Matroidal Approximations of Independence Systems

TL;DR

This paper explores a heuristic for approximating maximum weight bases in independence systems using matroidal structures, which can outperform classical greedy algorithms even with minimal prior knowledge.

Contribution

It introduces a new heuristic based on inner matroids for independence systems, improving approximation guarantees over traditional methods.

Findings

Heuristic can outperform greedy algorithms in worst-case scenarios.

Performance improves even with no additional knowledge (${\mathcal O}={\mathcal I}$).

Provides theoretical analysis of the heuristic's approximation guarantees.

Abstract

Milgrom (2017) has proposed a heuristic for determining a maximum weight basis of an independence system ${\mathcal I}$ given that we want an approximation guarantee only for sets in a prescribed ${\mathcal O}\subseteq {\mathcal I}$. This ${\mathcal O}$ reflects prior knowledge of the designer about the location of the optimal basis. The heuristic is based on finding an `inner matroid', one contained in the independence system. We show that even in the case ${\mathcal O}={\mathcal I}$ of zero additional knowledge the worst-case performance of this new heuristic can be better than that of the classical greedy algorithm.

Equations

$$V^{\ast }(\mathcal{I},v)=\underset{S\in \mathcal{I}}{\max }v(S),V^{\ast }(\mathcal{O},v)=\underset{S\in \mathcal{O}}{\max }v(S),\text{ and }{V}_g^{\ast }(\mathcal{I},v)=\text{value of the greedy solution}.$$

$$\rho (\mathcal{I},\mathcal{O},\mathcal{J})\text{raise}0.40903pt:=\underset{S\in \mathcal{O}\setminus \{\varnothing \}}{\min }\underset{S^{\prime }\in {\mathcal{B}}_{\mathcal{J}}(S)}{\min }\frac{|S^{\prime }|}{|S|};$$

$$\rho (\mathcal{I},\mathcal{O},\mathcal{J})=\underset{S\in \mathcal{O}\setminus \{\varnothing \}}{\min }\frac{l_{\mathcal{J}}(S)}{|S|}.$$

$$\rho (\mathcal{I},\mathcal{O},\mathcal{M})=\underset{S\in \mathcal{O}\setminus \{\varnothing \}}{\min }\frac{r_{\mathcal{M}}(S)}{|S|}=\underset{S\in \mathcal{O}\setminus \{\varnothing \}}{\min }\underset{S^{\prime }\in \mathcal{M}:S^{\prime }\subseteq S}{\max }\frac{|S^{\prime }|}{|S|},$$

$$\rho (\mathcal{I},\mathcal{I},{\mathcal{M}}^{\prime })=\frac{1}{k_2}=q(\mathcal{I}).$$

$$\rho \left(\mathcal{M},\left(\genfrac{}{}{0ex}{}{E_2}{k_2}\right),{\mathcal{M}}_2\right)=1>\frac{1}{k_2}=q(\mathcal{I}).$$

$$W\text{raise}0.40903pt:=\{w\in {\mathbb{R}}_{+}^E:1^{\top }w=1\}\cap \{w\in {\mathbb{R}}_{+}^E:\varnothing \ne \mathcal{O}\cap {\mathrm{argmax}}_{I\in \mathcal{I}}w(I)\}$$

$$\{w\in {\mathbb{R}}_{+}^E:\varnothing \ne \mathcal{O}\cap {\mathrm{argmax}}_{I\in \mathcal{I}}w(I)\}$$

$$\rho (\mathcal{I},\mathcal{O},\mathcal{M})=\underset{v\in W}{\min }\frac{V^{\ast }(\mathcal{M},v)}{V^{\ast }(\mathcal{O},v)}.$$

$$\rho (\mathcal{I},\mathcal{O},\mathcal{M})>\underset{v\in W}{\min }\frac{V^{\ast }(\mathcal{M},v)}{V^{\ast }(\mathcal{O},v)}.$$

$$v^{\ast }\in {\mathrm{argmin}}_{v\in W:\mathrm{supp}(v)\le \ell }\frac{V^{\ast }(\mathcal{M},v)}{V^{\ast }(\mathcal{O},v)},$$

$$\underset{S\in \mathcal{O}\setminus \{\varnothing \}}{\min }\frac{r_{\mathcal{M}}(S)}{|S|}>\frac{V^{\ast }(\mathcal{M},v^{\ast })}{V^{\ast }(\mathcal{O},v^{\ast })}.$$

$$\underset{S\in {\mathcal{O}}^{\prime }\setminus \{\varnothing \}}{\min }\frac{r_{{\mathcal{M}}^{\prime }}(S)}{|S|}\ge \underset{S\in \mathcal{O}\setminus \{\varnothing \}}{\min }\frac{r_{\mathcal{M}}(S)}{|S|},$$

$$\frac{V^{\ast }(\mathcal{M},v^{\ast })}{V^{\ast }(\mathcal{O},v^{\ast })}\ge \frac{V^{\ast }({\mathcal{M}}^{\prime },v^{\ast })}{V^{\ast }({\mathcal{O}}^{\prime },v^{\ast })}.$$

$$\underset{S\in {\mathcal{O}}^{\prime }\setminus \{\varnothing \}}{\min }\frac{r_{{\mathcal{M}}^{\prime }}(S)}{|S|}=\rho ({\mathcal{I}}^{\prime },{\mathcal{O}}^{\prime },{\mathcal{M}}^{\prime })>\frac{V^{\ast }({\mathcal{M}}^{\prime },v^{\ast })}{V^{\ast }({\mathcal{O}}^{\prime },v^{\ast })}.$$

$${\hat{v}}^{\alpha }=\frac{1}{{\sum }_{i\in E}{v}_i^{\alpha }}{v}^{\alpha }\text{ and observe }\frac{{\hat{v}}^{\alpha }(S)}{{\hat{v}}^{\alpha }(T)}=\frac{v^{\alpha }(S)}{v^{\alpha }(T)}\forall S,T:v(T)>0.$$

$${\hat{v}}_j^{\alpha }\text{raise}0.40903pt:=\{\begin{array}{ll}\bar{v}+\alpha & \text{ if }{v}_j^{\ast }=\bar{v},\\ v_j^{\ast }& \text{ otherwise.}\end{array}$$

$$f(\alpha )\text{raise}0.40903pt:=\frac{V^{\ast }(\mathcal{M},{\hat{v}}_j^{\alpha })}{V^{\ast }(\mathcal{O},{\hat{v}}_j^{\alpha })}.$$

$$f(\alpha )\text{raise}0.40903pt:=\frac{V^{\ast }(\mathcal{M},{\hat{v}}_j^{\ast })+\alpha |\{i\in X_{\mathcal{M}}:\bar{v}=v_i^{\ast }\}|}{V^{\ast }(\mathcal{O},{\hat{v}}_j^{\alpha })+\alpha |\{i\in X_{\mathcal{O}}:\bar{v}=v_i^{\ast }\}|}.$$

$${\rho }^M(\mathcal{I},\mathcal{O})\text{raise}0.40903pt:=\underset{\mathcal{M}\subseteq \mathcal{I}:\mathcal{M}\text{ is a matroid}}{\max }\rho (\mathcal{I},\mathcal{O},\mathcal{M}).$$

$$r(X)\le |X|$$

$$\text{If }X\subseteq Y,\text{ then }r(X)\le r(Y)$$

$$r(X\cup Y)+r(X\cap Y)\le r(X)+r(Y)$$

$$\begin{array}{rlrl}{\rho }^M(\mathcal{I},\mathcal{O})=\underset{\delta }{\max }\underset{S\in \mathcal{O}}{\min }(1-\frac{{\delta }_S}{r(S)})& & & \\ \text{s.t. }0\le {\delta }_S& \le r(S)& & \forall S\subseteq E\\ [r(S)-{\delta }_S]& \le [r(T)-{\delta }_T]& & \forall S\subset T\subseteq E\\ [r(S\cup j\cup k)-{\delta }_{S\cup j\cup k}]-[r(S\cup j)-{\delta }_{S\cup j}]& \le [r(S\cup k)-{\delta }_{S\cup k}]-[r(S)-{\delta }_S]& & \forall j,k\notin S\subseteq E\\ {\delta }_S& \text{ integral }& & \forall S\subseteq E\end{array}$$

$${\rho }^M(\mathcal{I},\mathcal{I})\ge 1/2>2/5=q(\mathcal{I}).$$

$$\mathcal{I}=\left\{I\subseteq E:\text{ either }I=\{-3,-2,-1\}\text{ or }|I\cap \{-3,-2,-1\}|\le 2\right\}.$$

$$q(F)=3/7.$$

$$\mathcal{M}=\left\{I\subseteq E:|I\cap \{-3,-2,-1\}|\le 2\right\}.$$

$${\rho }^M(\mathcal{I},\mathcal{I})=2/3.$$

Full text

††journal:

1 Introduction

Given a ground set $E=\{1,\ldots,n\}$ and a family $\mathcal{I}$ of subsets of $E$, the pair $(E,\mathcal{I})$ is a called an independence system if $\emptyset\in\mathcal{I}$ and for all $B\subseteq A\in\mathcal{I}$ we have $B\in\mathcal{I}$ as well. Elements of $\mathcal{I}$ are called independent sets. For any $A\subseteq E$, a set $B\subseteq A$ is called a basis of $A$ if $B\in\mathcal{I}$ and $B\cup\{j\}\not\in\mathcal{I}$ for all $j\in A\setminus B$. For $A\subseteq E$ denote by $\mathcal{B}_{\mathcal{I}}(A)$ (or $\mathcal{B}(A)$ when no ambiguity) the set of bases of $A$ (with respect to $\mathcal{I}$). The bases with respect to $E$ are denoted by $\mathcal{B}_{\mathcal{I}}$ or just $\mathcal{B}$. Minimally dependent sets are called circuits. If we associate weights $v_{i}\in{\mathbb{R}}_{+}$ with each $i\in E$, the problem of finding a maximum weight basis of $(E,\mathcal{I})$ can be expressed as $\max_{A\in\mathcal{B}}\sum_{i\in A}v_{i}$. For convenience we will write $\sum_{i\in S}v_{i}$ as $v(S)$ for all $S\subseteq E$. The problem is NP-hard by reduction to Hamiltonian path in a directed graph [4].

An independence system $(E,\mathcal{I})$ with $\{i\}\in\mathcal{I}$ for all $i\in E$ is called normal (for graphs or matroids, the term loopless is more common). A non-normal independence system can be made normal by deleting the elements $e\in E$ with $\{e\}\notin\mathcal{I}$ without changing the solution to the problem of finding a maximum weight basis.

Milgrom [5] proposed a heuristic for determining a maximum weight basis given prior information about the optimal solution. This prior information is expressed as a subset ${\mathcal{O}}$ of $\mathcal{I}$. The collection ${\mathcal{O}}$ is to be interpreted to be the set of likely candidates for an optimal basis. The performance of Milgrom’s heuristic is evaluated with respect to ${\mathcal{O}}$. It is based on finding an ‘inner matroid’, one contained in the independence system. We compare the worst-case performance of this heuristic with the well known greedy algorithm for the same problem. We show that the worst-case performance of Milgrom’s heuristic even in the absence of additional assumptions on ${\mathcal{O}}$ may be better than that of the classical greedy algorithm. Additionally, as illustrated by example when one exploits ${\mathcal{O}}$, there are aspects of Milgrom’s proposal that bear further investigation. We defer a discussion of these matters till the end. In the next section we describe the greedy algorithm and state its worst case performance.

2 Greedy Algorithm

We describe the greedy algorithm (see algorithm 1 next page) for finding a basis of $E$ with possibly large weight

Remark*.*

If some coefficients in the objective function are equal, the order of the $v_{i}$’s is not unique and therefore the algorithm’s outcome is not necessarily unique. To ensure uniqueness assume an exogenously given tie breaking rule.

A worst case bound on the quality of the greedy solution in terms of the rank quotient of an independence system can be found in [3, 4].

Definition 1**.**

Let $(E,\mathcal{I})$ be an independence system. The rank of $F\subseteq E$ is defined by $r(F)\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\max\{|B|\,:\,B\text{ basis of }F\}$ and the lower rank of $F$ is defined by $l(F)\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\min\{|B|\colon B\text{ basis of }F\};$ both map $2^{E}$ into the nonnegative integers. The rank quotient of $\mathcal{I}$ is denoted by $q(\mathcal{I})\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\min_{F\subseteq E,r(F)>0}\,\frac{l(F)}{r(F)}.$

Following an axiomatization of matroids by [2], we define a matroid to be an independence system $(E,\mathcal{I})$ with $q(\mathcal{I})=1$. There are several equivalent characterizations, some of which we will use later:

Basis exchange:

For every pair of bases $B^{1},B^{2}\in\mathcal{B}$ and $i\in B^{2}\setminus B^{1}$ there exists a $j\in B^{1}\setminus B^{2}$ such that $(B^{1}\cup\{i\})\setminus\{j\}$ is a basis.

Augmentation property:

For every pair of independent sets $I,J\in\mathcal{I}$ with $|I|<|J|$ there exists $j\in J\setminus I$ such that $I\cup\{j\}$ is an independent set.

Rank axioms:

see (R1)-(R3) on page R1.

For proofs of these and more see [6].

If $(E,\mathcal{I})$ is a normal independence system, then, $q(\mathcal{I})\in\left\{\frac{a}{b}\mid a\in\{1,2,\dots,l(\mathcal{I})\}\text{ and }b\in\{1,2,\dots,r(\mathcal{I})\}\right\}.$ Hence, for normal independence systems, $q(\mathcal{I})\geq 1/r(\mathcal{I}).$ However, a better bound is known:

Theorem 1** (2).**

Let $(E,\mathcal{I})$ be an independence system. If, for any $A\in\mathcal{I}$ and $e\in E$ the set $A\cup\{e\}$ contains at most $p$ circuits, then $q(\mathcal{I})\geq 1/p$. If $(E,\mathcal{I})$ is the intersection of $p$ matroids, then, $q(\mathcal{I})\geq 1/p$.

Theorem 2** (3, 4).**

For an independence system $\mathcal{I}$ on $E$ with objective function $v\gneqq\mathbf{0}$, let $I_{g}$ be the solution returned by Algorithm (1) and $I_{o}$ be a maximum weight basis. Then, $q(\mathcal{I})\leq\frac{v(I_{g})}{v(I_{o})}\leq 1.$ Also, for every independence system there are weights $v\in\{0,1\}^{E}$, such that the first inequality holds with equality.

The only attempt to improve upon this bound we are aware of involves incorporating a partial enumeration stage into the greedy algorithm, see [2].

3 Inner Matroid

Milgrom [5] proposes an alternative to the greedy algorithm with two parts. First, is the introduction of a-priori information on where the optimal basis may lie. Formally, let ${\mathcal{O}}\subseteq\mathcal{I}$ be a collection of independent sets one of which is conjectured to be an optimal weight basis. ${\mathcal{O}}$ need not satisfy the hereditary property (so $J\subset I\in{\mathcal{O}}$ does not imply $J\in{\mathcal{O}}$) and may exclude a basis that is in fact optimal (see Example 11). Call ${\mathcal{O}}$ the acceptable set. The second part defines a matroid ‘inside’ the independence system (but not necessarily containing all of ${\mathcal{O}}$) and applies the greedy algorithm to find an optimal weight basis of that matroid. We describe this approach here.

Given an independence system $(E,\mathcal{I}),$ an acceptable set ${\mathcal{O}}$, and a weight vector $v\in{\mathbb{R}}_{+}^{E}$ let

[TABLE]

Independence system $(E,\mathcal{M})$ is a matroid by the previous definition involving rank quotient and Theorem (2) if and only if $V^{*}(\mathcal{M},v)=V_{g}^{*}(\mathcal{M},v)$ for all $v\in{\mathbb{R}}_{+}^{E}.$

Call an independence system $(E,\mathcal{J})$ contained in $(E,\mathcal{I})$, an inner independence system, if $\mathcal{J}\subseteq\mathcal{I}$. If $(E,\mathcal{J})$ is a matroid it is called an inner matroid.

The approximation quality of $\mathcal{J}$ for $\mathcal{I}$ with respect to the acceptable set ${\mathcal{O}}\subseteq\mathcal{I}$ is given by

[TABLE]

so this picks the element from ${\mathcal{O}}$ that is least well approximated by a basis of $\mathcal{J}$. In terms of the lower rank $l_{\mathcal{J}}$ of $(E,\mathcal{J})$ this yields

[TABLE]

If $\mathcal{M}$ is an inner matroid, then $r_{\mathcal{M}}=l_{\mathcal{M}}$ and

[TABLE]

which is the form [5] originally chose.

It is natural to ask when $\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})=1$.

Theorem 3**.**

If $(E,\mathcal{I})$ is a normal independence system with acceptable set ${\mathcal{O}}\subseteq\mathcal{I}$, then, there exists a matroid $\mathcal{M}$ with $\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})=1$ if and only if $\mathcal{I}$ has an inner matroid containing ${\mathcal{O}}^{\dagger}\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\bigcup_{F\subseteq B\in{\mathcal{O}}}\{F\}$.

Proof.

If $\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})=1$ then ${\mathcal{O}}\subseteq\mathcal{M}$. For $I\subseteq J\in{\mathcal{O}}$ follows $J\in\mathcal{M}$ and by hereditary property of matroid $I\in\mathcal{M}.$   ∎

We give three examples to suggest that nothing stronger is possible.

Definition 2**.**

Let $U^{k}_{n}$ denote the uniform matroid of all subsets of an $n$-element set of cardinality at most $k$, where $0\leq k\leq n$. If we want to specify the ground set $E$ of $n$ elements explicitly we can also write $U^{k}_{E}.$

Example 4**.**

In all cases the ground set is $E=\{1,2,3,4\}$.

• •

Let $\mathcal{I}=\{\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{3,4\}\}$. $(E,\mathcal{I})$ is not a matroid as the basis exchange axiom is violated by $\{1,2\}$ and $\{3,4\}$. If ${\mathcal{O}}=\{\{1\},\{2\},\{3\},\{4\}\}$, then, $\rho(\mathcal{I},{\mathcal{O}},U^{1}_{4})=1$. This demonstrates, that the condition “${\mathcal{O}}$* is a matroid” is too strong. However, in this case ${\mathcal{O}}^{\dagger}$ is a matroid.*

• •

Let $\mathcal{I}$ be as above, but ${\mathcal{O}}=\{\{1,2\},\{3,4\}\}$. Then, $\rho(\mathcal{I},{\mathcal{O}},U^{1}_{4})=1/2$. In this case ${\mathcal{O}}^{\dagger}$ is not a matroid.

• •

Let $\mathcal{I}=\{\emptyset\{1\},\{2\},\{3\},\{4\},\{1,2\},\{2,3\},\{3,4\},\{1,3\},\{1,4\},\{2,4\},\{1,2,3\}\}.$ $(E,\mathcal{I})$ is not a matroid, since $\{3,4\}$ is a independent set that can not be augmented with an element from the larger independent set $\{1,2,3\}$. If ${\mathcal{O}}=\{\{2,3\},\{3,4\}\}$ the smallest inner matroid of $\mathcal{I}$ containing ${\mathcal{O}}$ is $\mathcal{M}=U^{2}_{\{2,3,4\}}$; and $\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})=1.$ However ${\mathcal{O}}^{\dagger}$ is not a matroid, since it lacks the basis $\{2,4\}$ that would be required by basis exchange. So requiring “${\mathcal{O}}^{\dagger}$* to be a matroid” is too strong.*

Remark*.*

Clearly $\rho(\mathcal{I},{\mathcal{O}},\mathcal{J})\geq\rho(\mathcal{I},{\mathcal{O}}^{\dagger},\mathcal{J}),$ since the minimum on the right hand side is determined over a larger set.

However, $\rho(\mathcal{I},{\mathcal{O}},\mathcal{J})>\rho(\mathcal{I},{\mathcal{O}}^{\dagger},\mathcal{J})$ is possible, as the following example demonstrates:

Example 5**.**

Consider $E=\{1,2,3,4,5,6\}$ and the independence system $\mathcal{I}=2^{\{1,2,3,4\}}\cup 2^{\{3,4,5,6\}}.$ Clearly $\{1\},\{5,6\}\in\mathcal{I}$. The augmentation property requires that $\{1,5\}$ or $\{1,6\}$ has to be independent, which is not the case. Therefore $\mathcal{I}$ is not a matroid. Let ${\mathcal{O}}=\{\{1,2,3,4\},\{3,4,5,6\}\}$ be our acceptable set and let $\mathcal{M}=2^{\{1,2,3,4\}}$. Now, let us determine $\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})=\min_{S\in{\mathcal{O}}\setminus\{\emptyset\}}\frac{r_{\mathcal{M}}(S)}{|S|}.$ With $r_{\mathcal{M}}(\{1,2,3,4\})=4$ and $r_{\mathcal{M}}(\{3,4,5,6\})=2$ we deduce that $\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})=\frac{2}{4}=\frac{1}{2}.$

Now, ${\mathcal{O}}^{\dagger}$ contains $O=\{4,5,6\}$ with $r_{\mathcal{M}}(O)=1$. Even worse $O=\{6\}\in{\mathcal{O}}^{\dagger}$ with $r_{\mathcal{M}}(O)=0$ which demonstrates $\rho(\mathcal{I},{\mathcal{O}}^{\dagger},\mathcal{M})=0<\frac{1}{2}=\rho(\mathcal{I},{\mathcal{O}},\mathcal{M})$.

Corollary 1**.**

The example demonstrates, that $\rho(\mathcal{I},{\mathcal{O}}^{\dagger},\mathcal{M})>0$ is only possible if $e\in\bigcup_{O\in{\mathcal{O}}}O$ implies $\{e\}\in\mathcal{M}.$

Example 6** (twin-peaks).**

Consider two disjoint sets $E_{1}$ and $E_{2}$ and integers $k_{1},k_{2}$ such that $k_{1}<|E_{1}|<k_{2}<|E_{2}|.$ Let $\mathcal{M}_{1}=U^{k_{1}}_{E_{1}}$ and $\mathcal{M}_{2}=U^{k_{2}}_{E_{2}}$.

Define $(E,\mathcal{I})$ to be an independence system on $E\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}E_{1}\dot{\cup}E_{2}$ where a set $A\subseteq E$ is independent if and only if $A$ is an independent set in $\mathcal{M}_{1}$ or $\mathcal{M}_{2}$, thus $\mathcal{I}=\mathcal{M}_{1}\cup\mathcal{M}_{2}$. Call this a ‘twin-peaks’ independence system. Choose an $F\subseteq E$ which consists of one element from $E_{1}$ and $k_{2}$ elements from $E_{2}$. Now $l(F)=1$ and $r(F)=k_{2}$. Since larger ratios between sets in this independence system are impossible, $q(\mathcal{I})=\frac{1}{k_{2}}$.

To determine the best matroid approximating ${\mathcal{O}}=\mathcal{I}$ contained in $\mathcal{I}$, we must examine all matroids $\mathcal{M}^{\prime}$ contained in $\mathcal{I}:$

Suppose first that there exists $i\in E_{1},j\in E_{2}$ such that $\{i\},\{j\}\in{{\mathcal{M}}^{\prime}}$. We show that the rank of ${{\mathcal{M}}^{\prime}}$ is 1, i.e. the largest bases of ${{\mathcal{M}}^{\prime}}$ are of size one. If not, there exists a set $F\in{{\mathcal{M}}^{\prime}}$ of cardinality two. By construction of $\mathcal{I}$ the set $F$ must be a subset of $E_{1}$ or $E_{2}$, wlog. let $F\subseteq E_{1}$; since $F$ and $\{j\}$ are independent sets in ${{\mathcal{M}}^{\prime}}$ there has to be an element in $F$ with which we could augment $\{j\}$ to be an independent set $F^{\prime}$ in ${{\mathcal{M}}^{\prime}}$. But $F^{\prime}$ contains elements from $E_{1}$ and $E_{2}$ and therefore is not independent in $\mathcal{I}$ and therefore not in ${{\mathcal{M}}^{\prime}}.$

The largest matroid of this kind contained in $\mathcal{I}$ is $U^{1}_{E}$. Hence, every independent set from $\mathcal{I}$ is approximated by an arbitrary contained singleton and we obtain the worst case bound of $1/\max(k_{1},k_{2})=1/k_{2}.$ 2. 2.

On the other hand, if $r({{\mathcal{M}}^{\prime}})>1$ then we can conclude that either all independent sets of ${\mathcal{M}}^{\prime}$ are contained in $E_{1}$ or they are contained in $E_{2}$ and the inclusion-wise largest matroids contained in $E_{1}$ and $E_{2}$ are $\mathcal{M}_{1}$ and $\mathcal{M}_{2},$ respectively.

For ${{\mathcal{M}}^{\prime}}=\mathcal{M}_{1}$, approximating sets $F\in\mathcal{M}_{2}\setminus\{\emptyset\}$ is only possible with the empty set which yields a quotient of $0/|F|$. For ${{\mathcal{M}}^{\prime}}=\mathcal{M}_{2}$, approximating sets approximating nonempty sets $F\in\mathcal{M}_{1}\setminus\{\emptyset\}$ is only possible with the empty set which yields a quotient of $0/|F|.$

In summary we obtain the best approximation in the first case and can conclude

[TABLE]

However, $\rho$ can be much better if we exploit additional knowledge. As an illustration, suppose that ${\mathcal{O}}=\binom{E_{2}}{k_{2}}$, i.e., all subsets of size at most $k_{2}$ of $E_{2}$. This might be justified by a situation, where we know that the optimal solutions which occur in our environment, have to live in $E_{2}$, maybe since $k_{1}\ll k_{2}$. Unsurprisingly, the matroid $\mathcal{M}_{2}$ gives us the best conceivable guarantee of $1$:

[TABLE]

Milgrom [5] proposes that the maximum weight basis of an inner matroid $\mathcal{M}$ be used as a solution to the problem of finding a maximum weight basis in $(E,\mathcal{I})$. The objective function value of this solution will be $V^{*}(\mathcal{J},v)$. [5] gives a bound on the quality of this solution in terms of $\rho(\mathcal{I},{\mathcal{O}},\mathcal{J})$.

Definition 3**.**

For a given independence system $(E,\mathcal{I})$ and $e\in E$ define a new independence system $\mathcal{I}\setminus\{e\}$ on $E\setminus\{e\}$ by $\mathcal{I}\setminus\{e\}\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\{I\in\mathcal{I}\colon e\notin I\}.$ Additionally, for $\{e\}\in\mathcal{I}$ define the independence system $\mathcal{I}\mathbin{/}\{e\}$ on $E\setminus\{e\}$ by $\mathcal{I}/\{e\}\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\{I\setminus\{e\}\colon e\in I\in\mathcal{I}\}.$

Remark*.*

It is easy to see, that $(E\setminus\{e\},\mathcal{I}\setminus\{e\})$ is an independence system and $\mathcal{I}\mathbin{/}\{e\}$ is an independence system if $\{e\}\in\mathcal{I}$. Furthermore, if $(E,\mathcal{I})$ is a matroid, so is $(E\setminus\{e\},\mathcal{I}\setminus\{e\})$.

Lemma 7**.**

Given an independence system $(E,\mathcal{I})$ and a set ${\mathcal{O}}$ the set

[TABLE]

is compact.

Proof.

Clearly the set $\{w\in{\mathbb{R}}^{E}_{+}\colon\mathbf{1}^{\top}w=1\}$ is compact. Also,

[TABLE]

where each set on the right is a closed cone. As ${\mathcal{O}}$ is finite the right hand side is closed. Therefore $W$ is compact.   ∎

Theorem 8** (5).**

Let $(E,\mathcal{M})$ be an inner matroid and $W$ be the above defined set of non-negative and non-trivial weight vectors $v$ such that at least one member of ${\mathcal{O}}$ is a maximum weight basis of $(E,\mathcal{M})$. Then,

[TABLE]

As a convenience for the reader we provide here an independent, fuller proof of the result than [5].

Proof.

If the theorem is false, there must exist an independence system $(E,\mathcal{I})$, a set ${\mathcal{O}}$ and an inner matroid $(E,\mathcal{M})$ such that

[TABLE]

Among all such counterexamples choose one where $|E|$ is minimal. Among all such counterexamples choose one, where the support of $v$ is smallest. Denote the size of the support by $\ell$.

Let

[TABLE]

such that it is a vector with as few distinct values as possible. Choose $X_{\mathcal{O}}\in\operatorname{arg\,max}_{S\in{\mathcal{O}}}v^{*}(S),$ and $X_{\mathcal{M}}\in\operatorname{arg\,max}_{S\in\mathcal{M}}v^{*}(S)$ such that $|X_{\mathcal{O}}\cap X_{\mathcal{M}}|$ is maximal. In case of ties, choose a tie-breaking rule, so that greedy yields $X_{\mathcal{M}}$. By the minimum support assumption, $v^{*}_{j}=0$ for all $j\not\in X_{\mathcal{M}}\cup X_{{\mathcal{O}}}$. Additionally:

Claim**.**

$v^{*}_{j}>0$* for all $j\in E\setminus X_{\mathcal{O}}.$*

Proof of claim.

Suppose there exists $j\in E\setminus X_{\mathcal{O}}$ with $v^{*}_{j}=0.$ Let ${{\mathcal{I}}^{\prime}}=\mathcal{I}\setminus\{j\}$, $E^{\prime}=E\setminus\{j\}$, ${{\mathcal{M}}^{\prime}}=\mathcal{M}\setminus\{j\}$ and ${{\mathcal{O}}^{\prime}}=\{O\in{\mathcal{O}}\colon j\notin O\}.$ Because $j\notin X_{\mathcal{O}}\in{{\mathcal{O}}^{\prime}}$ we have ${{\mathcal{O}}^{\prime}}\neq\emptyset$.

For our counterexample we have by (1) and (2):

[TABLE]

Now, since the minimum on the left below is taken over a smaller set and $r_{{\mathcal{M}}^{\prime}}(S)=r_{\mathcal{M}}(S)$ for all $S\in{{\mathcal{O}}^{\prime}}$,

[TABLE]

On the other hand, $V^{*}(\mathcal{M},v^{*})\geq V^{*}({{\mathcal{M}}^{\prime}},v^{*})$ and $V^{*}({\mathcal{O}},v^{*})=V^{*}({{\mathcal{O}}^{\prime}},v^{*})>0$ yield

[TABLE]

Together

[TABLE]

Therefore, we obtain a counterexample with $|E^{\prime}|<|E|$ in contradiction to our assumption of minimality of the counterexample.   $\boxminus$

Claim**.**

$v^{*}_{k}=0\,\,\forall k\in X_{\mathcal{M}}\setminus X_{\mathcal{O}}$.

Proof of claim.

Suppose a $k\in X_{\mathcal{M}}\setminus X_{\mathcal{O}}$ with $v^{*}_{k}>0$ exists. Let $e^{k}$ denote the $k$-th unit vector. Consider the valuations $v^{\alpha}=v^{*}-\alpha e^{k}$ for $0\leq\alpha<v^{*}_{k}$ and their normalization

[TABLE]

Now $\hat{v}^{\alpha}$ differs from $v^{*}$ in two ways. First, the weight of one element outside $X_{\mathcal{O}}$ was reduced by $\alpha$. Second, the weights were scaled. Hence, $X_{\mathcal{O}}$ is optimal with respect to $v^{\alpha}$ for all $\alpha\geq 0$.

If there exists no $\epsilon>0$ such that $X_{\mathcal{M}}$ remains optimal for $\mathcal{M}$ with respect to $v^{\alpha}$ for $0\leq\alpha\leq\epsilon$, then it is the case that $X_{\mathcal{M}}$ is not optimal for any $v^{\epsilon}$ with $\epsilon>0.$ This requires, that the run of the greedy algorithm changes if $\epsilon>0$. However, the greedy algorithm does not depend on the exact weights of $v^{\epsilon}$ but only on their ordering with respect to decreasing weight. The only change in ordering is, that with respect to $v^{\epsilon}$ for $\epsilon>0$ that item $k$ will occur later. So before the greedy algorithm for $v^{\epsilon}$ considers $k$ it picks up some item $k^{\prime}$ that is not part of $X_{\mathcal{M}}.$ Now, as $k^{\prime}$ was not choosen for $v^{0}$, item $k^{\prime}$ must be dependent with respect to the previously choosen items and $k$. Hence there exists a circuit in $X_{\mathcal{M}}\cup k^{\prime}$ that contains $k$ and $k^{\prime}.$ Now, as this happens for arbitrarily small $\epsilon$ it has to be the case, that $v_{k}=v_{k^{\prime}}$ (and all elements choosen inbetween have that same weight). But this yields, that $X_{\mathcal{M}}\cup k^{\prime}$ is dependent while $X_{\mathcal{M}}\setminus\{k\}\cup\{k^{\prime}\}$ is another basis and $k^{\prime}\in X_{\mathcal{O}}\setminus X_{\mathcal{M}}$ which contradicts our assumption, that $X_{\mathcal{M}}$ and $X_{\mathcal{O}}$ were chosen to have a maximal intersection.

Therefore, such an $\epsilon>0$ has to exist so that $X_{\mathcal{M}}$ stays optimal for $v^{\alpha}$ with $0<\alpha<\epsilon$. Now, observe $v^{\alpha}(X_{\mathcal{M}})$ and $\hat{v}^{\alpha}(X_{\mathcal{M}})/\hat{v}^{\alpha}(X_{\mathcal{O}})$ are decreasing for $\alpha$ increasing from [math] to $\epsilon$. As $|E|$ and $|\operatorname{supp}v|$ do not change this contradicts our choice of $v^{*}$ in (2).   $\boxminus$

Together with the previous claim we obtain:

Claim**.**

$X_{\mathcal{M}}\setminus X_{\mathcal{O}}=\emptyset$, therefore $X_{\mathcal{M}}\subseteq X_{\mathcal{O}}$ and $X_{\mathcal{O}}=E$.

Claim**.**

For every $i\in X_{\mathcal{M}}$ there exists an $j\in X_{\mathcal{O}}\setminus X_{\mathcal{M}}$ such that $v^{*}_{i}=v^{*}_{j}$ and vice-versa.

Proof of claim.

Consider a valuation where $v^{*}_{i}$ is slightly decreased. If $X_{\mathcal{O}}$ and $X_{\mathcal{M}}$ do not change, then a smaller quotient (1) for that valuation would result, which is a contradiction. As $X_{\mathcal{O}}=E$ that cannot change either; so $X_{\mathcal{M}}$ has to change for arbitrarily small changes to $v^{*}_{i}$, and it can change only, by dropping $i$ and picking up some new element $j\in X_{\mathcal{O}}\setminus X_{\mathcal{M}}.$ But this requires $v^{*}_{i}=v^{*}_{j}$.

Similarly, consider a valuation, where $v^{*}_{j}$ is slightly increased. If $X_{\mathcal{O}}$ and $X_{\mathcal{M}}$ do not change, then a smaller quotient for that valuation would result, which contradicts assumption. Again, $X_{\mathcal{O}}=E$ ensures that $X_{\mathcal{O}}$ remains optimal during these slight increases. However, it could be, that $j$ immediately enters $X_{\mathcal{M}}$ but this requires, that just as immediately some $i^{\prime}\in X_{\mathcal{M}}$ exits $X_{\mathcal{M}}$. But this requires $v^{*}_{i^{\prime}}=v^{*}_{j}$.   $\boxminus$

Let $\bar{v}=\max_{i\in E}v^{*}_{i}$, and $\dbarv\mathrel{\raise 0.40903pt\hbox{:}\mkern-4.2mu=}\max_{i:v^{*}_{i}\neq\bar{v}}v^{*}_{i},$ and set

[TABLE]

If $\dbarv=0$ then $v^{*}$ is a multiple of the all ones vector, and the proof would be done, as such a $v^{*}$ could not provide a counterexample.

Otherwise, if $\dbarv>0$ for $|\alpha|\leq\bar{v}-\dbarv$, $X_{\mathcal{O}}$ and $X_{\mathcal{M}}$ remain optimal for $V^{*}({\mathcal{O}},\hat{v}_{j}^{\alpha})$ and $V^{*}(\mathcal{M},\hat{v}_{j}^{\alpha})$ respectively, since $X_{\mathcal{O}}=E$ and $v\geq 0$ implies, that $X_{\mathcal{O}}$ is best possible. Regarding the optimality of $X_{\mathcal{M}}$, notice, that the greedy algorithm finds an optimal solution, but for $|\alpha|\leq\bar{v}-\dbarv$ the ordering of the elements does not change, hence $X_{\mathcal{M}}$ remains optimal. Let

[TABLE]

By our analysis we deduce

[TABLE]

If $f^{\prime}(0)<0$, then $f(\epsilon)<f(0)=\frac{V^{*}(\mathcal{M},v^{*})}{V^{*}({\mathcal{O}},v^{*})}$ (for small $\epsilon$) contrary to the assumption, that $v^{*}$ was chosen best possible among those with minimal $|E|$ and support of size $\ell$. For small $\epsilon$ neither $E$ nor $\operatorname{supp}(v^{\epsilon})$ have changed.

If $f^{\prime}(0)>0$, then $f(-\epsilon)<f(0)=\frac{V^{*}(\mathcal{M},v^{*})}{V^{*}({\mathcal{O}},v^{*})}$ (for small $\epsilon$) contrary to the assumption, that $v^{*}$ was chosen best possible among those with minimal $|E|$ and support of size $\ell$. For small $\epsilon$ neither $E$ nor $\operatorname{supp}(v^{-\epsilon})$ have changed.

Finally, if $f^{\prime}(0)=0$, it is straightforward to show that $f$ is a constant function for $\alpha\in[\dbarv-\bar{v},\bar{v}-\dbarv]$, hence $f(\dbarv-\bar{v})=f(0)$ but the number of distinct values of $\hat{v}^{\bar{\bar{v}}-\bar{v}}$ has decreased by $1$ contrary to the assumption that $v^{*}$ was chosen to have the minimum number of distinct values.

Since in all three cases we got a contradiction, we conclude that $v^{*}$ is just a multiple of the all-ones vector.   ∎

Remark*.*

Uniqueness of $X_{\mathcal{O}}$ and $X_{\mathcal{M}}$ is important for the proof. Consider a twin-peaks independence system consisting of $\{1\}$ and of $\{2,3\}$. The bases are $\{1\}$ and $\{2,3\}$. Let $2,1,1$ be the coefficients of the objective function. Optimal are both $X_{1}=\{1\}$ and $X_{2}=\{2,3\}.$

If we choose $X_{2}$ for the optimal solution and $\alpha$ increases $v_{1}$ by [math] upwards then the objective function is $\alpha|X_{1}|$ which does not relate to $X_{2}$.

If we choose $X_{1}$ for the optimal solution and $\alpha$ decreases $v-1$ by [math] downwards, then, the objective function is constantly $|X_{2}|$ which does not relate to $X_{1}$.

Theorem 8 about approximating an independence system by a matroid motivates the search for a matroid that yields a best approximation. To this end, Milgrom [5] defines the following substitutability index:

[TABLE]

Milgrom [5] gives no algorithm for determining the best inner matroid $\mathcal{M}^{\prime}$. We describe an integer program for finding it.

Let $r$ be the rank function of the independence system $(E,\mathcal{I})$. We use the fact that matroids are completely characterized by their rank functions. A function $r:2^{E}\to{\mathbb{Z}}_{+}$ is the rank function of matroid $(E,\mathcal{I})$, with $\mathcal{I}=\{F\subseteq E\colon r(F)=|F|\}$ if and only if for all $X,Y\subseteq E$:

[TABLE]

We formulate the problem of finding the inner matroid as the problem of finding a suitable rank function, $r^{\prime}(S)=r(S)-\delta_{S}$, with $\delta_{S}$, integral, must be chosen to ensure that $r^{\prime}$ is a matroid rank function. Hence,

[TABLE]

Obviously, this is not an efficient means to find an optimal inner matroid, since it is an integer-program Additionally, the usually given independence oracle is of no use either. However, if one faces a single independence system with possibly different objective functions but always, for reasons of fundamental value, the same acceptable set ${\mathcal{O}}$, then finding a good inner matroid is just preprocessing.

4 Possible advantages over Greedy

In this section we compare the outcome of the algorithm in Milgrom [5] with greedy in the zero knowledge case of ${\mathcal{O}}=\mathcal{I}$. Example 6 investigated a case, in which under zero knowledge the usual rank-quotient bound could not be improved. This does not exclude the possibility that there exist independence systems $(E,\mathcal{I})$ where $\rho^{M}(\mathcal{I},\mathcal{I})>q(\mathcal{I})$.

Theorem 9**.**

There exists an independence system $(E,\mathcal{I})$ such that $\rho^{M}(\mathcal{I},\mathcal{I})>q(\mathcal{I})$.

Proof.

Let $E=\{-2,-1,1,2,3,4\}$ and $\mathcal{I}=\left\{I\subseteq E\colon\text{ either }I=\{-2,-1\}\text{ or }|I\cap\{-2,-1\}|\leq 1\right\}.$

On the one hand the set $\{-1,1,2,3,4\}$ is independent and maximally independent, hence a basis of $E$. On the other hand the set $\{-2,-1\}$ is independent and cannot be augmented; hence it is a basis of size $2$. Notice, that all two element sets are independent in $\mathcal{I}$. Therefore we can conclude $l(E)=2$ and $r(E)=5$ and $q(\mathcal{I})\leq 2/5$.

By the earlier remark we made for normal independence systems $q(\mathcal{I})\in\left\{\frac{a}{b}\mid a\in\{1,2,\dots,l(\mathcal{I})\}\right.$ and $\left.b\in\{1,2,\linebreak\dots,r(\mathcal{I})\}\right\}$. So if the rank quotient were smaller than $2/5$ for some set $F$ we would need $r(F)>2$. Hence $|F|\geq 3$. Recall that all pairs of elements are independent, hence the lower rank of any set of size at least $2$ is at least $2$. Therefore, $q(\mathcal{I})=2/5.$

Now consider the almost free matroid $\mathcal{M}$ on set $E$ given by $\mathcal{M}=\left\{I\subseteq E\colon|I\cap\{-2,-1\}|\leq 1\right\}.$ As it it contains one ‘OR’-clause less than $\mathcal{I}$, clearly, $\mathcal{M}\subseteq\mathcal{I}.$ Additionally we note that $\mathcal{M}=U^{1}_{\{-2,-1\}}\oplus U^{4}_{\{1,2,3,4\}}$, hence $\mathcal{M}$ is a matroid, in fact an inner matroid of $\mathcal{I}$.

Further, by construction $\mathcal{M}=\mathcal{I}\setminus\{\{-2,-1\}\}$, i.e., all sets of $\mathcal{I}$ but $\{-2,-1\}$ are independent in $\mathcal{M}$ and therefore perfectly approximated. Now, clearly the best approximations of the set $\{-2,-1\}$ from $\mathcal{I}$ are $\{1\}$ or $\{2\}$ both of approximation quality $1/2$. Therefore $\rho(\mathcal{I},\mathcal{I},\mathcal{M})=1/2$ and $\rho^{M}(\mathcal{I},\mathcal{I})\geq\rho(\mathcal{I},\mathcal{I},\mathcal{M})=1/2.$222To argue that $\rho^{M}(\mathcal{I},\mathcal{I})=1/2$ we prove, that there exists no matroid ${\mathcal{M}}^{\prime}$ with $\rho(\mathcal{I},\mathcal{I},{{\mathcal{M}}^{\prime}})>1/2$. For a contradiction suppose, inner matroid ${{\mathcal{M}}^{\prime}}\subset\mathcal{I}$ fulfills $\rho(\mathcal{I},\mathcal{I},{{\mathcal{M}}^{\prime}})>1/2$. As it has to approximate the set $L=\{-2,-1\}$ better than $1/2$, we can conclude that $L\in{{\mathcal{M}}^{\prime}}$. By ratio assumption and $I=\{-1,1,2,3,4\}\in\mathcal{I}$ we know that ${\mathcal{M}}^{\prime}$ contains a set of $3$ elements of $I$. Wlog. we can assume that that set is either of form $J_{1}=\{-1,1,2\}$ or $J_{2}=\{2,3,4\}.$ Now applying the augmentation property of matroids to $L$ of size $2$ and $J_{1}$ or $J_{2}$ of size $3$ shows that ${{\mathcal{M}}^{\prime}}$ contains an independent set $M$ of size $3$ containing $-2,-1$ and one element of $1,2,3,4$. But by definition, such $B$ is not contained in $\mathcal{I}$ and therefore note in ${{\mathcal{M}}^{\prime}}.$ This demonstrates

[TABLE]

Hence, a well chosen inner matroid (following Milgrom) can provide a better performance guarantee than direct application of the greedy algorithm to an independence system.   ∎

This independence system may appear contrived. This is not the case. It is the independence system of a knapsack problem with capacity $8$ and a weight of $4$ for items $-2,-1$ and weight of $1$ for items $1,2,3,4$.

There is a classic approximation algorithm for the knapsack problem. It generates two solutions and picks the best of them. One of those solution is a greedy solution (according to value per weight), and has a performance guarantee of $1/2$. Theorem 9 is no stronger than the classic approximation algorithm for knapsack. Therefore, we provide a knapsack instance and an inner matroid of its independence system with an approximation quality of better than $1/2$.

Theorem 10**.**

For the independence system $(\mathcal{I},E)$ induced by the knapsack problem on items $E=\{-3,-2,-1,\linebreak 1,2,3,4,5\}$ with weights $w_{i}=5$ if $i<0$ and $w_{i}=1$ if $i>0$ and capacity $15$ holds $\rho^{M}(\mathcal{I},\mathcal{I})=2/3>1/2>3/7=q(\mathcal{I})$. Therefore, there exists a matroid that permits $2/3$-approximation and is hence better than the knapsack-greedy algorithm.

Proof.

Consider the knapsack problem on items $E=\{-3,-2,-1,1,2,3,4,5\}$ with weights $w_{i}=5$ if $i<0$ and $w_{i}=1$ if $i>0$ and capacity $15$. It induces the independence system

[TABLE]

On the one hand the set $\{-2,-1,1,2,3,4,5\}$ is independent and maximally independent, hence a basis of $E$. On the other hand the set $\{-3,-2,-1\}$ is independent and cannot be augmented; hence it is a basis of size $3$. Notice, that any triple of elements is independent in $\mathcal{I}$. Therefore we can conclude $l(E)=3$ and $r(E)=7$ and $q(\mathcal{I})\leq 3/7$.

By earlier remark we had for normal independence systems $q(\mathcal{I})\in\left\{\frac{a}{b}\mid a\in\{1,2,\dots,l(\mathcal{I})\}\right.$ and $\left.b\in\{1,2,\dots,r(\mathcal{I})\}\right\}$; together with the observation that all triplets are independent in this $\mathcal{I}$ it reduces to $q(\mathcal{I})\in\left\{\frac{a}{b}\mid a\in\{1,2,\dots,3\}\right.$ and $\left.b\in\{1,2,\dots,7\}\right\}.$ If there were $F\subseteq E$ with $q(F)=l(F)/r(F)=a/b<3/7$ then because of $b\leq 7$ we need $a\leq 2$. But as triplets are independent, $b=l(F)\leq 2$ requires $|F|\leq 2$ which yields $a=r(F)\leq 2$. But this rules out $q(F)<3/7.$ Therefore

[TABLE]

Now consider the almost free matroid $\mathcal{M}$ on set $E$ given by

[TABLE]

As it it contains all independent sets of $\mathcal{I}$ except for $\{-3,-2,-1\}$, clearly $\mathcal{M}\subseteq\mathcal{I}.$ Additionally we note that $\mathcal{M}=U^{2}_{\{-3,-2,-1\}}\oplus U^{5}_{\{1,2,3,4,5\}}$, hence $\mathcal{M}$ is a matroid, in fact an inner matroid of $\mathcal{I}$.

Further, by construction $\mathcal{M}=\mathcal{I}-\{\{-3,-2,-1\}\}$. Therefore, all sets of $\mathcal{I}$ but $\{-3,-2,-1\}$ are independent in $\mathcal{M}$ and therefore perfectly approximated. Now, clearly the best approximations of the set $\{-3,-2,-1\}$ from $\mathcal{I}$ are all of its two-element subsets $\{1,2\},\{2,3\},\{1,3\}$ of approximation quality $2/3$. Therefore $\rho(\mathcal{I},\mathcal{I},\mathcal{M})=2/3$ and $\rho^{M}(\mathcal{I},\mathcal{I})\geq\rho(\mathcal{I},\mathcal{I},\mathcal{M})=2/3.$

Finally, we want to argue that $\rho^{M}(\mathcal{I},\mathcal{I})=2/3$. Otherwise, there would exist a matroid ${{\mathcal{M}}^{\prime}}\subseteq\mathcal{I}$ with $\rho(\mathcal{I},\mathcal{I},{{\mathcal{M}}^{\prime}})>2/3$. Since $\{-3,-2,-1\}\in\mathcal{I}$ and the approximation-ratio is assumed to be greater than $2/3$ we can conclude that $\{-3,-2,-1\}\in{{\mathcal{M}}^{\prime}}.$ Similarly, $\{1,2,3,4,5\}\in\mathcal{I}$ implies that a subset of it of size greater $5*2/3$, has to be contained in ${{\mathcal{M}}^{\prime}}$. Without loss of generality suppose $\{1,2,3,4\}\in{{\mathcal{M}}^{\prime}}.$ As ${{\mathcal{M}}^{\prime}}$ is a matroid and contains independent sets $\{-3,-2,-1\}$ and $\{1,2,3,4\}$ of different size, there has to exist an element, say $1$, from the second set, so that $\{-3,-2,-1,1\}\in{{\mathcal{M}}^{\prime}}$. However, because of $\{-3,-2,-1,1\}\notin\mathcal{I}$ this is impossible. This demonstrates

[TABLE]

Therefore, when running the plain greedy algorithm on the constructed matroid $\mathcal{M}$ for arbitrary objective functions, we obtain a performance guarantee of $2/3$ which is strictly better than the $1/2$-guarantee of the usual knapsack-greedy algorithm.   ∎

The previous two examples, in particular the one from Theorem 9, raise the question of how to easily find a good inner matroid. We have already described an integer program (IP) to do so, which is clearly impractical. Therefore, we outline a heuristic that could, in some cases, find a good (but not best) inner matroid, by further restricting the circuits of the independence system.

Remark*.*

Consider again $E=\{-2,-1,1,2,3,4\}$ and $\mathcal{I}=\left\{I\subseteq E\colon\text{ either }I=\{-2,-1\}\text{ or }|I\cap\{-2,-1\}|\leq 1\right\}$ from the proof of Theorem 9. The circuits of $\mathcal{I}$ are $\mathcal{C}=\{\{-2,-1,i\}\colon i\in\{1,2,3,4\}\}.$ If $\mathcal{I}$ were a matroid, then, strong circuit elimination should hold. I.e., strong circuit elimination for $\{-2,-1,1\}$ and $\{-2,-1,2\}$ would require that there exits circuits in $\{-2,-1,1,2\}$

• containing $1$ and avoiding $-2$,

• containing $2$ and avoiding $-2$,

• containing $1$ and avoiding $-1$, and

• containing $2$ and avoiding $-1$.

In comparison with $\mathcal{C}$ we see, that no circuit in $\{-2,-1,1,2\}$ contains $1$ and avoids $-2$ and so on. To get “closer” to an inner matroid, we could either drop one of the problematic circuits, say $\{-2,-1,1\}$ but this would enlarge the independence system, the ‘opposite’ to an inner matroid, or we could

add missing circuits:

for example above we could add circuits $\{1,-1\},\{2,-1\},\{1,-2\},\{2,-2\}$; this way we would get closer to an inner matroid, while some of the original circuits we started with, like $\{-2,-1,1\}$ would remain dependent, but would no longer be minimal dependent sets, as they contain some of the newly minted circuits, like $\{1,-2\}$. If we would add all circuits that could be derived in this fashion, we would, indeed end with an inner matroid, that turns out to be $U^{1}_{E}.$ However, it provides only a Milgrom approximation guarantee of $1/5$ worse than the rank-quotient of $1/2$. Or we could

“destroy” the offending circuits, by adding circuits contained in them:

an obvious way to make the offending circuits $\{-2,-1,1\}$ and $\{-2,-1,2\}$ go away is to add a circuit $\{-2,-1\}$. In fact all previous circuits contain this one, so that the resulting circuit system would be $\mathcal{C}^{\prime}=\{\{-2,-1\}\}$. Indeed, the independence system induced by this single circuit is just the inner matroid $\mathcal{M}=U^{1}_{\{-2,-1\}}\oplus U^{4}_{\{1,2,3,4\}}$ from the proof of Theorem 9 realizing the approximation guarantee of $1/2$.

Remark*.*

One might wonder, what else can be said about the independence system $\mathcal{I}$ of Theorem 9. As it turn out $\mathcal{I}\mathbin{/}\{-2\}$ is one of the previously defined twin-peaks independence systems with ground sets of size $1$ and $4$.

The zero knowledge assumption is weaker than necessary. If $v\geq\mathbf{0}$, we know that a maximum weight independent set must be, unsurprisingly, a basis of $E$. Thus, a more reasonable knowledge assumption is ${\mathcal{O}}=\mathcal{B}_{\mathcal{I}}$. This provides another possibility to improve over simple greedy for independence systems.

Suppose $v\geq\mathbf{0}$ and ${\mathcal{O}}\subset\mathcal{B}_{\mathcal{I}}$. In this case, prior information rules out some basis from being optimal. To ensure an apples with apples comparison between the greedy algorithm and the algorithm in Milgrom [5] we must allow the greedy algorithm to make use of the information contained in ${\mathcal{O}}$ as well. This may require modifying the greedy algorithm so much that it become an entirely different algorithm altogether. A straightforward way to do this is to apply the greedy algorithm to an independence system $(E,{\mathcal{O}}^{\dagger})$ of sets contained in elements of ${\mathcal{O}}$. The example we describe below shows that the greedy algorithm will be dominated by Milgrom’s heuristic. Under one interpretation of ${\mathcal{O}}$ this is not surprising. Suppose, that $\mathcal{I}\setminus{\mathcal{O}}$ is the collection of sets that can never be optimal. Then, ${\mathcal{O}}^{\dagger}$ would include sets that should definitely not be selected.

Example 11**.**

Consider the stable set problem (largest set of pairwise nonadjacent vertices) on a path of $9$ vertices:

123456789

The underlying independence system has $E=\{1,2,3,4,5,6,7,8,9\}$ and the stable sets form the set-system $\mathcal{I}$. The maximum cardinality basis of $\mathcal{I}$ is the set $\{1,3,5,7,9\}$ of cardinality $5$. An example of a small basis of $E$ would be $\{2,5,8\}$ of size $3$. For the Korte-Hausmann bound, consider the set $\{1,2,3\}$ that demonstrates $q\leq\frac{1}{2}$.

Suppose ${\mathcal{O}}$ is the collection of stable sets $S$ of size at most $4$ with $|S\cap\{3,6,9\}|\leq 1$. Notice, $\{1,3,5,7\}\in{\mathcal{O}}$.

Consider the set $\{1,2,3\}$. Assuming only subsets of elements of ${\mathcal{O}}$, are independent, this set has a rank quotient of $1/2$, i.e., $q({\mathcal{O}}^{\dagger})=1/2$. Hence, the greedy algorithm applied to the independence system consisting of the elements of ${\mathcal{O}}^{\dagger}$ will have a worst case bound of at most $1/2$.

The following matroid $\mathcal{M}=U^{1}_{\{1,2\}}\oplus U^{1}_{\{4,5\}}\oplus U^{1}_{\{7,8\}}$ is contained in $\mathcal{I}$ since every independent set in $\mathcal{M}$ is a stable set from $\mathcal{I}$. It is straightforward to verify that $\rho^{M}(\mathcal{I},{\mathcal{O}})=3/4>1/2$. Interestingly, the largest set of $\mathcal{I}$ is not independent in the matroid $\mathcal{M}$.

This example highlights the essential difference between Milgrom’s heuristic and the greedy algorithm. The first is looking for a basis that has large overlap with each set in ${\mathcal{O}}$. The greedy algorithm seeks a basis that has a large overlap with every set that is contained in ${\mathcal{O}}^{\dagger}$.

5 Discussion

Our reading of [5] suggests that Milgrom proposed the existence of a ‘good’ inner matroid as an explanation for the success of the greedy algorithm in some settings. He writes:

“In practice, procedures based on greedy algorithms often perform very well.”

Our analysis shows that existence of a good inner matroid by itself is not sufficient to explain the practical observation. It is possible for there to be no good inner matroid approximation, yet the greedy algorithm performs well. Further, there are examples where the inner matroid approximation will dominate that of the greedy algorithm.

Certainly, the prior knowledge encoded in ${\mathcal{O}}$ is beneficial. The idea of incorporating prior information about the optimal solution into an optimization problem is not new. There are three approaches we are aware of: probabilistic, uncertainty sets and stability. The first encodes the prior information in terms of a probability distribution over the possible objective value coefficients. (This is sometimes relaxed to a class of distributions sharing common moments.) The second, assumes that the objective coefficients are drawn from some set given a-priori. Stability assumes that the optimal solution of the instance under examination does not change under perturbations to the objective function (see [1]). The novelty of Milgrom’s proposal is to encode the prior information in a description of the set of possible optima.

However, as Theorem 9 makes clear, even in the absence of prior knowledge the inner matroid approximation can outperform greedy. This is because the inner matroid approximation exploits global information about the structure of the independence system. Roughly speaking, the inner matroid approximation ‘kills’ off low rank bases so as to prevent greedy from being stuck there. If the optimal basis is not to be found among these low rank sets, no harm is done. If not, precisely because they are low rank, one can approximate the optimal objective function value well using only a subset of the elements of these low rank sets.

Acknowledgements

The authors want to thank a referee for helpful comments and finding a gap in a previous proof, that moved us to state and prove Theorem 9.