that $v_1$
has at most $\dfrac{k^2+k-4}{2}$ neighbors in
$\varOmega\cup\bigcup\limits_{i=1}^{k-1}V(C_i)$ and that $v_1$ has
at least $\dfrac{k^2+3k}{2}-\dfrac{k^2+k-4}{2}=k+2$ neighbors in
$G''$, a contradiction. So as we claimed, vertex $v_1$ has more than
$\dfrac{k}{2}$ neighbors in $C_{k-2}$, and similarly $v_s$ has more
than $\dfrac{k}{2}$ neighbors in $C_{k-2}$. But now it can be observed that the vertices of $C_{k-2}$ and the vertices of $P'$ form a
cycle of length at least $k+2$, and, together with the cycles $C_i$ 
(where $1\leq i\leq k-1$ and $i\neq k-2$) and an appropriate cycle of
length $k$ of $G\lbrack \Omega\rbrack$, we get $k$ disjoint cycles
of different lengths. This concludes this part of the proof.

\medskip

\textbf{b)} Let $G$ be a complete bipartite graph with two partite
sets of size $\dfrac{k(k+3)}{2}-1$ and $n-\dfrac{k(k+3)}{2}+1$,
respectively, where $n$ is large enough. Then
$\delta(G)=\dfrac{k(k+3)}{2}-1$. Note that $G$ does not have $k$
vertex-disjoint cycles of different lengths. So the bound
on $\delta(G)$ in our statement is sharp.
\end{proof}

While the bound on $\delta$ in Theorem~\ref{ThLargeOrder} is best possible,
we have no clue about how far from optimal the bound on $n$ is.
We believe it would be interesting to investigate this aspect further.


\subsection{Large graphs with bounded maximum degree}

We conclude this section by pointing out that if we restrict ourselves to large graphs with bounded maximum degree, then we may
further refine Theorem \ref{ThLargeOrder}. 
In particular, we believe the following conjecture should be true.

\begin{conjecture}\label{ConjMaximumDegree}
For every two integers $k$ and $D$, there is an integer $n_0$ such that every
graph $G$ of order at least $n_0$ with
$k+1\leq\delta(G)\leq\Delta(G)\leq D$ has $k$ vertex-disjoint
cycles of different lengths.
\end{conjecture}

\noindent We here make a first step towards Conjecture~\ref{ConjMaximumDegree} by proving it for $k=2$, see Theorem~\ref{ThConjMaximumDegree}.

\medskip

Before presenting the proof, we need to introduce a few preliminary results concerning graphs having only one type of cycles.
In the upcoming results, we use $n(G)$ to denote the order of some graph $G$. For any integer
$i$, we denote by $N_i(G)$ the set of vertices of $G$ with degree
$i$, and set $n_i(G)=|N_i(G)|$. 

\begin{lemma}\label{LeCycleTriangle}
Let $G$ be a graph with $\delta(G)\geq 2$. If all cycles of $G$ have length~$3$, then $n_2(G)\geq \dfrac{n(G)}{3}+2$.
\end{lemma}

\begin{proof}
We first claim that every block (maximal $2$-connected subgraph) of $G$ is either a $K_2$ or a
triangle. Let $B$ be an arbitrary block of $G$. If $B\neq K_2$, then
$B$ is 2-connected. If $B$ has at least four vertices, then, by
Theorem~\ref{theorem:dirac-degree},
it contains a cycle of order at least 4, which is
not in our assumption. So we conclude that $B$ has exactly three
vertices. Thus $B$ is a triangle.

Now we prove the lemma by induction on the number of blocks of $G$. If
$G$ has only one block, then $G$ is a triangle and the assertion is
trivial. If $G$ is disconnected, then we can complete the proof by
applying the induction hypothesis to each component of $G$. Now we
assume that $G$ is connected and separable.

Let $B$ be an end-block of $G$, and $x_0$ be the cut-vertex of $G$
contained in $B$. Recall that $B$ is a triangle. If
$d_{G-B}(x_0)\geq 2$, then let $G'=G-(B-x_0)$. By the induction
hypothesis, we have $n_2(G')\geq \dfrac{n(G')}{3}+2$. Note that
$n(G)=n(G')+2$ and $n_2(G)\geq n_2(G')+1$. We then have $$n_2(G)\geq
n_2(G')+1\geq\frac{n(G')}{3}+2+1=\frac{n(G)-2}{3}+3\geq\frac{n(G)}{3}+2.$$
Now we assume that $d_{G-B}(x_0)=1$. Let $P=(x_0,x_1,\cdots,x_t)$ be
a path of $G$ such that $d(x_i)=2$ for $1\leq i\leq t-1$ and
$d(x_t)\geq 3$. Let $G'=G-B-(P-x_t)$. By the induction hypothesis, we have
$n_2(G')\geq \dfrac{n(G')}{3}+2$. Note that $n(G)=n(G')+t+2$ and
$n_2(G)\geq n_2(G')+t$. Since $t\geq 1$, we now have $$n_2(G)\geq
n_2(G')+t\geq\frac{n(G')}{3}+2+t=\frac{n(G)-t-2}{3}+t+2\geq
\frac{n(G)}{3}+2,$$ which concludes the proof.
\end{proof}

\begin{lemma}\label{LeCycleQuadrangle}
Let $G$ be a graph with $\delta(G)\geq 2$. If all cycles of $G$ have length~$4$, then $n_2(G)\geq \dfrac{n(G)}{5}+2$.
\end{lemma}

\begin{proof}
We first claim that every block of $G$ is either a $K_2$ or a
complete bipartite graph $K_{2,s}$ with $s\geq 2$. Let $B$ be an
arbitrary block of $G$. If $B\neq K_2$, then $B$ is 2-connected.
Thus $B$ contains a cycle, which is a $C_4$ by our assumption. If
$B$ is not bipartite, then $B$ has an odd cycle, which is not a
$C_4$, a contradiction. This implies that $B$ is bipartite. Let
$X,Y$ be the two partite sets of $B$ and let
$C=(x_1,y_1,x_2,y_2,x_1)$ be a $C_4$ of $G$, where $x_1,x_2\in X$
and $y_1,y_2\in Y$. Suppose that $X$ has a third vertex $x_3$. Since
$B$ is 2-connected, there is a path $P$ with two end-vertices in $C$
such that $x_3\in V(P)$ and all internal vertices of $P$ are not in
$C$. If $P$ has length more than 2, then $B$ contains a cycle longer
than $C$, which is not a $C_4$, a contradiction. Thus $P$ has length
2 and $x_3y_1,x_3y_2\in E(B)$. If $Y$ has a third vertex $y_3$, then
by a similar analysis as above, we can see that $x_1y_3,x_2y_3\in
E(G)$. But in this case $C'=(x_1,y_3,x_2,y_2,x_3,y_1,x_1)$ is a
cycle being not a $C_4$, a contradiction. This implies that either
$|X|=2$ or $|Y|=2$. We suppose without loss of generality that
$|X|=2$. Since $B$ is 2-connected, vertices $x_1,x_2$ are adjacent to every
vertex in $Y$. Thus $B$ is a complete bipartite graph $K_{2,s}$ with
$s\geq 2$.

Now we prove the lemma by induction on the block number of $G$. If
$G$ has only one block, then $G=K_{2,s}$ with $s\geq 2$ and the
assertion is trivial. If $G$ is disconnected, then we can complete
the proof by applying the induction hypothesis to each component of
$G$. Now we assume that $G$ is connected and separable.

Let $B$ be an end-block of $G$, and $x_0$ be the cut-vertex of $G$
contained in $B$. Note that $B=K_{2,s}$ with $s\geq 2$. If
$d_{G-B}(x)\geq 2$, then let $G'=G-(B-x)$. By the induction hypothesis,
$n_2(G')\geq \dfrac{n(G')}{5}+2$. Note that $n(G)=n(G')+s+1$ and
$$n_2(G)\geq\left\{\begin{array}{ll}
n_2(G')+2,  & \mbox{if } s=2;\\
n_2(G')+s-2,    & \mbox{if } s\geq 3.
\end{array}\right.$$ One can compute that $n_2(G)\geq \dfrac{n(G)}{5}+2$. Now we
assume that $d_{G-B}(x)=1$. Let $P=(x_0,x_1,\cdots, x_t)$ be a path
of $G-(B-x_0)$ such that $d(x_i)=2$ for $1\leq i\leq t-1$ and
$d(x_t)\geq 3$. Let $G'=G-B-(P-x_t)$. By the induction hypothesis,
we have $n_2(G')\geq \dfrac{n(G')}{5}+2$. Note that $n(G)=n(G')+s+t+1$ and
$$n_2(G)\geq\left\{\begin{array}{ll}
n_2(G')+t+1,  & \mbox{if } s=2;\\
n_2(G')+s+t-3,    & \mbox{if } s\geq 3,
\end{array}\right.$$
where $t\geq 1$. This yields $n_2(G)\geq
\dfrac{n(G)}{5}+2$.
\end{proof}

We are now ready to prove Conjecture~\ref{ConjMaximumDegree} for $k=2$.

\begin{theorem}\label{ThConjMaximumDegree}
For every $D\geq 3$, every graph $G$ of order more than
$20D-46$ with $3 \leq \delta(G) \leq \Delta(G) \leq D$ has two vertex-disjoint cycles of different lengths.
\end{theorem}

\begin{proof}
Suppose that $G$ does not contain two
vertex-disjoint cycles of different lengths. We will show that
$n(G)$ is bounded above by $20D-46$. Let $g$ denote the girth of $G$. We
distinguish two cases.

\medskip

\noindent\textbf{Case 1.} $g=3$ or $g=4$.

Let $C$ be a shortest cycle of $G$. Then $C$ is either a $C_3$ or a
$C_4$. Let $H$ be an induced subgraph of $G$ defined by a sequence
of vertex sets $U_0,U_1,\cdots,U_s$ such that:

\begin{enumerate}
    \item $U_0=V(C)$ and $U_s=V(H)$;

    \item for every $i$, where $0\leq i\leq s-1$, there is a vertex $x_i\in
V(G)\backslash U_i$ such that $d_{U_i}(x_i)>\dfrac{d(x_i)}{2}$ and
$U_{i+1}=U_i\cup\{x_i\}$;

    \item for every vertex $x\in V(G)\backslash U_s$, we have $d_{U_s}(x)\leq
\dfrac{d(x)}{2}$.
\end{enumerate}

Note that $n(H)=n(C)+s$. We first show that $s$ is bounded above by a
constant (depending on $D$). Note that $e(U_0,V(G)\backslash
U_0)=\sum_{v\in U_0}(d(v)-2)$ is bounded above by $4D-8$. Observe that $e(U_{i+1},V(G)\backslash U_{i+1})\leq e(U_i,V(G)\backslash
U_i)-1$. This implies that $e(U_s,V(G)\backslash U_s)\leq 4D-8-s$
and $s\leq 4D-8$.

Let $F=G-H$. Recall that for every vertex $x\in V(F)$, we have
$d_F(x)\geq\left\lceil \dfrac{d(x)}{2}\right\rceil\geq 2$. Thus
$\delta(F)\geq 2$. If $F$ contains a cycle of length different from that of
$C$, then we are done. So we assume that either every cycle
of $F$ is a triangle, or every cycle of $F$ is a $C_4$. Note that
every vertex in $N_2(F)$ has a neighbor in $H$. Furthermore, we have $n_2(F)\leq
e(V(H),V(F))\leq 4D-8-s$. Now, applying Lemmas~\ref{LeCycleTriangle} and~\ref{LeCycleQuadrangle},
we get $$n(F)\leq 5(n_2(F)-2)\leq 5(4D-10-s),$$ and,
thus,
$$n(G)=n(H)+n(F)\leq 4+s+5(4D-10-s)\leq 20D-46.$$

\medskip

\noindent\textbf{Case 2.} $g\geq 5$.

Let $C=(x_1,\ldots,x_g,x_1)$ be a cycle of $G$ of order $g$. 
We may assume that any vertex of $V(G)\setminus V(C)$ has at most one
neighbor in $C$ (for otherwise we would have a cycle of length less
than $g$). So, the graph $G'=G-C$ is of minimum degree at least $2$
and since $G'$ cannot be of minimum degree at least $3$ (for otherwise $G'$
would contain two cycles of different lengths, and then one of these
cycles and $C$ would be disjoint and of different lengths), actually 
$G'$ is of minimum degree exactly $2$. Let $P=(y_1,\ldots, y_t)$ be a
longest path in $G'$. By maximality of $t$, all neighbors of $y_1$
in $G'$ are in $P$. If $y_1$ has three neighbors in $G'$, clearly we
are done. It follows that $y_1$ has exactly two neighbors in $P$ and
then, necessarily, the neighbor of $y_1$ in $P$ distinct from $y_2$
is $y_g$. So $C'=(y_1,\ldots, y_g, y_1)$ is a cycle of $G'$ of
length $g$. We claim that $d_{G'}(y_2) = 2$. Indeed suppose that
$y_2$ has a neighbor in $P$ distinct from $y_1$ and $y_3$.
Necessarily we have $t\geq g+1$ and the neighbor of $y_2$ is
$y_{g+1}$. But then $(y_1,y_g,y_{g+1},y_2,y_1)$ is a $4$-cycle,
which is impossible. Suppose now that $y_2$ has a neighbor $y$ in
$V(G')\setminus V(P)$. Then $(y, y_2,\ldots, y_t)$ is a longest path
in $G'$. By similar arguments as earlier, $y_g$ is a neighbor of $y$. But then
$(y_1,y_2,y,y_g,y_1)$ is a $4$-cycle, which is again impossible.
So, each of the vertices $y_1$ and $y_2$ has one neighbor in $C$.
When $g \geq 7$, it can be noted that $y_1$, $y_2$ and a
shortest path between their neighbors in $C$ form a cycle of length
less than $g$, which is impossible. So we are done when $g \geq 7$.

Assume now that $g=5$. Recall that each of the vertices $y_1$, $y_2$
has one neighbor in $C$ and clearly the neighbors of $y_1,y_2$ in
$C$ are distinct and not adjacent (for otherwise we would have a
cycle of length less than 5). So vertices $y_1$, $y_2$ and a path
between their neighbors in $C$ of length 3 form a cycle $C_1$ of
length $6$. Let us consider the graph $G_1=G-(V(C)\cup \lbrace y_1,
y_2\rbrace)$. Clearly every vertex of $G_1$ distinct from $y_3$ and
$y_5$ is of degree at least $2$ in $G_1$. If one of the vertices
$y_3$ and $y_5$ is of degree at least $2$ in $G_1$, then, by
considering a longest path in $G_1$, we deduce that $G_1$
contains a cycle, and then this cycle and one of the cycles $C$ and
$C_1$ would form two vertex-disjoint cycles of different lengths.
Suppose now that  $y_3$ and $y_5$ are both of degree $1$ in $G_1$.
Since $(y_4, y_3, y_2,y_1,y_5,\ldots, y_t)$ is a path of length $t$,
by the previous arguments $y_4$ is of degree $2$ in $G'$. Since
$n\geq 15$, we have $V(G)\backslash(V(C)\cup V(C'))\neq\emptyset$, 
and the graph $G_2=G-V(C)\cup V(C')$ is of minimum
degree at least $2$. It follows that  $G_2$ contains a cycle, and
then, as previously, we are done. This concludes the case $g=5$.

Suppose now that $g=6$. As $P'=(y_5,\ldots, y_1, y_6,\ldots, y_t)$
is also longest path of $G'$, by the previous reasoning, vertices $y_5$
and $y_4$ are of degree $2$ in $G'$. It follows that each of the
vertices $y_1$, $y_2$, and $y_4$ has one neighbor in $C$. Then the subgraph $G_1$ of $G$ induced by
$V(C)\cup\lbrace y_1, y_2, y_3, y_4 \rbrace$ contains two cycles of
different lengths. It is clear that $C''=(y_t, y_{t-1},\ldots,
y_{t-5}, y_t)$ is a $6$-cycle of $G$. If $t\geq 10$, clearly
$V(C'')$ is vertex-disjoint with $V(G_1)$
and we are done. It cannot be the case that $t=9$ or $t=7$ as
$y_4$ and $y_2$ are of degree $2$ in $G'$. Suppose now that $t=6$.
Since $n\geq 15$, we have $V(G)\backslash(V(C)\cup V(C'))\neq\emptyset$ and
we may assume that the graph $G-V(C)\cup V(P)$ does not contain
cycles (for otherwise we are done). So there is a vertex $z$ in
$V(G)\backslash(V(C)\cup V(C'))$ that has a neighbor $y_i$ (where
$i=3$ or $i=6$). But then the vertices of $C$ and $z$ would form a
path of order $7$, a contradiction to the maximality of $t$.

Suppose now that $t=8$. Clearly, vertices $y_7$ and $y_8$ are of
degree $2$ in $G'$. Since $n\geq 15$, subset $V(G)\setminus
(V(C)\cup V(P))$ is non-empty, and the graph $G-V(C)\cup V(P)$ is of
minimum degree at most $1$. Then there exists a vertex $z$ of
$G-V(C)\cup V(P)$ having a neighbor in $V(P)$. Since $y_1$, $y_2$,
$y_4$, $y_5$, $y_7$ and $y_8$ are of degree $2$ in $G'$, this
neighbor is either $y_3$ or $y_6$. Without loss of generality, we
may suppose that this neighbor is $y_3$. Suppose that $z$ has
another neighbor in $P$. Necessarily this other neighbor is $y_6$ --
but then $(z,y_3,y_4,y_5,y_6,z)$ would be a $5$-cycle, which is
impossible. So the only neighbor of $z$ in $P$ is $y_3$. Then $z$
has a neighbor $u$ in $G-V(C)\cup V(P)$. Suppose that $u$ has a
neighbor $v$ in $G-V(C)\cup V(P)$. Then $(v,u,z,y_3,\ldots,y_8)$
would be a path of $G'$ of length $9$, which by maximality of $t=8$
is impossible. It follows that $u$ has a neighbor in $P$, and
necessarily this neighbor is $y_6$ (because $G'$ does not contain
triangles). Note that $z$ and $u$ are of degree $2$ in
$G'$. So the eight vertices of $\varOmega=\lbrace
y_1,y_2,y_4,y_5,y_7,y_8, z,u\rbrace$ are of degree $2$ in $G'$ and
then each vertex of $\varOmega$ has exactly one neighbor in $C$
(which has order $6$). It follows that there exists a vertex $w$ of
$C$ having two neighbors $a$ and $b$ in $\varOmega$. 
Since it can be checked that $d_{G'}(a,b)\leq 3$, we eventually get a cycle of length at
most $5$, which is not possible. So the case $g=6$ is settled and
the result is proved for $g\geq 5$. This concludes the whole proof.
\end{proof}

Taking $D=3$ in Theorem \ref{ThConjMaximumDegree}, we get that every
cubic graph of order more than $20D-46=14$ has two vertex-disjoint
cycles of different lengths.

\begin{theorem}\label{ThCubicGraph}
Every cubic graph of order at least $15$ contains two
vertex-disjoint cycles of different lengths.
\end{theorem}

\begin{figure}[t]
\begin{center}
\begin{tikzpicture}
   \tikzstyle{VertexStyle}=[shape = circle, fill=black, minimum size = 4pt, inner sep=0pt]
   \SetVertexNoLabel	
   \tikzstyle{EdgeStyle}=[line width=1pt]
   \grHeawood[RA=2.5]
\end{tikzpicture}
\end{center}
\caption{The Heawood graph.}
\label{figure:heawood}
\end{figure}

\noindent We remark that the Heawood graph
($(3,6)$-cage, see Figure~\ref{figure:heawood}) of order 14 does not contain two vertex-disjoint
cycles of different lengths. Thus the bound on the
order in Theorem~\ref{ThCubicGraph} is sharp.


\section{Disjoint cycles of different lengths in digraphs} \label{section:directed}

The results from this section are motivated by the following conjecture raised by Bermond and Thomassen in 1981~\cite{BT81}.

\begin{conjecture}[Bermond and Thomassen~\cite{BT81}] \label{conjecture:bt}
For every $k \geq 2$, every digraph of minimum out-degree at least $2k-1$ contains at least $k$ vertex-disjoint directed cycles.
\end{conjecture}

\noindent If true, Conjecture~\ref{conjecture:bt} would be best possible because of the bidirected complete graph on $2k$ vertices.
Towards Conjecture~\ref{conjecture:bt}, one could more generally wonder whether for every $k$ there is a smallest finite function $f(k)$ such that every digraph of minimum out-degree at least $f(k)$ contains at least $k$ vertex-disjoint directed cycles (so $f(k)$ should be equal to $2k-1$ according to Conjecture~\ref{conjecture:bt}). Thomassen first proved in~\cite{Tho83} that $f(k)$ exists for every $k \geq 1$. Later, Alon improved the value of $f(k)$ to $64k$~\cite{Alo96}. For $k=2$, Thomassen proved in~\cite{Tho83} that $f(k)=3$, which confirms Conjecture~\ref{conjecture:bt} for $k=2$. Later on, Lichiardopol, P\'or and Sereni proved that for $k=3$ the best value for $f(k)$ is $5$, again confirming Conjecture~\ref{conjecture:bt} for $k=3$~\cite{LPS09}. This apart, Conjecture~\ref{conjecture:bt} is still open, though some more partial results may be found in literature (see \textit{e.g.} \cite{BLS09}).

Motivated by Conjecture~\ref{conjecture:bt} and in the same flavour as in Section~\ref{section:undirected}, one can wonder about the existence of a (smallest) finite function $g(k)$ such that every digraph with minimum out-degree at least~$g(k)$ contains $k$ (vertex-) disjoint (directed) cycles of distinct lengths. In this context, the following was conjectured by Lichiardopol~\cite{Lic14}:

\begin{conjecture}[Lichiardopol~\cite{Lic14}] \label{conjecture:digraph-lengths}
For every $k \geq 2$, there exists an integer $g(k)$ such that every digraph of minimum out-degree at least $g(k)$ contains $k$ vertex-disjoint directed cycles of distinct lengths.
\end{conjecture}

\noindent It is worth pointing out that a similar function $h(k)$ for the existence of $k$ disjoint cycles of the same length does not exist, 
as Alon proved that there exist digraphs of arbitrarily large minimum out-degree having no two (not necessarily disjoint) cycles of the same length~\cite{Alo96}.
Lichiardopol proved Conjecture~\ref{conjecture:digraph-lengths} for $k=2$ in~\cite{Lic14}, solving a question of Henning and Yeo~\cite{HY12}. We note that $g(k)$ should in general be a quadratic function of $k$; for an illustration of this statement, consider a complete bidirected digraph $D$ on $g(k)+1$ vertices. Since $\delta^+(D)=g(k)$, there exist $k$ disjoint cycles of different lengths in $D$. It follows that $$g(k)+1\geq 2+\cdots +k+1,$$ hence $g(k)+1\geq \frac{(k+1)(k+2)}{2}-1$, which yields $g(k)\geq \frac{k^2+3k-2}{2}$ in general.

\medskip

In this section, we verify Conjecture~\ref{conjecture:digraph-lengths} in several contexts. 
We first verify it for tournaments in Section~\ref{section:tournaments}.
Using the probabilistic method, we then verify it, in Sections~\ref{section:regular} and~\ref{section:small},
for some regular digraphs,
and digraphs where the order is a polynomial function of the minimum out-degree.
Some concluding remarks are gathered in Section~\ref{section:ccl-digraphs}.


\subsection{Tournaments} \label{section:tournaments}

We here verify Conjecture~\ref{conjecture:digraph-lengths} for tournaments. 
More precisely, for every $k \geq 1$, we study the smallest finite function $g_t(k)$ such that
every tournament of minimum out-degree at least $g_t(k)$ has $k$ vertex-disjoint directed cycles of different lengths.
We exhibit both an upper bound and a lower bound on $g_t(k)$ for every $k$.

In order to prove our upper bound, we need to introduce the following result first.

\begin{lemma} \label{lemma:tournaments}
Every tournament of minimum out-degree $\delta\geq 1$ contains a directed cycle of order at least $2\delta+1$. 
\end{lemma}

\begin{proof}
We proceed by induction on the order $n\geq 2\delta+1$ of a tournament $T$ of minimum out-degree $\delta$. We claim that the assertion is true for $n=2\delta+1$. Indeed, in this case $T$ is a regular tournament of degree $\delta$. So $T$ is strongly-connected, and then, by Theorem~\ref{theorem:camion}, we get that $T$ is Hamiltonian. So the vertices of $T$ form a directed cycle of order $2\delta+1$, and we are done. Suppose now that the assertion is true up to the row $n-1$, where $n\geq 2\delta+2$, and let us study it for $n$. So $T$ is a tournament of minimum out-degree $\delta$ and of order $n\geq 2\delta+2$. If $T$ is strongly-connected, then $T$ is Hamiltonian and again we are done. Suppose thus that $T$ is not strongly-connected. 
In that case, we consider a strongly-connected component $B$ of $T$ that is terminal (in the decomposition into strongly-connected components of $T$),
\textit{i.e.} with no arcs out-going to the rest of $T$.
Then $B$ has minimum out-degree at least~$\delta$.
So, by the induction hypothesis, we deduce that $B$ (and therefore $T$) contains a directed cycle of order at least $2\delta+1$. So the assertion is true for $n$, which concludes the proof.
\end{proof}

We are now ready to exhibit an upper bound on $g_t(k)$, and hence to confirm Conjecture~\ref{conjecture:bt} for tournaments.

\begin{theorem}  \label{theorem:tournament}
For every $k\geq 1$, we have $g_t(k) \leq \frac{k^2+4k-3}{2}$. 
\end{theorem}

\begin{proof}
We proceed by induction on $k$. The assertion is clearly true for $k=1$. So suppose that the assertion is true up to row $k-1$ (where $k\geq 2$), and let us study it for $k$. Let $T$ be a tournament of minimum out-degree at least $\frac{k^2+4k-3}{2}$. By the induction hypothesis $T$ contains $k-1$ disjoint cycles $C_1,\ldots, C_{k-1}$ of different lengths.
From Theorem~\ref{theorem:moon} (applied on the $C_i$'s),
we now deduce that $T$ has a collection $C_1', ..., C_{k-1}'$ of disjoint cycles with $\lvert V(C_i')\rvert =i+2$, and therefore of different lengths.
 
We have $$\lvert V(C_1')\cup\cdots\cup V(C_{k-1}')\rvert =3+\cdots + k+1 =\frac{k^2+3k-4}{2}<\frac{k^2+4k-3}{2}.$$ It thus follows that the tournament $T'=T-V(C_1')\cup\cdots\cup V(C_{k-1}')$ is of positive order and of minimum out-degree at least $\frac{k^2+4k-3}{2}-\frac{k^2+3k-4}{2}=\frac{k+1}{2}$. According to Lemma~\ref{lemma:tournaments}, we know that $T'$ contains a cycle $C'_k$ of length at least $k+2$ and, together with the directed cycles $C_1', ..., C_{k-1}'$, we get then $k$ disjoint cycles of different lengths. Consequently, the assertion is true for $k$, and the result is proved. 
\end{proof}

We now deduce a lower bound on $g_t(k)$.

\begin{observation} \label{observation:tournaments}
For every $k \geq 1$, we have $g_t(k)\geq \frac{k^2+5k-2}{4}$.
\end{observation}

\begin{proof}
Let $T$ be a regular tournament of degree $g_t(k)$. 
Then $T$ contains $k$ disjoint cycles of different lengths. 
Since the order of $T$ is $2g_t(k)+1$, it follows that $$2g_t(k)+1\geq 3+\cdots +k+2,$$ 
hence $2g_t(k)+1\geq \frac{(k+2)(k+3)}{2}-3$, which yields $g_t(k)\geq \frac{k^2+5k-2}{4}$.
\end{proof}

So, in the context of tournaments, according to Theorem~\ref{theorem:tournament} and Observation~\ref{observation:tournaments} we get the following.

\begin{corollary} \label{corollary:tournaments}
For every $k \geq 1$, we have $$\frac{k^2+5k-2}{4} \leq g_t(k) \leq \frac{k^2+4k-3}{2}.$$
\end{corollary}


\subsection{Regular digraphs} \label{section:regular}

We now use the probabilistic method to prove, in the current section and upcoming Section~\ref{section:small}, 
Conjecture~\ref{conjecture:digraph-lengths} in two new contexts.
To this aim, we first need to introduce a few tools and lemmas.
The first two are classic tools of the probabilistic method, namely 
Chernoff's Inequality and the Lov\'asz Local Lemma (see \textit{e.g.}~\cite{MR02}).

\begin{proposition}[Chernoff's Inequality] \label{prop: Chernoff}
Let $X$ be a binomial random variable $BIN(n,p)$. Then, for any $0 \leq t \leq np$,
we have $Pr[X-np < -t] \leq e^{-t^2/3np}$.
\end{proposition}

\begin{proposition}[Lov\'{a}sz Local Lemma -- Symmetric version] \label{prop: Lovasz}
Let $A_1,..., A_n$ be a finite set of events in some probability space $\Omega$ such that each $A_i$ occurs with probability at most $p$, where each $A_i$ is mutually independent of all but at most $d$ other events. If $4pd \leq 1$, then $Pr[\cap_{i=1}^{n} \overline{A_i}] > 0$.
\end{proposition}

We will also be needing the following fact on the existence of $k$ (not necessarily disjoint) cycles with distinct lengths in a digraph with minimum out-degree at least~$k$.

\begin{proposition} \label{lemma:k-cycle-lengths}
For every $k \geq 1$, every digraph $D$ with minimum out-degree at least $k$ contains $k$ directed cycles of distinct lengths.
\end{proposition}

\begin{proof}
Consider the out-neighbours of the last vertex of a longest directed path in $D$.
\end{proof}

We are now ready to prove the main result of this section. By a \textit{$r$-regular digraph}, we refer to a digraph whose all vertices have in- and out-degree $r$.

\begin{theorem} \label{theorem:regular}
Let $k \geq 500$ and $r \geq \frac{k^2}{2}(1+7(\frac{\ln k}{k})^\frac{1}{3})$. Then every $r$-regular digraph contains at least $k$ vertex-disjoint directed cycles of distinct lengths.
\end{theorem}

\begin{proof}
Let $D$ be a simple $r$-regular digraph, and assume $r \geq \frac{k^2}{2}(1+7(\frac{\ln k}{k})^\frac{1}{3})$. 
The main idea is to prove, using several probabilistic tools, that we 
can partition the vertex set of $D$ into $k$~parts $V_1,...,V_k$ such that each $V_i$ induces a digraph 
of minimum out-degree at least~$i$. With such a partition in hand, one can then get the~$k$ desired 
disjoint cycles by just considering each of the $V_i$'s successively, and picking, in each of the digraphs 
they induce, one cycle whose length is different from the lengths of the previously picked cycles. This is 
possible according to Proposition~\ref{lemma:k-cycle-lengths} due to the out-degree property of the 
partition $V_1,...,V_k$.

We first introduce some notation and assumptions we use throughout (and further) to deal with 
our computations. Every parameter has to be thought of as a function of $k$. By writing $o(1)$,
we refer to a term tending to $0$ as $k$ tends to infinity. Given two terms $a$ and $b$, we write 
$a \thicksim b$ if $a/b$ tends to $1$, and $a \ll b$ if $a/b$ tends to $0$ as $k$ tends to infinity. 
Let $k':=k+\lfloor k^{2/3}(\ln k)^{1/3}\rfloor$ and $s$ be the sum of the first $k'$ integers, that 
is $s:=\frac{k'(k'+1)}{2}$. It is assumed throughout that $r \thicksim s \thicksim \frac{k^2}{2}$; so 
$r$ and $s$ will sometimes freely be replaced by $\frac{k^2}{2}$ to simplify 
some computation (assuming this has no impact on the computation).

We now show that, under our assumptions, the desired partition $V_1,...,V_k$ of $V(D)$ 
exists with non-zero probability. For this purpose, let us just randomly $k$-color 
the vertices of $D$, where assigning color~$j$ to some vertex means that we put it into $V_j$.
All colors are not assigned uniformly, but in such a way that, for every color $j \in \{1,...,k\}$, 
the probability $p_j$ that some vertex is colored~$j$ is:
\begin{displaymath}
p_j := \frac{j+\lfloor k^{2/3}(\ln k)^{1/3}\rfloor}{s}.
\end{displaymath}
Note that $\sum_{j=1}^k{p_j}<1$, therefore a vertex gets no color with probability $1-\sum_{j=1}^k{p_j}$.

Let $X_v^j$ denote the number of out-neighbors of $v$ colored $j$ by the random process. Clearly $X_v^j \thicksim BIN(r,p_j)$. So that we can later apply the Lov\'asz Local Lemma, let us define our set of bad events. Let $A_v$ be the event $$A_v := \bigcup_{j=1}^{k}(v \textrm{ is colored } j \textrm{ and } X_v^j<j ).$$ Second, let $B$ be the event that at least one of the~$k$ colors does not appear among a fixed subset $\{u_1, u_2, ..., u_r\}$ of $r$ vertices of $D$. It should be clear that any two events $A_v$ and $A_u$ are dependent if $(u \cup N^+(u)) \cap (v \cup N^+(v)) \ne \emptyset$. So, since $D$ is $r$-regular, each $A_v$ depends on at most $r^2$ other bad events $A_u$. Since the event $B$ only depends on the colors of~$r$ fixed vertices, similarly $B$ depends on at most $r^2$ other bad events. 
To apply the Lov\'asz Local Lemma, every bad event $A \in (\cup_{v}A_v)\cup B$ must hence fulfill $4r^2\Pr(A)\leq 1$.

Concerning $B$, we have:
\begin{align*}
\Pr(B) \leq \sum_{j=1}^{k}(1-p_j)^r &\leq k \left(1-\frac{\lfloor k^{2/3} (\ln k)^{1/3} \rfloor+1}{s} \right)^r \\
&\leq ke^{-\frac{k^{2/3} (\ln k)^{1/3} r}{s}} \\
&\leq k^{- \left(\frac{k}{\ln k} \right)^{2/3} \cdot \frac{r}{k^2}  + 1},
\end{align*}

\noindent where we used the fact that $s \leq k^2$ for $k\geq 500$. Therefore, we have that:

\begin{equation*}
4r^2 \Pr(B) \leq 4r^2 k^{- \left(\frac{k}{\ln k} \right)^{2/3} \cdot \frac{r}{k^2}  + 1}\leq 1,
\end{equation*}
since $r \geq k^2/2$ and $k \geq 500$.

 Now consider the $A_v$'s. Since
$$ s=\frac{(k')^2}{2}\left(1+\frac{1}{k'}\right) \leq \frac{k^2}{2}\left(1+\left(\frac{\ln k}{k}\right)^\frac{1}{3}\right)^2\left(1+\frac{2}{k}\right),  $$
\noindent then
$$ \left(1-4\left(\frac{\ln k}{k}\right)^\frac{1}{3}\right)rp_j 
\geq \left(1-4\left(\frac{\ln k}{k}\right)^\frac{1}{3}\right)\frac{rj}{s} \geq j   $$
since $k \geq 500$.
Applying Chernoff's Inequality, we get:
\begin{align*}
\Pr( X_v^j<j ) &\leq \Pr\left( X_v^j<\left(1-4\left(\frac{\ln k}{k}\right)^\frac{1}{3}\right)rp_j \right) \\
&\leq e^{-16\left(\frac{\ln k}{k}\right)^\frac{2}{3}\frac{rp_j}{3}}.
\end{align*}

\noindent Then:
\begin{displaymath}
\Pr(A_v) \leq \sum_{j=1}^{k}{p_j e^{-16\left(\frac{\ln k}{k}\right)^\frac{2}{3}\frac{rp_j}{3}}}.
\end{displaymath}

We hence want every term in the sum to be smaller than $\frac{1}{4r^2k}$.
Thus, it suffices
to show that, for each $j$, we have $$\frac{\left\lfloor k^{2/3}(\ln k)^{1/3}\right\rfloor+j}{s} 
e^{-16\left(\frac{\ln k}{k}\right)^\frac{2}{3}\frac{r}{3}\frac{\left\lfloor k^{2/3}(\ln k)^{1/3}\right\rfloor+j}{s}} < 
\frac{1}{4r^2k}.$$
Since $k^2/2 \leq s \leq k^2$ for $k \geq 500$ and $r \geq s$, this reduces to showing that we have
$$ \left(\left\lfloor k^{2/3}(\ln k)^{1/3}\right\rfloor+j\right) k^{-\frac{16r}{3s}} \leq \frac{k}{8r^2}.$$
Since $k$ is sufficiently large and $j \leq k$, this, in turn, amounts to showing that 
$k^{-\frac{16r}{3s}} \leq \frac{1}{16r^2}$ or equivalently, that $\frac{k^{\frac{16r}{3s}}}{16r^2} \geq 1$.
Since $k$ is sufficiently large, $r \geq s$ and $r \geq k^2 /2$, the function on the left is increasing;
hence it suffices to show that the inequality holds for $r = k^2 /2$. Since $r \geq s$,
we only need to show that $k^{16/3} \geq 4 k^4$, which is clearly true when $k \geq 500$.

Under all these conditions, the requirements for applying the Lov\'asz Local Lemma are thus met; 
we can hence deduce the desired partition $V_1,...,V_k$, thus the claimed $k$ disjoint cycles of distinct lengths.
\end{proof}


\subsection{Small digraphs} \label{section:small}

We now prove Conjecture~\ref{conjecture:digraph-lengths} for digraphs whose order can be expressed as some particular function of the minimum out-degree.

\begin{theorem} \label{theorem:order}
Let $k \geq 1$. Then every simple digraph of order at most $c^{r^d}$ (where $c$ and $d$ are two constants satisfying $c>1$ and $0<d <1$) and minimum out-degree $r \geq c_0\max\{k^{\frac{1}{1-d}}, k^2\}$ contains at least $k$ vertex-disjoint directed cycles of distinct lengths, where $c_0=\max\{2,(24\ln c)^\frac{1}{1-d}\}$.
\end{theorem}

\begin{proof}
Let $D$ be such a digraph with order $n$ where we assume the out-degree of every vertex is exactly~$r$. We herein reuse the terminology introduced in the proof of Theorem~\ref{theorem:regular}.

The proof is similar to that of Theorem~\ref{theorem:regular}, except that the random $k$-coloring of the vertices of $D$ is this time performed uniformly (all $k$ colors have the same probability to be assigned). Given any vertex $v$ of $D$, let $X_v$ denote the number of out-neighbors of $v$ being assigned the same color by the random coloring. Clearly $X_v \thicksim BIN(r,1/k)$. Our two kinds of bad events are the following. First $A_v$ is, for every vertex $v$, the event that $X_v<k$. Second, let $B$ be the event that at least one of the $k$ colors does not appear at all.

We first assume the following:

\begin{equation} \label{E:2:4}
\frac{r}{2k} \geq k \Leftrightarrow r\geq 2k^2.
\end{equation}

\noindent Therefore, applying Chernoff's Inequality, we have:
$$ \Pr(A_v) = \Pr(X_v<k) \leq \Pr\left(X_v< \frac{r}{2k}\right) \leq e^{-\frac{r}{12k}}.$$

\noindent By the Union Bound, we deduce:
\begin{align} \label{eq1}
\Pr\left(\bigcup_v{A_v} \bigcup B\right) \leq \sum_v{\Pr(A_v)} + \Pr(B) & \leq ne^{-\frac{r}{12k}} + k \left(1-\frac{1}{k}\right)^n \nonumber \\
 & \leq ne^{-\frac{r}{12k}} + k\left(1-\frac{1}{k}\right)^r \nonumber \\
 & \leq ne^{-\frac{r}{12k}} + ke^{-\frac{r}{k}}.
\end{align}

\noindent Since we want this obtained sum to be smaller than $1$, we would like each of its two terms to be strictly smaller that $1/2$. Then, concerning the first term:

\begin{equation*}
ne^{-\frac{r}{12k}} < \frac{1}{2} \Leftrightarrow \frac{r}{12k} > \ln(2n).
\end{equation*}

\noindent Since $n\leq c^{r^d}$, we have $\ln(2n) \leq 2r^d\ln c$. So we need a stronger inequality:

\begin{equation} \label{E:2:5}
\frac{r}{12k} \geq 2r^d\ln c \Leftrightarrow r \geq (24k\ln c)^{\frac{1}{1-d}}.
\end{equation}

\noindent Concerning the second term of Inequality~(\ref{eq1}), we want:

\begin{equation} \label{E:2:6}
ke^{-\frac{r}{k}} < \frac{1}{2} \Leftrightarrow r > k\ln(2k).
\end{equation}

Suppose now that Inequalities~(\ref{E:2:5}) and (\ref{E:2:6}) hold. Then there exists a partition $V_1, ..., V_k$ of $V(D)$ such that the out-degree of each vertex in the part containing it is at least $k$. We are now able to successfully pick a cycle from each of these parts in such a way that all picked cycles have distinct lengths (by applying Proposition~\ref{lemma:k-cycle-lengths}).

So that all conditions of Inequalities~(\ref{E:2:4}), (\ref{E:2:5}) and (\ref{E:2:6}) are met, we then just need $r \geq c_0\max\{k^{\frac{1}{1-d}}, k^2\}$, where $c_0=\max\{2,(24\ln c)^\frac{1}{1-d}\}$, as claimed.
\end{proof}


\subsection{Concluding remarks} \label{section:ccl-digraphs}

In Sections~\ref{section:regular} and~\ref{section:small}, we have proved that Conjecture~\ref{conjecture:digraph-lengths} holds 
for some regular digraphs (Theorem~\ref{theorem:regular}) and digraphs with bounded order (Theorem~\ref{theorem:order}). About these two results, let us mention the following:

\begin{enumerate}
    \item In the proofs of Theorems~\ref{theorem:regular} and~\ref{theorem:order}, if we require the parts of the partition $V_1,...,V_k$ to include more vertices, then we can deduce $k$ distinct cycles whose lengths are 'more than just distinct'. Possible such additional properties are \textit{e.g.} the cycles to be of even or odd lengths only, or to be of lengths divisible by some fixed integer, etc.
    
    \item Let us mention that the bound on~$r$ in the statement of Theorem~\ref{theorem:regular} is asymptotically best possible.
	This is the main reason why the random process in the proof is not uniform,
	and thus why the result requires $k$ to be large enough.
	As a consequence, the proof could probably be adapted so that it works for all values of $k$,
	but, on the other hand, this would require to consider $r$-regular digraphs with $r$ being a much bigger function of $k$.

    \item In the statement of Theorem~\ref{theorem:order}, it is worth mentioning that we can lower the requirement on the out-degree if we assume a smaller upper bound on $n$. For example, if $n$ is bounded above by a polynomial function of $r$ ($n < r^d$ for some constant $d > 0$), then we can require the digraph to have minimum out-degree at least $k^2(\ln k)^3$ only (which sticks closer to what is stated in Conjecture~\ref{conjecture:digraph-lengths}).

    \item In case Conjecture~\ref{conjecture:digraph-lengths} is false and some counterexamples exist, Theorem~\ref{theorem:order} gives a lower bound on the order of these counterexamples. These should be of large order.
\end{enumerate}


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