A new lower bound for online strip packing

European Journal of Operational Research 250 (2016) 754–759

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European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor

Discrete Optimization

A new lower bound for online strip packing✩ Guosong Yu a,∗, Yanling Mao b, Jiaoliao Xiao b a b

Department of Mathematics, Nanchang University, Nanchang 330031, PR China Department of Management Science and Engineering, Nanchang University, Nanchang 330031, PR China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 12 January 2015 Accepted 6 October 2015 Available online 22 October 2015

In this paper, we consider the online strip packing problem, in which a list of online rectangles has to be packed without overlap or rotation into a strip of width 1 and infinite length √ so as to minimize the required height of the packing. We derive a new improved lower bound of (3 + 5)/2 ≈ 2.618 for the competitive ratio for this problem. This result improves the best known lower bound of 2.589.

Keywords: Packing Strip packing Online algorithm Competitive ratio

© 2015 Elsevier B.V. and Association of European Operational Research Societies (EURO) within the International Federation of Operational Research Societies (IFORS). All rights reserved.

1. Introduction In this paper, we consider online strip packing of rectangles. Rectangles arrive one by one in an online fashion and have to be packed into a strip of width 1 and infinite length without known any information about future rectangles. The rectangles must be packed without overlap and rotation and couldn’t be moved when they are already packed. The objective is to minimize the total height of the packing. The strip packing problem was first considered by Baker, Coffman, and Rivest (1980). They showed that this problem is NP−Hard. Strip packing has many real-world applications in manufacturing, logistics, and computer science, e.g., VLSI layout design, stock cutting problem. To evaluate the performance of an online algorithm we adopt competitive analysis. For any list of rectangles L, the height of a strip used by algorithm A and by the optimal solution is denoted by A(L) and OPT(L), respectively. The (absolute) competitive ratio of A, denoted by RA , is given by



RA = sup L



A(L) . OPT (L)

The asymptotic competitive ratio R∞ of A is defined by A



R∞ A

= lim sup sup n→∞

L





A(L)  OPT (L) = n . OPT (L) 

✩ This work was supported by the project of National Natural Science Foundation of China (no. 71263038) and MOE project of Humanities and Social Sciences Foundation (no. 12YJA630091). ∗ Corresponding author. Tel.: +86 13576089185. E-mail address: [email protected] (G. Yu).

For the offline strip packing problem, Coffman, Garey, Johnson, and Tarjan (1980) presented the algorithms NFDH and FFDH with asymptotic competitive ratios of 2 and 1.7, respectively. An AFPTAS was given by Kenyon and Rémila (2000). Sleator (1980) presented an approximation algorithm with an absolute competitive ratio of 2.5. This was independently improved by Schiermeyer (1994) and Steinberg (1997) with algorithms of absolute competitive ratio 2. Harren and van Stee (2009) first broke the barrier of 2 and presented an algorithm with a absolute competitive ratio of 1.936. Then this bound was improved to 5/3 + ε for any ε > 0 by Harren, Jansen, Prädel, and van Stee (2014). For the online strip packing problem, Baker and Schwarz (1983) showed the first fit shelf algorithm has absolute competitive ratio √ of 6.99. The upper bound was improved to 7/2 + 10 ≈ 6.6623 by Ye, Han, and Zhang (2009) and Hurink and Paulus (2007) independently. Regarding the lower bound on the competitive ratio for online strip packing, Brown, Baker, and Katseff (1982) derived a lower bound ρ ≥ 2 on the competitive ratio of any online algorithm by constructing adversary sequences (BBK sequences). Then Johannes (2006) and Hurink and Paulus (2008) improved the bound to 2.25 and 2.43 by studying BBK sequences. Kern and Paulus (2013) finally showed that the BBK sequences can√be packed by providing matching upper and lower bounds of 3/2 + 33/6 ≈ 2.457. The current best known result ρ ≥ 2.589 was presented by Harren and Kern (2015) by constructing modified BBK sequences. A related problem is the multiple-strip packing problem. Zhuk (2006) first considered this problem and showed that there is no approximation algorithm with absolute competitive ratio better than 2 unless P = NP even if there are only two strips. Latter Ye, Han, and Zhang (2011) presented a nearly optimal algorithm with an absolute competitive ratio of 2 + ε for any ε > 0. Then

http://dx.doi.org/10.1016/j.ejor.2015.10.012 0377-2217/© 2015 Elsevier B.V. and Association of European Operational Research Societies (EURO) within the International Federation of Operational Research Societies (IFORS). All rights reserved.

G. Yu et al. / European Journal of Operational Research 250 (2016) 754–759

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P3 R2

F2 y2

b2

P2 R1

R2

P2

si P1

x2

a2

y1

sk Fig. 2. An optimal packing.

P1 x1

R1 si

Fig. 1. The list of Ln .

Bougeret, Dutot, Jansen, Otte, and Trystram (2010) presented an approximation algorithm with absolute competitive ratio of 2, which is the best possible unless P = NP. For the online multiple-strip packing problem, Ye et al. (2011) designed both randomized and deterministic online algorithms with competitive ratios better than the previous bound 10 presented by Zhuk (2006). 2. The instance construction In this section, we first describe the construction of new modified BBK sequences Ln = ( J1 , . . . , Ji , . . . ) in order to get a new lower bound for the competitive ratio of online strip packing, where each item Ji denotes a rectangle of height Ji and width w( Ji ). In the end of this section, we will give the reason why we design this type of our sequences. Assume √ that there exists a ρ −competitive online algorithm A with ρ < (3 + 5)/2. Since Brown et al. (1982) derived a lower bound 2, we suppose that ρ > 2 in the following. We can define Ln as the list of items

(P1 , s1 , . . . , sk , R1 , P2 , F2 , R2 , . . . , Pn−1 , Fn−1 , Rn−1 , Pn ). The sequences Ln consist of four types of rectangles:Pi , si , Ri , Fi . For any i, Pi and Ri are thin items with small widths. Ri is special in Ln , because it may not appear in Ln . For any i, Ri is included in the sequence only if the online algorithm A satisfies some conditions which we will state in the following. For any i, si is a rectangle with small height and near-full width. After the online algorithm A packed the rectangle P1 , s1 , . . . , si , . . . arrived online and may be packed either below P1 or above P1 . When the first rectangle si (denoted by sk ) is packed above P1 , new si doesn’t arrive. For any i ≥ 2, Fi is a block item with the full width of 1. For convenience, sk is also said to be F1 , i.e. sk = F1 . We can describe the sequences Ln (see Fig. 1) with • • • • • •

F2 sk

P1 = 1, w(P1 ) = δ0 si = θ + ε , w(si ) = 1 − δi for i = 1, . . . , k; (ρ −1)(x +P +y )

i i i Ri = + ε for i = 1, . . . , n − 1; ρ Pi = (xi−1 + Pi−1 + yi−1 ) + ε for i = 2, . . . , n;  P2 P2 F2 = max{min{x1 − k−1 i=1 si , ρ }, min{y1 , ρ }, x2 } + ε ;

Pi

Fi = max{min{yi−1 , ρ }, Fi−1 , xi } + ε for i = 3, . . . , n − 1.

ρ −1 The value ε is a small positive value and θ = −2ρ(ρ+3 (Note that −1)2 √ θ > 0 if ρ < (3 + 5)/2). The value δ 0 is a positive number no more 2

δ

1 i−1 than 2n−1 and δi = 4n for i = 1, . . . , k. For i ≥ 2, w(Ri−1 ) = w(Pi ) = 2δk and w(Fi ) = 1. The value x1 denotes the distance between P1 and the bottom of the strip, and xi denotes the distance between Fi−1 and Pi for i = 2, . . . , n. The value y1 denotes the distance between P1 and sk ,and yi denotes the distance between Pi and Fi for i = 2, . . . , n − 1. For i = 1, R1 is included in the sequence if the following condition  P2 P2 holds: either x1 − k−1 i=1 si > ρ or y1 > ρ (Note that the condition inP

P

1 .). For i ≥ 2, Ri is included dicates that x1 + y1 > ρ2 , i.e. x1 + y1 > ρ −1 Pi in the sequence if xi + yi > ρ −1 . In the next section, we will show that Ri must √ be packed below Fi if the competitive ratio of A is less than (3 + 5)/2. The sole function of the positive number ε is to ensure the structure of any online packing. For convenience, we assume that ε is small enough to be omitted from the analysis. We now show that Fi should be packed above Pi for i ≥ 2. For i = 2, if R1 is included in the sequence which means either  P2 P2 P2 x1 − k−1 i=1 si > ρ or y1 > ρ , then F2 ≥ ρ . Thus x1 + P1 + y1 − R1 =

x1 +P1 +y1

P

= ρ2 ≤ F2 which means F2 couldn’t be packed below sk . Since F2 ≥ x2 , the online algorithm A must pack F2 above P2 . If R1 P is not included in the sequence, then x1 − (s1 + · · · + sk−1 ) ≤ ρ2  P and y1 ≤ ρ2 . Thus, F2 = max{x1 − k−1 i=1 si , y1 , x2 } which means F2 should be packed above P2 . For i ≥ 3, if Ri−1 is included in the sequence, let bi−1 be the distance between Ri−1 and ρ

Fi−1 , then bi−1 ≤ xi−1 + Pi−1 + yi−1 − Ri−1 =

xi−1 +Pi−1 +yi−1

P

= ρi . So Pi Fi = max{min{yi−1 , ρ }, Fi−1 , xi } ≥ max{min{yi−1 , bi−1 }, Fi−1 , xi } which means Fi should be packed above Pi . If Ri−1 is not included in P

ρ

P

the sequence, then yi−1 ≤ ρi (note that xi−1 + yi−1 ≤ ρi−1 −1 indicates P P yi−1 ≤ ρi ). So Fi = max{min{yi−1 , ρi }, Fi−1 , xi } = max{yi−1 , Fi−1 , xi } which also means Fi should be packed above Pi . For the list ( J1 , . . . , Ji ), let A(Ji ) denote the height of the packing by the algorithm A and OPT(Ji ) denote the height of the optimal off-line packing. It is not difficult to determine OPT(Jj ). We list them in the following (see Fig. 2): • • • • • •

OPT (P1 ) = 1;  OPT (si ) = P1 + ij=1 s j for i = 1, . . . , k;  OPT (R1 ) = P1 + kj=1 s j or OPT (R1 ) = R1 + sk ;  OPT (Pi ) = Pi + i−1 j=1 Fj for i ≥ 2; i OPT (Fi ) = Pi + j=1 Fj for i ≥ 2;  OPT (Ri ) = Ri + ij=1 Fj for i ≥ 2.

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G. Yu et al. / European Journal of Operational Research 250 (2016) 754–759

Now we will show the reason why we design this type of sequences after we give two definitions. We define Gi through

A(Pi ) + Gi = ρ · OPT (Pi ).

(1)

R1

With similar methods as in the literature (Harren & Kern, 2015), we define a potential function i to obtain a contradiction which is defined by

i =

xi + G i . Pi

z1

(2)

In the process of studying i , we found that i has no good rule for the basic BBK sequences (i.e. only Pi and Fi are included in the sequences Ln ). But, if 2 and Fi are small enough, then i has a good rule, that is, i+1 − i is less than a negative constant. In order to ensure that 2 is small enough, we need to reduce F1 if it is relatively large. We use a series of si of small height and near-full width instead of F1 in our new BBK sequences. The special structure of si guarantees the effect of sk (i.e. the first rectangle si packed above P1 ) is similar to that of F1 (So sk is also said to be F1 in this paper) and sk has the advantage that sk is smaller than F1 . Now we need to reduce Fi for i = 2, . . . , n − 1 if it is relatively large. If a suitable rectangle(i.e. Ri ) with height larger than Pi can be packed between Fi−1 and Fi (F0 refers to the bottom of the strip), then Fi+1 may be smaller than its original value. In order to make sure that Ri can be packed between Fi−1 and (ρ −1)(xi +Pi +yi ) Fi , we design the proper value of Ri = . This value can ρ

also ensure that the gap between Ri and Fi−1 and the gap between Ri

sk

y1 P1 x1

si Fig. 3. If R1 is packed above sk .

Lemma 3.4. If R1 is included in the sequence, then it must be packed below sk . Proof. If the online algorithm A packs R1 above Sk , then OPT (R1 ) = P1 + s1 + · · · + sk or OPT (R1 ) = R1 + sk . Let z1 denote the distance between sk and R1 , we have (see Fig. 3)

and Fi will not exceed (xi + Pi + yi ) − Ri = i ρi i = i+1 ρ . This is the reason why Fi (i = 2, . . . , n − 1) has the above definition.

x1 + P1 + y1 + sk + z1 + R1 ≤ ρ · OPT (R1 ).

3. Lower bound on the competitive ratio

1 the bottom of the strip. If y1 > ρ2 = 1 ρ1 1 , then y1 > ρ −1 (x1 + (ρ −1)(x1 +P1 +y1 ) ρ −1 1 P1 ). Thus R1 = > ρ (x1 + P1 + ρ −1 (x1 + P1 )) = P1 + ρ

x +P +y

P

For the case OPT (R1 ) = P1 + s1 + · · · + sk , then R1 ≤ P1 + s1 + · · · + sk−1 ≤ P1 + x1 since si (i = 1, . . . , k − 1) is placed between P1 and P

In this section, we will show that the potential function i decreases to be negative contradicting the fact that i ≥ 0.

x +P +y

P

Lemma 3.1. yi ≤ Gi + (ρ − 1)Fi for i ≥ 2.

x1 , contradicting R1 ≤ P1 + x1 . So y1 ≤ ρ2 . Since R1 is not included P P in the sequence if x1 − (s1 + · · · + sk−1 ) ≤ ρ2 and y1 ≤ ρ2 , we have

Proof. The online algorithm A is by assumption ρ −competitive after packing rectangle Fi , which means

x1 + P1 + y1 + sk + z1 + R1 ≤

x1 − (s1 + · · · + sk−1 ) > ρ2 . Then P

A(Fi ) ≤ ρ · OPT (Fi ).

ρ · (P1 + s1 + · · · + sk−1 + sk ) P2

<

ρ · (P1 + x1 −

A(Pi ) + yi + Fi ≤ ρ · (OPT (Pi ) + Fi ),

=

ρ · (P1 + x1 −

i.e., yi ≤ Gi + (ρ − 1)Fi . 

=

(ρ − 1)(P1 + x1 ) − y1 + ρ · sk .

Thus

Lemma 3.2. For i ≥ 2,

x1 + P1 + y1

ρ

+ sk )

z1 ≤ (ρ − 2)(P1 + x1 ) − 2y1 − R1 + (ρ − 1)sk

Proof. Since Gi+1 = ρ · OPT (Pi+1 ) − A(Pi+1 ), A(Pi+1 ) = A(Pi ) + yi + Fi + xi+1 + Pi+1 and OPT (Pi+1 ) = OPT (Pi ) + Fi + (Pi+1 − Pi ), we have

≤ (ρ − 2 −

xi+1 + Gi+1 = xi+1 + ρ · OPT (Pi+1 ) − A(Pi+1 )

=

= xi+1 + ρ · [OPT (Pi ) + Fi + (Pi+1 − Pi )] =



By (2) and Lemma 3.2, we get the following conclusion. Lemma 3.3. For i ≥ 2,

i+1

+ sk )

Hence, we have

xi+1 + Gi+1 = Gi + (ρ − 1)Fi + (ρ − 1)Pi+1 − ρ Pi − yi .

− [A(Pi ) + yi + Fi + xi+1 + Pi+1 ] (ρ · OPT (Pi ) − A(Pi )) + (ρ − 1)Fi + (ρ − 1)Pi+1 − ρ Pi − yi = Gi + (ρ − 1)Fi + (ρ − 1)Pi+1 − ρ Pi − yi .

ρ

Gi + (ρ − 1)Fi − ρ Pi − yi = (ρ − 1) + . Pi+1

Now we show that Ri couldn’t be packed above Fi by the following lemmas.

ρ −1 )(P1 + x1 ) + (ρ − 1)sk ρ

ρ 2 − 3ρ + 1 (P1 + x1 ) + (ρ − 1)sk ρ ρ 2 − 3ρ + 1 −ρ 2 + 3ρ − 1 ≤ · 1 + (ρ − 1) ρ 2(ρ − 1)2 2 (ρ − 2)(ρ − 3ρ + 1) = < 0. 2ρ(ρ − 1) For the case OPT (R1 ) = R1 + sk , with R1 = ρρ−1 (x1 + P1 + y1 ) >

ρ −1 1 ρ ( ρ −1 P1 + P1 ) = P1 = 1, we have

z1 ≤ (ρ − 1)R1 − (x1 + P1 + y1 ) + (ρ − 1)sk = (ρ − 1)R1 − =

ρ R + (ρ − 1)sk ρ −1 1

ρ 2 − 3ρ + 1 R1 + (ρ − 1)sk ρ −1

G. Yu et al. / European Journal of Operational Research 250 (2016) 754–759

757

Fig. 4. If Ri is packed above Fi .

Fig. 5. 2 ≤ (ρ − 2) + α .

ρ 2 − 3ρ + 1 −ρ 2 + 3ρ − 1 · 1 + (ρ − 1) ρ −1 2(ρ − 1)2 2 ρ − 3ρ + 1 = < 0. 2(ρ − 1) <

Then we derive a contradiction to z1 ≥ 0.



In the following, we denote α = (ρ − 1)θ =

−ρ 2 +3ρ −1 . 2(ρ −1)

Lemma 3.5. For i ≥ 2, if Ri is included in the sequence and i ≤ (ρ − F 2) + α and Pi ≤ (ρ − 2) + α , then Ri must be packed between Fi−1 and i Fi . 1 Proof. Since Ri = ρρ−1 (xi + Pi + yi ) > ρρ−1 ( ρ −1 Pi + Pi ) = Pi , then Ri couldn’t be packed below Fi−1 . We will show that Ri couldn’t be packed above Fi either. If Ri is packed above Fi , let zi denote the distance between Fi and Ri , then A(Pi ) + yi + Fi + zi + Ri ≤ ρ · (OPT (Pi ) + Fi + Ri − Pi ) (see Fig. 4). Thus

zi ≤ (ρ · OPT (Pi ) − A(Pi )) + (ρ − 1)Fi + (ρ − 1)Ri − ρ Pi − yi

ρ −1 (xi + Pi + yi ) − ρ Pi − yi ρ (ρ − 1)2 < Gi + (ρ − 1)Fi + xi − (ρ − 1)Pi as <1 ρ = (xi + Gi ) + (ρ − 1)Fi − (ρ − 1)Pi ≤ (ρ − 2 + α)Pi + (ρ − 1)(ρ − 2 + α)Pi − (ρ − 1)Pi = (ρ 2 − 3ρ + 1 + ρα)Pi −ρ 2 + 3ρ − 1 = (ρ 2 − 3ρ + 1 + ρ · )Pi 2(ρ − 1) (ρ − 2)(ρ 2 − 3ρ + 1) = Pi < 0. 2(ρ − 1) = Gi + (ρ − 1)Fi + (ρ − 1) ·

Then we derive a contradiction to zi ≥ 0.  Lemma 3.6.

2 ≤ (ρ − 2) + α . Proof. With A(P2 ) + G2 = ρ · OPT (P2 ) and P2 = x1 + P1 + y1 , we have x1 + P1 + y1 + sk + x2 + P2 + G2 = ρ · (P2 + sk ) (see Fig. 5). Then x2 +

G2 = (ρ − 2)P2 + (ρ − 1)sk = (ρ − 2)P2 + (ρ − 1)θ . Since P2 ≥ P1 = )θ ≤ (ρ − 2) + (ρ − 1)θ = (ρ − 2) + 1, we get 2 = (ρ − 2) + (ρ −1 P2 α.  Lemma 3.7.

F2 P2

≤ ρ − 2 + α and

Proof. If F2 = x2 , then

F2 P2

=

x2 P2

F2 P3

≤ ρ1 .

≤ 2 ≤ ρ − 2 + α and

F2 P3

=

x2 x2 +P2 +y2

α 1 2 2 ≤  ·P2 +P =  +1 ≤ ρρ −2+ −1+α = ρ 2 −ρ +1 < ρ . 2 2 2 2 k−1 P2 P If F2 = x2 , then F2 = max{min{x1 − i=1 si , ρ }, min{y1 , ρ2 }} ≤ P2 F2 F2 F2 1 1 ρ . Thus, P ≤ ρ < ρ − 2 + α and P ≤ P ≤ ρ . 

≤

 ·P

x2 x2 +P2



ρ 2 −3ρ +3

2

3

2

Lemma 3.8. 3 < ρ − 2. Proof. By lemma 3.3 we get

G2 + (ρ − 1)F2 − ρ P2 − y2 P3 G2 + (ρ − 1)F2 − ρ P2 − y2 = (ρ − 1) + . x2 + P2 + y2

3 = (ρ − 1) +

Consider the derivative with respect to y2 we get

∂ 3 (ρ − 1)P2 − (x2 + G2 + (ρ − 1)F2 ) = >0 ∂ y2 (x2 + P2 + y2 )2 as x2 + G2 ≤ (ρ − 2 + α)P2 and

F2 P2

≤ ρ − 2 + α.

Thus, we have

3 ≤ (ρ − 1) +

G2 + (ρ − 1)F2 − ρ P2 − (G2 + (ρ − 1)F2 ) x2 + P2 + (G2 + (ρ − 1)F2 )

ρ P2 x2 + G2 + (ρ − 1)F2 + P2 ρ = (ρ − 1) − 2 + (ρ − 1) PF22 + 1 = (ρ − 1) −

≤ (ρ − 1) −

ρ ρ − 2 + α + (ρ − 1)(ρ − 2 + α) + 1

758

G. Yu et al. / European Journal of Operational Research 250 (2016) 754–759

= (ρ − 1) −

ρ (ρ − 1)2 + ρα

< ρ − 2.

Fi

The last step holds as α =

−ρ 2 +3ρ −1 2(ρ −1)

Lemma 3.9. For i ≥ 3, if i ≤ ρ − 2 and and

Fi Pi+1

2 ρ +1 ≤ ρ1 , where β = ρ 2−3 ρ −1 < 0.

<

−ρ 2 +3ρ −1

Fi−1 Pi

ρ

.



≤ ρ1 , then i+1 − i ≤ β

Pi

Proof. By Lemma 3.3 we get

i+1 − i = (ρ − 1) +

xi

Gi + (ρ − 1)Fi − ρ Pi − yi − i . xi + Pi + yi

Fi−1

Consider the derivative with respect to yi we get

bi−1 ≤

yi−1

∂ (ρ − 1)Pi − (xi + Gi + (ρ − 1)Fi ) ( − i ) = ∂ yi i+1 (xi + Pi + yi )2 ∂

as ∂ y i = 0 for yi does not affect i . i Now we show that ∂∂y (i+1 − i ) > 0 and i+1 − i ≤ β for i ≥

xi−1

i

ai−1 ≤

i

Hence

Fig. 6.

Gi + (ρ − 1)Fi − ρ Pi − (Gi + (ρ − 1)Fi ) − i xi + Pi + (Gi + (ρ − 1)Fi ) ρ Pi = (ρ − 1) − − i xi + Gi + (ρ − 1)Fi + Pi ρ Pi = (ρ − 1) − − i ρ xi + Gi + Pi ρ Pi ≤ (ρ − 1) − − i ρ(xi + Gi ) + Pi

i+1 − i ≤ (ρ − 1) +

ρ − i ρ i + 1 ρ − (ρ − 2) ≤ (ρ − 1) − ρ(ρ − 2) + 1 ρ 2 − 3ρ + 1 = . (ρ − 1)2 ≤ (ρ − 1) −

Pi

=

Fi−1 Pi

≤ ρ1 . If Fi = Fi−1 , then

Fi Pi

P

=

min{yi−1 , ρi } Pi

≤

ρ Pi

= ρ1 . So we have xi + Gi + (ρ − 1)Fi = i Pi + (ρ − 1)Fi ≤ (ρ − 2)Pi + ρρ−1 Pi < (ρ − 1)Pi since i ≤ ρ − 2. Thus ∂∂y (i+1 − i ) > 0 i

when Fi = xi . Hence

Gi + (ρ − 1)Fi − ρ Pi − (Gi + (ρ − 1)Fi ) − i xi + Pi + (Gi + (ρ − 1)Fi ) ρ Pi = (ρ − 1) − − i xi + Gi + (ρ − 1)Fi + Pi

i+1 − i ≤ (ρ − 1) +

ρ

= (ρ − 1) −

i + (ρ − 1) PFii + 1 ρ − i ≤ (ρ − 1) − i + ρρ−1 + 1 = (ρ − 1) −

ρ2

ρ i + 2ρ − 1 ρ2 ≤ (ρ − 1) − 2ρ − 1

Fi−2

≤ ρ1 .

ρ 2 − 3ρ + 1 . 2ρ − 1

ρ The inequality (∗2) holds as (ρ − 1) − ρ  +2 ρ −1 − i attains i 2

ρ its maximum when i = 0 since ∂ ∂ ((ρ − 1) − ρ  +2 ρ −1 − i ) = i i 2

3 ρ3 − 1 ≤ (2ρρ−1)2 − 1 < 0. (ρ i +2ρ −1)2 2 2 ρ +1 ρ 2 −3ρ +1 ρ +1 , 2ρ −1 } = ρ 2−3 Thus i+1 − i ≤ max{ ρ(ρ−3 ρ −1 < 0. −1)2

xi

Pi+1 1

≤

xi xi +Pi

≤

xi +Gi xi +Gi +Pi

F

F

1 i i Pi+1 ≤ ρ (see Fig. 6). If Fi = xi , then Pi+1 = i F Fi 1 i =  +1 ≤ ρρ −2 −1 < ρ . If Fi = xi , then Pi+1 ≤ Pi ≤ i

Lemma 3.7 , Lemma 3.8 and Lemma 3.9 show that the potential

Case 2: Fi = xi .

Fi Pi

Fi Pi+1

Pi ρ

ρ. 

The inequality (∗1) holds as i ≤ ρ − 2 and ∂ ∂ ((ρ − 1) − i ρ ρ2 ρ2 ρ 2 − 1 > 0. −  ) = − 1 ≥ − 1 = ( ) i ρ i +1 (ρ i +1)2 (ρ(ρ −2)+1)2 (ρ −1)2 If Fi = Fi−1 , then

=

It remains to show that (∗1)

Ri−1

Pi−1

3 by case distinction on the way that Fi is determined. Case 1: Fi = xi . Since i ≤ ρ − 2, we have xi + Gi + (ρ − 1)Fi = ρ xi + Gi ≤ ρ(xi + ∂ Gi ) = ρ i Pi ≤ ρ(ρ − 2)Pi < (ρ − 1)Pi . Then ∂ y (i+1 − i ) > 0.

Pi ρ

− i

− i

i (i ≥ 3) decreases by a constant in every step until i becomes a negative number. Then we derive a contradiction to i ≥ 0 for any i. Thus we have Theorem 3.10. There exists no online algorithm for strip packing with competitive ratio

ρ<

√ 3+ 5 = 2.618 . . . . 2

4. The limitation of using the list Ln Theorem 4.1. With the √ list Ln , no lower bound on the competitive ratio larger than ρ = (3 + 5)/2 can be obtained for online strip packing. Proof. We only need to find an algorithm that its competitive ratio is ρ when presented with our modified BBK sequences Ln . Consider the online algorithm A: if the list of the online rectangles is ( J1 , . . . , Ji , . . . ), then the online algorithm A packs J1 above the bottom of the strip and packs Ji above Ji−1 for i ≥ 2; Let c1 denote the distance between J1 and the bottom of the strip and ci denote the distance between Ji and Ji−1 for i ≥ 2, the online algorithm A sets J

(∗2)

c2i−1 = ρ2i−1 −1 and c2i = 0. P1 With our list Ln , the online algorithm A chooses x1 = ρ −1 , y1 = 0 P1 and places s1 above P1 (i.e. k = 1). Since x1 + y1 ≤ ρ −1 (i.e. x1 + y1 ≤

G. Yu et al. / European Journal of Operational Research 250 (2016) 754–759 P2

ρ ), then R1 is not included in the sequence. And so on, for i ≥ 2, the

A(Pi ) − ρ · OPT (Pi ) = 2ρ ·

Ln = (P1 , s1 , P2 , F2 , . . . , Pn−1 , Fn−1 , Pn ). Since P1 = 1 and Pi = xi−1 + Pi−1 + yi−1 , we obtain Pi = ρρ−1 Pi−1 = P

•

•

•

i−1

1 A(P1 ) = x1 + P1 = ρ −1 + P1 = ρρ−1 and OPT (P1 ) = P1 = 1, then A(P1 ) < ρ · OPT(P1 ). P2 −1 A(P2 ) = x1 + P1 + s1 + x2 + P2 = x2 + 2P2 = ρ −1 + 2P2 = 2ρρ−1 P2 and √OPT (P2 ) = P2 + s1 = P2 (Note that s1 = θ = 0 for ρ = −1 (3 + 5)/2), then A(P2 ) = ρ · OPT (P2 ) since 2ρρ−1 = ρ. for i ≥ 3,

P

A(Pi ) = x1 + x2 + · · · + xi + P1 + · · · + Pi + s1 + F2 + · · · + Fi−1 = x1 + x2 + · · · + xi + P1 + · · · + Pi + θ + x2 + · · · + xi−1 = x1 + 2(x2 + · · · + xi−1 ) + xi + P1 + · · · + Pi =

1 + ρ −1

2ρ + ··· + (ρ − 1)2

2ρ i−2 + (ρ − 1)i−1

ρ i−1 (ρ − 1)i

 ρ i−1 ρ −1 ρ −1   i−1 1 1 ρ ρ +2 = − + ρ −1 ρ −1 ρ −1 ρ −1  ρ i−1   ρ i−1 · + ρ· − (ρ − 1) ρ −1 ρ −1   ρ i−1 1 +ρ = 2+ ρ −1 ρ −1  1  2ρ − − (ρ − 1) + ρ −1 ρ −1  ρ i−1 1 = 2ρ · + ( − 2ρ + 1) as ρ −1 ρ −1 = ρ − 2 and ρ 2 − 3ρ + 1 = 0, +1+

ρ

+ ··· +

and

OPT (Pi ) = Pi + s1 + F2 + · · · + Fi−1 = Pi + θ + x2 + · · · + xi−1

ρi ρ ρ i−2 + + ··· + (ρ − 1)i (ρ − 1)2 (ρ − 1)i−1  ρ i−1  ρ i−1 ρ ρ = + − · ρ −1 ρ −1 ρ −1 ρ −1  ρ i−1 =ρ· + ( − ρ + 1), ρ −1 =

then



ρ i−1

+ ( − 2ρ + 1)

ρ −1   ρ i−1 −ρ· ρ· + ( − ρ + 1) ρ −1  ρ i−1 = ρ(2 − ρ) + ρ( − 3ρ + 1) ρ −1

P

i and yi = 0, places Fi above Pi , online algorithm A chooses xi = ρ −1 Pi and Ri is not included in the sequence(since xi + yi ≤ ρ −1 ). So

i · · · = ( ρρ−1 )i−1 P1 = ( ρρ−1 )i−1 and xi = ρ −1 = ρ i . It is easy to see (ρ −1) that xi > xi−1 , then Fi = xi for i ≥ 2. We show that this algorithm is ρ −competitive when presented with Ln in the following way:

759

< 0. •

for any i, A(Fi ) = A(Pi ) + Fi and OPT (Fi ) = OPT (Pi ) + Fi , then A(Fi ) < ρ · OPT(Fi ). 

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