Beating Competitive Ratio 4 for Graphic Matroid Secretary

TL;DR

A competitive ratio arbitrarily close to e is shown, showing a competitive ratio arbitrarily close to e even for constant girth values, supporting the strong matroid secretary conjecture.

Abstract

One of the classic problems in online decision-making is the *secretary problem* where to goal is to maximize the probability of choosing the largest number from a randomly ordered sequence. A natural extension allows selecting multiple values under a combinatorial constraint. Babaioff, Immorlica, Kempe, and Kleinberg (SODA'07, JACM'18) introduced the *matroid secretary conjecture*, suggesting an $O(1)$-competitive algorithm exists for matroids. Many works since have attempted to obtain algorithms for both general matroids and specific classes of matroids. The ultimate goal is to obtain an $e$-competitive algorithm, and the *strong matroid secretary conjecture* states that this is possible for general matroids. A key class of matroids is the *graphic matroid*, where a set of graph edges is independent if it contains no cycle. The rich combinatorial structure of graphs makes them a natural first step towards solving a problem for general matroids. Babaioff et al. (SODA'07, JACM'18) first studied the graphic matroid setting, achieving a $16$-competitive algorithm. Subsequent works have improved the competitive ratio, most recently to 4 by Soto, Turkieltaub, and Verdugo (SODA'18). We break this $4$-competitive barrier, presenting a new algorithm with a competitive ratio of $3.95$. For simple graphs, we further improve this to $3.77$. Intuitively, solving the problem for simple graphs is easier since they lack length-two cycles. A natural question is whether a ratio arbitrarily close to $e$ can be achieved by assuming sufficiently large girth. We answer this affirmatively, showing a competitive ratio arbitrarily close to $e$ even for constant girth values, supporting the strong matroid secretary conjecture. We also prove this bound is tight: for any constant $g$, no algorithm can achieve a ratio better than $e$ even when the graph has girth at least $g$.

Figures (1)

• Figure 1: The described "worst-case" example for Algorithm \ref{['alg:graph']}. At step $t$, we are presented with an edge $e_t$ in the optimum solution. $e_t$ may be blocked from being taken due to an earlier step $j$. We suppose that the outgoing edge $e^u$ from $u$ in $T^{\textnormal{opt}}_j$ goes to $v$ while the outgoing edge $e^v$ from $v$ in $T^{\textnormal{opt}}_j$ goes to a third vertex $y$; these are the only possible values of $e_j$ with a potential to block $e_t$. All of $e_t, e^u, e^v$ are depicted. Additionally depicted is an edge $e_i$ outgoing from the endpoint of $e_j$ selected as $b$ -- this edge being presented at step $i$ would block $e_j$ from being taken. Each of the four images depicts one equally likely possibility for the random choice of $(a, b)$ in the case of $e_t$ as well as in the case of $e_j$, where $e_j$ is assumed to be the edge outgoing from $b$ in $T^{\textnormal{opt}}_j$ (as this is the only case relevant for an increased probability arising from $e_j$ not being taken). Only in one case is $e_t$ guaranteed to be taken (assuming steps other than those depicted do not block $e_t$ from being taken).

Theorems & Definitions (65)

• Theorem 1
• Theorem 2
• Theorem 3
• Theorem 4
• Lemma 5
• proof
• Definition 6
• Definition 7
• Definition 8
• Lemma 9