Byzantine Game Theory: Sun Tzus Boxes

TL;DR

The paper defines the Byzantine Selection Problem, where an organizer must select $\ell$ agents from $n$ with known values $v_i$, while up to $t$ agents may be Byzantine and report zero. Deterministically, choosing the top-$\ell$ values is optimal, but when $t\ge\ell$ the adversary can nullify the chosen set; hence randomized mechanisms are studied to maximize the expected payoff against worst-case byzantine sets, under an oblivious adversary. For $\ell=1$, the authors derive a closed-form optimal structure: the solution concentrates on a prefix with $v_1 p_1 = \dots = v_i p_i$ and $p_{i+1..n}=0$, with $p^i_j \propto 1/v_j$, and value $\frac{i-t}{\sum_{j=1}^i 1/v_j}$, computable in linear time. For general $\ell$, the problem is reduced to marginals $p'_i$ in $\Delta_\ell([n])$ and solved via a water-filling interpretation (nice pseudo-distributions) and a linear-time two-pointer algorithm over breakpoints, enabling linear-time sampling from the optimal distribution. The work bridges Byzantine fault tolerance with centralized selection problems, yielding practical, robust mechanisms for auctions, hiring, or committee selection under adversarial reporting.

Abstract

We introduce the Byzantine Selection Problem, living at the intersection of game theory and fault-tolerant distributed computing. Here, an event organizer is presented with a group of $n$ agents, and wants to select $\ell < n$ of them to form a team. For these purposes, each agent $i$ self-reports a positive skill value $v_i$, and a team's value is the sum of its members' skill values. Ideally, the value of the team should be as large as possible, which can be easily achieved by selecting agents with the highest $\ell$ skill values. However, an unknown subset of at most $t < n$ agents are byzantine and hence not to be trusted, rendering their true skill values as $0$. In the spirit of the distributed computing literature, the identity of the byzantine agents is not random but instead chosen by an adversary aiming to minimize the value of the chosen team. Can we still select a team with good guarantees in this adversarial setting? As it turns out, deterministically, it remains optimal to select agents with the highest $\ell$ values. Yet, if $t \geq \ell$, the adversary can choose to make all selected agents byzantine, leading to a team of value zero. To provide meaningful guarantees, one hence needs to allow for randomization, in which case the expected value of the selected team needs to be maximized, assuming again that the adversary plays to minimize it. For this case, we provide linear-time randomized algorithms that maximize the expected value of the selected team.

Figures (1)

• Figure 1: Consider an example with $n = 7, t = 1, \ell = 5$ and $\mathbf{v} = (12, 8, 8, 6, 4, 3, 2).$ This is depicted in \ref{['fig:water-level-1']} by rectangles with heights given by $\mathbf{v},$ each of area 1. One can understand pseudo-distributions $p$ using a water-filling metaphor: each rectangle corresponds to a container comprising one unit of volume and each value $p_i \in [0, 1]$ corresponds to pouring $p_i$ units of water into the $i$-th container. By the choice of widths, pouring $p_i$ units of water into the $i$-th container makes the water rise to height $h_i := v_i \cdot p_i$ inside the container. Given $(t, \ell),$ a pseudo-distribution is $(E, i)$-nice if: (0) it uses at most $\ell$ units of water; (1) the water rises to level $E$ in the first $t + 1$ containers; (2) each container $k$ among the first $i$ is saturated for $E$, i.e., either the water rises to height $E$ in $k$ or $E < v_k$ and $k$ is full; (3) container $i + 1$ (if it exists) is not saturated for $E$; (4) all subsequent containers are empty. Given $E$, there can exist at most one maximal $E$-nice pseudo-distribution: fill in the first $t + 1$ containers to level $E,$ if this exceeds the water budget $\ell,$ then no solution exists, otherwise continue in order through the next containers, saturating them until there is not enough water left to saturate current container. This is demonstrated for $E = 7$ in \ref{['fig:water-level-2']}: the first $3 \geq t + 1$ containers rise to level $7$, the next two containers have $p_4 = p_5 = 1$ (hence $i = 5$), the following container is not saturated: $p_6 = \frac{2}{3}$, and the last container is empty: $p_7 = 0.$ Overall, $\mathbf{p} = (\frac{7}{12}, \frac{7}{8}, \frac{7}{8}, 1, 1, \frac{2}{3}, 0),$ whose entries sum up to $\ell = 5.$

Theorems & Definitions (13)

• theorem 1
• theorem 2
• theorem 3
• lemma 4
• lemma 5
• lemma 6
• lemma 6
• lemma 7
• definition 1
• lemma 8