100 lines
6.2 KiB
TeX
100 lines
6.2 KiB
TeX
\documentclass[a4paper,11pt]{article}
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\usepackage{chao}
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\DeclareMathOperator{\supp}{supp}
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\title{Cogirth of Perturbed Graphic Matroids}
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\author{}
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\date{}
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\begin{document}
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\maketitle
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We want to use basepacking on perturbed graphic matroid. Basepacking works for deletion closed matroid classes with constant cogirth-packing gap.
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PGMs are closed under deletion. It remains to show PGMs have constant gap.
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Geelen and Kapadia reduce the cogirth problem on PGMs to finding the cogirth of $(1,t)$-signed grafts.
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We work on binary matroid $M$ defined on the following binary matrix:
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\[
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\begin{array}{ccc}
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& \begin{array}{cc} E(G) & \tau \end{array} \\
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\begin{array}{r} V(G) \\ \omega \end{array}
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&
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\begin{pmatrix}
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A(G) & T \\
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\sigma & \alpha
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\end{pmatrix}
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\end{array}
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\]
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where $G$ is a connect graph and $A(G)$ is the incidence matrix of $G$.
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The goal is to find the cogirth of $M/\tau$.
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This matroid can be built from a graphic matroid via a series of extension, contraction and co-extension. We first show that $M([A,T])/T$ has constant gap, and then prove a constant gap of $M/\tau$.
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\begin{lemma}
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Let $M$ be a binary matroid with binary representation $B\in \F_2^{n\times m}$. If $M$ has constant gap, then $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ has constant gap for any row vector $\sigma\in \F_2^{m}$.
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\end{lemma}
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\begin{proof}
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Let $M'$ be $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ and let the ground set be $E$.
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$\lambda(M')\leq \lambda(M)$ (row space).
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Let $F^*\in \argmin_{F\subset E} \frac{|E-F|}{r'(E)-r'(F)}$.
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\begin{equation*}
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\begin{aligned}
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\sigma(M') &= \frac{|E-F^*|}{r'(E)-r'(F^*)}\\
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&\geq \frac{|E-F^*|}{r(E)+1-r(F^*)}\\
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&\geq \frac{|E-F^*|}{2(r(E)-r(F^*)}\\
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&\geq \frac{\sigma(M)}{2}
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\end{aligned}
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\end{equation*}
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So the gap $\frac{\lambda(M')}{\sigma(M')}\leq \frac{2\lambda(M)}{\sigma(M)}$.
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\end{proof}
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\begin{lemma}
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Let $M$ be a binary matroid with representation $A\in \F_2^{n\times m}$. Let $A'$ be the binary matrix $[A,T]$ for any $T\in \F_2^n$. If $M$ and deletion minors of $M$ have constant gap, then $M(A')/T$ also has constant gap.
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\end{lemma}
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\paragraph{Graphic case}
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Before proving this lemma consider a easier case where $M$ is graphic. The new element $T$ identifies a vertex set $T$ (abusing notations). The minimum cocircuit of $M'/T$ is exactly the minimum even cut in $(G,T)$. Consider the fundamental circuits $C(B,T)$. $C(B,T)\cap E$ for any spanning tree $B$ is exactly the $T$-join in $B$. What is a base of $M'/T$? $B-e$ for all spanning tree $B$ and $e\in C(B,T)\setminus \set{T}$. This base hitting set of $M'/t$ is the even cut.
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We want to prove the followings:
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\[
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\lambda(M'/t)\leq \lambda_2(G)\leq 2\sigma_2(G)\leq 2(|J|+\sigma(M'/T)|_{E\setminus J}),
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\]
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where $\lambda_2(G)$ and $\sigma_2(G)$ are the integral and fractional 3-cut in $G$, $J$ is the minimum $T$-cut. To see the last inequality, consider for any base $B$ the size of $J\cap B$. If this is at least 2, then RHS satisfies the 2-cut constraint for $B$; otherwise, $B\setminus J$ should be a base in $M'/T$ since $T$-cut intersects every $T$-join, and we have $\sigma(M'/T)|_{B\setminus J}\geq 1$ as a constraint in the hitting set.
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Then we discuss the connections in the minimum cut of $G$, the minimum $T$-cut of $G$ and the minimum even cut of $G$. It is easy to see that the mincut is either an odd cut or an even cut.
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\begin{enumerate}
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\item $J$ is mincut. $\lambda(M'/t)\leq 2(\lambda(G)+\sigma(M'/t))\leq 6\sigma(M'/t)$
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\item $\lambda(M'/t)$ is the mincut. $\lambda(M'/t)=\lambda(G)\leq 2\sigma(G)\leq 2\sigma(M'/t)$
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\end{enumerate}
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Now we generalize this idea to binary matroids. Note that we need constant gap for all deletion minors of $M$ since we are going to use the gap of 2-cocycle LP of $M$.
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Instead of $T$-cut we find the minimum hitting set of $C(B,T)$ for all base $B$. An useful lemma is the following.
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\begin{lemma}
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Let $M$ be a binary matroid and let $M'$ be $(M+f)/f$ for a new element $f$ ($+$ is extension on binary representation).
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The cogirth of $M$ is either the minimum hitting set of $\{C(B,f)|\forall B\}$ or the cogirth of $M'$.
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\end{lemma}
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\begin{proof}
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For binary matroid $M$, finding the cogirth of $M$ is equivalent to finding the smallest nonempty set in the cocircuit space.
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Let $A$ be the binary representation of $M$. Note that the cocircuit space of $M$ is the same as the row space of $A$. So we work on matrix instead. Any cocircuit of $M$ have a indicator vector $yA$ and thus we can use a row vector $y$ to indicate any cocircuit. $f$-extension can be also seen as adding a new column $v$ to the binary matrix. This new element gives a dichotomy on the cocircuit space:
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\begin{enumerate}
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\item even cocycle. $yv=0$ (Note that cocycle here only means that the set is in the cocircuit space.)
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\item odd cocycle. $yv=1$
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\end{enumerate}
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Now we consider the hitting sets. It follows from definitions that the minimum hitting set of $\{B-e |\forall B ,\forall e\in B\cap C(B,f)\}$ is exactly the minimum cocircuit of $(M+f)/f$, which is the minimum set in the cocircuit space with $yv=0$. So the minimum hitting set of $B-e$ is always an even cocycle.
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Consider a set $X$ which fails to hit all $\{C(B,f)|\forall B\}$. Then there must be a set $F\subseteq E\setminus X$ that span the new element $f$. On matrix, this is equivalent to the existence of $\chi_F$ that $A\chi_F=v$.
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% Using Farkas' lemma on $\mathbb F_2$, this implies that there dose not exists $y$ that $\supp(yA)\subset X\land yv=1$. \note{chatGPT says one can use Farkas lemma on finite field like this, need to verify.}
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\begin{lemma}[Farkas' lemma on $\F_2$]
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Let $A$ be a $n\times m$ binary matrix and $b$ be a $n$-dimensional binary vector. Either there is a $x\in \F_2^m$ such that $Ax=b$, or there is $y\in \F_2^n$ such that $y^TA=0$ and $y^Tb=1$.
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\end{lemma}
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Then $X$ fails to be a hitting set is equivalent to the fact that there does not exist $y$ satisfying $\supp(yA)\subset X\land yv=1$.
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So $X$ hits all $\{C(B,f)|\forall B\}$ iff there is such a $y$. Minimizing $X$ pushes it to $\supp(yA)$ which is an odd cocycle. Hence, the minimum hitting set for $C(B,f)$ is always an odd cocycle.
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The theorem then follows directly from the fact that any cocycle is either odd or even.
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\end{proof}
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\end{document} |