A tight analysis of BrownBakerKatseff sequences for online strip packing
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DOI: 10.1007/s1087801294631
 Cite this article as:
 Kern, W. & Paulus, J.J. J Comb Optim (2013) 26: 333. doi:10.1007/s1087801294631
Abstract
We study certain adversary sequences for online strip packing which were first designed and investigated by Brown, Baker and Katseff (Acta Inform. 18:207–225) and determine the optimal competitive ratio for packing such BrownBakerKatseff sequences online. As a byproduct of our result, we get a new lower bound of \(\rho \ge 3/2+\sqrt{33}/6 \approx 2.457\) for the competitive ratio of online strip packing.
Keywords
Online algorithm Strip packing Competitive ratio1 Introduction
In the twodimensional strip packing problem a number of rectangles have to be packed without rotation or overlap into a strip such that the height of the strip used is minimum. The width of the rectangles is bounded by 1 and the strip has width 1 and infinite height. Baker et al. (1980) show that this problem is NPhard.
We study the online version of this packing problem. In the online version the rectangles are given to the online algorithm one by one from a list, and the next rectangle is given as soon as the current rectangle is irrevocably placed into the strip. To evaluate the performance of an online algorithm we employ competitive analysis. For a list of rectangles L, the height of a strip used by online algorithm A and by the optimal solution is denoted by A(L) and OPT(L), respectively. The optimal solution is not restricted in any way by the ordering of the rectangles in the list. Competitive analysis measures the absolute worstcase performance of online algorithm A by its competitive ratio sup_{ L }{A(L)/OPT(L)} (cf. Pruhs et al. 2004).
Regarding the upper bound on the competitive ratio for online strip packing, recent advances have been made by Ye et al. (2009) and Hurink and Paulus (2008). Independently they show that a modification of the wellknown shelf algorithm yields an online algorithm with competitive ratio \(7/2+\sqrt{10}\approx6.6623\). We refer to these two papers for a more extensive overview of the literature.
In the early 80’s, Brown et al. (1982) derived a lower bound ρ≥2 on the competitive ratio of any online algorithm by constructing certain (adversary) sequences in a fairly straightforward way (cf. Sect. 2). These sequences were further studied by Johannes (2006) and Hurink and Paulus (2008), who derived improved lower bounds of 2.25 and 2.43, resp. (Both results are computer aided and presented in terms of online parallel machine scheduling, a closely related problem.) The paper of Hurink and Paulus (2008) also presents an upper bound of ρ≤2.5 for packing such “BrownBakerKatseff sequences”. The purpose of our present paper is to close the gap between 2.43 and 2.5 by presenting a tight analysis, showing that BrownBakerKatseff sequences can be packed online with competitive ratio \(\rho = 3/2 +\sqrt{33}/6\) and that this is best possible. As a byproduct, we obtain a new lower bound ρ≈2.457 for online strip packing.
The result of this paper has been presented at the CTW 2010 (cf. Kern 2010). Meanwhile, in a joint work with R. Harren, we tried to analyze a modified version of BrownBakerKatseff sequences, yielding a slightly better lower bound (but no exact analysis as here), cf. Harren and Kern (2011).
2 The instance construction
In this section we describe the construction of BrownBakerKatseff sequences L _{ n } according to Brown et al. (1982). In addition, we present an online algorithm for packing the sequences L _{ n } online with ratio \(\hat{\rho}= 3/2+\sqrt{33}/6\). For convenience, let throughout this note \(\hat{\rho}=3/2+\sqrt{33}/6\).
By definition of the rectangles’ heights and widths, an online algorithm can only pack the rectangles one above the other in the same order as the rectangles appear in the list L _{ n }. An optimal offline packing is obtained by first packing the rectangles q _{ i } on top of each other and then pack all p _{ i } next to each other on top of the qrectangles. The sole function of the positive term ϵ is to ensure this structure on any online packing. From now on we assume that ϵ is small enough to be omitted from the analysis.
We start with the (simpler) upper bound:
Theorem 1
Each list L _{ n } can be packed online with competitive ratio \(\hat{\rho}=\frac{3}{2}+\frac{\sqrt{33}}{6}\).
Proof
Consider the online algorithm A that chooses \(\beta_{0}=\hat{\rho}1\), \(\alpha_{2}=1/(\hat{\rho}1)\), and all other gaps equal to 0. So \(p_{0}=1,q_{1}=q_{2}=\hat{\rho}1, p_{1}=\hat{\rho}\), and p _{2} can be computed from p _{2}=p _{1}+α _{2} p _{2}: We get \(p_{2}=\hat{\rho}/(1\frac{1}{\hat{\rho}1})=\frac{(\hat{\rho}1)\hat{\rho}}{\hat{\rho}2}\), or \(p_{2}= (\hat{\rho}1)(3\hat{\rho}2)=4\hat{\rho}2\) by our choice of \(\hat{\rho}\).

After packing p _{0}=1 we have \(A(L_{0})=\hat{\rho}\) and OPT(L _{0})=1. Thus, the competitive ratio is exactly \(\hat{\rho}\) at this point.

After packing rectangle q _{1} the online and optimal packing increase by the same amount. Thus the competitive ratio decreases.

After \(p_{1}=\hat{\rho}\) we have \(A(L_{1})=\beta_{0} p_{0} +p_{0}+q_{1}+p_{1}=\hat{\rho}1+1+\hat{\rho} 1 +\hat{\rho}\), while OPT(L _{1})=q _{1}+p _{1}=2ρ−1. Hence \({A(L_{1})}/{\mathit{OPT}(L_{1})}=(3\hat{\rho}1)/(2\hat{\rho}1)<\hat{\rho}\).

After q _{2} we have \(A(L_{1}q_{2})=A(L_{1})+q_{2}+\alpha_{2}p_{2}=3\hat{\rho}1+\hat{\rho}1+3\hat{\rho}2=7\hat{\rho}4\), while OPT(L _{1} q _{2})=p _{1}+q _{1}+q _{2}=3ρ−2. So \({A(L_{1}q_{2})}/{\mathit{OPT}(L_{1}q_{2})}=(7\hat{\rho}4)/(3\hat{\rho}2)=\hat{\rho}\). Again the competitive ratio is exactly \(\hat{\rho}\) at this point (by definition of \(\hat{\rho}\)).

After p _{2} we have \(A(L_{2})=A(L_{1}q_{2})+p_{2}=7\hat{\rho}4+4\hat{\rho}2=11\hat{\rho}6\), while \(\mathit{OPT}(L_{2})=q_{1}+q_{2}+p_{2}=2\hat{\rho}2+4\hat{\rho}2=6\hat{\rho}4\). Thus \({A(L_{2})}/{\mathit{OPT}(L_{2})}=(11\hat{\rho}6)/(6\hat{\rho}4)<\hat{\rho}\).

For i≥3 there are no more gaps introduced by online algorithm A. In particular, \(p_{2}=p_{3}=p_{4}=\cdots=(\hat{\rho}1)(3\hat{\rho}2) \) and \(q_{i}=\alpha_{2}p_{2}=3\hat{\rho}2\) for all i≥3. When packing q _{ i }, the online and optimal packing increase by the same amount and, thus, ρcompetitiveness is not violated. After packing p _{ i+1}, however, we have OPT(L _{ i+1})=OPT(L _{ i })+q _{ i+1} and \(A(L_{i+1})=A(L_{i})+q_{i+1}+p_{i+1}=A(L_{i})+\hat{\rho}q_{i+1}\). The height of the online packing grows exactly \(\hat{\rho}\) times as fast as the optimal packing.
3 Lower bound on the competitive ratio
In this section we prove a lower bound of \(\hat{\rho}={3}/{2}+{\sqrt{33}}/{6}\) on the competitive ratio for online packing BrownBakerKatseff sequences—and hence for online strip packing in general. The outline of the proof is as follows.
Assume that there exists a ρcompetitive online algorithm A with \(\rho<\hat{\rho}\). We present this algorithm with the list L _{ n }, with n arbitrarily large. To obtain a contradiction we define a potential function Φ_{ i } on the state of the online packing after packing rectangle p _{ i }. We argue that this potential function is both bounded from below and that it decreases to −∞, giving us the required contradiction.
We admit that the potential function looks rather involved. Its exact form is simply motivated by the technical analysis that follows. In particular, the potential function is designed such that a certain “shift invariance” (cf. Lemma 2 below) holds, which helps a lot in simplifying the analysis. (Yet, in Harren and Kern 2011 we analyze “extended” BrownBakerKatseff sequences with a simpler potential Φ_{ i }=γ _{ i }+β _{ i } and considerable technical effort.)
The values of α _{ i } and β _{ i } are nonnegative by definition and γ _{ i } is nonnegative by the ρcompetitiveness of online algorithm A. Observe that shifting p _{ i }, say, upward, increases β _{ i } and decreases γ _{ i } by the same amount, so that shifting p _{ i } has actually no effect on Φ_{ i }. In Lemma 2 we will see that the same holds w.r.t. shifting q _{ i }. Thus the decisions of the online algorithm in “phase i” have no effect on Φ_{ i }, but rather on subsequent potential values. (This phenomenon is observed also elsewhere, cf., e.g. Fuchs et al. 2005 or Harren and Kern 2011.) The crucial result is Lemma 6, stating that \(\rho <\hat{\rho}\) forces a significant decrease of the potential in each step. Thus Φ_{ i+1}→−∞ for i→∞. On the other hand, in Lemma 5 we show among other things that \(\alpha_{i} < 1/(\hat{\rho}1)\). As a consequence, Φ_{ i }>−1 for all i. This contradiction finally implies \(\rho \ge \hat{\rho}\), the main result of this note:
Theorem 2
The best possible ratio for packing BrownBakerKatseff sequences is \(\hat{\rho}=3/2+\sqrt{33}/6\). This value also provides a lower bound for online strip packing.
The remainder of this note is concerned with the proofs of the afore mentioned lemmata. We start by providing some basic properties (Lemmas 1 and 2 below) of the potential function Φ_{ i }. In the following, ρ is assumed to be suitably close to \(\hat{\rho}\) whenever this is necessary.
Lemma 1
If Φ_{ i }≤ρ−1 then γ _{ i }+β _{ i }+α _{ i }≤ρ−1.
Lemma 2
The potential Φ_{ i } is invariant under shifting p _{ i } and/or q _{ i }.
Proof
Shifting rectangle p _{ i }, say, upward by one unit does not affect OPT(L _{ i }) nor α _{ i }. Furthermore, A(L _{ i }) increases by one unit and, hence, γ _{ i } p _{ i } decreases by one unit. At the same time β _{ i } p _{ i } increases by one unit, so that γ _{ i }+β _{ i } remains constant and Φ_{ i } is invariant under shifting p _{ i }.
Clearly, shifting p _{ i } and/or q _{ i } does have an effect on subsequent values like p _{ i+1} and Φ_{ i+1}. Furthermore, even when we are only interested in packing L _{ i }, shifting up p _{ i } or shifting down the concatenated q _{ i } p _{ i } may result in an infeasible packing. (Indeed, as we shift q _{ i } p _{ i } down, p _{ i } decreases and after a while the packing may cease to be ρcompetitive!) As long as we are only interested in the value of Φ_{ i }, however, we may well shift p _{ i } and q _{ i } as we like, disregarding the competitiveness constraints. We will make extensive use of this observation further on.
Lemma 3
Proof
Lemma 4
If q _{ i+1}=max{α _{ i } p _{ i },q _{ i }}, then we may assume w.l.o.g. that β _{ i }=0.
Proof
Shifting rectangle p _{ i } down decreases the distance β _{ i } p _{ i } and increases α _{ i+1} p _{ i+1}. However, when we keep all other distances equal it does not affect p _{ j } with j>i. Due to the increase in α _{ i+1} p _{ i+1} some q _{ j } with j>i may increase, but this is only in favor of the online algorithm since the optimal value increases by exactly the same amount. So the alternative online algorithm that schedules p _{ i } earlier and leaves all other distances unchanged is also feasible. □
Lemma 5
For i≥0, Φ_{ i }≤ρ−1, and for i≥1, \(\alpha_{i}\leq 1/(\hat{\rho}1)\) and \(q_{i}/p_{i}\leq 1(\hat{\rho}1)\). In case q _{ i }=α _{ i−1} p _{ i−1} or q _{ i }=q _{ i−1}, we even have \(q_{i}/p_{i}\leq (1\alpha_{i})/(\hat{\rho}1)\).
Proof
 Case 1:

q _{ i+1}=α _{ i } p _{ i }.
By Lemma 4 we may assume β _{ i }=0. By (1), this further implies p _{ i }=(1−α _{ i+1})p _{ i+1}. Henceby induction.$$\frac{q_{i+1}}{p_{i+1}}=\frac{\alpha_i p_i}{p_i+\alpha_{i+1}p_{i+1}} = \frac{\alpha_i(1\alpha_{i+1})p_{i+1}}{(1\alpha_{i+1})p_{i+1}+\alpha_{i+1}p_{i+1}} = (1\alpha_{i+1})\alpha_i \leq \frac{1\alpha_{i+1}}{\hat{\rho}1}$$The online algorithm A is by assumption ρcompetitive after packing rectangle q _{ i+1}, which means that the distance between rectangles q _{ i+1} and p _{ i } is not too large, i.e., α _{ i+1} p _{ i+1}≤γ _{ i } p _{ i }+(ρ−1)q _{ i+1}=γ _{ i } p _{ i }+(ρ−1)α _{ i } p _{ i }. Together with Lemma 1 this gives for \(\rho \le \hat{\rho}\). (This upper bound for α _{ i+1}, obtained by setting γ _{ i }=0 and α _{ i }=ρ−1 (cf. Lemma 1) is rather weak, but sufficient for our purposes.)  Case 2:

q _{ i+1}=β _{ i } p _{ i }.
By Lemma 1 we have β _{ i }≤ρ−1 and thus, which is less than \(1/(\hat{\rho}1)\) for \(\rho \le \hat{\rho}\).The online algorithm A is by assumption ρcompetitive after packing rectangle q _{ i+1}, which means that the distance between rectangles q _{ i+1} and p _{ i } is not too large, i.e. α _{ i+1} p _{ i+1}≤γ _{ i } p _{ i }+(ρ−1)q _{ i+1}=γ _{ i } p _{ i }+(ρ−1)β _{ i } p _{ i }. This, together with β _{ i }+γ _{ i }≤ρ−1 (by Lemma 1) gives for \(\rho \le \hat{\rho}\). (In the upper bound computation for α _{ i } above we assumed that γ _{ i } is as large as possible and β _{ i } is as small as possible. This is justified by the fact that \(\frac{\rho1}{\rho}<\frac{1}{\hat{\rho}1}\).)For the potential, Lemma 3 gives which is strictly less than ρ−1 for ρ sufficiently close to \(\hat{\rho}\). (Again, for the last inequality in (3), note that increasing β _{ i } as much as possible instead of γ _{ i } is justified: If we let \(f(\beta,\gamma)=\frac{\gamma+2(\rho1)\beta 1}{1+\beta}\), thenThe latter, however, is true as we assume β+γ≤ρ−1.)$$\frac{\partial f}{\partial \gamma}=\frac{1}{1+\beta } < \frac{\partial f}{\partial \beta }=\frac{2(\rho 1)\gamma +1}{(1+\beta )^2} \quad \Leftrightarrow \quad \beta + \gamma < 2 \rho 1.$$  Case 3:

q _{ i+1}=q _{ i }.
Induction yields$$\frac{q_{i+1}}{p_{i+1}}=\frac{q_i}{p_{i+1}}\le (1\alpha_{i+1})\frac{q_i}{p_i}\leq\frac{1\alpha_{i+1}}{\hat{\rho}1}.$$To argue that \(\alpha_{i+1}\leq 1/(\hat{\rho}1)\) we shift the concatenated rectangles q _{ i },p _{ i } down as far as possible, i.e., until either γ _{ i }=0 or α _{ i }=0 (thereby increasing α _{ i+1}). By shifting q _{ i },p _{ i } down, the length of p _{ i } decreases, therefore γ _{ i } can become 0. At the same time p _{ i+1} increases, causing the optimal and online solution to increase by the same amount. So the online algorithm is still ρcompetitive after this shift.
If γ _{ i }=0, then α _{ i+1} p _{ i+1}≤(ρ−1)q _{ i+1}=(ρ−1)q _{ i }≤p _{ i }. Thus \(\alpha_{i+1}\leq 1/2 \leq 1/(\hat{\rho}1)\).
If α _{ i }=0, the rectangles p _{ i−1},q _{ i },p _{ i } are concatenated. To show \(\alpha_{i+1}\leq 1/(\hat{\rho}1)\) for this case, we distinguish three subcases. Case 3a:

q _{ i+1}=q _{ i }=α _{ i−1} p _{ i−1}.
 Case 3b:

Since α _{ i }=0 and β _{ i }=0 we have Φ_{ i }=γ _{ i }≤(2(ρ−1)^{2}−1)/ρ (cf. (3)). Thus, The latter is increasing in ρ, so we conclude (after multiplying the enumerator and denominator with ρ) thatby definition of \(\hat{\rho}\). (Recall that \(3\hat{\rho}^{2}9\hat{\rho}+4=0\).)$$\alpha_{i+1} \le \frac{(\hat{\rho}1)^2+2((\hat{\rho}1)^21)}{\hat{\rho}+(\hat{\rho}1)^2+2((\hat{\rho}1)^21)}=\frac{1}{\hat{\rho}1}$$
 Case 3c:

q _{ i+1}=q _{ i }=q _{ i−1}. By Lemma 4 we may assume β _{ i−1}=0, so that actually q _{ i−1},p _{ i−1},q _{ i },p _{ i } are concatenated and p _{ i }=p _{ i−1}. Since γ _{ i } p _{ i }=γ _{ i−1} p _{ i−1}+(ρ−1)q _{ i }−p _{ i }<γ _{ i−1} p _{ i−1}, i.e., the improvement of A upon ρcompetitiveness decreases, the value of α _{ i+1} is smaller than α _{ i } could at most be, thus, in particular, less than \(1/(\hat{\rho}1)\).
Lemma 6
\(\Phi_{i+1} \leq \Phi_{i}  \frac{\hat{\rho}\rho}{\hat{\rho}1}\).
Proof
 Case 1:

q _{ i+1}=α _{ i } p _{ i }.
By Lemma 4 we can assume β _{ i }=0. Furthermore, Lemma 2 allows us to shift p _{ i+1} resp. the concatenated q _{ i+1} p _{ i+1} down until α _{ i+1}=β _{ i+1}=0 without affecting Φ_{ i+1} (although this might result in a negative value of γ _{ i+1} in case Φ_{ i+1} is negative). Summarizing, let us assume that q _{ i },p _{ i },q _{ i+1},p _{ i+1} are concatenated.
We seek to analyze how Φ_{ i } and Φ_{ i+1} vary as the concatenated q _{ i },p _{ i },q _{ i+1},p _{ i+1} are shifted upward. By Lemma 2, Φ_{ i } remains unchanged. As to Φ_{ i+1}, observe that shifting q _{ i },p _{ i },q _{ i+1},p _{ i+1} upward by one unit will increase α _{ i } p _{ i } and hence q _{ i+1} as well as p _{ i } and p _{ i+1} by one unit each. Hence A(L _{ i+1}) increases by 4. On the other hand, OPT(L _{ i+1}) increases by 2, so that γ _{ i+1} p _{ i+1} increases by (2ρ−4). Hence shifting q _{ i },p _{ i },q _{ i+1},p _{ i+1} upward will increase Φ_{ i+1} as long as Φ_{ i+1}<2ρ−4. So we are led to distinguish the following two cases: Φ_{ i+1}>2ρ−4:

In this case, Φ_{ i+1} will increase as we shift q _{ i },p _{ i },q _{ i+1},p _{ i+1} downward. Doing so, we decrease q _{ i+1}. So we eventually end up in a situation where either q _{ i+1}=q _{ i } (which will be treated in Case 3 below) or γ _{ i } becomes zero (revealing that Φ_{ i }=−(ρ−2)α/(1−α) must be negative). But γ _{ i }=0 implies So Φ_{ i+1}=γ _{ i+1}=(ρ−1)α _{ i }−1<0, contradicting our assumption that Φ_{ i+1}>2ρ−4.
 Φ_{ i+1}≤2ρ−4:

In this case Φ_{ i+1} increases as we shift q _{ i },p _{ i },q _{ i+1},p _{ i+1} upward until either q _{ i } gets tight in the sense that A(L _{ i }−p _{ i })=ρ OPT(L _{ i }−p _{ i }) or Φ_{ i+1}=2ρ−4 is reached. The latter is impossible: Since γ _{ i+1} p _{ i+1}=γ _{ i } p _{ i }+(ρ−1)q _{ i+1}−p _{ i+1}, we conclude (using Lemma 1) that contradicting our assumption that Φ_{ i+1}=2ρ−4.
Finally, assume that q _{ i } gets tight. In this case, γ _{ i } p _{ i }=ρ(p _{ i }−p _{ i−1})−p _{ i }≥ρα _{ i } p _{ i }−p _{ i }. So γ _{ i }≥ρα _{ i }−1. Similarily, γ _{ i+1} p _{ i+1}=γ _{ i } p _{ i }+(ρ−1)q _{ i+1}−p _{ i+1}. Dividing by p _{ i+1}=p _{ i }, we obtain γ _{ i+1}=γ _{ i }+(ρ−1)α _{ i }−1. Henceis maximized when γ _{ i } is as small as possible, i.e. γ _{ i }=ρα _{ i }−1. Thus$$\Delta\Phi= \Phi_{i+1}\Phi_i= \gamma_{i+1}\Phi_i= \gamma_i+(\rho1)\alpha_i1\frac{\gamma_i(\rho2)\alpha_i}{1\alpha_i}$$This shows that ΔΦ is indeed strictly negative since$$\Delta\Phi \le 2\rho\alpha_i\alpha_i2\frac{2\alpha_i1}{1\alpha_i}= (2\rho1)\alpha_i \frac{1}{1\alpha_i}.$$for \(\rho \le \hat{\rho}\).$$\alpha_i(1\alpha_i) \le \frac{1}{2}\cdot\frac{1}{2} =\frac{1}{4} < \frac{1}{2\rho1}$$  Case 2:

q _{ i+1}=β _{ i } p _{ i }.
By Lemma 3 we have$$\Delta\Phi=\Phi_{i+1}\Phi_i=\frac{\gamma_i+2(\rho1)\beta_i1}{1+\beta_i}\frac{\gamma_i+\beta_i(\rho2)\alpha_i}{1\alpha_i}.$$The derivative with respect to γ _{ i } of the above is 1/(1+β _{ i })−1/(1−α _{ i })≤0, so ΔΦ is decreasing in γ _{ i }. Thus we may choose γ _{ i }=0. Additionally, we have α _{ i }≤β _{ i }, otherwise we are not in this case. With γ _{ i }=0 and under the constraints α _{ i }≤β _{ i }≤ρ−1,α _{ i }+β _{ i }≤ρ−1 we haveIndeed, the function$$\Phi_{i+1}\Phi_i = \frac{2(\rho1)\beta_i1}{1+\beta_i} \frac{\beta_i(\rho2)\alpha_i}{1\alpha_i}<0.04.$$has partial derivative Δ_{ α }=0 for β=ρ−2. Plugging this value into Δ, we find that the resulting Δ is independent of α and equals \(\rho 2\frac{1}{\rho1} < 0.1\). On the boundary α=0 the function Δ=Δ(β) is less than −0.04 for β∈[0,ρ−1] and on the boundary α=β the function Δ=Δ(β) is bounded from above by −0.2 for \(\beta \in [0,\frac{\rho1}{2}]\). (Note that α=β implies \(\beta \le \frac{\rho1}{2}\).) We omit the details.$$\Delta=\Delta(\alpha,\beta)=\frac{2(\rho1)\beta1}{1+\beta}\frac{\beta(\rho2)\alpha}{1\alpha}$$  Case 3:

q _{ i+1}=q _{ i }.
By definition of q _{ i+1}, we have α _{ i } p _{ i }≤q _{ i+1}=q _{ i }. Thus, Lemma 5 implies \(\alpha_{i} \le q_{i}/p_{i} \le \frac{1\alpha_{i}}{\hat{\rho}1}\) or, equivalently, \(\alpha_{i} \le 1/\hat{\rho}\). Thus we conclude thatfor \(\rho \le \hat{\rho}\).$$\Phi_i = \frac{\gamma_i(\rho2)\alpha_i}{1\alpha_i}\ge \frac{(\rho2)\alpha_i}{1\alpha_i} \ge \frac{(\rho2)/\hat{\rho}}{11/\hat{\rho}} > \hat{\rho}3 $$(4)Now, as in Case 1, let us assume that β _{ i }=β _{ i+1}=0 and shift the concatenated q _{ i+1} p _{ i+1} down until α _{ i+1} becomes zero as well. This leaves Φ_{ i+1} (and, of course, also Φ_{ i }) invariant, but may result in a negative value of γ _{ i+1} in case Φ_{ i+1} (=γ _{ i+1} after the shift) is negative. (As we are only interested in ΔΦ, we do not care about negative values of γ _{ i+1} here.) Thus assume that q _{ i },p _{ i },q _{ i+1},p _{ i+1} are concatenated.
We claim that shifting the concatenated q _{ i },p _{ i },q _{ i+1},p _{ i+1} down increases Φ_{ i+1} (while leaving Φ_{ i } unchanged). Indeed moving down decreases both p _{ i } and p _{ i+1} by one unit, so that in total A(L _{ i+1}) decreases by 3 units, while OPT(L _{ i+1}) decreases by only one unit. Thus γ _{ i+1} p _{ i+1} increases with 3−ρ units while p _{ i+1} decreases with one unit. This shows that moving q _{ i },p _{ i },q _{ i+1},p _{ i+1} down increases Φ_{ i+1}=γ _{ i+1} whenever Φ_{ i+1}>ρ−3.
To show that Φ decreases, we may assume that Φ_{ i+1}>ρ−3—else a significant decrease of at least \(\hat{\rho}\rho\) follows already immediately from (4). But then, as we have seen above, we may also assume w.l.o.g. that q _{ i },p _{ i },q _{ i+1},p _{ i+1} is shifted down as far as possible, i.e., until α _{ i }=0. (Note that this may even result in a negative value of γ _{ i }, but we do not care, as we are only interested in Φvalues.) When α _{ i }=0, however, thenis significantly negative since γ _{ i+1} p _{ i+1}=γ _{ i } p _{ i }+(ρ−1)q _{ i+1}−p _{ i+1}, so that \(\gamma_{i+1} \le \gamma_{i} +\frac{\rho1}{\hat{\rho}1} 1\).$$\Delta\Phi = \Phi_{i+1}\Phi_i = \gamma_{i+1} \gamma_i$$
4 Conclusions
We have solved the (30 year old) problem of analysing BrownBakerKatseff sequences and provided an optimal online algorithm for these sequences. Up to now all previous lower bounds for online strip packing were based on BrownBakerKatseff sequences. A natural question to ask is whether (or to what extent) these sequences are worst case sequences for online strip packing. In Harren and Kern (2011) we have tried to analyze more generally sequences of “thin” items p _{ i } and “blocking” items q _{ j }, which are not necessarily alternating. We suceeded in proving lower and upper bounds close to 2.6 for such sequences. Determining the exact value appears to be involved, but in any case, our results show that to close the gap between the currently best lower bound of 2.6 and the currently best upper bound of 6.6, completely new ideas are needed.
Acknowledgement
Part of this research has been funded by the Dutch BSIK/BRICKS project.
Open Access
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