Consider the following game played by two players, called Waiter and Client, on the edges of $K_n$ (where $n$ is divisible by 3). Initially, all the edges are unclaimed. In each round, Waiter picks two yet unclaimed edges. Client then chooses one of these two edges to be added to Waiter’s graph and one to be added to Client’s graph. Waiter wins if she forces Client to create a $K_3$-factor in Client’s graph at some point, while if she does not manage to do that, Client wins.

It is not difficult to see that for large enough $n$, Waiter has a winning strategy. The question considered by Clemens et al. is how long the game will last if Waiter aims to win as soon as possible, Client aims to delay her as much as possible, and both players play optimally. Denote this optimal number of rounds by ${\tau }_{WC}({\mathcal{F}}_{n,K_3-\text{fac}},1)$. Clemens et al. proved that $\frac{13}{12}n\le {\tau }_{WC}({\mathcal{F}}_{n,K_3-\text{fac}},1)\le \frac{7}{6}n+o\left(n\right)$, and conjectured that ${\tau }_{WC}({\mathcal{F}}_{n,K_3-\text{fac}},1)=\frac{7}{6}n+o\left(n\right)$. In this note, we verify their conjecture.

考虑以下两个由两个服务员（称为“服务员”和“客户”）玩的游戏， ${ķ}_{ñ}$ （在哪里 $ñ$可被3整除。最初，所有边缘均无人认领。在每个回合中，侍者挑选两个尚未领取的边缘。然后，客户选择这两个边之一添加到服务员的图，并选择一个添加到客户的图。如果服务员强迫客户创建一个服务员，则该服务员将获胜。${ķ}_3$有时会影响客户图表中的因素，而如果客户未能做到这一点，则客户会获胜。

不难看出，足够大 $ñ$，侍者有制胜法宝。克莱门斯等人考虑的问题。如果Waiter希望尽快赢得比赛，游戏将持续多长时间，Client希望尽可能延迟她的比赛，并且双方玩家都将发挥最佳状态。用以下方式表示此最佳轮数${\tau }_{\mathrm{w\; ^}C}（{\mathcal{F}}_{ñ，{ķ}_3-\text{事实}}，\mathrm{1个}）$。Clemens等。证明$\frac{13}{12}ñ\le {\tau }_{\mathrm{w\; ^}C}（{\mathcal{F}}_{ñ，{ķ}_3-\text{事实}}，\mathrm{1个}）\le \frac{7}{6}ñ+Ø（ñ）$，并认为 ${\tau }_{\mathrm{w\; ^}C}（{\mathcal{F}}_{ñ，{ķ}_3-\text{事实}}，\mathrm{1个}）=\frac{7}{6}ñ+Ø（ñ）$。在本说明中，我们验证了他们的猜想。