论文标题
完整的四边形图的交叉数$ k_ {1,1,m,n} $
The crossing number of the complete 4-partite graph $K_{1,1,m,n}$
论文作者
论文摘要
令$ \ textrm {cr}(g)$表示图$ g $的交叉数。众所周知的Zarankiewicz的猜想(ZC)主张$ \ textrm {Cr}(k_ {m,n})$ 1954。1971年,Harborth在$ \ textrm {cr}上给出了一个猜想(hc)(k_ cr}(k__ {k_ {x_1,...,...,...,...,...,x_n})$。如果Ho等人的ZC为TRUE,则$ K_ {1,M,N} $上的HC。在2021年。在本文中,我们显示了以下结果:如果$ m $和$ n $均匀,则\ [\ textrm {cr}(k_ {1,1,m,n})\ geq \ frac {1} {2}(\ textrm {cr}(k_ {m+1,n+3})+\ textrm {cr}(k_ {m+3,n+1}) - mn- \ frac {1} \ [\ textrm {cr}(k_ {1,1,m,n})\ geq \ frac {1} {2}(\ textrm {cr}(k_ {1,m+1,n+1})+\ textrm {cr}(k_ {2,m,n}) - \ frac {1} {1} {1} {4} {4} {4} {4}(m+1)(m+1)(n+1)(n+1)(n+1) \ begin {equation} \ nonumber \ begin {split} \ textrm {cr}(k_ {1,1,m,n})&\ geq \ frac {1} {4}(\ textrm {cr}(k_ {m+1,n+2})+\ textrm {cr}(k_ {m+3,n+2})+2 \ 2 \ 2 \ textrm {cr}(k_ {2,m,m,n}) \\& - m(n+1) - \ frac {1} {4}(n+1)^2)。 \ end {split} \ end {equation}我们结果中的下限意味着,如果$ m $和$ n $都是均匀的,并且zc是TRUE的,则HC在$ k_ {1,1,m,n} $ holds上;如果$ m $和$ n $中的至少一个是奇数的,而$ k_ {2,m,n} $上的zc和hc都是true的,则hc on $ k_ {1,1,m,n} $ holds。
Let $\textrm{cr}(G)$ denote the crossing number of a graph $G$. The well-known Zarankiewicz's conjecture (ZC) asserted $\textrm{cr}(K_{m,n})$ in 1954. In 1971, Harborth gave a conjecture (HC) on $\textrm{cr}(K_{x_1,...,x_n})$. HC on $K_{1,m,n}$ is verified if ZC is true by Ho et al. in 2021. In this paper, we showed the following results: If both $m$ and $n$ are even, then \[\textrm{cr}(K_{1,1,m,n})\geq \frac{1}{2}(\textrm{cr}(K_{m+1,n+3})+\textrm{cr}(K_{m+3,n+1})-mn-\frac{1}{4}(m^2+n^2));\] If both $m$ and $n$ are odd, then \[\textrm{cr}(K_{1,1,m,n})\geq \frac{1}{2}(\textrm{cr}(K_{1,m+1,n+1})+\textrm{cr}(K_{2,m,n})-\frac{1}{4}(m+1)(n+1)+1);\] If $m$ is even and $n$ is odd, then \begin{equation}\nonumber \begin{split} \textrm{cr}(K_{1,1,m,n})&\geq \frac{1}{4}(\textrm{cr}(K_{m+1,n+2})+\textrm{cr}(K_{m+3,n+2})+2\textrm{cr}(K_{2,m,n}) \\&-m(n+1)-\frac{1}{4}(n+1)^2). \end{split} \end{equation} The lower bounds in our result imply that if both $m$ and $n$ are even and ZC is true, then HC on $K_{1,1,m,n}$ holds; if at least one of $m$ and $n$ is odd and both ZC and HC on $K_{2,m,n}$ are true, then HC on $K_{1,1,m,n}$ holds.