Fully-Dynamic and Kinetic Conflict-Free Coloring of Intervals with Respect to Points

TL;DR

This paper studies the problem of maintaining conflict-free colorings of intervals with respect to points in a dynamic setting, providing bounds on recoloring costs and number of colors, including for kinetic scenarios.

Contribution

It introduces the fully-dynamic conflict-free coloring problem for intervals, establishes bounds on recoloring costs versus number of colors, and explores kinetic case solutions.

Findings

Lower bound on recolorings related to number of colors

Algorithms using O(log n) colors with O(log n) recolorings

Kinetic setting with 4-color coloring and tight bounds

Abstract

We introduce the fully-dynamic conflict-free coloring problem for a set $S$ of intervals in $\mathbb{R}^1$ with respect to points, where the goal is to maintain a conflict-free coloring for$S$ under insertions and deletions. A coloring is conflict-free if for each point $p$ contained in some interval, $p$ is contained in an interval whose color is not shared with any other interval containing $p$. We investigate trade-offs between the number of colors used and the number of intervals that are recolored upon insertion or deletion of an interval. Our results include: - a lower bound on the number of recolorings as a function of the number of colors, which implies that with $O(1)$ recolorings per update the worst-case number of colors is $\Omega(\log n/\log\log n)$, and that any strategy using $O(1/\varepsilon)$ colors needs $\Omega(\varepsilon n^{\varepsilon})$ recolorings; - a coloring strategy that uses $O(\log n)$ colors at the cost of $O(\log n)$ recolorings, and another strategy that uses $O(1/\varepsilon)$ colors at the cost of $O(n^{\varepsilon}/\varepsilon)$ recolorings; - stronger upper and lower bounds for special cases. We also consider the kinetic setting where the intervals move continuously (but there are no insertions or deletions); here we show how to maintain a coloring with only four colors at the cost of three recolorings per event and show this is tight.

Equations

$$|R^{*}_{i}|\geqslant\left\{\begin{array}[]{l l}\dfrac{\left\lfloor|R^{*}_{i-1}|/4r\right\rfloor}{2c}&\text{if }i>1,\\[10.0pt] \dfrac{n_{1}}{c}&\text{if }i=1.\end{array}\right.$$

$$|R_i^{\ast }|⩾\frac{\lfloor |R_{i-1}^{\ast }|/4r\rfloor }{2c}⩾\frac{1}{2c}\cdot \left(\left(\frac{n_1}{(8rc{)}^{i-1}}-1\right)/4r-1\right)>\frac{n_1}{(8rc{)}^i}-1.$$

$$|R_{i}|\geqslant\left\{\begin{array}[]{l l}\left\lfloor\dfrac{|R_{i-1}|}{r+2}\right\rfloor&\text{if }i>1,\\[20.0pt] \dfrac{n}{2}&\text{if }i=1.\end{array}\right.$$

$$\begin{array}[]{lll}|R_{i}|&\geqslant&\left\lfloor|R_{i-1}|/(r+2)\right\rfloor\\[5.69054pt] &\geqslant&\left\lfloor\left(\frac{n}{(r+2)^{i-1}}-2\right)/(r+2)\right\rfloor\\[5.69054pt] &\geqslant&\left(\left(\frac{n}{(r+2)^{i-1}}-2\right)/(r+2)-1\right)\\[5.69054pt] &=&\frac{n}{(r+2)^{i}}-\frac{2}{r+2}-1\\[5.69054pt] &>&\frac{n}{(r+2)^{i}}-2.\end{array}$$

$$\frac{n/2}{(r+2{)}^{\rho }}-2⩽|R_{\rho }|<r+1,$$

$$G_1,\dots ,G_7:=\{I_1\},\{I_1,I_2\},\{I_1,I_2,I_3,\},\{I_1,I_2,I_3,I_4\},\{I_1,I_3,I_4\},\{I_3,I_4\},\{I_3\}.$$

Full text

• •

a lower bound on the number of recolorings as a function of the number of colors, which implies that with $O(1)$ recolorings per update the worst-case number of colors is $\Omega(\log n/\log\log n)$, and that any strategy using $O(1/\varepsilon)$ colors needs $\Omega(\varepsilon n^{\varepsilon})$ recolorings;

• •

a coloring strategy that uses $O(\log n)$ colors at the cost of $O(\log n)$ recolorings, and another strategy that uses $O(1/\varepsilon)$ colors at the cost of $O(n^{\varepsilon}/\varepsilon)$ recolorings;

• •

stronger upper and lower bounds for special cases.

We also consider the kinetic setting where the intervals move continuously (but there are no insertions or deletions); here we show how to maintain a coloring with only four colors at the cost of three recolorings per event and show this is tight.

1 Introduction

Consider a set $S$ of fixed base stations that can be used for communication by mobile clients. Each base station has a transmission range, and a client can potentially communicate via that base station when it lies within the transmission range. However, when a client is within reach of several base stations that use the same frequency, the signals will interfere. Hence, the frequencies of the base stations should be assigned in such a way that this problem does not arise. Moreover, the number of used frequencies should not be too large. Even et al. [11] and Smorodinsky [16] introduced conflict-free colorings to model this problem, as follows. Let $S$ be a set of disks in the plane, and for a point $q\in\mathbb{R}^{2}$ let $S(q)\subseteq S$ denote the set of disks containing the point $q$. A coloring of the disks in $S$ is conflict-free if, for any point $q\in\mathbb{R}^{2}$ with non-empty $S(q)$, the set $S(q)$ has at least one disk with a color that is unique among the disks in $S(q)$. Even et al. [11] proved that any set of $n$ disks in the plane admits a conflict-free coloring with $O(\log n)$ colors, and this bound is tight in the worst case.

The concept of conflict-free colorings can be generalized and extended in several ways, giving rise to a host of challenging problems. Below we mention some of them, focussing on the papers most directly related to our work. A more extensive overview is given by Smorodinsky [17]. One obvious generalization is to work with types of regions other than disks. For instance, Even et al. [11] showed how to find a coloring with $O(\log n)$ colors for a set of translations of any single centrally symmetric polygon. Har-Peled and Smorodinsky [13] extended this result to regions with near-linear union complexity. One can also consider the dual setting, where one wants to color a given set $P$ of $n$ points in the plane, such that any disk—or rectangle, or other range from a given family—contains at least one point with a unique color (if it contains any point at all). This too was studied by Even et al. [11] and they show that this can be done with $O(\log n)$ colors when the ranges are disks or scaled translations of a single centrally symmetric convex polygon.

The results mentioned above deal with the static setting, in which the set of objects to be colored is known in advance. This may not always be the case, leading Fiat et al. [12] to introduce the online version of the conflict-free coloring problem. Here the objects to be colored arrive one at a time, and each object must be colored upon arrival. There is also a lot of work on online coloring of other types (e.g. [15, 10]), but we restrict our discussion to conflict-free colorings. Fiat et al. [12] show that when coloring points in the plane with respect to disks, $n$ colors may be needed in the online version. Hence, they turn their attention to the 1-dimensional problem of online coloring points with respect to intervals. They prove that this can be done deterministically with $O(\log^{2}n)$ colors and randomized with $O(\log n\log\log n)$ colors with high probability. Later Chen [8] gave a randomized algorithm that uses $O(\log n)$ colors with high probability. In the same paper, similar results were obtained for conflict-free colorings of points with respect to halfplanes, unit disks and axis-aligned rectangles of almost the same size. In these cases the colorings use $O(\mathrm{polylog}\ n)$ colors with high probability. Bar-Noy, Cheilaris, and Smorodinsky [3] discussed several versions of the deterministic one-dimensional variant. Furthermore, Abam et al. [1] studied the dual version of coloring intervals on a line with respect to points, providing an algorithm that use $O(\log^{3}n)$ colors in general or $O(\log n)$ colors when all intervals contain a specific point. Later, Bar-Noy et al. [2] considered the case where recolorings are allowed for each insertion. They prove that for coloring points in the plane with respect to halfplanes, one can obtain a coloring with $O(\log n)$ colors in an online setting at the cost of $O(n)$ recolorings in total. More recent variants include strong conflict-free colorings [7, 14], where we require several unique colors, and conflict-free multicolorings [4], which allow assigning multiple colors to a point. Even more variants of online conflict-free colorings can be found in the survey [17].

Our contributions.

We introduce a variant of the conflict-free coloring problem where the objects to be colored arrive and disappear over time. This fully-dynamic conflict-free coloring problem models a scenario where new base stations may be deployed (to deal with increased capacity demands, for example) and existing base stations may break down or be taken out of service (either permanently or temporarily). We also define the semi-dynamic conflict-free coloring problem as the online variant where recolorings are allowed (or the fully-dynamic variant without deletions). Note that when we talk about the dynamic variant, we mean fully-dynamic. These natural variants have, to the best of our knowledge, not been considered so far. It is easy to see that, unless one maintains a coloring in which any two intersecting objects have distinct colors, there is always a sequence of deletions that invalidates a given conflict-free coloring. Hence, recolorings are needed to ensure that the new coloring is conflict-free. This leads to the question: how many recolorings are needed to maintain a coloring with a certain number of colors? We initiate the study of fully-dynamic conflict-free colorings by considering the problem of coloring intervals with respect to points. In this variant, we are given a (dynamic) set $S$ of intervals in $\mathbb{R}^{1}$, which we want to color such that for any point $q\in\mathbb{R}^{1}$ the set $S(q)$ of intervals containing $q$ contains an interval with a unique color. This version of the problem can be used to model the case where the base stations are located along a highway, for instance, and 1-dimensional range and frequency assignment problems have already been studied in various settings [2, 7, 12]. Moreover, the lower bounds that we prove hold for the 2-dimensional problem as well. In the static setting, coloring intervals is rather easy: a simple procedure yields a conflict-free coloring with three colors. The dynamic version turns out to be much more challenging.

In Section 2 we prove lower bounds on the possible tradeoffs between the number of colors used and the worst-case number of recolorings per update: for any algorithm that maintains a conflict-free coloring on a sequence of $n$ insertions of intervals with at most $c(n)$ colors and at most $r(n)$ recolorings per insertion, we must have $r(n)>n^{1/(c(n)+1)}/(8c(n))$. This implies that for $O(1/\varepsilon)$ colors we need $\Omega(\varepsilon n^{\varepsilon})$ recolorings per updated, and with only $O(1)$ recolorings per update we must use $\Omega(\log n/\log\log n)$ colors.

In Section 3 we then present several algorithms that achieve bounds close to our lower bound. All bounds are worst-case, unless specifically stated otherwise. First, we present two algorithms for the case where the interval endpoints come from a universe of size $U$. One algorithm uses $O(\log U)$ colors and two recolorings per update; the other uses $O(\log_{t}U)$ colors and $O(t)$ recolorings per update in the worst case, where $2\leqslant t\leqslant U$ is a parameter. We then extend the second algorithm to an unbounded universe, leading to two results: we can use $O(\log_{t}n)$ colors and perform at most $O(t\log_{t}n)$ recolorings per update for any fixed $t\geqslant 2$, or we can use $O(1/\varepsilon)$ colors and $O(n^{\varepsilon}/\varepsilon)$ recolorings, amortized, for any fixed $\varepsilon>0$.

Next, in Section 4 we consider the online setting, where intervals arrive one by one and recolorings are not allowed. Abam et al. [1] observed that for this setting there is a lower bound of $\log n$ on the number of required colors. They also provide an upper bound of $O(\log^{3}n)$ for general instances, as well as an upper bound of $\log n$ when all intervals are nested and contain the origin. We provide an upper bound of $\log n$ on nested intervals even if not all of them contain the origin. We also use the result of Bar-Noy et al. [2] to provide a non-deterministic algorithm on any instance (non-nested) with $O(\log n)$ colors with high probability.

Finally, in Section 5 we turn our attention to kinetic conflict-free colorings. Here the intervals do not appear or disappear, but their endpoints move continuously on the real line. At each event where two endpoints of different intervals cross each other, the coloring may need to be adapted so that it stays conflict-free. One way to handle this is to delete the two intervals involved in the event, and re-insert them with the new endpoint order. We show that a specialized approach is much more efficient: we show how to maintain a conflict-free coloring with four colors at the cost of three recolorings per event. We also show that on average $\Theta(1)$ recolorings per event are needed in the worst case when using only four colors.

Preliminaries: the chain method.

We first present the so called chain method that colors intervals statically with three colors. This method is used later, but we present it here to give the reader a sense of CF-colorings. Consider the intervals $I_{1},\ldots,I_{n}$ and suppose without loss of generality that their union is an interval. Let $I_{i}$ be the interval with the leftmost left endpoint in this component. We color $I_{i}$ red. Then, of all intervals whose left endpoint lies inside $I_{i}$ we select the interval $I_{j}$ with the rightmost right endpoint and color it blue, of all intervals whose left endpoint lies inside $I_{j}$ we select the interval $I_{k}$ sticking out furthest to the right and color it red, and so on. This way we create a chain of intervals that are alternatingly colored red and blue. We then color all the remaining intervals using the dummy color, see Fig. 1.

2 Lower bounds for semi-dynamic conflict-free colorings

In this section we present lower bounds on the semi-dynamic (insertion only) conflict-free coloring problem for intervals. More precisely, we present lower bounds on the number of recolorings necessary to guarantee a given upper bound on the number of colors. We prove a general lower bound and a stronger bound for so-called local algorithms. The general lower bound uses a construction where the choice of segments to be added depends on the colors of the segments already inserted. This adaptive construction is also valid for randomized algorithms, but it does not give a lower bound on the expected behavior.

Theorem 2.1**.**

Let alg be a deterministic algorithm for the semi-dynamic conflict-free coloring of intervals. Suppose that on any sequence of $n>0$ insertions, alg uses at most $c(n)$ colors and $r(n)$ recolorings per insertion, where $r(n)>0$. Then $r(n)>n^{1/(c(n)+1)}/(8c(n))$.

Proof.

We first fix a value for $n$ and define $c:=c(n)$ and $r:=r(n)$. Our construction will proceed in rounds. In the $i$-th round we insert a set $R_{i}$ of $n_{i}$ disjoint intervals—which intervals we insert depends on the current coloring provided by alg. After $R_{i}$ has been inserted (and colored by alg), we choose one of the colors used by alg for $R_{i}$ to be the designated color for the $i$-th round. We denote this designated color by $c_{i}$. We will argue that in each round we can pick a different designated color, so that the number of rounds, $\rho$, is a lower bound on the number of colors used by alg. We then prove a lower bound on $\rho$ in terms of $n,c$, and $r$, and derive the theorem from the inequality $\rho\leqslant c$.

To describe our construction more precisely, we need to introduce some notation and terminology. Let $R_{i}:=\{I_{1},\ldots,I_{n_{i}}\}$, where the intervals are numbered from left to right. (Recall that the intervals in $R_{i}$ are disjoint.) To each interval $I=I_{j}$ we associate the set $I^{e}:=(a,b)$, where $a$ is the right endpoint of $I$, and $b$ is the left endpoint of $I_{j+1}$ if $j<n_{i}$ and $+\infty$ if $j=n_{i}$, that is, $I^{e}$ represents the empty space to the right of $I$. We call $(I,I^{e})$ an $i$-brick. We define the color of a brick $(I,I^{e})$ to be the color of $I$, and we say a point or an interval is contained in this brick if it is contained in $I\cup I^{e}$. Recall that each round $R_{i}$ has a designated color $c_{i}$. We say that an $i$-brick $B:=(I,I^{e})$ is living if:

• •

$I$ has the designated color $c_{i}$;

• •

if $i>1$ then both $I$ and $I^{e}$ contain living $(i-1)$-bricks.

A brick that is not alive is called dead and an event such as a recoloring that causes a brick to become dead is said to kill the brick. By recoloring an interval $I$, alg can kill the brick $B=(I,I^{e})$ and the death of $B$ may cause some bricks containing $B$ to be killed as well.

We can now describe how we generate the set $R_{i}$ of intervals we insert in the $i$-th round and how we pick the designated colors. (Note that the designated color of a round is fixed once it is picked; it is not updated when recolorings occur.) We denote by $R_{i}^{*}$ the subset of intervals $I\in R_{i}$ such that $(I,I^{e})$ is a living $i$-brick. Note that $R_{i}^{*}$ can be defined only after the $i$-th round, when we have picked the designated color $c_{i}$.

The set $R_{1}$ contains the $\frac{n}{2}$ intervals $[0,1],[2,3],\ldots,[n-2,n-1]$, and the designated color $c_{1}$ of the first round is the color used most often in the coloring produced by alg after insertion of the last interval in $R_{1}$. 2. 2.

To generate $R_{i}$ for $i>1$, we proceed as follows. Partition $R^{*}_{i-1}$ into groups of $4r$ consecutive intervals. (If $|R^{*}_{i-1}|$ is not a multiple of $4r$, the final group will be smaller than $4r$. This group will be ignored.) For each group $G:=I_{1},\ldots,I_{4r}$ we put an interval $I_{G}$ into $R_{i}$, which starts at the left endpoint of $I_{1}$ and ends slightly before the left endpoint of $I_{2r+1}$; see Fig. 2 for an illustration.

The designated color $c_{i}$ is picked as follows. Consider the coloring after the last interval of $R_{i}$ has been inserted, and let $C(i)$ be the set of colors assigned by alg to intervals in $R_{i}$ and that are not a designated color from a previous round—we argue below that $C(i)\neq\emptyset$. Then we pick $c_{i}$ as the color from $C(i)$ that maximizes the number of living $i$-bricks.

We continue generating sets $R_{i}$ in this manner until $|R^{*}_{i}|<4r$, at which point the construction finishes. Below we prove that in each round alg must introduce a new designated color, and we prove a lower bound on the number of rounds in the construction.

**Claim 1. **Let $B=(I,I^{e})$ be a living $i$-brick. Then for any $j\in\{1,\ldots,i\}$ there is a point $q_{j}\in I\cup I^{e}$ that is contained in a single interval of color $c_{j}$ and in no other interval from $\bigcup_{\ell=1}^{i-1}R_{\ell}$. Moreover, there is a point $q_{j}\in I\cup I^{e}$ not contained in any interval from $\bigcup_{\ell=1}^{i-1}R_{\ell}$.

Proof of claim. We prove this by induction on $i$. For $i=1$ the statement is trivially true, so suppose $i>1$. By definition, both $I$ and $I^{e}$ contain living $(i-1)$-bricks, $\overline{B}$ and $\overline{B}^{e}$. Using the induction hypothesis we can now select a point $q_{j}$ with the desired properties: for $j=i$ we use that $\overline{B}$ contains a point that is not contained in any interval from $\bigcup_{\ell=1}^{i-1}R_{\ell}$, for $j<i$ we use that $\overline{B}^{e}$ contains a point in an interval of color $c_{j}$ and in no other interval from $\bigcup_{\ell=1}^{i-1}R_{\ell}$, and to find a point not contained in any interval from $\bigcup_{\ell=1}^{i-1}R_{\ell}$ we can also use $\overline{B}^{e}$. $\square$

Now consider the situation after the $i$-th round, but before we have chosen the designated color $c_{i}$. We say that a color $c$ is eligible (to become $c_{i}$) if $c\neq c_{1},\ldots,c_{i-1}$, and we say that an $i$-brick $(I,I^{e})$ is eligible if its color is eligible and $I$ and $I^{e}$ both contain living $(i-1)$-bricks (if $i>1$). Note that due to some recolorings, some of the newly inserted intervals might not contain any living brick and hence can never be living no matter the designated color; the next claim shows that at most half intervals inserted this round are eligible.

**Claim 2. **Immediately after the $i$-th round, at least half of the $i$-bricks are eligible.

Proof of claim. Consider an $i$-brick $(I,I^{e})$. At the beginning of the $i$-th round, before we have actually inserted the intervals from $R_{i}$, both the interval $I$ and its empty space $I^{e}$ contain $2r$ living $(i-1)$-bricks. As the intervals from $R_{i}$ are inserted, alg may recolor certain intervals from $R_{1}\cup\ldots\cup R_{i-1}$, thereby killing some of these $(i-1)$-bricks. Now suppose that alg recolored at most $2r-1$ of the intervals from $R_{1}\cup\ldots\cup R_{i-1}$ that are contained in $I\cup I^{e}$. Then both $I$ and $I^{e}$ still contain a living $(i-1)$-brick. By the previous claim and the definition of a conflict-free coloring, this implies alg cannot use any of the colors $c_{j}$ with $j<i$ for $I$. Hence, the color of $I$ is eligible and the $i$-brick $(I,I^{e})$ is eligible as well.

It remains to observe that alg can do at most $rn_{i}$ recolorings during the $i$-th round. We just argued that to prevent an $i$-brick from becoming eligible, alg must do at least $2r$ recolorings inside that brick. Hence, alg can prevent at most half of the $i$-bricks from becoming eligible. $\square$

Recall that after the $i$-th round we pick the designated color $c_{i}$ that maximizes the number of living $i$-bricks. In other words, $c_{i}$ is chosen to maximize $|R^{*}_{i}|$. Next we prove a lower bound on this number. Recall that $\rho$ denotes the number of rounds.

**Claim 3. **For all $1\leqslant i\leqslant\rho$ we have $|R^{*}_{i}|\geqslant n_{1}/(8rc)^{i}-1$.

Proof of claim. Since alg can use at most $c$ colors, we have $|R^{*}_{1}|\geqslant n_{1}/c$. Moreover, for $i>1$ the number of intervals we insert is $\left\lfloor|R^{*}_{i-1}|/4r\right\rfloor$. By the previous claim at least half of these are eligible. The eligible intervals have at most $c$ different colors, so if we choose $c_{i}$ to be the most common color among them we see that $|R^{*}_{i}|\geqslant\left\lfloor|R^{*}_{i-1}|/4r\right\rfloor/2c$. We thus obtain the following recurrence:

[TABLE]

We can now prove the result using induction.

[TABLE]

$\square$

Finally we can derive the desired relation between $n,c$, and $r$. Since $n_{1}=n/2$ and $n_{i+1}<n_{i}/2$ for all $i=1,\ldots,\rho-1$, the total number of insertions is less than $n$. The construction finishes when $|R^{*}_{i}|<4r$. Hence, $\rho$, the total number of rounds, must be such that $n/(2(8rc)^{\rho})-1\leqslant|R^{*}_{\rho}|<4r$, which implies $\rho>\log_{8rc}(n/(8r+2))>\log_{8rc}n-1$. The number of colors used by alg is at least $\rho$, since every round has a different designated color. Thus $c>\log_{8rc}n-1$ and so $n\leqslant(8rc)^{c+1}$, from which the theorem follows. ∎

Two interesting special cases of the theorem are the following: with $r=O(1)$ we will have $c=\operatorname{\Omega}\left(\log n/\log\log n\right)$, and for $c=O(1/\varepsilon)$ (for some small fixed $\varepsilon>0$) we need $r=\operatorname{\Omega}\left(\varepsilon n^{\varepsilon}\right)$. Note that the theorem requires $r>0$. Obviously the $\operatorname{\Omega}\left(\log n/\log\log n\right)$ lower bound on $c$ that we get for $r=1$ also holds for $r=0$. For the special case of $r=0$—this is the standard online version of the problem—we can prove a stronger result, however: here we need at least $\lfloor\log n\rfloor+1$ colors. This bound even holds for a nested set of intervals, that is, a set $S$ such that $I\subset I^{\prime}$, $I^{\prime}\subset I$, or $I\cap I^{\prime}=\emptyset$ for any two intervals $I,I^{\prime}\in S$. We also show in Section 4 that a greedy algorithm achieves this bound for nested intervals.

2.1 Local algorithms

We now prove a stronger lower bound for so-called local algorithms. Intuitively, these are deterministic algorithms where the color assigned to a newly inserted interval $I$ only depends on the structure and the coloring of the connected component where $I$ is inserted—hence the name local. More precisely, local algorithms are defined as follows.

Suppose we insert an interval $I$ into a set $S$ of intervals that have already been colored. The union of the set $S\cup\{I\}$ consists of one or more connected components. We define $S(I)\subseteq S$ to be the set of intervals from $S$ that are in the same connected component as $I$. (In other words, if we consider the interval graph induced by $S\cup\{I\}$ then the intervals in $S(I)$ form a connected component with $I$.) Order the intervals in $S(I)\cup\{I\}$ from left to right according to their left endpoint, and then assign to every interval its rank in this ordering as its label. (Here we assume that all endpoints of the intervals in $S(I)\subseteq S$ are distinct. It suffices to prove our lower bound for this restricted case.) Based on this labelling we define a signature for $S(I)\cup\{I\}$ as follows. Let $\lambda_{1},\ldots,\lambda_{k}$, where $k:=|S(I)|+1$, be the sequence of labels obtained by ordering the intervals from left to right according to their right endpoint. Furthermore, let $c_{i}$ be the color of the interval labeled $\lambda_{i}$, where $c_{i}=\mbox{\sc nil}$ if the interval labeled $i$ has not yet been colored. Then we define the signature of $S(I)\cup I$ to be the sequence $\operatorname{\mbox{sig}}(I):=\langle\lambda_{1},\ldots,\lambda_{k},c_{1},\ldots,c_{k}\rangle$; see Fig. 3.

We now define a semi-dynamic algorithm alg to be local if upon insertion of an interval $I$ the following holds: (i) alg only performs recoloring in $S(I)$, and (ii) the color assigned to $I$ and the recolorings in $S(I)$ are uniquely determined by $\operatorname{\mbox{sig}}(I)$, that is, the algorithm is deterministic with respect to $\operatorname{\mbox{sig}}(I)$. Note that randomized algorithms are not local. Also note that when a set of disjoint intervals is inserted into an initially empty set, a local algorithm will give each interval the same color. Thus the algorithms we present in the next section are not local, as the coloring depends on a global tree structure.

To strengthen Theorem 2.1 for the case of local algorithms, it suffices to observe that the intervals inserted in the same round must all receive the same color. Hence, the factors $c$ in the denominators of Inequality (1) disappear, leading to the theorem below. Note that for $r(n)=O(1)$, we now get the lower bound $c(n)=\operatorname{\Omega}(\log n)$.

Theorem 2.2**.**

Let alg be a local algorithm for the semi-dynamic conflict-free coloring of intervals. Suppose that on any sequence of $n>0$ insertions, alg uses at most $c(n)$ colors and $r(n)$ recolorings per insertion, where $r(n)>0$. Then $r(n)\geqslant n^{1/(c(n)+2)}-2$.

Proof.

We first fix a value for $n$ and define $c:=c(n)$ and $r:=r(n)$. Similar to the previous construction, this construction also proceeds in rounds. In the $i$-th round we insert a set $R_{i}$ of $n_{i}$ disjoint intervals. We will argue that in each round alg picks a different color, so that the number of rounds, $\rho$, is a lower bound on the number of colors used by alg. We then prove a lower bound on $\rho$ in terms of $n,c$, and $r$, and derive the theorem from the inequality $\rho\leqslant c$.

In the $i$-th round, we insert the following intervals:

The set $R_{1}$ contains the $\frac{n}{2}$ intervals $[0,1],[2,3],\ldots,[2n-2,2n-1]$. Since all intervals are disjoint, the local information alg can use for each interval consists only of that single interval. Hence, since alg is deterministic, it colors all intervals in $R_{1}$ with the same color. 2. 2.

To generate $R_{i}$ for $i>1$, we proceed as follows. Partition $R_{i-1}$ into groups of $r+2$ consecutive intervals. (If $|R_{i-1}|$ is not a multiple of $r+2$, the final group will be smaller than $r+2$. This group will be ignored.) For each group $G:=I_{1},\ldots,I_{r+2}$ we put an interval $I_{G}$ into $R_{i}$, which starts at the left endpoint of $I_{1}$ and ends slightly before the left endpoint of $I_{r+2}$. Since local information for alg for each of these intervals is the same and alg is deterministic, it colors all intervals of $R_{i}$ using the same color and the output signature of $S(I)$ is the same for all $I\in R_{i}$.

We continue generating sets $R_{i}$ in this manner until $|R^{*}_{i}|<r+1$, at which point the construction finishes. Note that if $R_{\rho-1}$ consists of exactly $r+1$ intervals, we cannot apply the above construction as it is. In this case, $R_{\rho}$ consists of a single interval which starts at the left endpoint of $I_{1}$ and ends at $2n-1$, that is the new interval contains every other interval.

By construction, for each interval $I$ in $R_{i}$, for $i\in\{1,\ldots,\rho\}$, and for each $j\in\{1,\ldots,i\}$ there is a point $q$ that is contained in a single interval of $R_{j}$ and in no other interval from $R_{1}\cup\cdots\cup R_{i}$. Moreover, the rightmost point of $I$ is not contained in any interval from $R_{1}\cup\cdots\cup R_{i}$. Furthermore, since for each newly inserted interval $I$ alg can only recolor $r$ of the intervals that $I$ intersects, there exists a set of intervals $I_{1},\cdots,I_{i-1}$ such that for every $I_{j}$ in this set, $I_{j}\in R_{j}$, $I_{j}$ is contained in $I_{j+1}$ ($I_{i}=I$), and $I_{j}$ is not recolored by alg. Hence, these properties remain valid during the insertion process.

**Claim 4. **For all $1\leqslant i\leqslant\rho$ we have $|R_{i}|\geqslant n/(r+2)^{i}-2$.

Proof of claim. Since alg colors all intervals of $R_{1}$ using the same color, we have $|R_{1}|=n/2$. Moreover, since for $i>1$ the number of intervals we insert is $\left\lfloor|R_{i-1}|/(r+2)\right\rfloor$, we have that $|R_{i}|\geqslant\left\lfloor|R_{i-1}|/(r+2)\right\rfloor$. We thus obtain the following recurrence:

[TABLE]

We can now prove the result using induction.

[TABLE]

$\square$

Finally we can derive the desired relation between $n,c$, and $r$. Since $|R_{1}|=n/2$ and $|R_{i+1}|<|R_{i}|/2$ for all $i=1,\ldots,\rho-1$, the total number of insertions is less than $n$. The construction finishes when $|R_{i}|<r+1$. Hence, $\rho$, the total number of rounds, must be such that

[TABLE]

which implies $\rho>\log_{r+2}(n/(2r+6))>\log_{r+2}n-2$. The number of colors used by alg is at least $\rho$, since every round uses a different color. Thus $c>\log_{r+2}n-2$ and so $n\leqslant(r+2)^{c+2}$, from which the theorem follows. ∎

3 Upper bounds for fully-dynamic conflict-free colorings

Next we present algorithms to maintain a conflict-free coloring for a set $S$ of intervals under insertions and deletions. The algorithms use the same structure, which we describe first. From now on, we use $n$ to denote the current number of intervals in $S$.

Let $P$ be the set of $2n$ endpoints of the intervals in $S$. (To simplify the presentation we assume that all endpoints are distinct, but the solution is easily adapted to the general case.) We will maintain a B-tree on the set $P$. A B-tree of minimum degree $t$ on a set of points in $\mathbb{R}^{1}$ is a multi-way search tree in which each internal node has between $t$ and $2t$ children (except the root, which may have fewer children) and all leaves are at the same level; see the book by Cormen et al. [9, Chapter 18] for details. Thus each internal or leaf node stores between $t-1$ and $2t-1$ points from $P$ (again, the root may store fewer points). We denote the set of points stored in a node $v$ by $P(v):=\{p_{1}(v),\ldots,p_{n_{v}}(v)\}$, where $n_{v}:=|P(v)|$ and the points are numbered from left to right. For an internal node $v$ we denote the $i$-th subtree of $v$, where $1\leqslant i\leqslant n_{v}+1$, by $\mathcal{T}_{i}(v)$. Note that the search-tree property guarantees that all points in $\mathcal{T}_{i}(v)$ lie in the range $(p_{i-1}(v),p_{i}(v))$, where $p_{0}=-\infty$ and $p_{n_{v}+1}=\infty$.

We now associate each interval $I\in S$ to the highest node $v$ such that $I$ contains at least one of the points in $P(v)$, either in its interior or as one of its endpoints. Thus our structure is essentially an interval tree [5, Chapter 10] but with a B-tree as underlying tree structure. We denote the set of intervals associated to a node $v$ by $S(v)$. Note that if $\mathrm{level}(v)=\mathrm{level}(w)=\ell$, for some nodes $v\neq w$, and $I\in S(v)$ and $I^{\prime}\in S(w)$, then $I$ and $I^{\prime}$ are separated by a point $p_{i}(z)$ of some node $z$ at level $m<\ell$. Hence, $I\cap I^{\prime}=\emptyset$.

We partition $S(v)$ into subsets $S_{1}(v),\ldots,S_{n_{v}}(v)$ such that $S_{i}(v)$ contains all intervals $I\in S(v)$ for which $p_{i}(v)$ is the leftmost point from $P(v)$ contained in $I$. From each subset $S_{i}(v)$ we pick at most two extreme intervals: the left-extreme interval $I_{i,\mathrm{left}}(v)$ is the interval from $S_{i}(v)$ with the leftmost left endpoint, and the right-extreme interval $I_{i,\mathrm{right}}(v)$ is the interval from $S_{i}(v)$ with the rightmost right endpoint. Since all intervals from $S_{i}(v)$ contain the point $p_{i}(v)$, every interval from $S_{i}(v)$ is contained in $I_{i,\mathrm{left}}(v)\cup I_{i,\mathrm{right}}(v)$. Note that it may happen that $I_{i,\mathrm{left}}(v)=I_{i,\mathrm{right}}(v)$. Finally, we define $S_{\mathrm{extr}}(v):=\bigcup_{i=1}^{n_{v}}\{I_{i,\mathrm{left}}(v),I_{i,\mathrm{right}}(v)\}$ to be the set of all extreme intervals at $v$.

Our two coloring algorithms both maintain a coloring with the following properties.

(A.1)

For each level $\ell$ of the tree $\mathcal{T}$, there is a set $C(\ell)$ of colors such that these color sets are disjoint between different levels.

(A.2)

For each node $v$ at level $\ell$ in $\mathcal{T}$, the intervals from $S_{\mathrm{extr}}(v)$ are colored locally conflict-free using colors from $C(\ell)$ and a universal dummy color. Here locally conflict-free means that the coloring of $S_{\mathrm{extr}}(v)$ is conflict-free if we ignore all other intervals.

(A.3)

All non-extreme intervals receive a universal dummy color, which is distinct from any of the other colors used, that is, the dummy color is not in $C(\ell)$ for any level $\ell$.

The two coloring algorithms that we present differ in the size of the sets $C(\ell)$ and in which local coloring algorithm is used for the sets $S_{\mathrm{extr}}(v)$. It is not hard to show that the properties above guarantee a conflict-free coloring.

Lemma 3.1**.**

Any coloring with properties (A.1)–(A.3) is conflict free and uses at most $1+\sum_{\ell}|C(\ell)|$ colors.

Next we describe two algorithms based on this general framework: one for the easy case where the interval endpoints come from a finite universe, and one for the general case.

Solutions for a polynomially-bounded universe.

The framework above uses a B-tree on the interval endpoints. If the interval endpoints come from a universe of size $U$—for concreteness, assume the endpoints are integers in the range $0,\ldots,U-1$—then we can use a B-tree on the set $\{0,\ldots,U-1\}$. Thus the B-tree structure never has to be changed.

Theorem 3.2**.**

Let $S$ be a dynamic set of intervals whose endpoints are integers in $\{0,\ldots,U-1\}$.

1. (i)

We can maintain a conflict-free coloring on $S$ that uses $O(\log U)$ colors and that performs at most two recolorings per insertion and deletion. 2. (ii)

For any $t$ with $2\leqslant t\leqslant U$, we can maintain a conflict-free coloring on $S$ that uses $O(\log_{t}U)$ colors and performs $O(t)$ recolorings per insertion and deletion.

Proof.

For both results we use the general framework described above.

1. (i)

We pick $|C(\ell)|=4t-2$ for all levels $\ell$. Since a B-tree of minimum degree $t$ on $U$ points has height $O(\log_{t}U)$, the total number of colors is $O(t\log_{t}U)$. With $4t-2$ colors, we can give each interval in $S_{\mathrm{extr}}(v)$ its own color, so the coloring is locally conflict free. To get $O(\log U)$ colors we now pick $t=2$. To insert an interval $I$, we find the set $S_{i}(v)$ into which $I$ should be inserted, and check if $I$ should replace the current left-extreme and/or right-extreme interval in $S_{i}(v)$. Thus an insertion requires at most two recolorings. Deletions can be handled in a similar way: if $I$ is an extreme interval, then we have to recolor at most two intervals that become extreme after the deletion of $I$. 2. (ii)

We pick $|C(\ell)|=2$ for all levels $\ell$, yielding $O(\log_{t}U)$ colors in total. We now color $S_{\mathrm{extr}}(v)$ using the chain method with $C(\mathrm{level}(v))$ and the global dummy color. After coloring every connected component of $S_{\mathrm{extr}}(v)$ in this manner we have a locally conflict-free coloring of $S_{\mathrm{extr}}(v)$. Insertions and deletions into $S(v)$ are simply handled by updating $S_{\mathrm{extr}}(v)$ and recoloring $S_{\mathrm{extr}}(v)$ from scratch. Since at most two intervals start or stop being extreme and $|S_{\mathrm{extr}}(v)|\leqslant 4t-2$, we use $O(t)$ recolorings.

∎

When $U$ is polynomially bounded in $n$—that is, $U=O(n^{k})$ for some constant $k$—this gives very efficient coloring schemes. In particular, we can then get $O(\log n)$ colors with at most two recolorings using method (i), and we can get $O(1/\varepsilon)$ colors with $O(n^{\varepsilon})$ recolorings (for any fixed $\varepsilon>0$) by setting $t=U^{\varepsilon/k}$ in method (ii).

Note finally that we do not need to explicitly store the whole tree as it is enough to compute the location of any node when needed, yielding a linear space complexity.

A general solution.

If the interval endpoints do not come from a bounded universe then we cannot use a fixed tree structure. Next we explain how to deal with this when we apply the method from Theorem 3.2(ii), which colors the sets $S_{\mathrm{extr}}(v)$ using the chain method: we take the interval with the leftmost left endpoint, and color it blue. Then, among all intervals whose left endpoint lies in the blue interval, we pick the one with the rightmost right endpoint and color it red. We then repeat this process, alternating between blue and red, until we reach the rightmost interval. Finally, we color all uncolored intervals grey.

Suppose we want to insert a new interval $I$ into the set $S$. We first insert the two endpoints of $I$ into the B-tree $\mathcal{T}$. Inserting an endpoint $p$ can be done in a standard manner. The basic operations for an insertion are (i) to split a full node and (ii) to insert a point into a non-full leaf node.

Splitting a full node $v$ (that is, a node with $2t-1$ points) is done by moving the median point into the parent of $v$, creating a node containing the $t-1$ points to the left of the median and another node containing the $t-1$ points to the right of the median. Note that this means that some intervals from $S(v)$ may have to be moved to $S(\mathit{parent}(v))$. Thus splitting a node $v$ involves recoloring intervals in $S(v)$ and $S(\mathit{parent}(v))$. Observe that an interval only needs to be recolored if it was extreme before or after the change. Hence, we recolor $O(t)$ intervals when we split a node $v$.

Since an insertion splits only nodes on a root-to-leaf path and the depth of $\mathcal{T}$ is $O(\log_{t}n)$, the total number of recolorings due to node splitting is $O(t\log_{t}n)$. Moreover, inserting a point into a non-full leaf node only takes $O(t)$ recolorings. We conclude that an insertion performs $O(t\log_{t}n)$ recolorings in total. For deletions the argument is similar. Since recoloring at a single node induces $O(t)$ recolorings, the total number of recolorings is $O(t\log_{t}n)$.

Theorem 3.3**.**

Let $S$ be a dynamic set of intervals.

1. (i)

For any fixed $t\geqslant 2$ we can maintain a conflict-free coloring on $S$ that uses $O(\log_{t}n)$ colors and that performs $O(t\log_{t}n)$ recolorings per insertion and deletion, where $n$ is the current number of intervals in $S$. In particular, we can maintain a conflict-free coloring with $O(\log n)$ colors using $O(\log n)$ recolorings per update. 2. (ii)

For any fixed $\varepsilon>0$ we can maintain a conflict-free coloring on $S$ that uses $O(1/\varepsilon)$ colors and that performs $O(n^{\varepsilon}/\varepsilon)$ recolorings per insertion or deletion. The bound on the number of recolorings is amortized.

The idea behind part (ii) is to use a $t$ with $n^{\varepsilon}/2\leqslant t\leqslant 2n^{\varepsilon}$. This causes the bound in (ii) to be amortized, since now we need to change $t$ when $n$ has halved or doubled.

We have not been able to efficiently generalize the first method of Theorem 3.2 to an unbounded universe. The problem is that splitting a node $v$ may require many intervals in $S_{\mathrm{extr}}(v)$ to be recolored, since many intervals may be moved to $\mathit{parent}(v)$. Hence, the method would use the same number of recolorings as the chain method, but more colors.

Bounded-length intervals.

Next we present a simple method that allows us to improve the bounds when the intervals have length between $1$ and $L$ for some constant $L>1$.

Theorem 3.4**.**

Let $S$ be a dynamic set of intervals with lengths in the range $[1,L)$ for some fixed $L>1$. Suppose we have a dynamic conflict-free coloring algorithm for a general set of intervals that uses at most $c(n)$ colors and at most $r(n)$ recolorings for any insertion or deletion. Then we can obtain a dynamic conflict-free coloring algorithm on $S$ that uses at most $2\cdot c(2L)+1$ colors and at most $2\cdot r(2L)+1$ recolorings for any insertion or deletion.

Proof.

For each pair of integers $i,j$ so that $i\geqslant 0$ and $0\leqslant j<L$ let $x_{i,j}:=iL+j$, and for each $i\geqslant 0$ define $X_{i}:=\{x_{i,j}:0\leqslant j<L\}$. Note that the point sets $X_{i}$ form a partition of the non-negative integer points in $\mathbb{R}^{1}$ into subsets of $L$ consecutive points. We assign each interval $I\in S$ to the leftmost point $x_{i,j}$ contained in it, and we denote the subset of intervals assigned to a point in $X_{i}$ by $S_{i}$. We will color each subset $S_{i}$ separately. Since an interval in $S_{i}$ can intersect only intervals in $S_{i-1}\cup S_{i}\cup S_{i+1}$, we can use the same color set for all $S_{i}$ with $i$ even, and the same color set for all $S_{i}$ with $i$ odd. It remains to argue that we can maintain a coloring for $S_{i}$ using $c(2L)$ colors (in addition to a global dummy color) and $2\cdot r(2L)+1$ recolorings.

Let $S_{\mathrm{extr}}(x_{i,j})$ contain the two extreme intervals assigned to $x_{i,j}$: the interval sticking out furthest to the left and, among the remaining intervals, the one sticking out furthest to the right. We give each non-extreme interval the dummy color and maintain a conflict-free coloring for $S_{\mathrm{extr}}(i):=\bigcup_{j}S_{\mathrm{extr}}(x_{i,j})$. When we insert a new interval into $S_{i}$ it replaces at most one extreme interval, so we do at most one insertion and one deletion in $S_{\mathrm{extr}}(i)$. Thus we do at most $2\cdot r(2L)+1$ recolorings. (The +1 is because the interval that becomes non-extreme is recolored to the dummy color.) For deletions the argument is similar. ∎

For instance, by applying Theorem 3.3(i) we can maintain a coloring with $O(\log L)$ colors and $O(\log L)$ recolorings. We can also plug in a trivial dynamic algorithm with $c(n)=n$ and $r(n)=0$ to obtain $4L+1$ colors with only 1 recoloring per update; when $L$ is sufficiently small this gives a better result.

4 Online conflict-free colorings

In this section we consider recolorings to be too expensive and hence forbid them all together. In this setting, any conflict-free coloring must be a proper coloring if we allow deletions: if any point is contained in two intervals of the same color, then deleting all other intervals would result in a conflict. Thus we would need $n$ colors in the worst case. We therefore only allow insertions, that is, we study the online version. Note that Theorem 2.1 does not cover the case $r=0$. However, Abam et al. [1] proved that there is a logarithmic lower bound on the number of colors for the online case. For completeness we provide the proof.

Theorem 4.1** (Abam et al. [1]).**

For each $n>0$, there is a sequence of $n$ intervals for which any online conflict-free coloring requires at least $\lfloor\log_{2}n\rfloor+1$ colors.

Proof.

We reduce the static conflict-free coloring problem for $n$ points with respect to intervals in  $\mathbb{R}^{1}$ to the online problem for coloring intervals with respect to points. The former problem asks for a coloring of the $n$ given points, which we can assume to be the integers $1,\ldots,n$, such that for any two points $p,q$, the subset of consecutive points between $p$ and $q$ has a unique color. Since this problem requires $\lfloor\log_{2}n\rfloor+1$ colors [11], this reduction proves the theorem.

We map the instance $\mathcal{I}\{1,\ldots,n\}$ to an instance $\mathcal{I}^{\prime}$ of the online conflict-free coloring of intervals as follows: for each $i=1,\ldots,n$, we insert the interval $I_{i}:=[-i,i]$ at time $i$. See Fig. 4 for an illustration.

We now prove that there is an interval containing exactly the subset $S\subseteq\mathcal{I}$ if and only if there is a point $q$ and a time step $t$ such that the point $q$ is contained in exactly $S^{\prime}:=\{I_{i}|p_{i}\in S\}$ at time $t$ in $\mathcal{I}^{\prime}$. Let $i,j$ be such that $S=\{i,\ldots,j\}$. We claim that the query point $q=i-\frac{1}{2}$ at time $j$ is contained in exactly $S^{\prime}$; see Fig. 5.

First, since $q$ is not contained in $I_{i-1}=[-i+1,i-1]$, and $I_{i-1}$ contains all intervals with lower index, $q$ is not contained in $I_{\ell}$ for $\ell<i$. Moreover, since $q$ is contained in $I_{i}=[-i,i]$ and $I_{i}$ is contained in any higher indexed interval, $q$ is also contained in those. Since we are at time $j$, only intervals $I_{1},\ldots,I_{j}$ are present. Thus, at time $j$, the point $q$ is indeed contained in exactly the intervals of $S^{\prime}$, which concludes the proof. ∎

Notice that the intervals in the lower-bound construction are nested, that is, any two intervals are either disjoint or one contains the other. We show that for inputs restricted to such nested intervals a simple greedy algorithm requires only $\lfloor\log_{2}n\rfloor+1$ colors. For the case where all nested intervals cover a specific point, Abam et al. [1] provided an algorithm that uses $O(\log n)$ colors.

We label the colors as $0,1,2,3,\ldots$ in the order in which they are introduced. For each newly inserted interval, the greedy algorithm gives it the dummy color [math] if it is contained in another interval; otherwise, it gets the smallest color $c\geqslant 1$ that keeps the coloring conflict-free. Next we prove that this greedy algorithm with a dummy color is optimal on nested intervals.

Theorem 4.2**.**

The greedy algorithm with a dummy color uses at most $\lfloor\log_{2}n\rfloor+1$ colors on nested instances.

Proof.

We prove the following statement:

For any ordered set $S$ of $n$ nested intervals and for any color $i\geqslant 1$ the following holds: if the greedy algorithm assigns color $i$ to an interval $I\in S$, then $I$ contains least $2^{i-1}-1$ other intervals from $S$.

If this statement holds, then the algorithm indeed uses at most $\lfloor\log_{2}n\rfloor+1$ colors. We prove the statement by induction on the number $n$ of intervals.

The case $n=1$ is obvious, so now consider the case $n>1$ and assume the statement holds on any instance of less than $n$ intervals. For any $1\leqslant j\leqslant n$, define $S_{j}:=\{I_{1},\ldots,I_{j}\}$ to be the first $j$ intervals in $S$. Let $i$ be the color of $I_{n}$, the last interval inserted. Let $S(I_{n})\subseteq S_{n-1}$ be the set of intervals contained in $I_{n}$. We can assume without loss of generality that $S(I_{n})$ does not contain any interval $I_{j}$ using color $i$ or higher otherwise we can apply the induction hypothesis to $S_{j}$.

Since the algorithm did not use colors $1,\ldots,i-1$ for $I_{n}$, there is a query point $q$ contained in exactly one interval $I_{j}$ of color $i-1$. Note that $I_{j}\subsetneq I_{n}$. Let $S(I_{j})\subseteq S$ be the set of intervals contained in $I_{j}$ (excluding $I_{j}$ itself). By the induction hypothesis, $|S(I_{j})|\geqslant 2^{i-2}-1$. We now prove that the color of any interval in $S(I_{n})\setminus\{S(I_{j})\cup I_{j}\}$ is not influenced by $S(I_{j})\cup\{I_{j}\}$.

**Claim 5. **Any interval in $S(I_{n})\setminus\{S(I_{j})\cup\{I_{j}\}\}$ of color $k<i$ in the full instance is also colored with $k$ in the instance where only $S(I_{n})\setminus\{S(I_{j})\cup\{I_{j}\}\}$ is inserted.

Proof of claim. We show the claim using the following invariant: each interval of $S(I_{n})\setminus\{S(I_{j})\cup\{I_{j}\}\}$ being colored receives the same color in the restricted instance as it receives in the full instance. The invariant obviously holds when no interval is inserted. Suppose now that the invariant holds, and let $I_{\ell}\in S(I_{n})\setminus\{S(I_{j})\cup\{I_{j}\}\}$ be the next inserted interval. Let $k<i$ be the color of $I_{\ell}$ in the full instance. If $I_{\ell}\cap I_{j}=\emptyset$, the invariant obviously holds after the insertion of $I_{\ell}$ since it holds before the insertion of $I_{\ell}$ and since no interval in $S(I_{j})\cup\{I_{j}\}$ intersects $I_{\ell}$. Suppose therefore that $I_{\ell}\supset I_{j}$. In this case, $k\neq i-1$ since we assumed there is a query point $q\in I_{j}$ such that $I_{j}$ is the only interval of color $i-1$ containing $q$. Therefore, $I_{j}$ has no influence on the color of $I_{\ell}$. Moreover, no interval in $S(I_{j})$ can have color $i-1$: suppose there is an interval $I_{m}\in S(I_{j})$ of color $i-1$. Then, for any point $p\in I_{m}$, there must be an interval of unique color smaller than $i-1$, and therefore one of these intervals, say $I^{\prime}_{m}$ colored with $i^{\prime}<i-1$, must contain $I_{m}$. However, in that case, the color $i^{\prime}$ was available when coloring $I_{m}$ which contradicts the greedy choice. Thus, no interval in $S(I_{j})$ has color $i-1$ and hence the color of $I_{\ell}$ is only determined by intervals in $S(I_{n})\setminus\{S(I_{j})\cup I_{j}\}$ and therefore the invariant is maintained for $I_{\ell}$ too. $\square$

We now have two cases.

By removing $S(I_{j})\cup\{I_{j}\}$, the interval $I_{n}$ still cannot use color $i-1$ (by the claim, all the other intervals keep the same colors). Then there is an interval $I_{\ell}$ not in $S(I_{j})\cup\{I_{j}\}$, that uses color $i-1$ and is contained in $I_{n}$, which by the induction hypothesis contains at least $2^{i-2}-1$ other intervals. Since $I_{\ell}$ cannot contain $I_{j}$—this follows from the definition of $I_{j}$—the intervals contained in $I_{j}$ and the ones contained in $I_{\ell}$ are distinct. Hence, $I_{n}$ contains at least $(2^{i-2}-1)+(2^{i-2}-1)+2=2^{i-1}$ intervals. 2. 2.

By removing $S(I_{j})\cup\{I_{j}\}$, the interval $I_{n}$ can use color $i-1$ (by the claim, all the other intervals keep the same colors). By the induction hypothesis, $I_{n}$ contains at least $2^{i-2}-1$ intervals in the instance without $S(I_{j})\cup\{I_{j}\}$, see Fig. 6 for an illustration. By adding back $S(I_{j})$ and $I_{j}$, we have that $I_{n}$ contains at least $(2^{i-2}-1)+(2^{i-2}-1)+1=2^{i-1}-1$ intervals.

∎

Non-deterministic.

Our main interest in this paper is in deterministic algorithms. For the online case, we note that there is a randomized algorithm that uses $O(\log n)$ colors with high probability. This follows from the general technique of Bar-Noy et al. [2], who present an online algorithm for so-called $k$-degenerate hypergraphs that uses $O(k\log n)$ colors with high probability. A hypergraph $H=(V,E)$ is called $k$-degenerate if for every subset $V^{\prime}\subseteq V$ of vertices, and for every ordering $v_{1},\ldots,v_{i}$ of $V^{\prime}$, the following holds: $\sum_{j=1}^{i}\deg_{D[1,\ldots,j]}(v_{j})\leqslant k|V^{\prime}|$, where $\deg_{D[1,\ldots,j]}(v_{j})$ denotes the degree of $v_{j}$ in the Delaunay graph of $H[v_{1},\ldots,v_{j}]$ (the Delaunay graph of a hypergraph is the subhypergraph on all vertices containing the hyperedges of size 2). To apply their result it suffices to show that the hypergraph generated by intervals is 2-degenerate, which can be done using an amortized counting of the number of new Delaunay edges at each insertion as follows.

Theorem 4.3**.**

Given an online sequence of intervals we can maintain a conflict free coloring of this sequence using $O(\log n)$ colors with high probability.

Proof.

To apply the result of Bar-Noy et al. Let $I_{1},\ldots,I_{i}$ be a sequence of intervals inserted in that order. Each time an interval is inserted, each endpoint $p$ is labelled (or relabelled) 0, 1, or 2, depending if the region of the real line immediately to the right of $p$ is covered by zero, one, or two or more intervals respectively. Now it suffices to notice two things: first, the label can only increase over time, second, each time a Delaunay edge appears, at least one endpoint label becomes 2 (either a 1 is turned into a 2, or one of the endpoints of the new interval is immediately labelled 2). Therefore, there cannot be more Delaunay edges created than the number of endpoints. ∎

5 Kinetic conflict-free colorings

In this section we consider conflict-free colorings of a set of intervals in $\mathbb{R}^{1}$ whose endpoints move continuously. Note that we allow the endpoints of an interval to move independently, that is, we allow the intervals to expand or shrink over time but we do not allow the right endpoint of an interval to cross the left endpoint of the same interval. We show that by using only three recolorings per event—an event is when two endpoints cross each other—we can maintain a conflict-free coloring consisting of only four colors. Our recoloring strategy is based on the chain method discussed in the introduction. This method uses three colors: two colors for the chain and one dummy color. To be able to maintain the coloring in the kinetic setting without using many recolorings, we relax the chain properties and we allow ourselves three colors for the chain. Next we describe the invariants we maintain on the chain and its coloring, and we explain how to re-establish the invariants when two endpoints cross each other. In the remainder we assume that the endpoints of the chains are in general position except at events, and that events do not coincide (that is, we never have three coinciding endpoints and we never have two events at the same time). These conditions can be removed by using consistent tie-breaking.

The chain invariants.

Let $S$ be the set of intervals to be colored, where all interval endpoints are distinct. (Recall that we assumed this to be the case except at event times.) Consider a subset $\mathcal{C}\subseteq S$ and order the intervals in $\mathcal{C}$ according to their left endpoint. We denote the predecessor of an interval $I\in\mathcal{C}$ in this order by $\mathit{pred}_{\mathcal{C}}(I)$, and we denote its successor by $\mathit{succ}_{\mathcal{C}}(I)$. A chain (for $S$) is defined as a subset $\mathcal{C}$ with the following three properties.

(C1)

Any interval $I\in\mathcal{C}$ can intersect only two other intervals in $\mathcal{C}$, namely $\mathit{pred}_{\mathcal{C}}(I)$ and $\mathit{succ}_{\mathcal{C}}(I)$.

(C2)

Any interval $I\in S\setminus\mathcal{C}$ is completely covered by the intervals in $\mathcal{C}$.

(C3)

No interval $I\in\mathcal{C}$ is fully contained in any other interval $I^{\prime}\in S$.

Now consider a set $S$ and a chain $\mathcal{C}$ for $S$. We maintain the following color invariant: each interval $I\in\mathcal{C}$ has a non-dummy color and this color is different from the color of $\mathit{succ}(I)$, and each interval in $S\setminus\mathcal{C}$ has the dummy color.

Lemma 5.1**.**

Let $S$ be a set of intervals and $\mathcal{C}$ be a chain for $S$. Then any coloring of $S$ satisfying the color invariant is conflict-free.

Proof.

Property (C2) implies that any point contained in at least one interval of $S$ is contained in a chain interval. Furthermore, from property (C1) we know that any point $x\in\mathbb{R}^{1}$ is contained in at most two intervals of $\mathcal{C}$, and that if $x$ is contained in two chain intervals they must be consecutive. The color invariant guarantees that two consecutive chain intervals have different colors, so the (at most two) chain intervals containing $x$ provide a unique color for $x$. Hence, the coloring is conflict-free. ∎

Handling events.

Our kinetic coloring algorithm maintains a chain $\mathcal{C}$ for $I$ and a coloring with three colors (excluding the dummy color) satisfying the color invariant. Later we show how to re-establish the color invariant at each event, but first we show how to update the chain by adding at most one interval to the chain and removing at most two. We distinguish several cases.

• •

Case A: The right endpoints of two intervals $I$ and $I^{\prime}$ cross.

Assume without loss of generality that $I$ is shorter than $I^{\prime}$. We have two subcases.

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Subcase A.1: Interval $I$ is contained in $I^{\prime}$ before the event. In this case $I$ was not a chain interval before the event. If after the event $I$ is still fully covered by the chain intervals, then there is nothing to do: we can keep the same chain. Otherwise, property (C2) is violated after the event. We now proceed as follows. First we add $I$ to the chain. If $I$ intersects $\mathit{pred}_{\mathcal{C}}(I^{\prime})$—note that $I^{\prime}$ must be a chain interval if (C2) is violated—then we remove $I^{\prime}$ from the chain.

• –

Subcase A.2: Interval $I$ is contained in $I^{\prime}$ after the event. If $I$ was not a chain interval, there is nothing to do. Otherwise property (C3) no longer holds after the event, and we have to remove $I$ from the chain. If $I^{\prime}$ is also a chain interval then this suffices. Otherwise we add $I^{\prime}$ to the chain, and remove $\mathit{pred}_{\mathcal{C}}(I)$ if $\mathit{pred}_{\mathcal{C}}(\mathit{pred}_{\mathcal{C}}(I))$ intersects $I^{\prime}$.

• •

Case B: The left endpoints of two intervals $I$ and $I^{\prime}$ cross.

This case is symmetric to Case A.

• •

Case C: The right endpoint of an interval $I$ crosses the left endpoint of an interval $I^{\prime}$.

Again we have two subcases.

• –

Subcase C.1: Intervals $I$ and $I^{\prime}$ start intersecting. Note that properties (C2) and (C3) still hold after the event. The only possible violation is in property (C1), namely when both $I$ and $I^{\prime}$ are chain intervals and there is a chain interval $I^{\prime\prime}$ with $\mathit{pred}_{\mathcal{C}}(I^{\prime\prime})=I$ and $\mathit{succ}_{\mathcal{C}}(I^{\prime\prime})=I^{\prime}$. In this case we simply remove $I^{\prime\prime}$ from the chain.

• –

Subcase C.2: Intervals $I$ and $I^{\prime}$ stop intersecting. First note that this cannot violate properties (C1) and (C3). The only possible violation is property (C2), namely when both $I$ and $I^{\prime}$ are chain intervals and there is at least one non-chain interval containing the common endpoint of $I$ and $I^{\prime}$ at the event. Of all such non-chain intervals, let $I^{\prime\prime}$ be the interval with the leftmost left endpoint. Note that $I^{\prime\prime}$ is not contained in any other interval, so we can add $I^{\prime\prime}$ to the chain without violating (C3). After adding $I^{\prime\prime}$ we check if we have to remove $I$ and/or $I^{\prime}$: if $I^{\prime\prime}$ intersects $\mathit{pred}_{\mathcal{C}}(I)$ then we remove $I$ from the chain, and if $I^{\prime\prime}$ intersects $\mathit{succ}_{\mathcal{C}}(I^{\prime})$ then we remove $I^{\prime}$ from the chain.

It is easy to check that in each of these cases the new chain that we generate has the chain properties (C1)–(C3). Next we show that each case requires at most three recolorings and summarize the result.

Lemma 5.2**.**

In each of the above cases, the changes to the chain require at most three recolorings to re-establish the color invariant.

Proof.

An interval that is a non-chain interval before and after the event need not be recolored. We have seen that we add at most one interval to the chain and remove at most two. Removing an interval from the chain requires recoloring it to the dummy color. Adding an interval requires giving it a color that is different from its predecessor and its successor in the chain, which we can do since we have three colors available for the chain. It remains to check if any other chain intervals need to be recolored. This can only happen in case C.1, when $I$ and $I^{\prime}$ are chain intervals that are not removed. In this case it suffices to recolor $I$ with a color different from the color of $I^{\prime}$ and from the color of $\mathit{pred}_{\mathcal{C}}(I)$. In all cases, the number of recolorings is at most three. ∎

Theorem 5.3**.**

Let $S$ be a kinetic set of intervals in $\mathbb{R}^{1}$. We can maintain a conflict-free coloring for $S$ with four colors at the cost of at most three recolorings per event, where an event is when two interval endpoints cross each other.

A lower bound.

Now consider the simple scenario where the intervals are rigid—each interval keeps the same length over time—and each interval is either stationary or moves with unit speed to the right. Our coloring algorithm may perform recolorings whenever two endpoints cross, which means that we do $O(n^{2})$ recolorings in total. We show that even in this simple setting, this bound is tight in the worst case if we use at most four colors.

Consider four intervals $I_{1},I_{2},I_{3},I_{4}$ where $I_{i}=(a_{i},b_{i})$, with $a_{i}<b_{i}$ as shown in Figure 8. Here $I_{2}\subset I_{1}$, $I_{4}\subset I_{3}$, the right endpoints of $I_{1}$ and $I_{2}$ are contained in $I_{3}\cap I_{4}$, and the left endpoints of $I_{3}$ and $I_{4}$ are contained in $I_{1}\cap I_{2}$. The exact locations of the endpoints with respect to each other is not important and we focus on the different overlap sets of the gadget. Specifically within a gadget there is a point contained in each of the following sets,

[TABLE]

Based on these sets we can show that no coloring for crossing gadgets exists that provides a valid conflict-free coloring for each combination of intersection sets between the two gadgets. The proof relies on the following lemma.

Lemma 5.4**.**

Let $G=\{I_{1},I_{2},I_{3},I_{4}\}$ and $H=\{J_{1},J_{2},J_{3},J_{4}\}$ be two gadgets, with overlap sets $G_{1},\ldots,G_{7}$ and $H_{1},\ldots,H_{7}$ as defined above. There is no 4-coloring for $G$ and $H$ such that all sets $\{G_{1},\ldots,G_{7}\}\cup\{H_{1},\ldots,H_{7}\}\cup\{G_{i}\cup H_{i}|\ 1\leqslant i,j\leqslant 7\ \}$ are conflict-free.

Proof.

We can assume that not both $I_{1},I_{2},I_{3},I_{4}$ and $J_{1},J_{2},J_{3},J_{4}$ use all four colors, otherwise $G_{4}\cup H_{4}=\{I_{1},I_{2},I_{3},I_{4},J_{1},J_{2},J_{3},J_{4}\}$ is not conflict-free. It is also not possible to use at most two colors, since each gadget by itself needs to be conflict-free. Hence, suppose that there are exactly three colors among $I_{1},I_{2},I_{3},I_{4}$ (the other case is symmetric), say two are red, one is blue, and one is green. We define $\operatorname{col}(G_{i})$, respectively $\operatorname{col}(H_{i})$, to be the multiset of the colors used by the intervals in $G_{i}$, respectively $H_{i}$. Then $\operatorname{col}(G_{4})=\{\mbox{red},\mbox{red},\mbox{blue},\mbox{green}\}$ and without loss of generality, $\operatorname{col}(G_{2})=\{\mbox{red},\mbox{blue}\}$ and $\operatorname{col}(G_{6})=\{\mbox{red},\mbox{green}\}$. We now have two cases.

One interval among $J_{1},J_{2},J_{3},J_{4}$ uses the fourth color, say yellow. If $J_{1}$ or $J_{2}$ is yellow, then either $\operatorname{col}(H_{6})=\{\mbox{red},\mbox{blue}\}$, implying that $G_{2}\cup H_{6}$ is not conflict-free; or $\operatorname{col}(H_{6})=\{\mbox{red},\mbox{green}\}$ implying that $G_{6}\cup H_{6}$ is not conflict-free; or $\operatorname{col}(H_{6})=\{\mbox{blue},\mbox{green}\}$ implying that $G_{4}\cup H_{6}$ is not conflict-free. A similar argument holds when  $J_{3}$ or $J_{4}$ is yellow. 2. 2.

Two intervals among $J_{1},J_{2},J_{3},J_{4}$ use yellow. It follows that $H_{4}$ contains two yellow intervals and the remaining two intervals are colored either $\{\mbox{red},\mbox{blue}\}$, implying that $G_{2}\cup H_{4}$ is not conflict-free; or $\{\mbox{red},\mbox{green}\}$, implying that $G_{6}\cup H_{4}$ is not conflict-free; or $\{\mbox{blue},\mbox{green}\}$, implying that $G_{4}\cup H_{4}$ is not conflict-free.

∎

Now we place $\Omega(n)$ of these gadgets in two groups and for simplicity assume a gadget has width of 1. The gadgets in the first group are spaced with distance 2 between them, so a gadget from the second group can fit between any two consecutive gadgets. In the second group the gadgets are spaced with distance $3n$ between them, so that all gadgets of the first group fit between them. All gadgets of the first group then move at the same speed, starting somewhere to the left of the second group and moving to the right. The gadgets of the second group remain stationary. These motions ensure that each gadget of first group will cross each gadget of the second group, generating $\Omega(n^{2})$ crossing events, each of which results in at least one recoloring by Lemma 5.4.

Theorem 5.5**.**

For any $n>0$, there is a set of $8n$ intervals, each of which is either stationary or moves with unit speed to the right, so that when coloring the intervals using four colors at least $n^{2}$ recolorings are required to maintain a conflict-free coloring.

6 Conclusions

We introduced the fully-dynamic conflict-free coloring problem, where we want to maintain a conflict-free coloring for a set of geometric objects under insertions and deletions, and we presented a number of lower bounds and algorithms for the 1-dimensional version of the problem. There are several open problems.

First of all, there is still a gap between our upper and lower bounds. Recall that Theorem 2.1 gives a lower bound in the semi-dynamic (insertion-only) case. It would be interesting to see if a stronger lower bound can be obtained by mixing insertions and deletions.

Second, our lower bounds give a bound on the number of colors needed as a function of the worst-case number of re-colorings per insertion. It would also be interesting to investigate lower bounds as a function of the amortized number of re-colorings and/or to develop better algorithms in this setting.

Finally, an obvious area of future research is to consider the problem in two- or higher-dimensional space. In particular the problem of maintaining a conflict-free coloring for a set of disks in the plane under insertions and deletions—which comes much closer to the application that motivated conflict-free colorings, namely frequency assignment in wireless networks—is of interest. We note that very recently De Berg and Markovic [6] obtained several results on the two-dimensional problem.