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\title{Cogirth of Perturbed Graphic Matroids}
\author{}
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\maketitle

\section{Introduction}
Geelen and Kapadia design randomized polynomial time algorithms for computing girth and cogirth of perturbed graphic matroids \cite{geelen_computing_2018}. They leave an open problem that if there are deterministic polynomial time algorithms. We solve the cogirth part using base packing techniques.

% We want to use basepacking on perturbed graphic matroid. Basepacking works for deletion closed matroid classes with constant cogirth-packing gap. 
A binary matroid is a low rank perturbed graphic matroid (PGM) if it has a binary representation $A+P$, where $A$ is the incidence matrix of a graph and $P$ is a binary matrix with rank at most a constant $r$.
For fixed $r$, low rank perturbed graphic matroids are closed under minors. PGMs play a central role in the following conjecture.\footnote{Details can be found in the introduction of \cite{geelen_computing_2018}.}
\begin{conjecture}
For any proper minor closed class $\mathcal M$ of binary matroids, there is a polynomial time algorithm for computing the girth of matroids in $\mathcal M$.
\end{conjecture}

Geelen and Kapadia reduce the cogirth problem on low rank PGMs to the one on $(1,t)$-signed-grafts.
% We prove that matroids on $(1,t)$-signed-grafts have constant cogirth-packing gap.
Let $G$ be a graph and let $A(G)$ be its incidence matrix.
The incidence matrix of a $(1,t)$-signed-graft $(G,\set{s},T,B,C,D)$ is the binary matrix
\[
A=
\begin{array}{ccc}
      & \begin{array}{cc} E(G) & T \end{array} \\
    \begin{array}{r} V(G) \\ \set{s} \end{array}
      &
        \begin{pmatrix}
          A(G)  &   B \\
          C     &   D
        \end{pmatrix}
\end{array},
\]
where $T$ indexes $t$ new columns and $\set{s}$ indexes a new row.
The matroid $M(A)$ is the linear matroid on the matrix $A\in \F_2^{(V(G)+S)\times (E(G)+T)}$.

\subsection{previous works}
The cogirth problem on $(1,t)$-signed-grafts can be considered as variation of graph min-cut under congruency constraints.
\begin{problem}[$t$-dimensional even cut, \cite{geelen_computing_2018}]
\label{prob:tdimevencut}
Let $G=(V,E)$ be a graph and let $\ell:V\to \F_2^{t}$ be a $t$-dimensional coloring on vertices. Given a edge set $C\subset E$ and a coloring $D\in  \F_2^{1\times t}$, find a non-empty vertex set $X\subset V$ that minimizes the smaller value of the following two:
\begin{enumerate}
\item the minimum $|\delta(X)|$ such that $\sum_{v\in X}\ell(v)=0$;
\item the minimum $|\delta(X)\setsymdiff C|$ such that $\sum_{v\in X}\ell(v)=D$.
\end{enumerate}
\end{problem}

Consider a special case of \autoref{prob:tdimevencut} that $\ell=D=0$ for all vectices and $C=E$. Then the $\delta(X)$ achieving the minimum value of case 2 is exactly the max-cut of $G$.
This observation suggests that one cannot deal with the two cases separately.

Another interesting special case is that $C=\emptyset$ and $D=0$. The problem becomes graph min-cut with congruency constrants, 
which is a special case of submodular function minimization under congruency constraints (SFMC) studied by Nägele \etal \cite{nagele_submodular_2019}.
They show that SFMC with constant number of constraints can be solved in polynomial time if the modular is prime. However, the objective in case 2 of \autoref{prob:tdimevencut} is not submodular so their method does not generalize.

\subsection{proof outline}

\section{Proof of constant gap}
\begin{lemma}
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 $\sigma\in \F_2^{m}$.
\end{lemma}
\begin{proof}
Let $M'$ be $M\left(\begin{bmatrix}B\\ \sigma\end{bmatrix}\right)$ and let the ground set be $E$.
$\lambda(M')\leq \lambda(M)$ (row space). 
Let $F^*\in \argmin_{F\subset E} \frac{|E-F|}{r'(E)-r'(F)}$.

\begin{equation*}
\begin{aligned}
\sigma(M')  &=      \frac{|E-F^*|}{r'(E)-r'(F^*)}\\
            &\geq   \frac{|E-F^*|}{r(E)+1-r(F^*)}\\
			&\geq   \frac{|E-F^*|}{2(r(E)-r(F^*)}\\
			&\geq   \frac{\sigma(M)}{2}
\end{aligned}
\end{equation*}
So the gap $\frac{\lambda(M')}{\sigma(M')}\leq \frac{2\lambda(M)}{\sigma(M)}$.
\end{proof}

\begin{lemma}\label{contraction_gap}
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.
\end{lemma}

\paragraph{Graphic case}
Before proving this lemma consider an 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.
We want to prove the followings:
\[
    \lambda(M'/t)\leq \lambda_2(G)\leq 2\sigma_2(G)\leq 2(|J|+\sigma(M'/T)|_{E\setminus J}),
\]
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.

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.
\begin{enumerate}
\item $J$ is mincut. $\lambda(M'/t)\leq 2(\lambda(G)+\sigma(M'/t))\leq 6\sigma(M'/t)$
\item $\lambda(M'/t)$ is the mincut. $\lambda(M'/t)=\lambda(G)\leq 2\sigma(G)\leq 2\sigma(M'/t)$
\end{enumerate}

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 $M$.

Instead of $T$-cut we find the minimum hitting set of $C(B,T)$ for all base $B$. An useful lemma is the following.
\begin{lemma}\label{TcutTjoin}
Let $M$ be a binary matroid and let $M'$ be $(M+f)/f$ for a new element $f$ ($+$ is extension on binary representation).
The cogirth of $M$ is either the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$ or the cogirth of $M'$.
\end{lemma}

\begin{proof}
For binary matroid $M$, finding the cogirth of $M$ is equivalent to finding the smallest nonempty set in the cocircuit space.

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:

\begin{enumerate}
\item   even cocycle. $yv=0$ (Note that cocycle here only means that the set is in the cocircuit space.)
\item   odd cocycle. $yv=1$
\end{enumerate}

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.

Consider a set $X$ which fails to hit all $\{C(B,f)\setminus f|\forall B\}$. 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 a indicator vector $\chi_F$ such that $A\chi_F=v$. 
% 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.}
\begin{lemma}[Farkas' lemma on $\F_2$]
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 $y^Tb=1$.
\end{lemma}
Then $X$ fails to be a hitting set is equivalent to the fact that there does not exist $y$ satisfying $\supp(yA)\subset X\land yv=1$.
So $X$ hits all $\{C(B,f)\setminus f|\forall B\}$ iff there is such a $y$. Minimizing $X$ pushes it to $\supp(yA)$ which is an odd cocycle. Hence, the minimum hitting set for $\{C(B,f)\setminus f|\forall B\}$ is always an odd cocycle.

The theorem then follows directly from the fact that any cocycle is either odd or even.
\end{proof}

Now we prove \autoref{contraction_gap}.
\begin{proof}
We follow the same framework as graphic case. Let $c$ be the 2-hitting set gap and let $c'$ be the cogirth-packing gap of $M$. We have
\begin{equation}\label{eq1}
\lambda(M'/t)\leq \lambda_2(G)\leq c\sigma_2(G)\leq c(|J|+\sigma(M'/T)|_{E\setminus J})
\end{equation}

where $J$ is the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$.

We then apply \autoref{TcutTjoin}. If $\lambda(M'/t)=\lambda(M)$, then we have $\lambda(M'/t)= \lambda(M)\leq c \sigma(M)\leq c\sigma(M'/t)$, since the optimal solution to $\sigma(M'/t)$ is feasible to $\sigma(M)$. Otherwise, we extend \autoref{eq1} and get $\lambda(M'/t)\leq c(|J|+\sigma(M'/T))=c(\lambda(M)+\sigma(M'/T))\leq c(c'+1)\sigma(M'/T)$.
\end{proof}

The constant gap for matroids on $(1,t)$-signed grafts then follows.
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