1. Introduction
The minimum weight basis problem on a matroid is, given an assignment of real weights to the ground elements, to compute the minimum weight of a basis, the latter being the sum of weights of the basis elements. Thanks to classical results of Rado [10], Gale [6] and Edmonds [4], the algorithmic aspect of this problem is well understood: the minimum weight basis problem on a downward closed set system can be solved by the greedy algorithm precisely when the system forms the family of independent sets of a matroid.
In this paper, we are interested in structural aspects of the minimum weight basis problem. Given a weighting of ground elements, the following questions naturally arise.

How does the optimal value change when an element is contracted or deleted?

How does the optimal value change when the weight of an element is changed?

By how much can the weight of a single element be changed without changing optimal bases?

What simultaneous changes of the weights preserve optimality of optimal bases?

What ground elements belong to all, to none or to some but not to all of the optimal bases?
Question 3 is a special case of Question 4 and was already answered by Tarjan [11] (for graphic matroids) and by Libura [8] (for general matroids) in terms of fundamental circuits and cuts relative to a given optimal basis. Question 5 was answered by Cechlárova and Lacko [3] in terms of the rank function of the underlying matroid. But to our best knowledge, no answers (in any terms) to Questions 1, 2 and 4 were known so far.
It turns out that all five questions can be answered using the concept of minmax weight of a ground element , which we define as the minimum, over all circuits containing , of the maximum weight of an element in the independent set :
Further, we call
the bottleneck weight of . The answers to the aforementioned questions – are given by the corresponding Theorems 4, 3, 2, 5 and 1 in the next section. All necessary matroid concepts are recalled in Section 3.
2. Results
Let be a loopless matroid, that is, no element belongs to all bases and no singleton set is dependent. All our results concern the minimum weight basis problem on . Given a weighting , a basis is optimal (or simply optimal if the weighting is clear from the context) if its weight is minimal among all bases. The weight of such a basis, that is, the number , is the optimal value under the weighting .
Contraction and deletion
Given a ground element , the independent sets of the matroid , obtained by contracting the element , are all sets with and , while those of the matroid , obtained by deleting the element , are all sets with . Since the matroid is loopless, each of these two matroids contains at least one nonempty independent set. Note that the set of ground elements of both matroids and is . For every basis of , either is a basis of (if ), or is a basis of (if ). This gives us a known recursion
But what about the opposite direction: if we already know the optimal value in the matroid , what are the optimal values and in the two submatroids and ? Our main result (Theorem 1) gives the answer.
Theorem 1 (Tropical Kirchhoff’s formula).
Let be a loopless matroid, and be a ground element. For every weighting , the following equalities hold:

;

.
In particular, (a) and (b) yield the equality . Thus, the minmax weight of an element in the matroid is determined by the minimum weights of bases in the submatroids and .
Remark 1.
In the special case of graphic matroids (see creftype 1 in Section 3), Theorem 1(a) gives us the tropical version of the classical arithmetic effective conductance formula for electrical networks proved by Kirchhoff [7] already in 1847. The spanning tree polynomial of an undirected connected graph is , where the sum is over all spanning trees of . Kirchhoff’s formula (see also [12, Theorem 8] for a detailed exposition) states that, when the edges of are interpreted as electrical resistors and their weights as electrical conductances (reciprocals of electrical resistances), then the effective conductance between the endpoints of any edge is exactly the ratio , where is the graph obtained from by contracting the edge .
In the tropical semifield “addition” means taking the minimum, “multiplication” means adding the numbers, and “division” turns into subtraction. In particular, the spanning tree polynomial of a graph turns into the tropical polynomial . The function computed by is the wellknown minimum weight spanning tree problem. The ratio of polynomials in Kirchhoff’s formula turns into the difference of their tropical versions. So, a natural question arises: what is the tropical analogue of the effective conductance between the endpoints of an edge ? Theorem 1(a) gives the answer (even in general matroids): this is exactly the bottleneck weight of .
Postoptimality
If we change the weight of a single ground element , what is the optimal value under the new weighting ? We show that the difference (of the new and the old optimal values) is determined by the bottleneck weights of the element under the old and the new weighting.
Theorem 2 (Postoptimality).
Let and let be weightings that differ only in the weights given to the element . Then
Since the minmax weight does not depend on the weight of the element itself, we have and, hence, . Thus, when a weighting and an element are fixed, and a new weighting gives weight to the element , then the behavior of the function
only depends on whether or not. If , then for the constant . If , then is constant or constant , whichever of these two numbers is larger.
If all weights are nonnegative, and if the new weighting gives weight to the element , then , and Theorem 2 directly yields the following consequence.
Corollary 1.
If the weights are nonnegative, and if the weight of a single ground element is dropped down to zero, then the minimum weight of a basis decreases by exactly the bottleneck weight of this element.
creftypecap 1 allows us to compute the optimal value under an arbitrary nonnegative weighting by computing the bottleneck weights of the elements of any (fixed in advance) basis.
Corollary 2.
Let be a basis. Given a nonnegative weighting , consider the sequence of weightings , where , and each next weighting is obtained from by setting the weights of the elements to zero. Then
Proof.
creftypecap 1 gives us the recursion which rolls out into . Since the weighting gives weight to all elements of the basis , we have . Since the weights are nonnegative, the basis is of minimum weight under the weighting . Hence, . ∎
Sensitivity
Given a weighting the sensitivity question is: what changes of the weights preserve optimality of optimal bases. That is, if is a new weighting, under what conditions do optimal bases remain optimal? In the case when only changes the weight of a single element , this question was answered by Libura [8] in terms of fundamental circuits and cuts relative to an optimal basis (see Remark 7 in Section 7); in the case of graphic matroids (where bases are spanning trees, see creftype 1 in Section 3), the same answer was given earlier by Tarjan [11]. We answer this question in terms of the minmax weight of . Namely, we associate with every element its tolerance under the (old) weighting:
Theorem 3 (Sensitivity, local change).
Let and let be weightings that only differ in the weights given to the element , and be an optimal basis.

If , then is optimal if and only if .

If , then is optimal if and only if .

If , then is optimal.
By Theorem 3(c), the weight of a single element can be changed by while preserving optimality. In contrast, if we allow the weights of two or more elements to be changed, then only changes by at most preserve optimality.
Theorem 4 (Sensitivity, global change).
Let be a weighting, and be an optimal basis.

If a weighting satisfies for all , then the basis is optimal.

For every there is a weighting such that holds for all elements but the basis is not optimal.
Claim (b) shows that the upper bound in claim (a) is tight.
Persistency
Given a weighting , the set of ground elements is split into three (not necessarily nonempty) subsets:

= elements belonging to all optimal bases;

= elements not belonging to any optimal basis;

= elements that belong to some but not to all optimal bases.
The persistency problem is to determine this partition. Elements are called persistent. Knowing which ground elements belong to which of these three subsets may be helpful when constructing an optimal basis. Namely, we can contract all elements of (that is, include them into the solution), remove all elements of , and try to extend our partial solution to an optimal basis by only treating the elements of .
Cechlárová and Lacko [3] characterized the sets and in terms of the rank function of the underlying matroid: iff removing from the set of all elements not heavier than decreases the rank of this set, and iff adding to the set of all elements lighter than leaves the rank of this set unchanged. We characterize these sets in terms of minmax weights of ground elements.
Theorem 5 (Persistency).
Let be a weighting, and be a ground element.

if and only if ;

if and only if ;

if and only if .

If all weights are distinct, then is the unique optimal basis.
Organization
3. Preliminaries
We use standard matroid terminology as, for example, in Oxley’s book [9]. A matroid on a finite set of ground elements is a pair , where is a nonempty downward closed collection of subsets of , called independent sets, with the augmentation property: whenever and are independent sets of cardinalities , there is an element such that the set is independent; as customary, we abbreviate to and write for .
Bases and circuits
An independent set is a basis if it is contained in no other independent set. The augmentation property implies that all bases have the same cardinality. This property also yields the basis exchange axiom: if and are bases, then for every element there is an element such that is a basis. The following two important refinements of the basis exchange axiom are known as the symmetric basis exchange and the bijective basis exchange.
Proposition 1 (Brualdi [1], Brylawski [2]).
Let and be bases.

For every there is an such that both and are bases.

There is a bijection such that the set is a basis for every .
A subset of is dependent if it is not independent. A circuit is a dependent set whose proper subsets are all independent. For a ground element , an circuit is a circuit containing . An element is a loop if the set is dependent, and is a coloop if belongs to all bases. To avoid pathological situations, we assume that our matroid is loopless: no ground element is a loop or a coloop. We only need this assumption to ensure two properties: every circuit contains at least two elements, and for every ground element at least one circuit exists.
Fundamental paths and cuts
Let be a basis, and a ground element. If , then the set must contain at least one circuit, because is independent but is dependent. An important fact, shown by Brualdi [1, Lemma 1] (see also Oxley [9, Proposition 1.1.4]) and known as the unique circuit property, is that the set contains a unique circuit . Since is independent, this circuit is an circuit (that is, contains ). This unique circuit is known as the fundamental circuit of relative to . Motivated by graphic matroids (cf. creftype 1), we call the independent set
the fundamental path of relative to . If is a basis element, then the set
is known as the fundamental cut of relative to . Note that and . Also note the duality: if and , then
if and only if . 
The unique circuit property yields the following equivalent definition of and .
Proposition 2.
Let be a basis, and a ground element.

If , then .

If , then .
That is, if , then consists of all elements from outside the basis that can replace in . If , then consists of all basis elements that can be replaced by in .
Proof.
Claim (a) follows from claim (b) and the aforementioned duality. To show claim (b), let and ; hence, . Take an arbitrary element . If , then cannot be a basis because it contains the circuit . If , then is a basis because is removed from the unique circuit contained in (the set is independent and has the same cardinality as ). ∎
Remark 2.
If , then is nonempty, because is not a loop (the set is independent). If , then Proposition 2 implies that the set is also nonempty. Indeed, since is not a coloop (does not belong to all bases), holds for some basis . By the basis exchange axiom, there is some element such that is a basis. By Proposition 2(b), the element belongs to the set ; hence, .
Proposition 3.
Let be a basis, and an circuit. Then .
Proof.
Let be an circuit. Since the set is independent, it lies in some basis , and holds since is already dependent. By Proposition 1(a), there is an such that both sets and are bases. By Proposition 2, this is equivalent to and . Thus, . Since is a basis, and both circuits and lie in , the uniqueness of fundamental circuits yields . Hence, , as claimed. ∎
Minmax and bottleneck weights of elements
A weighting is an assignment of real weights to the ground elements. The weight of a set is the sum of the weights of its elements. The minimum weight basis problem on a matroid is, given a weighting , to determine the minimum weight of a basis:
We call the number the optimal value (under the weighting ), and call a basis optimal (or just optimal, if the weighting is clear from the context) if holds, that is, if the basis is of minimal weight.
Remark 3.
Note that in the context of the minimum weight basis problem our assumption that no ground element is a loop or a coloop is quite natural. If is a loop, then it belongs to none of the bases, and the element contributes nothing to the optimal value . If is a coloop, then it belongs to all bases, and the contribution of the element to the optimal value is predetermined.
The minmax weight of an element under a weighting is the minimum, over all circuits , of the maximum weight of an element in the (independent) set :
Since is not a loop (the set is independent), the set is nonempty for every circuit . Moreover, since is not a coloop, at least one circuit exists. So, the minmax weight is welldefined. Note that the minmax weight of does not depend on the weight of the element itself: it only depends on the weights of the remaining elements. So, all three relations , and are possible. We call
the bottleneck weight of .
Remark 4.
Thanks to a classical “bottleneck extrema” result of Edmonds and Fulkerson [5], the minmax weight of a ground element has an equivalent definition as the “maxmin” weight in terms of cocircuits:
(1) 
A set is a cocircuit if it intersects every basis, and no proper subset has this property. In other words, cocircuits in a matroid are circuits of the dual matroid ; bases in are complements of the bases of . An cocircuit is a cocircuit containing the element . To obtain Eq. 1 from the main theorem of Edmonds and Fulkerson [5], one has only to verify that the family is a blocking family for the family , i.e., that members of are minimal (under set inclusion) subsets of intersecting all members of . This follows from a wellknown fact that holds for every circuit and every cocircuit (see, for example [9, Proposition 2.1.11]). In this paper, we will not use this equivalent “maxmin” definition Eq. 1 of : we only mention it for interested readers.
Example 1.
The graphic matroid (or cycle matroid) determined by an undirected connected graph has edges of as its ground elements. Independent sets are forests, bases are spanning trees of , and circuits are simple cycles in . A loop is an edge with identical endpoints, and a coloop is an edge whose deletion destroys the connectivity of (such edges are also called bridges). The minmax weight of an edge is the minimum, over all simple paths in of length at least two between the endpoints of , of the maximum weight of an edge in this path. The bottleneck weight of an edge is also known as the bottleneck distance between the endpoints of . If is a spanning tree of and is an edge of , then the set consists of all edges of the unique path in between the endpoints of . If , then consists of all edges of lying between the two trees of , except the edge itself.
For the rest of the paper, let be an arbitrary loopless matroid.
4. Main lemma
The following lemma (illustrated in Figure 1) is our main technical tool.
Lemma 1 (Main lemma).
Let be a weighting, a ground element, and an optimal basis.

If , then

the minimum weight of an element in ;

the minimum weight of a basis avoiding the element is .


If , then

the maximum weight of an element in ;

the minimum weight of a basis containing the element is .

Given a weighting and an element , by an circuit witnessing the minmax weight of an element we will mean an circuit on which the minmax weight of the element is achieved, that is, for which is the maximum weight of an element .
Proof of Lemma 1(a).
Let be a weighting, a ground element, and an optimal basis. Assume that , and let be a lightest element in . Our goal is to show that

, and

the set is a lightest basis among all bases avoiding the element .
(i) To show the inequality , suppose for a contradiction that holds. Since , Proposition 2 implies that the set is a basis. But then its weight is smaller than that of , contradicting the optimality of .
To show the inequality , let be an circuit witnessing the minmax weight of the element . Hence, , where is a heaviest element of . By Proposition 3, there is an element in the intersection . Then because and is a heaviest element of , and because and is a lightest element of . Hence, .
To show the opposite inequality , consider the fundamental circuit of the element relative to the basis . Since , we have . Thus, both and belong to the same circuit . Let be a heaviest element in . Since in the definition of the minmax weight we take the minimum over all circuits containing , we have . So, it remains to show that holds. Suppose for a contradiction that we have a strict inequality . Then (clearly) and, hence, . By Proposition 2, the set is a basis. But the weight of this basis is , contradicting the optimality of the basis . Thus, , as desired.
(ii) Let be the family of all bases avoiding the element ; hence, . Since belongs to , Proposition 2 implies that the set is a basis, and this basis belongs to . Our goal is to show that this specific basis has the smallest weight among all bases in .
So, let be a lightest basis in ; hence, . By Proposition 1(b), there is a bijection such that the set is a basis for every element . Since the basis is optimal, this yields for every . Consider the basis where . Its weight is
Since is a basis, Proposition 2 implies that . Since both elements and belong to , and since is a lightest element of , we have . So , meaning that is a lightest basis in , as claimed. ∎
Remark 5.
The inequality in part (i) also follows from the equivalent definition Eq. 1 of the minmax weight (see Remark 4). For this, it is enough to verify that the set is a cocircuit. The set intersects every basis with : by the basis exchange axiom, is a basis for some ; hence, . Moreover, no proper subset of has this property: for every , the set does not intersect the basis .
Proof of Lemma 1(b).
Let be a weighting, a ground element, and an optimal basis. Assume that , and let be a heaviest element in . Our goal is to show that

, and

the set is a lightest basis among all bases containing the element .
(i) To show the inequality , suppose for a contradiction that . Since , Proposition 2 implies that the set is a basis. But then its weight is smaller than that of , contradicting the optimality of .
The inequality holds because is an circuit, and takes the minimum (of the maximum weights) over all circuits. To show the opposite inequality , suppose for a contradiction that we have a strict inequality , and let be an circuit witnessing . Hence, holds for all . Since , Proposition 2 implies that is also a basis. Since and is an circuit, Proposition 3 implies that some element belongs to . So, by Proposition 2, the set is a basis. But since , we have , contradicting the optimality of .
(ii) Let be the family of all bases containing the element ; hence, . Since belongs to , Proposition 2 implies that the set is a basis, and this basis belongs to . Our goal is to show that this specific basis has the smallest weight among all bases in .
So, let be a lightest basis in ; hence, . By Proposition 1(b), there is a bijection such that the set is a basis for every element . Since the basis is optimal, this yields for every . Consider the basis where . Its weight is
Since is a basis, Proposition 2 implies that . Since both elements and belong to , and since is a heaviest element of , we have . So , meaning hat is a lightest basis in , as claimed. ∎
Remark 6 (From optimal bases to bottleneck weights).
Proposition 4.
Let be a weighting and be an optimal basis with . If , then , where is the smallest index for which the set is dependent.
Proof.
For , let be the set of the lightest elements of , and let . The set is independent because is not a loop, and the set is dependent, because is a basis and . So, there is a unique index such that the set is independent but is dependent. Our goal is to show that holds for this .
Since is independent but is dependent, the set contains an circuit . Since , the uniqueness of fundamental circuits yields ; hence, . Since the set is independent, the last element of must be contained in . Since and since is a heaviest element of , is also a heaviest element of . Thus, Lemma 1(b) gives . ∎
5. Proof of Theorem 1
Let, as before, be a loopless matroid, and be a ground element. Recall that the independent sets of the matroid , obtained by contracting the element , are all sets with and , while those of the matroid , obtained by deleting the element , are all sets with . Our goal is to show that, for every weighting , the following equalities hold:

;

.
Take an arbitrary optimal basis of ; hence, . In the proof of both equalities (a) and (b), we distinguish two cases depending on whether our element belongs to or not.
6. Proof of Theorem 2
Let and let be weightings that differ only in the weights given to the element . Our goal is to show the equality .
7. Proof of Theorem 3
Fix a ground element , and let