5.9 KiB
title
| title |
|---|
| Counterexample Search Notes |
Goal
Try to find a counterexample to the rainbow-base-cover conjecture:
If a rank-
rmatroid on2relements decomposes into two disjoint bases, and the elements are colored byrcolors 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:
- whether
Mhas two disjoint bases; - for each pairing, which of the
2^rtransversals are actual bases; - whether some 3 rainbow bases cover all
2relements; - whether some 4 rainbow bases cover every element exactly twice.
This was implemented in 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 = 1:1pairingr = 2:3pairingsr = 3:15pairingsr = 4:105pairings
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 <= 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 <= 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:
((0,5),(1,4),(2,3),(6,7))
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:
1graph 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
((0,1),(0,2),(0,3),(0,4),(1,2),(1,3),(1,4),(2,3))
and pairing
((0,3),(1,5),(2,4),(6,7)).
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:
- exhaustive or semi-exhaustive graphic search on 10 edges and 6 vertices;
- sparse paving matroids of rank 5 on 10 elements;
- biased random linear constructions, especially sparse or highly structured matrices rather than dense uniform random ones;
- 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 ⊆ {0,1}^r. Then:
- 3-cover means there exist
x,y,z in Fsuch that in every coordinate, not all ofx_i,y_i,z_iare equal; - exact 4-double-cover means there exist
x_1,x_2,x_3,x_4 in Fsuch that every coordinate has exactly two0s and two1s.
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-
rmatroids on2relements withr <= 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).