Seymour’s 1-flowing conjecture III

A long time ago, I gave some background to the definition of a 1-flowing matroid. Assume $M$ is a matroid on the ground set $E$, and $e$ is an element in $E$. Take $c_{x}$ to be a non-negative integral capacity assigned to each element, $x$, in $E-e$. A flow is an assignment of a non-negative real, $f_{C}$, to each circuit, $C$, that contains $e$, with the constraint that if $x$ is in $E-e$, then the sum of $f_{C}$ over all circuits that contain $e$ and $x$ does not exceed $c_{x}$. We say that $M$ is $e$-flowing if, for every assignment of capacities, there is a flow whose value (sum over all circuits containing $e$) is equal to
\[\min\left\{\sum_{x\in C^{*}-e} c_{x}\mid C^{*}\ \text{is a cocircuit containing}\ e\right\}.\]
If $M$ is $e$-flowing for every $e$ in $E$, then $M$ is $1$-flowing.

We are motivated to study $1$-flowing matroids because problems that should really be intractable, magically become easy to solve when we consider such matroids. Consider the case of a graph with distinct vertices $s$ and $t$. We might ask for the largest number of edge-disjoint paths from $s$ to $t$. Now this resembles a difficult problem: we are asking to find a largest-possible pairwise-disjoint subfamily from a family of sets (the edge-sets of $s-t$ paths). Finding the maximum number of pairwise-disjoint sets is famously an NP-hard problem. However, as is well known, the problem of finding edge-disjoint $s-t$ paths can be solved using network-flow techniques by assigning a capacity of one to every edge. This really reflects the fact (discussed earlier) that graphic matroids are $1$-flowing. The fact that the maximum value of a flow through a network is equal to the minimum capacity of a cut gives enough structure to the problem to render it feasible.

In my most recent post, we can see another intimation that difficult problems may become tractable when dealing with $1$-flowing matroids. Let $M$ be a matroid on the ground set $E$, where $e$ is an element in $E$. We assign a real-valued variable $a_{x}$ to each element $x\in E-e$, and we insist that $a_{x}$ takes only non-negative values. In addition, for each circuit, $C$, that contains $e$, we require that the sum of variables, taken over all elements in $C-e$, must be at least one. Consider the set, $P$, of points in the Euclidean space $\mathbb{R}^{E-e}$ corresponding to assignments of values to variables such that these constraints are satisfied. Since $P$ is an intersection of half-spaces, it is a polyhedron. Now $M$ is $e$-flowing if and only if the $0$-dimensional extreme points (that is, the vertices) of $P$ have entirely integral coordinates. This means that integer programming problems associated with $P$ are in fact amenable to linear programming techniques. In other words, if there is a linear objective function which assigns a cost to each point in $P$, and we wish to find lowest-cost point in $P$ with integer coordinates, then we can relax the integrality constraint, since running the unconstrained linear program will return an integral point in any case. As integer programming is $NP$-hard, and linear progamming is solvable in polynomial time, this gives us some intuition as to why ostensibly difficult problems are easy for $1$-flowing matroids.

Proposition. The class of $1$-flowing matroids is closed under minors.

Proof. Let $M$ be a $1$-flowing matroid on the ground set $E$, and let $e$ be an element in $E$. Let $f$ be in $E-e$. We will show that $M\backslash f$ is $e$-flowing. Assume that each element $x\in E(M\backslash f)-e$ has been assigned a capacity $c_{x}$. We assign exactly the same capacities to these elements in $M$, and in addition we assign $f$ a capacity of $0$. Now every cocircuit of $M\backslash f$ is of the form $C^{*}-f$ where $C^{*}$ is a cocircuit of $M$, or is equal to a cocircuit of $M$. It follows that
\min\left\{\sum_{x\in C^{*}-e} c_{x}\mid e\in C^{*}\in\mathcal{C}^{*}(M)\right\}\leq\min\left\{\sum_{x\in C^{*}-e} c_{x}\mid e\in C^{*}\in\mathcal{C}^{*}(M\backslash f)\right\}.
If this is a strict inequality, then there is some cocircuit $C^{*}$ in $M$ that contains $e$, where the sum of capacities in $C^{*}-e$ is less than the minimum such sum in $M\backslash f$. Obviously $C^{*}$ is not a cocircuit of $M\backslash f$, so $f$ is in $\operatorname{cl}^{*}(C^{*})$. Thus some cocircuit $C^{*}_{0}$ of $M$ satisfies $f\in C^{*}_{0}\subseteq C^{*}\cup f$. But then $C^{*}_{0}-f$ is a cocircuit of $M\backslash f$ with total capacity at most the total capacity of $C^{*}$, a contradiction. Therefore the two minima are equal.

Since $M$ is $e$-flowing, there is a flow with value equal to these minima. We construct a flow for $M\backslash f$ with the same value by simply omitting all circuits that contains $e$ and $f$. This shows $M\backslash f$ is $e$-flowing.

Next we consider $M/f$. Assume that we assign the capacity $c_{x}$ to each $x$ in $E(M/f)-e$. Let $d$ be
\min\left\{\sum_{x\in C^{*}-e} c_{x}\mid e\in C^{*}\in\mathcal{C}^{*}(M/f)\right\}.
In $M$ we assign exactly the same capacities, and we also assign $f$ the capacity of $d$. It follows immediately that the minimum sum of capacities, taken over all cocircuits that contain $e$, is exactly the same in both matroids. Since $M$ is $1$-flowing, there is a flow whose value is equal to this minimum. For every circuit, $C$, in $M$ that contains $e$, there is a circuit, $\theta(C)$, of $M/f$ such that $\theta(C)\subseteq C$ and $e\in\theta(C)$. To construct a flow in $M/f$ we give each circuit $\theta(C)$ the sum of the values given to all pre-images of $C$ under $\theta$. Every other circuit receives a value of zero. This is enough to demonstrate that $M/f$ is $e$-flowing. The rest follows by easy induction. $\square$

Now we can hope to characterise the class of $1$-flowing matroids via its excluded minors. We might also note that the class is closed under duality. (This proof uses linear programming duality, and can be constructed from Theorem 1.17 in [Cor2001] and (3.1) in [Sey1977].) Thus the set of excluded minors will be closed under duality.

Let $M$ be isomorphic to $U_{2,4}$ and let $e$ be in $E$. We assign a capacity of one to the elements in $E-e$. Each cocircuit containing $e$ has a total capacity of two. Can we find a flow with this value? This would mean assigning values, $f_{1}$, $f_{2}$, and $f_{3}$, to the three circuits containing $e$, in such a way that $f_{1}+f_{2}+f_{3}=2$. However, the capacity constraints require that the sum of any two of $f_{1}$, $f_{2}$, and $f_{3}$ is at most one. A moment’s thought shows that this means $2(f_{1}+f_{2}+f_{3})\leq 3$, so there can be no solution to this system. Therefore $U_{2,4}$ is not $1$-flowing, and hence every $1$-flowing matroid is binary. The converse does not hold, as $AG(3,2)$ is not $1$-flowing (and is an excluded minor for the class). Seymour discovered two more excluded minors, $T_{11}$ and its dual. $T_{11}$ is the lift matroid of the even-cycle graph obtained from $K_{5}$ by making every edge negative.

Now we have arrived at the titular, and extremely lovely, conjecture by Seymour:

Conjecture (Seymour 1981). The set of excluded minors for the class of $1$-flowing matroids is $U_{2,4}$, $AG(3,2)$, $T_{11}$, and $T_{11}^{*}$.

[Cor2001] G. Cornuéjols, Combinatorial optimisation: Packing and covering. CBMS-NSF Regional Conference Series in Applied Mathematics, 74. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2001.

[Sey1977] P. D. Seymour, The matroids with the max-flow min-cut property. J. Combinatorial Theory Ser. B 23 (1977) 189-222.

[Sey1981] P. D. Seymour, Matroids and multicommodity flows. European J. Combin. 2 (1981) 257-290.

4 thoughts on “Seymour’s 1-flowing conjecture III

  1. T11 is a single-element extension of R10. Lists of binary matroids have been available for a long time (see our paper “On Matroid Generation” joint with Robert Kingan and Wendy Myrvold).

    The 3-connected rank 5 and 11 element matroids are only 32 in number. Have all 32 been checked to support Seymour’s conjecture that T11 is the only 11-element rank-5 minimal excluded minor. There’s only about 20 of the 32 that have to be checked since the others have AG(3,2) minor and won’t be a minimal excluded minor.

    For example Gordon Royle shows how to check AG(3,2) here

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