rainbow_base_cover/notes/counterexample-search-2026-03-26.md

---
title: Counterexample Search Notes
---

# Goal

Try to find a counterexample to the rainbow-base-cover conjecture:

> If a rank-$r$ matroid on $2r$ elements decomposes into two disjoint bases, and the elements are colored by $r$ colors each used twice, then three rainbow bases cover the ground set.

I also tracked the stronger special case of the double-cover conjecture in the partition-matroid setting:

> For the rainbow bases, are there always four of them covering each element exactly twice?

# Search Design

## Encoding

Fix a pairing of the $2r$ elements into color classes. A rainbow basis chooses exactly one element from each pair, so for fixed coloring there are at most $2^r$ rainbow candidates.

For a matroid $M$, I tested:

1. whether $M$ has two disjoint bases;
2. for each pairing, which of the $2^r$ transversals are actual bases;
3. whether some $3$ rainbow bases cover all $2r$ elements;
4. whether some 4 rainbow bases cover every element exactly twice.

This was implemented in [search_rainbow_counterexample.py](/Users/congyu/rainbow_base_cover/search_rainbow_counterexample.py).

## Exhaustive part

Sage's $AllMatroids(2r, r)$ is available up to $r = 4$, so I checked all unlabeled rank-$r$ matroids on $2r$ elements for $r = 1,2,3,4$, and all pairings of the ground set:

- $r = 1$: $1$ pairing
- $r = 2$: $3$ pairings
- $r = 3$: $15$ pairings
- $r = 4$: $105$ pairings

## Additional rank-5 probes

Since Sage does not exhaust all rank-5 matroids on 10 elements, I also sampled random linear matroids over small fields, and exhaustively checked some graphic families:

- random rank-$5$ linear matroids over $GF(2), GF(3), GF(4), GF(5)$;
- all simple graphic matroids coming from 8-edge graphs on 5 vertices.

# Results

## Exhaustive search for $r \le 4$

No counterexample appeared.

| rank $r$ | matroids checked | qualifying matroids | max observed minimum cover | any 3-cover failure? | any 4-double-cover failure? |
|---|---:|---:|---:|---|---|
| 1 | 2 | 1 | 2 | no | no |
| 2 | 7 | 3 | 2 | no | no |
| 3 | 38 | 17 | 3 | no | no |
| 4 | 940 | 730 | 3 | no | no |

So:

- the rainbow-cover conjecture holds for every matroid in Sage's complete database with $2r \le 8$;
- in the same range, the stronger partition-matroid double-cover statement also holds.

## Explicit rank-4 witness requiring 3 bases

The first rank-$4$ example I found with minimum cover number exactly $3$ is

- matroid: all_n08_r04_#493
- pairing: $\bigl((0,5),(1,4),(2,3),(6,7)\bigr)$

For this pairing there are exactly $8$ rainbow bases:

$(0,1,3,6)$, $(0,1,3,7)$, $(0,2,4,6)$, $(0,2,4,7)$, $(1,2,5,6)$, $(1,2,5,7)$, $(3,4,5,6)$, $(3,4,5,7)$.

This instance still has:

- minimum rainbow cover size $= 3$;
- a 4-rainbow exact double cover.

So the search really is reaching the sharp bound $3$, not just easy cases with cover number $2$.

## Graphic search

The known lower-bound example $K_4$ is reproduced computationally:

- all simple graphs on $4$ vertices with $6$ edges: $1$ graph checked;
- maximum minimum cover number: $3$;
- no 3-cover failure;
- no 4-double-cover failure.

I also checked all simple 8-edge graphs on 5 vertices:

- graphs checked: $45$;
- connected graphs: $45$;
- qualifying graphic matroids: $45$;
- maximum minimum cover number: $3$;
- no 3-cover failure;
- no 4-double-cover failure.

One concrete simple graphic rank-$4$ witness with minimum cover $3$ is the graph with edge set

$\bigl((0,1),(0,2),(0,3),(0,4),(1,2),(1,3),(1,4),(2,3)\bigr)$

and pairing

$\bigl((0,3),(1,5),(2,4),(6,7)\bigr)$.

## Random rank-5 linear search

No sampled rank-$5$ linear matroid produced a counterexample.

Finished runs:

| field | samples | distinct matroids checked | qualifying matroids | max observed minimum cover | any 3-cover failure? | any 4-double-cover failure? |
|---|---:|---:|---:|---:|---|---|
| $GF(2)$ | 200 | 200 | 92 | 3 | no | no |
| $GF(2)$ | 500 | 500 | 245 | 3 | no | no |
| $GF(3)$ | 200 | 200 | 180 | 3 | no | no |
| $GF(4)$ | 200 | 200 | 99 | 3 | no | no |
| $GF(5)$ | 200 | 200 | 200 | 2 | no | no |

# Observations

## Small-rank evidence is strong

Up through rank 4, the search is exhaustive, not heuristic. In that range I found no obstruction even to the stronger four-basis exact double-cover property.

## Rank 4 already has nontrivial tight examples

The bound $3$ is still best possible in rank $4$: there are pairings where $2$ rainbow bases do not suffice, but $3$ do.

## The partition-matroid special case looks robust

At least computationally, the partition-matroid case of the double-cover conjecture behaves better than expected:

- exhaustive success for all matroids on 8 elements of rank 4;
- no failures in the rank-5 linear samples over four small fields;
- no failures in the checked graphic families.

# What I would try next

## More targeted rank-5 search

The next most plausible places to look are:

1. exhaustive or semi-exhaustive graphic search on 10 edges and 6 vertices;
2. sparse paving matroids of rank 5 on 10 elements;
3. biased random linear constructions, especially sparse or highly structured matrices rather than dense uniform random ones;
4. direct search for the stronger four-basis exact double-cover failure, since that might break before the three-cover statement does.

## Structural reformulation

For a fixed pairing, the rainbow bases form a subset $F \subseteq \{0,1\}^r$. Then:

- $3$-cover means there exist $x,y,z \in F$ such that in every coordinate, not all of $x_i,y_i,z_i$ are equal;
- exact $4$-double-cover means there exist $x_1,x_2,x_3,x_4 \in F$ such that every coordinate has exactly two $0$s and two $1$s.

This reformulation may be a better starting point for a structural attack than thinking directly in matroid language.

# Current Status

I did not find a counterexample.

The strongest completed evidence from this turn is:

- exhaustive verification for all rank-$r$ matroids on $2r$ elements with $r \le 4$;
- exhaustive verification for simple graphic rank-4 instances on 8 edges;
- no sampled failure among several hundred rank-$5$ linear matroids over $GF(2), GF(3), GF(4), GF(5)$.