Add proof attempt and fix math markup in notes

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

@@ -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;
+1. whether $M$ has two disjoint bases;
-2. for each pairing, which of the `2^r` transversals are actual bases;
+2. for each pairing, which of the $2^r$ transversals are actual bases;
-3. whether some 3 rainbow bases cover all `2r` elements;
+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
+- $r = 1$: $1$ pairing
-- `r = 2`: `3` pairings
+- $r = 2$: $3$ pairings
-- `r = 3`: `15` pairings
+- $r = 3$: $15$ pairings
-- `r = 4`: `105` 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`
+- matroid: all_n08_r04_#493
-- pairing: `((0,5),(1,4),(2,3),(6,7))`
+- 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;
+- all simple graphs on $4$ vertices with $6$ edges: $1$ graph checked;
-- maximum minimum cover number: `3`;
+- 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`;
+- graphs checked: $45$;
-- connected graphs: `45`;
+- connected graphs: $45$;
-- qualifying graphic matroids: `45`;
+- qualifying graphic matroids: $45$;
-- maximum minimum cover number: `3`;
+- 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 |
+| $GF(2)$ | 200 | 200 | 92 | 3 | no | no |
-| `GF(2)` | 500 | 500 | 245 | 3 | no | no |
+| $GF(2)$ | 500 | 500 | 245 | 3 | no | no |
-| `GF(3)` | 200 | 200 | 180 | 3 | no | no |
+| $GF(3)$ | 200 | 200 | 180 | 3 | no | no |
-| `GF(4)` | 200 | 200 | 99 | 3 | no | no |
+| $GF(4)$ | 200 | 200 | 99 | 3 | no | no |
-| `GF(5)` | 200 | 200 | 200 | 2 | 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;
+- $3$-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.
+- 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 $GF(2), GF(3), GF(4), GF(5)$.

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

@@ -0,0 +1,263 @@
+---
+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.