The problem of scheduling tasks on $p$ processors so that no task ever gets too far behind is often described as a game with cups and water. In the $p$-processor cup game on $n$ cups, there are two players, a filler and an emptier, that take turns adding and removing water from a set of $n$ cups. In each turn, the filler adds $p$ units of water to the cups, placing at most $1$ unit of water in each cup, and then the emptier selects $p$ cups to remove up to $1$ unit of water from. The emptier's goal is to minimize the backlog, which is the height of the fullest cup. The $p$-processor cup game has been studied in many different settings, dating back to the late 1960's. All of the past work shares one common assumption: that $p$ is fixed. This paper initiates the study of what happens when the number of available processors $p$ varies over time, resulting in what we call the \emph{variable-processor cup game}. Remarkably, the optimal bounds for the variable-processor cup game differ dramatically from its classical counterpart. Whereas the $p$-processor cup has optimal backlog $\Theta(\log n)$, the variable-processor game has optimal backlog $\Theta(n)$. Moreover, there is an efficient filling strategy that yields backlog $\Omega(n^{1 - \epsilon})$ in quasi-polynomial time against any deterministic emptying strategy. We additionally show that straightforward uses of randomization cannot be used to help the emptier. In particular, for any positive constant $\Delta$, and any $\Delta$-greedy-like randomized emptying algorithm $\mathcal{A}$, there is a filling strategy that achieves backlog $\Omega(n^{1 - \epsilon})$ against $\mathcal{A}$ in quasi-polynomial time.
翻译:将水排到 $ p$ 的处理器上, 使任何任务不会太落后的问题通常被描述为 杯子和水的游戏。 在 杯子上 $ p$ 的处理器杯游戏中, 有两个玩家, 一个填充器和一个空格, 轮流将水从一套 $ 美元 的杯子上加起来并去除。 在每转, 填充器在杯子上增加 $ 单位, 在每个杯子上最多放1美元, 然后由空格选择 $ 的杯子从杯子上去除最多1美元 美元 。 在牌子杯子上, 以美元 的高度 。 $ 美元 的处理器已经在许多不同的场合里进行了研究。 所有过去的工作都有一个共同的假设: 美元是固定的 。 本文开始研究当可用的处理器数量在任何时间上变化时会发生什么, 导致我们称之为 emp - cal- 美元 平价 的杯子游戏 战略 。 最优性 。