Packing a Knapsack of Unknown Capacity

For $j\in\{k,k+1\}$, let $r(j)\in\{1,\dots,n\}$ be the index of the iteration in which Universal inserts $\Pi_{j}$ into $\Pi$. We distinguish two cases.

If $r(k)<r(k+1)$, then the item $\Pi_{k+1}$ cannot be a swap item, since it would appear in front of the item $\Pi_{k}$ if it was. As each non-swap item is inserted into $\Pi$ such that all items left of it are larger with respect to $\tilde{d}$, the claim follows.

If $r(k)>r(k+1)$, since it is not a swap item, $\Pi_{k}$ is put in front of $\Pi_{k+1}$ because it has a higher density. ∎

Since $\Pi_{k}$ is a swap item, it stands in front of all items inserted earlier into $\Pi$. Hence, all items that appear in front of $\Pi_{k}$ in $\Pi$ have been inserted in a later iteration of Universal. Since Universal processes items in order of non-decreasing sizes, we have $l(\Pi_{j})\geq l(\Pi_{k})$. ∎

We distinguish three cases.

First case: $\Pi_{j}$ is a swap item and $d(\Pi_{j})\geq d(\Pi_{k})$. By Lemma 4, we have $l(\Pi_{j})\geq l(\Pi_{k})$, and the claim trivially holds.

Second case: $\Pi_{j}$ is a swap item and $d(\Pi_{j})<d(\Pi_{k})$. Since $\Pi_{j}$ is a swap item, there is a capacity $C\geq l(\Pi_{j})$ such that

In particular, for $C=l(\Pi_{j})$ we obtain

Since, by Lemma 4, $l(\Pi_{j})\geq l(\Pi_{k})$, the item $\Pi_{k}$ is included in the set on the right hand side of (1). We conclude that $v(\Pi_{j})>v(\Pi_{k})$.

Third case: $\Pi_{j}$ is not a swap item. Let $\Pi_{j^{\prime}}$ be the first swap item after $\Pi_{j}$ in $\Pi$, i.e.,

Note that the minimum is attained as $\Pi_{k}$ is a swap item. The analysis of the first two cases implies that $v(\Pi_{j^{\prime}})\geq v(\Pi_{k})$. By Lemma 3 we have $d(\Pi_{j})\geq d(\Pi_{j+1})\geq\dots\geq d(\Pi_{j^{\prime}})$, and by Lemma 4 we have $l(\Pi_{j})\geq l(\Pi_{j^{\prime}})$. Hence, $v(\Pi_{j})\geq v(\Pi_{j^{\prime}})\geq v(\Pi_{k})$. ∎

We distinguish whether $\Pi_{k}$ is a swap item, or not.

If $\Pi_{k}$ is a swap item, by definition, $\Pi_{k}$ is worth more than the greedy set for some capacity $C\geq l(\Pi_{k})$. Thus,

Since items whose size is strictly larger than $l(\Pi_{k})$ are inserted into $\Pi$ at a later iteration of Universal, they can only end up behind $\Pi_{k}$ if they are smaller with respect to $\tilde{d}$. Hence,

and thus $v(\Pi_{k})>v(\{\Pi_{j}\,|\,j>k\text{ and }\tilde{d}(\Pi_{j})\succ\tilde{d}(\Pi_{k})\})$, as claimed.

If, on the other hand, $\Pi_{k}$ is not a swap item, let $\Pi_{k^{\prime}}$ be the first swap item after it in $\Pi$. If no such item exists, the claim holds by Lemma 3, since

Otherwise, by Lemma 3, we obtain $\tilde{d}(\Pi_{k})\succ\tilde{d}(\Pi_{k+1})\succ\dots\succ\tilde{d}(\Pi_{k^{\prime}})$ and hence

Consequently, and by the argument above for swap items,

Finally, by Lemma 5, we have $v(\Pi_{k})\geq v(\Pi_{k^{\prime}})\geq v(\{\Pi_{j}\,|\,j>k\text{ and }\tilde{d}(\Pi_{j})\succ\tilde{d}(\Pi_{k})\})$. ∎

We show that for every set of items $\mathcal{I}$, the permutation $\Pi=\textsc{Universal}(\mathcal{I})$ satisfies $v(\textsc{Opt}(\mathcal{I},C))\leq 2v(\Pi(C))$ for every capacity $C\leq l(\mathcal{I})$. By Theorem 1, it suffices to show $v(\Pi(C))\geq v(\textsc{MGreedy}(\mathcal{I},C))$ for all capacities. We distinguish between swap capacities and capacities where MGreedy outputs a greedy set.

First, assume that $C$ is a swap capacity, and let $\{\Pi_{k}\}=\textsc{MGreedy}(\mathcal{I},C)$ be the swap item returned by the modified greedy algorithm. Then, $\Pi(C)$ contains at least one item $\Pi_{j}$ with $j\leq k$. By Lemma 5 we have

Now assume that $C$ is not a swap capacity. Let $G^{+}=\textsc{MGreedy}(\mathcal{I},C)\setminus\Pi(C)$ be the set of items in the greedy set for capacity $C$ that are not packed by the permutation $\Pi$. Similarly, let $U^{+}=\Pi(C)\setminus\textsc{MGreedy}(\mathcal{I},C)$. If $G^{+}=\emptyset$, then $v(\Pi(C))\geq v(\textsc{MGreedy}(\mathcal{I},C))$ and we are done. Suppose now that $G^{+}\neq\emptyset$. Then, also $U^{+}\neq\emptyset$. For all items $i\in U^{+}$, we have $l(i)\leq C$ and $i\notin\textsc{MGreedy}(\mathcal{I},C)$. Since $C$ is not a swap capacity, $\textsc{MGreedy}(\mathcal{I},C)$ is the greedy set for capacity $C$, and thus $\tilde{d}(i)\prec\tilde{d}(i^{\prime})$ for all $i\in U^{+}$ and $i^{\prime}\in G^{+}$. By definition of $\Pi(C)$ and since $U^{+}\neq\emptyset$, we also have $k=\min\{j\mid\Pi_{j}\in U^{+}\}<\min\{k^{\prime}\mid\Pi_{k^{\prime}}\in G^{+}\}$, i.e., the first item $\Pi_{k}\in U^{+}$ in $\Pi$ is encountered before every item from $G^{+}$. It follows that

Using $v(U^{+})\geq v(\Pi_{k})$ and $v(\Pi_{k})\geq v(G^{+})$ (Lemma 6) we get

∎

We first argue how all swap items can be determined in time $\Theta(n\log n)$. We use that an item is a swap item if and only if it is worth more than all smaller items of higher density combined. This is true, because every item $i$ that is worth more than all smaller items of higher density is a swap item for capacity $l(i)$. Conversely, a swap item $i$ for capacity $C\geq l(i)$ is worth more than all items of higher density that are smaller than $C$.

We maintain a balanced search tree for items that is ordered by size and stores the total value of the items of both subtrees in the corresponding root. Inserting an item into this tree as well as determining whether an item is worth more than all smaller items in the tree both takes time $\Theta(\log n)$. To determine the set of swap items, we iterate over all items in order of decreasing densities and insert items one by one into the search tree. After each insertion, we query whether the newly inserted item is worth more than all smaller items in the tree. This is true if and only if the item is worth more than all smaller items of higher density, i.e., if and only if the item is a swap item. Including the initial sorting by density, we can determine all swap items in time $\Theta(n\log n)$.

We construct the output permutation $\Pi$ by iterating over the items in order of increasing size, as in Algorithm 2. We maintain a list $L$ of balanced search trees, each ordered by density. Except for the last tree in $L$, every tree contains exactly one swap item, which is the item of smallest density in the tree. The density of a tree is the density of this swap item (or 0 if the tree has no swap item). Each tree stores the items in $\Pi$ to the left of the corresponding swap item (if it exists) and to the right of the swap item of the preceding tree in $L$ (if it exists). We start with a list containing a single tree with no corresponding swap item, which eventually holds all non-swap items that end up behind the last swap item in $\Pi$. Whenever we encounter a new swap item, we add a new tree consisting of only this swap item to the front of $L$. For each non-swap item, we have to find the correct tree to insert it into. Once we know the tree, we can determine the position at which to insert the item into the tree, and thus in $\Pi$, in time $\Theta(\log n)$ simply by searching the tree.

To complete the proof, we need an efficient way to find the correct tree in $L$ for a non-swap item. For this purpose, we maintain a sublist $L^{\prime}$ of $L$ that contains only those trees that are needed for the remainder of the algorithm. Whenever a new swap item $s$ adds a tree to the front of $L$, we also add the tree to the front of $L^{\prime}$. Observe that from this point on no items are inserted into trees of a higher density than $s$. Hence, before inserting the tree of $s$ to $L^{\prime}$, we may remove trees of higher density from the front of $L^{\prime}$. This guarantees that $L^{\prime}$ remains sorted by density. We can thus implement $L^{\prime}$ as a balanced search tree order by density. This way, we can find the correct tree for each non-swap item in time $\Theta(\log n)$. Since every tree is removed at most once from $L^{\prime}$, the amortized cost for maintaining the sublist is constant for each swap item.

Since Universal requires $n$ iterations, the total running time is $\Theta(n\log n)$. ∎

We give a family of instances, one for each size $n\geq 3$. We ensure that for every item $i$ of the instance of size $n$, there is a capacity $C$, such that packing item $i$ first can only lead to a solution that is worse than $\textsc{Opt}(\mathcal{I},C)$ by a factor of at least $(2-4/n)$. This completes the proof, as the factor approaches $2$ for increasing values of $n$.

The instance of size $n$ is given by $\mathcal{I}=\{1,2,\dots,n\}$ with

where $F_{i}$ denotes the $i$-th Fibonacci number ($F_{1}=1,F_{2}=1,F_{3}=2,\dots$).

We need to show that, no matter which item is tried first (i.e., no matter which item is the root of the policy), there is a capacity for which this choice ruins the solution. Observe that both values and sizes of the items are strictly increasing. Assume that item $i\geq 3$ is packed first. Since the smallest item has size $l(1)=F_{n}$, for capacity $C_{i}=2F_{n}+F_{i}-2<2F_{n}+F_{i}-1=l(1)+l(i)$, no additional item fits the knapsack. However, the unique optimum solution in this case is $\textsc{Opt}(\mathcal{I},C_{i})=\{i-1,i-2\}$. These two items fit the knapsack, as $l(i-1)+l(i-2)=2F_{n}+F_{i-1}+F_{i-2}-2=2F_{n}+F_{i}-2=C_{i}$. By definition,

Hence, policies that first pack item $i\geq 3$ do not achieve a robustness factor $\alpha<2-3/n$.

Now, assume that one of the two smallest items is packed first. For capacity $C_{1,2}=l(n)=2F_{n}-1<2F_{n}=l(1)+l(2)$, no additional item fits the knapsack. The unique optimum solution, however, is to pack item $n$. It remains to compute the ratios

Hence, policies that first pack item 1 or item 2 do not achieve a robustness factor $\alpha<2-4/n$. ∎

Given an instance $\mathcal{I}$ of the oblivious knapsack problem with unit densities and any capacity $C\leq v(\mathcal{I})$, we compare the packing $\Pi(C)$ that results from the solution $\Pi=\textsc{UniversalUD}(\mathcal{I}$) with an optimal packing $\textsc{Opt}(\mathcal{I},C)$. We define the set $M$ of items in $\Pi(C)$ for which at least one smaller item is not in $\Pi(C)$, i.e., more precisely, let $M=\{i\in\Pi(C)\mid\exists j\in\mathcal{I}\backslash\Pi(C):j\prec i\}$.

We first consider the case that $M\neq\emptyset$ and set $i=\min_{\prec}M$ to be the smallest item in $M$ with respect to ‘$\prec$’. Consider the iteration $r$ of UniversalUD in which $i$ is inserted into $\Pi$, i.e., $i=L_{r}$. By definition of $M$, there is an item $j\prec i$ with $j\notin\Pi(C).$ Let $j$ be the first such item in $\Pi$. Since $j\prec i$, we have $j\in\Pi^{(r)}$. From $i\in\Pi(C)$ and $j\notin\Pi(C)$, it follows that $i$ precedes $j$ in $\Pi$ (and thus in $\Pi^{(r)}$). Let $i^{\prime}$ be the item directly preceding $j$ in $\Pi^{(r)}$. If $i^{\prime}=i$, $i$ was compared with $j$ when it was inserted into $\Pi^{(r)}$, with the result that $v(i)\geq\varphi v(j)$ and thus $v(\Pi(C))\geq\varphi v(j)$. If $i^{\prime}\neq i$, by definition of $j$, we still have $i^{\prime}\in\Pi(C)$. Also, either $i^{\prime}\succ j$ and thus $v(i^{\prime})\geq v(j)$, or $j$ was compared with $i^{\prime}$ when it was inserted into $\Pi$ in an earlier iteration of UniversalUD, with the result that $v(i^{\prime})>\frac{1}{\varphi}v(j)$. Again, $v(\Pi(C))\geq v(i)+v(i^{\prime})>v(j)+\frac{1}{\varphi}v(j)=\varphi v(j)$.

In both cases it follows from $j\notin\Pi(C)$ that $v(\textsc{Opt}(\mathcal{I},C))\leq C<v(\Pi(C))+v(j)$, and using $v(j)\leq\frac{1}{\varphi}v(\Pi(C))$ we get

Now, assume that $M=\emptyset$. Intuitively, this means that $\Pi(C)$ consists of a prefix of $L$ (the smallest items). Let $i_{1}\succ\dots\succ i_{k}$ be the items in $\Pi(C)\setminus\textsc{Opt}(\mathcal{I},C)$, and let $j_{1}\succ\dots\succ j_{l}$ be the items in $\textsc{Opt}(\mathcal{I},C)\setminus\Pi(C)$. As $\Pi(C)$ consists of a prefix of $L$, we have $|\Pi(C)|\geq|\textsc{Opt}(\mathcal{I},C)|$ and thus $k\geq l$. If $k=0$, the claim trivially holds. Otherwise, since $M$ is empty, we have $j_{l}\succ i_{1}$. Is suffices to show $v(j_{h})\leq\varphi v(i_{h})$ for all $h\leq l$. To this end, we consider any fixed $h\leq l$. From $v(\{i_{1},\dots,i_{h-1}\})\leq v(\{j_{1},\dots,j_{h-1}\})$ it follows that

This implies that $j_{h}$ cannot precede all items of $\{i_{h},\dots,i_{k}\}$ in $\Pi$, as $j_{h}\notin\Pi(C)$. Hence, there is an item $i\in\{i_{h},\dots,i_{k}\}$ that precedes $j_{h}$ in $\Pi$. Since $j_{h}\succ i$, in the iteration when UniversalUD inserted $j_{h}$ into $\Pi$, $i$ was already present. From the fact that $i$ ended up preceding $j_{h}$ it follows that $j_{k}$ was compared with $i$ and thus $v(j_{h})<\varphi v(i)\leq\varphi v(i_{h})$. We obtain

which implies the result. ∎

To improve the running time from the naïve $\Theta(n^{2})$, we maintain a balanced search tree $T$ that stores a subset of the items in $\Pi$ sorted decreasingly by their sizes. Whenever an item gets inserted to the front of $\Pi$, and only then, we also insert it into $T$. This way, the items in $T$ remain sorted by their positions in $\Pi$ throughout the execution of the algorithm. We need an efficient way of finding, in each iteration $r$ of UniversalUD (Algorithm 3), the first item $i$ in $\Pi^{(r)}$ for which $v(L_{r})\geq\varphi v(i)$, or detecting that no such item exists. We claim that, if such an item exists, it is stored in $T$ and can thus be found in time $\Theta(\log n)$.

It suffices to show that for every item $i\in T$ and its predecessor $j$ in $T$ we have that none of the items that precede $i$ in $\Pi$ are smaller than $j$. To see this, we argue that none of the items between $j$ and $i$ in $\Pi$ are smaller than $j$. We can then repeat the argument for $j$ and its predecessor $j^{\prime}$, etc. For the sake of contradiction, let $i^{\prime}$ be the first item between $j$ and $i$ with $v(i^{\prime})<v(j)$. None of the items between $j$ and $i^{\prime}$ are smaller than $j$, hence both $j$ and $i^{\prime}$ are inserted into $\Pi$ earlier than all of them. Let $r$ be the iteration in which $j$ is inserted into $\Pi$. Since $i^{\prime}$ is inserted earlier into $\Pi$, and since $j$ is inserted to the front of $\Pi^{(r)}$, $i^{\prime}$ is at the front of $\Pi^{(r-1)}$. This is a contradiction to $i^{\prime}$ not being in $T$. ∎

Consider an instance of the oblivious knapsack problem with five items of unit density and values equal to $v_{1}=1+\varepsilon,v_{2}=1+\varepsilon,v_{3}=2/\varphi,v_{4}=1+1/\varphi^{2},v_{5}=\varphi$, for sufficiently small $\varepsilon>0$. We show that no algorithm achieves a robustness factor of $\varphi-\delta$ for this instance. To this end we consider an arbitrary algorithm $\mathscr{A}$ and distinguish different cases depending on which item the algorithm tries to pack first.

1. (a)

   If $\mathscr{A}$ tries item 1 or item 2 first, it cannot fit any additional item for a capacity equal to $v_{5}=\varphi$, as even $v_{1}+v_{2}>\varphi$. For this capacity $\mathscr{A}$ is worse by a factor of $\varphi/(1+\varepsilon)>\varphi-\delta$ than the optimum solution, which packs item 5.

2. (b)

   If $\mathscr{A}$ tries item 3 first, it cannot fit any additional item for a capacity equal to $v_{1}+v_{2}=2+2\varepsilon$, as even $v_{3}+v_{1}>2+2\varepsilon$. For this capacity $\mathscr{A}$ is worse by a factor of $(1+\varepsilon)\varphi>\varphi-\delta$ than the optimum solution which packs items 1 and 2.

3. (c)

   If $\mathscr{A}$ tries item 4 first, it cannot fit any additional item for a capacity equal to $v_{2}+v_{3}=1+2/\varphi+\varepsilon$, as even $v_{4}+v_{1}=2+1/\varphi^{2}+\varepsilon>1+2/\varphi+\varepsilon$. For this capacity $\mathscr{A}$ is worse by a factor of $\frac{1+2/\varphi+\varepsilon}{1+1/\varphi^{2}}>\frac{\varphi+1/\varphi}{1+1/\varphi^{2}}=\varphi>\varphi-\delta$ than the optimum solution which packs items 2 and 3.

4. (d)

   If $\mathscr{A}$ tries item 5 first, it cannot fit any additional item for a capacity equal to $v_{3}+v_{4}=\varphi+1$, as even $v_{5}+v_{1}=\varphi+1+\varepsilon>\varphi+1$. For this capacity $\mathscr{A}$ is worse by a factor of $\frac{\varphi+1}{\varphi}=\varphi>\varphi-\delta$ than the optimum solution which packs items 3 and 4.