A Combinatorial Metrical Task System Problem Under the Uniform Metric

Abstract

We consider a variant of the metrical task system (MTS) problem under the uniform metric, where each decision corresponds to some combinatorial object in a fixed set (e.g., the set of all s-t paths of a fixed graph). Typical algorithms such as Marking algorithm are not known to solve this problem efficiently and straightforward implementations takes exponential time for many classes of combinatorial sets. We propose a modification of Marking algorithm, which we call Weighted Marking algorithm. We show that Weighted Marking algorithm still keeps $O(\log n)$ competitive ratio for the standard MTS problem with n states. On the other hand, combining with known sampling techniques for combinatorial sets, Weighted Marking algorithm works efficiently for various classes of combinatorial sets.

A On Assumption 1

As is well known as a folklore, we can assume without loss of generality that the loss vectors $\varvec{\ell }_t$ are small enough, so that Assumption 1 is satisfied.

Assumption 1

Whenever the previous state $\varvec{c}_{t-1}$ satisfies $\varvec{c}_{t-1} \cdot \varvec{L}< L$, where $\varvec{L}$ is the cumulative loss vectors up to round $t-1$ in the current phase, and the phase did not end at round $t-1$, i.e., $\min _{\varvec{c}^* \in C} \varvec{c}^* \cdot \varvec{L}< 1$, then $\varvec{\ell }_t$ satisfies the two conditions:

• 1. $\varvec{c}_{t-1} \cdot (\varvec{L}+ \varvec{\ell }_t) \le L$, and

• 2. $\min _{\varvec{c}^* \in C} \varvec{c}^* \cdot (\varvec{L}+ \varvec{\ell }_t) \le 1$.

This is because, when $\varvec{\ell }_t$ violates the assumption, then we can replace $\varvec{\ell }_t$ by a sequence of non-negative loss vectors $\varvec{\ell }_{t_1}, \varvec{\ell }_{t_2}, \ldots , \varvec{\ell }_{t_k}$ so that $\varvec{\ell }_t = \varvec{\ell }_{t_1} + \cdots + \varvec{\ell }_{t_k}$ and the new sequence of loss vectors satisfy the assumption in the following way:

• 1. If the first condition is violated, i.e., $\varvec{c}_{t-1} \cdot (\varvec{L}+ \varvec{\ell }_t) = a > L$, then we let

  $$\begin{aligned} \alpha _1 = \frac{L-\varvec{c}_{t-1} \cdot \varvec{L}}{a - \varvec{c}_{t-1} \cdot \varvec{L}}. \end{aligned}$$

  Otherwise, we let $\alpha _1 = 1$. In the former case, we can easily verify that $0< \alpha _1 < 1$ and $\varvec{c}_{t-1} \cdot (\varvec{L}+ \alpha _1 \varvec{\ell }_t) = L$.

• 2. If the second condition is violated, i.e., $\min _{\varvec{c}^* \in C} \varvec{c}^* \cdot (\varvec{L}+\varvec{\ell }_t) > 1$, then we let $0< \alpha _2 < 1$ be such that $\min _{\varvec{c}^* \in C} \varvec{c}^* \cdot (\varvec{L}+\alpha _2 \varvec{\ell }_t) = 1$. Otherwise, we let $\alpha _2 = 1$. Note that, in the former case, we can find such $\alpha _2$ efficiently by binary search.

• 3. Let $\alpha = \min \{\alpha _1,\alpha _2\}$ and $\varvec{\ell }_{t_1} = \alpha \varvec{\ell }_t$ and $\varvec{\ell }_{t_2} = (1 - \alpha ) \varvec{\ell }_t$. Then, clearly $\varvec{\ell }_{t_1}$ satisfies Assumption 1. If $\varvec{\ell }_{t_2}$ still violates the assumption, then repeat the same procedure for $\varvec{\ell }_{t_2}$ recursively.