Here we provide the remained part of the complete proof for the tight ratio ofF F. We will use the results of Chapter 4, so we continue here with the analysis of the case when the last common bin is small.
Suppose that the size of the last common bin is smaller than 2/3. For the rest of the upper bound proof, fixx > 0so that s(Cγ) = 23 −2x. Lemma 4.2.2(ii) implies s(Cγ) > 1/2and thus x <1/12.
Since now the regular weight of the last bin is smaller than0.8, we need to compensate for this.
This is indeed possible due to the fact that now Lemma 4.2.1(iii) implies that the inner common bins are larger than2/3 +xand this allows us to improve the bounds of Lemma 4.2.5 by an amount proportional tox.
Note that Ci, i > 1, cannot be a 5+-bin: Sinces(C1) < 5/6, all items in Ci have size larger than1/6and five of them would add up to more than5/6, contradicting the assumption thatCi is a common bin.
Case 1: Ciis a2-bin. ThenCi contains at least one itemcof size larger than1/3as otherwise s(Ci−1) ≤ 2/3 contradicting Lemma 4.2.1(iii) together with the assumption that S(Cγ) < 2/3.
The other itemc0inCi satisfiesc0 > 16 +y. Thus Case 2: Ci is a3-bin. Suppose thatCicontains an itemc >1/3. Then the remaining two items in Cihave size at least 16 +yand we obtain
r(Ci−1) +v(Ci)≥1− 6 5y+3
5(y+y) + 0.1 = 1.1≥1 + 3 5x
sincex <1/12. Otherwise all three items inCihave size at most1/3. We claim that one of them has size at least 16+x, as otherwise, usingx <1/12, we haves(Ci)<3(16+x) = 12+ 3x < 23+x,
Case 3: SupposeCi is a 4-bin. All items in Ci are small (i.e. have sizes between1/6and1/3), as otherwise s(Ci) ≥ 13 + 3· 16 = 56, contradicting the assumption that Ci is a common bin. As
2 Let γk denote the number of k-bins that do not contain an exceptional item among the inner common bins, i.e., amongC2, . . . ,Cγ−1. Letα= 2(γ2+γ3) +γ4.
Lemma 8.3.2 Suppose thats(Cγ)<2/3. The following holds:
(i) Ifα≥8then the total weight of the common bins is at leastw(C)≥γ−0.2.
(ii) Ifα≥4then the total weight of the common bins is at leastw(C)≥γ−0.3.
Proof. We apply Lemma 8.3.1 for any i = 1, . . . , γ −2 such that Ci+1 does not contain an exceptional item. Otherwise, i.e., ifCi+1 contains an exceptional item and also fori =γ −1we apply Lemma 4.2.5. Summing all the resulting bounds onr(Ci) +v(Ci+1)andr(Cγ) = 0.8−125x we obtain that the total weight of the common bins is
s(C)≥γ−1 + (γ2+γ3)3
5x+γ4 3
10x+ (0.8−12
5 x) =γ−0.2 + (3α−24)x 10 .
Forα≥8we have3α≥24and (i) follows. Forα≥4we usex <1/12, which gives(3α−24)x≥
−12x≥ −1and (ii) follows. 2
Theorem 8.3.1 For any instance of bin packing,F F ≤1.7·OP T.
Proof. Ifs(Cγ)≥ 2/3then the theorem follows by Theorem 4.2.1. Thus assumes(Cγ)< 23 and F F ≥ 1.7·OP T + 0.1. We distinguish several cases and in each we derive a contradiction or prove the theorem statementF F ≤1.7·OP T, leading to an indirect proof as well.
Case 1: OP T ≥ 21. By Lemma 4.2.2(iv) we haveγ ≥ 12. Thus there are at least 10 inner common bins and at most 2 of them have an exceptional item. Thus α ≥ γ2 +γ3 +γ4 ≥ 8and w(C)≥γ −0.2by Lemma 8.3.2(i). Now Lemma 4.2.6 and Proposition 4.2.1 imply the theorem.
Case 2: OP T ≥8,OP T 6≡4(mod 10), andOP T 6≡7(mod 10). ThenF F ≥1.7·OP T+0.3, thus we can use Lemma 4.2.2(iv) withτ = 3and we obtainγ ≥6. There are no exceptional items, sinceOP T 6≡ 7(mod 10), and thusα ≥ 4. Lemma 8.3.2(ii) impliesW > β + (γ −0.3) +δ = F F −0.3≥1.7·OP T, a contradiction.
Case 3: OP T = 14. ThenF F = 24. Lemma 4.2.2(iv) withτ = 2 impliesγ ≥ 9. There are no exceptional items, thusγ2 +γ3+γ4 ≥7. Ifγ2 +γ3 ≥ 1thenα ≥ 8and the theorem follows by Lemma 8.3.2(i). In the remaining caseγ4 ≥7, thus there are seven4-bins among the common bins. Using Lemma 4.2.1(v) for five of these common4-bins, Lemma 4.2.1(iv) for some four of the remainingγ−5≥4common bins and Lemma 4.2.1(i) for the remaining24−9 = 15bins we get
S > 4 + 4· 2
3+ 15· 1
2 >14 = OP T , a contradiction.
Case 4: OP T = 17. Then F F = 29. Lemma 4.2.2(iv) gives γ ≥ 10. Thus γ2 + γ3 + γ4 ≥ 6. If γ4 ≤ 4then α ≥ 2(6 −γ4) +γ4 ≥ 8 and the theorem follows by Lemma 8.3.2(i).
Otherwise there are five4-bins among the common bins. Using Lemma 4.2.1(v) for these five 4-bins, Lemma 4.2.1(iv) for some five of the remainingγ−5≥5common bins and Lemma 4.2.1(i) for the remaining29−10 = 19bins we get
S > 4 + 5· 2
3+ 19· 1
2 >17 = OP T ,
a contradiction.
Case 5: OP T = 7. ThenF F = 12. First we claim thatδ = 7. OtherwiseS >6·23 + 6· 12 = 7, a contradiction. It follows that there are at most three 2-bins in the F F packing, since by Lemma 4.2.7 noOP T-bin can contain two 2-items. Next we claim that no twoF F-bins have total size greater than or equal to3/2. Otherwise there remain at least three2+-bins in theF F packing andS > 32+ 3·23+ 7·12 = 7, a contradiction. Since there are five2+-bins and at most three2-bins, there have to be at least two 3+-bins. LetC be the last 3+-bin and B some bin before it. Then C contains three items of size larger than1−s(B)and s(B) +s(C) ≥ s(B) + 3(1−s(B)) = 3−2·s(B). Since no two bins have total size3/2or more, this impliess(B)≥3/4. Furthermore, this implies that there is a single bin beforeC, as otherwise there would again be two bins with total size at least3/2. I.e.,Bis the first bin,Cis the second bin and there are exactly three2-bins.
C has at least three items and they are packed into differentOP T-bins by Assumption 4.2.1. We claim that one of these three bins contains both a2-itemcand a D-itemdwith sized >1/2: Each OP T-bin contains a D-item and there is at most oneDitem of size at most1/2; furthermore, there is at most oneOP T-bin not containing a2-item, as there are six2-items in the three2-bins. Thus the condition excludes at most twoOP T-bins. Fixc0 to be an item fromC packed with such ac anddin the sameOP T-bin. Note thatcanddare in laterF F-bins thanC, asBandCare the first bins and they are3+-bins. We havec0+c <1/2as they are packed withd >1/2in anOP T-bin.
On the other hand we claim thats(C)−c0 <1/2: otherwise we note thatc0 >1−s(B), asc0was not packed inB and thuss(B) +s(C) > s(B) +c0 + 1/2 > s(B) + (1−s(B)) + 1/2 = 3/2, contradicting the first claim in the proof. Thuss(C) +c= (s(C)−c0) + (c0+c)<1/2 + 1/2 = 1 andF F should have packedcintoC, which is the final contradiction. 2 We note that the last case ofOP T = 7is also covered in the manuscript [65], we have included it for completeness.