rainbow_base_cover/notes/proof-attempt-2026-03-26.md

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Proof Attempt and the Strongly Base Orderable Case

Status

I do not have a proof of the full conjecture.

I do have a clean proof of a strong special case:

If the matroid is strongly base orderable, then two rainbow bases already suffice.

So in that class the conjectured bound 3 is true with room to spare.

Setup

Let M be a rank-r matroid on E, and suppose

• E = A \sqcup B,
• A and B are bases,
• the ground set is partitioned into r color classes of size 2.

Write

• A = \{a_1,\dots,a_r\},
• B = \{b_1,\dots,b_r\}.

Later I will assume that b_i = \varphi(a_i) for a bijection \varphi : A \to B.

A Useful Reformulation

Fix any bijection \varphi : A \to B, and write b_i = \varphi(a_i).

For each subset I \subseteq [r], define the transversal

[ T_I := {a_i : i \notin I} \cup {b_i : i \in I}. ]

So T_I picks exactly one element from each pair \{a_i,b_i\}.

The question is:

• when is T_I rainbow?
• when is T_I a basis?

The second question depends on the matroid. The first depends only on the coloring.

The Color Graph

From the coloring, build a multigraph G on vertex set [r] as follows.

For each color pair:

• if the color pair is \{a_i,b_i\}, add a loop at i;
• if it is \{a_i,b_j\} or \{b_i,a_j\} with i \neq j, add an edge ij with label 0;
• if it is \{a_i,a_j\} or \{b_i,b_j\} with i \neq j, add an edge ij with label 1.

Because each element of E appears in exactly one color pair, each vertex i contributes exactly two half-edges, one coming from a_i and one from b_i. Hence every vertex has degree 2, so each connected component of G is a cycle or a loop.

Lemma: The Parity System Is Always Consistent

For the graph G, consider the system over \mathbf F_2

[ x_i + x_j = \lambda(ij) ]

for every non-loop edge ij, where \lambda(ij) \in \{0,1\} is its label, and for a loop the equation is x_i + x_i = 0.

Proof

It is enough to show that on every cycle, the sum of the labels is 0 mod 2.

Take one cycle component C. Let

• p be the number of $A$-A color pairs on C,
• q be the number of $B$-B color pairs on C,
• m be the number of cross pairs ($A$-B or $B$-A) on C.

Then:

• the number of $A$-elements used by the color pairs of C is 2p + m,
• the number of $B$-elements used by the color pairs of C is 2q + m.

But each vertex in C contributes exactly one element from A and one from B, so these two numbers are equal. Therefore 2p + m = 2q + m, hence p=q.

The label-1 edges are exactly the $A$-A and $B$-B pairs, so the number of label-1 edges on C is

[ p+q = 2p, ]

which is even.

So every cycle has even total label, and the system is consistent.

The Strongly Base Orderable Case

Recall the definition.

Definition

A matroid M is strongly base orderable if for every two bases B_1,B_2 there exists a bijection \varphi : B_1 \to B_2 such that for every subset X \subseteq B_1,

[ (B_1 \setminus X) \cup \varphi(X) ]

is again a basis.

Theorem

Let M be a rank-r matroid on E, and suppose

• E = A \sqcup B with A,B bases,
• \varphi : A \to B is a bijection such that (A \setminus X) \cup \varphi(X) is a basis for every X \subseteq A,
• the elements of E are colored by r colors, each used exactly twice.

Then there exist two rainbow bases whose union is E.

In particular, every strongly base orderable matroid satisfies the rainbow-cover conjecture, and in fact needs at most 2 rainbow bases.

Proof

Write b_i = \varphi(a_i) and build the labeled multigraph G above.

By the lemma, the parity system

[ x_i + x_j = \lambda(ij) ]

has a solution x = (x_1,\dots,x_r) \in \{0,1\}^r.

Let

[ I := {i \in [r] : x_i = 1} ]

Define

[ R := (A \setminus {a_i : i \in I}) \cup \varphi({a_i : i \in I}) = {a_i : x_i = 0} \cup {b_i : x_i = 1}. ]

By hypothesis, R is a basis of M.

I claim that R is rainbow.

Take any color pair.

1. If the pair is \{a_i,b_i\}, then R contains exactly one of them by construction.
2. If the pair is \{a_i,b_j\} or \{b_i,a_j\}, then the edge ij has label 0, so x_i = x_j. Hence exactly one of a_i and b_j lies in R.
3. If the pair is \{a_i,a_j\} or \{b_i,b_j\}, then the edge ij has label 1, so x_i \neq x_j. Hence exactly one of the two same-side elements lies in R.

Thus R contains exactly one element of each color, so R is rainbow.

Now take the complementary solution x' = 1-x. It also satisfies the same parity system, because the equations are linear over \mathbf F_2.

The corresponding set is

[ R' := {a_i : x_i = 1} \cup {b_i : x_i = 0} = E \setminus R. ]

Again, by hypothesis, R' is a basis, and by the same argument it is rainbow.

Finally,

[ R \cup R' = E. ]

So R and R' are two rainbow bases covering the whole ground set.

Corollary for the Double-Cover Variant

In the strongly base orderable case, the partition-matroid special case of the double-cover conjecture also holds:

• the two complementary rainbow bases R and R' are common bases of M and the color partition matroid;
• therefore R,R,R',R' are four common bases covering each element exactly twice.

So this special case is stronger than the original $3$-cover statement.

Why This Does Not Yet Prove the Full Conjecture

The proof above splits the problem into two parts:

1. the coloring contributes a parity system on a 2-regular graph;
2. the matroid decides which transversals T_I are actually bases.

The first part is completely benign: for every bijection \varphi : A \to B, the parity system is always solvable.

The real difficulty is the second part.

In the strongly base orderable case, every T_I is a basis, so the parity solution immediately gives a rainbow basis. In a general matroid, the parity-compatible sets T_I need not be bases at all.

Failed General Approaches

Attempt 1: Replace strong base orderability by multiple symmetric exchange

Take disjoint bases A,B. A standard exchange theorem says that for each subset X \subseteq A, there exists some subset Y \subseteq B with |Y|=|X| such that

[ (A \setminus X) \cup Y ]

is a basis.

This is not enough here. The coloring prescribes a very specific subset of B once one chooses X; call it Y_{\mathrm{col}}(X).

The exchange theorem only gives existence of some Y, not the prescribed Y_{\mathrm{col}}(X). That is exactly the gap.

Attempt 2: Work component-by-component on the color graph

The color graph is a disjoint union of cycles. Each cycle has two natural local rainbow states.

This suggests an induction on the cycle components.

I could not make this work, because matroid basis choices do not decompose independently over those color components. Local feasible choices on two different color cycles need not glue to a global basis.

Attempt 3: Use the common-base polytope

Let P be the partition matroid of the coloring. Then both M and P have rank r, and E decomposes into two bases in both. Hence the vector (1/2)\mathbf 1_E lies in the intersection of the two base polytopes.

If one could always write (1/2)\mathbf 1_E as the average of four common-base vertices, one would get the exact double-cover statement immediately.

I do not see a mechanism forcing such a $4$-vertex decomposition. General Carathéodory bounds are far too weak.

What Seems Most Promising

The strongly base orderable proof suggests the right intermediate object.

Fix complementary bases A,B and a bijection \varphi : A \to B. The coloring always singles out a nonempty affine subspace of \{0,1\}^r consisting of the parity-compatible transversals T_I.

So a possible route to the full conjecture is:

Find a structural reason that, for some good choice of A,B,\varphi, at least three of those parity-compatible transversals are bases.

Strongly base orderable matroids are the extreme case where all of them are bases. For a general matroid, I do not currently know how to force even one parity-compatible transversal to be a basis from a fixed bijection, let alone three.

Bottom Line

I could not prove the full conjecture.

I did prove the following self-contained special case:

If M is strongly base orderable and E is the disjoint union of two bases, then for every coloring of E into r pairs there exist two rainbow bases covering E.

That is the strongest proof-level progress I found in this turn.