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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}
We prove that if $M(A)$ is a binary matroid on $(1,t)$-signed-graft and $t$ is a constant then $M(A)/T$ has constant cogirth-packing gap,
thus showing a deterministic polynomial-time algorithm for computing the cogirth of low rank PGMs.

We first prove that for any binary matroid with constant gap, adding another row to it still keeps the gap constant. Then we can work on the incidence matrix of graphs with $t$ additional columns.

Contracting a set $T$ from a matroid $M$ can be done by contracting elements in $T$ one by one, $M/e_1/\dots/e_t$. We fix an arbitrary order for elements in $T$ and let $T_i$ be the set of first $i$ elements. We prove that if $(M/T_i)|E(G)$ has constant gap then so does $(M/T_{i+1})|E(G)$.

\section{Proof of constant gap}
We first show that adding one additional row keeps the gap constant.
\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*}
Thus we have $\frac{\lambda(M')}{\sigma(M')}\leq \frac{2\lambda(M)}{\sigma(M)}$.
\end{proof}

Now we show that if $(M/T_i)|E(G)$ and its deletion minors have constant gap then $(M/T_{i+1})|E(G)$ also has constant gap.

\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 $[A,\tau]$ for any binary vector $\tau\in \F_2^n$. 
If $M$ and deletion minors of $M$ have constant gap, then $M(A')/\tau$ has constant gap.
\end{lemma}

\paragraph{Graphic case}
Before proving this lemma consider an easier case where $M$ is graphic. The new element $\tau$ identifies a vertex set $T$. 
A cut $\delta(X)$ is even if $|X\cap T|$ is even. The minimum cocircuit of $M(A')/\tau$ is the minimum even cut in $(G,T)$.

For a spanning tree $B$, let $C(B,\tau)$ be the fundamental circuit in $B\cup \set{\tau}$. Then $C(B,\tau)\cap E$ is a $T$-join, since it induces a subgraph in which the set of vertices with odd degree is exactly $T$.

% Now we try to characterize the set of bases of $M(A')/\tau$.
% A graph contains a $T$-join if and only if every component contains an even number of vertices in $T$.
One can see that the set of bases is $\set{B-e}$ for all spanning tree $B$ and all $e\in C(B,\tau)\cap B$. 
Note that the minimum base hitting set of $M(A')/t$ is the minimum even cut of $(G,T)$.


To show that gap we prove the followings,
\[
    \lambda(M(A')/\tau)\leq \lambda_2(G)\leq 2\sigma_2(G)\leq 2(|J|+\sigma(M(A')/\tau)|_{E\setminus J}),
\]
where $\lambda_2(G)$ and $\sigma_2(G)$ are the integral and fractional 3-cut in $G$, and $J$ is the minimum $T$-cut. 

Any 3-cut in $G$ hits every spanning tree at least twice. So any 3-cut is a hitting set for the bases of $M(A')/\tau$. It follows that $\lambda(M(A')/\tau)\leq \lambda_2(G)$.
The second inequality holds since the integrality gap for graph $k$-cut is 2.

For the last inequality, we claim that $x=\chi_J + \sigma(M(A')/\tau)|_{E\setminus J}$ is a feasible solution to 3-cut LP. First it is easy to see that $x(e)\leq 1$ for any edge $e$. 
Consider the size of $J\cap B$ for a spanning tree $B$. If this is at least 2, then $x$ satisfies the 3-cut constraint for $B$; Otherwise, $B\setminus J$ should be a base of $M'/\tau$ since $T$-cut intersects every $T$-join, and we have $\sigma(M'/T)|_{B\setminus J}\geq 1$ as a constraint in the base hitting set LP of $M(A')/\tau$.

To finish the proof consider the connections in the minimum cut of $G$, the minimum $T$-cut of $G$ and the minimum even cut of $G$. Note that if $\delta(X)$ is a $T$-cut then $|X\cap T|$ is odd. Then the minimum cut is either an $T$-cut or an even cut.
\begin{enumerate}
\item If $J$ is the mincut. $\lambda(M'/t)\leq 2(\lambda(G)+\sigma(M'/t))\leq 2(2\sigma(G)+\sigma(M'/t)) \leq 6\sigma(M'/t)$
\item If $\lambda(M'/t)$ is the mincut. $\lambda(M'/t)=\lambda(G)\leq 2\sigma(G)\leq 2\sigma(M'/t)$
\end{enumerate}
For either case, we have constant gap.

Now we generalize this idea to binary matroids. Before the proof we need some useful lemmas.

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 $M$.
\begin{lemma}[$(1,d)$-good implies $(k,kd)$-good]
	...
\end{lemma}

Instead of $T$-cut we find the minimum hitting set of $C(B,\tau)$ for all base $B$. An useful lemma is the following. We write $M+f$ for an one element extension on the binary representation of $M$.
\begin{lemma}\label{TcutTjoin}
Let $M$ be a binary matroid and let $M'$ be $(M+f)/f$ for a new element $f$.
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 $A\in \F_2^{r\times n}$ be the binary representation of $M$. Note that the cocircuit space of $M$ is the same as the row space of $A$. So we will work on binary matrices. 
$f$-extension can be seen as adding a new column $v$ to the binary matrix $A$. None-empty sets in the cocircuit spaces are called \emph{cocycles}. 
Any cocycle has an indicator vector $yA$ and thus we can use a row vector $y$ to represent any cocycle. 
We say a cocycle is even if $yv=0$ and odd if $yv=1$.

First consider the cogirth of $M'$. 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 $M'$, which is the minimum set in the cocircuit space with $yv=0$. So the cogirth of $M'$ is the minimum even cocycle.

On the other hand, 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 $A\chi_F=v$. Now we apply the following lemma which is easy to prove.
% 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 the minimum 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 the 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'/\tau)\leq \lambda_2(M)\leq c\sigma_2(M)\leq c(|J|+\sigma(M'/T)|_{E\setminus J})
\end{equation}

where $J$ is the minimum hitting set of $\{C(B,f)\setminus f|\forall B\}$ and $\lambda_2(M)$ ($\sigma_2(M)$) is the integral (fractional) 2-hitting set of bases of $M$.

We then apply \autoref{TcutTjoin}.
If $\lambda(M'/\tau)=\lambda(M)$, then we have $\lambda(M'/\tau)= \lambda(M)\leq c \sigma(M)\leq c\sigma(M'/\tau)$, since the optimal solution to $\sigma(M'/\tau)$ is feasible to $\sigma(M)$;
Otherwise, we extend \autoref{eq1} and get $\lambda(M'/\tau)\leq c(|J|+\sigma(M'/\tau))=c(\lambda(M)+\sigma(M'/\tau))\leq c(c'+1)\sigma(M'/\tau)$.
\end{proof}

To finish the proof of constant gap of $M/T$ we note that $(M/T_t)|E(G)=M/T$ and $(M/T_{i+1})|E(G)=[((M/T_{i})|E(G))+e_{i+1}]/e_{i+1}$ and the constant gap for deletion minors can be proven by wrapping the entire proof into an induction on the size of $E(G)$.

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