Add alternating-path conjecture and cographic proof

Introduce a "Known results" section discussing β(M1∩M2), state the
Aharoni–Berger conjectures and the Davies–McDiarmid theorem, and add
three bibliography entries (emlektabla16_2024, rainbow bases SIAM,
berczi_partitioning_2024).

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

@@ -1,12 +1,12 @@
 ---
-title: Counterexample Search Notes
+title: codex - 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.
+> 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 $GF(2), GF(3), GF(4), GF(5)$.

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

@@ -0,0 +1,263 @@
+---
+title: codex - 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.

notes/main.md

@@ -2,11 +2,13 @@
 title: Matroid Rainbow Base Cover
 ----
 
+This problem is proposed in [@emlektabla16_2024].
+
 Let $M=(E,\mathcal B)$  be a rank-$r$ matroid whose ground set decomposes into two disjoint bases. Furthermore,
 assume that $E$ is colored by $r$ colors, each color appearing exactly twice. A basis of $M$ is called rainbow if
 it does not contain two elements of the same color.
 
-::: Problem
+::: {.Problem #prob-rainbowbasecover}
 What is the minimum number of rainbow bases needed to cover $E$?
 :::
 
@@ -19,8 +21,92 @@ Currently known bounds:
 
 ::: {.Conjecture #conj-doublecover}
 Let $M_1$ and $M_2$ be two matroid on the same ground set $E$. Assume that $E$ decomposes into two bases in both of them.
-Then $M_1$ and $M_2$ has four common bases that cover each element exactly twice.
+Then $M_1$ and $M_2$ has four common bases (allow repetition) that cover each element exactly twice.
 :::
 
 Let $M_1$ be the partition matroid of colors and let $M_2$ be $M$.
-If [@conj-doublecover] is true, then 3 common bases will be enough to cover $E$.
+If [@conj-doublecover] is true, then 3 common bases will be enough to cover $E$.
+
+# Known results
+
+Let $\beta(M)$ be the covering number of matroid $M$. Given two matroids $M_1,M_2$ on the same ground set, $\beta(M_1\cap M_2)$ is the minimum number of common independent sets needed to cover the ground set.
+[@conj-doublecover] is basically asking for $\beta(M_1\cap M_2)$ when the ground set $E$ is partitioned into 2 common bases of $M_1$ and $M_2$.
+
+## Results and conjectures on $\beta(M_1\cap M_2)$
+
+One can see that $\beta(M_1\cap M_2)\geq \min \set{\beta(M_1),\beta(M_2)}$. Aharoni and Berger showed that $\beta(M_1\cap M_2)\leq 2 \max \set{\beta(M_1),\beta(M_2)}$
+
+::: {.Conjecture title="Aharoni and Berger"}
+Let $M_1$ and $M_2$ be two matroids on the same ground set.
+
+1. If $\beta(M_1)\neq \beta(M_2)$, then $\beta(M_1\cap M_2)=\max \set{\beta(M_1),\beta(M_2)}$.
+2. If $\beta(M_1)= \beta(M_2)$, then $\beta(M_1\cap M_2)\leq \max \set{\beta(M_1),\beta(M_2)}+1$.
+:::
+
+Cases that have been verified:
+
+- If $\beta(M_1)=\beta(M_2)=2$ then $\beta(M_1\cap M_2)=3$.^[Note that this case does not give an answer to [@prob-rainbowbasecover] since the covering in $\beta(M_1\cap M_2)$ uses common independent sets instead of common bases.]
+- If $\beta(M_1)=2,\beta(M_2)=3$, then $\beta(M_1\cap M_2)\leq 4$.
+- If $\beta(M_1)=2,\beta(M_2)=k\geq 4$, then $\beta(M_1\cap M_2)\leq 2\ceil{k/2}+2$.
+
+::: {.Theorem title="Davies and McDiarmid"}
+Let $M_1$ and $M_2$ be strongly base orderable matroids on the same ground set.
+Then $\beta(M_1\cap M_2)= \max \set{\beta(M_1),\beta(M_2)}$.
+:::
+
+See [@berczi_partitioning_2024] for refs. Bérczi and Schwarcz generalized SBO using the following definition.
+
+::: {.Definition title="strongly base reorderable"}
+A matroid is called **strongly base reorderable**(SBRO) if for any pair $A,B$ of bases, there exists bases $A',B'$ and a bijection $\phi:A'\setminus B'\to B'\setminus A'$ such that $A'\cap B'=A\cap B,A'\cup B'=A\cup B$, and $(A'\setminus X)\cup \phi(X)$ is a basis for every $X\subset A'\setminus B'$.
+:::
+
+::: Lemma
+The class of SRBO matroids is complete(closed under taking minors, duals, direct sums, truncations and induction by directed graphs).
+:::
+
+They also showed that SBRO is sufficient for $\beta(M_1\cap M_2)=\max \set{\beta(M_1),\beta(M_2)}$.
+
+::: Theorem
+Let $M_1$ and $M_2$ be strongly base reorderable matroids on the same ground set.
+Then $\beta(M_1\cap M_2)= \max \set{\beta(M_1),\beta(M_2)}$.
+:::
+
+## Circuit cover
+
+Let $M=(E,\mathcal B)$ be a matroid and let $A,B\in \mathcal B$ be two bases of $M$.
+Given a graph $G=(A\Delta B, F)$, we say $G$ covers a circuit $C\subset A\Delta B$ if the induced subgraph $G[C]$ contains an edge.
+For a class of circuits $\mathcal C$ in $A\Delta B$, we say $G$ covers $\mathcal C$ if $G$ covers every circuit in $\mathcal C$.
+
+
+::: {.Conjecture #conj-alterpath}
+Let $A,B$ be bases of a matroid $M$. Then there exists a path that alternates between $A\setminus B$ and $B\setminus A$ and covers $\mathcal C[A\cup B]$.
+:::
+
+Bérczi and Schwarcz [@berczi_partitioning_2024] verified this conjecture for graphic matroids, paving matroids and spikes. They also showed an useful lemma:
+
+::: {.Lemma title="informal" #lem-reduction}
+To verify [@conj-alterpath] for a minor closed class of matroids, one only need to consider matroids in this class whose ground set can be partitioned into two disjoint bases.
+:::
+
+
+# [@conj-alterpath] on cographic matroid
+
+::: Lemma
+[@conj-alterpath] holds on cographic matroids.
+:::
+
+::: Proof
+By [@lem-reduction] it suffices to show the following: Let $A,B$ be two spanning trees on vertex set $V$. One can find a sequence of edges in $A$ and $B$ such that
+
+1. no consecutive edges belong to $A$ or $B$, and that
+2. every cut in $(V,A\cup B)$ contains some consecutive edge pair in the sequence.
+
+We prove by induction on $|V|$.
+
+- If $|V|=2$, there is only one non-empty cut.
+- Suppose that the lemma holds for $|V|<k$. Let $G=(V,A\cup B)$ be the union of two spanning trees on $k$ vertices. The average degree of $G$ is $4-4/k$ and the degree of any vertex must be at least 2. Thus there is always a vertex $v$ with $\deg(v)\in \set{2,3}$.
+    - If $\deg(v)=2$, then $v$ must be a leaf in both spanning trees, so $G-v$ is the union of two spanning trees. Let $a,b$ be the edges incident to $v$ in $G$. By induction hypothesis there is a desired edge sequence $\mathcal S$ for the graph $G-v$. We append $a,b$ (or $b,a$, depending on which tree the last element of $\mathcal S$ belongs to) to $\mathcal S$. Then the only cut that is not covered by $\mathcal S$ is $\delta(v)$, which contains the pair $a,b.$
+    - If $\deg(v)=3$, then assume $v$ is a leaf in $A$ and a deg-2 vertex in $B$. Let $a,b_1,b_2$ be the edges incident to $v$ and let $x,y$ be the neighbors of $v$ in $B$. We construct a new graph $G'$ by deleting $v$ from $G$ and adding a new edge $d=(x,y)$. Then $G'$ is the union of two spanning trees and there is a desired sequence $\mathcal S'$ for $G'$. We construct a new sequence $\mathcal S$ by replacing $d$ in $\mathcal S'$ with 3 edges $b_1,a,b_2$. For $X\subsetneq V\setminus \set{v}$, the cut $\delta(X)$ is always covered by $\mathcal S'$ but possibly use $d$. However, any cut separating $x$ and $y$ must contain $b_1$ or $b_2$ in $G$, which keeps $\mathcal S$ feasible. The remaining case is $\delta(v)$ which is covered by $b_1,a,b_2$.
+:::
+
+# Common base cover

reference.bib

@@ -0,0 +1,30 @@
+@misc{emlektabla16_2024,
+  title        = {16th Eml{\'e}kt{\'a}bla Workshop},
+  year         = {2024},
+  howpublished = {\url{https://users.renyi.hu/~emlektab/emlektabla16.pdf}},
+  note         = {Workshop PDF; accessed 2026-03-26}
+}
+
+@article{noauthor_rainbow_nodate,
+	title = {Rainbow {Bases} in {Matroids}},
+	url = {https://epubs.siam.org/doi/10.1137/22M1516750},
+	abstract = {Abstract. Recently, it was proved by Bérczi and Schwarcz that the problem of factorizing a matroid into rainbow bases with respect to a given partition of its ground set is algorithmically intractable. On the other hand, many special cases were left open. We first show that the problem remains hard if the matroid is graphic, answering a question of Bérczi and Schwarcz. As another special case, we consider the problem of deciding whether a given digraph can be factorized into subgraphs which are spanning trees in the underlying sense and respect upper bounds on the indegree of every vertex. We prove that this problem is also hard. This answers a question of Frank. In the second part of the article, we deal with the relaxed problem of covering the ground set of a matroid by rainbow bases. Among other results, we show that there is a linear function such that every matroid that can be factorized into bases for some can be covered by rainbow bases if every partition class contains at most 2 elements.},
+	language = {en},
+	urldate = {2026-03-26},
+	journal = {SIAM Journal on Discrete Mathematics},
+}
+
+@article{berczi_partitioning_2024,
+	title = {Partitioning into common independent sets via relaxing strongly base orderability},
+	volume = {202},
+	issn = {00973165},
+	url = {https://linkinghub.elsevier.com/retrieve/pii/S0097316523000857},
+	doi = {10.1016/j.jcta.2023.105817},
+	language = {en},
+	urldate = {2026-03-26},
+	journal = {Journal of Combinatorial Theory, Series A},
+	author = {Bérczi, Kristóf and Schwarcz, Tamás},
+	month = feb,
+	year = {2024},
+	pages = {105817},
+}