From 10c8dccad8922c2f8584b7aa2ad60adaa052ea50 Mon Sep 17 00:00:00 2001 From: Yu Cong Date: Wed, 5 Nov 2025 00:02:13 +0800 Subject: [PATCH] Update notes.tex --- notes.tex | 9 +++++++-- 1 file changed, 7 insertions(+), 2 deletions(-) diff --git a/notes.tex b/notes.tex index c47e720..d818dba 100644 --- a/notes.tex +++ b/notes.tex @@ -86,8 +86,13 @@ Let $A$ be the binary representation of $M$. Note that the cocircuit space of $M 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 $\{C(B,f)|\forall B\}$. 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 $A\chi_F=v$. Using Farkas' lemma on $\mathbb F_2$, this implies that there dose not exists $y$ that $\supp(yA)\in X\land yv=1$. (chatGPT says one can use Farkas lemma on finite field like this, need to verify). -So $X$ hits all $\{C(B,f)|\forall B\}$ iff there is such a $y$. Minimizing $X$ pushes $\supp(yA)$ which is an odd cocycle. Hence, the minimum hitting set for $C(B,f)$ is always an odd cocycle. +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 $A\chi_F=v$. +% 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 $\{C(B,f)|\forall B\}$ 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. The theorem then follows directly from the fact that any cocycle is either odd or even.