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Yu Cong
2025-11-26 17:25:14 +08:00
parent 6a024d43f9
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notes.tex
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@@ -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. 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 $yv=1$. 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. 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.} % 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 $y^Tb=1$. 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} \end{lemma}
Then $X$ fails to be a hitting set is equivalent to the fact that there does not exist $y$ satisfying $\supp(yA)\subset X\land yv=1$. 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. The theorem then follows directly from the fact that any cocycle is either odd or even.
\end{proof} \end{proof}