We define the problem as a two-player game between Algorithm and Builder. The game is played in rounds. Each round, Builder presents an interval that is neither contained in nor contains any previously presented interval. Algorithm immediately and irrevocably assigns the interval a color that has not been assigned to any interval intersecting it. The set of intervals form an interval representation for a unit interval graph and the colors form a proper coloring of that graph. For every positive integer $\omega$, we define the value $R(\omega)$ as the maximum number of colors for which Builder has a strategy that forces Algorithm to use $R(\omega)$ colors with the restriction that the unit interval graph constructed cannot contain a clique of size $\omega$. In 1981, Chrobak and \'{S}lusarek showed that $R(\omega)\leq2\omega -1$. In 2005, Epstein and Levy showed that $R(\omega)\geq\lfloor{3\omega/2\rfloor}$. This problem remained unsolved for $\omega\geq 3$. In 2022, Bir\'o and Curbelo showed that $R(3)=5$. In this paper, we show that $R(4)=7$

翻译：我们将该问题定义为算法与构建者之间的双人博弈。游戏以回合制进行。每回合中，构建者给出一个既不包含于先前任何区间，也不包含任何先前区间的区间。算法立即且不可撤销地为该区间分配一种颜色，该颜色不得与任何与之相交的区间已分配的颜色相同。这些区间集合构成一个单元区间图的区间表示，而颜色集合构成该图的一个正常着色。对于每个正整数 $\omega$，我们定义值 $R(\omega)$ 为构建者拥有一种策略所能迫使算法使用的最大颜色数，同时限制所构建的单元区间图不能包含大小为 $\omega$ 的团。1981年，Chrobak 和 \'{S}lusarek 证明了 $R(\omega)\leq2\omega -1$。2005年，Epstein 和 Levy 证明了 $R(\omega)\geq\lfloor{3\omega/2\rfloor}$。对于 $\omega\geq 3$，该问题一直悬而未决。2022年，Bir\'o 和 Curbelo 证明了 $R(3)=5$。在本文中，我们证明 $R(4)=7$。