Klaus Jansen and Rob van Stee- On Strip Packing With Rotations

8/3/2019 Klaus Jansen and Rob van Stee- On Strip Packing With Rotations

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On Strip Packing With Rotations

Klaus Jansen

Institut fur Informatik

Universitat zu KielGermany

[email protected]

Rob van Stee

Fakultat fur Informatik

Universitat KarlsruheGermany

[email protected]

ABSTRACT

We present an asymptotic fully polynomial time approxi-mation scheme for two-dimensional strip packing with rota-tions. In this problem, a set of rectangles need to be packedinto a rectangle (strip) of fixed width and minimum height,and these rectangles can be rotated by 90. Additionally, wepresent a simple asymptotic polynomial time approximationscheme, and give an improved algorithm for two-dimensional

bin packing with rotations.

Categories and Subject Descriptors

F.2 [Analysis of algorithms and problem complexity]:Nonnumerical algorithms and problems

General Terms

Algorithms

Keywords

strip packing, rotations, approximation scheme

1. INTRODUCTIONIn strip packing, we receive a list L of items p1, p2, . . . , pn.

Each item is given by its width wi and height hi, and eachitem pi must be assigned to a position (xi, yi), where xi 0,yi 0, and xi + wi 1. Further, the positions must beassigned in such a way that no two items in the strip overlap.All items are placed aligned with the coordinate axes. Thegoal is to minimize the maximum height that is reached by

Research supported by EU Project APPOL, Approxi-mation and Online Algorithms, IST-2001-32012 and EUProject CRESCCO, Critical resource sharing for coopera-tion in complex systems, IST-2001-33135.Work performed while the author was at CWI, The Nether-lands. Research supported by the Netherlands Organizationfor Scientific Research (NWO), project number SION 612-061-000.

Permission to make digital or hard copies of all or part of this work forpersonal or classroom use is granted without fee provided that copies arenot made or distributed for profit or commercial advantage and that copiesbear this notice and the full citation on the first page. To copy otherwise, torepublish, to post on servers or to redistribute to lists, requires prior specificpermission and/or a fee.STOC05, May 22-24, 2005, Baltimore, Maryland, USA.Copyright 2005 ACM 1-58113-960-8/05/0005 ...$5.00.

any item. This problem has many applications, for instancecutting objects out of a strip of material in such a way thatthe amount of material wasted is minimized.

In this paper, we consider the version of the problemwhere an item may be rotated by 90 before it is packed.If an item is rotated, we require xi + hi 1 instead ofxi+wi 1. This may allow for a better packing, but it com-plicates the problem since the number of possible packingsis now a factor of 2n larger. In the above-mentioned appli-cation, allowing rotations corresponds to assuming that thematerial used for cutting is featureless (i.e., the orientationof the items on the strip does not matter). In practice, itis often important that cutting takes place along horizontalor vertical lines. We therefore focus on the case where only90 rotations are allowed.

The standard measure of algorithm quality for strip pack-ing is the asymptotic performance ratio, which we now de-fine. For a given input sequence , let A() be the maxi-mum height used by algorithm A on . Let opt() be theminimum possible height needed to pack . The asymptoticperformance ratio for an algorithm A is defined to be

RA = lim suph

sup( A()opt()

opt() = h).Coffman, Garey, Johnson and Tarjan [2] presented the al-

gorithms NFDH (Next Fit Decreasing Height) and FFDH(First Fit Decreasing Height) for strip packing without ro-tations and showed that their asymptotic performance ratioare 2 and 1.7, respectively.

The ratio of 2 is also obtained by BLWD (Bottom Left-most Decreasing Width) which was presented by Baker,Coffman and Rivest [1]. An asymptotic fully polynomialtime approximation scheme for this problem was given byKenyon and Remila [8].

The proofs for the upper bounds of NFDH and BLDW de-pend only on area arguments and therefore also hold in thiscase. An algorithm with asymptotic performance bound of

1.613 was presented by Miyazawa and Wakabayashi [11]. Asimple algorithm with an upper bound of 3/2 was presentedby Epstein and van Stee [3].

Our main result is the following. We denote by opt(L)the height of the optimal strip packing of L which is a listof rotatable items.

Theorem 1. There is an algorithm alg which, given alist L of n rectangles with side lengths at most 1 that canbe rotated 90, and a positive number < 1/2, produces apacking of L in a strip of width 1 and height alg(L) such


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that

alg(L) (1 + )opt(L) + O

1

2

.

The time complexity ofalg is polynomial in n and 1/.

This algorithm works in three phases: a linear program-ming relaxation, a rounding phase and a strip packing phase(without allowed rotations). The AFPTAS uses ideas from[5, 6, 7, 8].

Additionally, we first present a simple asymptotic poly-nomial time approximation scheme. For this we use thealgorithm from [8] as a subroutine. Basically we do a pre-processing step to obtain a modified list of items with onlya constant number of different widths and heights. Thenthe number of orientations for this modified set of items ispolynomial in n. For any fixed orientation, we can apply thealgorithm from [8]. The crucial point of the proof is to showthat we do not lose too much in the preprocessing step, andtherefore have a good approximation. The additive constantfor this APTAS is O(1/3). It is possible to get an additiveconstant of O(1/2) also in this case, but at the cost of asignificantly more complex preprocessing step.

These results immediately imply an asymptotic (2 + )-

approximation algorithm for two-dimensional bin packingwith rotations (that runs also in time polynomial in 1/).Previously Miyazawa and Wakabayashi [11] showed an up-per bound of 2.64 for this problem, which was improved to9/4 by Epstein and van Stee [3].

We begin by describing some of the methods used in [8]which are also important in the current paper in Section2. We then describe our APTAS in detail in Section 3 andanalyze it in Section 4. After that we describe our AFPTASin Section 5. Finally in Section 6, we describe the relationto the two-dimensional bin packing problem with rotations.

2. METHODSWe describe some of the methods used in [8] which are also

important in the current paper. In this section, we assumethat items cannot be rotated (the setting from [8]).

Fractional strip packing

A fractional strip packing of L is a packing of any list L

obtained from L by subdividing some of its rectangles byhorizontal cuts: each rectangle (wi, hi) is replaced by a

sequence of rectangles (wi, h1i ), (wi, h

2i ), . . . , (wi, h

kii ) such

thatPkij=1 hji = hi.

In the case that L contains only items ofm distinct widthsin (, 1], where > 0 is some constant, it is possible tofind a fractional strip packing of L which is within 1 of theoptimal fractional strip packing fsp(L) in polynomial time.Moreover, it is possible to turn this packing into a regularstrip packing at the loss of only an additive constant 2m.Denote the height of the optimal strip packing for L byopt(L). We conclude that we find a packing with height atmost

fsp(L) + 1 + 2m opt(L) + 2m + 1. (1)

Modified NFDH (Next Fit Decreasing Height)

This is a method for adding narrow items (items of width atmost ) to a packing of wide items such as described above.Such a packing leaves empty rectangles at the right side ofthe strip.

Each of these rectangles is packed with narrow items us-ing NFDH (starting with the highest narrow item in the firstrectangle). Wenn all rectangles have been used, the remain-ing items (if any) are packed above the packing using NFDHon the entire width of the strip. For further details, we referto [8].

Grouping and rounding

This method is a variation on the linear rounding defined byFernandez de la Vega and Lueker [4]. It works as follows.

We stack up the rectangles ofL by order of non-increasingwidths to obtain a left-justified stack of total height h(L).We define m 1 threshold rectangles, where a rectangle isa threshold rectangle if its interior or lower boundary inter-sects some line y = ih(L)/m for some i {1, . . . , m1}. Wecut these threshold rectangles along the lines y = ih(L)/m.This creates m groups of items that have height exactlyh(L)/m.

First, the widths of the rectangles in the first group arerounded up to 1, and the widths of the rectangles in eachsubsequent group are rounded up to the widest width in thatgroup. This defines L+.

Second, the widths of the rectangles in each group are

rounded down to the widest width of the next group (downto 0 for the last group). This defines L.It is easy to find a strip packing for L using the methods

from [8] (reduction to fractional strip packing). Moreover,it can be seen that the stack associated with L+ is exactlythe union of a bottom part of width 1 and height h(L)/mand the stack associated with L. Thus

fsp(L) fsp(L+) = fsp(L) +h(L)

m. (2)

Partial ordering

We say that L L if the stack associated to L (used forthe grouping above), viewed as a region of the plane, iscontained in the stack associated to L. Note that L

L

implies that fsp(L) fsp(L

). As an example, in thegrouping above we have

L L L+.

3. A SIMPLE APTAS

3.1 DefinitionsSince the input rectangles may be rotated, we define ar-

bitrarily that the width is the shorter side of the rectangle.(It may be that the item is packed such that the width isthe vertical size.) Thus for each rectangle, 0 < wi hi 1.

We define

HL

=

n

Xi=1 hi

.

This is the height of a stack of the items in L when all theseitems are placed with hi as their vertical dimension, and isan upper bound for the height of any stack of items in L.

Let A(L) denote the total area of the items in L. This isindependent of rotations.

We need to pack the items in L into a strip of width 1and unbounded height. We will use the following constantsin our algorithm: which is the cutoff width for smallrectangles, and m = (1/)3 which is the number of groups


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used in the rounding described in the previous section. Thevalue will be determined by the desired approximationaccuracy .

3.2 Few and wide itemsWe are given n items with m1 different heights and m2

different widths. All widths (and heights) are at least andat most 1. There are at most m1m2 groups of items, whereeach group contains identical items. Denote the number ofitems in group i by ni (i = 1, . . . , m1m2). Then for group i,there are at most ni+1 possibilities with regard to the rota-tion. If the items in this group are squares, the orientationof the items does not matter (there is only one orientation).Otherwise, we can place ki items such that the width is thevertical dimension and ni ki items such that the height isthe vertical dimension, for each ki = 0, 1, . . . , ni.

In total, this implies we have

m1m2Yi=1

(ni + 1)

distinct possibilities for the orientations of the items. Eachsuch possibility is called an orientation of the input list L.It can be represented by a vector v with m1m2 coordinates.

After an orientation of the list has been fixed, we canfind a nearly optimal solution for packing these items us-ing a reduction to fractional strip packing as in [8]. SincePm1m2i=1 ni = n, the number of orientations is polynomial in

the number of items: it is at mostn

m1m2+ 1

m1m2which is polynomial in n for given (fixed) m1 and m2. Thuswe can find the (nearly) best possible packing for these itemsin polynomial time by solving all the fractional strip packingproblems.

Equation (1) implies that for a given, fixed orientation vof the items, this gives us a packing with height at most

fsp(L, v) + 2m(v) + 1 opt(L, v) + 2m(v) + 1, (3)

where fsp(L, v) and opt(L, v) are the optimal fractionalstrip packing and the optimal strip packing for the inputlist L using orientation v, respectively, and m(v) m1m2is the number of distinct widths in orientation v.

3.3 Rounding procedureFirst, the input list Lgeneral will be partitioned into two

sublists. Lnarrow will contain all items with width at most

and L will contain all other items. The items in Lnarrow willbe ignored for the time being. The dimensions of the itemsin L are going to be rounded up according to a procedurewhich we will now describe.

We are going to start by applying a geometric roundingto both heights and widths. In this way, we obtain an in-put instance with a limited number of distinct heights andwidths, of which the total area is still close to the total areaof the original input instance.

However, this is by itself not enough to ensure that thesolution of the (fractional) strip packing problem for thismodified problem instance is close to the real solution. Sup-pose we fix an orientation and the problem instance containsmany items of horizontal size just smaller than 1 /2. If thesewidths get rounded up to a value which is just larger than

1/2, it is clear that the height of the optimal strip packingmay change by an arbitrarily high amount.

In the original paper [8], this problem was solved by trans-forming adjustments of the widths (horizontal sizes) into avertical adjustment of the problem instance, for which itwas trivial to calculate the effect on the height of the op-timal strip packing. To enable us to do this in the currentproblem, we are going to do something similar: in fact ourrounding is going to be a mixture of geometric rounding and

the grouping and rounding method from [8]. We formalizethis idea below.

Let I be the set L rotate(L). That is, for each rectangle(wi, hi) in L, I contains both (wi, hi) and (hi, wi). Thus|I| = 2|L|.

From I we can obtain a modified list I+ such that I I+,as described in section 2. We use m 1 threshold rectanglesat heights y = ih(I)/m for i = 1, . . . , m 1.

Consider a rectangle (wi, hi) in L. Suppose that in I, thefirst corresponding rectangle (wi, hi) has been rounded to(wi, hi) while the other rectangle, (hi, wi), has been roundedto (hi, wi). We then round the original rectangle (wi, hi) to

(min(wi, wi), min(hi, hi)),

where x is the first power of 1/(1 +

) greater than or equalto x.

Property 1. The number of groups created in the round-ing procedure is bounded by a constant.

Proof. The rounding is only applied to items that haveheight and width larger than . Thus the highest class that

can contain items is m which is the solution of (1+)m

=. We find

m = log

log(1 + )

1

log

1

h, we have

hf A(Lgeneral)

1

1 + +

4

()2m

+

102

25m + 1. (7)


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We are now ready to prove Theorem 2. Consider theorientation of the items in L in the optimal strip packing,i.e. the list M. At some point, our algorithm will try an ori-entation of L which corresponds to this orientation. Fromhere on, we only consider the height of the packing that ouralgorithm finds for this particular orientation. Clearly, theheight that it outputs in Step 4 cannot be higher than this.

By (7), we have either alg(L) h or

alg(L)

1 + + 4()2m

A(Lgeneral)

1 + 102

25m + 1

where h is the height of the strip packing produced in Step3(a) of the algorithm. By (3) and Lemma 1, using as inLemma 2 that the number of different widths is at most5150

m in any orientation, we find

h fsp(L) +51

25m + 1

1 + +

4

()2m

fsp(M) +

51

25m + 1

1 + +

4

()2m

opt(M) +

51

25m + 1

1 + + 4()2m

opt(Lgeneral) + 51

25m + 1.

Since A(Lgeneral) opt(Lgeneral), this implies

alg(L)

1 + +

4

()2m

opt(Lgeneral)

1 +

102

25m + 1.

Taking m = (1/)3 and = /(6 + ), this implies that

alg(Lgeneral) (1 + )opt(Lgeneral) +1121

3+ 1

for < 1/2. This proves Theorem 2.

5. THE AFPTAS

5.1 LP relaxationLet Lgeneral be the set of all rectangles {p1, . . . , pn}. A

configuration C is given by two disjoint sets (A, B): a set ofnon-rotated rectangles A {p1, . . . , pn} and a set of ro-tated rectangles B {p1, . . . , pn} that have total widthPpjA

wj +PpjB

hj 1. Let C be the set of all con-

figurations.We use the following linear program relaxation of strip

packing with rotations:

MinXCC

xC

s.t.1

hj XC=(A,B):pjA

xC

+1

wj

XC=(A,B):pjB

xC 1 j = 1, . . . , n ,

xC 0, C C.

(8)

The variable xC in the linear program indicates the totalspace reserved for configuration C C. The objective valuePCC xC is the total height of the packing (where we im-

plicitly allow to rotate a rectangle and to cut a rectangle intoseveral pieces). The inequality for rectangle pj can be inter-preted as follows: the first term 1/hj

PC=(A,B):pjA

xC and

the second term 1/wjPC=(A,B):pjB

xC is the non-rotated

and rotated fraction of the rectangle pj.Let LINR(Lgeneral) be the minimum objective value of

the linear program (8).We note that LINR(Lgeneral) OP T(Lgeneral) where

OP T(Lgeneral) is the height of an optimum strip packingwith allowed 90 degrees rotations. To see this consider anoptimum solution of this strip packing problem (with corre-sponding values xC). In this solution each rectangle is eitherrotated or not. Ifpj is non-rotated, then

PC=(A,B):pjA

xC

hj . In the other case (ifRj is rotated)PC=(A,B):pjB

xC wj . Therefore the inequalities are satisfied for all rectan-gles, and this implies the lower bound.

The linear program (8) can be transformed into a max-min resource sharing problem with one block and n cover-ing constraints [6, 7]. Using the Lagrangian decompositionmethod the max-min resource sharing problem can be solvedapproximately with accuracy in O(n(2 + ln n)) iterationsor coordination steps [5], each requiring for a given pricevector y = (y1, . . . , yn) and

= /6 an (1 ) approxi-mate solution of a multi-choice knapsack problem with twochoices per item:

MaxnXj=1

yjhj

yj,0 + yj

wjyj,1

s.t.nXj=1

wjyj,0 + hjyj,1 1,

yj,0 + yj,1 1, j = 1, . . . , n ,yj,0, yj,1 {0, 1}, j = 1, . . . , n .

(9)

Lawler [9] showed that an (1 )-approximate solutionof this Knapsack Problem can be computed in O(n2/)time. The overhead (i.e. the numerical calculations) periteration in [5] can be bounded by O(n lnln(n/)) that isless than the running time required by the knapsack sub-routine. This implies that the linear program (8) can besolved approximately with multiplicative accuracy (1+ ) in

O(n3(2 + ln n)1) time (where = 3/20 and 1).

5.2 Rounding the LP solutionLet hmax = maxj max{hj , wj}. For each rectangle pj and

each choice {0, 1} we define two fractions

xj,0 =1

hj

XC=(A,B)C:pjA

xC

xj,1 =1

wj

XC=(A,B)C:pjB

xC.

W.l.o.g. we suppose that xj,0 + xj,1 = 1. Then we con-struct an instance I of a strip packing problem with rectan-gles (wj , xj,0 hj) of height xj,0hj and width wj (ifxj,0 > 0)

and rectangles (hj, xj,1 wj) of height xj,1wj and width hj(if xj,1 > 0).

Let Lnarrow = {(x, y) I|x , j = 1, . . . , n} and L =

I Lnarrow. For the set L of wide rectangles we constructa left-justified stack of height h(L) and split the stack intom = (1/)2 groups of equal height h(L)/m (here we divideagain the threshold rectangles into two pieces). Let ai, bibe the smallest and largest width of group i. Let Wj,i {hj , wj} be the set of widths in group i corresponding torectangle pj (i.e. Wj,i stores implicitly the rotated or non-rotated rectangle pieces corresponding to pj). For the set


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Lnarrow we compute the total area

A(Lnarrow) =nXj=1

24 Xwj

xj,0hjwj +Xhj

xj,1hjwj

35

and place the corresponding rectangles in group 0.For each group i {0, . . . , m} and rectangle pj let zj,i =

Pw:wWj,i yj,i(w) be the fraction of the original rectangle pjplaced in group i (where yj,i(w) is the fraction of rectanglepj with width w {wj , hj} placed in group i). Consider thecase that we have either a non-rotated or a rotated piece ofpj in group i (i.e. Wj,i = {wj} or Wj,i = {hj}). In this casewe set Hj,i = hj (if pj is non-rotated) and Hj,i = wj (if pjis rotated).

The goal is now to get an unique orientation for almostall rectangles. To obtain such an assignment we do the fol-lowing steps:

(1) for each group i {1, . . . , m} and rectangle pj withtwo pieces (rotated and non-rotated) in group i : re-place them by one rectangle

(max(hj, wj), zj,imin(hj , wj))

of width max(hj , wj) and height zj,imin(hj , wj) . Inthis case Hj,i = min(hj , wj), zj,i = yj,i(wj) + yj,i(hj)and Wj,i = max(hj , wj).

(2) for each rectangle pj with two choices in group 0: re-place them by one rectangle (wj, zj,0 hj).

(3) round all tasks over the groups using a general assign-ment problem:

nXj=1

zj,0hjwj A(Lnarrow)

n

Xj=1zj,iHj,i h(L)/m i = 1, . . . , m

mXi=0

zj,i = 1 j = 1, . . . , n

zj,i 0 j = 1, . . . , n ,i = 1, . . . , m .

(10)

This problem formulation is closely related to schedulingtasks on unrelated machines; i.e. with m + 1 machines andmakespan as objective function [10]. One can round nowthe variables zj,i such that there are at most m fractionalvariables. Let F be the set of rectangles pj with fractionalvalues zj,i (0, 1) after the rounding. After removing theserectangles we have an unique orientation (either rotated ornon-rotated) for each remaining rectangle. The rectanglesin F can be placed on top of the packing with additional

height bounded by mhmax m (using hmax 1). Wehave obtained now a strip packing instance with a set L ={(Wj,i, Hj,i) : z

j,i = 1, i 1} of wide rectangles and a set

Lnarrow = {(wj , hj) : zj,0 = 1} of narrow rectangles.Our algorithm computes (L)+ by producing a new stack

packing. In this way an instance of the strip packing prob-lem with at most m = (1/)2 distinct widths is generated.Finally we apply the approximation algorithm for strip pack-ing [8] for the instance (L)+ L

narrow (with multiplicative

accuracy (1 + ) in the fractional strip packing phase andm = (1/)2 groups).

5.3 AlgorithmOur approximation algorithm works now as follows:

Input: a list Lgeneral ofn rectangles pj with height hj andwidth wj where hj , wj (0, 1],

(0) set = min(1, /8), = /( + 2), and m = (1/)2,

(1) compute a (1 + ) approximate solution of the linear

program relaxation (8) and compute the values

xj,0 =1

hj

XC=(A,B):pjA

xC

xj,1 =1

wj

XC=(A,B):pjB

xC

for each rectangle pj ,

(2) construct an instance I with rectangles (wj , xj,0 hj)and (hj, xj,1 wj) (for xj, > 0), and let L = {(x, y) I : x > } and Lnarrow = {(x, y) I : x

},

(3) use the rounding technique to obtain a strip packinginstance (with a set L of wide and a set Lnarrow of

narrow rectangles) and a set F of rectangles pj withfractional values zj,i (0, 1),

(5) construct instance (L)+ Lnarrow and apply the ap-proximation algorithm for strip packing by Kenyonand Remila [8],

(6) place the rectangles in F on top of the packing.

5.4 Analysis of the AFPTASTo prove the upper bound for the total height of the pack-

ing we use the following inequalities:

(1) L (L)+

(2) A(L+) A(L)1 + 1m

(3) F SP(L+) F SP(L)

1 +

1

m

(4) F SP(L) LINR(Lgeneral)(1 + )

(5) A(L Lnarrow) LINR(Lgeneral)(1 + )

where F SP(L) and LINR(Lgeneral) is the optimal objectivevalue of the fractional strip packing for L and of the linearprogram relaxation for Lgeneral, respectively.

The fractional strip packing algorithm for (L)+ computesa solution of height (1 + )F SP((L)+) with at most m =(1/)2 non-zero variables. This solution is transformed intoan integral strip packing of height

hwide (1 + )F SP((L)+) + 2m

(1 + )2

1 +1

m

2LINR(Lgeneral) + 2m.

In a similar way, we have for the total area

A((L)+ Lnarrow) (1 + )

1 +

1

m

2LINR(Lgeneral).

The integral strip packing (see algorithm by [8]) constructesa nice packing for the wide rectangles and leaves a well struc-tured free space for the narrow rectangles. Therefore the


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final height hfinal (after placing the narrow rectangles andadding the rectangles in F) is at most

max

"(1 + )2

1 +

1

m

2LINR(Lgeneral) + 3m,

A((L)+ Lnarrow)

1 + 5m + 1

(1 + )OP T(Lgeneral) + O(1/

2)

using = /( + 2), = min(1, /8) and m = (1/)2 =O(1/2). This proves Theorem 1.

6. TWO-DIMENSIONAL BIN PACKING

WITH ROTATIONSWe conclude by giving an upper bound of 2 + for two-

dimensional bin packing with rotations.Take = /2. Suppose our algorithm with as input pa-

rameter gives a strip packing with a height ofH for a givenset of items. Denote the height of the optimal strip packingby H. We have

H (1 + )H + O 12 .

Given this strip packing, we can define cuts at all integerheights. Then starting from the cut at height 1, we canallocate the items to square bins as follows: put all the itemsbelow this cut into a separate bin (the upper edge of someof these items may coincide with the cut), and put all theitems that intersect the cut into another bin. Continue withthe next higher cut until all items are packed into bins.

Using the above method, we put the items into at most2H bins. Additionally, there does not exist a packing intobins which uses less than H bins, because such a packingcould trivially be turned into a strip packing that uses aheight at most H 1, which is a contradiction.

Thus the number of bins that we use to pack the items

is at most 2 + 2 = 2 + times the optimal number of binsrequired for these items, plus an additive constant.

Acknowledgment

The authors wish to thank Claire Kenyon who defined therounding procedure for the APTAS as presented above andthereby substantially simplified the proof.

7. REFERENCES[1] Brenda S. Baker, Edward G. Coffman, and Ronald L.

Rivest. Orthogonal packings in two dimensions. SIAMJournal on Computing, 9:846855, 1980.

[2] Edward G. Coffman, Michael R. Garey, David S.Johnson, and Robert E. Tarjan. Performance boundsfor level oriented two-dimensional packing algorithms.SIAM Journal on Computing, 9:808826, 1980.

[3] Leah Epstein and Rob van Stee. This side up! InProceedings of the 2nd Workshop on Approximationand Online Algorithms (WAOA 2004), LNCS 3351,pages 4860, 2005.

[4] Wenceslas Fernandez de la Vega and George S.Lueker. Bin packing can be solved within 1 + inlinear time. Combinatorica 1(4):349355 (1981).

[5] Mike D. Grigoriadis, Leonid G. Khachiyan, LorantPorkolab and Jorge Villavicencio. Approximatemax-min resource sharing for structured concaveoptimization. SIAM Journal on Optimization,11:1081-1091, 2001.

[6] Klaus Jansen. Scheduling malleable parallel tasks: anasymptotic fully polynomial time approximationscheme. Algorithmica, 39:5981, 2004.

[7] Klaus Jansen and Lorant Porkolab. On preemptiveresource constrained scheduling: polynomial-timeapproximation schemes. In Proceedings of the 9thInternational Conference on Integer Programming andCombinatorial Optimization (IPCO 2002), LNCS2337, pages 329349, 2002.

[8] Claire Kenyon and Eric Remila. A near-optimalsolution to a two-dimensional cutting stock problem.Mathematics of Operations Research, 25:645-656, 2000.

[9] Eugene Lawler. Fast approximation algorithms forknapsack problems, Mathematics of OperationsResearch, 4:339-356, 1979.

[10] Jan K. Lenstra, David B. Shmoys, and Eva Tardos.Approximation algorithms for scheduling unrelated

parallel machines. Mathematical Programming,24:259-272, 1990.

[11] Flavio Keidi Miyazawa and Yoshiko Wakabayashi.Packing problems with orthogonal rotations. InProceedings of the 6th Latin American TheoreticalInformatics Symposium (LATIN 2004), LNCS 2976,pages 359368, 2004.

Klaus Jansen and Rob van Stee- On Strip Packing With Rotations

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