Abstract
Let f(n) denote the maximum possible number of edges in an n-vertex graph containing no cycles of equal length. The problem of determining f(n) was posed by Erdos in 1975. This paper provides a lower bound for f(n).
Full Text
Preamble
On the Number of Edges in Some Graphs
Chunhui Lai
School of Mathematics and Statistics, Minnan Normal University, Zhangzhou, Fujian, P.R. China
Abstract
In 1975, P. Erdős proposed the problem of determining the maximum number $f(n)$ of edges in a graph with $n$ vertices in which any two cycles have different lengths. The sequence $(c_1, c_2, \cdots, c_n)$ represents the cycle length distribution of a graph $G$ with $n$ vertices, where $c_i$ is the number of cycles of length $i$ in $G$. Let $f(a_1, a_2, \cdots, a_n)$ denote the maximum possible number of edges in a graph satisfying $c_i \leq a_i$, where $a_i$ is a nonnegative integer. In 1991, Shi posed the problem of determining $f(a_1, a_2, \cdots, an)$, which extended Erdős's problem. It is clear that $f(n) = f(1, 1, \cdots, 1)$. Let $g(n, m) = f(a_1, a_2, \cdots, a_n)$, where $a_i = 1$ if $i/m$ is an integer, and $a_i = 0$ otherwise. It is clear that $f(n) = g(n, 1)$.
We prove that
$$\liminf_{n\to\infty} \frac{g(n,m)-n}{\sqrt{n}} \geq \sqrt{2 + \frac{7654}{99}} \approx 2.444$$
for all even integers $m$, which improves upon previous bounds (Lai, 2017). We show that
$$\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} \geq \sqrt{2 + \frac{7654}{19071}}$$
which is better than the previous bounds: $\sqrt{2}$ (Shi, 1988) and earlier results. We make the following conjecture:
$$\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} = \sqrt{2 + \frac{7654}{19071}}.$$
Key words: Graph, cycle, number of edges
AMS 2000 MSC: 05C38, 05C35
Funding: Project supported by the National Science Foundation of China (No. 61379021; No. 11401290), NSF of Fujian (2015J01018; 2018J01423), Fujian Provincial Training Foundation for "Bai-Quan-Wan Talents Engineering", Project of Fujian Education Department (JZ160455), the Institute of Meteorological Big Data-Digital Fujian and Fujian Key Laboratory of Data Science and Statistics, Project supported by Minnan Normal University.
Email address: laich2011@msn.cn; laichunhui@mnnu.edu.cn (Chunhui Lai)
Introduction
Let $f(n)$ be the maximum number of edges in a graph with $n$ vertices in which no two cycles have the same length. In 1975, Erdős raised the problem of determining $f(n)$ (see Bondy and Murty \cite{BondyMurty1976}, p. 247, Problem 11). Shi \cite{Shi1988} proved the following lower bound:
Theorem 1 (Shi \cite{Shi1988}). $f(n) \geq n + \left\lfloor\left(\sqrt{8n-23}+1\right)/2\right\rfloor$ for $n \geq 3$.
Additional related results were obtained by Chen, Lehel, Jacobson and Shreve \cite{ChenLehelJacobsonShreve1998}, Jia \cite{Jia1996}, Lai \cite{Lai1993, Lai2001, Lai2003, Lai2017}, and Shi \cite{Shi1992, Shi1994}. Boros, Caro, Füredi and Yuster \cite{BorosCaroFurediYuster2001} proved the following upper bound:
Theorem 2 (Boros, Caro, Füredi and Yuster \cite{BorosCaroFurediYuster2001}). For $n$ sufficiently large, $f(n) < n + 1.98\sqrt{n}$.
Lai \cite{Lai2017} improved Shi's lower bound as follows:
Theorem 3 (Lai \cite{Lai2017}). Let $t = 1260r + 169$ ($r \geq 1$), then
$$f(n) \geq n + \frac{t^2 + 87978t + 15957}{4t + 1}$$
for $n \geq \frac{2119}{4}t^2 + \frac{26399}{2}t + 6932215$.
Lai \cite{Lai1993} proposed the following conjecture:
Conjecture 4 (Lai \cite{Lai1993}).
$$\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} = \sqrt{2}.$$
It would be nice to prove that $\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} > \sqrt{2}$. Survey papers on this problem can be found in Tian \cite{Tian1986}, Zhang \cite{Zhang2007}, and Lai and Liu \cite{LaiLiu2014}. The progress on all 50 problems in \cite{BondyMurty1976} can be found in Locke \cite{Locke}.
The sequence $(c_1, c_2, \cdots, c_n)$ is the cycle length distribution of a graph $G$ with $n$ vertices, where $c_i$ is the number of cycles of length $i$ in $G$. Let $f(a_1, a_2, \cdots, a_n)$ denote the maximum possible number of edges in a graph which satisfies $c_i \leq a_i$, where $a_i$ is a nonnegative integer. Shi \cite{Shi1991} posed the problem of determining $f(a_1, a_2, \cdots, a_n)$, which extended the problem due to Erdős. It is clear that $f(n) = f(1, 1, \cdots, 1)$. Let $g(n, m) = f(a_1, a_2, \cdots, a_n)$, where $a_i = 1$ if $i/m$ is an integer, and $a_i = 0$ otherwise. It is clear that $f(n) = g(n, 1)$.
In this paper, we obtain the following results:
Theorem 5. Let $m$ be even, $s_1 > s_2$, $s_1 + 3s_2 > k$, then
$$g(n, m) \geq n + (k + s_1 + 2s_2 + 1)t - 1$$
for $n \geq \left(\frac{3}{4}mk^2 + \frac{1}{2}mks_1 + \frac{3}{4}mk + \frac{1}{2}mks_2 + \frac{1}{2}ms_1s_2 + \frac{9}{4}ms_2 - k - s_1 - 2s_2 + \frac{1}{2}m - 1\right)t + 1$.
Theorem 6. Let $t = 1260r + 169$ ($r \geq 1$), then for $n \geq \frac{1309}{2}t^2 - \frac{1349159}{3}t + 6932215$,
$$f(n) \geq n + \frac{119}{3}t - 26399.$$
Proof of Theorem 5
Proof. Let
$$n_t = \left(\frac{3}{4}mk^2 + \frac{1}{2}mks_1 + \frac{3}{4}mk + \frac{1}{2}mks_2 + \frac{1}{2}ms_1s_2 + \frac{9}{4}ms_2 - k - s_1 - 2s_2 + \frac{1}{2}m - 1\right)t + 1,$$
where $m$ is even, $s_1 > s_2$, $s_1 + 3s_2 > k$, and $n \geq n_t$. It suffices to show that there exists a graph $G$ on $n$ vertices with $n + (k + s_1 + 2s_2 + 1)t - 1$ edges such that all cycles in $G$ have distinct lengths and all cycle lengths are multiples of $m$.
We construct the graph $G$ consisting of several subgraphs $B_i$ for $0 \leq i \leq s_1t$, $i = s_1t + j$ ($1 \leq j \leq s_2t$), and $i = s_1t + s_2t + j$ ($1 \leq j \leq t$). These subgraphs all share a single common vertex $x$, and are otherwise vertex-disjoint.
For $1 \leq i \leq s_2t$, let the subgraph $B_{s_1t+i}$ consist of a cycle
$$C_{s_1t+i} = xa_1^i a_2^i \cdots a_{ms_1t+2ms_2t+mi-1}^i x$$
and a path
$$P_{s_1t+i} = a_{ms_1t+ms_2t+mi}^i \cdots a_{ms_1t+2ms_2t+mi-2}^i.$$
Based on this construction, $B_{s_1t+i}$ contains exactly three cycles of lengths $ms_1t + mi$, $ms_1t + ms_2t + mi$, and $ms_1t + 2ms_2t + mi$.
For $1 \leq i \leq t$, let the subgraph $B_{s_1t+s_2t+i}$ consist of a cycle
$$C_{s_1t+s_2t+i} = xy_1^i \cdots y_{ms_1t+3ms_2t+mk(k+1)t+mi-1}^i x$$
and $k$ paths sharing the common vertex $x$, with their other endpoints on the cycle $C_{s_1t+s_2t+i}$:
$$P_{s_1t+s_2t+i,p} = y_{ms_1t+3ms_2t-mkt+m(p-1)t+mi}^i \cdots y_{ms_1t+3ms_2t+mk(2p-1)t+m(p-1)t+mi}^i \quad (p = 1, 2, \ldots, k).$$
As a cycle with $k$ chords contains exactly $\binom{k+2}{2}$ distinct cycles, $B_{s_1t+s_2t+i}$ contains cycles of lengths $ms_1t + 3ms_2t + mkht + (h + j - 1)mt + mi$ for $j \geq 1$, $h \geq 0$, and $k + 1 \geq j + h$.
$B_0$ is a path with one endpoint at $x$ and length $n - n_t$. All other $B_i$ are simply cycles of length $mi$.
Therefore, $g(n, m) \geq n + (k + s_1 + 2s_2 + 1)t - 1$ for $n \geq n_t$. This completes the proof.
From Theorem 5, we have
$$\liminf_{n\to\infty} \frac{g(n,m)-n}{\sqrt{n}} \geq \sqrt{\frac{k + s_1 + 2s_2 + 1}{\frac{3}{4}mk^2 + \frac{1}{2}mks_1 + \frac{3}{4}mk + \frac{1}{2}mks_2 + \frac{1}{2}ms_1s_2 + \frac{9}{4}ms_2 + \frac{1}{2}m}}$$
for all even integers $m$.
Let $s_1 = 28499066$, $s_2 = 4749839$, $k = 14249542$. Then
$$\liminf_{n\to\infty} \frac{g(n,m)-n}{\sqrt{n}} \geq \sqrt{2.444}$$
for all even integers $m$.
Proof of Theorem 6
Proof. Let $t = 1260r + 169$ with $r \geq 1$, and let
$$n_t = \frac{1309}{2}t^2 - \frac{1349159}{3}t + 6932215.$$
For $n \geq n_t$, it suffices to show that there exists a graph $G$ on $n$ vertices with $n + \frac{119}{3}t - 26399$ edges such that all cycles in $G$ have distinct lengths.
We construct the graph $G$ consisting of several subgraphs $B_i$ for $0 \leq i \leq 22t$, $i = 22t + j$ ($1 \leq j \leq \frac{5t-8}{3}$), $i = 23t + \frac{2t-2}{3} + j$ ($1 \leq j \leq \frac{5t-8}{3}$), and $i = 32t + j - 60$ ($58 \leq j \leq t - 742$). These subgraphs all share a single common vertex $x$, and are otherwise vertex-disjoint.
For $1 \leq i \leq \frac{5t-8}{3}$, let the subgraph $B_{22t+i}$ consist of a cycle
$$C_{22t+i} = xa_1^i a_2^i \cdots a_{28t+\frac{2t-2}{3}+2i-3}^i x$$
and a path
$$P_{22t+i} = a_{56t-2}^i \cdots a_{76t-4}^i.$$
Based on this construction, $B_{22t+i}$ contains exactly three cycles of lengths $22t + i$, $25t + \frac{t-1}{3} + i - 1$, and $28t + \frac{2t-2}{3} + 2i - 2$.
For $1 \leq i \leq \frac{5t-8}{3}$, let the subgraph $B_{23t+\frac{2t-2}{3}+i}$ consist of a cycle
$$C_{23t+\frac{2t-2}{3}+i} = xb_1^i b_2^i \cdots b_{28t+\frac{2t-2}{3}+2i-2}^i x$$
and a path
$$P_{23t+\frac{2t-2}{3}+i} = b_{11t-1}^i \cdots b_{76t-4}^i.$$
Based on this construction, $B_{23t+\frac{2t-2}{3}+i}$ contains exactly three cycles of lengths $23t + \frac{2t-2}{3} + i$, $27t + i - 1$, and $28t + \frac{2t-2}{3} + 2i - 1$.
For $58 \leq i \leq t - 742$, let the subgraph $B_{32t+i-60}$ consist of a cycle
$$C_{32t+i-60} = xy_1^i \cdots y_{137t+11i+890}^i x$$
and ten paths sharing the common vertex $x$, with their other endpoints on the cycle $C_{32t+i-60}$:
\begin{align}
& P_{32t+i-60,1} = y_{11t-2}^i \cdots y_{21t-59+i}^i, \
& P_{32t+i-60,2} = y_{12t-2}^i \cdots y_{31t-53+2i}^i, \
& P_{32t+i-60,3} = y_{12t-2}^i \cdots y_{41t+156+3i}^i, \
& P_{32t+i-60,4} = y_{13t-2}^i \cdots y_{51t+155+4i}^i, \
& P_{32t+i-60,5} = y_{13t-2}^i \cdots y_{61t+155+5i}^i, \
& P_{32t+i-60,6} = y_{14t-2}^i \cdots y_{71t+154+6i}^i, \
& P_{32t+i-60,7} = y_{14t-2}^i \cdots y_{81t+153+7i}^i, \
& P_{32t+i-60,8} = y_{15t-2}^i \cdots y_{91t+147+8i}^i, \
& P_{32t+i-60,9} = y_{15t-2}^i \cdots y_{101t+149+9i}^i, \
& P_{32t+i-60,10} = y_{16t-2}^i \cdots y_{111t+151+10i}^i.
\end{align}
As a cycle with $d$ chords contains exactly $\binom{d+2}{2}$ distinct cycles, $B_{32t+i-60}$ contains cycles of lengths: $32t + i - 60$, $33t + i + 4$, $34t + i + 207$, $35t + i - 3$, $36t + i - 2$, $37t + i - 3$, $38t + i - 3$, $39t + i - 8$, $40t + i$, $41t + i$, $42t + i + 739$, $43t + 2i - 54$, $43t + 2i + 213$, $45t + 2i + 206$, $45t + 2i - 3$, $47t + 2i - 3$, $47t + 2i - 4$, $49t + 2i - 9$, $49t + 2i - 6$, $51t + 2i + 2$, $51t + 2i + 741$, $53t + 3i + 155$, $54t + 3i + 212$, $55t + 3i + 206$, $56t + 3i - 4$, $57t + 3i - 4$, $58t + 3i - 10$, $59t + 3i - 7$, $60t + 3i - 4$, $61t + 3i + 743$, $64t + 4i + 154$, $64t + 4i + 212$, $66t + 4i + 205$, $66t + 4i - 5$, $68t + 4i - 10$, $68t + 4i - 8$, $70t + 4i - 5$, $70t + 4i + 737$, $74t + 5i + 154$, $75t + 5i + 211$, $76t + 5i + 204$, $77t + 5i - 11$, $78t + 5i - 8$, $79t + 5i - 6$, $80t + 5i + 736$, $85t + 6i + 153$, $85t + 6i + 210$, $87t + 6i + 198$, $87t + 6i - 9$, $89t + 6i - 6$, $89t + 6i + 735$, $95t + 7i + 152$, $96t + 7i + 204$, $97t + 7i + 200$, $98t + 7i - 7$, $99t + 7i + 735$, $106t + 8i + 146$, $106t + 8i + 206$, $108t + 8i + 202$, $108t + 8i + 734$, $116t + 9i + 148$, $117t + 9i + 208$, $118t + 9i + 943$, $127t + 10i + 150$, $127t + 10i + 949$, $137t + 11i + 891$.
$B_0$ is a path with one endpoint at $x$ and length $n - n_t$. All other $B_i$ are simply cycles of length $i$.
Therefore, $f(n) \geq n + \frac{119}{3}t - 26399$ for $n \geq n_t$. This completes the proof.
From Theorem 6, we have
$$\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} \geq \sqrt{2 + \frac{7654}{19071}},$$
which is better than the previous bounds: $\sqrt{2}$ (see \cite{Shi1988}) and $\sqrt{2 + \frac{7654}{19071}}$ (see \cite{Lai2017}). Combining this with Boros, Caro, Füredi and Yuster's upper bound (Theorem 2), we obtain
$$1.98 \geq \limsup_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} \geq \liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} \geq \sqrt{2 + \frac{7654}{19071}}.$$
From the proof of Theorem 6, we have
$$\liminf_{n\to\infty} \frac{g(n,m)-n}{\sqrt{n}} \geq \sqrt{2 + \frac{7654}{19071}}$$
for all integers $m$.
If $m = 1$ and $1 \leq i \leq t$, we could construct a subgraph similar to $B_{s_1t+s_2t+i}$ consisting of a cycle $C_{s_1t+s_2t+i}$ and $k$ paths sharing a common vertex $x$, with their other endpoints on the cycle $C_{s_1t+s_2t+i}$, such that all cycles in $B_{s_1t+s_2t+i}$ have distinct lengths. This would yield $\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} \geq \sqrt{2.444} > \sqrt{2 + \frac{7654}{19071}}$. However, we have only constructed such a subgraph for $m = 1$ and $58 \leq i \leq t - 742$, using a cycle with ten paths sharing a common vertex $x$ to obtain $\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} \geq \sqrt{2 + \frac{7654}{19071}}$.
Since the liminf for $\frac{g(n,m)-n}{\sqrt{n}}$ for even $m$ is $\sqrt{2.444}$, it is reasonable to suspect that such a lower bound also holds for $\frac{f(n)-n}{\sqrt{n}}$. We make the following conjecture:
Conjecture 7.
$$\liminf_{n\to\infty} \frac{f(n)-n}{\sqrt{n}} = \sqrt{2.444}.$$
Acknowledgment
The author would like to thank Professor Endre Boros, Yair Caro, Ronald J. Gould, Gyula O.H. Katona, and Raphael Yuster for their advice. The author would also like to thank the referees for their many valuable comments and suggestions.
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