Single-source shortest paths in the CONGEST model with improved bounds

Abstract

Computing shortest paths from a single source is one of the central problems studied in the CONGEST model of distributed computing. After many years in which no algorithmic progress was made, Elkin [STOC ‘17] provided the first improvement over the distributed Bellman-Ford algorithm. Since then, several improved algorithms have been published. The state-of-the-art algorithm for weighted directed graphs (with polynomially bounded non-negative integer weights) requires $\tilde{O}(\min \{\sqrt{n}D^{1/2} ,\sqrt{n}D^{1/4} + n^{3/5} + D\})$ rounds [Forster and Nanongkai, FOCS ‘18], which is still quite far from the known lower bound of $\tilde{\Omega }(\sqrt{n} + D)$ rounds [Elkin, STOC ‘04]; here D is the diameter of the underlying network and n is the number of vertices in it. For the $(1+o(1))$-approximate version of this problem and the same class of graphs, Forster and Nanongkai [FOCS ‘18] obtained a better upper bound of $\tilde{O}(\sqrt{n}D^{1/4} + D)$ rounds. In the same paper, they stated that achieving the same bound for the exact case remains a major open problem. In this paper we resolve the above mentioned problem by devising a new randomized algorithm for computing shortest paths from a single source in $\tilde{O}(\sqrt{n}D^{1/4} + D)$ rounds. Our algorithm is based on a novel weight-modifying technique that allows us to compute approximate distances that preserve a certain form of the triangle inequality for the edges in the graph.

A missing proofs and procedures

Lemma A.1

For every $v \in V$, we must have $d(s,v,H_v) \ge d(s,v,G)$.

Proof

Let $v \in V$. It is sufficient to show that $w_v(x,y) \ge d(x,y,G)$ holds for every edge $(x,y) \in E(H_v)$ as then every path from s to v in $H_v$ has weight at least d(s, v, G) which clearly implies that $d(s,v,H_v) \ge d(s,v,G)$. Consider such an edge $(x,y) \in E(H_v)$. Note that, by definition, we have $w_v(x,y) = d_3(x,y) = \min \big (\{d_1(x,y)\} \cup \{d_2(x,z) + d_1(z,y) \mid z \in S\}\big )$. Observe that by construction we have that $d_1(x,y) \ge d(x,y,G)$ and $d_1(z,y) \ge d(z,y,G)$ for every $z \in S$. Therefore, to conclude that $w_v(x,y) \ge d(x,y,G)$ we only have to show that $d_2(x,z) \ge d(x,z,G)$ for every $z \in S$. To this end, note that $d_2(x,z) \ge d(x,z,G_{sk})$ for every $z \in S$, and so it is enough to show that $d(x,z,G_{sk}) \ge d(x,z,G)$ for every $z \in S$. This clearly holds as the weight of each edge $(x',y')$ in $G_{sk}$ is at least $d_1(x',y')$ which at least $d(x',y',G)$. $\square$

Lemma A.2

For every $x' \in S'$ and $v \in V$ such that $d(x',v,G) \le 2\lceil n / D^{1/2} \rceil$, the probability that $d_3(x',v) \le d(x',v,G)$ is at least $1 - 1/n^6$.

Proof

Let $x' \in S'$ and $v \in V$ be such that $d(x',v,G) \le 2\lceil n / D^{1/2} \rceil$. If we actually have $d(x',v,G) \le 3\lceil \sqrt{n}D^{1/4} \rceil = h_1$, then $d_1(x',v) = d(x',v,G)$ and so $d_3(x',v) \le d_1(x',v) = d(x',v,G)$ always holds.

Otherwise, we have $d(x',v,G) > 3\lceil \sqrt{n}D^{1/4} \rceil$. Let $\pi$ be some shortest path from $x'$ to v in G. Partition $\pi$ into k consecutive subpaths $\pi _1,\ldots ,\pi _k$ such that each of the subpaths $\pi _1,\ldots ,\pi _{k-1}$ contains exactly $\lceil \sqrt{n}D^{1/4} \rceil$ vertices and $\pi _k$ contains at most $\lceil \sqrt{n}D^{1/4}\rceil$ vertices (note that $k \ge 4$).

We first show that if $V(\pi _i) \cap S \ne \varnothing$ holds for every $i \in \{2,\ldots ,k-1\}$, then $d_3(x',v) \le d(x',v,G)$. Let $m = \lfloor k/2 - 1\rfloor$. Set $x_0 = x'$ and for each $i \in \{1,\ldots ,m\}$ let $x_i$ be some vertex in $V(\pi _{2i+1}) \cap S$. Note that for each $i \in \{0,\ldots ,m-1\}$, the subpath of $\pi$ that goes from $x_i$ to $x_{i+1}$ contains the subpath $\pi _{2i+2}$ (as $x_i$ belongs to $\pi _{2i+1}$ and $x_{i+1}$ to $\pi _{2i+3}$) and so it must contain at least $\lceil \sqrt{n}D^{1/4} \rceil$ edges. Moreover, this subpath is also a subpath of $\pi _{2i+1} \circ \pi _{2i+2} \circ \pi _{2i+3}$ and so it can contain at most $3\lceil \sqrt{n}D^{1/4} \rceil$ edges. We conclude that $\lceil \sqrt{n}D^{1/4} \rceil \le d(x_i, x_{i+1},G) \le 3\lceil \sqrt{n}D^{1/4} \rceil = h_1$ holds for every $i \in \{0,\ldots ,m-1\}$, and so also that $w_{G_{sk}}(x_i,x_{i+1}) = d_1(x_i,x_{i+1}) = d(x_i,x_{i+1},G)$. It follows that

$$\begin{aligned} d(x',x_m,G_{sk})= & {} d(x_0,x_m,G_{sk}) \\\le & {} w_{G_{sk}}(x_0,\ldots ,x_m) = \sum \nolimits _{i=0}^{m-1}{d_1(x_i,x_{i+1})} \\= & {} \sum \nolimits _{i=0}^{m-1}{d(x_i,x_{i+1},G)} \\= & {} d(x_0,x_m,G) = d(x',x_m,G) \end{aligned}$$

Since $d(x',x_m,G_{sk}) \le d(x',x_m,G) \le d(x',v,G) \le 2\lceil n / D^{1/2} \rceil$, we get that $d_2(x',x_m) = d(x',x_m,G_{sk})$ and so $d_2(x',x_m) \le d(x',x_m,G)$. As $2m + 1 \ge k-2$, it must be that the subpath of $\pi$ that goes from $x_m$ to v is also a subpath of $\pi _{k-2} \circ \pi _{k-1} \circ \pi _{k}$ and so $d(x_m,v,G) \le 3\lceil \sqrt{n}D^{1/4}\rceil$. It follows that $d_1(x_m,v) = d(x_m,v,G)$, and so $d_3(x',v) \le d_2(x',x_m) + d_1(x_m,v) \le d(x',x_m,G) + d(x_m,v,G) = d(x',v,G)$.

We claim that with probability at most $1/n^{6}$ there exists an $i \in \{2,\ldots ,k-1\}$ for which $V(\pi _i) \cap S = \varnothing$. Indeed, for a given $i \in \{2,\ldots ,k-1\}$, the probability that $V(\pi _i) \cap S = \varnothing$ is

$\left( 1 - \frac{\alpha }{n} \right) ^{|V(\pi _i) |} = \left( 1 - \frac{7\log _2{n}}{\sqrt{n}D^{1/4}} \right) ^{\lceil \sqrt{n}D^{1/4} \rceil } \le e^{-7\log _2{n}} \le n^{-7}$

It follows, by the union bound, that the probability that there exists such an i is at most $k / n^7 \le 1 / n^6$. $\square$

Lemma A.3

For every $v \in V$, the probability that $d(s,v,H_v) \le d(s,v,G)$ is at least $1- 1/n^4$.

Proof

Let $v \in V$. If $d(s,v,G) = \infty$, then $d(s,v,H_v) \le d(s,v,G)$ always holds. If $d(s,v,G) \le 2\lceil n / D^{1/2} \rceil$, then, by Lemma A.2, we have $d_3(s,v) \le d(s,v,G)$ with probability at least $1 - 1/n^6 \ge 1 - 1/n^4$. As $w_v(s,v) = d_3(s,v)$, we get that $d(s,v,H_v) \le w_v(s,v) = d_3(s,v)$, and so $d(s,v,H_v) \le d(s,v,G)$ holds with probability at least $1 - 1/n^4$.

We are left with the case $\infty> d(s,v,G) > 2\lceil n / D^{1/2} \rceil$. Let $\pi$ be some shortest path from s to v in G. Partition $\pi$ into k consecutive subpaths $\pi _1,\ldots ,\pi _k$ such that each of the subpaths $\pi _1,\ldots ,\pi _{k-1}$ contains exactly $\lceil n / D^{1/2} \rceil$ vertices and $\pi _k$ contains at most $\lceil n / D^{1/2} \rceil$ vertices (note that $k \ge 3$).

We first show that if $V(\pi _i) \cap S' \ne \varnothing$ holds for every $i \in \{2,\ldots ,k-1\}$, then $d(s,v,H_v) \le d(s,v,G)$ holds with probability at least $1-1/n^5$. Set $x_1 = s$, $x_k = v$ and for each $i \in \{2,\ldots ,k-1\}$ let $x_i$ be some vertex in $V(\pi _{i}) \cap S'$. Note that for each $i \in \{1,\ldots ,k-1\}$, the subpath of $\pi$ that goes from $x_i$ to $x_{i+1}$ is also a subpath of $\pi _{i} \circ \pi _{i+1}$ and so contains at most $2\lceil n / D^{1/2} \rceil$ edges. We get that $d(x_i,x_{i+1},G) \le 2\lceil n / D^{1/2} \rceil$ for all $i \in \{1,\ldots ,k-1\}$.

By Lemma A.2, the probability that $d_3(x_i,x_{i+1}) > d(x_i,x_{i+1},G)$ holds for a given $i \in \{1,\ldots ,k-1\}$ is at most $1/n^6$. We get, by applying the union bound, that with probability at most $1/n^5$ there exists an $i \in \{1,\ldots ,k-1\}$ for which $d_3(x_i,x_{i+1}) > d(x_i,x_{i+1},G)$. We conclude that with probability at least $1 - 1/n^5$ we have $d_3(x_i,x_{i+1}) \le d(x_i,x_{i+1},G)$ for all $i \in \{1,\ldots ,k-1\}$. Given that this event has occurred, we must have

$$\begin{aligned} d(s,v,H_v)= & {} d(x_0,x_k,H_v) \\\le & {} w_v(x_0,\ldots ,x_k) = \sum \nolimits _{i=0}^{k-1}{d_3(x_i,x_{i+1})} \\\le & {} \sum \nolimits _{i=0}^{k-1}{d(x_i,x_{i+1},G)} = d(x_0,x_k,G) \\= & {} d(s,v,G) \end{aligned}$$

It follows that $d(s,v,H_v) \le d(s,v,G)$ holds with probability at least $1 - 1/n^5$.

We claim that with probability at most $1/n^{5}$ there exists an $i \in \{2,\ldots ,k-1\}$ for which $V(\pi _i) \cap S' = \varnothing$. Indeed, for a given $i \in \{2,\ldots ,k-1\}$, the probability that $V(\pi _i) \cap S' = \varnothing$ is

$$\begin{aligned} \left( 1- pp'\right) ^{|V(\pi _i) |}= & {} \left( 1 - \frac{\alpha }{n} \frac{\alpha '}{6\alpha } \right) ^{|V(\pi _i) |}\\= & {} \left( 1 - \frac{6\sqrt{D} \log {n}}{n} \right) ^{\lceil n / D^{1/2} \rceil }\\\le & {} 2^{-6\log {n}} = n^{-6} \end{aligned}$$

It follows, by the union bound, that the probability that there exists such an i is at most $k / n^6 \le 1 / n^5$. Now, the probability that we have $d(s,v,H_v) \le d(s,v,G)$ is at least $(1-1/n^5)^2 \ge 1 - 1/n^4$. $\square$

Proof of Claim 4.2

Let $x \in V {\setminus } \{v\}$ and assume that $\hat{d}(v,x) \ne \infty$. For every $i \in \{0,\ldots ,\beta +1\}$, let $q_i$ be the path that is obtained from $p_{(0,x)}$ after i unfolding steps (that is, $q_i$ is the value of the path p in the Procedure UnfoldPath after applying i iterations of the loop). We will show by induction on i that the path $q_i$ must be simple for every $i \in \{0,\ldots ,\beta +1\}$.

In the base case ($i = 0$), we have $q_0 = p_{(0,x)}$, and so, by construction, $q_0$ must be a simple path unless it starts with a shortcut edge and v occurs more than once in that path. But, the latter can never be the case, as this would imply that $p_{(0,x)}$ is not the path in $P_{(0,x)}$ with the smallest possible weight. Indeed, in such a case, the subpath of $q_0$ that goes from the second occurrence of v to x must also belong to $P_{(0,x)}$, and so the smallest possible weight of a path in $P_{(0,x)}$ is 0, whereas the weight of the path $p_{(0,x)}$ is $> 0$ as it contains a shortcut edge.

Assume now that the claim holds for some $i \in \{0,\ldots ,\beta \}$, and prove it for $i+1$. First, note that if the first edge of $q_i$ is not a shortcut edge then the claim is trivial as $q_i$ and $q_{i+1}$ are the same path. So assume that $q_i$ starts with a shortcut edge e that goes from v to u, and let $(x_1,\ldots ,x_m)$ be the sequence of vertices that the path $q_{i}$ traverses (note that $x_1 = v$, $x_2 = u$ and $x_m = x$). Recall that $q_{i+1}$ is obtained from $q_{i}$ by replacing the edge e with the path $p_{(i+1,u)}$. Let $(y_1,\ldots ,y_{m'})$ be the sequence of vertices that the path $p_{(i+1,u)}$ traverses (and, again, note that $y_1 = v$ and $y_{m'} = u$). It follows that the sequence of vertices that the path $q_{i+1}$ traverses is $(y_1,\ldots ,y_{m'-1},x_{2},\ldots ,x_{m})$.

Now, by the induction hypothesis, we know that $q_i$ is a simple path, and so we must have $x_{i_1} \ne x_{i_2}$ for every two different indices $i_1$ and $i_2$ in $\{2,\ldots ,m\}$. It is also very easy to see that $p_{(i+1,u)}$ must be a simple path (as otherwise it would not be a path with the smallest weight in $P_{(i+1,u)}$), thus we must also have $y_{i_1} \ne y_{i_2}$ for every two different indices $i_1$ and $i_2$ in $\{1,\ldots ,m'-1\}$. So to conclude that $q_{i+1}$ is simple, we are left to show that $y_{i_1} \ne x_{i_2}$ for every $i_1 \in \{1,\ldots ,m'-1\}$ and $i_2 \in \{2,\ldots ,m\}$. Let $i_1$ and $i_2$ be two such indices, and assume towards a contradiction that $y_{i_1} = x_{i_2}$. If $i_1 = 1$, then we must have $x_{i_2} = y_{1} = v$ and so v appears more then once in $q_i$ which is a contradiction to the assumption that $q_i$ is simple. Otherwise, $i_1 > 1$. Let $p'$ be the subpath of $p_{(i+1,u)}$ that goes from $y_1$ to $y_{i_1}$. It is easy to see that $w_{H_{i+1}}(p') < w_{H_{i+1}}(p_{(i+1,u)})$ and $p' \in P_{(i+1,y_{i_1})}$. Therefore, $H_i$ must contain a shortcut edge $e'$ from $y_1 = v$ to $y_{i_1} = x_{i_2}$ of weight $\le w_{H_{i+1}(p')} < w_{H_{i+1}}(p_{(i+1,u)})$. Let $p''$ be the subpath of $q_i$ that goes from $x_{i_2}$ to $x_{m} = x$. The path $(e') \circ p''$ exists in $H_i$. It clearly belongs to $P_{(i,x)}$, but its weight in $H_i$ is smaller than that of $q_i$ which is impossible. $\square$

Proof of Lemma 4.5

Let $p = (x_1,\ldots ,x_m)$ be the unfolded path of x from w and let $(z_\beta ,\ldots ,z_1,z_0)$ be the associated vertices. Let $x_i$ be some vertex on this path which is different from x and w. Let $z \in \{z_\beta ,\ldots ,z_1,z_0\}$ be the closest vertex to $x_i$ in p among the vertices in $\{x_1,\ldots ,x_i\}$. At the ith iteration, the algorithm (that starts from w) detects the subpath of p from z to $x_i$ and so we have $\hat{d}(w,x_i) \le d_{i+1}(w,z_i) + w_{H_i}(z_i,\ldots ,x_i)$. Now, let $z' \in \{z_\beta ,\ldots ,z_1,z_0\}$ be the closest vertex to $x_i$ in p among the vertices in $\{x_i,\ldots ,x_m\}$. In the i-th iteration of the algorithm that starts from $x_i$, the algorithm considers the subpath of p from $x_i$ to $z'$ and then continues as in the algorithm that starts from w. It follows that $\hat{d}(w,x_i) + \hat{d}(x_i,x) \le \hat{d}(w,x)$. $\square$