Axiomatic and Erdős-Moon approaches to tournament rankings

TL;DR

The paper studies how to rank players in tournaments under fairness axioms by merging the Erdős–Moon injective-minimization approach with axiomatic constraints. It defines the Erdős–Moon number $\mathsf{EMN}(\mathcal{A})$ as the worst-case minimum backward-arc proportion across all finite tournaments for rankings satisfying $\mathcal{A}$, and proves $\mathsf{EMN}(\mathsf{Cop})=\mathsf{EMN}(\mathsf{sCop})=3/4$, with several axioms remaining open. A fixed-point recalculation framework on simplicial rankings is developed to produce $\mathcal{A}$-fair rankings, and it is shown that every finite tournament admits a linear fair ranking (hence a spectral fair ranking) via appropriate recalculations. The exact value for the Copeland-related axioms is established as tight, using constructive lower-bound examples that approach $3/4$, which highlights a fundamental trade-off between axiomatic fairness and backward-arc minimization in ranking tournaments.

Abstract

Tournament ranking is a function that assigns each vertex of a tournament (i.e., a directed graph without loops, in which each pair of different vertexes is connected by exactly one arc) a number called the rank of the vertex. One of approaches to constructing tournament rankings suggests choosing a ranking that satisfies a fixed set of axioms. In another approach, proposed by Erdős and Moon, only injective rankings are considered, and among them, one that minimises the number of backward arcs is selected (an arc $x\to y$ is called backward iff the rank of $x$ is less than the rank of $y$). We combine these two approaches as follows: among the rankings that satisfy a fixed set of axioms, we choose one that minimises the number of backward arcs. The Erdős-Moon approach naturally leads to the question of how small the proportion of backward arcs can be guaranteed when using injective rankings. Erdős and Moon showed that the answer to this question is $1/2$. A similar question arises in our approach: how small the proportion of backward arcs can be guaranteed when using rankings that satisfy a set of axioms $\mathcal{A}$? We call this number the Erdős-Moon number of $\mathcal{A}$. We prove that the Erdős-Moon number of the Copeland axiom equals $3/4$.

Figures (6)

• Figure 1: Example of a tournament and its ranking
• Figure 2: Even case. $\text{A}\leqslant\text{B}$ because $\text{C}=\text{D}$ and $\text{A+C}\leqslant\text{D+B}$
• Figure 3: Even case. $\text{E}+\text{A}+\text{F}\leqslant\frac{{l}({l}-1)}{2} + \frac{{l}^2}{2} + \frac{{l}({l}-1)}{2}$
• Figure 4: Odd case. ${\tilde{\text{A}}}\leqslant\tilde{\text{B}}$ because $\tilde{\text{C}}=\tilde{\text{D}}$ and $\tilde{\text{A}}+\tilde{\text{C}}\leqslant\tilde{\text{D}}+\tilde{\text{B}}$
• Figure 5: Odd case. $2\tilde{\text{G}}=\tilde{\text{I}}+\tilde{\text{H}}+\tilde{\text{A}}+\tilde{\text{J}} \leqslant{l}+(\tilde{\text{H}}+\tilde{\text{B}})+{l}={l} + ({l}\,{+}\,{1}){l} + {l}$

Theorems & Definitions (24)

• Definition 2.2
• Definition 2.3
• Example 2.4
• Example 2.5
• Definition 2.6
• Remark 2.7
• Definition 2.9
• Definition 2.10
• Definition 2.11
• Proposition 2.12