The Bidirected Cut Relaxation for Steiner Tree has Integrality Gap Smaller than 2

TL;DR

This work studies the Steiner tree problem through the Bidirected Cut Relaxation (BCR) and proves an upper bound of $1.9988$ on its integrality gap, improving upon the longstanding bound of $2$. The authors develop a novel scale-or-contract framework that blends a primal-dual perspective with component contractions, and crucially construct dual solutions to BCR that may be non-laminar, a departure from classical primal-dual methods. A key technical tool is the notion of locally $\gamma$-MST-optimal instances, which enables either the identification of an improving component or the guarantee that the terminal MST cost remains closely tied to the BCR value, yielding the improved bound. The approach also involves an explicit dual-growth procedure on a subdivided graph to ensure feasibility, and it yields a polynomial-time procedure that could be turned into an algorithm by approximating the contracting step. Overall, the results offer a scalable path toward near-2 approximations for Steiner tree and have potential implications for prize-collecting Steiner tree and related network-design problems.

Abstract

The Steiner tree problem is one of the most prominent problems in network design. Given an edge-weighted undirected graph and a subset of the vertices, called terminals, the task is to compute a minimum-weight tree containing all terminals (and possibly further vertices). The best-known approximation algorithms for Steiner tree involve enumeration of a (polynomial but) very large number of candidate components and are therefore slow in practice. A promising ingredient for the design of fast and accurate approximation algorithms for Steiner tree is the bidirected cut relaxation (BCR): bidirect all edges, choose an arbitrary terminal as a root, and enforce that each cut containing some terminal but not the root has one unit of fractional edges leaving it. BCR is known to be integral in the spanning tree case [Edmonds'67], i.e., when all the vertices are terminals. For general instances, however, it was not even known whether the integrality gap of BCR is better than the integrality gap of the natural undirected relaxation, which is exactly 2. We resolve this question by proving an upper bound of 1.9988 on the integrality gap of BCR.

Figures (13)

• Figure 1: An instance of the Steiner tree problem that arises from a minimum-cost spanning tree problem by subdividing edges. Terminals are shown as squares and Steiner nodes as circles; all edges have cost $1$. On the left, we see in blue the dual solution $\overline{y}$ resulting from the dual solution $y$ computed by the primal-dual algorithm by omitting the set containing the root $r$. Scaling up $\overline{y}$ by a factor of $2$ does not yield a feasible dual solution to \ref{['eq:bcr-tree']} because the constraints corresponding to outgoing edges of the two terminals in $R\setminus \{r\}$ are already tight with respect to $\overline{y}$. On the right, we see in blue the dual solution $z$ resulting from our dual growth process for \ref{['eq:bcr-tree']}. Scaling up $z$ by a factor of $2$ yields an optimum dual solution for \ref{['eq:bcr-tree']}. Notice that our construction does not provide a dual solution with laminar support.
• Figure 2: The left part of the figure shows a known instance of the Steiner tree problem where \ref{['eq:bcr-tree']} is not integral; see e.g. Vicari20. The numbers next to the edges show the edge costs and terminals are shown as squares. A terminal MST (in the metric closure of the instance) has cost $8$ and this is also the cost of an optimum Steiner tree. On the right we see the dual solution $(1+\delta)z$ that our dual growth procedure yields for $\delta=\frac{7}{8}$, which in this example is an optimum dual solution. In this example, for every time $t\in [0, t_{\max}) = [0,2)$, the partition $\mathcal{S}^t$ consists of all singleton sets $\{s\}$ with $s\in R$. The blue sets are sets $U_{\{s_1\}}$ (at different times of the algorithm) and the green ones sets $U_{\{s_2\}}$. For $\delta > \frac{7}{8}$, the sets $U_{\{s_1\}}$ and $U_{\{s_2\}}$ would for some time $t < t_{\max}$ contain the root vertex $r$ .
• Figure 3: Consider the Steiner tree instance $(G,R)$ that arises as the metric closure of the depicted graph where all shown edges have cost $1$. The terminal set $R$ is shown by squares. Then $\mathrm{opt}=k+q+1$ and $\mathrm{mst}(G[R]) = 2k + q$. For any $\gamma > 0$, we can choose $q$ large enough compared to $k$ to obtain $\mathrm{mst}(G[R]) < (1+\gamma) \cdot \mathrm{opt}$. However, if we run our dual growth procedure for any value of $\delta > \frac{1}{k}$, we would grow dual variables corresponding to sets $U_S$ containing the root $r$ (for $S=\{s_i\}$ with $i\in \{1,\dots,k\}$).
• Figure 4: Consider the Steiner tree instance $(G,R)$ that arises as the metric closure of the left depicted graph where all shown edges have cost $1$. The terminal set $R$ is shown by squares. Then $\mathrm{opt}=k+2q$ and $\mathrm{mst}(G[R]) = 2k + 2q$. For any $\gamma > 0$, we can choose $q$ large enough compared to $k$ to obtain $\mathrm{mst}(G[R]) < (1+\gamma) \cdot \mathrm{opt}$. However, if we run our dual growth procedure for any value of $\delta > \frac{1}{k+1}$, we would grow dual variables corresponding to sets $U_S$ containing the root $r$. Note that here, for any time $t \ge \frac{k+2}{(k+1) \cdot (1+\delta)}$, even all dual variables $U_S$ we grow would contain the root $r$. Thus, if we stopped growing those dual variables containing the root (or simply set them to zero at the end) to ensure feasibility, then the dual solution we obtain is not sufficiently good to prove an integrality gap below two. (We obtain a lower bound of $\frac{k+2}{k+1}\cdot (k+q)$ on the value of \ref{['eq:bcr-tree']} and can choose $k$ and $\frac{q}{k}$ arbitrarily large.) We address this issue by first identifying an improving component to contract and only then applying our dual growth procedure. For example, we might consider a component $K$ with vertex set $\{s_1, v_1, r, \tilde{s}_1, \dots, \tilde{s}_k\}$. Then contracting the set $X$ of terminals connected by $K$, we obtain the (metric closure of) the instance shown on the right.
• Figure 5: The top part (a) of the figure shows the terminal sets that contributed to an $S$-tight path $P$ that starts at a terminal $s\in S$. The sets $S$ with $s\in S$ are shown in black and dashed. All other sets contributing to $P$ can be partitioned into chains $\mathcal{C}^1$, $\mathcal{C}^2$, and $\mathcal{C}^3$, where any two sets belonging two different chains $\mathcal{C}^i$ are disjoint. The bottom part (b) of the figure illustrates the path $P$ with subpaths $P_1$, $P_2$, and $P_3$. The sets from a chain $\mathcal{C}^i$ contribute only on edges of $P_i$. The sets containing the terminal $s$ can contribute to any edge of $P$ (including the edges of $P_1$, $P_2$, and $P_3$).

Theorems & Definitions (55)

• Theorem 1.1
• Definition 2.1
• Definition 2.2
• Theorem 2.3
• Theorem 3.1
• proof
• Definition 3.1
• Lemma 3.1
• Definition 4.1
• Definition 4.5