# Online Coloring of Short Intervals

We study the online graph coloring problem restricted to the intersection graphs of intervals with lengths in [1,σ]. For σ=1 it is the class of unit interval graphs, and for σ=∞ the class of all interval graphs. Our focus is on intermediary classes. We present a (1+σ)-competitive algorithm, which beats the state of the art for 1 < σ < 2. For σ = 1 our algorithm matches the performance of FirstFit, which is 2-competitive for unit interval graphs. For σ=2 it matches the Kierstead-Trotter algorithm, which is 3-competitive for all interval graphs. On the lower bound side, we prove that no algorithm is better than 5/3-competitive for any σ>1, nor better than 7/4-competitive for any σ>2, nor better than 5/2-competitive for arbitrarily large values of σ.

## Authors

• 2 publications
• 1 publication
• 2 publications
• 13 publications
• 3 publications
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• ### Online Maximum k-Interval Coverage Problem

We study the online maximum coverage problem on a line, in which, given ...
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• ### On the deficiency of complete multipartite graphs

An edge-coloring of a graph G with colors 1,...,t is an interval t-color...
12/03/2019 ∙ by Armen R. Davtyan, et al. ∙ 0

• ### Distributed recoloring of interval and chordal graphs

One of the fundamental and most-studied algorithmic problems in distribu...
09/13/2021 ∙ by Nicolas Bousquet, et al. ∙ 0

• ### Online Coloring and a New Type of Adversary for Online Graph Problems

We introduce a new type of adversary for online graph problems. The new ...
05/21/2020 ∙ by Yaqiao Li, et al. ∙ 0

• ### L(p,q)-Labeling of Graphs with Interval Representations

We provide upper bounds on the L(p,q)-labeling number of graphs which ha...
09/12/2019 ∙ by Mehmet Akif Yetim, et al. ∙ 0

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## 1. Introduction

In the online graph coloring

problem the input graph is presented to the algorithm vertex by vertex, along with all the edges adjacent to the already presented vertices. Each vertex must be assigned a color, different than any of its neighbors, immediately and irrevocably at the moment it is presented, without any knowledge of the remaining part of the graph. The objective is to minimize the number of used colors. The problem and its variants attract much attention, both for theoretical properties and practical applications in network multiplexing, resource allocation, and job scheduling.

The standard performance measure, used to analyze online algorithms, is the competitive ratio, i.e., the worst-case guarantee on the ratio of the solution given by an online algorithm to the optimal offline solution (see Section 1.1 for a formal definition).

In the general case, of online coloring of arbitrary graphs there is no hope for any algorithm with a constant competitive ratio. The best known algorithm [7] uses colors for -vertex -colorable graphs, i.e. it is -competitive, and there is a lower bound [8] showing that no online graph coloring algorithm can be -competitive. It is thus common to study the problem restricted to particular graph classes.

Having in mind the applications in scheduling, one of the important special cases is the class of interval graphs, i.e. intersection graphs of intervals on the real line. The classic result is by Kierstead and Trotter [11], who designed a -competitive algorithm and proved a matching lower bound. However, in the special case of unit interval graphs, i.e. intersection graphs of intervals of a fixed (unit, w.l.o.g.) length, already the simple greedy FirstFit algorithm is -competitive [4].

A natural question arises, what happens in between the interval and unit interval graph classes. In particular, we ask about the optimal competitive ratio of online coloring algorithms for intersection graphs of intervals of length restricted to a fixed range. Formally, let us introduce the -interval coloring problem.

###### Definition 1.

For , the -interval coloring problem asks: Given a sequence of closed intervals , such that for every , find a sequence of colors, , such that

 ∀i≠j ([li,ri]∩[lj,rj]≠∅)⇒(ci≠cj),

minimizing the number of distinct colors .

We study the problem in the online setting, i.e., intervals are presented one by one, in an arbitrary order, and each interval has to be colored immediately and irrevocably after it is presented.

Note that we choose to include the interval representation in the input, instead of presenting the mere graph. It seems a plausible modelling choice given the scheduling applications. Moreover, it lets algorithms exploit geometric properties of the input, and not only structural graph properties. Naturally, any lower bound obtained for this variant of the problem transfers to the harder variant with no interval representation in the input.

### 1.1. Our Results

Before we state our results, let us give a formal definition of the competitive ratio. In this paper we focus on the asymptotic competitive ratio.

###### Definition 2.

Let be an online graph coloring algorithm, and let denote the maximum number of colors uses to color any graph which can be colored offline using -colors (i.e. its chromatic number is at most ). We say that has the asymptotic competitive ratio (or that is -competitive, for short), if .

Another popular performance measure for online algorithms is the absolute competitive ratio, which requires that holds for all (and not only in the limit). The choice of the asymptotic, instead of absolute, competitive ratio for our analysis makes things easier for the algorithm and harder for the lower bounds. In our algorithm, sadly, we do not know how to get rid of a constant additive overhead, which vanishes only with growing . The good side is, our lower bounds for the asymptotic competitive ratio imply the identical lower bounds for the absolute competitive ratio.

#### 1.1.1. Algorithm.

Our positive result is the existence of a -competitive algorithm.

###### Theorem 3.

For every , , there is an algorithm for online -interval coloring with asymptotic competitive ratio.

Note that for every -interval coloring algorithm is also a correct -interval coloring algorithm, with the same upper bound on its competitive ratio. Therefore, for Theorem 3 yields an online -interval coloring algorithm with a competitive ratio arbitrarily close to . This distinction between rational and irrational values of becomes somewhat less peculiar in the light of the results of Fishburn and Graham [5], who proved, among other things, that the classes of graphs with interval representation with lengths in are right-continuous exactly at irrational .

Until now, the state-of-the art was the -competitive FirstFit algorithm [4] for and the -competitive Kierstead-Trotter algorithm [11] for . Thus, our algorithm matches the performance of FirstFit for , and beats the Kierstead-Trotter algorithm up until .

#### 1.1.2. Lower Bounds.

In order to prove lower bounds for online problems, it is often convenient to look at the problem as a combinatorial game between two players, Algorithm and Presenter. In our case, in each round Presenter reveals an interval, and Algorithm immediately and irrevocably assigns it a color. While Algorithm tries to minimize the number of different colors it assigns, the Presenter’s goal is to force Algorithm to use as many colors as possible. A strategy for Presenter implies a lower bound on the competitive ratio of any algorithm solving the problem.

Our negative results include a series of strategies for Presenter with the following consequences.

###### Theorem 4.

For every there is no online algorithm for -interval coloring with the asymptotic competitive ratio less than .

###### Theorem 5.

For every there is no online algorithm for -interval coloring with the asymptotic competitive ratio less than .

###### Theorem 6.

For every there is such that there is no online algorithm for -interval coloring with the asymptotic competitive ratio .

The following, more illustrative, statement is a direct corollary of Theorem 6.

###### Corollary 7.

There is no online algorithm that works for all and uses at most colors for -colorable graphs (for any function ).

### 1.2. Methods

Our algorithm is inspired by the recent result for online coloring of unit disk intersection graphs [9]. We cover the real line with overlapping blocks, grouped into a constant number of classes. Each class gets a private set of available colors. When an interval is presented, the algorithm chooses a block in a round-robin fashion, and greedily assigns a color from its class.

All our lower bounds can be considered as generalizations of the lower bound for online coloring of unit interval graphs by Epstein and Levy [4]. In particular, we heavily use their separation strategy, which also appears in [1]. The novel technique is the recursive composition of strategies, materialized in Lemmas 12, 15, 17, and 21. Our lower bound borrows also from the work of Kierstead and Trotter [11]. However, in order to control the length of intervals independently of the number of colors, we cannot simply use the pigeonhole principle, as they did. Instead, we develop Lemmas 19 and 20, which let us overcome this issue, at a cost of a worse bound for the competitive ratio, i.e.  instead of .

### 1.3. Related Work

Interval graphs have been intensively studied since the sixties [2, 12], and, in particular, they are known to be perfect, i.e. the chromatic number of an interval graph always equals the size of the largest clique (see, e.g., [6]). To construct an optimal coloring offline it is sufficient to color the graph greedily in a nondecreasing order of the left ends of the intervals.

For the most basic approach for online coloring, that is the FirstFit algorithm, the competitive ratio for interval graphs is unknown. After a series of papers, the most recent results state that FirstFit is at least - and at most -competitive [10, 13]. Kierstead and Trotter [11] designed a more involved online coloring algorithm, which uses at most colors for -colorable interval graphs, and proved that there exists a strategy that forces any online coloring algorithm to use exactly that number of colors. The same lower and upper bounds were obtained independently by Chrobak and Ślusarek [3, 14]. For intersection graphs of intervals of unit length any online coloring algorithm uses at least colors, and FirstFit uses at most colors [4].

It seems a natural question to ask if it is possible to improve the bound of by assuming that interval lengths belong to a fixed range. The study of interval graphs with bounded length representations was initiated by Fishburn and Graham [5]. However, it focused mainly on the combinatorial structure, and not its algorithmic applications.

Kierstead and Trotter [11] give, for every , a strategy for Presenter to construct an -colorable set of intervals while forcing Algorithm to use at least colors. However, the lengths of presented intervals increase with the increasing . For this reason, with the interval length restricted to , their lower bound is only for the absolute competitive ratio and does not exclude, say, an algorithm that always uses at most colors. On the contrary, in Theorem 6 we rule out the existence of such an algorithm.

## 2. Algorithm

###### Theorem (Reminder of Theorem 3).

For every , , there is an algorithm for online -interval coloring with asymptotic competitive ratio.

###### Proof.

Let us present an algorithm which, in principle, works for any real , however only for a rational it achieves the declared competitive ratio. The algorithm has a positive integer parameter . Increasing the parameter brings the asymptotic competitive ratio closer to at the cost of increasing the additive constant. More precisely, given an -colorable set of intervals our algorithm colors it using at most colors, and thus its competitive ratio is . For a rational , in order to obtain exactly the declared asymptotic competitive ratio it is sufficient to set to the smallest possible denominator of a simple fraction representation of . Let . The algorithm will use colors from the set .

Now, let us consider the partition of the real line into small blocks. For , the -th small block occupies the interval . Moreover, we define large blocks. The -th large block occupies the interval . See Fig. 1.

Let us point out certain properties of the blocks, which will be useful in the further analysis. Each large block is the sum of consecutive small blocks, and each small block is a subset of consecutive large blocks. Further, length of a large block is , and for any two intervals of length in that both have the left endpoint in the same large block, the two intervals intersect. Thus, the intervals whose left endpoints belong to a fixed large block form a clique. Finally, if the indices of two large blocks differ by at least , then any two intervals – one with the left endpoint in one block, the other with the left endpoint in the other – do not intersect.

With each small block the algorithm associates a small counter, and with each large block the algorithm associates a large counter. Let denote the small counter of the -th small block, and denote the large counter of the -th large block. Initially, all the small and large counters are set to zero.

To assign a color to an interval, the algorithm proceeds as follows:

1. Let be the index of the small block containing the left endpoint of the interval.

2. Let be the index of the large block containing the left endpoint of the interval such that . Note that there is exactly one such .

3. Assign to the interval the color .

4. Increase the small counter by one.

5. Increase the large counter by one.

First let us argue that the algorithm outputs a proper coloring. Consider any two intervals which were assigned the same color. Let and denote the indices of the large blocks selected for these intervals by the algorithm. Since the colors of the two intervals have the same first coordinates, we have that . However, since the second coordinates, which are determined by large counters, are also the same, and must be different, and thus they differ by at least . Hence the left endpoints of the large blocks and are at least apart, and the two considered intervals do not intersect, thus the coloring is proper.

It remains to bound the number of colors in terms of the clique number . Let be the index of the maximum large counter . Clearly, the algorithm used at most colors in total. Let denote the set of intervals with the left endpoints in the -th large block and colored with a color in . Observe that . Let denote the number of intervals in which have the left endpoint in the -th small block. Recall that the -th large block is the sum of small blocks – indexed , , …, – and thus . Now, observe that, because of the round robin selection in the step 2 of the algorithm,

 Sk⩾b⋅(xk−1)+1.

Let denote the set of all intervals with the left endpoints in the -th large block. We can bound the number of intervals in

 |D| = j+b−1∑k=jSk ⩾ j+b−1∑k=j(b⋅(xk−1)+1) = b⋅(Lj−b)+b.

Recall that is a clique and thus the clique number of the input graph is at least the size of . Therefore , and the algorithm used at most

 ⌈b⋅(1+σ)⌉⋅(ωb+b−1)

colors. ∎

## 3. Lower Bounds

In this section we show several families of strategies for Presenter that force Algorithm to use many colors while the introduced set of intervals is colorable with a smaller number of colors, and contains only short intervals. We start with a short, informal presentation of these strategies.

Epstein and Levy [4] give the following strategy for Presenter in online coloring of unit intervals (see the proof of Lemma 12 for all the details, Figure 2 may help visualize this strategy).

1. Presenter plays a clique of initial intervals. Algorithm has to use different colors, let denote the set of these colors.

2. To the left of the initial intervals, Presenter plays a clique of separation intervals so that all intervals colored with colors in are slightly shifted to the left of all intervals colored with colors not in .

3. Presenter plays a clique of final intervals that intersect all the initial intervals, and right-most separation intervals.

Algorithm has to use at least colors in total.

In Section 3.1 we generalize this strategy. We observe, that instead of presenting a clique in the first step, Presenter can use an arbitrary strategy that requires slightly shorter intervals. For -interval coloring we can construct a recursive strategy that applies this trick roughly times. Using this simple observations we obtain a family of strategies for different . Corollary 13 gives that, for example, there is no online algorithm with asymptotic competitive ratio for -interval coloring (for any ).

Now, consider the following strategy for Presenter in online coloring of -intervals (see the proof of Lemma 15 for all the details, Figure 3 may help visualize this strategy).

1. Presenter plays a clique of initial intervals of length . Algorithm has to use different colors, let denote the set of these colors.

2. To the left of the initial intervals, Presenter plays a clique of left separation intervals of length so that all intervals colored with colors in are slightly shifted to the left of all intervals colored with colors not in . Let denote the set of colors of right-most left separation intervals.

3. To the right of initial intervals, Presenter plays a clique of right separation intervals of length so that all intervals colored with colors in are slightly shifted to the right of all intervals colored with colors not in .

4. Presenter plays a clique of final intervals of length that intersect all the initial intervals, right-most left separation intervals, and left-most right separation intervals.

We get that there is no online algorithm with asymptotic competitive ratio for -interval coloring (for any ), i.e. we’ve sketched an informal proof of Theorem 4.

In Section 3.2 we use the above strategy for -intervals as a recursive step that can be used to obtain strategies for larger ’s. We get another family of strategies, where for -interval coloring we can apply the recursive step roughly times. Corollary 16 gives that, for example, there is no online algorithm with asymptotic competitive ratio for -interval coloring (for any ).

Now, consider the following strategy for Presenter in online coloring of -intervals (see the proof of Lemma 17 for all the details, Figures 4 and 5 may help visualize this strategy).

1. Presenter plays a clique of left initial intervals of length . Algorithm has to use different colors, let denote the set of these colors.

2. To the right of the left initial intervals, Presenter plays a clique of right initial intervals of length . Algorithm has to use different colors, let denote the set of these colors.

3. To the left of the left initial intervals, Presenter plays a clique of left separation intervals of length so that all intervals colored with colors in are slightly shifted to the left of all intervals colored with colors not in . Let denote the set of colors of right-most left separation intervals.

4. To the right of the right initial intervals, Presenter plays a clique of right separation intervals of length so that all intervals colored with colors in are slightly shifted to the right of all intervals colored with colors not in . Let denote the set of colors of left-most right separation intervals.

5. Let , and .

1. If , Presenter plays a clique of final intervals of length that intersect all the initial intervals, right-most left separation intervals, and left-most right-separation intervals. In total, Algorithm has to use at least colors in this case.

2. If (which implies ), Presenter plays a clique of pre-final intervals of length that intersect all the right initial intervals, and left-most right separation intervals. Then, Presenter plays a clique of final intervals of length that intersect all the left initial intervals, right-most left separation intervals, and all the pre-final intervals. The sets of colors of final and pre-final intervals are disjoint, and moreover do not intersect with . A short calculation shows that in this case Algorithm also has to use at least colors.

Thus, we get that there is no online algorithm with asymptotic competitive ratio for -interval coloring (for any ), i.e. we’ve sketched an informal proof of Theorem 5.

In Section 3.3 we use the above strategy for -intervals as a recursive step that can be used to obtain strategies for larger ’s. We get another family of strategies, where for -interval coloring we can apply the recursive step roughly times. Corollary 18 gives that, for example, no online algorithm with asymptotic competitive ratio for -interval coloring (for any ).

To get a lower bound better than we combine our method with some ideas from the lower bound by Kierstead and Trotter [11]. In Section 3.4 we prove Corollary 22, which gives that, for example, there is no -competitive online algorithm for -interval coloring, there is no -competitive online algorithm for -interval coloring, and there is no -competitive online algorithm for -interval coloring (for any ).

Table 1 summarizes the aforementioned strategies, and illustrates the growth of the interval length required to prove better and better lower bounds.

To properly capture asymptotic properties of the introduced strategies we give the following formal definitions.

###### Definition 8.

For and , an -strategy is a strategy for Presenter that forces Algorithm to use at least colors subject to the following constraints:

1. the set of introduced intervals is -colorable,

2. every introduced interval has length at least and at most ,

3. every introduced interval is contained in the interval .

We are interested in providing strategies that achieve the biggest possible ratio for large . This motivates the following definition.

###### Definition 9.

An -schema is a set of -strategies for all such that .

The term in the above definition accounts for the fact that sometimes in a proof we would like to introduce, say,

-clique. Then, for odd

’s a rounding is required, which results in small inaccuracies we need to control.

###### Remark 10.

Note that the existence of an -schema implies a lower bound of for the asymptotic competitive ratio of any online algorithm solving the -interval coloring problem.

To put the above definitions in context, note that Kierstead and Trotter [11] give, for all , an -strategy. However, their family of strategies does not yield an -schema, because the length of the presented intervals grows with .

###### Example 11 (⟨1,1,1⟩-schema).

For any , a strategy that introduces the interval in every round is an -strategy. The set of these strategies is a -schema.

In this section we show a series of constructions that use an existing schema to create another schema with different parameters. The -schema given above is the initial step for those constructions.

Let be an -strategy. We will say that Presenter uses strategy in the interval to denote that Presenter plays according to , presenting intervals shifted by , until Algorithm uses colors.

### 3.1. Warm-up

Our first construction is a natural generalization of the strategy for unit intervals given by Epstein and Levy [4]. It is surpassed by more involved strategies coming later, but it serves as a gentle introduction to our framework.

###### Lemma 12.

If there is an -schema, then there is a -schema for every .

###### Proof.

To prove the lemma we need to provide an -strategy for every . Let us fix an , and let . The -schema contains an -strategy for some . The strategy for Presenter consists of three phases (see Fig. 2). In the first phase, called the initial phase, Presenter uses strategy inside the interval . Let and let denote the set of colors used by Algorithm in the initial phase.

The second phase, borrowed from [4, 1], is called the separation phase. In this phase, Presenter plays the following separation strategy for rounds. Let and . In the -th round of the separation phase Presenter introduces the interval . If Algorithm colors the interval with one of the colors in , let and , which means that the next interval will be shifted slightly to the right. Otherwise, let and , which means that the next interval will be shifted slightly to the left. Observe that all intervals introduced in the separation phase have length and . Thus, every interval introduced in the separation phase is contained in and any two of those intervals intersect. Furthermore, the choice of ’s and ’s guarantees that for any two intervals , introduced in the separation phase, colored with a color in , and colored with a color not in , we have that the left endpoint of is to the left of the left endpoint of . Let be the set of right-most intervals introduced in the separation phase. Let be the set of colors used by Algorithm on the intervals in . As , we get that sets of colors and are disjoint.

For the last phase, called the final phase, let be the left-most right endpoint of an interval in . In the final phase Presenter introduces times the same interval . This interval intersects all intervals introduced in the initial phase, all intervals in , and no other interval introduced in the separation phase. Thus, Algorithm must use colors in the final phase that are different from the colors in both and . Let denote the set of colors used by Algorithm in the final phase.

The presented set of intervals is clearly -colorable and Algorithm used at least many colors. The longest interval presented has length , and all intervals are contained in . Thus, we have constructed a -schema. ∎

###### Corollary 13.

There is an -schema, for every and every , where is the -th Fibonacci number (, ).

###### Proof.

Starting with a -schema and repeatedly applying Lemma 12 one can generate111Knowing the desired target values of and , one needs to properly adjust the value for each application of Lemma 12, e.g., it is sufficient to set it to . a family of -schemas, such that , , and . Solving the recurrence equations we get , , and . ∎

Note that this method cannot give a lower bound with the multiplicative factor better than . However, we can get arbitrarily close to this bound. That is, for every there is a and such that for each there is a strategy for Presenter to present intervals of length at most and force Algorithm to use colors on an -colorable set of intervals.

###### Observation 14.

There is no online algorithm that works for all and uses at most colors for -colorable graphs (for any function ).

### 3.2. The 5/3 Lower Bound

###### Lemma 15.

If there is an -schema, then there is a -schema for every .

###### Proof.

The proof of this lemma is very similar to the proof of Lemma 12, but now we have two separation phases instead of just one, see Fig. 3. Let us fix an , and let . Let be an -strategy for some .

In the initial phase, Presenter uses inside interval , and forces Algorithm to use colors. Let denote the set of those colors.

In the separation phase, Presenter plays the separation strategy two times. First, Presenter plays the separation strategy for rounds in the region pushing to the right colors not in . Let be the set of right-most intervals from this first separation. Let denote the set of colors used by Algorithm to color . Then, Presenter plays the separation strategy for rounds in the region pushing to the left colors not in . Let be the set of left-most intervals from this second separation. Let denote the set of colors used by Algorithm to color .

Let be the left-most right endpoint of an interval in . Let be the right-most left endpoint of an interval in . In the final phase Presenter introduces times the same interval .

The presented set of intervals is clearly -colorable and Algorithm used at least many colors. The longest interval presented has length at most , and all intervals are contained in . Thus, we have constructed a -schema. ∎

###### Corollary 16.

There is an -schema, for every and every , where

###### Proof.

The argument is similar to Corollary 13, but now we solve the recurrence equation , . ∎

Note that, similarly to Observation 14, one could already use Corollary 16 to get a lower bound arbitrarily close to for the asymptotic competitive ratio of any online algorithm that work for all . Nonetheless in Section 3.4 we prove a stronger lower bound.

###### Theorem (Reminder of Theorem 4).

For every there is no online algorithm for -interval coloring with the asymptotic competitive ratio less than .

###### Proof.

Assume for contradiction that for some there exists an online algorithm for -interval coloring with the asymptotic competitive ratio , for some . By the definition of the asymptotic competitive ratio, there is an such that for every the algorithm colors every -colorable set of intervals using at most colors.

Observe that, for , Corollary 16 gives a -schema. By the definition of schema, there is an such that for every there is a strategy for Presenter to present an -colorable set of intervals, of length in , and force Algorithm to use colors. For we reach a contradiction. ∎

### 3.3. The 7/4 Lower Bound

###### Lemma 17.

If there is an -schema, then there is a -schema for every .

###### Proof.

The proof of this lemma is a bit more complicated than the previous ones, as we now have two initial phases, two separation phases and a strategy branching, see Fig. 4 and Fig. 5. Let us fix an , and let . Let be an -strategy for some .

In the initial phase, Presenter uses strategy twice: first, inside interval , and then inside interval . Algorithm uses colors in each of these games. We get a set of colors used by Algorithm in the first game, and a set of colors used by Algorithm in the second game. Note that might be non-empty.

In the separation phase, Presenter plays the separation strategy two times. First, Presenter plays the separation strategy for rounds in the region pushing to the right colors not in . Let be the set of right-most intervals from the first separation phase. Let denote the set of colors used by Algorithm to color . Then, Presenter plays the separation strategy for rounds in the region pushing to the left colors not in . Let be the set of left-most intervals from the second separation phase. Let denote the set of colors used by Algorithm to color . Let be the left-most right endpoint of an interval in . Let be the right-most left endpoint of an interval in .

There are two cases in the final phase. Let , and analogously . We have that .

Case 1: If , then Presenter introduces times the same interval .

Each interval introduced in the final phase intersects with all intervals from both initial phases and all intervals in . Thus, Algorithm is forced to use colors in total.

Case 2: If , then Presenter introduces intervals, all of them having endpoints . Let be the set of colors used by Algorithm in this pre-final phase. We have , and we assumed that , thus we have , and now we are in Case 1 with , see Fig. 5.

The longest interval introduced by Presenter in both cases has length strictly less than , and the whole game is played in the region . ∎

###### Corollary 18.

There is an -schema, for every and every , where

###### Proof.

The argument is similar to Corollaries 13 and 16, but now we solve the recurrence equations , , and , , . ∎

Note that, similarly to Observation 14, one could already use Corollary 18 to get a lower bound arbitrarily close to for the asymptotic competitive ratio of any online algorithm that work for all . Nonetheless in Section 3.4 we prove a stronger lower bound.

###### Theorem (Reminder of Theorem 5).

For every there is no online algorithm for -interval coloring with the asymptotic competitive ratio less than .

###### Proof.

Observe that, for and , Corollary 18 gives a -schema. Then proceed analogously to the proof of Theorem 4. ∎

### 3.4. The 5/2 Lower Bound

To prove our main negative result we need two simple combinatorial lemmas.

###### Lemma 19.

Let . For every four sets , each of size , if their intersection is small: , their sum is large: .

###### Proof.

Each element which belongs to the sum but does not belong to the intersection can belong to at most three sets. Thus, we have

and so

 3⋅∣∣ ∣∣4⋃i=1Xi∣∣ ∣∣⩾4k−∣∣ ∣∣4⋂i=1Xi∣∣ ∣∣⩾(3+γ)⋅k.

###### Lemma 20.

Let , and be a family of sets, each of size . Then, either

 ∣∣ ∣∣4n⋃i=1Xi∣∣ ∣∣⩾(3+γ3)nk,

or the sequence can be covered with four disjoint intervals , , , , such that for the intersection of ’s is large:

 |Y1∩Y2∩Y3∩Y4|⩾(1−γ)⋅k.
###### Proof.

Consider families of sets defined as follows: for every , and for every and . See Fig. 6.

If for some we have , then we are done. Thus, we assume that . Let . We prove that , by induction on . For the statement is obvious because . For and arbitrary , let . By the induction hypothesis . We may ignore some elements of those sets and assume that , moreover we assumed that , where and . We apply Lemma 19 and get