Linear-in-$\varDelta$ lower bounds in the LOCAL model

Abstract

By prior work, there is a distributed graph algorithm that finds a maximal fractional matching (maximal edge packing) in $O(\varDelta )$ rounds, independently of $n$; here $\varDelta$ is the maximum degree of the graph and $n$ is the number of nodes in the graph. We show that this is optimal: there is no distributed algorithm that finds a maximal fractional matching in $o(\varDelta )$ rounds, independently of $n$. Our work gives the first linear-in-$\varDelta$ lower bound for a natural graph problem in the standard $\mathsf{LOCAL }$ model of distributed computing—prior lower bounds for a wide range of graph problems have been at best logarithmic in $\varDelta$.

Appendices

Appendix A: Derandomisation

As discussed in Sect. 5.2, unbounded outputs require special care. In this appendix we note that even though Naor and Stockmeyer [25] assume bounded outputs, their result on derandomising local algorithms applies in our setting, too.

Recall that a randomised algorithm is an $\mathsf{ID }$-algorithm such that each node has in addition access to a source of random bits. Let $\mathcal {A}$ be a randomised $t(\varDelta )$-time algorithm that computes a maximal FMon graphs of maximum degree $\varDelta$ or possibly fails with some small probability. Given an assignment of random bit strings $\rho :V(G)\rightarrow \{0,1\}^*$ to the nodes of a graph $G$, denote by $\mathcal {A}^\rho$ the deterministic algorithm that computes as $\mathcal {A}$, but uses $\rho$ for randomness.

The proof of Theorem 5.1 in [25] is using the following fact whose proof we reproduce here for convenience.

Lemma 6

For every $n$, there is an $n$-set $S_n\subseteq \mathbb {N}$ of identifiers and an assignment $\rho _n:S_n\rightarrow \{0,1\}^*$ such that $\mathcal {A}^{\rho _n}$ is correct on all graphs that have identifiers from $S_n$.

Proof

Denote by $k=k(n)$ the number of graphs $G$ with $V(G)\subseteq \{1,\ldots ,n\}$. Let $X_1,\ldots ,X_q\subseteq \mathbb {N}$ be any $q$ disjoint sets of size $n$. Suppose for the sake of contradiction that the claim is false for each $X_i$. That is, for any assignment $\rho :X_i\rightarrow \{0,1\}^*$ of random bits, $\mathcal {A}^\rho$ fails on at least one of the $k$ many graphs $G$ with $V(G)\subseteq X_i$. By averaging, this implies that for each $i$ there is a particular graph $G_i$, $V(G_i)\subseteq X_i$, on which $\mathcal {A}$ fails with probability at least $1/k$. Consider the graph $G$ that is the disjoint union of the graphs $G_1,\ldots ,G_q$. Since $\mathcal {A}$ fails independently on each of the components $G_i$, the failure probability on $G$ is at least $1-(1-1/k)^q$. But this probability can be made arbitrarily close to $1$ by choosing a large enough $q$, which contradicts the correctness of $\mathcal {A}$. $\square$

The deterministic algorithms $\mathcal {A}^{\rho _n}$ allow us to again obtain a $t(\varDelta )$-time $\mathsf{OI }$-algorithm, which establishes the $\varOmega (\varDelta )$ lower bound for $\mathcal {A}$. Only small modifications to Sect. 5.4 are needed:

• Step (i). Instead of the infinite set $I\subseteq \mathbb {N}$ as previously provided by Lemma 3, we can use the finite Ramsey’s theorem to find arbitrarily large sets $I_n\subseteq S_n$ (i.e., $|I_n|\rightarrow \infty$ as $n\rightarrow \infty$) with the property that $\mathcal {A}^{\rho _n}$ fully saturates the nodes of a loopy $\mathsf{OI }$-neighbourhood that has identifiers from $I_n$ (Lemma 4).

• Step (ii). Then, passing again to sufficiently sparse subsets $J_n\subseteq I_n$, we can reprove Lemma 5 and Corollary 1, which only require that $J$ is large enough.

This concludes the lower bound proof for randomised $\mathsf{LOCAL }$ algorithms.

Appendix B: Combinatorial proof of Lemma 2

In $T$ there is a unique simple directed path $x\!\rightsquigarrow \!y$ between any two nodes $x,y\in V(T)$. We use $V(x\!\rightsquigarrow \!y)$ and $E(x\!\rightsquigarrow \!y)$ to denote the nodes and edges of the path. Also, we set $V_\mathsf{in }(x\!\rightsquigarrow \!y) := V(x\!\rightsquigarrow \!y){\setminus } \{x,y\}$. We will assign to each path $x\!\rightsquigarrow \!y$ an integer value, denoted $[\![x\!\rightsquigarrow \!y]\!]$, which will determine the relative order of the endpoints.

By definition, in the $\mathsf{PO }$ model, we are given the following linear orders:

• Each node $v \in V(T)$ has a linear order $\prec _v$ on its incident edges.

• Each edge $e \in E(T)$ has a linear order $\prec _e$ on its incident nodes.

For notational convenience, we extend these relations a little: for $v \in V_\mathsf{in }(x\!\rightsquigarrow \!y)$ we define $x\prec _v y \iff e\prec _v e'$, where $e$ is the last edge on the path $x\!\rightsquigarrow \!v$ and $e'$ is the first edge on the path $v\!\rightsquigarrow \!y$; similarly, for $e\in E(x\!\rightsquigarrow \!y)$, we define $x\prec _e y\iff x'\prec _e y'$, where $e=\{x',y'\}$ and $x'$ and $y'$ appear on the path $x\!\rightsquigarrow \!y$ in this order.

For any statement $P$, we will use the following type of Iverson bracket notation:

$$\begin{aligned}{}[P] := {\left\{ \begin{array}{ll} +1&{} \text {if }P \text { is true}, \\ -1&{} \text {if }P \text { is false}. \end{array}\right. } \end{aligned}$$

We can now define

$$\begin{aligned}{}[\![x\!\rightsquigarrow \!y]\!]\, := \sum _{e\in E(x\rightsquigarrow y)} [x\prec _e y]\quad +\ \sum _{v\in V_\mathsf{in }(x\rightsquigarrow y)} [x\prec _v y]. \end{aligned}$$

(6)

In particular, $[\![x\!\rightsquigarrow \!x]\!]= 0$. The linear order $\prec$ on $V(T)$ is now defined by setting

$$\begin{aligned} x \prec y \iff [\![x\!\rightsquigarrow \!y ]\!]> 0. \end{aligned}$$

See Fig. 12. Next, we show that this is indeed a linear order.

Fig. 12

In this example, $[\![u\!\rightsquigarrow \!v ]\!]= +1$, $[\![v\!\rightsquigarrow \!u ]\!]= -1$, and hence $u \prec v$

Antisymmetry and totality. Since $[x\prec _v y] = -[y\prec _v x]$ and $[x\prec _e y] = -[y\prec _e x]$, we have the property that

$$\begin{aligned}{}[\![x\!\rightsquigarrow \!y]\!]= -[\![y\!\rightsquigarrow \!x]\!]. \end{aligned}$$

Moreover, if $x\ne y$, the first sum in (6) is odd iff the second sum in (6) is even. Therefore $[\![x\!\rightsquigarrow \!y]\!]$ is always odd; in particular, it is non-zero. These properties establish that either $x \prec y$ or $y \prec x$ (but never both).

Transitivity. Let $x,y,z \in V(T)$ be three distinct nodes with $x \prec y$ and $y \prec z$; we need to show that $x\prec z$. Denote by $v \in V(T)$ the unique node in the intersection of the paths $x\!\rightsquigarrow \!z$, $z\!\rightsquigarrow \!y$, and $y\!\rightsquigarrow \!x$.

Viewing the path $x\!\rightsquigarrow \!z$ piecewise as $x\!\rightsquigarrow \!v \!\rightsquigarrow \!z$ we write

$$\begin{aligned}{}[\![x\!\rightsquigarrow \!z]\!]= [\![x\!\rightsquigarrow \!v]\!]+ [x\prec _v z] + [\![v\!\rightsquigarrow \!z]\!], \end{aligned}$$

where it is understood that $[x\prec _v z] := 0$ in the degenerate cases where $v\in \{x,z\}$. Similar decompositions can be written for $z\!\rightsquigarrow \!y$ and $y\!\rightsquigarrow \!x$. Indeed, it is easily checked that

$$\begin{aligned}&[\![x\!\rightsquigarrow \!z]\!]+ [\![z\!\rightsquigarrow \!y]\!]+ [\![y\!\rightsquigarrow \!x]\!]\\&\quad = [x\prec _v z] + [z\prec _v y] + [y\prec _v x]. \end{aligned}$$

By assumption, $[\![z\!\rightsquigarrow \!y]\!], [\![y\!\rightsquigarrow \!x]\!]\le -1$, so we get

$$\begin{aligned}{}[\![x\!\rightsquigarrow \!z]\!]\ge 2 + [x\prec _v z] + [z\prec _v y] + [y\prec _v x]. \end{aligned}$$

The only way the right hand side can be negative is if

$$\begin{aligned} = [z\prec _v y]= [y\prec _v x] = -1,\end{aligned}$$

but this is equivalent to having $z\prec _v x \prec _v y \prec _v z$, which is impossible. Hence $[\![x\!\rightsquigarrow \!z]\!]\ge 0$. But since $x\ne z$ we must have in fact that $x \prec z$.