Abstract 1 Introduction 2 Generalized Min Sum Set Cover 3 2-approximation for Min Latency Set Cover 4 A Tail Bound for the Sum of Bernoulli Random Variables References

A 4.509-Approximation Algorithm for Generalized Min Sum Set Cover

Amey Bhangale ORCID Department of Computer Science and Engineering, University of California, Riverside, CA, USA    Yezhou Zhang111Corresponding author ORCID Department of Computer Science and Engineering, University of California, Riverside, CA, USA
Abstract

We study the generalized min-sum set cover (GMSSC) problem, where given a collection of hyperedges E with arbitrary covering requirements {ke∈ℤ+:e∈E}, the objective is to find an ordering of the vertices that minimizes the total cover time of the hyperedges. A hyperedge e is considered covered at the first time when ke of its vertices appear in the ordering.

We present a 4.509-approximation algorithm for GMSSC, improving upon the previous best-known guarantee of 4.642 [2, SODA’21]. Our approach retains the general LP-based framework of Bansal, Batra, Farhadi, and Tetali [2] but provides an improved analysis that narrows the gap toward the lower bound of 4-approximation assuming P≠NP. Our analysis takes advantage of the constraints of the linear program in a nontrivial way, along with new lower-tail bounds for the sums of independent Bernoulli random variables, which could be of independent interest.

Keywords and phrases:
Generalized Min Sum Set Cover, Approximation Algorithm, Min latency set cover, Linear programming, Knapsack cover inequalities
Category:
Track A: Algorithms, Complexity and Games
Funding:
Amey Bhangale: Supported by the Hellman Fellowship award and NSF CAREER award 2440882.
Yezhou Zhang: Supported by the Hellman Fellowship award and NSF CAREER award 2440882.
Copyright and License:
[Uncaptioned image] © Amey Bhangale and Yezhou Zhang; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation → Approximation algorithms analysis
Related Version:
Full Version: https://arxiv.org/abs/2605.10031
Editors:
Sayan Bhattacharya, Danupon Nanongkai, Michael Benedikt, and Gabriele Puppis

1 Introduction

In this paper, we study approximation algorithms for the following related problems.

Min Sum Set Cover Problem.

The min sum set cover problem (MSSC) takes a universe of n elements, denoted as [n], as input and a collection of it subsets. Given a permutation of all elements in [n], a subset S is considered covered at the earliest time an element from S appears in the permutation. The goal of the min sum set cover problem is to find a permutation of [n] that minimizes the sum of the cover time of all given subsets. Alternatively, this problem can be described on a hypergraph: given a hypergraph G=(V,H) with n vertices and a set of hyperedges e∈H. For a schedule (permutation) of n vertices, the cover time is defined as the earliest time that one of v∈e appears in the schedule. The goal is to find a schedule that minimizes the sum of the cover time of all hyperedges. This problem was formally defined in [6], which also introduced a 4-approximation greedy algorithm and a matching hardness result.

Generalized Min Sum Set Cover Problem.

The generalized min sum set cover problem (GMSSC), also known as the multiple intents re-ranking problem generalized MSSC by assigning each hyperedge an individual cover requirement ke≤|e|, such that the hyperedge is considered at the earliest time when ke of its vertices appear in the schedule. The problem was first introduced by Azar, Gamzu, and Yin in [1], which also provided an O⁢(log⁡(max⁡ke))-approximation algorithm for the problem. Bansal, Gupta, and Krishnaswamy first provided a constant, 485-approximation in [3] with a linear program relaxation, which included strengthening constraints known as KC (Knapsack Cover) inequalities from [5]. Skutella and Williamson later improved the approximation factor to 28 in [11] by applying the α-point rounding algorithm to the LP relaxation, and Im, Sviridenko, and Zwaan further improved the factor to 12.4 in [9]. Recently, Bansal, Batra, Farhadi, and Tetali gave a 4.642-approximation in [2] by adding a linear transformation, or the kernel, to the LP fractional optimal solution and rounding based on the adjusted solution. This remains the best-known approximation to date. The hardness of GMSSC is 4−ϵ, derived from MSSC in [6].

Min Latency Set Cover Problem.

The min latency set cover problem, first formally studied in [7], is a special case of GMSSC where each hyperedge has a cover requirement ke=|e|. It is a well-studied problem, also known as the special case of the classic precedence-constrained scheduling problem, and has many 2-approximation algorithms (see, e.g, [10] for a survey). Bansal and Khot proved the Unique-Game hardness of this problem is 2−ϵ in [4].

1.1 Our Results

Our main theorem improves the 4.624-approximation for GMSSC given by Bansal, Batra, Farhadi, and Tetali in [2].

Theorem 1.

There exists a 4.509-approximation algorithm for the generalized min sum set cover (GMSSC) problem.

In [2], the authors gave an approximation algorithm for GMSSC and showed that the algorithm achieves an approximation ratio of 4.624. They also remarked that the numerical analysis suggests that the approximation factor is no more than 4.5232, even though they were unable to prove this analytically. Finally, they showed that getting better than a 4.509 factor approximation has a natural bottleneck for their approach. Towards proving the main theorem above, we provide a tighter analysis of their algorithm and show that it indeed achieves a 4.509 approximation ratio. Our analysis leverages the linear program’s constraints in a non-trivial manner along with new lower-tail bounds for the sums of independent Bernoulli random variables, which could be of independent interest (see Section 1.2 for more details).

Our second theorem gives a 2-approximation algorithm for the min latency set cover problem using the framework of Bansal, Batra, Farhadi, and Tetali [2], improving upon the e=2.718⁢… approximation algorithm given in the same paper.

Theorem 2.

There exists a 2-approximation algorithm for the min latency set cover problem.

Recall that a 2-approximation algorithms were already known before [10]. We designed a new kernel which, when plugged into the framework of [2], gives the optimal approximation algorithm for the min latency set cover problem.

In [2], the authors already showed that their framework gave an optimal 4-approximation algorithm for MSSC (all ke=1). Theorem 2 now shows that the framework provides an optimal approximation algorithm (assuming the Unique Games Conjecture) for min sum latency set cover (all ke=|e|). It would be interesting to see if it gives an optimal approximation algorithm for GMSSC (arbitrary ke).

1.2 Techniques

This section provides an overview of the LP-rounding based algorithmic techniques employed in this paper.

The foundation of our approach lies in the work of [3], which defined a commonly used linear program (LP) for MSSC and GMSSC. This LP returns a fractional solution x where xv,t is an indicator of whether vertex v is scheduled at time t. The core LP constraint is given by

∑v∈e∑t′<txv,<t≥∑t′<txe,t′,∀e,t,

where ∑t′<txe,t′ indicates the fraction of edge e covered at time t. However, a simple randomized rounding algorithm, rounding based on ∑t′≤txv,t′, proves ineffective even for MSSC. This is because if the LP fractionally covers a hyperedge of n vertices by scheduling each vertex to the extent of 1/n at time t, this approach leaves this hyperedge uncovered with probability (1−1/n)n≈1/e≫0 for any time after t. For a long period after t, the LP solver may consider this hyperedge covered, but yet no vertex might be scheduled in an integral solution, leading to an arbitrarily large approximation ratio. To overcome this, a linear transform, or kernel K, was introduced and applied to the fractional optimal solution x from the LP in [2]. The standard α-point randomized rounding was then applied to Z=K⁢x to obtain a feasible integral solution. Using a kernel K⁢(t,t′)=β/t⁢(t′≤t), for β=2, in [2], the authors achieved a 4-approximation for MSSC, which corresponded to the known hardness result. This was accomplished by analyzing the expected cover time of a hyperedge in the fractional optimal solution x, denoted cx⁢(e), and its corresponding expected time in the integral solution z, denoted cz⁢(e), along with a random tie-breaking rule for the final integral solution. While cx⁢(e) was obtained from the optimal fractional solution, cz⁢(e) was related to the sum of random Bernoulli variables. The approximation factor was then formulated as a convex optimization problem involving the ratio cz⁢(e)/cx⁢(e). Through convex optimization techniques, it was proven that the worst-case scenario occurs when the LP solver assigns the extent 1/n to all vertices. Nevertheless, the β/t kernel effectively increases the probability of scheduling v∈e over time, leading to a feasible solution.

For the GMSSC problem, the best known hardness is 4, and all previous approximations follow the extended analysis of MSSC. In [3], an unbounded integrality gap is demonstrated for the natural extended LP for GMSSC, and a set of strengthening constraints, KC (Knapsack Cover) inequalities from [5], was introduced for GMSSC.

∑v∈e∖S∑t′<txv,t′≥(ke−|S|)⁢∑t′<txe,t′,∀e,t,S⊆e.

The KC inequalities eliminated the unbounded integrality gap and enabled the proof of several constant-factor approximation algorithms. However, applying the same algorithm and analysis to GMSSC in [2] yielded only a 4.642-approximation factor. This result left a gap between GMSSC and MSSC, and the method to eliminate this gap remained unknown. For instance, if ke−1 vertices are already scheduled in the integral solution (let S be the set of these vertices), the above KC inequality effectively transforms into a constraint in the LP for MSSC. This suggests that GMSSC could potentially be reduced to MSSC after a certain time.

Our paper aims primarily to eliminate this gap between GMSSC and MSSC after a certain time in the scheduling. To this end, we address a key analytical challenge in GMSSC: for a hyperedge e, the remaining cover requirement at time t, ke⁢(t), potentially decreases over time as vertices are getting scheduled. This dynamic nature makes the probability of covering e difficult to accurately express with a single function. We establish a novel upper bound Pke⁢(t) on the probability that a hyperedge e remains uncovered at time t, explicitly involving ke⁢(t). Furthermore, by leveraging multiple KC inequalities, we demonstrate that the core KC inequality (after applying the kernel) ∑v∈e∖Ae⁢(t)zv,<t≥ke⁢(t)⁢ze,<t can be more accurately expressed as:

∑v∈e∖Ae⁢(t)zv,<t≈ke⁢(t)⁢ze,<t+δ⁢(t),

where Ae⁢(t) is the set of vertices guaranteed to be scheduled in the integral solution before time t, δ⁢(t) is a crucial gap extracted from other KC inequalities.

Consequently, we derive a more precise expression for cz⁢(e), the expected cover time in the integral solution. More precisely, since a vertex v∈e∖Ae⁢(t) is scheduled at time t with probability zv,<t in the rounding algorithm, the probability that e is not covered at time t is precisely the probability that at most ke⁢(t)−1 vertices from e∖Ae⁢(t) are scheduled before time t. If we let Pk⁢(x) to denote an upper bound on Pr⁡[S≤k−1] if k⁢x=𝔼[S] where S is the sum of Bernoulli random variables, then

cz(e)≈te+∑t>tePke⁢(t)(ze,<t+δ(t)/ke(t))), (1)

where te is a special time extracted from the LP solution. Compared to previous results, where cz⁢(e)≤te+∑t>tePc⁢(ze,<t) (here, the quantity Pc⁢(ze,<t) is an upper bound on the probability if e is not covered at time t, independent of ke⁢(t), proved in [2]), our expression more precisely reflects the probability of covering the hyperedge assuming that ke⁢(t) is known. This allows for a more fine-grained analysis in terms of the change of ke⁢(t) over time. Accordingly, we present several extended convex optimization results to argue that the worst-case behavior of GMSSC is similar to that of MSSC. Crucially, the existence of δ⁢(t) and the properties of Pke⁢(t) allow us to prove that the worst-case scenario always involves ke⁢(t) reducing to 1 prior to te. On a more technical side, we give a refined upper bound on Pk⁢(x) in Theorem 4, and use these bounds to relate various terms in the expression (1) for cz⁢(e) as t increases. This more refined analysis yielded a 4.509-approximation for GMSSC.

Beyond GMSSC, we also study the min latency set cover problem. While 2-approximation algorithms are already known for this problem, we explore whether this LP framework can yield a matching approximation. In the min latency set cover problem, the LP solver needs to schedule all vertices of a hyperedge before it is considered covered. However, a simple randomized algorithm again proves insufficient. If the LP assigns an extent of 1−1/n to all vertices of a hyperedge, making the edge almost scheduled fractionally, the simple randomized algorithm covers this hyperedge at time t with probability (1−1/n)n≈1/e, which is much less than 1. Consequently, we adopt a kernel:

K⁢(t,t′)=α⁢t′t⁢(t+1),

which retains the property of increasing the probability of scheduling a vertex over time, but at a faster rate, as a divergent kernel is not required here. This approach, using a similar analysis, yields a 2-approximation. Although this result for the min latency set cover is not novel, it highlights the robustness and power of this LP-rounding framework and offers insights that could potentially lead to further improvements in the approximation factor for GMSSC.

1.3 Organization

Section 2 presents our primary analysis regarding the Generalized Minimum Sum Set Cover (GMSSC) problem. Specifically, we begin by briefly introducing the linear program, its relaxation, and the rounding algorithm from [2] in Sections 2.1 and 2.2, respectively. Our main contribution, detailed in Section 2.3, is the achievement of a 4.509-approximation factor within this LP-rounding based framework.

Section 3 focuses on the min latency set cover problem, where we prove that this algorithm achieves a 2-approximation using a distinct kernel compared to the MSSC/GMSSC approach. Finally, Section 4 establishes a new tail bound on the sum of Bernoulli random variables, which is crucial for the analysis presented in Section 2.3.

2 Generalized Min Sum Set Cover

2.1 Linear Program

Assume that time is discrete and that time t∈{1,2,…}, refers to the time interval (t−1,t]. e refers to an edge and a hyperedge interchangeably. For each vertex v and time t, there is a variable xv,t, a 0-1 indicator of whether v is assigned at time t. For each edge e and time t, there is a variable ue,t intended to be 1 if e is uncovered at the beginning of t. The following LP program and relaxation for MSSC was introduced in [3]:

Minimize ∑e,tue,ts.t.
∑vxv,t≤1, ∀t, (2)
ue,t+∑v∈e∑t′<txv,t′≥1, ∀e,t, (3)
ue,t,xv,t≥0, ∀e,v,t. (4)

The constraint (2) ensures that at most one vertex is scheduled simultaneously. The constraint (3) guarantees that ue,t can be set to 0 only if some v∈e is scheduled strictly before t.

For GMSSC, the demands ke are arbitrary, so the natural extension of (3),

ke⁢ue,t+∑v∈e∑t′<txv,t′≥ke,∀e,t,

becomes extremely weak. For example, if a hyperedge has 100 vertices to cover, and the LP solver assigns ∑t′<txv,t′=1 to 99 vertices but sets all other vertices to 0. In such a case, the natural extension of (3) allows the LP solver to set ue,t=0.01, and e is almost covered in the fractional solution. However, since there are only 99 vertices with ∑t′<txv,t′≠0, we are unable to obtain a valid integral solution in any case. Consequently, the Knapsack Cover (KC) Inequalities were introduced to replace (3) and strengthen the linear program in [3].

(ke−|S|)⁢ue,t+∑v∈e∖S∑t′<txv,t′≥(ke−|S|),∀e,t,S⊆e,|S|<ke. (5)

The KC inequalities guarantee that the LP solver can only consider this hyperedge e covered when ke many vertices are almost scheduled.

2.2 Rounding algorithm

Once the LP solver provides the optimal fractional solution x, we shall apply a rounding algorithm to x to generate the integral solution. However, directly rounding ∑t′<txv,t′ may not guarantee a feasible integral solution. For example, when ke=1, the LP solver can set ∑t′≤txv,t′=1/n→0 for all v∈e. The challenge is that we may not schedule any vertex for a long period in the rounding procedure, since the probability is negligible. Therefore, we apply an additional transformation to x to ensure that the integral solution gives a feasible schedule after the LP solver covers the edge. For all v and t, we apply the following transformation,

zv,t=∑t′K⁢(t,t′)⁢xv,t′. (6)

The new solution z=K⁢x will no longer satisfy the constraints in the original linear program, but z can be rounded to a feasible schedule if K is chosen properly. The rounding algorithm is as follows:

Algorithm 1 Rounding Algorithm from [2].

After the rounding procedure, the approximation algorithm gives the desired integral solution σ.

2.3 Analysis

For a vector xv, define

xv,<t:=∑t′<txv,t′.

For the GMSSC problem, we select the kernel,

K⁢(t,t′)=βt⋅𝟙⁢[t′≤t]. (7)

Note that β=2.043 in our final setting. Correspondingly,

zv,t=βt⁢∑t′≤txv,t′andzv,<t=β⁢∑t′′<txv,≤t′′t′′=β⁢∑t′≤txv,t′⁢∑t′≤t′′t1t′′≥β⁢∑t′≤txv,t′⁢ln⁡tt′. (8)

This kernel ensures that even if xv,<t→0, zv,<T eventually reach 1 for some large T.

Fix an edge e and denote the cover time of e in the LP as cx⁢(e)

cx⁢(e):=∑tue,t.

Let xe,t=ue,t−ue,t+1, then cx⁢(e) can be written as

cx⁢(e)=∑tue,t=∑tt⁢xe,t.

Additionally, define Covσ⁢(e) as the cover time of e in σ. Similarly, Covτ⁢(e) is the cover time of e in τ.

According to the rounding procedure, each vertex v∈e will be scheduled with probability min⁡(zv,<t,1) at time t. Let yv,t=min⁡(1,zv,<t), then for each vertex v, the event whether v will be scheduled can be considered as a Bernoulli random variable with success probability Yv,t∼B⁢(yv,t). Furthermore, the event that edge e is scheduled at time t is determined by the sum of |e| independent Bernoulli random variables. Let pt⁢(e) be the probability that e is not scheduled at time t in τ. By the definition of Covτ⁢(e) and the above analysis,

𝔼[Covτ⁢(e)]=∑tpt⁢(e)=∑tPr⁡[∑v∈eYv,t≤ke−1]. (9)

Let S=∑v∈eYv,t, then 𝔼[S]=∑v∈eyv,t. However, it is a challenge to analyze all vertices in e simultaneously. The Hoeffding inequality, in [8], gives an upper bound on (9).

Lemma 3.

If S is sum of n i.i.d Bernoulli random variables and 𝔼[S]=n⁢p, then

Pr⁡[S≤c]≤ ∑i=0c(ni)⁢pi⁢(1−p)n−i, if 0≤c≤n⁢p−1,
Pr⁡[S≤c]≤ max0≤s≤c⁢∑i=0c−s(n−si)⁢(n⁢p−sn−s)i⁢(1−n⁢p−sn−s)n−s−i, if n⁢p−1<c<n⁢p,
Pr⁡[S≤c]≤ 1, if n⁢p≤c≤n.

This lemma allows us to describe the sum of Bernoulli random variables using 𝔼[S]. Remark that the bounds in Lemma 3 can be further bounded by some exponential bound derived from Poisson distributions.

Up to this point, the above analysis is virtually identical to the framework in [2]. However, there are some gaps in the framework that need to be addressed. Let ze=K⁢xe, and consider the following KC inequalities

(ke−|S|)⁢ue,t+∑v∈e∖Sxv,<t≥ke−|S|.

Apply the kernel K to both sides, we have

∑v∈e∖Szv,<t≥(ke−|S|)⁢ze,<t.

Since zv,<t corresponds to yv,t and ze,<t corresponds to xe,t, the KC inequalities potentially allow us to bound 𝔼[S] with ze,t or xe,t. However, if zv,<t>1 for some vertices, this approach fails as ∑v∈ezv,<t≠∑v∈eyv,t. Thus, fix a time t, define Ae⁢(t)={v|zv,<t≥1}, Be⁢(t)=e∖Ae⁢(t), and let ke⁢(t)=|Be⁢(t)|. Note that the vertices in Ae⁢(t) are definitely scheduled before time t in the schedule τ. We have

∑v∈e∖Ae⁢(t)zv,<t≥(ke−|Ae⁢(t)|)⁢ze,<t.

Or equivalently,

∑v∈Be⁢(t)zv,<t≥ke⁢(t)⋅ze,<t. (10)

The challenge arises from the fact that ke⁢(t) will decrease over time, so even the bound in Lemma 3 changes. In [2], the authors gave a common upper bound on Pr⁡[S≤k−1] for all k as a function of 𝔼[S]k. Since it is independent of k, it is always “safe” as ke⁢(t) changes. However, such a safe bound is not tight for any ke⁢(t) and introduces some loss. To improve, we need to process the KC inequalities with caution and extract more information from them.

First, for any fixed ke⁢(t), we show the exponential upper bound for pt⁢(e). We prove the following lemma that further upper bounds Lemma 3.

Theorem 4.

Let S=∑Yv, for any integer k≥1, then function

Pk⁢(x)={e−k⁢(x−k−1k),if x>1,1,if 0≤x≤1.

is a non-increasing and convex upper bound on the probability Pr⁡[S≤k−1] if k⁢x=𝔼[S] where S is the sum of Bernoulli random variables.

Proof.

We show a detailed proof of this bound in Section 4. ◀ We note on some important properties of Pk(.) here.

  1. 1.

    Pk⁢(x) is non-increasing and convex on [1,∞) for all k≥1.

  2. 2.

    P1⁢(1)<P2⁢(1)<P3⁢(1)<… and P1⁢(2)>P2⁢(2)>P3⁢(2)>….

  3. 3.

    Pk⁢(x)≤1 for all x and k.

Now Theorem 4 can serve as an upper bound on pt⁢(e), i.e.,

pt⁢(e)≤Pr⁡[∑v∈Bt⁢(e)Yv,t≤ke⁢(t)−1]≤Pke⁢(t)⁢(∑v∈Be⁢(t)yv,tke⁢(t)).

Define te as the earliest time such that ze,≤te≥1, and let

cz⁢(e)=te+∑t>tept⁢(e).

Intuitively, we only study the probability pt⁢(e) when t>te, and assume that for any time t≤te the edge e will not be covered (i.e., pt⁢(e)=1 for all t≤te). Hence, by definition, cz⁢(e) is an upper bound on 𝔼[Covτ⁢(e)], that is, 𝔼[Covτ⁢(e)]≤cz⁢(e).

Using (10), we can apparently state that

pt⁢(e)≤Pke⁢(t)⁢(ze,<t),andcz⁢(e)≤te+∑t>tePke⁢(t)⁢(ze,<t).

However, consider an illustrative example with ke=3, we draw P3⁢(x), P2⁢(x), and P1⁢(x) in Figure 1. The black curve is the upper bound on pt⁢(e) in [2] as a reference.

Figure 1: Function graph of Pi⁢(x).

The worst case for cz⁢(e) seems to be the case that probabilities are taken subsequently from P3⁢(x), P2⁢(x), and P1⁢(x), corresponding to the uppermost curve for any x, which implies that we switch between different Pk⁢(x) twice. The problem is whether such “jumps” are “lossless” and yield the maximum value of cz⁢(e). If we consider a scenario in which the jump from P2(.) to P1(.) occurs at ze,<t=2.5, we observe a probability pt⁢(e) rise as P1⁢(2.5)>P2⁢(2.5). By the definition of pt⁢(e), its value never increases when zv,<t increases. Hence, there must be some additional terms to strengthen (10) that we can extract from other KC inequalities. Next, we use multiple KC inequalities to approximate ∑v∈Be⁢(t)zv,<t.

Lemma 5.

cz⁢(e)≤e1β⁢(1+1β−1⁢e−1)⁢cx⁢(e).

Proof.

Define a series of time slots {t1,t2,…⁢tke−1} such that ti is the earliest time that there are ke−i vertices, denoted Qi={qke,…,qi+1}, satisfying zv,<ti≥1. In other words, Qi=Ae⁢(ti), |Qi|=ke−i and Qi∖Qi+1=qi+1. From time ti+1 to ti, the cover requirement for edge e changes from i+1 to i. Therefore, for t∈(ti+1,ti),

pt⁢(e)≤Pi+1⁢(∑v∈e∖Qi+1yv,ti+1)

Assume that te≤tke−1≤⋯≤t2≤t1, i.e., all these vertices are not guaranteed to be scheduled by te. In general cz⁢(e) is at most,

cz⁢(e) ≤te+∑t=tetke−1−1Pke⁢(∑v∈eyv,tke) (11)
+∑t=tke−1tke−2−1Pke−1⁢(∑v∈e∖Qke−1yv,tke−1)+⋯+∑t>t1P1⁢(∑v∈e∖Q1yv,t). (12)

Now, we carefully use multiple KC inequalities to evaluate ∑yv,t with ze,<t. For any ti>te, assume ze,<ti=1+δi>1, and consider the following KC inequality at time ti with the set S being Qi+1,

∑v∈e∖Qi+1zv,<ti≥(i+1)⋅ze,<ti, (13)
⟹ z{qi+1},<ti+∑v∈e∖Qizv,<ti≥(i+1)⋅ze,<ti, (14)

Using the KC inequality for x, for any time t, ∑v∈e∖Qixv,<t≥i⋅xe,<t. Therefore, for any time t>ti,

∑v∈e∖Qizv,t≥i⋅ze,t,

which leads to

∑v∈e∖Qizv,<t =∑v∈e∖Qizv,<ti+∑v∈e∖Qi∑t′=tit−1zv,t′
≥i⋅ze,<ti+(ze,<ti−zqi+1,<ti)+i⁢∑t′=tit−1ze,t′
≥i⋅ze,<t+δi,

where we used the fact that zqi+1,<ti<1 and ze,<ti=1+δi.

Hence, as ti≥te, there exists a non-decreasing gap between ∑v∈e∖Qizv,<t and i⁢ze,<t. Moreover, these gaps can accumulate over time. For example, at any time t>tke−1, we have

∑v∈e∖{qke−1}zv,<t≥(ke−1)⁢ze,<t+δke−1.

At time t≥tke−2, using the above inequality, we obtain

zqke−2,<t+∑v∈e∖Qke−2zv,<t≥(ke−1)⁢ze,<t+δke−1.

Applying the argument above, we get

∑v∈e∖Qke−2zv,<t≥(ke−2)⁢ze,<t+δke−2+δke−1.

Therefore, for any time ti+1≤t<ti, we use the following inequality to approximate the probability that e is not covered before time t.

pt⁢(e)≤Pi+1⁢(ze,<t+∑j=i+1ke−1δji+1)

Hence, the upper bound (11) can be written as

𝔼[Covτ⁢(e)] ≤te+∑t=tetke−1−1Pke⁢(ze,<t)+⋯+∑t>t2t1P2⁢(ze,<t+∑j=2ke−1δj2) (15)
+∑t>t1P1⁢(ze,<t+∑j=1ke−1δj).

Doing similar calculations as in (8), but for e instead of v, we get a similar formula for ze,<t,

ze,<t≥β⁢∑t′≤txe,t′⁢ln⁡tt′.

Let t∗ be the (fractional) time such that

β⁢∑t′≤⌊t∗⌋xe,t⁢ln⁡t∗/t′=1.

Naturally, ze,<t⁣∗≥1. Since te is the earliest time that ze,≤te≥1, we have ⌊t∗⌋≥te. Let γt=β⁢(∑t′≤tat′⁢ln⁡(t/t′)) where at corresponds to xe,t in our analysis for GMSSC. Using the non-increasing property of Pk, substituting ze,<t with γt only increases cz⁢(e). Therefore, we formulate cz⁢(e)/cx⁢(e) as the following optimization problem. Recall that cx⁢(e)=∑tt⁢xe,t.

(ℱ)Maximize⁢⌊t∗⌋+∑t=⌊t∗⌋+1tke−1−1Pke⁢(γt)+⋯+∑t1P1⁢(γt+∑jδj)∑tt⁢at (16)
s.t.∥a∥1=1,γt∗=1,(zqi+1,<ti<1, and ⁢zqi+1,≤ti>1)⁢…

For simplicity, we denote ∑t>⌊t∗⌋Pke⁢(t)⁢(γt+∑j=ke⁢(t)ke−1δj/ke⁢(t)) as ∑t>t∗P⁢(γt), where P⁢(x)≤1, P′⁢(x)≤0 and P′′⁢(x)≥0.

The following fact uses the convexity of a function to conclude its maximum value on a compact region.

Fact 6 ([2] Fact 8).

Let f:Rn→R be a convex and non-negative function on some compact region H, and g:Rn→R is a linear and positive function on H, then the maximum of maxx∈H⁡f⁢(x)/g⁢(x) is attained at an extreme point of H.

Proof.

Consider any x,y∈H and λ∈[0,1],

max⁡(f⁢(x)g⁢(x),f⁢(y)g⁢(y))≥λ⁢f⁢(x)+(1−λ)⁢f⁢(y)λ⁢g⁢(x)+(1−λ)⁢g⁢(y)≥f⁢(λ⁢x+(1−λ)⁢y)g⁢(λ⁢x+(1−λ)⁢y).

◀

The following lemma describes the maximum scenario for ℱ.

Lemma 7.

If P⁢(x) is function such that P⁢(x)≤1, P′′⁢(x)≥0, and P′⁢(x)≤0 for all x≥1, then for the maximum problem

ℱ=⌊t∗⌋+∑t>t∗P⁢(γt)∑tt⁢at,

with at least m≥2 non-trivial constraints, it always attains its maximum value when some at=1.

Proof.

In [2], the authors proved this lemma when m=2. Let us consider the case where there are more non-zero variables. We focus on how to reduce the number of nonzero variables. Since each Pi is convex, ∑tPke⁢(t) is also a convex function. Using Fact 6, the max value of ℱ attains at some extreme point, where there are at most m non-zero variables.

Assume that there non-zero variables are c1≤c2≤⋯≤cm and corresponding values are a1,a2,…,am. The following two cases discuss the non-zero variables before t∗ or after t∗, respectively.

  1. 1.

    c1≤…⁢cl−1≤t∗.

    Suppose that ∑i=1l−1ai=D for some constant D≤1 and let c1=t∗⁢e−b1, c2=t∗⁢e−b2, up to cl−1=t∗⁢e−bl−1. The constraint γt∗=1 becomes

    β⁢∑i=1l−1ai⁢bi=1.

    For any time t≥t∗,

    γt=D⁢β⁢ln⁡t/t∗+β⁢∑i=1l−1ai⁢bi+∑i=lai⁢ln⁡t/ci=1+D⁢β⁢ln⁡t/t∗+∑i=lai⁢ln⁡t/ci.

    Which indicates that the numerator is independent of any individual ai or bi for i<l. Then ℱ becomes a minimization problem. In particular,

    Minimizet∗⁢β⁢(∑i=1l−1ai⁢e−bi)+∑i=lci⁢ais.t.β⁢(∑i=1l−1ai⁢bi)=1.

    Using the convexity of e−x, the extreme point is at b1=b2=⋯=bl−1. Consequently, we reduce the number of non-zero variables before t∗ to 1.

  2. 2.

    t∗≤cl≤⋯≤cm.

    Now we assume that b1=b2=⋯=bl−1=B, the constraint γt∗=1 becomes

    D⁢B⁢β=1,

    and is independent of cl,…,cm. Let cl=t∗⁢bl, cl+1=t∗⁢bl+1, up to cm=t∗⁢bm. For time t>t∗, the denominator is

    t∗⁢e−B⁢D⁢β+t∗⁢∑i=lbi⁢ai. (17)

    While the numerator is

    t∗+∑t=t∗+1c1P⁢(1+D⁢β⁢ln⁡t/t∗)+∑c1+1c2P⁢(1+D⁢β⁢ln⁡t/t∗+β⁢al⁢ln⁡t/t∗⁢a)
    +⋯+∑cm∞P⁢(1+D⁢β⁢ln⁡t/t∗+β⁢∑i=lmai⁢ln⁡t/t∗⁢bi).

    Let x=t/t∗, we can upper bound the numerator using the fact that all Pi are non-increasing and Pi⁢(γt)≤1. For convenience, we denote Am⁢(x)=1+D⁢β⁢ln⁡x+β⁢∑i=lmai⁢(ln⁡x−ln⁡bi). Hence,

    t∗⁢(1+∫1blP⁢(1+D⁢β⁢ln⁡x)⁢𝑑x+⋯+∫bm−1bmP⁢(Am−1⁢(x))⁢𝑑x)
    +t∗⁢(∫bm∞P⁢(Am−1⁢(x)+β⁢am⁢(ln⁡x−ln⁡bm))⁢𝑑x). (18)

    Notice that only the last two terms contain bm, so we first try to eliminate bm. Let (2) be f⁢(bm) and let (17) be g⁢(bm), we show that f⁢(bm)/g⁢(bm) always reaches its maximum value when bm=bm−1 and thus eliminate bm and cm.

Claim 8.

The function f⁢(bm)/g⁢(bm) achieves its maximum value at bm=bm−1.

Proof.

We rename bm as b here. To prove this claim, we show that (f⁢(b)/g⁢(b))′≤0 for all b≥bm−1. Define

h⁢(b)=f′⁢(b)⁢g⁢(b)−f⁢(b)′⁢g⁢(b).

and correspondingly,

h′⁢(b)=f′′⁢(b)⁢g⁢(b)−f⁢(b)⁢g′′⁢(b).

It suffices to show that h⁢(b)≤0 for all b≥bm−1. Note that g′⁢(b)=am, and g′′⁢(b)=0. Therefore, h′=f′′⁢g, and the sign of h′ depends only on f′′. We then compute f′⁢(b) and f′′⁢(b).

f′⁢(b) =P⁢(Am−1⁢(b))−P⁢(Am−1⁢(b)+ln⁡b−ln⁡b)
−βb⁢am⁢∫b∞P′⁢(1+D⁢β⁢ln⁡x+∑i=lmai⁢(ln⁡x−ln⁡bi))⁢𝑑x
substituting x=bt
=−β⁢am⁢∫1∞P′⁢(1+D⁢β⁢ln⁡t+D⁢β⁢ln⁡b+∑i=lm−1ai⁢(ln⁡b+ln⁡t−ln⁡bi)+am⁢ln⁡t)⁢𝑑t.

Hence,

f′′⁢(b)=−β2⁢amb⁢∫1∞P′′⁢(1+D⁢β⁢ln⁡t+D⁢β⁢ln⁡b+∑i=lm−1ai⁢(ln⁡b+ln⁡t−ln⁡bi)+am⁢ln⁡t)⁢𝑑t.

Because all Pi is convex on [1,∞), for all b≥bm−1

P′′⁢(1+D⁢β⁢ln⁡t+D⁢β⁢ln⁡b+∑i=lm−1ai⁢(ln⁡b+ln⁡t−ln⁡bi)+am⁢ln⁡t)≥0.

Thus, f′′⁢(b)≤0 and h′⁢(b)≤0 for all b≥bm−1.

Meanwhile, we need to show that h⁢(bm−1)≤0. However, proving h⁢(bm−1)≤0 is equivalent to solving a subproblem of this claim, particularly, exactly one less non-zero variable after t∗. Therefore, we can recursively apply the same argument and reduce the number of non-zero variables. Finally, we reach the case where there is one non-zero variable after t∗, and we prove h⁢(1)≤0. We list the proof of h⁢(1)≤0 here.

f⁢(1)=1+∫1∞P⁢(1+β⁢ln⁡x)⁢𝑑xandg⁢(1)=D⁢e−1/D⁢β,
f′⁢(1)=−∫1∞β⁢(1−D)⁢P′⁢(1+β⁢ln⁡x)⁢𝑑x⁢andg′⁢(1)=1−D.

Note that f′⁢(1)≥0 as P′⁢(x)≤0 and g⁢(1)≤1, we have

h⁢(1) =f′⁢(1)⁢g⁢(1)−f⁢(1)⁢g′⁢(1)≤f′⁢(1)−f⁢(1)⁢(1−D)
≤−(1−D)⁢(1+∫1∞β⁢P′⁢(1+β⁢ln⁡x)+P⁢(1+β⁢ln⁡x)⁢d⁢x)
=−(1−D)⁢(1+∫1∞(ey−1β⁢P⁢(y))′⁢𝑑y) (let 1+β⁢ln⁡x=y)
≤(1−D)⁢(1−P⁢(1))≤0.

Consequently, h⁢(b) is a non-increasing function on [bm−1,∞) and h⁢(b)≤h⁢(bm−1) for all b≥bm−1. ⊲

With Claim 8, we conclude that any m≥3 can be reduced to the m=1 case, where the remaining constraint is γt∗=1. Solving γt∗=1 when some au=1, we obtain u=t∗⋅e−1/β. ◀

Using Lemma 7, we conclude that the maximum value of ℱ occurs at time u=t∗⋅e−1/β such that xe,u=1, and we simplify ℱ as follow.

ℱ ≤1u⁢(t∗+∫t∗tke−1Pke⁢(β⁢ln⁡tu)⁢𝑑t+⋯+∫t1∞P1⁢(β⁢ln⁡tu+∑jδj)⁢𝑑t)
≤e1β+1β⁢(∫11+δke−1Pke⁢(x)⁢exβ⁢𝑑x+⋯) (19)
+1β⁢(∫1+δ21+δ1P2⁢(x+∑j=2ke−1δj2)⁢exβ⁢𝑑x+∫1+δ1∞P1⁢(x+∑j=1ke−1δj)⁢exβ⁢𝑑x). (let x=β⁢ln⁡(t/u) and change of variables)

Now, for the integral part, we argue that it is at most ∫1∞e−x⁢ex/β⁢𝑑x by demonstrating that any jump between different Pi only reduces the value of ℱ.

Figure 2: The loss of jumping between Pi.

Consider ke=2, and according to the analysis above, our goal is to show that

∫11+δ1P2⁢(x)⁢exβ⁢𝑑x+∫1+δ1∞P1⁢(x+δ1)⁢exβ⁢𝑑x≤∫1∞P1⁢(x)⁢exβ⁢𝑑x.

We draw three curves P2⁢(x)⁢exp⁡(x/β), P1⁢(x)⁢exp⁡(x/β) and P1⁢(x+δ1)⁢exp⁡(x/β) in Fig 2. Comparing LHS with RHS (gray area), we observe two differences: the blue area represents the gain from ∫11+δ1P2⁢(x)xβ⁢𝑑x, and the red area represents the loss due to the gap δ1. If we can prove that the red area is larger than the blue one, we can argue that the jump (δ1>0) introduces some loss, and the maximum value returned from ∫1∞P1⁢(x)xβ⁢𝑑x.

The following claim formally proves that the worst case is always P1(.) due to these gaps δi.

Claim 9.

For any b≥a≥0 and Δ≥0,

∫1+a1+bPi⁢(x+a+Δi)⁢exβ⁢𝑑x+∫1+b∞P1⁢(x+b+a+Δi−1)⁢exβ⁢𝑑x≤∫1+a∞P1⁢(x+a+Δi)⁢exβ⁢𝑑x.

Proof.

Let LHS be a function of b,

L⁢(b)=∫1+a1+bPi⁢(x+a+Δi)⁢exβ⁢𝑑x+∫1+b∞P1⁢(x+b+a+Δi−1)⁢exβ⁢𝑑x.

When b=a, as Pi is a non-increasing function,

L⁢(a)=∫1+a∞P1⁢(x+a+Δi−1+a)⁢𝑑x≤∫1+a∞P1⁢(x+Δi+a)⁢𝑑x=R⁢H⁢S.

Meanwhile when b→∞,

limb→∞L⁢(b) =∫1+a∞Pi⁢(x+Δ+ai)⁢exβ⁢𝑑x
=1i−1/β⁢exp⁡(−Δ+(i+1)⁢a+1i+1+aβ)
=1i−1/β⁢e−1i⁢e1+aβ⁢exp⁡(−Δ+(i+1)⁢ai).

Given the facts that

1i−1/β⁢e−1i≤11−1/β⁢e−1,andi+1i≥2.

We conclude that

limb→∞L⁢(b)≤ββ−1⁢e−1−2⁢a−Δi⁢e1+aβ=∫1+a∞P1⁢(x+a+Δi)⁢exβ⁢𝑑x.

The claim holds for both b=a and as b→∞. Then we check if L has any extreme point on [a,∞). Compute the derivative of L⁢(b),

L′⁢(b) =exp⁡(−i⁢(b+1+Δ+ai)+1+bβ)
−2⁢β−1β−1⁢exp⁡(−(1+2⁢b+Δ+ai−1)+1+bβ).

Let E=(2⁢β−1)/(β−1), and note that E>e. Solving for the extreme point b0 from L′⁢(b0)=0, we get

(i−2)⁢b0=1−ln⁡2⁢β−1β−1−1i+(1i−1−1i)⁢(Δ+a). (20)

We discuss the extreme point based on the value of i.

  1. 1.

    i>4. In this case, the RHS of (20) is a decreasing function of Δ and a, as 1/(i−1)<1/i for all i≥3. Hence,

    b0<−1i⁢(i−2)<0.

    Thus, L⁢(b) is monotone on [0,∞). The maximum of L⁢(b) is either L⁢(a) or L⁢(∞), and the claim holds.

  2. 2.

    i=4. Now L⁢(b) has no extreme points. Since L is monotone, the claim holds.

  3. 3.

    i=3. Potentially, b0>a≥0, but consider L′⁢(0),

    L′⁢(0) <e−1+Δ+ai+1β−e⋅e−1−Δ+ai−1+1β
    ≤e1β⁢(e−1i⁢e−Δ+ai−e−Δ+ai−1)<0.

    Hence, L⁢(b0) gives the minimum value. The maximum value is L⁢(a) or L⁢(∞), so the claim holds.

  4. 4.

    i=2. We compute the extreme point,

    b0=−12⁢(Δ+a+1)+12−2⁢ln⁡E.

    Let b0≥a so that there is an extreme point on [a,∞), we have,

    Δ+3⁢a+1≤22−2⁢ln⁡E.

    Consider L′⁢(a)

    L′⁢(a)=e1+aβ⁢(e−22⁢(1+3⁢a+Δ)−eln⁡E−1−Δ−3⁢a).

    The sign of L′⁢(a) depends on

    −22⁢(1+3⁢a+Δ)−(ln⁡E−Δ−1−3⁢a)=2−22⁢(1+3⁢a+Δ)−ln⁡E≤0.

    As a result, L′⁢(a)≤0, and L⁢(b0) again gives the minimum value. The maximum value is L⁢(a) or L⁢(∞), and the claim is valid.

Towards upper bounding (2.3), we set a=δi, b=δi−1, and Δ represents the sum of all the other δj. ⊲

Using Claim 9 recursively, we obtain

ℱ ≤e1β+1β⁢(⋯+∫1+δ21+δ1P2⁢(x+12⁢∑j=2ke−1δj)⁢𝑑x+∫1+δ1∞P1⁢(x+∑j=1ke−1δj)⁢exβ⁢𝑑x)
≤e1β+1β⁢(⋯+∫1+δ2∞P1⁢(x+12⁢∑j=3ke−1δj+δ2)⁢exβ⁢𝑑x)
≤e1β+1β⁢(⋯+∫1+δ3∞P1⁢(x+13⁢∑j=4ke−1δj+δ3)⁢exβ⁢𝑑x)
≤⋯
≤e1β+1β⁢(∫1∞P1⁢(x)⁢exβ⁢𝑑x)
=e1β⁢(1+1β−1⁢e−1).

On the other hand, if t1≤te, then cz⁢(e)≤te+∑tePke−1⁢(ze,<t)+…, which is equivalent to the case where there are ke−1 vertices remain to be covered, and allowing us to apply the same argument. Hence,

cz⁢(e)≤e1β⁢(1+1β−1⁢e−1)⁢cx⁢(e).

◀

Lemma 10.

𝔼[Covσ⁢(e)]≤β⁢cz⁢(e)

Proof.

Apply the kernel to (2),

∑vzv=∑vK⁢xv≤K⁢𝟏=β⁢𝟏.

Intuitively, the kernel allows the rounding algorithm to schedule at most β vertices at the same time, indicating that 𝔼[Covσ⁢(e)]≈β⁢𝔼[Covτ⁢(e)] due to the random tie-breaking rule. See [2] Lemma 19 for more details. ◀

Theorem 1. [Restated, see original statement.]

There exists a 4.509-approximation algorithm for the generalized min sum set cover (GMSSC) problem.

Proof.

With Lemma 5 and Lemma 10, it follows that the approximation ratio is at most

β⁢e1β⁢(1+1β−1⁢e−1).

The ratio is 4.509 when β=2.043. ◀

3 2-approximation for Min Latency Set Cover

We also study the min latency set cover problem, which is a special case of GMSSC where ke=|e| for all hyperedges e. Apply K to xe and xv, the KC inequalities become

zv,<t≥ze,<t,∀v∈e.

Hence, by the earliest time te such that ze,≤te≥1, all the vertices v∈e have been scheduled in τ.

For this problem, we apply a different kernel,

K⁢(t,t′)=α⁢t′t⁢(t+1)⁢𝟙⁢[t′≤t].

Similar to the GMSSC case, the following two lemmas suffice to yield the desired approximation.

Lemma 11.
cz⁢(e)≤αα−1⁢cx⁢(e).

Proof.

Under the kernel K=α⁢t′/t⁢(t+1), the relationship between x and z is

ze,<t =∑t′≤t∑t′′≤t′K⁢(t′,t′′)⁢xv,t′′=∑t′′<t∑t′=t′′t−1K⁢(t,t′′)=α⁢∑t′=1tt−t′t⁢xe,t′,

where we use ∑q=t′′t−1K⁢(q,t′′)=α⁢t′′⁢∑q=t′′t−1(1/q−1/(q+1))=α⁢(1−t′′/t).

Define t∗ as the time that α⁢∑t′=1t∗((t∗−t′)/t∗)⁢xe,t′=1. Again, ze,<t∗=1 indicates t∗≥te. Since cz⁢(e)≤te, the corresponding maximum problem becomes

maximizet∗∑tt⁢at,s.t.∥a∥1=1,α⁢∑t≤t∗at⁢t∗−tt∗=1.

This problem has two non-trivial constraints, implying it has at most 2 non-zero variables. Let au=s and av=1−s.

  1. 1.

    u<v≤t∗, the problem is

    Minimizes⁢u+(1−s)⁢v,
    s.t.s⁢t∗−ut∗+(1−s)⁢t∗−vt∗=1−s⁢u+(1−s)⁢vt∗=1α.

    In this case, s⁢u+(1−s)⁢v is a constant (t∗)⁢(1−1/α). Hence, the ratio in (3) is at most

    α⁢t∗t∗⁢(α−1)=αα−1.
  2. 2.

    u<t∗<v, then the problem is,

    Minimizesu+(1−s)v,s.t.st∗−ut∗=1α,

    This minimization problem is simply an increasing function of v, which leads to v=t∗.

◀

Lemma 12.
𝔼[Covσ⁢(e)]≤α2⁢cz⁢(e).

Proof.

Apply K to (2), we have

∑vzv=∑vK⁢xv≤K⁢𝟏.

Furthermore,

(K⁢𝟏)t=∑t′K⁢(t,t′)=∑t′≤tα⁢t′t⁢(t+1)=αt⁢(t+1)⋅t⁢(t+1)2=α2.

Hence, by time te, σ schedules at most (α/2)⁢te vertices and must include all the vertices v∈e, then

𝔼[Covσ⁢(e)]≤α2⁢te=α2⁢cz⁢(e).

◀

Theorem 2. [Restated, see original statement.]

There exists a 2-approximation algorithm for the min latency set cover problem.

Proof.

Combining the above two lemmas, we obtain the ratio for the min latency set cover problem,

α22⁢(α−1),α≥2,

which attains the minimum value 2 at α=2. ◀

▶ Remark 13.

Theorem 2, along with the 4-approximation algorithm for MSSC, which is produced by the same algorithm with a different kernel, indicates that achieving a 4-approximation for GMSSC within this framework using a universal kernel to all vertices is difficult.

4 A Tail Bound for the Sum of Bernoulli Random Variables

Now we prove Theorem 4. See 4

Proof.

Rewrite Lemma 3 as,

Pr⁡[S≤k−1]≤∑i=0k−1(ni)⁢(k⁢xn)i⁢(1−k⁢xn)n−i,if x≥1.
Lemma 14 ([2] Lemma 29).

For λ=n⁢p=𝔼[S],

∑i=0k−1(ni)⁢pi⁢(1−p)n−i≤e−λ⁢∑i=0k−1λii!.

This lemma is a stronger version of the Poisson Limit Theorem, and we use it to approximate the sum of binomial distributions. When k=1 this lemma directly gives e−x and the theorem is valid. Hence, it suffices to prove the following claim:

Claim 15.

For all k≥2, and x≥1,

exp⁡(−k⁢(x−k−1k))≥e−k⁢x⁢∑i=0k−1(k⁢x)ii!.

Proof.

Define η=(k−1)/k and the following function,

R⁢(x)=e−k⁢x⁢∑i=0k−1(k⁢x)ii!/e−k⁢(x−k−1k)=e−k⁢x+k⁢(x−k−1k)⁢∑i=0k−1(k⁢x)ii!.

If we can prove that R⁢(x)≤1 for all x≥η, we finish the proof. First, using the Maclaurin expansion of ex, we have

ex≥∑i=0k−1xii!,

Let x=k−1, we obtain ek−1≥∑i=0k−1(k−1)i/i!. Then at x=η,

R⁢(η)=e−(k−1)⁢∑i=0k−1(k−1)ii!≤1.

On the other hand, consider R⁢(∞),

limx→∞R⁢(x)=e−k−1k⁢limx→∞∑i=0k−1e(−k+k)⁢x⁢(k⁢x)ii!. (21)

As k≥2, −k+k<0,

limx→∞(k⁢x)ii!⋅e(k−k)⁢x=0.

Because k is a constant, we conclude that R⁢(∞)=0. It remains to analyze the extreme point. Since R⁢(η)≤1 and R⁢(∞)<1, we assume that there are some extreme points with the maximum value on [η,∞). Compute the derivative of R⁢(x),

R′⁢(x) =e−k⁢x+k⁢(x−k−1k)⁢((−k+k)⁢(∑i=0k−1(k⁢x)ii!)+k⁢∑i=0k−2(k⁢x)ii!)
=k⋅e−k⁢x+k⁢(x−k−1k)⁢(−k⁢(k⁢x)k−1(k−1)!+∑i=0k−1(k⁢x)ii!).

The extreme point x0 satisfies

k⁢(k⁢x0)k−1(k−1)!=∑i=0k−1(k⁢x0)ii!.

Hence, we can eliminate the summation in R⁢(x)

R⁢(x)≤R⁢(x0)=k⁢(k⁢x0)k−1(k−1)!⁢e(k−k)⁢x0−k−1k. (22)

Consider (22) as function of x0,

S⁢(t):=k⁢(k⁢t)k−1(k−1)!⁢e(k−k)⁢t−k−1k.

The corresponding derivative is,

S′⁢(t)=k(k−1)!⁢e−k−1k⁢e−k⁢t+k⁢t⁢(k⁢t)k−2⁢[(−k+k)⁢k⁢t+k⁢(k−1)].

The sign of S′⁢(t) is determined by (−k+k)⁢k⁢t+k⁢(k−1). Therefore, S′⁢(η)>0 and S′⁢(∞)<0. The extreme point t0=(k−1)/(k−k) gives the maximum value.

S⁢(t0) =(k)k⁢(k+1)k−1(k−1)!⁢e−k−1k−k+1 (23)
=k⁢(kk−1)k−1⁢(k−1)k−1(k−1)!⁢e−(k−1)⁢e−k−1k.

Using the Stirling approximation 2⁢π⁢n⁢(n/e)n≤n!,

S⁢(t0)≤(kk−1)k−1⁢k2⁢π⁢(k−1)⁢e−k+1k.

Consider (k/(k−1))k−1, substituting r=1/k and taking the logarithm,

(k−1)⁢ln⁡(kk−1) =−(1r2−1)⁢ln⁡(1−r)
=(1r2−1)⁢(∑j=1∞rjj) (Maclaurin series of ln⁡(1−r))
=(1r+12+r3+…)−(r+r22+…)
≤1r+12=k+12.

Therefore,

S⁢(t0)≤e12+1k⁢k(k−1)⁢2⁢π. (24)

This is a decreasing function of k. When k≥9, (24) is less than 1. Meanwhile, we can verify that (23) is also less than 1 for 2≤k≤8. Finally, we conclude that R⁢(x)≤R⁢(x0)≤S⁢(t0)≤1 for all k≥2 and finish the proof. ⊲ ◀

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