From 22037a5f2358e1d82759c218a87ffcf2c461e038 Mon Sep 17 00:00:00 2001 From: Yu Cong Date: Thu, 26 Mar 2026 13:26:07 +0800 Subject: [PATCH] Add proof attempt and fix math markup in notes --- notes/counterexample-search-2026-03-26.md | 88 ++++---- notes/proof-attempt-2026-03-26.md | 263 ++++++++++++++++++++++ 2 files changed, 307 insertions(+), 44 deletions(-) create mode 100644 notes/proof-attempt-2026-03-26.md diff --git a/notes/counterexample-search-2026-03-26.md b/notes/counterexample-search-2026-03-26.md index bb294dd..7c50d1d 100644 --- a/notes/counterexample-search-2026-03-26.md +++ b/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; -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 $GF(2), GF(3), GF(4), GF(5)$. diff --git a/notes/proof-attempt-2026-03-26.md b/notes/proof-attempt-2026-03-26.md new file mode 100644 index 0000000..8321954 --- /dev/null +++ b/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.