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notes.tex

@@ -49,7 +49,7 @@ Let $G=(V,E)$ be a graph and let $\ell:V\to \F_2^{t}$ be a $t$-dimensional color
 \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.
+This observation suggests that one cannot solve the two cases separately in polynomial time.
 
 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}.
@@ -64,6 +64,19 @@ We first prove that for any binary matroid with constant gap, adding another row
 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}
+
+\paragraph{notations and useful lemmas}
+Let $M=(E,\mathcal I)$ be a matroid. We write $\lambda(M)$ for the cogirth of $M$ and write $\sigma(M)$ for the strength of $M$.
+It is known that the strength of $M$ equals the maximum fractional base packing number and $\sigma(M)=\min_{F\subset E} \frac{|E-F|}{r(E)-r(F)}$.
+
+In the proof we frequently use the following result on binary matroids.
+The \emph{cocircuit space} of a binary matroid $M=(E,\mathcal I)$ is the subspace of $\F_2^{E}$ that are generated by the incidence vectors of cocircuits of $M$.
+\begin{lemma}[\cite{Oxley_2011}, Chapter 9]
+Let $M=(E,\mathcal I)$ be a binary matroid and let $A\in \F_2^{r(M)\times |E|}$ be its binary representation. 
+Then the row space of $A$ equals the cocircuit space of $M$.
+Any non-zero vector in the cocircuit space represents a dependent set of $M$.
+\end{lemma}
+
 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 $\sigma\in \F_2^{m}$.
@@ -77,10 +90,10 @@ We can assume that $r(E)-r(F^*)\geq 1$ since otherwise we have $r'(E)-r'(F^*)\le
 
 \begin{equation*}
 \begin{aligned}
-\sigma(M')  &=      \frac{|E-F^*|}{r'(E)-r'(F^*)}\\
+\sigma(M')  &=      \frac{|E-F^*|}{r'(E)-r'(F^*)}
-            &\geq   \frac{|E-F^*|}{r(E)+1-r(F^*)}\\
+            \geq   \frac{|E-F^*|}{r(E)+1-r(F^*)}\\
-			&\geq   \frac{|E-F^*|}{2(r(E)-r(F^*))}\\
+			&\geq   \frac{|E-F^*|}{2(r(E)-r(F^*))}
-			&\geq   \frac{\sigma(M)}{2}
+			\geq   \frac{\sigma(M)}{2}
 \end{aligned}
 \end{equation*}
 Thus we have $\frac{\lambda(M')}{\sigma(M')}\leq \frac{2\lambda(M)}{\sigma(M)}$.
@@ -95,14 +108,16 @@ If $M$ and deletion minors of $M$ have constant gap, then $M(A')/\tau$ has const
 
 \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 $(G,T)$.
+Note that for a cut $\delta(X)$ one can sum up the rows labelled by $v\in X$ in the incidence matrix and get an indicator vector of $\delta(X)$.
+A cut $\delta(X)$ is even if $|X\cap T|$ is even.
+Then the minimum cocircuit of $M(A')/\tau$ is the minimum even cut in $(G,T)$.
+Let $G=(V,E)$ be a graph and let $T\subset V$ be a vertex set with even size.
+An edge set $F\subset E$ is a \emph{$T$-join} if $X$ induces a subgraph in which the set of vertices with odd degree is exactly $T$.
+An cut $\delta(X)$ is a \emph{$T$-cut} if $|X\cap T|$ odd. Note that any $T$-cut hits every $T$-join. For proofs, we refer to Section~12.4 in \cite{Korte_Vygen_2018}.
 
 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 $(G,T)$.
+% 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,
@@ -117,14 +132,14 @@ 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.
+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 $G$. Recall that the minimum $T$-cut is the minimum cut $\delta(X)$ with $|X\cap T|$ odd. Then the minimum cut is either a $T$-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.
+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 $M$.
 \begin{lemma}[$(1,d)$-good implies $(k,kd)$-good]
@@ -134,7 +149,7 @@ Note that we need constant gap for all deletion minors of $M$ since we are going
 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 $M'$.
+The cogirth of $M$ is either the minimum hitting set of $\{C(B,f)\setminus f|\forall B\in\mathcal B(M)\}$ or the cogirth of $M'$.
 \end{lemma}
 
 \begin{proof}
@@ -145,7 +160,7 @@ $f$-extension can be seen as adding a new column $v$ to the binary matrix $A$. N
 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.
+First consider the cogirth of $M'$. It follows from definitions that the minimum hitting set of $\{B-e |\forall B\in\mathcal{B}(M),\forall e\in B\cap C(B,f)\}$ 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.}
@@ -164,7 +179,6 @@ We follow the same framework as the graphic case. Let $c$ be the 2-hitting set g
 \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}.

ref.bib

@@ -30,3 +30,5 @@
 	year = {2019},
 	pages = {1351--1386},
 }
+@book{Oxley_2011, title={Matroid Theory}, ISBN={9780198566946}, url={https://academic.oup.com/book/34846}, DOI={10.1093/acprof:oso/9780198566946.001.0001}, publisher={Oxford University Press}, author={Oxley, James}, year={2011}, month=feb }
+@book{Korte_Vygen_2018, address={Berlin, Heidelberg}, series={Algorithms and Combinatorics}, title={Combinatorial Optimization: Theory and Algorithms}, volume={21}, rights={https://www.springer.com/tdm}, ISBN={9783662560389}, url={https://link.springer.com/10.1007/978-3-662-56039-6}, DOI={10.1007/978-3-662-56039-6}, publisher={Springer Berlin Heidelberg}, author={Korte, Bernhard and Vygen, Jens}, year={2018}, collection={Algorithms and Combinatorics}, language={en} }