Bounds on Functionality and Symmetric Difference -- Two Intriguing Graph Parameters

TL;DR

This paper characterizes the functionality of random graphs $G(n,p)$ across all $p$, establishes bounds for all graphs, and explores the symmetric difference parameter, especially in interval and circular arc graphs, revealing new bounds and relationships.

Contribution

It provides the first comprehensive bounds on functionality for all $G(n,p)$ and general graphs, and clarifies the relationship between functionality and symmetric difference in specific graph classes.

Findings

Functionality of $G(n,p)$ is characterized for all $p$, with bounds matching up to a constant.

Maximum functionality is roughly $rac{1}{ oot 2 }$ at $p hicksim 1/ oot 2 $.

Symmetric difference in interval graphs is at most $O( oot 3 )$, with examples reaching $ heta( oot 4 )$.

Abstract

Functionality ($\mathrm{fun}$) is a graph parameter that generalizes graph degeneracy defined by Alecu et al. [JCTB, 2021]. They research the relation of functionality to many other graphs parameters (tree-width, clique-width, VC-dimension, etc.). Extending their research, we completely characterize the functionality of random graph $G(n,p)$ for all possible $p$. We provide matching (up to a constant factor) lower and upper bound for a large range of $p$. It follows from our bounds for $G(n,p)$, that the maximum functionality (roughly $\sqrt{n}$) is achieved for $p \approx 1/\sqrt{n}$. We complement this by showing that every graph $G$ on $n$ vertices have $\mathrm{fun}(G) \le O(\sqrt{ n \ln n})$ and we give a nearly matching $\Omega(\sqrt{n})$-lower bound provided by incident graphs of projective planes. Previously known lower bounds for functionality were only logarithmic in the number of vertices. Further, we study a related graph parameter symmetric difference ($\mathrm{sd}$), the minimum of $|N(u) ~\Delta~ N(v)|$ over all pairs of vertices of the ``worst possible'' induced subgraph. It was observed by Alecu et al. that $\mathrm{fun}(G) \le \mathrm{sd}(G)+1$ for every graph $G$. They asked whether the functionality of interval graphs is bounded. Recently, Dallard et al. [RiM, 2024] answered this positively and they constructed an interval graph $G$ with $\mathrm{sd}(G) = \Theta(\sqrt[4]{n})$ (even though they did not mention the explicit bound), i.e., they separate the functionality and symmetric difference of interval graphs. We show that $\mathrm{sd}$ of interval graphs is at most $O(\sqrt[3]{n})$ and we provide a different example of an interval graph $G$ with $\mathrm{sd}(G) = \Theta(\sqrt[4]{n})$. Further, we show that $\mathrm{sd}$ of circular arc graphs is $\Theta(\sqrt{n})$.

Equations

$$\underset{H\subseteq G}{\max }\underset{v\in V(H)}{\min }{\mathrm{fun}}_H(v),$$

$$\underset{H\subseteq G}{\max }\underset{u,v\in V(H)}{\min }{\mathrm{sd}}_H(u,v),$$

$$|P|\le (b+1)(k+1)-1+k-b=bk+2k\le k^2+k.$$

$$P(\exists u_1\in N(v)\setminus S,u_2\notin N(v)\cup S\cup \{v\}:S\cap PN(u_1,u_2)=\varnothing ).$$

$$\sum _{u_1,u_2}(1-p{)}^{|PN(u_1,u_2)|-1}\le n^2(1-p{)}^{d-1}$$

$$p^{|B|}+(1-p{)}^{|B|}\le 2\cdot (1-p{)}^{\frac{\sqrt{n}}{2}}\le 2\cdot e^{-p\cdot \frac{\sqrt{n}}{2}}\le 2^{-\frac{3}{2}\cdot {\log }^{2}n}.$$

$$\mathbb{E}[|E^{\prime }|]=p\cdot k\cdot |V^{\prime }|\le \frac{3}{2}\sqrt{n}\cdot {\log }^{3}n$$

$$p^{|B_0|}+(1-p{)}^{|B_0|}\le 2\cdot (1-p{)}^{\frac{2n}{3}}\le 2\cdot e^{-p\cdot \frac{2n}{3}}\le 2^{-2\cdot {\log }^{2}n}.$$

$$e^{-\sqrt{n}{\log }^{3}n}+2^{-2\cdot {\log }^{2}n}\le 2^{-\frac{3}{2}\cdot {\log }^{2}n}.$$

$$\mathrm{Pr}[v_1,v_2:N(v_1)\cap V^{\prime }=N(v_2)\cap V^{\prime }]=2^{-k}.$$

$$\mathrm{Pr}[\exists v_1,v_2\notin V^{\prime }:N(v_1)\cap V^{\prime }=N(v_2)\cap V^{\prime }]\le \frac{n^2}{2^k}=\frac{1}{n}.$$

$$|r|$$

$$\le (k-j_k)\cdot d+d\le \frac{d}{5}\cdot d+d\le \frac{n}{4}$$

$$S=\bigl{\{}[i,i+d]\mid i\in\{0,\dots,t-d\}\bigr{\}}.$$

$$L_{i}=\bigl{\{}[a,b]\mid a,b\equiv i\pmod{2d};\ell_{i}\leq b-a\leq\ell_{i}+2d^{2}\bigr{\}},$$

Full text

Further, we study a related graph parameter symmetric difference, the minimum of $|N(u)\Delta N(v)|$ over all pairs of vertices of the “worst possible” induced subgraph. It was observed by Alecu et al. that $\mathrm{fun}(G)\leq\mathrm{sd}(G)+1$ for every graph $G$. We compare $\mathrm{fun}$ and $\mathrm{sd}$ for the class $\mathrm{INT}$ of interval graphs and $\mathrm{CA}$ of circular-arc graphs. We let $\mathrm{INT}_{n}$ denote the $n$-vertex interval graphs, similarly for $\mathrm{CA}_{n}$.

Alecu et al. ask, whether $\mathrm{fun}(\mathrm{INT})$ is bounded. Dallard et al. answer this positively in a recent preprint. On the other hand, we show that $\Omega(\sqrt[4]{n})\leq\mathrm{sd}(\mathrm{INT}_{n})\leq O(\sqrt[3]{n})$. For the related class $\mathrm{CA}$ we show that $\mathrm{sd}(\mathrm{CA}_{n})=\Theta(\sqrt{n})$.

We propose a follow-up question: is $\mathrm{fun}(\mathrm{CA})$ bounded?

1 Introduction

Let $G=(V,E)$ be a graph and $v\in V$ its vertex. The set of neighbors of $v$ in $G$ is denoted by $N_{G}(v)$, we omit the subscript if the graph is clear from the context. An adjacency matrix $A_{G}$ of $G$ is [math]-$1$ matrix such that its rows and columns are indexed by vertices of $G$ and $A[u,v]=1$ if and only if $u$ and $v$ are connected by an edge. Now, we define a functionality and symmetric difference of a graph – two principle notions of this papers – as introduced by Alecu et al. [1], and implicitly also Atminas et al. [3].

A vertex $v$ of a graph $G=(V,E)$ is a function of vertices $u_{1},\dots,u_{k}\in V$ (different from $v$) if there exists a boolean function $f$ of $k$ variables such that for any vertex $w\in V\setminus\{v,u_{1},\dots,u_{k}\}$ holds that A[v,w]=f\bigl{(}A[v,u_{1}],\dots,A[v,u_{k}]\bigr{)}. Informally, we can determine if $u$ and $w$ are connected from adjacency of $v$ with $u_{i}$’s. A functionality $\mathrm{fun}_{G}(v)$ of a vertex $v$ in $G$ is the minimum $k$ such that $v$ is a function of $k$ vertices of $G$. We drop the subscript and write just $\mathrm{fun}(v)$ if the graph $G$ is clear from the context. Then, a functionality $\mathrm{fun}(G)$ of a graph $G$ is defined as

[TABLE]

where the maximum is taken over all induced subgraphs $H$ of $G$.

It is observed in [3] that if $\mathrm{fun}(G)\leq k$ then we can encode $G$ using $(k+1)n\log n$ bits. So if every graph $G$ in some graph class $\mathcal{G}$ has bounded functionality then $\mathcal{G}$ contains at most $2^{O(n\log n)}$ graphs on $n$ vertices. Such classes are said to be of factorial growth and include diverse classes of practical importance (interval graphs, line graphs, forests, planar graphs, more generally all proper minor-closed classes). So Alecu et al. introduce functionality as a tool to study graph classes of factorial growth, and the related Implicit graph conjecture (although this conjecture was recently disproved [5]). This was also our original motivation. Moreover, functionality is a natural generalization of graph coloring number, and deserves a study for its own sake.

Alacu et al. [1] research the relation of functionality to many other graph parameters: in particular they provide a linear upper bound in terms of clique-width and a lower bound in terms of some function of VC-dimension. They also give lower bound on functionality of hypercube that is linear in the dimension.

Another parameter related to functionality is the so-called symmetric difference. Given two vertices $u,v$ of $G$, let $\mathrm{sd}_{G}(u,v)$ (or just $\mathrm{sd}(u,v)$ when the graph is clear from the context) be the number of vertices different from $u$ and $v$ that are adjacent to exactly one of $u$ and $v$. The symmetric difference $\mathrm{sd}(G)$ of a graph $G$ is defined as

[TABLE]

where the maximum is again taken over all induced subgraphs. We may view $\mathrm{sd}(u,v)$ as the size of set $(N(u)\mathrm{\Delta}N(v))\setminus\{u,v\}$, which explains the term “symmetric difference”. It is noted by Alecu et al. [1] that $\mathrm{fun}(G)\leq\mathrm{sd}(G)+1$. However, even looser opposite inequality does not hold as there are graphs of constant functionality and polynomial symmetric difference – for example the interval graphs. This was shown by Theorem 3.2 of Dallard et al. [4] and by our Theorem 3.4, while Corollary 5.3 of [4] shows that $\mathrm{sd}(\mathrm{INT})$ is unbounded, without providing explicit lower bounds.

The intersection graph of $F$ is a graph $G=(V,E)$ where $V=\{v_{1},\dots,v_{n}\}$ and two vertices $v_{i}$ and $v_{j}$ are connected if and only if the corresponding sets $S_{i}$ and $S_{j}$ intersect. An interval graph is an intersection graph of $n$ intervals on a real line. A circular arc graph is an intersection graph of $n$ arcs of a circle. We let $\mathrm{INT}_{n}$ denote the family of all intersection graphs with $n$ vertices, $\mathrm{INT}$ the family of all interval graphs. In the same vein, we define $\mathrm{CA}_{n}$ and $\mathrm{CA}$ for circular arc graphs.

1.1 Our Results

In this paper, we show several lower and upper bounds for functionality and symmetric difference of various graph classes. As far as we know, there were only logarithmic lower bounds for functionality [3, 1, 4]. Thus, it would have been possible that the functionality is at most logarithmic (similarly to the VC-dimension [6]). However, we show that the functionality of the incidence graph of a finite projective plane of order $k$ is exactly $k+1$, i.e., roughly $\sqrt{n}$. We complement this result with an almost matching upper bound that functionality of any graph is at most $O(\sqrt{n\log n})$. Further, we show for $\frac{3\log^{2}n}{n}\leq p\leq 1-\frac{3\log^{2}n}{n}$ that a random graph $G(n,p)$ has at least logarithmic functionality with probability $1-o(1)$. Note that if $p\leq o\left(\frac{\log n}{n}\right)$ then $\mathrm{fun}(G(n,p))\leq o(\log n)$ with high probability because the minimum degree of $G(n,p)$ for such $p$ is $o(\log n)$ with high probability and $\mathrm{fun}(G)$ is bounded by the minimum degree of the graph $G$. Similarly for $p\geq 1-o\left(\frac{\log n}{n}\right)$, as the functionality of a graph $G$ and its complement $\bar{G}$ is the same, and complement of $G(n,p)$ is $G(n,1-p)$. Thus, our range of $p$ is almost optimal – up to a logarithmic factor. Further, we prove that any vertex of $G(n,\frac{1}{2})$ is function of at most $O(\log n)$ vertices with high probability. Unfortunately, it does not imply that functionality of $G(n,\frac{1}{2})$ is logarithmic as it still can contain an induced subgraph of higher functionality. Overall, it suggests that graphs with polynomial functionality are quite rare and should be very structured (like the case of finite projective plane).

Further, we study symmetric difference of interval and circular arc graphs. We show that symmetric difference of circular arc graph is $\Theta(\sqrt{n})$, i.e., we prove that any circular arc graph has symmetric difference at most $O(\sqrt{n})$ and we present a circular arc graph of symmetric difference $\Omega(\sqrt{n})$. Recently, it was shown that interval graphs have bounded functionality and unbounded symmetric difference [4]. Even though it was not explicitly mentioned, the construction given by Dallard et al. [4] leads to the lower bound $\Omega(\sqrt[4]{n})$ for the symmetric difference of interval graphs. We independently came up with different construction leading to the same lower bound, however analysis of our construction is simpler than the one from the previous work. For interval graphs, we also present upper bound $O(\sqrt[3]{n})$ for symmetric difference. Thus, we are leaving a gap between the lower and upper bound. However, we show the symmetric difference of interval graphs is polynomial and strictly smaller than symmetric difference of circular arc graphs.

2 Functionality

2.1 Finite Projective Planes

Recall that a finite projective plane is a pair $(X,\mathcal{L})$, where $X$ is a finite set and $\mathcal{L}\subseteq 2^{X}$, satisfying the following axioms: [7]

For every $p\neq q\in X$, there is exactly one subset of $\mathcal{L}$ containing $p$ and $q$. 2. 2.

For every $L\neq M\in\mathcal{L}$, we have $|L\cap M|=1$. 3. 3.

There exists a subset $Y\subseteq X$ of size 4 such that $|L\cap Y|\leq 2$ for every $L\in\mathcal{L}$.

Elements of $X$ are called points and elements of $\mathcal{L}$ are called lines. We use several standard properties of a finite projective plane of order $k$: Each line contains exactly $k+1$ points. Each point is incident to exactly $k+1$ lines. The total number of points is $k^{2}+k+1$. The total number of lines is also $k^{2}+k+1$.

We note that for every $k$ which is a power of a prime, a finite projective plane of order $k$ can be constructed. The incidence graph of a finite projective plane is a bipartite graph with one part $X$ and the second part $\mathcal{L}$. In this graph, $x\in X$ is adjacent to $\ell\in\mathcal{L}$ iff $x$ is incident to $\ell$.

The following theorem shows that the incidence graph of a finite projective plane has functionality approximately $\sqrt{n}$, where $n$ is the number of vertices. Alecu et al. [1] have shown that there exists a function $f$ such that for every graph $G$ we have $\mathrm{vc}(G)\leq f(\mathrm{fun}(G))$. Our result complements this inequality by showing that the functionality of a graph cannot be upper bounded by its VC-dimension as it is known that the VC-dimension of a finite projective plane of any order is 2 [2].

Theorem 2.1

Consider a finite projective plane of order $k$ and its incidence graph $G$. Then, $\mathrm{fun}(G)=k+1$. Moreover, for any proper induced subgraph $G^{\prime}\subset G$ we have $\mathrm{fun}(G^{\prime})\leq k$.

Proof

First, note that $G$ is a $(k+1)$-regular graph, thus $\mathrm{fun}(G)\leq k+1$. Since $G$ is connected, every proper subgraph $G^{\prime}\subset G$ contains a vertex of degree at most $k$. Thus, $\mathrm{fun}(G^{\prime})\leq k$.

It remains to prove that functionality of every vertex $v\in V(G)$ is at least $k+1$. Let $\ell$ be a line of the projective plane. Because of the point-line duality of finite projective planes, it is enough to show that the functionality of $\ell$ is at least $k+1$. Let $S=\{p_{1},\dots,p_{a},\ell_{1},\dots,\ell_{b}\}$ be a set of points $p_{i}$ and lines $\ell_{j}$ (distinct from $\ell$) such that $|S|=a+b\leq k$. We will prove that there exist vertices $u$ and $w$ satisfying:

The vertices $u$ and $w$ are not in $S$ and they are not adjacent to any vertices in $S$. They are also different from $\ell$. 2. 2.

The vertex $u$ is adjacent to $\ell$, i.e., $u$ is a point incident to $\ell$. 3. 3.

The vertex $w$ is not adjacent to $\ell$, i.e., $w$ is a point not incident to $\ell$ or it is a line.

Once we prove the existence of these two vertices, we are done as their existence implies that $\ell$ is not a function of $S$.

There are $k+1$ points on each line and every two lines intersect in one point. It follows that there is a point $q$ on $\ell$ which is distinct from each $p_{i}$ and it is not incident with any $\ell_{i}$. Thus, the vertex $q$ is not adjacent to any vertex in $S$ and it is adjacent to $\ell$. We set $u=q$.

Let $L=\{\ell,\ell_{1},\dots,\ell_{b}\}$. We will prove the existence of the vertex $w$ by considering two cases. The first case is $b=k$, i.e., the set $S$ consists of $k$ lines and no points. We may consider any line $\ell^{\prime}\not\in L$ as the vertex $w$. The vertex $w$ is not adjacent to any vertex in $S$ as it contains only lines and it is not adjacent to the line $\ell$ either. Thus, the vertex $w$ satisfies the sought properties.

The second case is $b\leq k-1$. Let $P$ be the set of points $p_{1},\ldots,p_{a}$ together with all the points that are incident to some line from $L$. We will show that $|P|\leq k^{2}+k$, which implies existence of a vertex $w\in V(G)\setminus P$ satisfying the sought properties. First, note that if $b=0$ (i.e., $S$ contains only points), then $|P|\leq k+1+a=2k+1$. Next, we suppose that $b\geq 1$ which means that $L$ contains at least two lines. Recall that each pair of line intersects in a point. Thus, there are at most $(b+1)(k+1)-1$ points incident to the lines in $L$ (the $-1$ comes from the fact that we have at least two lines). Moreover, we have $a$ points $p_{1},\dots,p_{a}$. Thus, $|P|\leq(b+1)(k+1)-1+a$. Since $a+b\leq k$, we have $a\leq k-b$. Therefore,

[TABLE]

This concludes the second case and also the whole proof.

2.2 Upper Bound for General Graphs

Theorem 2.2

If $G$ is a graph with $n$ vertices, then $\mathrm{fun}(G)\leq\sqrt{cn\log n}$ for any $c>2$, if $n$ is big enough.

Proof

We show that there is $v\in V(G)$ such that $\mathrm{fun}_{G}(v)\leq d(n):=\sqrt{cn\log n}$. As $d(n)$ is increasing, this suffices to show existence of such vertex also in a subgraph of $G$. We will write $d=d(n)$.

For vertices $u\neq v\in V(G)$ we define their private neighborhood $PN(u,v)$ as the symmetric difference $N(u)\mathrm{\Delta}N(v)$. If $|PN(u,v)|<d$ for some $u$, $v$ then we are done, as we can certify neighborhood relation to $v$ by the set $PN(u,v)\cup\{u\}$. Next, suppose all sets $PN(u,v)$ have at least $d$ vertices. In this case we choose an arbitrary vertex $v\in V(G)$. We also choose a random set $S\subseteq V(G)$ by independently putting each vertex of $G-v$ to $S$ with probability $p=d/n$. The probability that adjacency to $v$ is not certified by $S$ is equal to

[TABLE]

We estimate this using the union bound by

[TABLE]

The $-1$ in the exponent is caused by $v\in PN(u_{1},u_{2})$, but $v$ cannot be chosen to $S$. Thus, the probability that $S$ is “bad for $v$” is at most $n^{2}e^{-p(d-1)}$, which is $o(1)$ whenever $c>2$.

Clearly the expected size of $S$ is $p(n-1)=\frac{n-1}{n}d$. So, by Markov inequality, $P(|S|>d)\leq\frac{n-1}{n}$. This means that the with positive probability $|S|\leq d$ and $S$ certifies adjacency to $v$.

2.3 Random Graphs

In this section, we prove our results about functionality of random graphs.

Theorem 2.3

The functionality of random graph $G=G(n,p)$ is $\Omega(\log n)$ with probability at least $1-\frac{1}{n^{\log n}}$ for any $\frac{3\log^{2}n}{n}\leq p\leq 1-\frac{3\log^{2}n}{n}$.

Proof

Note that $\mathrm{fun}(G)=\mathrm{fun}(\bar{G})$, where $\bar{G}$ is the complement of $G$. Since $G(n,1-p)$ is the complement of $G(n,p)$, we can suppose that $p\leq\frac{1}{2}$. We will prove that functionality of $G$ is larger than $k=\frac{1}{2}\log n$ (w.h.p.). Let $v,u_{1},\dots,u_{k}$ be vertices of $G$. We will show that $v$ is function of $u_{1},\dots,u_{k}$ only with small probability.

First suppose that $\frac{3\log^{2}n}{\sqrt{n}}\leq p\leq\frac{1}{2}$. We divide the rest of vertices $V^{\prime}=V(G)\setminus\{v,u_{1},\dots,u_{k}\}$ into buckets according to the adjacency to $u_{1},\dots,u_{k}$, i.e., in each bucket $B$ there are vertices such that for each $u_{i}$ all vertices in $B$ are adjacent to $u_{i}$ or none of them are adjacent to $u_{i}$. There are at most $2^{k}$ buckets and therefore, there is a bucket $B$ containing at least $\frac{|V^{\prime}|}{2^{k}}\geq\frac{\sqrt{n}}{2}$ vertices. However by definition of functionality, that means $v$ is connected to all vertices in $B$ or to none of them. This event occurs only with a probability

[TABLE]

We have $n\cdot\binom{n-1}{k}<n^{\frac{\log n}{2}}$ possibilities how to choose $v,u_{1},\dots,u_{k}$. Thus by union bound, we have that $\mathrm{fun}(G)\leq k$ with probability at most $\frac{1}{n^{\log n}}$.

Now, suppose that $\frac{3\log^{2}n}{n}\leq p\leq\frac{3\log^{2}n}{\sqrt{n}}$. Let $E^{\prime}$ be edges between $u_{1},\dots,u_{k}$ and $V^{\prime}=V\setminus\{v,u_{1},\dots,u_{k}\}$. We bound the expected size of $E^{\prime}$ as follows.

[TABLE]

By Chernoff bound, we have that probability that $|E^{\prime}|\geq 3\sqrt{n}\cdot\log^{3}n=\ell$ is at most $e^{-\sqrt{n}\log^{3}n}$. Thus with high probability, there are at least $|V^{\prime}|-\ell\geq\frac{2n}{3}$ vertices in $V^{\prime}$ that there is no edge between them and $u_{1},\dots,u_{k}$. Denote such vertices as $B_{0}$. Now, we proceed similarly as in the previous case. Again, the vertex $v$ has to be connected to all vertices in $B_{0}$ or none of them that occurs only with the following probability.

[TABLE]

Thus, the probability that $v$ is function of $u_{1},\dots,u_{k}$ is at most

[TABLE]

By the same union bound as before, we have that $\mathrm{fun}(G)\leq k$ with probability at most $\frac{1}{n^{\log n}}$.

Lemma 1

Each vertex of random graphs $G=G(n,\frac{1}{2})$ is function of at most $3\log n$ vertices with probability at least $1-\frac{1}{n}$.

Proof

Let $k=3\log n$ and $V^{\prime}$ be a set of $k$ vertices. We will show that with high probability there are no two vertices $v_{1},v_{2}\not\in V^{\prime}$ such that $N(v_{1})\cap V^{\prime}=N(v_{2})\cap V^{\prime}$, i.e., they have the same neighborhood in $V^{\prime}$. Let $v_{1},v_{2}\not\in V^{\prime}$.

[TABLE]

By union bound, we have:

[TABLE]

Thus, each vertex outside of $V^{\prime}$ has a unique neighborhood to $V^{\prime}$. Therefore, $V^{\prime}$ determines adjacency for any vertex not in $V^{\prime}$.

3 Symmetric Difference

In this section, we will prove our lower and upper bounds for symmetric difference of interval and circular arc graphs.

3.1 Circular Arc Graphs

In this section, we prove that symmetric difference of circular arc graphs is $\Theta(\sqrt{n})$. More formally, we prove the following two theorems.

Theorem 3.1

Any circular arc graph $G\in\mathrm{CA}_{n}$ has symmetric difference at most $O(\sqrt{n})$.

Theorem 3.2

There is a circular arc graph $G\in\mathrm{CA}_{n}$ of symmetric difference at least $\Omega(\sqrt{n})$.

First, we prove the upper bound, i.e., that every circular arc graph has symmetric difference at most $O(\sqrt{n})$. We use the following notations for arcs. Let $a,b$ be two points of a circle. Then, an arc $r=[a,b]$ is an arc beginning in $a$ and going in clockwise direction to $b$. We call $a$ as the starting point of $r$ and $b$ as the ending point of $a$.

Proof (Proof of Theorem 3.1)

Let circle arc graph $G=(V,E)$ be an intersection graph of a set of arcs $R=\{r_{1},\dots,r_{n}\}$ of a circle $C$. Without loss of generality, we can suppose that circumference of $C$ is $2n$, endpoints of all arcs $r_{i}\in R$ are integer points and are different for all arcs. Thus, each integer point $\{0,\dots,2n-1\}$ of $C$ is an endpoint of exactly one arc in $R$.

Consider an arc $r=[a,b]\in R$. We represent the arc $r$ as a point $(a,b)$ in a plane $\mathbb{R}^{2}$. Note that all these points are in the square $S$ with corners in the points $(0,0)$ and $(2n-1,2n-1)$ (some points maybe on the border of $S$). We divide the square $S$ into subsquares of size $2\sqrt{n}\times 2\sqrt{n}$. Since the area of $S$ is strictly less than $4n^{2}$, we have strictly less than $n$ subsquares. Thus, there is at least one subsquare that contains two points representing arcs, say $r=[a,b]$ and $r^{\prime}=[a^{\prime},b^{\prime}]$. It follows that $|a-a^{\prime}|,|b-b^{\prime}|\leq 2\sqrt{n}$. Suppose that $a^{\prime}>a$ and $b^{\prime}>b$, other cases are analogous. Then, each arc counted in $\mathrm{sd}(r,r^{\prime})$ has to start or end in an integer point from the interval $[a,a^{\prime}-1]$ or $[b,b^{\prime}-1]$. Since there are at most $4\sqrt{n}$ integer points in these two intervals and each integer point of $C$ is an endpoint of exactly one arc of $R$, we conclude that $\mathrm{sd}(r,r^{\prime})\leq 4\sqrt{n}$.

Now, we give a construction of a circular arc graph of symmetric difference at least $\Omega(\sqrt{n})$. Let $n$ be a square of an integer, i.e., $n=d^{2}$ for some $d\in\mathbb{N}$. We consider a circle $C$ of circumference $n$ and a set $P$ of integer points of $C$, i.e., $P=\{0,\dots,n-1\}$ ordered in clockwise direction. The length $|r|$ of an arc $a=[a,b]$ is equal to $b-a\pmod{n}$. We say the arc $[a,b]$ is integral if both $a$ and $b$ are integers.

We will represent each point $p\in P$ as two integer indices $0\leq i,j<d=\sqrt{n}$ such that $p=i\cdot d+j$. Note that each point $p$ has a unique such representation and we denote it as $(i,j)_{d}$. Let $R$ be a set of arcs \bigl{[}(i,j)_{d},(j,i)_{d}\bigr{]} for all possible $i\neq j$ such that length of each arc in $R$ is at most $\frac{n}{4}$. Note that we require that $i\neq j$, thus we do not consider zero-length arcs consisting only of a point of a form $(i,i)_{d}$. See Figure 1 for an illustration. Let $G$ be an intersection graphs of arcs in $R$. We we will prove that $\mathrm{sd}(G)\geq\Omega(\sqrt{n})$. First, we prove two auxiliary lemmas. The first one asserts that each sufficiently long arc of $C$ contains a lot of starting and ending points of arcs in $R$. The seconds one states that for any two arcs $r$ and $r^{\prime}$ we will find a long arc $s$ that is a subarc of only one of the arcs $r$ and $r^{\prime}$ (say $r$). Thus by the first lemma, the arc $r$ intersects many arcs that go “away” from the arc $r^{\prime}$ and that is enough to imply Theorem 3.2.

Lemma 2

Let $s$ be an integral arc of $C$ of length at least $d-1$. Then, it contains at least $\frac{d}{5}$ integer points such that they are starting points of arcs in $R$. Similarly, it contains at least $\frac{d}{5}$ integer points such that they are ending points of arcs in $R$.

Proof

Since $s$ is integral and its length is at least $d-1$, it contains at least $d$ consecutive integer points. Thus,there is an integer $j$ such that $s$ contains all points of the set $\{(j,t)_{d},\dots,(j,d-1)_{d},(j+1,0)_{d},\dots,(j+1,t-1)_{d}\}$ for some $t\in[d]$. For any $\ell\in[d]$, we define $j_{\ell}=j$ if $\ell\geq t$ and $j_{\ell}=j+1$ otherwise. It follows that $s$ contains the point $(j_{\ell},\ell)_{d}$ for every $\ell\in[d]$. Consider a subset of these points S=\left\{\bigl{(}j_{k},k\bigr{)}_{d}\mid 1\leq(k-j\mod d)\leq\frac{d}{5}\right\}. Observe that $|S|=\frac{d}{5}$. We claim that each point of $S$ is a starting point of an arc in $R$. Let $r$ be an arc \bigl{[}(j_{k},k)_{d},(k,j_{k})_{d}\bigr{]} where $1\leq(k-j\mod d)\leq\frac{d}{5}$.

[TABLE]

Thus, $r\in R$. Analogously, the set E=\left\{\bigl{(}j_{k},k\bigr{)}_{d}\mid 1\leq(j+1-k\mod d)\leq\frac{d}{5}\right\} of size $\frac{d}{5}$ contains ending points of arcs in $R$.

Lemma 3

Let $r=[(i,j)_{d},(j,i)_{d}]$ and $r^{\prime}=[(i^{\prime},j^{\prime})_{d},(j^{\prime},i^{\prime})_{d}]$ be two arcs in $R$. Then, at least one of the arcs $r$ and $r^{\prime}$ contains an integral subarc $s$ of length $d-2$ that is disjoint from the other arc.

Proof

If the arcs $r$ and $r^{\prime}$ are disjoint then the existence of $s$ is trivial as each arc in $R$ has length at least $d-1$ (the arc in $R$ of length exactly $d-1$ are those of the form $[(k,k+1)_{d},(k+1,k)_{d}]$).

Thus, suppose that $r$ and $r^{\prime}$ intersect and without loss of generality suppose that the ending point of $r$ lies inside $r^{\prime}$, i.e. we read in clockwise order the points $(i^{\prime},j^{\prime})_{d}$, $(j,i)_{d}$ and $(j^{\prime},i^{\prime})_{d}$. Moreover, we can assume without loss of generality that neither of the arcs $r$ and $r^{\prime}$ contains the point [math], i.e., we have $i\leq j$, $i^{\prime}\leq j^{\prime}$ and $j\leq j^{\prime}$. Otherwise, we could cyclically shift the numbering of points by a suitable multiple of $d$.

Consider an arc $s^{\prime}=[(j,i)_{d},(j^{\prime},i^{\prime})_{d}]$. Note that arc $s^{\prime}$ is a subarc of $r^{\prime}$ and intersects $r$ only in the point $(j,i)_{d}$. Thus, if $|s^{\prime}|\geq d-1$, the arc $r^{\prime}$ would contain the sought integral subarc $s$. We have $|s^{\prime}|=(j^{\prime}-j)\cdot d+i^{\prime}-i$. Thus, $|s^{\prime}|\geq d-1$ if $j^{\prime}-j\pmod{d}\geq 2$ or $j^{\prime}-j\pmod{d}=1$ and $i^{\prime}\geq i-1$.

Suppose that $|s^{\prime}|<d-1$. It follows that $j^{\prime}=j$ (and thus $i^{\prime}\neq i$) or $j^{\prime}-j\pmod{d}=1$ and $i^{\prime}\leq i-2$. First, suppose that $i^{\prime}\leq i-2$ (and $j^{\prime}-j\pmod{d}=1$). Then, the start point of $r$ also belongs to $r^{\prime}$ and the arc $s^{\prime}=[(i^{\prime},j^{\prime})_{d},(i,j)_{d}]$ has length at least $d-1$ and again it is a subarc of $r^{\prime}$ and intersects the arc $r$ only in the single point $(i,j)_{d}$.

The last case is when $j^{\prime}=j$. This implies that $i\neq i^{\prime}$ as all endpoints are unique. Moreover, since the point $(j,i)_{d}$ precedes the point $(j^{\prime},i^{\prime})_{d}$ in clockwise order, we get that $i<i^{\prime}$. Then, the arc $s^{\prime}=[(i,j)_{d},(i^{\prime},j^{\prime})_{d}]=[(i,j)_{d},(i^{\prime},j)_{d}]$ has length at least $d$ and is a subarc of only the arc $r$ with the except of its ending point $(i^{\prime},j^{\prime})_{d}$.

Now, we are ready to prove Theorem 3.2.

Proof (Proof of Theorem 3.2)

Consider two arcs $r=[(i,j)_{d},(j,i)_{d}]$ and $r^{\prime}=[(i^{\prime},j^{\prime})_{d},(j^{\prime},i^{\prime})_{d}]$ in $R$. Recall that all arcs in $R$ has length at most $\frac{n}{4}$. Thus, we can suppose that the arc $s^{\prime}=[(j^{\prime},i^{\prime})_{d},(i,j)_{d}]$ has length at least $\frac{n}{4}$ and is disjoint from $r$ and $r^{\prime}$, except the endpoints $(j^{\prime},i^{\prime})_{d}$ and $(i,j)_{d}$.

By Lemma 3, we suppose that $r$ contains a subarc $s$ of length $d-2$ that is disjoint with $r^{\prime}$ (the other case when $s$ is subarc of $r^{\prime}$ is analogous). Let $L$ be a set of points in $s$ such that they are ending points of arcs in $R$ (distinct from $r$). By Lemma 2, we have that $|L|\geq\frac{d}{5}-1$.

Let $t=[a,b]$ be an arc with the ending point $b$ in $L$. The arc $t$ intersects the arc $r$ as the ending point $b$ is a point of $s\subseteq r$. Since the arc $s^{\prime}$ has length at least $\frac{n}{4}$ and $t\in R$, the starting point $a$ has to be a point of $s^{\prime}$ and thus, the arc $t$ is disjoint with $r^{\prime}$. Therefore, $\mathrm{sd}(r,r^{\prime})\geq|L|\geq\Omega(\sqrt{n})$.

3.2 Interval Graphs

In this section, we will prove that symmetric difference of interval graphs is still polynomial but strictly less than the symmetric difference of circular arc graphs. In particular, we will prove the following two theorems.

Theorem 3.3

Any interval graph $G\in\mathrm{INT}_{n}$ has symmetric difference at most $O(\sqrt[3]{n})$.

Theorem 3.4

There is an interval graph $G\in\mathrm{INT}_{n}$ of symmetric difference at least $\Omega(\sqrt[4]{n})$.

First, we prove the upper bound.

Proof (Proof of Theorem 3.3)

Let $G$ be an intersection graph of intervals $R=\{r_{1},\dots,r_{n}\}$ and let $r_{i}=[a_{i},b_{i}]$. The intervals are numbered in the order given by their starting points, i.e., for every two indices $i,j\in[n]$ we have $i<j$ if and only if $a_{i}<a_{j}$. For an interval $r_{i}$, we define two sets:

$A_{i}=\{j\mid a_{i}<b_{j}\}$, i.e., it contains the indices of intervals in $R$ that ends after the interval $r_{i}$ starts. 2. 2.

$B_{i}=\{j\mid a_{j}<b_{i}\}$, i.e., it contains the indices of intervals in $R$ that starts before the interval $r_{i}$ ends.

Note that $N(r_{i})=A_{i}\cap B_{i}$. Moreover, $\mathrm{sd}(r_{i},r_{j})\leq|N(r_{i})\mathrm{\Delta}N(r_{j})|\leq|A_{i}\mathrm{\Delta}A_{j}|+|B_{i}\mathrm{\Delta}B_{j}|$. Note that for each pair $i,j$ holds that $A_{i}\subseteq A_{j}$ or $A_{j}\subseteq A_{i}$, thus $A_{i}\mathrm{\Delta}A_{j}$ is $A_{i}\setminus A_{j}$ or $A_{j}\setminus A_{i}$ and analogously with $B_{i}$ and $B_{j}$. We will prove that there are two intervals $r_{i}$ and $r_{j}$ such that $|A_{i}\mathrm{\Delta}A_{j}|+|B_{i}\mathrm{\Delta}B_{j}|\leq O(\sqrt[3]{n})$.

Let $d\in\mathbb{N}$ be a parameter. We will prove that $\mathrm{sd}(G)\leq O(\max\{d,d+\frac{n}{d^{2}}\})$. Thus, if we set $d=\sqrt[3]{n}$ we would get $\mathrm{sd}(G)\leq O(\sqrt[3]{n})$. Let $D_{\ell}=A_{\ell}\setminus A_{\ell+1}$. In other words, the set $D_{\ell}$ contains intervals of $R$ that end after the interval $r_{\ell}$ starts but before the interval $r_{\ell+1}$ starts. Note that $B_{i}=\{1,\dots,\ell\}$ where $\ell$ is the index such that $i\in D_{\ell}$. Let $k$ be the largest index such that $\sum_{\ell\leq k}|D_{\ell}|\leq d+2$. Note that for any two indices $i,j\leq k$ holds that $|A_{i}\mathrm{\Delta}A_{j}|\leq\sum_{\ell\leq k}|D_{\ell}|$.

First, we prove that if $k\leq d^{2}$, then $\mathrm{sd}(G)\leq 2d+2$. Suppose that actually $\sum_{\ell\leq k}|D_{\ell}|\leq d$. Then, $|D_{k+1}|\geq 3$. Let $i,j\in D_{k+1}$ such that $i,j\leq k$. Then, $B_{i}=B_{j}$ and $|A_{i}\mathrm{\Delta}A_{j}|\leq d$, therefore $\mathrm{sd}(G)\leq d$.

From now, we suppose that $d\leq\sum_{\ell\leq k}|D_{\ell}|\leq d+2$. Let $p\leq k$ be an index such that $|D_{p}|\geq 2$ (if such $p$ exists) and $i,j\in D_{p}=A_{p}\setminus A_{p+1}$. Thus, $B_{i}=B_{j}$. Since $i,j\leq p\leq k$, then $|A_{i}\mathrm{\Delta}A_{j}|\leq d+2$ and $\mathrm{sd}(G)\leq d+2$. Note that so far, we did not use the assumption that $k\leq d^{2}$.

Now, suppose that for all $\ell\leq k$ it holds that $|D_{\ell}|\leq 1$. Since $k\leq d^{2}$ there exists two indices $p<q\leq k$ such that $D_{p},D_{q}\neq\emptyset$ and $q-p\leq d$ (there are at least $d$ indices $\ell\leq k$ such that $|D_{\ell}|=1$). Let $i\in D_{p}$ and $j\in D_{q}$. Since $B_{i}=\{1,\dots,p\},B_{j}=\{1,\dots,q\}$, we have $|B_{j}\setminus B_{i}|\leq d$. Further, since $i,j\leq k$, we have $|A_{i}\mathrm{\Delta}A_{j}|\leq d+2$. Thus, $\mathrm{sd}(G)\leq 2d+2$.

Now, we suppose that $k>d^{2}$. It follows there are two indices $i,j\leq k$ such that $i\in D_{p}$ and $j\in D_{q}$ such that $|p-q|\leq\frac{n}{d^{2}}$. Therefore, $|A_{i}\mathrm{\Delta}A_{j}|\leq d+2$ and $|B_{i}\mathrm{\Delta}B_{j}|=|p-q|=\frac{n}{d^{2}}$ and $\mathrm{sd}(G)\leq d+2+\frac{n}{d^{2}}$.

Now, we give a construction of interval graph of symmetric difference $\Omega(\sqrt[4]{n})$. Let $d\in\mathbb{N}$. We construct $\Theta(d^{4})$ intervals on a line segment $[0,t]$ for $t=20d^{3}$ such that the corresponding intersection graph $G$ will have $\mathrm{sd}(G)\geq d$. There will be intervals of two types – short and long. See Figure 2 for an illustration. Short intervals have length $d$ and they start in each point $0,\dots,t-d$, i.e.,

[TABLE]

Long intervals will have various lengths. For $i\geq 0$, let $\ell_{i}=4d^{2}\cdot(i+1)$. For $0\leq i\leq 2d-1$, we define the $i$-th class of long intervals as

[TABLE]

i.e., the set $L_{i}$ contains intervals such that they start and end in points congruent to $i$ module $2d$ and their length is between $\ell_{i}$ and $\ell_{i}+2d^{2}$. Let $I=S\cup\bigcup_{0\leq i\leq 2d-1}L_{i}$ be a set of all constructed intervals. We start with two observations about $I$.

Observation 3.5

Any interval $[a,b]$ (for $a,b\in\mathbb{N}$) of the line segment $[0,t]$ of length $2d$ contains $d$ short intervals.

Observation 3.6

There are at most $O(d^{4})$ intervals in $I$.

Proof

Clearly, there are $t-d+1=O(d^{3})$ intervals in $S$. Note that for any $a\leq\frac{t}{2},a\equiv i\pmod{2d}$, there are exactly $d$ intervals of a form $[a,b]$ in $L_{i}$ because of the length constraints of the long interval. Analogously, for any $b\geq\frac{t}{2},a\equiv i\pmod{2d}$, there are exactly $d$ intervals of a form $[a,b]$ in $L_{i}$. We remark this indeed holds even for $L_{2d-1}$ as we set $t$ to be large enough. There are no other intervals in $L_{i}$. Since there are $O(d^{2})$ points $p\in[0,t]$ such that $p\equiv i\pmod{2d}$, it follows that $|L_{i}|\leq O(d^{3})$. Therefore, there are at most $O(d^{4})$ long intervals as there are $O(d)$ classes of long intervals.

The graph $G$ is an intersection graph of $I$. Now, we are ready to proof Theorem 3.4, i.e., $\mathrm{sd}(G)\geq\Omega(\sqrt[4]{n})$.

Proof (Proof of Theorem 3.4)

Let $r=[a,b]$ and $r^{\prime}=[a^{\prime},b^{\prime}]$ be two intervals in $I$. Without loss of generality let $a\leq a^{\prime}$. First, suppose the $r,r^{\prime}\in S$, i.e., both of them are short. In this case it holds that $a<a^{\prime}$ and $b<b^{\prime}$. Suppose that $b^{\prime}\leq\frac{t}{2}$. Then as already observed, there are $d$ intervals of a form $[b^{\prime},c]$ in $L_{i}$ for $b^{\prime}\equiv i\pmod{2d}$. Since $b^{\prime}$ is not in $r$, we have that $N(r)\mathrm{\Delta}N(r^{\prime})\geq d$. If $b^{\prime}>\frac{t}{2}$, then analogously there are $d$ intervals in $L_{j}$ ending in $a$ (for $a\equiv j\pmod{2d}$).

Now, suppose that $r,r^{\prime}\in L_{i}$ for some $i$. In this case $a,b,a^{\prime},b^{\prime}\equiv i\pmod{2d}$. Since $r\neq r^{\prime}$, it follows that $|a-a^{\prime}|\geq 2d$ or $|b-b^{\prime}|\geq 2d$. Thus, at least one of the intervals $r$ and $r^{\prime}$ has a private subinterval of length at least $2d$ and by Observation 3.5, we have that $N(r)\mathrm{\Delta}N(r^{\prime})\geq d$.

Let $k$ be a difference of length of $r$ and $r^{\prime}$. For the remaining cases we will prove that $k\geq 4d$. Then, at least one of the intervals $r$ and $r^{\prime}$ contains a private subinterval of length at least $2d$ and again by Observation 3.5, we have that $N(r)\mathrm{\Delta}N(r^{\prime})\geq d$. There are two remaining cases:

$r\in S,r^{\prime}\in L_{i}:$ Then, $|r|=d$ and $|r^{\prime}|\geq 4d^{2}$. 2. 2.

$r\in L_{i},r^{\prime}\in L_{j}$ for $i<j$: Then, $|r|\leq 4d^{2}\cdot(i+1)+2d^{2}$ and $|r^{\prime}|\geq 4d^{2}\cdot(j+1)$. It follows that $k=4d^{2}\cdot(j-i)-2d^{2}\geq 2d^{2}$.

Thus, in both cases we have that $k\geq 4d$ for $d\geq 2$.

We showed that $N(r)\mathrm{\Delta}N(r^{\prime})\geq d$ for all cases. Thus by Observation 3.6, we conclude that $\mathrm{sd}(G)\geq\Omega(\sqrt[4]{n})$.

3.2.1 Acknowledgements

The research presented in this paper has been started during the KAMAK workshop in 2021. We are grateful to the organizers of this wonderful event.