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@@ -6,7 +6,7 @@ title: Counterexample Search Notes
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.
> 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:
@@ -16,40 +16,40 @@ I also tracked the stronger special case of the double-cover conjecture in the p
## 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.
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:
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;
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:
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
- $r = 1$: $1$ 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)`;
- 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`
## 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? |
| 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 |
@@ -58,66 +58,66 @@ No counterexample appeared.
So:
- the rainbow-cover conjecture holds for every matroid in Sage's complete database with `2r <= 8`;
- 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
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))`
- 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:
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)`.
$(0,1,3,6)$, $(0,1,3,7)$, $(0,2,4,6)$, $(0,2,4,7)$, $(1,2,5,6)$, $(1,2,5,7)$, $(3,4,5,6)$, $(3,4,5,7)$.
This instance still has:
- minimum rainbow cover size `= 3`;
- 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`.
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:
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`;
- 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`;
- 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
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))`
$\bigl((0,1),(0,2),(0,3),(0,4),(1,2),(1,3),(1,4),(2,3)\bigr)$
and pairing
`((0,3),(1,5),(2,4),(6,7))`.
$\bigl((0,3),(1,5),(2,4),(6,7)\bigr)$.
## Random rank-5 linear search
No sampled rank-5 linear matroid produced a counterexample.
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 |
| $GF(2)$ | 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
@@ -127,7 +127,7 @@ Up through rank 4, the search is exhaustive, not heuristic. In that range I foun
## 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 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
@@ -150,10 +150,10 @@ The next most plausible places to look are:
## Structural reformulation
For a fixed pairing, the rainbow bases form a subset `F ⊆ {0,1}^r`. Then:
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.
- $3$-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.
@@ -163,6 +163,6 @@ 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 <= 4`;
- 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)`.
- no sampled failure among several hundred rank-$5$ linear matroids over $GF(2), GF(3), GF(4), GF(5)$.

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---
title: 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.