The Matroid Secretary Problem

In this post, I am going to discuss the matroid secretary problem, which is a very nice problem introduced by Babaioff, Immorlica, and Kleinberg [1].  I will try to give an up-to-date account, but am far from an expert in this area, so please feel free to comment if I miss or muddle anything.

Let’s warm up with the classical secretary problem, which has the following setup.  We want to hire a secretary.  There is a set $X$ of $n$ candidates, and each $x \in X$ has a competency $w(x)$.  We know that there are $n$ candidates, but we do not know the competency function.  The secretaries are then presented to us in a random order.  Once a secretary is presented to us, we interview him and discover his competence.  We then have to make an irrevocable decision to hire him or not.  If we hire him on the spot, the process ends.  If not, then due to a hyper-competitive labour market, he will be hired by another firm, and we must move on to the next candidate.

Let OPT be the maximum competency of all secretaries.  Our goal is to devise an algorithm that hires a secretary with competency close to OPT.  Say that an algorithm $\mathcal A$ is $\alpha$-competitive if OPT / $\mathbb{E} (\mathcal A) \leq \alpha$, where $\mathbb{E} (\mathcal A)$ is the expected competency outputted by $\mathcal A$ for a random permutation $\sigma$ of $X$.

Here is a $4$-competitive algorithm for the classical secretary problem.  We reject each of first $\frac{n}{2}$ candidates but we keep track of the highest competency, say $h$, of the first $\frac{n}{2}$ candidates.  We then hire the first candidate with competency at least $h$ (if none exists, we just hire the last secretary).  This algorithm is $4$-competitive because the probability that the most competent secretary is in the second half and the second most competent secretary is in the first half is more than $\frac{1}{4}$.  Interestingly, if we instead reject $\frac{1}{e}$ of the initial candidates, we obtain an $e$-competitive algorithm, which turns out to be best possible over all possible algorithms [2].

The matroid secretary problem has the same setup as above, except that there is a matroid $\mathcal M$ on the underlying set $X$ of secretaries.  We now want to hire a set of secretaries, subject to the condition that our hired set is independent in $\mathcal M$.  Again, there is an unknown weight function $w:X \to \mathbb R_+$, and the secretaries are presented to us in a random order.  After interviewing $x$, we must make an irrevocable decision to add or not add $x$ to our currently constructed independent set $I$.  Again, the goal is to devise an algorithm whose expected output is close to the maximum weight independent set.

Note that we recover the classical secretary problem when $\mathcal M$ is a rank-$1$ uniform matroid.  The generalization to matroids might seem like abstract nonsense, but the matroidal version is actually relevant in the theory of auctions, and has garnered a lot of interest.  The big (and still open) problem is the following conjecture.

Conjecture 1 (Babaioff, Immorlica, and Kleinberg [1]).  For every matroid $\mathcal M$, there is a $O(1)$-competitive algorithm.

Using a clever modification of the threshold price algorithm that we discussed for the classical secretary problem, Babaioff, Immorlica, and Kleinberg [1] proved that there is a $O(\log r)$-competitive algorithm, where $r$ is the rank of $\mathcal M$.  In a breakthrough paper, Chakraborty and Lachish [3] devised a $O(\sqrt{ \log r})$-competitive algorithm.  Their algorithm is a patchwork of a few different algorithms, and is quite complicated.  Finally, the state-of-the-art is a $O( \log \log r)$-competitive algorithm of Lachish [4].  Using different tools, Feldman, Svensson, and Zenklusen [5] have also obtained a $O( \log \log r)$-competitive algorithm for general matroids.

Another line of research is to weaken the problem slightly, in the hope of obtaining a constant-competitive algorithm.  Perhaps the most important modification is known as the random assignment model.  Note that in the original matroid secretary problem, we may as well assume that the weight function is chosen by an adversary.  In the random assignment model, we weaken the adversary, by only allowing her to choose the set of weights.  The weights are then randomly assigned to the set of secretaries.  Soto [6] proved that there is a constant-competitive algorithm for all matroids in the random assignment model.

Theorem 2 (Soto [6]). For each matroid $\mathcal M$, there is a $O(1)$-competitive algorithm in the random assignment model.

Soto’s algorithm uses the classic decomposition of a matroid into its principal minors.  See the recent paper of Fujishige [7] for a survey of this theory.  The principal minors of a matroid are uniformly dense ($\mathcal M$ is uniformly dense if $\max_{A \subseteq E} \frac{|A|}{r(A)}$ is obtained by $E$). Using the decomposition, one can reduce to the case of uniformly dense matroids, and Soto proved that there is indeed a $O(1)$-competitive algorithm for uniformly dense matroids. Note that his proof crucially assumes that we are in the random assignment model.

There are also many other interesting variants of the original matroid secretary problem. See Section 3 of Dinitz [8] for a nice overview.

I will finish by discussing the original matroid secretary problem for restricted classes of matroids.  In [1], it is proved that Conjecture 1 holds for graphic matroids, uniform matroids, partition matroids, and bounded-degree transversal matroids.  The bounded-degree condition for transversal matroids was later removed by Dmitrov and Plaxton [9].

Here is another class of matroids for which Conjecture 1 holds.  Let $\mathcal F$ be a laminar family and $c: \mathcal F \to \mathbb{N}$.  If we let $\mathcal I$ be the set of all $I$ such that $|I \cap F| \leq c(F)$ for all $F \in \mathcal F$, then $\mathcal I$ is the set of independent sets of a matroid.  Matroids arising in this way are called laminar matroids.  Note that the class of laminar matroids contains the partition matroids.  Im and Wang [10] showed that there is a constant-competitive matroid secretary algorithm for the class of laminar matroids.

Finally, Dinitz and Kortsarz gave a $O(1)$-competitive algorithm for the matroid secretary problem on regular matroids.  Their proof uses Seymour’s decomposition theorem for regular matroids [11].  Note that graphic matroids, cographic matroids, and $R_{10}$ all have $O(1)$-competitive matroid secretary algorithms.

Using the structure theorem for $\mathbb F$-representable matroids proved by Geelen, Gerards and Whittle [12], it may be possible to prove the following special case of Conjecture 1.

Conjecture 3.  For every finite field $\mathbb F$, there is a $O(1)$-competitive algorithm for the matroid secretary problem on the class of $\mathbb F$-representable matroids.

I suspect this might be tricky since one basic class of $\mathbb F$-representable matroids are those representable over a subfield $\mathbb F’$ of $\mathbb F$, and I don’t see how to get a $O(1)$-competitive matroid secretary algorithm for  $\mathbb F’$-representable matroids unless we are doing some cunning induction.

On the other hand, the structure theorem for binary matroids is slightly simpler (due to the absence of subfields), so the following special case of Conjecture 3 may be doable.

Conjecture 4. There is a $O(1)$-competitive algorithm for the matroid secretary problem on binary matroids.


[1] Moshe Babaioff, Nicole Immorlica, and Robert Kleinberg. Matroids, secretary problems, and online mechanisms. In Proc. SODA, pages 434–443, 2007.

[2] E. B. Dynkin. The optimum choice of the instant for stopping a Markov process. Soviet Math. Dokl, 4, 1963.

[3] Sourav Chakraborty and Oded Lachish. Improved competitive ratio for the matroid secretary problem. In Proc. SODA, pages 1702–1712, 2012.

[4] Lachish, Oded. O(log log rank)-competitive ratio for the matroid secretary problem. Foundations of Computer Science (FOCS), 2014 IEEE 55th Annual Symposium on. IEEE, 2014.

[5] Feldman, Moran, Ola Svensson, and Rico Zenklusen.  A simple O(log log (rank))-competitive algorithm for the matroid secretary problem.  Proceedings of the Twenty-Sixth Annual ACM-SIAM Symposium on Discrete Algorithms. SIAM, 2015.

[6] José A. Soto. Matroid secretary problem in the random assignment model. In Proc. SODA, pages 1275–1284, 2011.

[7] S. Fujishige. Theory of principal partitions revisited. In Research Trends in Combinatorial Optimization, pages 127–162, 2009.

[8] Dinitz, Michael. Recent advances on the matroid secretary problem. ACM SIGACT News 44 (2013), no. 2, 126–142.

[9] Nedialko B. Dimitrov and C. Greg Plaxton. Competitive weighted matching in transversal matroids. In Proc. ICALP, pages 397–408, 2008.

[10] Sungjin Im and Yajun Wang. Secretary problems: laminar matroid and interval scheduling. In Proc. SODA, pages 1265–1274, 2011.

[11] P. D. Seymour. Decomposition of regular matroids. J. Combin. Theory Ser. B, 28(3):305–359, 1980.

[12] J. Geelen, B. Gerards, G. Whittle. Structure in minor-closed-classes of matroids. Surveys in Combinatorics, London Mathematical Society Lecture Note Series 409, 327–362, 2013.

2 thoughts on “The Matroid Secretary Problem

  1. Just a remark. The usual presentation of the problem (I think) is that you want to maximise the probability of hiring the best secretary, rather than the expected quality of the secretary you hire. (But of course, for adversarial weights there can be unboundedly large gaps between the weights, so they amount to the same thing.)

    I’m not sure how that works out for the matroid secretary problem?

    • For the matroid secretary problem the two notions are indeed different. For example, the constant competitive algorithm for cographic matroids works as follows. The edge set of every 3-edge connected graph can be covered by three bonds. Thus, there is a simple algorithm that just randomly selects one of the three bonds (regardless of the weights). Since each edge will be taken with probability at least 1/3, this algorithm is 3-competitive. However, it will not select the best bond with probability bounded away from zero.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.