Michael A. Bender · Sándor P. Fekete · Alexander Kröller · Vincenzo Liberatore · Joseph S. B. Mitchell · Valentin Polishchuk · Jukka Suomela

The minimum backlog problem

Theoretical Computer Science · volume 605, pages 51–61, 2015 · doi:10.1016/j.tcs.2015.08.027

authors’ version publisher’s version arXiv.org


We study the minimum backlog problem (MBP). This online problem arises, e.g., in the context of sensor networks. We focus on two main variants of MBP.

The discrete MBP is a 2-person game played on a graph $G=(V,E)$. The player is initially located at a vertex of the graph. In each time step, the adversary pours a total of one unit of water into cups that are located on the vertices of the graph, arbitrarily distributing the water among the cups. The player then moves from her current vertex to an adjacent vertex and empties the cup at that vertex. The player's objective is to minimize the backlog, i.e., the maximum amount of water in any cup at any time.

The geometric MBP is a continuous-time version of the MBP: the cups are points in the two-dimensional plane, the adversary pours water continuously at a constant rate, and the player moves in the plane with unit speed. Again, the player's objective is to minimize the backlog.

We show that the competitive ratio of any algorithm for the MBP has a lower bound of $\Omega(D)$, where $D$ is the diameter of the graph (for the discrete MBP) or the diameter of the point set (for the geometric MBP). Therefore we focus on determining a strategy for the player that guarantees a uniform upper bound on the absolute value of the backlog.

For the absolute value of the backlog there is a trivial lower bound of $\Omega(D)$, and the deamortization analysis of Dietz and Sleator gives an upper bound of $O(D\log N)$ for $N$ cups. Our main result is a tight upper bound for the geometric MBP: we show that there is a strategy for the player that guarantees a backlog of $O(D)$, independently of the number of cups.

We also study a localized version of the discrete MBP: the adversary has a location within the graph and must act locally (filling cups) with respect to his position, just as the player acts locally (emptying cups) with respect to her position. We prove that deciding the value of this game is PSPACE-hard.

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© Elsevier 2015

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