From affe3151638c3ecd6005eb28c4db13c0223bd4fe Mon Sep 17 00:00:00 2001 From: Yu Cong Date: Wed, 26 Nov 2025 17:25:14 +0800 Subject: [PATCH] fix --- notes.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/notes.tex b/notes.tex index 9204938..618d8cf 100644 --- a/notes.tex +++ b/notes.tex @@ -143,7 +143,7 @@ $f$-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 an even cocycle. +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.} @@ -151,7 +151,7 @@ On the other hand, consider a set $X$ which fails to hit all $\{C(B,f)\setminus 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 an odd cocycle. +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}