generated from congyu/work_with_codex
264 lines
8.3 KiB
Markdown
264 lines
8.3 KiB
Markdown
---
|
|
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.
|