Relations between capacity utilization, minimal bin size and bin number
Als Manuskript gedruckt
Technische Universit¨at Dresden Herausgeber: Der Rektor
Relations between capacity utilization, minimal bin size and bin number Torsten Buchwald,
Kirsten Hoffmann, Guntram Scheithauer
MATH-NM-03-2013 May 2013
Contents 1 Introduction: problem formulation and goal
1
2 Related work
2
3 The FFDH1 - and FFDW1 -heuristic
4
4 The Bin Packing Problem
7
5 Conclusions and outlook
20
Relations between capacity utilization, minimal bin size and bin number Torsten Buchwald, Kirsten Hoffmann, Guntram Scheithauer Technische Universit¨at Dresden May 17, 2013 Abstract We consider the two-dimensional bin packing problem (BPP): given a set of rectangular items, find the minimal number of rectangular bins needed to pack all items. Rotation of the items is not permitted. We show for any integer k ≥ 3 that at most k − 1 bins are needed to pack all items if every item fits into a bin and if the total area of items does not exceed k/4-times the bin area. Moreover, this bound is tight. Furthermore, we show that only two bins are necessary to pack all items if the total area of items is not larger than the bin area, and if the height of each item is not larger than a third bin height and the width of every item does not exceed a half bin width.
Key words: Cutting and Packing, Bin Packing Problem, Strip Packing Problem
1
Introduction: problem formulation and goal
In this paper, we consider the two-dimensional bin packing problem (BPP) with nonrotatable rectangular items: Let L := (R1 , . . . , Rn ) denote a list of small rectangles Ri (items) of width wi ≤ W and height hi ≤ H, i ∈ K := {1, . . . , n}. The items have to be packed into a minimum number of rectangular bins of given width W and height H such that the items do not overlap each other, and each of them fits completely in one of the bins. Furthermore, the strip packing problem (SPP) is of interest where the height H of the bin is variable and has to be minimized. Both, BPP and SPP, are known to be NP-hard optimization problems (cf. [GJ79]). Because of their high applicability to practical problems, a lot of work is done in this field concerning heuristic approaches (leading to upper bounds of the optimal value), lower bounds and exact solution methods (cf. [DF92], [LMM02]). In most cases the solvability of the problems is out of question if natural demands are met (each item fits into a bin). For another type of packing problems, so-called orthogonal (decision) packing problem (OPP), the verification whether a feasible packing pattern of all items in a single bin exists 1
2
Relations between capacity utilization, minimal bin size and bin number
or not, is the main goal. Several algorithmic approaches have been published within the last 10 years (cf. [FS04], [CCM07], [BKRS09]). But, to the best of our knowledge, there exists only few work concerning theoretical aspects. In this paper, we improve known results and contribute in particular the statement that any set of rectangular items (each of them fits into a single bin) can be packed into at most k − 1 bins if the total area of items is not larger than k/4-times of bin area, for an integer k ≥ 3. In case that the total area of items does not exceed the bin area, we also show that only two bins are needed if some weak conditions are fulfilled. The next section reviews related work and motivates our investigations. Section 3 contains some results concerning SPP, whereas the BPP is considered in Section 4. In the last section we give some conclusions and an outlook of future research. In this paper, we consider various packing problems with rectangular items to be packed into one or a minimum number of rectangular containing regions. In any case, a list L = (R1 , . . . , Rn ) of rectangular items is given where rectangle Ri has width wi and height hi , i ∈ K := {1, . . . , n}. Rotation is not permitted. In the Strip Packing Problem (SPP) all items have to be packed into a single strip of given width W and minimum height H, whereas in the Bin Packing Problem (BPP) the goal is to find the minimum number of bins of size W × H needed to pack all items. Throughout this paper, for a list L = (R1 , . . . , Rn ) of rectangles, also represented by indexset K, let A(L) = A(K) denote the total area, i. e. X A(L) = A(K) := wi hi , i∈K
Abin := W H the bin area, and OP T (L) the optimal value of BPP (minimum number of bins) or SPP (minimal height), respectively. Furthermore, let X X e := e := e ⊆ K, w(K) wi , h(K) hi , ∀ K e i∈K
e := max{wi : i ∈ K}, e wmax (K)
e i∈K
e := max{hi : i ∈ K}, e hmax (K)
wmax := wmax (K),
hmax := hmax (K).
e ⊆ K, ∀K
We always assume that wi ≤ W/p and hi ≤ H/q for all i ∈ K holds true with some p ∈ [1, W ] ∩ N and q ∈ [1, H] ∩ N (N = {1, 2, . . . }). If not stated otherwise, p = q = 1. As usual, let (x)+ := max{0, x}.
2
Related work
To the best of our knowledge, the smallest known number of bins needed to pack any list L of items with A(L) ≤ Abin is 4. This number results from a theorem of Martello and Vigo [MV98] for the two-dimensional bin packing problem. Theorem 1 (Martello/Vigo (1998)). For any list L of rectangles with total area A(L) and bins of size W × H, we have A(L) OP T (L) ≤ 4 . WH
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
3
As a consequence of Theorem 1, at most four bins are needed to pack all items having total area not larger than the bin area. Some other known results suggest that without any or only with weak assumptions the packability of a list L of rectangular items into a single bin can be guaranteed if A(L) is not too large but near to 50 %. In 1968, Meir and Moser [MM68] published the following result: Theorem 2 (Meir/Moser (1968)). Any list L of rectangles with wi ≥ hi for all i ∈ K and total area A(L) can be packed into any rectangle of size W × H if W ≥ wmax and Abin = W H ≥ 2A(L) + W 2 /8. This result is best possible. As consequence of Theorem 2, we obtain the following corollary, e. g. used by Li and Cheng [LC90]. Corollary 3 (Meir/Moser (1968)). A list L of rectangles with wi ≥ hi for all i ∈ K can be packed into a unit square if wmax ≤ 1 and A(L) ≤ 7/16. A very useful result on packability of a list of rectangles was given in 1997 by Steinberg [Ste97]: Theorem 4 (Steinberg (1997)). If for a list L of rectangles and a bin of size W × H the following inequalities hold, wmax ≤ W,
hmax ≤ H,
2A(L) ≤ Abin − (2wmax − W )+ · (2hmax − H)+ ,
then it is possible to pack all items into the bin. Note that in case of wmax ≤ W/2 or hmax ≤ H/2 all items can be packed if their total area is at most 50% of the bin area. Consequently, if A(L) ≤ W H, then all items fit in a rectangular bin of size 2W × H, or W × 2H, but we cannot conclude that all items can be packed into two bins of size W × H, since the guillotine property is not guaranteed. For instance, three items of size ( 12 W + ε) × ( 21 H + ε) cannot be packed into two bins if ε > 0. Jansen and Zhang [JZ07] improved the result of Steinberg for the special case that there are no wide items in K, i. e. {i ∈ K : wi > W/2, hi ≤ H/2} = ∅. Theorem 5 (Jansen and Zhang (2007)). If A(K) ≤ 21 W H, and if the set K does not contain wide items, then the items of K can be packed in a single bin. Note that the theorem also holds true if K contains a large item (one item i with wi > W/2 and hi > H/2).
4
3
Relations between capacity utilization, minimal bin size and bin number
The FFDH1 - and FFDW1-heuristic
In this section we review some known results and give some new for the well known NFDH(Next-Fit Decreasing Height)- and FFDH-(First-Fit, Decreasing Height)-heuristic for SPP and for their modifications FFDH1- and FFDW1 -heuristic, which result when the FFDHheuristic is applied to a single bin. Within this section, we always assume that a strip of fixed width W is given and a list L = L of rectangular items with wmax ≤ W/p and hmax ≤ H/q where p ∈ [1, W ] ∩ N, q ∈ [1, H] ∩ N. Lemma 6. If p ≥ 2 and the total area A(L) of items does not exceed the bin area, i. e. A(L) ≤ W H, then the NFDH-heuristic yields a strip packing of height not larger than p 1 H. + p−1 q A proof of this lemma can be found in [HS12]. A similar result with stronger bound holds for the FFDH-heuristic. Let e hj , j = 1, 2, . . . , se denote the heights of the (horizontal) strips obtained by the FFDHheuristic applied for SPP with strip width W . Because of decreasing heights of items, we have e hj ≥ e hj+1 for j = 1, 2, . . . , se − 1. Furthermore, let e hse+1 := 0. Theorem 7. Let p ∈ [1, W ] ∩ N, and let wi ≤ W/p for i ∈ K. Then the following relation holds for the total area of items and strip height obtained by the FFDH-heuristic: s e X p e W hj+1. A(K) ≥ p + 1 j=1
Its proof is part of the proof of Theorem 3 in [CGJT80], and can also be found in [BS12]. Corollary 8. As a consequence of this theorem, the total height obtained by the FFDHheuristic for SPP to pack a set K of items with A(K) ≤ W H is not greater than p+1 1 H. + p q Proof: Using Theorem 7 we obtain: Pse e p+1 p+1 e j=1 hj ≤ pW A(K) + h1 ≤ pW W H +
H q
=
p+1 p
+
1 q
H.
In the following theorems, we apply, in general, the well-known FFDH-heuristic in order to pack a single bin, denoted as FFDH1-heuristic, which works as follows: Let again e hj , j = 1, 2, . . . , se denote the heights of the (horizontal) strips obtained by the FFDH-heuristic applied for SPP with strip width P W . In case that not all items can be packed into Psa single s e e bin by the FFDH-heuristic, then j=1 hj > H follows. We define s := arg max { j=1 e hj ≤ 1 1 H} as the number of strips packed by the FFDH -heuristic. The FFDH -heuristic stops just before the first item has to be packed into the (s + 1)st strip. Let K denote the item set packed by the FFDH1 -heuristic.
5
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Additionally to the FFDH1-heuristic, we also consider the First-Fit Decreasing-Width1heuristic (FFDW1 , transposed“ FFDH1-heuristic) where all items are sorted by non” increasing width, and are then packed sequentially in the left-most vertical strip in which they fit until the first item cannot be packed into the bin. Pn Lemma P 9. Let q ∈ N, L/q ≥ ℓ1 ≥ ℓ2 ≥ · · · ≥ ℓn > 0 and i=1 ℓi > L. Let s := s arg max { i=1 ℓi ≤ L}. Then s X q L≤ ℓj ≤ L. q+1 j=1 Ps q Proof: Let us assume j=1 ℓj < q+1 L. Then ℓs+1 > L/(q + 1). On the other hand, because of P assumption, s ≥ q and due to the non-increasing sorting of ℓi -values, we have P s s q L j=1 q+1 ≥ q+1 L, which leads to a contradiction. j=1 ℓj > P Because of the non-increasing strip heights e hj , j = 1, . . . , s with s := arg max { sj=1 e hj ≤ 1 H}, generated by the FFDH -heuristic, from Lemma 9 follows: Corollary 10. If not all items of K are packed by the FFDH1 -heuristic, then we have for the total height used: s X q e H≤ hj ≤ H. q+1 j=1 A similarPresult holds for the FFDW1-heuristic: Let w ej , j = 1, . . . , s with s := s arg max { j=1 w ej ≤ W } be the widths of the vertical strips obtained by the FFDW1heuristic. In case not all items are packed, we have for the total width used: X p W ≤ w ej ≤ W. p+1 j=1 s
In the FFDH1-heuristic we need to consider regular and non-regular items. If item r is packed into strip j and strip j is the highest non-empty strip at that time when item r is packed, then r is called a regular item, otherwise a non-regular or fallback item. Let Kij , j = 1, . . . , i, i = 1, . . . , s, denote the set of items packed into strip j at the time before strip i + 1 has to be opened (or all items are packed), i. e. when all regular items have been packed into strip i, andS probably some fallback items have been placed into strips below strip i. Hence, K = sj=1 Ksj . In case of K 6= K the first item not packed into the bin, is denoted by k0 . Corresponding to Kij , j = 1, . . . , i, i = eij := w(Kij ) = P P1, . . . , s, we define the used width w j∈Kij wj and area Aij := A(Kij ) = j∈Kij wj hj . Furthermore, let ∆i := max{0,
p W −w eii }, p+1
i = 1, . . . , s.
Theorem 11. Let A(K) > 21 W H. We assume that not all items are packed by the FFDH1heuristic, i. e. K 6= K, and k0 is the first item not packed. If there does not exist any subset e ⊆ K ∪ {k0 } with A(K) e ≥ 1 W H, that can be packed into the bin, then K 2 hk0
0 we distinguish two subcases:
1 W. (i) Strip i does not contain any item k ∈ Kii with width wk < p+1 1 Since the width of the first (regular) item in strip i is at least p+1 W +∆i−1 , and since at least p regular items are packed into strip i, we have w eii ≥ p W + ∆i−1 . Since inequality (1) Pi Pi−1 p+1 p is valid for i − 1 we obtain eij ≥ w eii + j=1 w ei−1,j ≥ p+1 W · i + ∆i−1 − ∆i−1 . j=1 w Consequently, formula (1) is also fulfilled for i.
1 W. (ii) Strip i contains a regular item t0 with width wt0 < p+1 This can only happen when further fallback items with total width larger than ∆i−1 are packed into strip i − 1 before item t0 is packed into strip i. Hence, w ei,i−1 − w ei−1,i−1 ≥ ∆i−1 . 1 Since the first item of strip i has width at least p+1 W +∆i−1 we obtain, because of induction, Pi P p W · i − ∆i . eij ≥ w eii + i−1 ei−1,j + w ei,i−1 − w ei−1,i−1 ≥ p+1 j=1 w j=1 w P p esj ≥ p+1 Consequently, inequality (1) is fulfilled, and we obtain sj=1 w W · s − ∆s . Ps p p W ·s. Otherwise, if ∆s > 0, then we have w ess = p+1 W −∆s , If ∆s = 0, then j=1 w esj ≥ p+1 Ps−1 p p and hence j=1 w esj ≥ p+1 W · s − ∆s − w ess = p+1 W · (s − 1). Ps p W · s we replace strip s by another one esj ≥ p+1 In order to construct a packing with j=1 w as follows: Sort the items of Kss ∪ {k0 } according to non-increasing width and pack them in this sequence into the s-th strip. Because of Corollary 10 the new packing Ps−1of strip s has p width w es ≥ p+1 W . Consequently, the obtained new bin packing fulfills j=1 w esj + w es ≥ p W · s. p+1
H. Since at least q strips are The rest of the proof is done by contradiction. Let hk0 ≥ p+1 2pq packed into the bin, for the total area A of the packed items follows P P P P P p A ≥ si=1 j∈Ksi wj hj ≥ si=1 j∈Ksi wj p+1 H ≥ si=1 w esi p+1 H ≥ s · p+1 W · p+1 H 2pq 2pq 2pq ≥q·
p W p+1
·
p+1 H 2pq
= 12 W H,
which contradicts the assumption. Corollary 12. Let A(K) > 12 W H. We assume that not all items are packed by the FFDW1-heuristic, i. e. K 6= K, and k0 is the first item not packed. If there does not exist e ⊆ K ∪ {k0 } with A(K) e ≥ 1 W H, that can be packed into the bin, then it holds any subset K 2 w k0
21 W H. If all items of K can be packed into a single bin, or if e ⊂ K with A(K) e ≥ 1 W H, which can be packed hk0 ≥ p+1 H, then there exists a subset K 2pq 2 into a single bin. Proof: Since at least one of the assumptions of Theorem 11 cannot be fulfilled, either all e with A(K) e ≥ 1 W H, which items of K can be packed into a bin, or there exists a subset K 2 1 can be packed into one bin. Since A(K) > 2 W H, the statement follows. Corollary 14. Let A(K) > 21 W H. If all items of K can be packed into a single bin, or if W , where k0 denotes the first item, which is not packed by the FFDW1-heuristic, wk0 ≥ q+1 2pq e ⊂ K with A(K) e ≥ 1 W H, which can be packed into a single then there exists a subset K 2 bin. Proof: The statement follows from the previous corollary applied on an instance with switched wi and hi , respectively W and H.
4
The Bin Packing Problem
In this section, we focus on relations between the maximum number of bins needed to pack all items and their total area. We state that any list L = L(K) of items with total area A(L) ≤ Abin can be packed into at most three bins. Then we improve a lemma of Bougeret et al. [BDJ+ 09], and investigate further conditions needed to pack all items into two bins. First of all, Steinberg’s theorem can be used to improve Theorem 1 as follows (suggested by an anonymous referee): Let an instance E of the two-dimensional bin packing problem be given with bins Pof size W × H and list L of rectangular items with wmax ≤ W , hmax ≤ H, and A(L) = i∈K wi hi ≤ W H = Abin . Because of Steinberg’s theorem, all items can be packed into a strip of width 2W and height H. Let (xi , yi ) denote the corresponding allocation point (lower left corner) of item i in such a packing. Then item i covers region Ri (xi , yi ) := {(x, y) : xi ≤ x < xi + wi , yi ≤ y < yi + hi } ⊂ [0, 2W ] × [0, H]. We define the following partition of K = {1, . . . , n}: K1 := {i ∈ K : xi + wi ≤ W },
K2 := {i ∈ K : xi ≥ W },
K3 := K\(K1 ∪ K2 ).
It is obvious, that all items of K1 can be packed into a single bin, and that all items of K2 can P be placed into a second bin. Since K3 = {i ∈ K : xi < W < xi + wi }, the total height i∈K3 hi of K3 -items does not exceed H. Hence, all items of K3 can also be packed into a single bin. Summarizing, we have Theorem 15. Consider bins P of size W × H and a list L of rectangular items with wmax ≤ W , hmax ≤ H. If A(L) = i∈K wi hi ≤ W H = Abin , then all items of L can be packed into at most three bins.
8
Relations between capacity utilization, minimal bin size and bin number
On the other hand, at least three bins are needed in following cases: Theorem 16. Let q ∈ [1, H] ∩ N. Then there exist lists L of items with wmax ≤ W (i. e. p = 1), hmax ≤ H/q, and A(K) ≤ W H such that at least three bins are needed to pack all items. Proof: The statement is a consequence of the following example. Consider 2q + 1 items 1 (H + ε) for some sufficient small positive ε and i ∈ K = with wi = 21 (W + ε) and hi = q+1 {1, . . . , 2q + 1}.P Then we have: A(K) = wi hi = (2q + 1) W2+ε · H+ε = 2q+1 (W H + ε(W + H + ε)) ≤ W H. q+1 2q+2 i∈K
Since only q items can be packed in one bin, at least three bins are necessary to pack all items.
A statement concerning the maximum number of bins needed to pack all items is given by Bougeret et al. [BDJ+ 09]. Lemma 17 (Bougeret at al. (2009)). Let k ≥ 3 and let L be a list of rectangles with widths and heights not greater than 1 and total area not greater than k/4. Then there exists a packing of L into k unit bins. Applying this lemma for k = 4 we obtain the result due to Theorem 1. Using the technique of Theorem 15, the Lemma of Bougeret et al. can be strengthened as follows: Lemma 18. Let k ≥ 3 and let L be a list of rectangles with wmax ≤ 1, hmax ≤ 1, and A(L) ≤ k/4. Then there exists a packing of L into k − 1 unit bins if k is even, and into k unit bins if k is odd. Proof: Let k = 2m, m ≥ 2, integer. Due to Steinberg’s theorem all items can be packed into a strip of width 2 · k/4 = k/2 = m and height 1. Using again the description of a packing by allocation points (xi , yi ) we can define m index sets Ip := {i ∈ I : p − 1 ≤ xi , xi + wi ≤ p}, p = 1, . . . , m, and m − 1 index sets Im+p := {i ∈ I : xi < p < xi + wi }, p = 1, . . . , m − 1. Obviously, the m − 1 = k − 1 index sets Ip form a partition of I, and each item set can be packed into a single bin. Now, for k = 4, the statement of Theorem 15 follows from Lemma 18: at most three bins are needed. In the following we show that Lemma 18 can further be strengthened. Lemma 19. Let L = L(K) be a list of rectangles with hmax ≤ H and wmax ≤ W . If w(K) ≤ 23 W , then the First Fit(FF) heuristic packs all items of K in at most two bins of size W × H. Proof: Let B1 and B2 denote sets of items packed by the FF-heuristic into the first two bins, respectively. Assume, there is an item j, which is not packed by the FF-algorithm into one of the first two bins. Then wj + w(B1 ) > W and wj + w(B2 ) > W , and therefore 2wj + w(B1 ) + w(B2 ) > 2W . Since wj + w(B1 ) + w(B2 ) ≤ 32 W we obtain wj > 21 W , and consequently, w(B1 ) + w(B2) < W . Therefore all items of B1 and B2 are packed by the FF-algorithm into the first bin, i. e. B2 = ∅ and item j fits within the second bin.
9
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Theorem 20. Let L = L(K) be a list of rectangles with hmax ≤ H and wmax ≤ W . If 3 A(L) ≤ W H, 4
(2)
then all items of L can be packed into at most two bins of size W × H. Proof: Let Ksmall := {i ∈ K : wi ≤ 12 W, hi ≤ 21 H}, Klarge := {i ∈ K : wi > 12 W, hi > 12 H}, Kwide := {i ∈ K : wi > 21 W, hi ≤ 12 H}, Ktall := {i ∈ K : wi ≤ 21 W, hi > 21 H}. W. l. o. g., we assume A(K) = 43 W H. The proof is done by a complete case-by-case analysis. Case 1: Let Ksmall = ∅, Kwide = ∅, Ktall ∪ Klarge 6= ∅. Because of hi > 21 H ∀i ∈ Ktall ∪ Klarge and area condition (2), we have w(Ktall ∪ Klarge ) < 3 W . Due to Lemma 19, all items are packed into two bins by the FF-algorithm. 2 Case 2: Let Ksmall = ∅, Kwide ∪ Klarge 6= ∅, Ktall = ∅. Similar to case 1 by interchanging wi with hi , and W with H. Case 3: Let Ksmall = ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge = ∅. There are three subcases to be considered: (i) Let w(Ktall ) ≤ W and h(Kwide ) ≤ H. The pieces of Ktall are packed side by side into the first bin; those of Kwide on top of each other into the second bin. (ii) Let w(Ktall ) > W . Sort the items of Ktall according to non-increasing height and pack them with the FFalgorithm. Due to Lemma 19 all items of Ktall are packed into two bins. Next, sort items of Kwide according to non-increasing width wi and pack them right-justified, from top to bottom into the second bin. There is no overlap since otherwise a contradiction to condition (2) arises (cf. Fig. 1): Let hk > 12 H be the height of an item of Ktall ,
Bin 1
Bin 2
Figure 1: Case 3: Packing of Ktall and Kwide which overlaps with an item of Kwide . Since item k is not packed into the first bin, we have A(Ktall ) > W hk . Because of overlap, h(Kwide ) > H − hk follows, and hence, A(Kwide ) > 12 W (H − hk ). Then, for the total area of items, we have A(K) = A(Ktall ∪ Kwide ) > W hk + 12 W (H − hk ) = 12 W (H + hk ) > 21 W (H + 12 H) = 43 W H.
10
Relations between capacity utilization, minimal bin size and bin number
(iii) Let h(Kwide ) > H. Similar to subcase (ii). Case 4: Let Ksmall = ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge 6= ∅. Two subcases can occur: (i) Let Klarge = {k}, ı. e. |Klarge | = 1. Pack item k bottom-left justified into the first bin. Then: (a) If w(Klarge ∪ Ktall ) ≤ W , then pack all items of Ktall besides item k. Since wk hk > 41 W H, we obtain A(Kwide ) < 1 W H because of condition (2). Since hi ≤ 12 H, all items of Kwide fit into the second bin 2 because of Steinberg’s theorem. (b) If w(Klarge ∪ Ktall ) > W , then, assuming that h(Klarge ∪ Kwide ) > H, we obtain a contradiction to condition (2) as follows: A(K) = A(Klarge ∪ Kwide ∪ Ktall ) > wk hk + 21 W (H − hk ) + (W − wk ) 12 H = wk (hk − 21 H) + wk 12 H + 21 W (H − hk ) + (W − wk ) 12 H > 21 W (hk − 21 H) + wk 21 H + 12 W (H − hk ) + (W − wk ) 21 H = 21 W (hk − 21 H + H − hk ) + (W − wk + wk ) 21 H = 12 W 12 H + W 12 H = 34 W H. Therefore, h(Klarge ∪ Kwide ) ≤ H, and all items of Kwide fit above item k. Since A(Ktall ) < 1 W H, all items of Ktall can be packed into the second bin because of Steinberg’s theorem. 2 (ii) Let Klarge = {j, k} (j 6= k), i. e. |Klarge | = 2. Because of A(j) > 41 W H and area condition (2), we have A({k} ∪ Kwide ) < 21 W H. Since wi > 21 H for i ∈ {k} ∪ Kwide , h({k} ∪ Kwide ) < H follows. Therefore, all these items can be packed on top of each other into the first bin. In a similar way, all items of {j} ∪ Ktall fit side by side within the second bin. Case 5: Let Ksmall 6= ∅, Kwide = ∅, Ktall = ∅, Klarge = ∅.
P Sort items of Ksmall according to non-increasing area. Let j := arg min { ji=1 wi hi ≥ P 1 W H}. Because of wi hi ≤ 14 W H for all i ∈ Ksmall , we have ji=1 wi hi < 21 W H. Due 4 to Steinberg’s theorem, all items of {1, . . . , j} can be packed into the first bin, and the remaining items into the second since their total area is (because of condition (2)) less than 12 W H. Case 6: Let Ksmall 6= ∅, Kwide = ∅, Ktall = ∅, Klarge 6= ∅. Pack one item of Klarge into the first bin. Then all other items fit within the second bin due to Theorem 5 since their total area is less than 21 W H. Case 7: Let Ksmall 6= ∅, Kwide = ∅, Ktall 6= ∅, Klarge = ∅. P Sort all items according to non-increasing area, and let j := arg max { ji=1 wi hi ≤ 21 W H}. Because of Lemma 9 (with q = 1, L = 12 W H), we have A({1, . . . , j}) ≥ 41 W H. Due to Steinberg’s theorem, all items of {1, . . . , j} can be packed into the first bin. Since the total area of non-packed items is at most 43 W H − 14 W H = 12 W H. Case 8: Let Ksmall 6= ∅, Kwide = ∅, Ktall 6= ∅, Klarge 6= ∅. Pack one item of Klarge into the first bin. Then all other items fit within the second bin due to Theorem 5 since their total area is less than 21 W H. Case 9: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall = ∅, Klarge = ∅. Similar to case 7.
11
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Case 10: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall = ∅, Klarge 6= ∅. Similar to case 8. Case 11: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge = ∅. We consider five subcases: (i) If w(Ktall ) ≤ W and A(Ktall ) ≥ 41 W H, then pack all items of Ktall into the first bin. All other items fit within the second bin because of Steinberg’s theorem since their total area is less than or equal to 12 W H and wi ≤ 12 W for all i. (ii) Let h(Kwide ) ≤ H, A(Kwide ) ≥ 41 W H. Similar to subcase (i) (iii) Let w(Ktall ) ≤ W , A(Ktall ) < 14 W H, h(Kwide ) ≤ H, A(Kwide ) < 14 W H. Let Iwide := Kwide and Itall := Ktall . We add successively all items of Ksmall to Iwide or Itall , respectively, such that their area does not exceed 21 W H. This approach works because of area condition (2) and since the area of a small item is not greater than 41 W H. We assume that we cannot add an item j to one of the sets without exceeding the area bound 21 W H. We obtain W H = 12 W H + 21 W H < A(Iwide ) + A(j) + A(Itall ) + A(j) ≤ 43 W H + A(j) ≤ 3 W H + 41 W H = W H, which is a contradiction. Consequently, all items can be added to 4 the sets Iwide and Itall and therefore all items can be packed into two bins due to Steinberg’s theorem. (iv) Let w(Ktall ) > W . Sort the items of Ktall according to non-increasing height hi and pack them with the FF algorithm side by side into two bins (cf. Lemma 19). Sort the items of Kwide according to non-increasing width wi and pack all of them right-justified from top to bottom also into the second bin. Similar to subcase (3) (ii) it can be shown that no overlap occur in bin two. We consider two subcases: (a) If A(B2 ) ≥ 14 W H, then all items, which are not packed into the second bin, fit within the first bin due to Steinberg’s theorem. (b) Let A(B2 ) < 41 W H. We define K tall := Ktall ∩ B2 and K wide := Kwide ∩ B2 . Then we have 14 W H > A(B2 ) > 1 W h(K wide ) 2
+ w(K tall ) 12 H. We obtain h(K wide )
W , we have A(Ktall ) > 12 W H, and therefore A(Ksmall ) < 14 W H. Next we show that all items of Ksmall can be packed additionally into the second bin: 2A(Ksmall ) < 2 · 14 W H = 21 W H H H − 21 w(K tall )H + w(K tall )2 W = 12 W H + w(K tall )H − 12 w(K tall )H − w(K tall )2 W H H ) − w(K tall )(H − ( 12 H + w(K tall ) W )) = (W − w(K tall ))( 12 H + w(K tall ) W 1 H = (W − w(K tall ))(H − ( 2 H − w(K tall ) W )) . H ))) − (2 · 21 W − (W − w(K tall )))(2 · 21 H − (H − ( 12 H − w(K tall ) W H )) − (2 · 21 W = (W − w(K tall ))(H − ( 12 H − w(K tall ) W H − (W − w(K tall )))+ (2 · 21 H − (H − ( 12 H − w(K tall ) W )))+ . Therefore, all items of Ksmall fit within the unused region in bin two of size (W −w(K tall ))× H (H − ( 12 H − w(K tall ) W )) (Fig. 2).
12
Relations between capacity utilization, minimal bin size and bin number
Bin 1
Bin 2
Figure 2: Case 11: Packing of Kwide , Ktall and Ksmall
(v) Subcase h(Kwide ) > H is similar to subcase (iv). Case 12: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge 6= ∅. We analyse again two subcases: (i) Let Klarge = {k}. Pack item k into the first bin. Further two subcases are considered: (a) If w(Klarge ∪ Ktall ) ≤ W , then all items of Ktall can also be placed into bin one besides item k. Since A(Kwide ∪ Ksmall ) < 21 W H these items fit within the second bin due to Steinberg’s theorem. (b) If w(Klarge ∪ Ktall ) > W , then similar to subcase (4)(i)(b) one can show that h(Klarge ∪ Kwide ) ≤ H holds true. Therefore, all items of Kwide ∪ Klarge fit within a bin, and all other within the second, similar to (a). (ii) Let Klarge = {j, k}. Pack items j and k bottom-left justified into the first, respectively the second bin. Similar to subcase (4)(ii) it can be shown that all items of Ktall can be placed besides item j in bin one, and that all items of Kwide fit above item k in bin two. We pack items of Kwide from top to bottom (top-justified). We define K small := {i ∈ Ksmall : wk + wi > W }, δk := wk − 21 W > 0 and θk := hk − 21 H > 0. We show that all items of K small fit between item k and items of Kwide within the second bin. Assuming h(K small ) + h(Kwide ) + hk > H we obtain a contradiction: A({k} ∪ K small ∪ Kwide ) > wk hk + (W − wk )(H − hk − h(Kwide )) + 12 W h(Kwide ) = wk hk + (W − wk )(H − hk ) + (wk − 21 W )h(Kwide ) = ( 21 W + δk )( 21 H + θk ) + ( 12 W − δk )( 21 H − θk ) + δk h(Kwide ) = 41 W H + 12 W θk + δk 21 H + δk θk + 41 W H − 12 W θk − δk 21 H + δk θk + δk h(Kwide ) = 12 W H + 2δk θk + δk h(Kwide ) > 12 W H. This contradicts area condition (2) since item j has area larger than 14 W H. Consequently, all items of K small can be packed between item k and items of Kwide into the second bin (Fig. 3). We obtain:
13
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
2A(Ksmall \K small ) = 2A(K) − 2A(B1 ) − 2A(B2 ) < 2 · 43 W H − 2A(j) − 2A(k) − 2A(Kwide ) < 32 W H − 21 W H − 2wk 12 H − 2 · 21 W h(Kwide ) < 32 W H − 21 W H − 2 · 12 W 12 H − 2δk 12 H − 2 · 21 W h(Kwide ) + 2δk h(Kwide ) = 12 W H − δk H − W h(Kwide ) + 2δk h(Kwide ) = ( 12 W − δk )(H − 2h(Kwide )) = ( 12 W − δk )(H − h(Kwide )) − ( 21 W − δk )(h(Kwide )) = ( 12 W − δk )(H − h(Kwide )) − ( 21 W − δk )+ (h(Kwide ))+ = ( 12 W − δk )(H − h(Kwide )) − (2 · ( 21 W − δk ) − ( 21 W − δk ))+ (2 · 12 H − (H − h(Kwide )))+ . Therefore all items of Ksmall \K small can be packed into the unused region of size ( 21 W − δk ) × (H − h(Kwide )) due to Steinberg’s theorem. W 2
j
k
Bin 1
Bin 2
Figure 3: Case 12: Packing of Klarge , Kwide , Ktall and Ksmall Theorem 21. Let L = L(K) be a list of rectangles with hmax ≤ H and wmax ≤ W . If A(L) ≤
k WH 4
for some k ≥ 3, k ∈ N,
then all items of L can be packed into at most k − 1 bins of size W × H. The statement is tight. Proof: The proof is done by reducing the general case to the special case of Theorem 20. Therefore, subsets of items with total area at least 41 W H are separated and packed into a single bin. We use again index sets Ksmall , Kwide , Ktall and Klarge as in proof of Theorem 20. Since wi hi > 12 W · 12 H = 41 W H for all items of Klarge , each item of Klarge can be packed into a single bin. In case of A(Kwide ) ≥ 14 W H, we obtain a desired reduction by the FF1 -algorithm (FFalgorithm applied to a single bin) as follows: If h(Kwide ) ≤ H, then all items fit in a single bin. Otherwise, the height obtained by the FF1 -algorithm when packing items of Kwide is at least 12 H since hi ≤ 12 H for all i ∈ Kwide so that the total area used is greater than 1 W H, and a further reduction can be made. 4 In a similar way, in case of A(Ktall ) ≥ 14 W H bin packings with area utilization at least 1 W H can be found. 4
14
Relations between capacity utilization, minimal bin size and bin number
Also in case of A(Ksmall ) ≥ 14 W H bin packings with area utilization at least 14 W H can be obtained by choosing subsets with total area not smaller than 14 W H but not larger than 1 W H (Theorem of Steinberg). 2 After that reduction process some items can remain in Kwide , Ktall and Ksmall but their total area is in each case smaller than 41 W H, and therefore is less than 34 W H. Hereby we proved that the general case can be reduced to that of Theorem 20. According to this theorem, two bins are needed for the remained items, and hence, in total at most k − 1 bins for all items. In order to show that the statement is tight, first we note that it is not fulfilled for k = 2 since items of size ε × H and W × ε cannot be packed into the same bin for ε > 0, and if ε is sufficient small, then their total area is less than 21 W H. Furthermore, for k ≥ 3, k − 1 items of size ( 21 W + ε) × ( 12 H + ε) with ε > 0 cannot be packed into k − 2 bins of size W × H. But, if ε is sufficient small the area condition is fulfilled, which shows that k − 1 is the optimal number of bins. Finally, for k ≥ 3, an instance of k items of size ( 21 W + ε) × ( 21 H + ε) with some sufficient small ε > 0 shows, that k4 W H is the maximum area bound for which all instances can be packed into k − 1 bins. In the following we are interested in statements how many bins are needed to pack all items of a list. Theorem 22. Let A(K) ≤ W H, wi ≤ W/3 and hi ≤ H/3 for all i ∈ K, i. e. p = q = 3. Then all items of L can be packed into at most two bins. Proof: Because of Corollary FFDH-heuristic for SPP packs all items of K with Pse8, the e total height not larger than j=1 hj ≤ 53 H. On the other hand, using the FFDH1-heuristic to fill a single bin covers at least 43 H of height because of Corollary 10. Therefore, at most two bins are needed to pack all items of K. Remark: A proof of Theorem 22, which is based on the NFDH-heuristic is given in [HS12]. In case of p = q = 3, Lemma 6 yields the NFDH-bound 11 H in contrast to the 6 FFDH-bound 53 H, so that a NFDH-based proof needs much more effort. Theorem 22 suggests the following Conjecture 23. Consider bins of size W × H and a list L of rectangular items with wmax ≤ W/2 and hmax ≤ H/2. If A(L) ≤ W H, then all items of L can be packed into at most two bins. In case of somewhat smaller items, we have: Theorem 24. Consider bins of size W × H and a list L of rectangular items. If wmax ≤ W/2,
hmax ≤ H/3,
(i. e. p = 2 and q = 3),
then all items of L can be packed into at most two bins.
A(K) ≤ W H,
15
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Proof: Without loss of generality, we may assume that A(K) = W H. We consider the following partition of the set of items: 1 1 K1 := {i ∈ K : wi ≤ W } and K2 := {i ∈ K : wi > W }. 3 3 Case 1: Let A(K1 ) ≤ 12 W H and A(K2 ) ≤ 12 W H. Because of Steinberg’s theorem, item set K1 as well as K2 can be packed into a bin so that at most two bins are needed to pack K. Case 2: Let A(K1 ) ≤ 12 W H and A(K2 ) > 21 W H. Pack items of K2 according to the FFDH1-heuristic into the first bin. Here, and in the following, let B1 denote those items packed into the first bin (either with FFDH1 , FFDW1 W for all items of K2 , from Corollary 14 follows or somehow else). Since wi > 13 W = q+1 2pq f2 ⊆ K2 with A(K f2 ) ≥ 1 W H, which can be packed into the that there exists a subset K 2 first bin. Because A(K\B1 ) = W H − A(B1 ) ≤ 12 W H, these items can be packed into the second bin using Steinberg’s theorem. Case 3: Let A(K1 ) > 21 W H and A(K2 ) ≤ 12 W H. Pack items of K1 according to the FFDH1-heuristic into the first bin,, now B1 ⊆ K1 , and let s denote again the number of (horizontal) strips. In case of A(B1 ) ≥ 12 W H, all remaining items can be packed into the second bin due to Steinberg’s theorem. In the following, let A(B1 ) < 12 W H and hi < Theorem 11 applied on K1 , i. e. p = q = 3.
p+1 H 2pq
=
2 H, 9
∀i ∈ K1 \B1 according to
e of K1 Suppose, the last condition is not fulfilled, then, because of Corollary 13, a subset K 1 e ≥ W H, which can be packed into a bin. Hence, all items of K would exist with A(K) 2 could be packed into two bins. 2 H 9
for all items
w ej+1 − ∆s w es+1 > 34 W (H − w e1 ) − ∆s w es+1
(3)
Because of sorting according to the FFDH1 -heuristic, we have hi < i ∈ K1 \B1 . With formula (1) and Theorem 11 we obtain A(B1 ) ≥ >
p W p+1
s P
j=1 3 2 W H − 4 3
( 43 W
−
2 W ) 92 H 3
=
1 WH 2
−
1 W H. 54
Now, sort the items of K2 according to non-increasing height, and pack them left-justified on top of the previous, as long as possible, until the bin height is reached, into the second bin (Fig. 4, left). Let H := h(K2 ). We distinguish three subcases: Subcase (i) Let H > H. Because of Corollary 10 the total height occupied in the second bin is at least 34 H. Since e := H − (H − 3 H) = 7H − H. A(K2 ) ≤ 12 W H and wi > 31 W , we have h(K2 ) < 23 H. Let H 4 4 Pack the remaining, non-packed items of K2 right-justified, starting at the bottom of the bin on top of each other into the second bin. Above these items it remains an unused e (Fig. 4, right). Moreover, A(K2 ) > 1 W h(K2 ) = 1 W H. rectangular area with size 21 W × H 3 3 We show that the items of K1 \B1 can be packed into the unused area of the second bin. For the total area of the non-packed items of K1 , we have:
A(K1 \B1 ) = A(K) − A(B1 ) − A(K2 ) < W H − ( 12 W H − W54H ) − W3 H = 21 W H + W54H − W3 H.
16
Relations between capacity utilization, minimal bin size and bin number
Figure 4: Packing of many items of K2 Therefore, 2A(K1 \B1 ) < 2( 21 W H + W54H − W3 H) = W H + W27H − 2W H 3 7W H 35 W 7H WH WH W W W = 8 + 27 − 6 H + 8 − 2 H = 216 W H − 6 H + 2 ( 4 − H). Two subcases have to be investigated: (a) Let 94 H − ( 47 H − H) ≤ 0, then 35 2A(K1 \B1 ) < 216 W H − W6 H + W2 ( 7H − H) 4 35 W W 7H = 216 W H − 6 H + 2 ( 4 − H) − (2wmax (K1 \B1 ) − W2 )+ ( 49 H − ( 47 H − H))+ < W2 ( 7H − H) − (2wmax (K1 \B1 ) − W2 )+ (2hmax (K1 \B1 ) − ( 47 H − H))+ 4 e − (2wmax (K1 \B1 ) − W )+ (2hmax (K1 \B1 ) − H) e +. = W2 H 2
Consequently, due to Steinberg’s theorem, the items of K1 \B1 can all be packed into a e rectangle of size 21 W × H. (b) Let 94 H − ( 74 H − H) > 0, then
35 2A(K1 \B1 ) < 216 W H − W6 H + W2 ( 7H − H) 4 35 W 7H W W = 216 W H − 6 H + 2 ( 4 − H) + 6 (2hmax (K1 \B1 ) − ( 47 H − H))+ − W6 (2hmax (K1 \B1 ) − ( 74 H − H))+ 35 ≤ 216 W H − W6 H + W2 ( 7H − H) + W6 (2 · 29 H − ( 47 H − H))+ − W6 (2hmax (K1 \B1 ) − ( 74 H − H))+ 4 35 = 216 W H − W6 H + W2 ( 7H − H) + W6 ( 49 H − 47 H + H) − W6 (2hmax (K1 \B1 ) − ( 47 H − H))+ 4 35 = 216 W H + W2 ( 7H − H) + W6 · −47 H − W6 (2hmax (K1 \B1 ) − ( 47 H − H))+ 4 36 < W2 ( 7H − H) − W6 (2hmax (K1 \B1 ) − ( 74 H − H))+ 4 W 7H = 2 ( 4 − H) − (2 · 13 W − W2 )+ (2hmax (K1 \B1 ) − ( 74 H − H))+ ≤ W2 ( 7H − H) − (2wmax (K1 \B1 ) − W2 )+ (2hmax (K1 \B1 ) − ( 74 H − H))+ 4 e − (2wmax (K1 \B1 ) − W )+ (2hmax (K1 \B1 ) − H) e +. = W2 H 2
Again, due to Steinberg’s theorem, the items of K1 \B1 can all be packed into a rectangle e of size 21 W × H.
Subcase (ii) Let H ≤ H and A(K2 ) ≥
WH 4
+
WH . 54
Because of H ≤ H all items of K2 are already packed on the left side of the second bin. For the total area of the not yet packed items of K1 it holds: A(K1 \B1 ) = A(K) − A(B1 ) − A(K2 ) ≤ W H − ( 21 W H − W54H ) − ( W4H + W54H ) = W4H . Because of wi ≤ 12 W ∀i ∈ K2 there exists a non-occupied area of size 12 W ×H in the second bin (cf. Fig. 4, left). Due to Steinberg’s theorem, all items of K1 \B1 can be packed into that rectangle.
17
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Subcase (iii) Let H ≤ H and A(K2 )
W H − We consider a partition of K1 into K1 := {i ∈ K1 : hi ≥
25 H} 81
and the following two subcases: f1 ) ≥ 1 W H. (a) Let A(K
WH . 54 ( W4H + W54H )
+
= 34 W H −
1 W H. 54
f1 := {i ∈ K1 : hi < and K
25 H}. 81
2
f1 using the FFDH1-heuristic Empty the first bin (i. e. set B1 := ∅) and pack items of K f1 ). In case not all items of K f1 are packed into the first into the first bin (now ∅ = 6 B1 ⊆ K bin, we obtain due to formula (1) and Theorem 7, P p A(B1 ) ≥ p+1 ej+1 − ∆s w es+1 > 34 W (H − w W sj=1 w e1 ) − ∆s w es+1 56 3 2 2 14 1 1 3 > 4 W 81 H − ( 4 W − 3 W ) 9 H = 27 W H − 54 W H = 2 W H.
Since A(B1 ) ≥ 12 W H all items of K\B1 can be packed into the second bin due to Steinberg’s theorem. f1 ) < 1 W H, (b) Let A(K 2 f1 ) > W H −( W H + W H )− 1 W H = W H − W H = 25 W H. then A(K1 ) = A(K)−A(K2 )−A(K 4 54 2 4 54 108 Let M = {1, . . . , m} := K1 \B1 (with B1 as defined at the beginning of Case 3) be sorted according to non-increasing item area. Because of Theorem 11, we have wi hi < 13 W 92 H = 2 W H for all i ∈ M. Furthermore, let B1 := B1 = K1 \M. Because of Theorem 11, all 27 items of K1 are packed in the first bin, i. e. K1 ⊆ B1 . Several subcases can occur: (1) Let w(K1 ) ≤ W . P 1 Let j ∈ M denote the smallest index with ji=1 wi hi ≥ 54 W H. Let M := {1, . . . , j}. P 2 Moreover, because of the assumed sorting in M, we have A(M) = ji=1 wi hi < 27 W H. Next we show that all items of M can also be packed into the first bin besides those of 1 1 W H + 54 W H = 21 W H, and consequently, the B1 . In that case, A(B1 ∪ M ) ≥ 21 W H − 54 remaining items can be packed into the second bin due to Steinberg’s theorem. In order to show this we distinguish two subcases: 1 (A) Let A(B1 ) ≤ 21 W H − 108 W H. Empty the first bin (i. e. B1 = ∅) and pack all items of K1 along the bottom side of the bin, i. e. B1 := K1 . Above these items a non-used rectangular area of size W × 23 H (Fig. 5, left) remains. It holds,
A(B1 \K1 ) + A(M) ≤ 12 W H −
1 WH 108
−
25 WH 108
+
2 WH 27
=
54−1−25+8 WH 108
= 13 W H.
Hence, all items of (B1 \K1 ) ∪ M can be packed into a rectangle of size W × 32 H . 1 (B) Let A(B1 ) > 21 W H − 108 W H. WH WH If A(K2 ) ≥ 4 + 108 , then a packing of K into two bins can be constructed similar to subcase (ii). Otherwise,
f1 ) > W H − ( W H + A(K1 ) = A(K) − A(K2 ) − A(K 4
WH ) 108
− 21 W H =
WH 4
−
Analogously to subcase (A), we obtain A(B1 \K1 ) + A(M) ≤ 12 W H −
26 WH 108
+
2 WH 27
=
54−26+8 WH 108
= 31 W H.
WH 108
=
26 W H. 108
18
Relations between capacity utilization, minimal bin size and bin number
k
Figure 5: Packing of items of K1 Therefore, also in this subcase, the items of M ∪ (B1 \K1 ) can be packed into a rectangle of size W × 32 H. (2) Let w(K1) > W . Sort the items of K1 according to non-increasing width and pack them left-justified side c1 denote by side along the bottom of the bin until the next item does not fit. Let K 3 c the set of packed items. Because of Corollary 10, we have w(K1 ) ≥ 4 W , and therefore c1 ) ≥ 25 W . Two cases are differentiated: A(K 108 c1 ) ≥ 7 W , (A) Let w(K 9 c1 ) ≥ 7 W 25 H = then A(K 9 81
175 W H. 729
M1 := {i ∈ M : wi hi ≤
We consider the following partition of M:
107 107 W H} and M2 := {i ∈ M : wi hi > W H}. 1458 1458
Two subcases are considered: 1 (α) Let A(M1 ) ≥ 54 W H. Let M1 = {1, . . . , m1 } beP sorted according to non-increasing item area. Let j1 ∈ M1 denote 1 1 the smallest index with ji=1 wi hi ≥ 54 W H. Let M1 := {1, . . . , j1 }. Moreover, because of P1 107 W H. Analogously to the the assumed sorting in M1 , we have A(M1 ) = ji=1 wi hi ≤ 1458 c previous case we show that all items of (B1 \K1 ) ∪ M1 can be packed into the unused area of size W × 32 H. For their total area, we have
c1 ) + A(M1 ) ≤ 1 W H − A(B1 \K 2
175 WH 729
+
107 WH 1458
=
729−350+107 WH 1458
=
486 WH 1458
= 13 W H.
Hence, A(B1 ) = A(B1 ∪ M1 ) ≥ 21 W H, and all remaining items can be packed into the second bin. 1 (β) Let A(M1 ) < 54 W H, then A(M2 ) = A(K) − A(K2 ) − A(B1 ) − A(M1 ) > W H − ( W4H + W54H ) − W2H − W54H 23 8 = 108−27−2−54−2 W H = 108 W H. Because of this and wi hi ≤ 31 W 29 H = 108 W H for all 108 i ∈ M2 , the set M2 contains at least three items. Pack three items of M2 side by side on c2 . Because of hi < 2 H for the top of the first bin (Fig. 6) and denote these items with M 9 all i ∈ M2 there exists an unused area of size W × 94 H. We show that there exists a set c1 ∪ M c2 ) with A(K c1 ∪ M c2 ∪ Kˇ1 ) ≥ 1 W H that can be packed into this unused Kˇ1 ⊆ K1 \(K 2 f = {1, . . . , m} c1 ∪ M c2 ). So we have A(M f) = A(K1 \(K c1 ∪ M c2 )) = area. Let M e := K1 \(K c1 ) − A(M c2 ) ≥ ( 3 − 1 − 1 − 2 )W H = 81−2−36−24 W H = 19 W H > 1 W H. A(K1 ) − A(K 4 54 3 9 108 108 9
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
19
f denote the smallest index with Pej wi hi ≥ 1 W H. Because of the sorting, we Let e j∈M i=1 9 Pej 2 obtain i=1 wi hi ≤ 9 W H. Hence, these items can be packed in the unused area and we 107 833 get A(B1 ) ≥ 175 W H + 3 · 1458 W H + 19 W H = 350+321+162 W H = 1458 W H > 12 W H. Due to 729 1458 Steinberg’s theorem all remaining items can be packed into the second bin.
c1 and M c2 Figure 6: Packing of items of K
c1 ) < 7 W . (B) Let w(K 9 c1 , wi > 2 W ∀i ∈ K c1 holds since otherwise a further item of Because of construction of K 9 c1 . K1 could be packed besides the items of K Empty again the first bin (i. e. B1 = ∅), and pack items of K1 using the FFDW1 -heuristic, into the bin. If A(B1 ) ≥ 12 W H, then all remaining items can be packed into the second bin. c1 are packed, i. e. K c1 ⊆ B1 . If A(B1 ) < 21 W H, then, because of Theorem 11, all items of K Let k denote the area-largest item of K1 \B1 . Because of Corollary 12 and the sorting of items with respect to non-increasing width one has wk < 29 W . We consider two subcases: 1 (α) Let wk hk ≥ 54 W H. Empty the first bin and pack items of K1 \{k} using the FFDH1 -heuristic into that bin. f1 ⊆ K1 \{k} denote those items now packed into the first bin. If A(B f1 ) ≥ 1 W H, then Let B 2 1 1 1 f f we are done. Otherwise, if A(B1 ) < 2 W H, then A(B1 ) ≥ 2 W H − 54 W H holds because of c1 are packed in the first inequality (3). Furthermore, because of Theorem 11, all items of K c1 ⊆ B f1 . We show that the items B f1 ∪ {k} can be packed into one bin. Empty bin, i. e. K c1 along the bottom side of bin one. Since the again the first bin and pack anew items of K 7 c1 . Then, for the total area total width is less than 9 W , item k fits besides the items of K f1 , we have of not yet packed items of B c1 ) ≤ 1 W H − 25 W H = 29 W H < 1 W H. f1 \K A(B 2
108
108
3
Therefore, these items can be packed into the unused area of size W × 23 H (Fig. 5, right). f1 ) + wk hk ≥ 1 W H. Because of inequality (3), the total area of this packing is A(B 2 Consequently, all items not packed into the first bin can be packed into the second due to Steinberg’s theorem.
1 (β) Let wk hk < 54 W H. Let again M = {1, . . . , m} := K1 \B1 be sorted according to non-increasing item area. Let Pj 1 W H. Let M := {1, . . . , j}. We j ∈ M denote the smallest index with i=1 wi hi ≥ 54 1 1 obtain 54 W H ≤ A(M) ≤ 27 W H. f1 denote Empty the first bin and pack items of K1 \M using the FFDH1 -heuristic. Let B
20
Relations between capacity utilization, minimal bin size and bin number
f1 ) ≥ 1 W H − 1 W H. If the set of items, which are packed now. Then we have A(B 2 54 f1 ) < 1 W H, empty again the first bin and pack items of K c1 side by side along the A(B 2 bottom of the first bin. Then f1 \K c1 ) + A(M ) ≤ 1 W H − 25 W H + 1 W H = 54−25+4 W H = 33 W H < 1 W H A(B 2 108 27 108 108 3 Hence, all these items can be packed into the first bin, and we have f1 ∪ M ) ≥ 1 W H. A(B1 ) = A(B 2 Therefore, all remaining non yet packed items fit in the second bin. Case 4: Let A(K1 ) > 21 W H and A(K2 ) > 12 W H.
Because of A(K) = A(K1 ) + A(K2 ) > 21 W H + 21 W H = W H W H is violated. Consequently, this case cannot occur.
the area condition A(K) ≤
Corollary 25. Consider bins of size W × H and a list L of rectangular items. If wmax ≤ W/3,
hmax ≤ H/2,
(i. e. p = 3 and q = 2),
A(K) ≤ W H,
then all items of L can be packed into at most two bins. Proof: The statement follows from the previous theorem applied on an instance with switched wi and hi , respectively W and H.
5
Conclusions and outlook
In this paper we showed that any list of rectangles can be packed into at most k − 1 bins if the total area of items does not exceed k/4-times the bin area for an integer k ≥ 3, which imprves the Lemma of Bougeret et al. [BDJ+ 09]. This result is tight. Furthermore, we investigated additional conditions to strengthen this statement. Especially, we showed that only two bins are necessary to pack all items if their total area is at most the bin area and if the height of each item is not larger than one-third of the bin height and if the width of every item does not exceed one-half of the bin width. It remains a lack between these two statements, which will be part of future research.
References [BDJ+ 09] M. Bougeret, P. F. Dutot, K. Jansen, C. Otte, and D. Trystram. Approximation algorithm for multiple strip packing. IT University of Copenhagen, Denmark, 2009. 7th WAOA. [BKRS09] G. Belov, V. Kartak, H. Rohling, and G. Scheithauer. One-dimensional relaxations and lp bounds for orthogonal packing (pages. International Transactions in Operational Research, 16:745–766, 2009. [BS12]
T. Buchwald and G. Scheithauer. A note on relations between capacity utilization and ” minimal bin number“. Preprint MATH-NM-05-2012, Technische Universit¨ at Dresden, 2012.
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
21
[CCM07] J. Carlier, F. Clautiaux, and A. Moukrim. New reduction procedures and lower bounds for the two-dimensional bin packing problem with fixed orientation. Computers & Oper. Res., 34:2223–2250, 2007. [CGJT80] E. G. Coffman, M. R. Garey, D. S. Johnson, and R. E. Tarjan. Performance bounds for level–oriented two–dimensional packing algorithms. SIAM J. on Computing, 9(4):808– 826, 1980. [DF92]
H. Dyckhoff and U. Finke. Cutting and packing in production and distribution. Physica Verlag, Heidelberg, 1992.
[FS04]
S. P. Fekete and J. Schepers. A general framework for bounds for higher-dimensional orthogonal packing problems. Math. Methods of Operations Research (MMOR), 60(2):311–329, 2004.
[GJ79]
M. R. Garey and D. S. Johnson. Computers and Intractability – A Guide to the Theory of NP-Completeness. Freeman, San Francisco, 1979.
[HS12]
K. Hoffmann and G. Scheithauer. Relations between capacity utilization and minimal bin number. Preprint MATH-NM-03-2012, Technische Universit¨ at Dresden, 2012.
[JZ07]
K. Jansen and G. Zhang. Maximizing the total profit of rectangles packed into a rectangle. Algorithmica, 47:323–342, 2007.
[LC90]
K. Li and K. H. Cheng. On three-dimensional packing. SIAM J. on Computing, 19(5):847–867, 1990.
[LMM02] A. Lodi, S. Martello, and M. Monaci. Two-dimensional packing problems: A survey. European Journal of Operational Research, 141(2):241–252, 2002. [MM68]
A. Meir and L. Moser. On packing of squares and cubes. J. of Combinatorial Theory, 5:126–134, 1968.
[MV98]
S. Martello and D. Vigo. Exact solution of the two-dimensional finite bin-packing problem. Management Science, 44:388–399, 1998.
[Ste97]
A. Steinberg. A strip-packing algorithm with absolute performance bound 2. SIAM J. on Computing, 26(2):401–409, 1997.
Technische Universit¨at Dresden Herausgeber: Der Rektor
Relations between capacity utilization, minimal bin size and bin number Torsten Buchwald,
Kirsten Hoffmann, Guntram Scheithauer
MATH-NM-03-2013 May 2013
Contents 1 Introduction: problem formulation and goal
1
2 Related work
2
3 The FFDH1 - and FFDW1 -heuristic
4
4 The Bin Packing Problem
7
5 Conclusions and outlook
20
Relations between capacity utilization, minimal bin size and bin number Torsten Buchwald, Kirsten Hoffmann, Guntram Scheithauer Technische Universit¨at Dresden May 17, 2013 Abstract We consider the two-dimensional bin packing problem (BPP): given a set of rectangular items, find the minimal number of rectangular bins needed to pack all items. Rotation of the items is not permitted. We show for any integer k ≥ 3 that at most k − 1 bins are needed to pack all items if every item fits into a bin and if the total area of items does not exceed k/4-times the bin area. Moreover, this bound is tight. Furthermore, we show that only two bins are necessary to pack all items if the total area of items is not larger than the bin area, and if the height of each item is not larger than a third bin height and the width of every item does not exceed a half bin width.
Key words: Cutting and Packing, Bin Packing Problem, Strip Packing Problem
1
Introduction: problem formulation and goal
In this paper, we consider the two-dimensional bin packing problem (BPP) with nonrotatable rectangular items: Let L := (R1 , . . . , Rn ) denote a list of small rectangles Ri (items) of width wi ≤ W and height hi ≤ H, i ∈ K := {1, . . . , n}. The items have to be packed into a minimum number of rectangular bins of given width W and height H such that the items do not overlap each other, and each of them fits completely in one of the bins. Furthermore, the strip packing problem (SPP) is of interest where the height H of the bin is variable and has to be minimized. Both, BPP and SPP, are known to be NP-hard optimization problems (cf. [GJ79]). Because of their high applicability to practical problems, a lot of work is done in this field concerning heuristic approaches (leading to upper bounds of the optimal value), lower bounds and exact solution methods (cf. [DF92], [LMM02]). In most cases the solvability of the problems is out of question if natural demands are met (each item fits into a bin). For another type of packing problems, so-called orthogonal (decision) packing problem (OPP), the verification whether a feasible packing pattern of all items in a single bin exists 1
2
Relations between capacity utilization, minimal bin size and bin number
or not, is the main goal. Several algorithmic approaches have been published within the last 10 years (cf. [FS04], [CCM07], [BKRS09]). But, to the best of our knowledge, there exists only few work concerning theoretical aspects. In this paper, we improve known results and contribute in particular the statement that any set of rectangular items (each of them fits into a single bin) can be packed into at most k − 1 bins if the total area of items is not larger than k/4-times of bin area, for an integer k ≥ 3. In case that the total area of items does not exceed the bin area, we also show that only two bins are needed if some weak conditions are fulfilled. The next section reviews related work and motivates our investigations. Section 3 contains some results concerning SPP, whereas the BPP is considered in Section 4. In the last section we give some conclusions and an outlook of future research. In this paper, we consider various packing problems with rectangular items to be packed into one or a minimum number of rectangular containing regions. In any case, a list L = (R1 , . . . , Rn ) of rectangular items is given where rectangle Ri has width wi and height hi , i ∈ K := {1, . . . , n}. Rotation is not permitted. In the Strip Packing Problem (SPP) all items have to be packed into a single strip of given width W and minimum height H, whereas in the Bin Packing Problem (BPP) the goal is to find the minimum number of bins of size W × H needed to pack all items. Throughout this paper, for a list L = (R1 , . . . , Rn ) of rectangles, also represented by indexset K, let A(L) = A(K) denote the total area, i. e. X A(L) = A(K) := wi hi , i∈K
Abin := W H the bin area, and OP T (L) the optimal value of BPP (minimum number of bins) or SPP (minimal height), respectively. Furthermore, let X X e := e := e ⊆ K, w(K) wi , h(K) hi , ∀ K e i∈K
e := max{wi : i ∈ K}, e wmax (K)
e i∈K
e := max{hi : i ∈ K}, e hmax (K)
wmax := wmax (K),
hmax := hmax (K).
e ⊆ K, ∀K
We always assume that wi ≤ W/p and hi ≤ H/q for all i ∈ K holds true with some p ∈ [1, W ] ∩ N and q ∈ [1, H] ∩ N (N = {1, 2, . . . }). If not stated otherwise, p = q = 1. As usual, let (x)+ := max{0, x}.
2
Related work
To the best of our knowledge, the smallest known number of bins needed to pack any list L of items with A(L) ≤ Abin is 4. This number results from a theorem of Martello and Vigo [MV98] for the two-dimensional bin packing problem. Theorem 1 (Martello/Vigo (1998)). For any list L of rectangles with total area A(L) and bins of size W × H, we have A(L) OP T (L) ≤ 4 . WH
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
3
As a consequence of Theorem 1, at most four bins are needed to pack all items having total area not larger than the bin area. Some other known results suggest that without any or only with weak assumptions the packability of a list L of rectangular items into a single bin can be guaranteed if A(L) is not too large but near to 50 %. In 1968, Meir and Moser [MM68] published the following result: Theorem 2 (Meir/Moser (1968)). Any list L of rectangles with wi ≥ hi for all i ∈ K and total area A(L) can be packed into any rectangle of size W × H if W ≥ wmax and Abin = W H ≥ 2A(L) + W 2 /8. This result is best possible. As consequence of Theorem 2, we obtain the following corollary, e. g. used by Li and Cheng [LC90]. Corollary 3 (Meir/Moser (1968)). A list L of rectangles with wi ≥ hi for all i ∈ K can be packed into a unit square if wmax ≤ 1 and A(L) ≤ 7/16. A very useful result on packability of a list of rectangles was given in 1997 by Steinberg [Ste97]: Theorem 4 (Steinberg (1997)). If for a list L of rectangles and a bin of size W × H the following inequalities hold, wmax ≤ W,
hmax ≤ H,
2A(L) ≤ Abin − (2wmax − W )+ · (2hmax − H)+ ,
then it is possible to pack all items into the bin. Note that in case of wmax ≤ W/2 or hmax ≤ H/2 all items can be packed if their total area is at most 50% of the bin area. Consequently, if A(L) ≤ W H, then all items fit in a rectangular bin of size 2W × H, or W × 2H, but we cannot conclude that all items can be packed into two bins of size W × H, since the guillotine property is not guaranteed. For instance, three items of size ( 12 W + ε) × ( 21 H + ε) cannot be packed into two bins if ε > 0. Jansen and Zhang [JZ07] improved the result of Steinberg for the special case that there are no wide items in K, i. e. {i ∈ K : wi > W/2, hi ≤ H/2} = ∅. Theorem 5 (Jansen and Zhang (2007)). If A(K) ≤ 21 W H, and if the set K does not contain wide items, then the items of K can be packed in a single bin. Note that the theorem also holds true if K contains a large item (one item i with wi > W/2 and hi > H/2).
4
3
Relations between capacity utilization, minimal bin size and bin number
The FFDH1 - and FFDW1-heuristic
In this section we review some known results and give some new for the well known NFDH(Next-Fit Decreasing Height)- and FFDH-(First-Fit, Decreasing Height)-heuristic for SPP and for their modifications FFDH1- and FFDW1 -heuristic, which result when the FFDHheuristic is applied to a single bin. Within this section, we always assume that a strip of fixed width W is given and a list L = L of rectangular items with wmax ≤ W/p and hmax ≤ H/q where p ∈ [1, W ] ∩ N, q ∈ [1, H] ∩ N. Lemma 6. If p ≥ 2 and the total area A(L) of items does not exceed the bin area, i. e. A(L) ≤ W H, then the NFDH-heuristic yields a strip packing of height not larger than p 1 H. + p−1 q A proof of this lemma can be found in [HS12]. A similar result with stronger bound holds for the FFDH-heuristic. Let e hj , j = 1, 2, . . . , se denote the heights of the (horizontal) strips obtained by the FFDHheuristic applied for SPP with strip width W . Because of decreasing heights of items, we have e hj ≥ e hj+1 for j = 1, 2, . . . , se − 1. Furthermore, let e hse+1 := 0. Theorem 7. Let p ∈ [1, W ] ∩ N, and let wi ≤ W/p for i ∈ K. Then the following relation holds for the total area of items and strip height obtained by the FFDH-heuristic: s e X p e W hj+1. A(K) ≥ p + 1 j=1
Its proof is part of the proof of Theorem 3 in [CGJT80], and can also be found in [BS12]. Corollary 8. As a consequence of this theorem, the total height obtained by the FFDHheuristic for SPP to pack a set K of items with A(K) ≤ W H is not greater than p+1 1 H. + p q Proof: Using Theorem 7 we obtain: Pse e p+1 p+1 e j=1 hj ≤ pW A(K) + h1 ≤ pW W H +
H q
=
p+1 p
+
1 q
H.
In the following theorems, we apply, in general, the well-known FFDH-heuristic in order to pack a single bin, denoted as FFDH1-heuristic, which works as follows: Let again e hj , j = 1, 2, . . . , se denote the heights of the (horizontal) strips obtained by the FFDH-heuristic applied for SPP with strip width P W . In case that not all items can be packed into Psa single s e e bin by the FFDH-heuristic, then j=1 hj > H follows. We define s := arg max { j=1 e hj ≤ 1 1 H} as the number of strips packed by the FFDH -heuristic. The FFDH -heuristic stops just before the first item has to be packed into the (s + 1)st strip. Let K denote the item set packed by the FFDH1 -heuristic.
5
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Additionally to the FFDH1-heuristic, we also consider the First-Fit Decreasing-Width1heuristic (FFDW1 , transposed“ FFDH1-heuristic) where all items are sorted by non” increasing width, and are then packed sequentially in the left-most vertical strip in which they fit until the first item cannot be packed into the bin. Pn Lemma P 9. Let q ∈ N, L/q ≥ ℓ1 ≥ ℓ2 ≥ · · · ≥ ℓn > 0 and i=1 ℓi > L. Let s := s arg max { i=1 ℓi ≤ L}. Then s X q L≤ ℓj ≤ L. q+1 j=1 Ps q Proof: Let us assume j=1 ℓj < q+1 L. Then ℓs+1 > L/(q + 1). On the other hand, because of P assumption, s ≥ q and due to the non-increasing sorting of ℓi -values, we have P s s q L j=1 q+1 ≥ q+1 L, which leads to a contradiction. j=1 ℓj > P Because of the non-increasing strip heights e hj , j = 1, . . . , s with s := arg max { sj=1 e hj ≤ 1 H}, generated by the FFDH -heuristic, from Lemma 9 follows: Corollary 10. If not all items of K are packed by the FFDH1 -heuristic, then we have for the total height used: s X q e H≤ hj ≤ H. q+1 j=1 A similarPresult holds for the FFDW1-heuristic: Let w ej , j = 1, . . . , s with s := s arg max { j=1 w ej ≤ W } be the widths of the vertical strips obtained by the FFDW1heuristic. In case not all items are packed, we have for the total width used: X p W ≤ w ej ≤ W. p+1 j=1 s
In the FFDH1-heuristic we need to consider regular and non-regular items. If item r is packed into strip j and strip j is the highest non-empty strip at that time when item r is packed, then r is called a regular item, otherwise a non-regular or fallback item. Let Kij , j = 1, . . . , i, i = 1, . . . , s, denote the set of items packed into strip j at the time before strip i + 1 has to be opened (or all items are packed), i. e. when all regular items have been packed into strip i, andS probably some fallback items have been placed into strips below strip i. Hence, K = sj=1 Ksj . In case of K 6= K the first item not packed into the bin, is denoted by k0 . Corresponding to Kij , j = 1, . . . , i, i = eij := w(Kij ) = P P1, . . . , s, we define the used width w j∈Kij wj and area Aij := A(Kij ) = j∈Kij wj hj . Furthermore, let ∆i := max{0,
p W −w eii }, p+1
i = 1, . . . , s.
Theorem 11. Let A(K) > 21 W H. We assume that not all items are packed by the FFDH1heuristic, i. e. K 6= K, and k0 is the first item not packed. If there does not exist any subset e ⊆ K ∪ {k0 } with A(K) e ≥ 1 W H, that can be packed into the bin, then K 2 hk0
0 we distinguish two subcases:
1 W. (i) Strip i does not contain any item k ∈ Kii with width wk < p+1 1 Since the width of the first (regular) item in strip i is at least p+1 W +∆i−1 , and since at least p regular items are packed into strip i, we have w eii ≥ p W + ∆i−1 . Since inequality (1) Pi Pi−1 p+1 p is valid for i − 1 we obtain eij ≥ w eii + j=1 w ei−1,j ≥ p+1 W · i + ∆i−1 − ∆i−1 . j=1 w Consequently, formula (1) is also fulfilled for i.
1 W. (ii) Strip i contains a regular item t0 with width wt0 < p+1 This can only happen when further fallback items with total width larger than ∆i−1 are packed into strip i − 1 before item t0 is packed into strip i. Hence, w ei,i−1 − w ei−1,i−1 ≥ ∆i−1 . 1 Since the first item of strip i has width at least p+1 W +∆i−1 we obtain, because of induction, Pi P p W · i − ∆i . eij ≥ w eii + i−1 ei−1,j + w ei,i−1 − w ei−1,i−1 ≥ p+1 j=1 w j=1 w P p esj ≥ p+1 Consequently, inequality (1) is fulfilled, and we obtain sj=1 w W · s − ∆s . Ps p p W ·s. Otherwise, if ∆s > 0, then we have w ess = p+1 W −∆s , If ∆s = 0, then j=1 w esj ≥ p+1 Ps−1 p p and hence j=1 w esj ≥ p+1 W · s − ∆s − w ess = p+1 W · (s − 1). Ps p W · s we replace strip s by another one esj ≥ p+1 In order to construct a packing with j=1 w as follows: Sort the items of Kss ∪ {k0 } according to non-increasing width and pack them in this sequence into the s-th strip. Because of Corollary 10 the new packing Ps−1of strip s has p width w es ≥ p+1 W . Consequently, the obtained new bin packing fulfills j=1 w esj + w es ≥ p W · s. p+1
H. Since at least q strips are The rest of the proof is done by contradiction. Let hk0 ≥ p+1 2pq packed into the bin, for the total area A of the packed items follows P P P P P p A ≥ si=1 j∈Ksi wj hj ≥ si=1 j∈Ksi wj p+1 H ≥ si=1 w esi p+1 H ≥ s · p+1 W · p+1 H 2pq 2pq 2pq ≥q·
p W p+1
·
p+1 H 2pq
= 12 W H,
which contradicts the assumption. Corollary 12. Let A(K) > 12 W H. We assume that not all items are packed by the FFDW1-heuristic, i. e. K 6= K, and k0 is the first item not packed. If there does not exist e ⊆ K ∪ {k0 } with A(K) e ≥ 1 W H, that can be packed into the bin, then it holds any subset K 2 w k0
21 W H. If all items of K can be packed into a single bin, or if e ⊂ K with A(K) e ≥ 1 W H, which can be packed hk0 ≥ p+1 H, then there exists a subset K 2pq 2 into a single bin. Proof: Since at least one of the assumptions of Theorem 11 cannot be fulfilled, either all e with A(K) e ≥ 1 W H, which items of K can be packed into a bin, or there exists a subset K 2 1 can be packed into one bin. Since A(K) > 2 W H, the statement follows. Corollary 14. Let A(K) > 21 W H. If all items of K can be packed into a single bin, or if W , where k0 denotes the first item, which is not packed by the FFDW1-heuristic, wk0 ≥ q+1 2pq e ⊂ K with A(K) e ≥ 1 W H, which can be packed into a single then there exists a subset K 2 bin. Proof: The statement follows from the previous corollary applied on an instance with switched wi and hi , respectively W and H.
4
The Bin Packing Problem
In this section, we focus on relations between the maximum number of bins needed to pack all items and their total area. We state that any list L = L(K) of items with total area A(L) ≤ Abin can be packed into at most three bins. Then we improve a lemma of Bougeret et al. [BDJ+ 09], and investigate further conditions needed to pack all items into two bins. First of all, Steinberg’s theorem can be used to improve Theorem 1 as follows (suggested by an anonymous referee): Let an instance E of the two-dimensional bin packing problem be given with bins Pof size W × H and list L of rectangular items with wmax ≤ W , hmax ≤ H, and A(L) = i∈K wi hi ≤ W H = Abin . Because of Steinberg’s theorem, all items can be packed into a strip of width 2W and height H. Let (xi , yi ) denote the corresponding allocation point (lower left corner) of item i in such a packing. Then item i covers region Ri (xi , yi ) := {(x, y) : xi ≤ x < xi + wi , yi ≤ y < yi + hi } ⊂ [0, 2W ] × [0, H]. We define the following partition of K = {1, . . . , n}: K1 := {i ∈ K : xi + wi ≤ W },
K2 := {i ∈ K : xi ≥ W },
K3 := K\(K1 ∪ K2 ).
It is obvious, that all items of K1 can be packed into a single bin, and that all items of K2 can P be placed into a second bin. Since K3 = {i ∈ K : xi < W < xi + wi }, the total height i∈K3 hi of K3 -items does not exceed H. Hence, all items of K3 can also be packed into a single bin. Summarizing, we have Theorem 15. Consider bins P of size W × H and a list L of rectangular items with wmax ≤ W , hmax ≤ H. If A(L) = i∈K wi hi ≤ W H = Abin , then all items of L can be packed into at most three bins.
8
Relations between capacity utilization, minimal bin size and bin number
On the other hand, at least three bins are needed in following cases: Theorem 16. Let q ∈ [1, H] ∩ N. Then there exist lists L of items with wmax ≤ W (i. e. p = 1), hmax ≤ H/q, and A(K) ≤ W H such that at least three bins are needed to pack all items. Proof: The statement is a consequence of the following example. Consider 2q + 1 items 1 (H + ε) for some sufficient small positive ε and i ∈ K = with wi = 21 (W + ε) and hi = q+1 {1, . . . , 2q + 1}.P Then we have: A(K) = wi hi = (2q + 1) W2+ε · H+ε = 2q+1 (W H + ε(W + H + ε)) ≤ W H. q+1 2q+2 i∈K
Since only q items can be packed in one bin, at least three bins are necessary to pack all items.
A statement concerning the maximum number of bins needed to pack all items is given by Bougeret et al. [BDJ+ 09]. Lemma 17 (Bougeret at al. (2009)). Let k ≥ 3 and let L be a list of rectangles with widths and heights not greater than 1 and total area not greater than k/4. Then there exists a packing of L into k unit bins. Applying this lemma for k = 4 we obtain the result due to Theorem 1. Using the technique of Theorem 15, the Lemma of Bougeret et al. can be strengthened as follows: Lemma 18. Let k ≥ 3 and let L be a list of rectangles with wmax ≤ 1, hmax ≤ 1, and A(L) ≤ k/4. Then there exists a packing of L into k − 1 unit bins if k is even, and into k unit bins if k is odd. Proof: Let k = 2m, m ≥ 2, integer. Due to Steinberg’s theorem all items can be packed into a strip of width 2 · k/4 = k/2 = m and height 1. Using again the description of a packing by allocation points (xi , yi ) we can define m index sets Ip := {i ∈ I : p − 1 ≤ xi , xi + wi ≤ p}, p = 1, . . . , m, and m − 1 index sets Im+p := {i ∈ I : xi < p < xi + wi }, p = 1, . . . , m − 1. Obviously, the m − 1 = k − 1 index sets Ip form a partition of I, and each item set can be packed into a single bin. Now, for k = 4, the statement of Theorem 15 follows from Lemma 18: at most three bins are needed. In the following we show that Lemma 18 can further be strengthened. Lemma 19. Let L = L(K) be a list of rectangles with hmax ≤ H and wmax ≤ W . If w(K) ≤ 23 W , then the First Fit(FF) heuristic packs all items of K in at most two bins of size W × H. Proof: Let B1 and B2 denote sets of items packed by the FF-heuristic into the first two bins, respectively. Assume, there is an item j, which is not packed by the FF-algorithm into one of the first two bins. Then wj + w(B1 ) > W and wj + w(B2 ) > W , and therefore 2wj + w(B1 ) + w(B2 ) > 2W . Since wj + w(B1 ) + w(B2 ) ≤ 32 W we obtain wj > 21 W , and consequently, w(B1 ) + w(B2) < W . Therefore all items of B1 and B2 are packed by the FF-algorithm into the first bin, i. e. B2 = ∅ and item j fits within the second bin.
9
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Theorem 20. Let L = L(K) be a list of rectangles with hmax ≤ H and wmax ≤ W . If 3 A(L) ≤ W H, 4
(2)
then all items of L can be packed into at most two bins of size W × H. Proof: Let Ksmall := {i ∈ K : wi ≤ 12 W, hi ≤ 21 H}, Klarge := {i ∈ K : wi > 12 W, hi > 12 H}, Kwide := {i ∈ K : wi > 21 W, hi ≤ 12 H}, Ktall := {i ∈ K : wi ≤ 21 W, hi > 21 H}. W. l. o. g., we assume A(K) = 43 W H. The proof is done by a complete case-by-case analysis. Case 1: Let Ksmall = ∅, Kwide = ∅, Ktall ∪ Klarge 6= ∅. Because of hi > 21 H ∀i ∈ Ktall ∪ Klarge and area condition (2), we have w(Ktall ∪ Klarge ) < 3 W . Due to Lemma 19, all items are packed into two bins by the FF-algorithm. 2 Case 2: Let Ksmall = ∅, Kwide ∪ Klarge 6= ∅, Ktall = ∅. Similar to case 1 by interchanging wi with hi , and W with H. Case 3: Let Ksmall = ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge = ∅. There are three subcases to be considered: (i) Let w(Ktall ) ≤ W and h(Kwide ) ≤ H. The pieces of Ktall are packed side by side into the first bin; those of Kwide on top of each other into the second bin. (ii) Let w(Ktall ) > W . Sort the items of Ktall according to non-increasing height and pack them with the FFalgorithm. Due to Lemma 19 all items of Ktall are packed into two bins. Next, sort items of Kwide according to non-increasing width wi and pack them right-justified, from top to bottom into the second bin. There is no overlap since otherwise a contradiction to condition (2) arises (cf. Fig. 1): Let hk > 12 H be the height of an item of Ktall ,
Bin 1
Bin 2
Figure 1: Case 3: Packing of Ktall and Kwide which overlaps with an item of Kwide . Since item k is not packed into the first bin, we have A(Ktall ) > W hk . Because of overlap, h(Kwide ) > H − hk follows, and hence, A(Kwide ) > 12 W (H − hk ). Then, for the total area of items, we have A(K) = A(Ktall ∪ Kwide ) > W hk + 12 W (H − hk ) = 12 W (H + hk ) > 21 W (H + 12 H) = 43 W H.
10
Relations between capacity utilization, minimal bin size and bin number
(iii) Let h(Kwide ) > H. Similar to subcase (ii). Case 4: Let Ksmall = ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge 6= ∅. Two subcases can occur: (i) Let Klarge = {k}, ı. e. |Klarge | = 1. Pack item k bottom-left justified into the first bin. Then: (a) If w(Klarge ∪ Ktall ) ≤ W , then pack all items of Ktall besides item k. Since wk hk > 41 W H, we obtain A(Kwide ) < 1 W H because of condition (2). Since hi ≤ 12 H, all items of Kwide fit into the second bin 2 because of Steinberg’s theorem. (b) If w(Klarge ∪ Ktall ) > W , then, assuming that h(Klarge ∪ Kwide ) > H, we obtain a contradiction to condition (2) as follows: A(K) = A(Klarge ∪ Kwide ∪ Ktall ) > wk hk + 21 W (H − hk ) + (W − wk ) 12 H = wk (hk − 21 H) + wk 12 H + 21 W (H − hk ) + (W − wk ) 12 H > 21 W (hk − 21 H) + wk 21 H + 12 W (H − hk ) + (W − wk ) 21 H = 21 W (hk − 21 H + H − hk ) + (W − wk + wk ) 21 H = 12 W 12 H + W 12 H = 34 W H. Therefore, h(Klarge ∪ Kwide ) ≤ H, and all items of Kwide fit above item k. Since A(Ktall ) < 1 W H, all items of Ktall can be packed into the second bin because of Steinberg’s theorem. 2 (ii) Let Klarge = {j, k} (j 6= k), i. e. |Klarge | = 2. Because of A(j) > 41 W H and area condition (2), we have A({k} ∪ Kwide ) < 21 W H. Since wi > 21 H for i ∈ {k} ∪ Kwide , h({k} ∪ Kwide ) < H follows. Therefore, all these items can be packed on top of each other into the first bin. In a similar way, all items of {j} ∪ Ktall fit side by side within the second bin. Case 5: Let Ksmall 6= ∅, Kwide = ∅, Ktall = ∅, Klarge = ∅.
P Sort items of Ksmall according to non-increasing area. Let j := arg min { ji=1 wi hi ≥ P 1 W H}. Because of wi hi ≤ 14 W H for all i ∈ Ksmall , we have ji=1 wi hi < 21 W H. Due 4 to Steinberg’s theorem, all items of {1, . . . , j} can be packed into the first bin, and the remaining items into the second since their total area is (because of condition (2)) less than 12 W H. Case 6: Let Ksmall 6= ∅, Kwide = ∅, Ktall = ∅, Klarge 6= ∅. Pack one item of Klarge into the first bin. Then all other items fit within the second bin due to Theorem 5 since their total area is less than 21 W H. Case 7: Let Ksmall 6= ∅, Kwide = ∅, Ktall 6= ∅, Klarge = ∅. P Sort all items according to non-increasing area, and let j := arg max { ji=1 wi hi ≤ 21 W H}. Because of Lemma 9 (with q = 1, L = 12 W H), we have A({1, . . . , j}) ≥ 41 W H. Due to Steinberg’s theorem, all items of {1, . . . , j} can be packed into the first bin. Since the total area of non-packed items is at most 43 W H − 14 W H = 12 W H. Case 8: Let Ksmall 6= ∅, Kwide = ∅, Ktall 6= ∅, Klarge 6= ∅. Pack one item of Klarge into the first bin. Then all other items fit within the second bin due to Theorem 5 since their total area is less than 21 W H. Case 9: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall = ∅, Klarge = ∅. Similar to case 7.
11
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Case 10: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall = ∅, Klarge 6= ∅. Similar to case 8. Case 11: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge = ∅. We consider five subcases: (i) If w(Ktall ) ≤ W and A(Ktall ) ≥ 41 W H, then pack all items of Ktall into the first bin. All other items fit within the second bin because of Steinberg’s theorem since their total area is less than or equal to 12 W H and wi ≤ 12 W for all i. (ii) Let h(Kwide ) ≤ H, A(Kwide ) ≥ 41 W H. Similar to subcase (i) (iii) Let w(Ktall ) ≤ W , A(Ktall ) < 14 W H, h(Kwide ) ≤ H, A(Kwide ) < 14 W H. Let Iwide := Kwide and Itall := Ktall . We add successively all items of Ksmall to Iwide or Itall , respectively, such that their area does not exceed 21 W H. This approach works because of area condition (2) and since the area of a small item is not greater than 41 W H. We assume that we cannot add an item j to one of the sets without exceeding the area bound 21 W H. We obtain W H = 12 W H + 21 W H < A(Iwide ) + A(j) + A(Itall ) + A(j) ≤ 43 W H + A(j) ≤ 3 W H + 41 W H = W H, which is a contradiction. Consequently, all items can be added to 4 the sets Iwide and Itall and therefore all items can be packed into two bins due to Steinberg’s theorem. (iv) Let w(Ktall ) > W . Sort the items of Ktall according to non-increasing height hi and pack them with the FF algorithm side by side into two bins (cf. Lemma 19). Sort the items of Kwide according to non-increasing width wi and pack all of them right-justified from top to bottom also into the second bin. Similar to subcase (3) (ii) it can be shown that no overlap occur in bin two. We consider two subcases: (a) If A(B2 ) ≥ 14 W H, then all items, which are not packed into the second bin, fit within the first bin due to Steinberg’s theorem. (b) Let A(B2 ) < 41 W H. We define K tall := Ktall ∩ B2 and K wide := Kwide ∩ B2 . Then we have 14 W H > A(B2 ) > 1 W h(K wide ) 2
+ w(K tall ) 12 H. We obtain h(K wide )
W , we have A(Ktall ) > 12 W H, and therefore A(Ksmall ) < 14 W H. Next we show that all items of Ksmall can be packed additionally into the second bin: 2A(Ksmall ) < 2 · 14 W H = 21 W H H H − 21 w(K tall )H + w(K tall )2 W = 12 W H + w(K tall )H − 12 w(K tall )H − w(K tall )2 W H H ) − w(K tall )(H − ( 12 H + w(K tall ) W )) = (W − w(K tall ))( 12 H + w(K tall ) W 1 H = (W − w(K tall ))(H − ( 2 H − w(K tall ) W )) . H ))) − (2 · 21 W − (W − w(K tall )))(2 · 21 H − (H − ( 12 H − w(K tall ) W H )) − (2 · 21 W = (W − w(K tall ))(H − ( 12 H − w(K tall ) W H − (W − w(K tall )))+ (2 · 21 H − (H − ( 12 H − w(K tall ) W )))+ . Therefore, all items of Ksmall fit within the unused region in bin two of size (W −w(K tall ))× H (H − ( 12 H − w(K tall ) W )) (Fig. 2).
12
Relations between capacity utilization, minimal bin size and bin number
Bin 1
Bin 2
Figure 2: Case 11: Packing of Kwide , Ktall and Ksmall
(v) Subcase h(Kwide ) > H is similar to subcase (iv). Case 12: Let Ksmall 6= ∅, Kwide 6= ∅, Ktall 6= ∅, Klarge 6= ∅. We analyse again two subcases: (i) Let Klarge = {k}. Pack item k into the first bin. Further two subcases are considered: (a) If w(Klarge ∪ Ktall ) ≤ W , then all items of Ktall can also be placed into bin one besides item k. Since A(Kwide ∪ Ksmall ) < 21 W H these items fit within the second bin due to Steinberg’s theorem. (b) If w(Klarge ∪ Ktall ) > W , then similar to subcase (4)(i)(b) one can show that h(Klarge ∪ Kwide ) ≤ H holds true. Therefore, all items of Kwide ∪ Klarge fit within a bin, and all other within the second, similar to (a). (ii) Let Klarge = {j, k}. Pack items j and k bottom-left justified into the first, respectively the second bin. Similar to subcase (4)(ii) it can be shown that all items of Ktall can be placed besides item j in bin one, and that all items of Kwide fit above item k in bin two. We pack items of Kwide from top to bottom (top-justified). We define K small := {i ∈ Ksmall : wk + wi > W }, δk := wk − 21 W > 0 and θk := hk − 21 H > 0. We show that all items of K small fit between item k and items of Kwide within the second bin. Assuming h(K small ) + h(Kwide ) + hk > H we obtain a contradiction: A({k} ∪ K small ∪ Kwide ) > wk hk + (W − wk )(H − hk − h(Kwide )) + 12 W h(Kwide ) = wk hk + (W − wk )(H − hk ) + (wk − 21 W )h(Kwide ) = ( 21 W + δk )( 21 H + θk ) + ( 12 W − δk )( 21 H − θk ) + δk h(Kwide ) = 41 W H + 12 W θk + δk 21 H + δk θk + 41 W H − 12 W θk − δk 21 H + δk θk + δk h(Kwide ) = 12 W H + 2δk θk + δk h(Kwide ) > 12 W H. This contradicts area condition (2) since item j has area larger than 14 W H. Consequently, all items of K small can be packed between item k and items of Kwide into the second bin (Fig. 3). We obtain:
13
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
2A(Ksmall \K small ) = 2A(K) − 2A(B1 ) − 2A(B2 ) < 2 · 43 W H − 2A(j) − 2A(k) − 2A(Kwide ) < 32 W H − 21 W H − 2wk 12 H − 2 · 21 W h(Kwide ) < 32 W H − 21 W H − 2 · 12 W 12 H − 2δk 12 H − 2 · 21 W h(Kwide ) + 2δk h(Kwide ) = 12 W H − δk H − W h(Kwide ) + 2δk h(Kwide ) = ( 12 W − δk )(H − 2h(Kwide )) = ( 12 W − δk )(H − h(Kwide )) − ( 21 W − δk )(h(Kwide )) = ( 12 W − δk )(H − h(Kwide )) − ( 21 W − δk )+ (h(Kwide ))+ = ( 12 W − δk )(H − h(Kwide )) − (2 · ( 21 W − δk ) − ( 21 W − δk ))+ (2 · 12 H − (H − h(Kwide )))+ . Therefore all items of Ksmall \K small can be packed into the unused region of size ( 21 W − δk ) × (H − h(Kwide )) due to Steinberg’s theorem. W 2
j
k
Bin 1
Bin 2
Figure 3: Case 12: Packing of Klarge , Kwide , Ktall and Ksmall Theorem 21. Let L = L(K) be a list of rectangles with hmax ≤ H and wmax ≤ W . If A(L) ≤
k WH 4
for some k ≥ 3, k ∈ N,
then all items of L can be packed into at most k − 1 bins of size W × H. The statement is tight. Proof: The proof is done by reducing the general case to the special case of Theorem 20. Therefore, subsets of items with total area at least 41 W H are separated and packed into a single bin. We use again index sets Ksmall , Kwide , Ktall and Klarge as in proof of Theorem 20. Since wi hi > 12 W · 12 H = 41 W H for all items of Klarge , each item of Klarge can be packed into a single bin. In case of A(Kwide ) ≥ 14 W H, we obtain a desired reduction by the FF1 -algorithm (FFalgorithm applied to a single bin) as follows: If h(Kwide ) ≤ H, then all items fit in a single bin. Otherwise, the height obtained by the FF1 -algorithm when packing items of Kwide is at least 12 H since hi ≤ 12 H for all i ∈ Kwide so that the total area used is greater than 1 W H, and a further reduction can be made. 4 In a similar way, in case of A(Ktall ) ≥ 14 W H bin packings with area utilization at least 1 W H can be found. 4
14
Relations between capacity utilization, minimal bin size and bin number
Also in case of A(Ksmall ) ≥ 14 W H bin packings with area utilization at least 14 W H can be obtained by choosing subsets with total area not smaller than 14 W H but not larger than 1 W H (Theorem of Steinberg). 2 After that reduction process some items can remain in Kwide , Ktall and Ksmall but their total area is in each case smaller than 41 W H, and therefore is less than 34 W H. Hereby we proved that the general case can be reduced to that of Theorem 20. According to this theorem, two bins are needed for the remained items, and hence, in total at most k − 1 bins for all items. In order to show that the statement is tight, first we note that it is not fulfilled for k = 2 since items of size ε × H and W × ε cannot be packed into the same bin for ε > 0, and if ε is sufficient small, then their total area is less than 21 W H. Furthermore, for k ≥ 3, k − 1 items of size ( 21 W + ε) × ( 12 H + ε) with ε > 0 cannot be packed into k − 2 bins of size W × H. But, if ε is sufficient small the area condition is fulfilled, which shows that k − 1 is the optimal number of bins. Finally, for k ≥ 3, an instance of k items of size ( 21 W + ε) × ( 21 H + ε) with some sufficient small ε > 0 shows, that k4 W H is the maximum area bound for which all instances can be packed into k − 1 bins. In the following we are interested in statements how many bins are needed to pack all items of a list. Theorem 22. Let A(K) ≤ W H, wi ≤ W/3 and hi ≤ H/3 for all i ∈ K, i. e. p = q = 3. Then all items of L can be packed into at most two bins. Proof: Because of Corollary FFDH-heuristic for SPP packs all items of K with Pse8, the e total height not larger than j=1 hj ≤ 53 H. On the other hand, using the FFDH1-heuristic to fill a single bin covers at least 43 H of height because of Corollary 10. Therefore, at most two bins are needed to pack all items of K. Remark: A proof of Theorem 22, which is based on the NFDH-heuristic is given in [HS12]. In case of p = q = 3, Lemma 6 yields the NFDH-bound 11 H in contrast to the 6 FFDH-bound 53 H, so that a NFDH-based proof needs much more effort. Theorem 22 suggests the following Conjecture 23. Consider bins of size W × H and a list L of rectangular items with wmax ≤ W/2 and hmax ≤ H/2. If A(L) ≤ W H, then all items of L can be packed into at most two bins. In case of somewhat smaller items, we have: Theorem 24. Consider bins of size W × H and a list L of rectangular items. If wmax ≤ W/2,
hmax ≤ H/3,
(i. e. p = 2 and q = 3),
then all items of L can be packed into at most two bins.
A(K) ≤ W H,
15
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Proof: Without loss of generality, we may assume that A(K) = W H. We consider the following partition of the set of items: 1 1 K1 := {i ∈ K : wi ≤ W } and K2 := {i ∈ K : wi > W }. 3 3 Case 1: Let A(K1 ) ≤ 12 W H and A(K2 ) ≤ 12 W H. Because of Steinberg’s theorem, item set K1 as well as K2 can be packed into a bin so that at most two bins are needed to pack K. Case 2: Let A(K1 ) ≤ 12 W H and A(K2 ) > 21 W H. Pack items of K2 according to the FFDH1-heuristic into the first bin. Here, and in the following, let B1 denote those items packed into the first bin (either with FFDH1 , FFDW1 W for all items of K2 , from Corollary 14 follows or somehow else). Since wi > 13 W = q+1 2pq f2 ⊆ K2 with A(K f2 ) ≥ 1 W H, which can be packed into the that there exists a subset K 2 first bin. Because A(K\B1 ) = W H − A(B1 ) ≤ 12 W H, these items can be packed into the second bin using Steinberg’s theorem. Case 3: Let A(K1 ) > 21 W H and A(K2 ) ≤ 12 W H. Pack items of K1 according to the FFDH1-heuristic into the first bin,, now B1 ⊆ K1 , and let s denote again the number of (horizontal) strips. In case of A(B1 ) ≥ 12 W H, all remaining items can be packed into the second bin due to Steinberg’s theorem. In the following, let A(B1 ) < 12 W H and hi < Theorem 11 applied on K1 , i. e. p = q = 3.
p+1 H 2pq
=
2 H, 9
∀i ∈ K1 \B1 according to
e of K1 Suppose, the last condition is not fulfilled, then, because of Corollary 13, a subset K 1 e ≥ W H, which can be packed into a bin. Hence, all items of K would exist with A(K) 2 could be packed into two bins. 2 H 9
for all items
w ej+1 − ∆s w es+1 > 34 W (H − w e1 ) − ∆s w es+1
(3)
Because of sorting according to the FFDH1 -heuristic, we have hi < i ∈ K1 \B1 . With formula (1) and Theorem 11 we obtain A(B1 ) ≥ >
p W p+1
s P
j=1 3 2 W H − 4 3
( 43 W
−
2 W ) 92 H 3
=
1 WH 2
−
1 W H. 54
Now, sort the items of K2 according to non-increasing height, and pack them left-justified on top of the previous, as long as possible, until the bin height is reached, into the second bin (Fig. 4, left). Let H := h(K2 ). We distinguish three subcases: Subcase (i) Let H > H. Because of Corollary 10 the total height occupied in the second bin is at least 34 H. Since e := H − (H − 3 H) = 7H − H. A(K2 ) ≤ 12 W H and wi > 31 W , we have h(K2 ) < 23 H. Let H 4 4 Pack the remaining, non-packed items of K2 right-justified, starting at the bottom of the bin on top of each other into the second bin. Above these items it remains an unused e (Fig. 4, right). Moreover, A(K2 ) > 1 W h(K2 ) = 1 W H. rectangular area with size 21 W × H 3 3 We show that the items of K1 \B1 can be packed into the unused area of the second bin. For the total area of the non-packed items of K1 , we have:
A(K1 \B1 ) = A(K) − A(B1 ) − A(K2 ) < W H − ( 12 W H − W54H ) − W3 H = 21 W H + W54H − W3 H.
16
Relations between capacity utilization, minimal bin size and bin number
Figure 4: Packing of many items of K2 Therefore, 2A(K1 \B1 ) < 2( 21 W H + W54H − W3 H) = W H + W27H − 2W H 3 7W H 35 W 7H WH WH W W W = 8 + 27 − 6 H + 8 − 2 H = 216 W H − 6 H + 2 ( 4 − H). Two subcases have to be investigated: (a) Let 94 H − ( 47 H − H) ≤ 0, then 35 2A(K1 \B1 ) < 216 W H − W6 H + W2 ( 7H − H) 4 35 W W 7H = 216 W H − 6 H + 2 ( 4 − H) − (2wmax (K1 \B1 ) − W2 )+ ( 49 H − ( 47 H − H))+ < W2 ( 7H − H) − (2wmax (K1 \B1 ) − W2 )+ (2hmax (K1 \B1 ) − ( 47 H − H))+ 4 e − (2wmax (K1 \B1 ) − W )+ (2hmax (K1 \B1 ) − H) e +. = W2 H 2
Consequently, due to Steinberg’s theorem, the items of K1 \B1 can all be packed into a e rectangle of size 21 W × H. (b) Let 94 H − ( 74 H − H) > 0, then
35 2A(K1 \B1 ) < 216 W H − W6 H + W2 ( 7H − H) 4 35 W 7H W W = 216 W H − 6 H + 2 ( 4 − H) + 6 (2hmax (K1 \B1 ) − ( 47 H − H))+ − W6 (2hmax (K1 \B1 ) − ( 74 H − H))+ 35 ≤ 216 W H − W6 H + W2 ( 7H − H) + W6 (2 · 29 H − ( 47 H − H))+ − W6 (2hmax (K1 \B1 ) − ( 74 H − H))+ 4 35 = 216 W H − W6 H + W2 ( 7H − H) + W6 ( 49 H − 47 H + H) − W6 (2hmax (K1 \B1 ) − ( 47 H − H))+ 4 35 = 216 W H + W2 ( 7H − H) + W6 · −47 H − W6 (2hmax (K1 \B1 ) − ( 47 H − H))+ 4 36 < W2 ( 7H − H) − W6 (2hmax (K1 \B1 ) − ( 74 H − H))+ 4 W 7H = 2 ( 4 − H) − (2 · 13 W − W2 )+ (2hmax (K1 \B1 ) − ( 74 H − H))+ ≤ W2 ( 7H − H) − (2wmax (K1 \B1 ) − W2 )+ (2hmax (K1 \B1 ) − ( 74 H − H))+ 4 e − (2wmax (K1 \B1 ) − W )+ (2hmax (K1 \B1 ) − H) e +. = W2 H 2
Again, due to Steinberg’s theorem, the items of K1 \B1 can all be packed into a rectangle e of size 21 W × H.
Subcase (ii) Let H ≤ H and A(K2 ) ≥
WH 4
+
WH . 54
Because of H ≤ H all items of K2 are already packed on the left side of the second bin. For the total area of the not yet packed items of K1 it holds: A(K1 \B1 ) = A(K) − A(B1 ) − A(K2 ) ≤ W H − ( 21 W H − W54H ) − ( W4H + W54H ) = W4H . Because of wi ≤ 12 W ∀i ∈ K2 there exists a non-occupied area of size 12 W ×H in the second bin (cf. Fig. 4, left). Due to Steinberg’s theorem, all items of K1 \B1 can be packed into that rectangle.
17
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
Subcase (iii) Let H ≤ H and A(K2 )
W H − We consider a partition of K1 into K1 := {i ∈ K1 : hi ≥
25 H} 81
and the following two subcases: f1 ) ≥ 1 W H. (a) Let A(K
WH . 54 ( W4H + W54H )
+
= 34 W H −
1 W H. 54
f1 := {i ∈ K1 : hi < and K
25 H}. 81
2
f1 using the FFDH1-heuristic Empty the first bin (i. e. set B1 := ∅) and pack items of K f1 ). In case not all items of K f1 are packed into the first into the first bin (now ∅ = 6 B1 ⊆ K bin, we obtain due to formula (1) and Theorem 7, P p A(B1 ) ≥ p+1 ej+1 − ∆s w es+1 > 34 W (H − w W sj=1 w e1 ) − ∆s w es+1 56 3 2 2 14 1 1 3 > 4 W 81 H − ( 4 W − 3 W ) 9 H = 27 W H − 54 W H = 2 W H.
Since A(B1 ) ≥ 12 W H all items of K\B1 can be packed into the second bin due to Steinberg’s theorem. f1 ) < 1 W H, (b) Let A(K 2 f1 ) > W H −( W H + W H )− 1 W H = W H − W H = 25 W H. then A(K1 ) = A(K)−A(K2 )−A(K 4 54 2 4 54 108 Let M = {1, . . . , m} := K1 \B1 (with B1 as defined at the beginning of Case 3) be sorted according to non-increasing item area. Because of Theorem 11, we have wi hi < 13 W 92 H = 2 W H for all i ∈ M. Furthermore, let B1 := B1 = K1 \M. Because of Theorem 11, all 27 items of K1 are packed in the first bin, i. e. K1 ⊆ B1 . Several subcases can occur: (1) Let w(K1 ) ≤ W . P 1 Let j ∈ M denote the smallest index with ji=1 wi hi ≥ 54 W H. Let M := {1, . . . , j}. P 2 Moreover, because of the assumed sorting in M, we have A(M) = ji=1 wi hi < 27 W H. Next we show that all items of M can also be packed into the first bin besides those of 1 1 W H + 54 W H = 21 W H, and consequently, the B1 . In that case, A(B1 ∪ M ) ≥ 21 W H − 54 remaining items can be packed into the second bin due to Steinberg’s theorem. In order to show this we distinguish two subcases: 1 (A) Let A(B1 ) ≤ 21 W H − 108 W H. Empty the first bin (i. e. B1 = ∅) and pack all items of K1 along the bottom side of the bin, i. e. B1 := K1 . Above these items a non-used rectangular area of size W × 23 H (Fig. 5, left) remains. It holds,
A(B1 \K1 ) + A(M) ≤ 12 W H −
1 WH 108
−
25 WH 108
+
2 WH 27
=
54−1−25+8 WH 108
= 13 W H.
Hence, all items of (B1 \K1 ) ∪ M can be packed into a rectangle of size W × 32 H . 1 (B) Let A(B1 ) > 21 W H − 108 W H. WH WH If A(K2 ) ≥ 4 + 108 , then a packing of K into two bins can be constructed similar to subcase (ii). Otherwise,
f1 ) > W H − ( W H + A(K1 ) = A(K) − A(K2 ) − A(K 4
WH ) 108
− 21 W H =
WH 4
−
Analogously to subcase (A), we obtain A(B1 \K1 ) + A(M) ≤ 12 W H −
26 WH 108
+
2 WH 27
=
54−26+8 WH 108
= 31 W H.
WH 108
=
26 W H. 108
18
Relations between capacity utilization, minimal bin size and bin number
k
Figure 5: Packing of items of K1 Therefore, also in this subcase, the items of M ∪ (B1 \K1 ) can be packed into a rectangle of size W × 32 H. (2) Let w(K1) > W . Sort the items of K1 according to non-increasing width and pack them left-justified side c1 denote by side along the bottom of the bin until the next item does not fit. Let K 3 c the set of packed items. Because of Corollary 10, we have w(K1 ) ≥ 4 W , and therefore c1 ) ≥ 25 W . Two cases are differentiated: A(K 108 c1 ) ≥ 7 W , (A) Let w(K 9 c1 ) ≥ 7 W 25 H = then A(K 9 81
175 W H. 729
M1 := {i ∈ M : wi hi ≤
We consider the following partition of M:
107 107 W H} and M2 := {i ∈ M : wi hi > W H}. 1458 1458
Two subcases are considered: 1 (α) Let A(M1 ) ≥ 54 W H. Let M1 = {1, . . . , m1 } beP sorted according to non-increasing item area. Let j1 ∈ M1 denote 1 1 the smallest index with ji=1 wi hi ≥ 54 W H. Let M1 := {1, . . . , j1 }. Moreover, because of P1 107 W H. Analogously to the the assumed sorting in M1 , we have A(M1 ) = ji=1 wi hi ≤ 1458 c previous case we show that all items of (B1 \K1 ) ∪ M1 can be packed into the unused area of size W × 32 H. For their total area, we have
c1 ) + A(M1 ) ≤ 1 W H − A(B1 \K 2
175 WH 729
+
107 WH 1458
=
729−350+107 WH 1458
=
486 WH 1458
= 13 W H.
Hence, A(B1 ) = A(B1 ∪ M1 ) ≥ 21 W H, and all remaining items can be packed into the second bin. 1 (β) Let A(M1 ) < 54 W H, then A(M2 ) = A(K) − A(K2 ) − A(B1 ) − A(M1 ) > W H − ( W4H + W54H ) − W2H − W54H 23 8 = 108−27−2−54−2 W H = 108 W H. Because of this and wi hi ≤ 31 W 29 H = 108 W H for all 108 i ∈ M2 , the set M2 contains at least three items. Pack three items of M2 side by side on c2 . Because of hi < 2 H for the top of the first bin (Fig. 6) and denote these items with M 9 all i ∈ M2 there exists an unused area of size W × 94 H. We show that there exists a set c1 ∪ M c2 ) with A(K c1 ∪ M c2 ∪ Kˇ1 ) ≥ 1 W H that can be packed into this unused Kˇ1 ⊆ K1 \(K 2 f = {1, . . . , m} c1 ∪ M c2 ). So we have A(M f) = A(K1 \(K c1 ∪ M c2 )) = area. Let M e := K1 \(K c1 ) − A(M c2 ) ≥ ( 3 − 1 − 1 − 2 )W H = 81−2−36−24 W H = 19 W H > 1 W H. A(K1 ) − A(K 4 54 3 9 108 108 9
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
19
f denote the smallest index with Pej wi hi ≥ 1 W H. Because of the sorting, we Let e j∈M i=1 9 Pej 2 obtain i=1 wi hi ≤ 9 W H. Hence, these items can be packed in the unused area and we 107 833 get A(B1 ) ≥ 175 W H + 3 · 1458 W H + 19 W H = 350+321+162 W H = 1458 W H > 12 W H. Due to 729 1458 Steinberg’s theorem all remaining items can be packed into the second bin.
c1 and M c2 Figure 6: Packing of items of K
c1 ) < 7 W . (B) Let w(K 9 c1 , wi > 2 W ∀i ∈ K c1 holds since otherwise a further item of Because of construction of K 9 c1 . K1 could be packed besides the items of K Empty again the first bin (i. e. B1 = ∅), and pack items of K1 using the FFDW1 -heuristic, into the bin. If A(B1 ) ≥ 12 W H, then all remaining items can be packed into the second bin. c1 are packed, i. e. K c1 ⊆ B1 . If A(B1 ) < 21 W H, then, because of Theorem 11, all items of K Let k denote the area-largest item of K1 \B1 . Because of Corollary 12 and the sorting of items with respect to non-increasing width one has wk < 29 W . We consider two subcases: 1 (α) Let wk hk ≥ 54 W H. Empty the first bin and pack items of K1 \{k} using the FFDH1 -heuristic into that bin. f1 ⊆ K1 \{k} denote those items now packed into the first bin. If A(B f1 ) ≥ 1 W H, then Let B 2 1 1 1 f f we are done. Otherwise, if A(B1 ) < 2 W H, then A(B1 ) ≥ 2 W H − 54 W H holds because of c1 are packed in the first inequality (3). Furthermore, because of Theorem 11, all items of K c1 ⊆ B f1 . We show that the items B f1 ∪ {k} can be packed into one bin. Empty bin, i. e. K c1 along the bottom side of bin one. Since the again the first bin and pack anew items of K 7 c1 . Then, for the total area total width is less than 9 W , item k fits besides the items of K f1 , we have of not yet packed items of B c1 ) ≤ 1 W H − 25 W H = 29 W H < 1 W H. f1 \K A(B 2
108
108
3
Therefore, these items can be packed into the unused area of size W × 23 H (Fig. 5, right). f1 ) + wk hk ≥ 1 W H. Because of inequality (3), the total area of this packing is A(B 2 Consequently, all items not packed into the first bin can be packed into the second due to Steinberg’s theorem.
1 (β) Let wk hk < 54 W H. Let again M = {1, . . . , m} := K1 \B1 be sorted according to non-increasing item area. Let Pj 1 W H. Let M := {1, . . . , j}. We j ∈ M denote the smallest index with i=1 wi hi ≥ 54 1 1 obtain 54 W H ≤ A(M) ≤ 27 W H. f1 denote Empty the first bin and pack items of K1 \M using the FFDH1 -heuristic. Let B
20
Relations between capacity utilization, minimal bin size and bin number
f1 ) ≥ 1 W H − 1 W H. If the set of items, which are packed now. Then we have A(B 2 54 f1 ) < 1 W H, empty again the first bin and pack items of K c1 side by side along the A(B 2 bottom of the first bin. Then f1 \K c1 ) + A(M ) ≤ 1 W H − 25 W H + 1 W H = 54−25+4 W H = 33 W H < 1 W H A(B 2 108 27 108 108 3 Hence, all these items can be packed into the first bin, and we have f1 ∪ M ) ≥ 1 W H. A(B1 ) = A(B 2 Therefore, all remaining non yet packed items fit in the second bin. Case 4: Let A(K1 ) > 21 W H and A(K2 ) > 12 W H.
Because of A(K) = A(K1 ) + A(K2 ) > 21 W H + 21 W H = W H W H is violated. Consequently, this case cannot occur.
the area condition A(K) ≤
Corollary 25. Consider bins of size W × H and a list L of rectangular items. If wmax ≤ W/3,
hmax ≤ H/2,
(i. e. p = 3 and q = 2),
A(K) ≤ W H,
then all items of L can be packed into at most two bins. Proof: The statement follows from the previous theorem applied on an instance with switched wi and hi , respectively W and H.
5
Conclusions and outlook
In this paper we showed that any list of rectangles can be packed into at most k − 1 bins if the total area of items does not exceed k/4-times the bin area for an integer k ≥ 3, which imprves the Lemma of Bougeret et al. [BDJ+ 09]. This result is tight. Furthermore, we investigated additional conditions to strengthen this statement. Especially, we showed that only two bins are necessary to pack all items if their total area is at most the bin area and if the height of each item is not larger than one-third of the bin height and if the width of every item does not exceed one-half of the bin width. It remains a lack between these two statements, which will be part of future research.
References [BDJ+ 09] M. Bougeret, P. F. Dutot, K. Jansen, C. Otte, and D. Trystram. Approximation algorithm for multiple strip packing. IT University of Copenhagen, Denmark, 2009. 7th WAOA. [BKRS09] G. Belov, V. Kartak, H. Rohling, and G. Scheithauer. One-dimensional relaxations and lp bounds for orthogonal packing (pages. International Transactions in Operational Research, 16:745–766, 2009. [BS12]
T. Buchwald and G. Scheithauer. A note on relations between capacity utilization and ” minimal bin number“. Preprint MATH-NM-05-2012, Technische Universit¨ at Dresden, 2012.
T. Buchwald, K. Hoffmann, G. Scheithauer. May 17, 2013
21
[CCM07] J. Carlier, F. Clautiaux, and A. Moukrim. New reduction procedures and lower bounds for the two-dimensional bin packing problem with fixed orientation. Computers & Oper. Res., 34:2223–2250, 2007. [CGJT80] E. G. Coffman, M. R. Garey, D. S. Johnson, and R. E. Tarjan. Performance bounds for level–oriented two–dimensional packing algorithms. SIAM J. on Computing, 9(4):808– 826, 1980. [DF92]
H. Dyckhoff and U. Finke. Cutting and packing in production and distribution. Physica Verlag, Heidelberg, 1992.
[FS04]
S. P. Fekete and J. Schepers. A general framework for bounds for higher-dimensional orthogonal packing problems. Math. Methods of Operations Research (MMOR), 60(2):311–329, 2004.
[GJ79]
M. R. Garey and D. S. Johnson. Computers and Intractability – A Guide to the Theory of NP-Completeness. Freeman, San Francisco, 1979.
[HS12]
K. Hoffmann and G. Scheithauer. Relations between capacity utilization and minimal bin number. Preprint MATH-NM-03-2012, Technische Universit¨ at Dresden, 2012.
[JZ07]
K. Jansen and G. Zhang. Maximizing the total profit of rectangles packed into a rectangle. Algorithmica, 47:323–342, 2007.
[LC90]
K. Li and K. H. Cheng. On three-dimensional packing. SIAM J. on Computing, 19(5):847–867, 1990.
[LMM02] A. Lodi, S. Martello, and M. Monaci. Two-dimensional packing problems: A survey. European Journal of Operational Research, 141(2):241–252, 2002. [MM68]
A. Meir and L. Moser. On packing of squares and cubes. J. of Combinatorial Theory, 5:126–134, 1968.
[MV98]
S. Martello and D. Vigo. Exact solution of the two-dimensional finite bin-packing problem. Management Science, 44:388–399, 1998.
[Ste97]
A. Steinberg. A strip-packing algorithm with absolute performance bound 2. SIAM J. on Computing, 26(2):401–409, 1997.
