On Solving Reachability in Grid Digraphs using a Psuedoseparator

TL;DR

This paper introduces a polynomial-time, space-efficient algorithm for the reachability problem in grid digraphs, utilizing a novel pseudoseparator to improve space bounds to approximately O(n^{1/4}).

Contribution

It presents a new divide-and-conquer approach using pseudoseparators, achieving better space complexity for grid digraph reachability.

Findings

Achieves O(n^{1/4 + ε}) space complexity for reachability.

Introduces the concept of pseudoseparator for divide-and-conquer algorithms.

Provides a polynomial-time algorithm with improved space bounds.

Abstract

The reachability problem asks to decide if there exists a path from one vertex to another in a digraph. In a grid digraph, the vertices are the points of a two-dimensional square grid, and an edge can occur between a vertex and its immediate horizontal and vertical neighbors only. Asano and Doerr (CCCG'11) presented the first simultaneous time-space bound for reachability in grid digraphs by solving the problem in polynomial time and $O(n^{1/2 + \epsilon})$ space. In 2018, the space complexity was improved to $\tilde{O}(n^{1/3})$ by Ashida and Nakagawa (SoCG'18). In this paper, we show that there exists a polynomial-time algorithm that uses $O(n^{1/4 + \epsilon})$ space to solve the reachability problem in a grid digraph containing $n$ vertices. We define and construct a new separator-like device called pseudoseparator to develop a divide-and-conquer algorithm. This algorithm works in a space-efficient manner to solve reachability.

Equations

$$c_l(v)=\{\begin{array}{ll}(x+1,y)& \text{ if }x<(i+1)t\text{ and }y=jt\\ (x,y+1)& \text{ if }x=(i+1)t\text{ and }y<(j+1)t\\ (x-1,y)& \text{ if }x>it\text{ and }y=(j+1)t\\ (x,y-1)& \text{ if }x=it\text{ and }y>jt\end{array}$$

$$S(m,h)=\{\begin{array}{ll}S(m,h^{1-\beta })+\tilde{O}(h^{1/2+\beta /2})& h>m^{1/8}\\ \tilde{O}(h)& \text{otherwise}.\end{array}$$

$$T(m,h)=\{\begin{array}{ll}q\cdot T(h^{1-\beta })+p& h>m^{1/8}\\ O(h)& \text{otherwise}.\end{array}$$

$$S(\hat{m})=\{\begin{array}{ll}S({\hat{m}}^{1-\alpha })+\tilde{O}(({\hat{m}}^{1+\alpha }{)}^{1/2+\beta /2})& \hat{m}>m^{1/8}\\ \tilde{O}({\hat{m}}^{1/4})& \text{otherwise}\end{array}$$

$$T(\hat{m})=\{\begin{array}{ll}q(\hat{m})\cdot T({\hat{m}}^{1-\alpha })+p(\hat{m})& \hat{m}>m^{1/8}\\ O(\hat{m})& \text{otherwise}.\end{array}$$

Full text

Asano et al. presented the first simultaneous time space bound for reachability in grid graphs by presenting an algorithm that solves the problem in polynomial time and $O(n^{1/2+\epsilon})$ space. In 2018, the space bound was improved to $\tilde{O}(n^{1/3})$ by Ashida and Nakagawa.

In this paper, we show that reachability in an $n$ vertex grid graph can be decided by an algorithm using $O(n^{1/4+\epsilon})$ space and polynomial time simultaneously.

1 Introduction

The problem of graph reachability is to decide whether there is a path from one vertex to another in a given graph. This problem has several applications in the field of algorithms and computational complexity theory. Reachability in directed and undirected graphs capture the complexity of nondeterministic and deterministic logarithmic space, respectively [11]. It is often used as a subroutine in various network related problems. Hence designing better for this problem is of utmost importance to computer scientists.

Standard graph traversal algorithms such as DFS and BFS give a linear time algorithm for this problem, but they require linear space as well. Savitch’s divide and conquer based algorithm can solve reachability in $O(\log^{2}n)$ space, but as a tradeoff, it requires $n^{O(\log n)}$ time [12]. Hence it is natural to ask whether we can get the best of both worlds and design an algorithm for graph reachability that runs in polynomial time and uses polylogarithmic space. Wigderson asked a relaxed version of this question in his survey - whether graph reachability can be solved by an algorithm that runs simultaneously in polynomial time and uses $O(n^{1-\epsilon})$ space [14]. In this paper, we address this problem for certain restricted classes of directed graphs.

Barnes et al showed that reachability in general graphs can be decided in simultaneously in $n/2^{\Theta(\sqrt{\log n})}$ space and polynomial time [6]. Although this algorithm gives a sublinear space bound, it still does not give a positive answer to Wigderson’s question.

However, for certain topologically restricted graph classes, we know of better space bounds simultaneously with polynomial time. Planar graphs are graphs that can be drawn on the plane such that no two edges of the graph cross each other at an intermediate point. Imai et al. showed that reachability in planar graph can be solved in $O(n^{1/2+\epsilon})$ space for any $\epsilon>0$ [9] . Later this space bound was improved to $\tilde{O}(n^{1/2})$ [4]. For graphs of higher genus, Chakraborty et al. gave an $\tilde{O}(n^{2/3}g^{1/3})$ space algorithm which additionally requires an embedding of the graph on a surface of genus $g$, as input [7]. They also gave an $\tilde{O}(n^{2/3})$ space algorithm for $H$ minor-free graphs which requires tree decomposition of the graph as an input and $O(n^{1/2+\epsilon})$ space algorithm for $K_{3,3}$-free and $K_{5}$-free graphs. For layered planar graphs, Chakraborty and Tewari showed that for every $\epsilon>0$ there is an $O(n^{\epsilon})$ space algorithm [8]. Stolee and Vinodchandran gave a polynomial time algorithm that, for any $\epsilon>0$ solves reachability in a directed acyclic graph with $O(n^{\epsilon})$ sources and embedded on the surface of genus $O(n^{\epsilon})$ using $O(n^{\epsilon})$ space [13]. For unique-path graphs, Kannan et al. give $O(n^{\epsilon})$ space and polynomial time algorithm [10].

Grid graphs are a subclass of planar graphs whose vertices are present at the integer lattice points of an $m\times m$ grid graph and edges can only occur between a vertex and its immediate vertical or horizontal neighbor. It has known that reachability in planar graphs can be reduced to reachability in grid graphs in logarithmic space [1]. The reduction, however, causes a quadratic blow-up in size with respect to the input graph. In this paper, we study the simultaneous time-space complexity of reachability in grid graphs.

Asano and Doerr presented a polynomial time algorithm that uses $O(n^{1/2+\epsilon})$ space for solving reachability in grid graphs [3]. Ashida and Nakagawa presented an algorithm with improved space complexity of $\widetilde{O}(n^{1/3})$ [5]. The latter algorithm proceeded by first dividing the input grid graph into subgrids. It then used a gadget to transform each subgrid into a planar graph, making the whole of the resultant graph planar. Finally, it used the planar reachability algorithm of Imai et al. [9] as a subroutine to get the desired space bound.

In this paper, we present a $O(n^{1/4+\epsilon})$ space and polynomial time algorithm for grid graph reachability, thereby significantly improving the space bound of Ashida and Nakagawa.

Theorem 1** (Main Theorem).**

For every $\epsilon>0$, there exists a polynomial time algorithm that can solve reachability in an $n$ vertex grid graph, using $O(n^{1/4+\epsilon})$ space.

To solve the problem, we divide the given grid graph into subgrids and recursively solve them such that the subgrids are smaller by a factor of $n^{\epsilon}$ and the depth of recursion is constant. To combine solutions, we replace paths in each grid with a single edge to get an auxiliary graph. Although this reduces the size of the graph by a factor polynomial in $n$ and preserves reachability as well, the auxiliary graph no longer remains planar. Since we are using a divide and conquer strategy on this graph, we define and use a new type of graph separator of the auxiliary graph, that we call as pseudoseparator. This separator again has size significantly smaller than the size of the auxiliary graph.

In Section 2, we state the definitions and notations that we use in this paper. In Section 3, we define the auxiliary graph and state various properties of it that we use later. In Section 4, we discuss the concept of a pseudoseparator. We give its formal definition and show how a pseudoseparator can be computed efficiently. In Section 5, we give the algorithm to solve reachability in an auxiliary graph and prove its correctness. Finally in Section 6 we use the algorithm of Section 5 to give an algorithm to decide reachability in grid graphs and thus prove Theorem 1.

2 Preliminaries

Let $[n]$ denote the set $\{0,1,2,\ldots,n\}$. We denote the vertex set of a graph $G$ by $V(G)$ and its edge set by $E(G)$. For a subset $U$ of $V(G)$, we denote the subgraph of $G$ induced by the vertices of $U$ as $G[U]$. For a graph $G$, we denote $\textsf{cc}(G)$ as the set of all connected components in the underlying undirected graph of $G$, where the undirected version is obtained by removing orientations on all edges of $G$. Henceforth, whenever we talk about connected components, we will mean the connected components of the underlying undirected graph.

In a drawing of a graph on a plane, each vertex is mapped to a point on the plane, and each edge is mapped to a simple arc whose endpoints coincide with the mappings of the end vertices of the edge. Moreover, the interior points of an arc corresponding to an edge does not intersect with any other vertex points. A graph is said to be planar if it can be drawn on the plane such that no two edges of the graph intersect except possibly at the endpoints. We will not be concerned with details about the representation of planar graphs. We note that the work of [2], and subsequently [11], implies a deterministic logarithmic space algorithm that decides whether or not a given graph is planar, and if it is, outputs a planar embedding. Hence when dealing with planar graphs, we will assume without loss of generality that a planar embedding is provided as well.

A $m\times m$ grid graph is a directed graph whose vertices are $[m]\times[m]=\{0,1,\ldots,m\}\times\{0,1,\ldots,m\}$ so that if $((i_{1},j_{1}),(i_{2},j_{2}))$ is an edge then $|i_{1}-i_{2}|+|j_{1}-j_{2}|=1$. It follows from definition that grid graphs are a subset of planar graphs.

3 Auxiliary Graph

Let $G$ be an $m\times m$ grid graph. We divide $G$ into $m^{2\alpha}$ subgrids such that each subgrid is a $m^{1-\alpha}\times m^{1-\alpha}$ grid. Formally, for $1\leq i,j\leq m^{\alpha}$, the $(i,j)$-th subgrid of $G$, denoted as $G[i,j]$ is the subgraph of $G$ induced by the set of vertices, $V(G[i,j])=\{(i^{\prime},j^{\prime})\mid(i-1)\cdot m^{1-\alpha}\leq i^{\prime}\leq i\cdot m^{1-\alpha}\textrm{ and }(j-1)\cdot m^{1-\alpha}\leq j^{\prime}\leq j\cdot m^{1-\alpha}\}$.

For $0<\alpha<1$ and $1\leq i,j\leq m^{\alpha}$, we define $\textsf{Aux}_{\alpha}(G)[i,j]$ as follows. The vertex set of $\textsf{Aux}_{\alpha}(G)[i,j]$ is $V(\textsf{Aux}_{\alpha}(G)[i,j])=\{(i^{\prime},j^{\prime})\mid i^{\prime}=k\cdot m^{1-\alpha}\textrm{ or }j=l\cdot m^{1-\alpha},\textrm{ such that }k\in\{i-1,i\}\textrm{ and }l\in\{j-1,j\}\}$. For two vertices $u,v$ in $\textsf{Aux}_{\alpha}(G)[i,j]$, $(u,v)$ is an edge in $\textsf{Aux}_{\alpha}(G)[i,j]$ if there is a path from $u$ to $v$ in the subgrid $G[i,j]$. In a drawing of $\textsf{Aux}_{\alpha}(G)[i,j]$, we use a straight line to represent the edge if $u$ and $v$ do not lie on a single side of $\textsf{Aux}_{\alpha}(G)[i,j]$, and an arc present inside the grid to represent it otherwise.

Now for $0<\alpha<1$, we define the $\alpha$-auxiliary graph, $\textsf{Aux}_{\alpha}(G)$ as follows. The vertex set of $\textsf{Aux}_{\alpha}(G)$, $V(\textsf{Aux}_{\alpha}(G))=\{(i,j)\mid i=k\cdot m^{1-\alpha}\textrm{ or }j=l\cdot m^{1-\alpha},\textrm{ such that }0\leq k,l\leq m^{\alpha}\}$. The edges of $\textsf{Aux}_{\alpha}(G)$ are the edges of $\textsf{Aux}_{\alpha}(G)[i,j]$ taken over all pairs $(i,j)$. Note that $\textsf{Aux}_{\alpha}(G)$ might have parallel edges, since an edge on a side of a block might be present in the adjacent block as well. In such cases we preserve both the edges, however in their different blocks of $\textsf{Aux}_{\alpha}(G)$ in the drawing of $\textsf{Aux}_{\alpha}(G)$ on the plane. Figure 1 contains an example of a grid graph partitioned into subgrids and its corresponding auxiliary graph. Since each block $\textsf{Aux}_{\alpha}(G)[i,j]$ contains $4m^{1-\alpha}$ vertices, the total number of vertices in $\textsf{Aux}_{\alpha}(G)$ would be at most $4m^{1+\alpha}$.

Our algorithm for reachability first constructs $\textsf{Aux}_{\alpha}(G)$ by solving each of the $m^{1-\alpha}\times m^{1-\alpha}$ grids recursively. It then uses a polynomial time subroutine to decide reachability in $\textsf{Aux}_{\alpha}(G)$. Note that we do not store the graph $\textsf{Aux}_{\alpha}(G)$ explicitly, since that would require too much space. Rather we solve a subgrid recursively whenever the subroutine queries for an edge in that subgrid of $\textsf{Aux}_{\alpha}(G)$.

Our strategy is to develop a polynomial time algorithm which solves reachability in $\textsf{Aux}_{\alpha}(G)$ using $\tilde{O}(\tilde{m}^{1/2+\beta/2})$ space where $\tilde{m}$ is the number of vertices in $\textsf{Aux}_{\alpha}(G)$. As discussed earlier, $\tilde{m}$ can be at most $4m^{1+\alpha}$. Hence, the main algorithm would require $\tilde{O}(m^{1/2+\beta/2+\alpha/2+\alpha\beta/2})$ space. For a fixed constant $\epsilon>0$, we can pick $\alpha>0$ and $\beta>0$ such that the space complexity becomes $O(m^{1/2+\epsilon})$.

3.1 Properties of the Auxiliary Graph

In the following definition, we give an ordered labeling of the vertices on some block of the auxiliary graph. The labeling is defined with respect to some vertex present in the block.

Definition 1**.**

Let $G$ be a $m\times m$ grid graph, $l=\textsf{Aux}_{\alpha}(G)[i,j]$ be a block of $\textsf{Aux}_{\alpha}(G)$ and $v=(x,y)$ be a vertex in $\textsf{Aux}_{\alpha}(G)[i,j]$. Let $t=m^{1-\alpha}$. We define the counter-clockwise adjacent vertex of $v$ with respect to the block $l$, $c_{l}(v)$ as follows:

[TABLE]

Similarly we also define the $r$-th counter-clockwise adjacent neighbour of $v$ with respect to the block $l$ inductively as follows. For $r=0$, $c_{l}^{r}(v)=v$ and otherwise we have $c_{l}^{r+1}(v)=c_{l}(c_{l}^{r}(v))$.

Note that for a block $l$ and vertices $v$ and $w$ in it, we write $v$ as $c_{l}^{p}(w)$ where $p$ is smallest non-negative integer for which $c_{l}^{p}(w)=v$. Next we formalize what it means to say that two edges of the auxiliary graph cross each other.

Definition 2**.**

Let $G$ be a grid graph and $l$ be a block of $\textsf{Aux}_{\alpha}(G)$. For two distinct edges $e$ and $f$ in the block, such that $e=(v,c_{l}^{p}(v)$ and $f=(c_{l}^{q}(v),c_{l}^{r}(v))$. We say that edges $e$ and $f$ cross each other if $\min(q,r)<p<\max(q,r)$.

Note the definition of cross given above is symmetric. That is, if edges $e$ and $f$ cross each other then $f$ and $e$ must cross each other as well. For an edge $f=(c_{l}^{q}(v),c_{l}^{r}(v))$, we define $\overleftarrow{f}=(c_{l}^{r}(v),c_{l}^{q}(v))$ and call it the reverse of $f$. We also note that if $e$ and $f$ cross each other, then $e$ and $\overleftarrow{f}$ also cross each other.

In Lemma 1 we state an equivalent condition of crossing of two edges, and in Lemmas 2 and 3 we state certain properties of the auxiliary graph that we use later.

Lemma 1**.**

Let $G$ be a grid graph and $l$ be a block of $\textsf{Aux}_{\alpha}(G)$. Let $w$ be an arbitrary vertex in the block $l$ and $e=(c_{l}^{p}(w),c_{l}^{q}(w))$ and $f=(c_{l}^{r}(w),c_{l}^{s}(w))$ be two distinct edges in $l$. Then $e$ and $f$ cross each other if and only if either of the following two conditions hold:

• •

$\min(p,q)<\min(r,s)<\max(p,q)<\max(r,s)$**

• •

$\min(r,s)<\min(p,q)<\max(r,s)<\max(p,q)$**

Proof.

We prove that if $\min(p,q)<\min(r,s)<\max(p,q)<\max(r,s)$ then $e$ and $f$ cross each other. We let $p<r<q<s$. Other cases can be proved by reversing appropriate edges. We thus have integers $r_{1}=r-p$, $q_{1}=q-p$ and $s_{1}=s-p$. Clearly, $r_{1}<q_{1}<s_{1}$. Let $v=c_{l}^{p}(w)$. Thus we have $e=(v,c_{l}^{q}(w))=(v,c_{l}^{q_{1}}(c_{l}^{p}(w)))=(v,c_{l}^{q_{1}}(v))$ and $f=(c_{l}^{r}(w),c_{l}^{s}(w))=(c_{l}^{r_{1}}(c_{l}^{p}(w)),c_{l}^{s_{1}}(c_{l}^{p}(w)))=(c_{l}^{r_{1}}(v),c_{l}^{s_{1}}(v))$ The proof for the second condition is similar.

Now, we prove that if $e=(c_{l}^{p}(w),c_{l}^{q}(w))$ and $f=(c_{l}^{r}(w),c_{l}^{s}(w))$ cross each other then either of the given two condition holds. We assume that $p$ is the smallest integer among $p$, $q$, $r$ and $s$. Other cases can be proved similarly. Now, let $v=c_{l}^{p}(w)$. We thus have integers $q_{1}=q-p,r_{1}=r-p$ and $s_{1}=s-p$ such that $e=(v,c_{l}^{q_{1}}(v))$ and $f=(c_{l}^{r_{1}}(v),c_{l}^{s_{1}}(v))$. Since $e$ and $f$ cross each other, we have $\min(r_{1},s_{1})<q_{1}<\max(r_{1},s_{1})$. Thus $\min(r_{1}+p,s_{1}+p)<q_{1}+p<\max(r_{1}+p,s_{1}+p)$. It follows that $\min(r,s)<q<\max(r,s)$. Since we assumed $p$ to be smallest integer among $p$, $q$, $r$ and $s$; we have $\min(p,q)<\min(r,s)<\max(p,q)<\max(r,s)$, thus proving the lemma. ∎

Definition 3**.**

Let $G$ be a grid graph and $l$ be a block of $\textsf{Aux}_{\alpha}(G)$. For a vertex $v$ and edges $f$, $g$ such that $f=(c_{l}^{q}(v),c_{l}^{r}(v))$ and $g=(c_{l}^{s}(v),c_{l}^{t}(v))$, we say that $f$ is closer to $v$ than $g$ if $\min(q,r)<\min(s,t)$.

Lemma 2**.**

Let $G$ be a grid graph and $e_{1}=(u_{1},v_{1})$ and $e_{2}=(u_{2},v_{2})$ be two edges in $\textsf{Aux}_{\alpha}(G)$. If $e_{1}$ and $e_{2}$ cross each other, then $\textsf{Aux}_{\alpha}(G)$ also contains the edges $(u_{1},v_{2})$ and $(u_{2},v_{1})$.

Proof.

Let $e_{1}=(v,c_{l}^{p}(v))$ and $e_{2}=(c_{l}^{q}(v),c_{l}^{r}(v))$ be two edges that cross each other in $\textsf{Aux}_{\alpha}(G)$. Let $l$ be the block of $\textsf{Aux}_{\alpha}(G)$ to which $e_{1}$ and $e_{2}$ belong. Consider the subgrid of $G$ which is solved to construct the block $l$. Since the edge $e_{1}$ exists in block $l$, there exists a path $P$ from $v$ to $c_{l}^{p}(v)$ in the underlying subgrid. This path $P$ divides the subgrid into two parts such that the vertices $c_{l}^{q}(v)$ and $c_{l}^{r}(v)$ belong to different parts of the subgrid. Thus, a path between $c_{l}^{q}(v)$ and $c_{l}^{r}(v)$ necessarly take a vertex of path $P$. Hence, there is a path from $v$ to $c_{l}^{r}(v)$ and a path from $c_{l}^{p}(v)$ to $c_{l}^{r}(v)$. Thus the lemma follows. ∎

Lemma 3**.**

Let $G$ be a grid graph and $e_{1}=(u_{1},v_{1})$ and $e_{2}=(u_{2},v_{2})$ be two edges in $\textsf{Aux}_{\alpha}(G)$. If $e_{1}$ and $e_{2}$ cross a certain edge $f=(x,y)$, and $e_{1}$ is $closer$ to $x$ than $e_{2}$, then the edge $(u_{1},v_{2})$ is also present in $\textsf{Aux}_{\alpha}(G)$.

Proof.

Let $e=(v,c_{l}^{p}(v))$, $f=(c_{l}^{q}(v),c_{l}^{r}(v))$ and $g=(c_{l}^{s}(v),c_{l}^{t}(v))$. If $c_{l}^{q}(v)=c_{l}^{s}(v)$ then the lemma trivially follows. Otherwise, we have two cases to consider:

• Case 1 ($f$ crosses $g$): In this case, we will have $(c_{l}^{q}(v),c_{l}^{t}(v))$ present in $\textsf{Aux}_{\alpha}(G)$ by Lemma 2.
• Case 2 ($f$ does not cross $g$): In this case, we have $\min(q,r)<\min(s,t)<p<\max(s,t)<\max(q,r)$. Since $f$ crosses $e$, we have the edge $(c_{l}^{q}(v),c_{l}^{p}(v))$ in $\textsf{Aux}_{\alpha}(G)$ by Lemma 2. This edge will cross $g$. Hence $(c_{l}^{q}(v),c_{l}^{t}(v))$ is present in $\textsf{Aux}_{\alpha}(G)$.

∎

4 Pseudoseparator in a Grid Graph

Imai et al. used a separator construction to solve the reachability problem in planar graphs [9]. A separator is a small set of vertices whose removal disconnects the graph into smaller components. An essential property of a separator is that, for any two vertices, a path between the vertices must contain a separator vertex if the vertices lie in two different components with respect to the separator.

Grid graphs are subclasses of planar and are known to have good separators. However, for a grid graph $G$, the graph $\textsf{Aux}_{\alpha}(G)$ might not have a small separator. Here we show that $\textsf{Aux}_{\alpha}(G)$ has a different kind of separator, which we call as a ${\rm PseudoSeparator}$ (see Definition 4). ${\rm PseudoSeparator}$ allows us to decide reachability in $\textsf{Aux}_{\alpha}(G)$, by using a divide and conquer strategy and obtain the claimed time and space bounds.

For a graph $H=(V_{1},E_{1})$ and a subgraph $C=(V_{2},E_{2})$ of $H$, define the graph $H\diamond C=(V_{3},E_{3})$ as $V_{3}=V_{1}\setminus V_{2}$ and $E_{3}=E_{1}\setminus\{e\in E_{1}\mid\exists f\in E_{2},e\textrm{ crosses }f\}$.

Definition 4**.**

Let $G$ be a grid graph and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$ with $h$ vertices. Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be a function A subgraph $C$ of $H$ is said to be an $f(h)$-${\rm PseudoSeparator}$ of $\textsf{Aux}_{\alpha}(G)$ if the size of every connected component in $\textsf{cc}(H\diamond C)$ is at most $f(h)$. The size of $C$ is the total number of vertices and edges of $C$ summed together.

For a vertex-induced subgraph $H$ of $\textsf{Aux}_{\alpha}(G)$, an $f(h)$-${\rm PseudoSeparator}$ is a subgraph $C$ of $H$ that has the property that, if we remove the vertices as well as all the edges which crosses one of the edges of ${\rm PseudoSeparator}$, the graph gets disconnected into small pieces. Moreover for every edge $e$ in $H$, if there exists distinct sets $U_{1}$ and $U_{2}$ in $\textsf{cc}(H\diamond C)$ such that one of the endpoints of $e$ is in $U_{1}$ and the other is in $U_{2}$, then there exists an edge $f$ in $C$ such that $e$ crosses $f$. Hence any path which connects two vertices in different components, must either contain a vertex of $C$ or must contain an edge that crosses an edge of $C$. We divide the graph using this ${\rm PseudoSeparator}$ and give an algorithm which recursively solves each subgraph and then combines their solution efficiently.

4.1 Constructing a Pseudoseparator

We briefly comment on how to construct a ${\rm PseudoSeparator}$ of a vertex induced subgraph $H$ of $\textsf{Aux}_{\alpha}(G)$. First, we pick a maximal subset of edges from $H$ so that no two edges cross (see Definition 5). Then we triangulate the resulting graph. This can be done in logspace. Next, we use Imai et al.’s algorithm to find a separator of the triangulated graph. Call the triangulated graph as $\widehat{H}$ and the separator vertices as $S$. The vertex set of ${\rm PseudoSeparator}$ of $H$ will contain all the vertices of $S$ and four additional vertices for each edge of $\widehat{H}[S]$ that is not present in $H$. The edge set of ${\rm PseudoSeparator}$ of $H$ will contain all the edges of $H$ which are also in $\widehat{H}[S]$ and four additional edges for each edge of $\widehat{H}[S]$ that is not present in $H$.

Definition 5**.**

Let $G$ be a grid graph and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$. We define $\textsf{planar}(H)$ as a subgraph of $H$. The vertex set of $\textsf{planar}(H)$ is same as that of $H$. For an edge $e\in H$, let $l$ be the block to which $e$ belongs and let $w$ be the lowest indexed vertex in that block. Then $e=(c^{i}_{l}(w),c^{j}_{l}(w))$ is in $\textsf{planar}(H))$ if there exists no other edge $f=(c^{x}_{l}(w),c^{y}_{l}(w))$ in $H$ such that $\min(x,y)<\min(i,j)<\max(x,y)<\max(i,j)$.

In Lemma 4 we show that the graph $\textsf{planar}(H)$ is indeed planar and prove a simple yet crucial property of this graph, that would help us to construct the ${\rm PseudoSeparator}$.

Lemma 4**.**

Let $G$ be a grid-graph and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$. No two edges of $\textsf{planar}(H)$ cross each other. Moreover, for any edge $e$ in $H$ that is not in $\textsf{planar}(H)$, there exists another edge in $\textsf{planar}(H)$ that crosses $e$.

Proof.

Let $l$ be a block of $\textsf{Aux}_{\alpha}(G)$ and $w$ be the smallest index vertex of $l$. Let $e=(c^{p}_{l}(w),c^{q}_{l}(w))$ and $f=(c^{r}_{l}(w),c^{s}_{l}(w))$ be two edges of $H$ that cross. We have, by lemma 1, that either $\min(p,q)<\min(r,s)<\max(p,q)<\max(r,s)$ or $\min(r,s)<\min(p,q)<\max(r,s)<\max(p,q)$. Hence, by our construction of $\textsf{planar}(H)$, atmost one of $e$ and $f$ belongs to it. Thus no two edges of $\textsf{planar}(H)$ can cross.

For the second part, we will prove by contradiction. Let us assume that there exists edges in $H$ which is not in $\textsf{planar}(H)$ and also not crossed by an edge in $\textsf{planar}(H)$. We pick edge $e=(c_{l}^{p}(w),c_{l}^{q}(w))$ from them such that $\min(p,q)$ of that edge is minimum. Since this edge is not present in $\textsf{planar}(H)$, we have by definition, an edge $f=(c_{l}^{r}(w),c_{l}^{s}(w))$ such that $\min(r,s)<\min(p,q)<\max(r,s)<\max(p,q)$. We pick the edge $f$ for which $\min(r,s)$ is minimum. Now, since this edge $f$ is not present in $\textsf{planar}(H)$, we have another edge $g=(c_{l}^{i}(w),c_{l}^{j}(w))$ in $\textsf{planar}(H)$ such that $\min(i,j)<\min(r,s)<\max(i,j)<\max(r,s)$. We pick $g$ such that $\min(i,j)$ is minimum and break ties by picking one whose $\max(i,j)$ is maximum. Now, we have the following cases:

• Case 1 ($i<r<j<s$): In this case, the edge $c_{l}^{i}(w),c_{l}^{s}(w)$ will be present in $H$. Since $i<p<s<q$, and $i<\min(r,s)$, this will contradict the way in which edge $f$ was chosen.

• Case 2 ($i<s<j<r$): In this case, the edge $(c_{l}^{r}(w),c_{l}^{j}(w))$ will be present in $H$. This edge will cross $e$ and hence not be present in $\textsf{planar}(H)$. Thus, we have an edge $g^{\prime}=(c_{l}^{i^{\prime}}(w),c_{l}^{j^{\prime}}(w))$ in $\textsf{planar}(H)$ such that $\min(i^{\prime},j^{\prime})<j<\max(i^{\prime},j^{\prime})<r$. We will thus have two subcases.

• Case 2a ($i<\min(i^{\prime},j^{\prime})$): Here, we will have $i<\min(i^{\prime},j^{\prime})<j<\max(i^{\prime},j^{\prime})$. Hence this edge will cross $g$ giving a contradiction to the first part of this lemma.

• Case 2b ($\min(i^{\prime},j^{\prime})\leq i$): Here, this edge should have been chosen instead of $g$ contradicting our choice of $g$.

The analysis of two remaining cases where $j<s<i<r$ and $j<r<i<s$ are similar to Cases $1$ and $2$ respectively. ∎

We next describe how to compute the triangulated planar graph $\widehat{H}$. Given $H$ as an input, we first find $\textsf{planar}(H)$ using Lemma 4. We then triangulate $\textsf{planar}(H)$ by first adding edges in the boundary of each block as follows: let $l$ be a block and $v$ be a vertex in $l$. Let $p$ be the smallest positive integer such that the vertex $c_{l}^{p}(v)$ is present in $H$. If the edge $(v,c_{l}^{p}(v))$ is not present in $\textsf{planar}(H)$, we add this in $\widehat{H}$. We add the edges in such a way that every edge of $l$ will be inside the boundary cycle. Finally, we triangulate the rest of the graph and add the triangulation edges to $\widehat{H}$. Note that this process can be done in logspace.

We will be using the following lemma, which was proven by Imai et al.

Lemma 5**.**

[9]** For every $\beta>0$, there exists a polynomial time and $\tilde{O}(h^{1/2+\beta/2})$ space algorithm that takes a $h$-vertex planar graph $P$ as input and outputs a set of vertices $S$, such that $|S|$ is $O(h^{1/2+\beta/2})$ and removal of $S$ disconnects the graph into components of size $O(h^{1-\beta})$.

We next show a property of triangulated graphs that we use in our construction of ${\rm PseudoSeparator}$. It is known that a simple cycle in a planar embedding of a planar graph divides the plane into two parts. We call these two parts the two sides of the cycle.

Lemma 6**.**

Let $G$ be a triangulated planar graph, and $S$ be a subset of its vertices. For every pair of vertex $u,v$ which belong to different components of $G\setminus S$, there exists a cycle in $G[S]$, such that $u$ and $v$ belong to different sides of this cycle.

Proof.

To prove the lemma, we first need some terminology. We call a set of faces an edge-connected region if it can be constructed in the following way:

• •

A set of a single face is an edge-connected region.

• •

If a set $F$ is an edge-connected region and $f$ is a face that shares an edge with one of the faces of $F$, then $F\cup\{f\}$ is an edge-connected region.

We can orient the edges of an undirected planar simple cycle to make it a directed cycle. This can help us identify the two sides of the cycle as interior (left-side) and exterior (right-side).

Let $C$ be a component of $G\setminus S$ and $S^{\prime}$ be the set of vertices of $S$ which are adjacent to at least one of the vertices of $C$ in $G$. Let $F$ be the set of triangle faces of $G$ to which at least one vertex of $C$ belongs. We first observe that for any face $f$ of $F$, the vertices of $f$ will either belong to $C$ or $S^{\prime}$. We see that $F$ is a region of edge-connected faces. Miller proved that we could write the boundary of the region of edge-connected faces as a set of vertex-disjoint simple cycles with disjoint exteriors [Miller]. These cycles will contain only the vertices of $S^{\prime}$. Hence the lemma follows. ∎

For a subgraph $H$ of $\textsf{Aux}_{\alpha}(G)$, we construct the graph $\textsf{psep}(H)$ in the following way.

Construct $\widehat{H}$ from $H$. 2. 2.

We then find a set $S$ of vertices in $\widehat{H}$ which divides it into components of size $O(n^{1-\beta})$ by applying Lemma 6. 3. 3.

Add each vertex of $S$ to the set $V(\textsf{psep}(H))$ and each edge of $\widehat{H}(S)$ which is also in $H$ to $E(\textsf{psep}(H))$. 4. 4.

Now let $e=(v,w)$ be a triangulation edge present in block $l$ of $\widehat{H}$ whose both endpoints are in $S$. Let $p$ and $q$ be integers such that $w=c_{l}^{p}(v)$ and $v=c_{l}^{q}(w)$ We add the following set of at most four vertices and four edges to $\textsf{psep}(H)$.

1. (a)

Let $p_{1}<p$ be the largest integer such that an edge $e_{1}$ with endpoints $v$ and $c_{l}^{p_{1}}(v)$ exists. 2. (b)

Let $p_{2}>p$ be the smallest integer such that an edge $e_{2}$ with endpoints $v$ and $c_{l}^{p_{2}}(v)$ exists. 3. (c)

Let $q_{1}<q$ be the largest integer such that an edge $e_{3}$ with endpoints $c_{l}^{q_{1}}(w)$ and $w$ exists. 4. (d)

Let $q_{2}>q$ be the smallest integer such that an edge $e_{4}$ with endpoints $c_{l}^{q_{2}}(w)$ and $w$ exists.

Note that the above edges could be directed either way. We add the vertices $c_{l}^{p_{1}}(v)$, $c_{l}^{p_{2}}(v)$, $c_{l}^{q_{1}}(w)$ and $c_{l}^{q_{2}}(w)$ to $V(\textsf{psep}(H))$. For $i=1,2,3,4$, we add the edges $e_{i}$ to $E(\textsf{psep}(H))$. We call these four edges as shadows of $e$.

In Lemma 7 we show that the set $\textsf{psep}(H)$ is a ${\rm PseudoSeparator}$ of $H$.

Lemma 7**.**

Let $G$ be a grid graph and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$. The graph $\textsf{psep}(H)$ is a $h^{1-\beta}$-${\rm PseudoSeparator}$ of $H$.

Proof.

Let $C=\textsf{psep}(H)$. Let $S$ be the set of vertices obtained from $\widehat{H}$ by using Lemma 5. We claim that if any two vertices $u$ and $v$ belong to different connected components of $\widehat{H}\setminus S$, then it belongs to diffent components of $\textsf{cc}(H\diamond C)$. We prove this by contradiction. Let us assume that it is not true. Then there is an edge from $e$ in $H$ and two distinct sets $U_{1}$ and $U_{2}$ of $\widehat{H}\setminus S$ such that one of the end point of $e$ is in $U_{1}$ and the other is in $U_{2}$. This edge $e$ does not cross any of the edge of $\textsf{psep}{H}$. Without loss of generality, let $e=(v,c_{l}^{p}(v))$, where $v\in U$ and $c_{l}^{p}(v)$ is not in $U$. (we pick the edge $e$ such that $p$ is minimum) Due to Lemma 6, it follows that there exists a triangulation edge $f$ such that $f=((c_{l}^{q}(v)),c_{l}^{r}(v))$ and that $e$ crosses $f$. We orient the triangulation edge so that $q<p<r$. Now, since $e$ is not present in $\textsf{planar}(H)$, by Lemma 4 there exists at least one edge that crosses it and is present in $\textsf{planar}(H)$. Let $g=(c_{l}^{s}(v),c_{l}^{t}(v))$ be one such edge such that $t-s$ is maximum. We thus have the following cases:

• Case 1 ($s<q<p<r<t$): In this case, since $g$ crosses $e$, by Lemma 2, we have that the edge $(c_{l}^{s}(v),c_{l}^{p}(v))$ is also present in $H$. This also crosses $f$ and thus should have been picked instead of $e$. Hence we have a contradiction.
• Case 2 ($q<t<p<s<r$): In this case, since $g$ crosses $e$, by Lemma 2, we have that the edge $(v,c_{l}^{t}(v))$ is also present in $H$. This also crosses $f$ and thus should have been picked instead of $e$. Hence we have a contradiction.
• Case 3 ($q<s<p<t<r$): In this case, the edge $(v,c_{l}^{t}(v))$ will also be present due to lemma 2. This edges also crosses $f$. Thus there will exists an edge in $\textsf{planar}(H)$ which crosses this edge. We pick edge $g^{\prime}=(c_{l}^{s^{\prime}}(v),c_{l}^{t^{\prime}}(v))$ such that $t^{\prime}-s^{\prime}$ is maximum. Clearly, $t^{\prime}<q$ and $s^{\prime}>r$. Now, since the edges $g$ and $g^{\prime}$ both crosses $e$ and $g^{\prime}$ is closer to $v$ than $g$, by lemma 3, the edge $(c_{l}^{s^{\prime}}(v),c_{l}^{t}(v))$ will also be present in the graph. This will again cross $f$. Now, any edge present in $\textsf{planar}(H)$ that crosses this edge will contradict the way in which $g$ or $g^{\prime}$ is chosen.
• Case 4 ($t<q<p<r<s$): In this case, the edge $(c_{l}^{s^{\prime}}(v),c_{l}^{p}(v))$ will also be present in the graph. This will cross $f$. Any edge present in $\textsf{planar}(H)$ that also crosses this edge will contradict the way $g$ is chosen.

In other cases, either $e$ would cross a $shadow$ edge of $f$ giving a contradiction, or $g$ would cross $f$ contradicting the fact that both of them are present in $\widehat{H}$. ∎

Summarizing Lemmas 5 and 7 we have Theorem 2.

Theorem 2**.**

Let $G$ be a grid graph and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$ with $h$ vertices. For any constant $\beta>0$, there exists an $\tilde{O}(h^{1/2+\beta/2})$ space and polynomial time algorithm that takes $H$ as input and outputs an $h^{1-\beta}$-${\rm PseudoSeparator}$ of size $O(h^{1/2+\beta/2})$.

5 Algorithm to Solve Reachability in Auxiliary Graph

In this section, we discuss the grid graph reachability algorithm. Let $G$ be a grid graph having $\widetilde{n}$ vertices. By induction we assume that we have access to a vertex induced subgraph $H$ of $\textsf{Aux}_{\alpha}(G)$, containing $h$ vertices. Below we describe a recursive procedure $\textsf{AuxReach}(H,x,y)$ that outputs true if there is a path from $x$ to $y$ in $H$ and outputs false otherwise.

5.1 Description of the Algorithm AuxReach

First we construct a $h^{1-\beta}$-${\rm PseudoSeparator}$ $C$ of $H$, using Theorem 2. We also ensure that $x$ and $y$ are part of $C$ (if not then we add them). Let $I_{1},I_{2},\ldots,I_{l}$ be the connected components of $H\diamond C$.

We maintain an array called visited of size $|C|$ to mark vertices or edges of the ${\rm PseudoSeparator}$ $C$. Each cell of visited corresponds to a distinct vertex or edge of $C$. For a vertex $v$ in $C$, we set $\textsf{visited}[v]:=1$ if there is a path from $x$ to $v$ in $H$, else it is set to 0. For an edge $e=(u,v)$ in $C$, we set $\textsf{visited}[e]:=u^{\prime}$ if (i) there is an edge $f=(u^{\prime},v^{\prime})$ that crosses $e$, (ii) there is a path from $x$ to $u^{\prime}$ in $H$ and (iii) $f$ is the closest such edge to $u$. Else $\textsf{visited}[e]$ is set to NULL. Initially, for all vertex $v\in C$, $\textsf{visited}[v]:=0$ and for all edges $e\in C$, $\textsf{visited}[e]:=\textsf{NULL}$. We say that a vertex $v$ is marked if either $\textsf{visited}[v]:=1$ or $\textsf{visited}[e]:=v$ for some edge $e$.

First set $\textsf{visited}[x]:=1$. We then perform an outer loop with $h$ iteration and in each iteration update certain entries of the array visited as follows. For every vertex $v\in C$, the algorithm sets $\textsf{visited}[v]:=1$ if there is a path from a marked vertex to $v$ such that the internal vertices of that path all belong to only one component $I_{i}$. Similarly, for each edge $e=(u,v)$ of $C$, the algorithm sets $\textsf{visited}[e]:=u^{\prime}$ if (i) there exists an edge $f=(u^{\prime},v^{\prime})$ which crosses $e$, (ii) there is a path from a marked vertex to $u^{\prime}$ such that the internal vertices of that path all belong to only one component $I_{i}$ and, (iii) $f$ is the closest such edge to $u$. Finally we output true if $\textsf{visited}[y]=1$ else output false. We use the procedure AuxReach recursively to check if there is a path between two vertices in a single connected component of $H\diamond C$. A formal description of AuxReach is given in Algorithm 1.

5.2 Proof of Correctness of AuxReach

Let $P$ be the shortest path from $x$ to $y$ in $H$. Suppose $P$ passes through the components $I_{\sigma_{1}},I_{\sigma_{2}},\ldots,I_{\sigma_{L}}$ in this order. The length of this sequence can be at most $\lvert H\rvert$. As the path leaves the component $I_{\sigma_{j}}$ and goes into $I_{\sigma_{j+1}}$, it can do in the following two ways only:

The path exits $I_{\sigma_{j}}$ through a vertex $w$ of ${\rm PseudoSeparator}$ as shown in Figure 2(a). In this case, Algorithm 1 would mark the vertex $w$. 2. 2.

The path exits $I_{\sigma_{j}}$ through an edge $(u,v)$ whose other endpoint is in $I_{\sigma_{j+1}}$. By Lemma 3, this edge will cross an edge $e=(x^{\prime},y^{\prime})$ of the ${\rm PseudoSeparator}$. In this case, Algorithm 1 would mark the vertex $u^{\prime}$, such that there is an edge $(u^{\prime},v^{\prime})$ that crosses $e$ as well and $(u^{\prime},v^{\prime})$ is closer than $(u,v)$ to $x^{\prime}$ and there is a path in $I_{\sigma_{j}}$ from a marked vertex to $u^{\prime}$. By Lemma 3, the edge $(u^{\prime},v)$ would be present in $H$ as well.

Thus after the $j$-th iteration, AuxReach would traverse the fragment of the path in the component $I_{\sigma_{j}}$ and either mark its endpoint or a vertex which is closer to the edge $e$ of $C$ which the path crosses. Finally, $t$ would be marked after $L$ iterations if and only if there is a path from $s$ to $t$ in $H$. We give a formal proof of correctness in Lemma 8. For a path $P=(u_{1},u_{2},\ldots u_{t})$, we define $\textsf{tail}(P):=u_{1}$ and $\textsf{head}(P):=u_{t}$.

Lemma 8**.**

Let $G$ be a grid graph and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$. Then for any two vertices $x,y$ in $H$, there is a path from $x$ to $y$ in $H$ if and only if $\textsf{AuxReach}(H,x,y)$ returns true.

Proof.

Firstly observe that a vertex is marked only if there is a path from some other marked vertex to that vertex in $H$. Hence if there is no path from $x$ to $y$ then $y$ is never marked by AuxReach and hence AuxReach returns false.

Now let $P$ be a path from $x$ to $y$ in $H$. We divide the path into subpaths $P_{1},P_{2},\ldots,P_{l}$, such that for each $i$, all vertices of $P_{i}$ belong to $U\cup V(C)$ for some connected component $U$ in $\textsf{cc}(H\diamond C)$ and either (i) $\textsf{head}(P_{i})=\textsf{tail}(P_{i+1})$, or (ii) $e_{i}=(\textsf{head}(P_{i}),\textsf{tail}(P_{i+1}))$ is an edge that crosses some edge $f_{i}\in C$. By Definition 4, we have that if condition (i) is true then $\textsf{head}(P_{i})$ is a vertex in $C$, and if condition (ii) is true then $\textsf{head}(P_{i})$ and $\textsf{tail}(P_{i+1})$ belong to two different components of $\textsf{cc}(H\diamond C)$ and $e_{i}$ is the edge between them.

We claim that after $i$-th iteration of loop in Line 1 of Algorithm 1, either of the following two statements hold:

$\textsf{head}(P_{i})$ is a vertex in $C$ and $\textsf{visited}[\textsf{head}(P_{i})]=1$. 2. 2.

There exists an edge $f_{i}=(u_{i},v_{i})$ of $C$ such that the edge $e_{i}=(\textsf{head}(P_{i}),\textsf{tail}(P_{i+1}))$ crosses $f_{i}$ and there is an edge $g_{i}=(u_{i}^{\prime},v_{i}^{\prime})$ which crosses $f_{i}$ as well, such that $g_{i}$ is closer to $u_{i}$ than $e_{i}$ and $\textsf{visited}[f_{i}]=u_{i}^{\prime}$.

We prove the claim by induction. The base case holds since $x$ is marked at the beginning. We assume that the claim is true after the $(i-1)$-th iteration. We have that $P_{i}$ belongs to $U\cup V(C)$ for some connected component $U$ in $\textsf{cc}(H\diamond C)$.

• Case 1 ($\textsf{head}(P_{i-1})=\textsf{tail}(P_{i})=w$(say)): By induction hypothesis $w$ was marked after the $(i-1)$-th iteration. If $\textsf{head}(P_{i})$ is a vertex in $C$ then it will be marked after the $i$-th iteration in Line 1. On the other hand if $e_{i}=(\textsf{head}(P_{i}),\textsf{tail}(P_{i+1}))$ is an edge that crosses some edge $f_{i}=(u_{i},v_{i})\in C$ then in the $i$-th iteration in Line 1, the algorithm marks a vertex $u_{i}^{\prime}$ such that, $g_{i}=(u_{i}^{\prime},v_{i}^{\prime})$ is the closest edge to $u_{i}$ that crosses $f_{i}$ and there is a path from $w$ to $u_{i}^{\prime}$.
• Case 2 ($e_{i-1}=(\textsf{head}(P_{i-1}),\textsf{tail}(P_{i}))$ is an edge that crosses some edge $f_{i-1}=(u_{i-1},v_{i-1})\in C$): By induction hypothesis, there is an edge $g_{i-1}=(u_{i-1}^{\prime},v_{i-1}^{\prime})$ which crosses $f_{i-1}$ as well, such that $g_{i-1}$ is closer to $u_{i-1}$ than $e_{i-1}$ and $\textsf{visited}[f_{i-1}]=u_{i-1}^{\prime}$. By Lemma 3 there is an edge in $H$ between $u_{i-1}^{\prime}$ and $\textsf{tail}(P_{i})$ as well. Now if $\textsf{head}(P_{i})$ is a vertex in $C$ then it will be marked after the $i$-th iteration in Line 1 by querying the graph $H[U\cup\{u_{i-1}^{\prime},\textsf{head}(P_{i})\}]$. On the other hand if $e_{i}=(\textsf{head}(P_{i}),\textsf{tail}(P_{i+1}))$ is an edge that crosses some edge $f_{i}=(u_{i},v_{i})\in C$ then in the $i$-th iteration in Line 1, AuxReach queries the graph $H[U\cup\{u_{i-1}^{\prime},u_{i}^{\prime}\}]$ and marks a vertex $u_{i}^{\prime}$ such that, $g_{i}=(u_{i}^{\prime},v_{i}^{\prime})$ is the closest edge to $u_{i}$ that crosses $f_{i}$ and there is a path from $u_{i-1}^{\prime}$ to $u_{i}^{\prime}$.

∎

Our subroutine would solve reachability in a subgraph $H$ (having size $h$) of $\textsf{Aux}_{\alpha}(G)$. We do not explicitly store a component of $\textsf{cc}(H\diamond C)$, since it might be too large. Instead, we identify a component with the lowest indexed vertex present in it and use Reingold’s algorithm on $H\diamond C$ to determine if a vertex is present in that component. We require $\widetilde{O}(h^{1/2+\beta/2})$ space to compute a $h^{1-\beta}$-${\rm PseudoSeparator}$ by Theorem 2. We can potentially mark all the vertex of the ${\rm PseudoSeparator}$ and for each edge of ${\rm PseudoSeparator}$ we mark at most one additional vertex. Since the size of ${\rm PseudoSeparator}$ is at most $O(h^{1/2+\beta/2})$, we require $\widetilde{O}(h^{1/2+\beta/2})$ space. The algorithm recurses on a graph with $h^{1-\beta}$ vertices. Hence the depth of the recursion is at most $3/(\log(1-\beta)^{-1})$, which is a constant.

Since the graph $H$ is given implicitly in our algorithm, there is an additional polynomial overhead involved in obtaining its vertices and edges. However, the total time complexity would remain a polynomial in the number of vertices since the recursion depth is constant.

Lemma 9**.**

Let $G$ be a grid graph on $m$ vertices and $H$ be a vertex induced subgraph of $\textsf{Aux}_{\alpha}(G)$ having $h$ vertices. For every $\beta>0$, AuxReach runs in $\widetilde{O}(\widetilde{h}^{1/2+\beta/2})$ space and polynomial time.

Proof.

Since the size of a component $U$ in $\textsf{cc}(H\diamond C)$ might be too large, we will not explicitly store it. Instead we identify a component by the lowest index vertex present in it and use Reingold’s algorithm on $H\diamond C$ to determine if a vertex is present in $U$. Let $S(m,h)$ and $T(m,h)$ denote the space and time complexity functions respectively of AuxReach, where $G$ is an $m\times m$ grid graph and $h$ is the number of vertices in the graph $H$. As noted earlier the depth of the recursion is at most $d:=3/(\log(1-\beta)^{-1})$.

Consider $S(m,h)$ for any $h>m^{1/8}$. By Theorem 2, we require $\widetilde{O}(h^{1/2+\beta/2})$ space to execute Line 1. We can potentially mark all the vertex of $C$ and for each edge $e$ of $C$ we store at most one additional vertex in $\textsf{visited}[e]$. Since the size of $C$ is at most $O(h^{1/2+\beta/2})$, we require $\widetilde{O}(h^{1/2+\beta/2})$ space to store $C$. By induction, a call to AuxReach in line 1 and 1 requires $S(h^{1-\beta})$ space which can be subsequently reused. Hence the space complexity satisfies the following recurrence. Then,

[TABLE]

Solving we get $S(m,h)=\widetilde{O}(h^{1/2+\beta/2}+m^{1/4})$.

Next we measure the time complexity of AuxReach. Consider the case when $h>m^{1/8}$. The number of steps in AuxReach minus the recursive calls to AuxReach is some polynomial in $h$, say $p$. Moreover AuxReach makes $q$ calls to AuxReach, where $q$ is some other polynomial in $h$. Hence $q(h)\leq p(h)$. Then,

[TABLE]

Solving the above recurrence we get $T(m,h)=O(p\cdot q^{d}+m^{1/4})=O(p^{2d}+m^{1/4})$. ∎

6 Solving Grid Graph

Let $G$ be an $m\times m$ grid graph. As mentioned in the introduction, our objective is to run Algorithm 1 on the graph $\textsf{Aux}_{\alpha}(G)$. By definition of $\textsf{Aux}_{\alpha}(G)$, for every pair of vertices $x,y$ in $\textsf{Aux}_{\alpha}(G)$, there is a path from $x$ to $y$ in $\textsf{Aux}_{\alpha}(G)$ if and only if there is a path from $x$ to $y$ in $G$. Hence it is sufficient to work with the graph $\textsf{Aux}_{\alpha}(G)$. However, we do not have explicit access to the edges of $\textsf{Aux}_{\alpha}(G)$. Note that we can obtain the edges of $\textsf{Aux}_{\alpha}(G)$ by solving the corresponding subgrid of $G$ to which that edge belongs. If the subgrid is small enough, then we use a standard linear space traversal algorithm. Otherwise, we use our algorithm recursively on the subgrid. Algorithm 2 outlines this method.

Consider an $\widehat{m}\times\widehat{m}$ grid graph $\widehat{G}$. Let $S(\widehat{m})$ be the space complexity and $T(\widehat{m})$ be the time complexity of executing GridReach on $\widehat{G}$. Note that the size of $\textsf{Aux}_{\alpha}(\widehat{G})$ is at most $\widehat{m}^{1+\alpha}$. For $\widehat{m}>m^{1/8}$, the space required to solve the grid-graph would be $S(\widehat{m})=S(\widehat{m}^{1-\alpha})+\widetilde{O}((\widehat{m}^{1+\alpha})^{1/2+\beta/2})$. This is because, a query whether $(u,v)\in\widehat{G}$ would invoke a recursion which would require $S(\widehat{m}^{1-\alpha})$ space and the main computation of ImplicitAuxReach could be done using $\widetilde{O}((\widehat{m}^{1+\alpha})^{1/2+\beta/2})$ space. Hence we get the following recurrence for space complexity.

[TABLE]

Similar to the analysis of AuxReach, for appropriate polynomials $p$ and $q$ ($q\leq p$), the time complexity would satisfy the following recurrence:

[TABLE]

Solving we get $S(m)=\widetilde{O}(m^{1/2+\beta/2+\alpha/2+\alpha\beta/2})$ and $T(m)=\poly(n)$. Since $\alpha$ and $\beta$ can be chosen arbitrarily, we can write the space complexity as $S(m)={O}(m^{1/2+\epsilon})$, for any constant $\epsilon>0$.