enumeration of the final solution.
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22
main.tex
22
main.tex
@@ -180,6 +180,17 @@ For general matroids, we want to show the following.
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Given a positive integer $k'<k$, then there exists a distribution on $k'$-cocycles such that for any base $B$ of $M'$, the expected size of intersection is $O(1)$ for fixed $k'$.
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\end{conjecture}
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\begin{theorem}
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If \autoref{conj:dist} is true for matroid $M$, then one can compute minimum $k$-cocycle in polynomial time.
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\end{theorem}
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\begin{proof}
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It follows from \autoref{conj:dist} that the minimum $k$-cocycle $C^*_k$ shares at most $h=O(1)$ elements with some base in the ideal base packing. The number of bases we need in the ideal base packing is polynomial (see next subsection).
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We enumerate all bases in this set and for each base $B$ enumerate all subsets with size in range $[r-h,r-k]$.
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Such a subset $I$ must be indenpendent and be contained in the flat $\overline{C^*_k}$. Then for such subset we further enumerate another subset $X$ such that $I\cup X$ is a rank-$r-k$ independent set. Thus, $\overline{\cl(I\cup X)}$ is a $k$-cocycle and we take the minimum one among all enumerations.
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One can check that each part of the enumeration can be done in polynomial time.
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\end{proof}
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Certainly \autoref{conj:dist} does not hold on any matroid.
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A counterexample would be uniform matroids $U_{2n,n}$.
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The size of every $k'$-cocycle is $k'+n$, and the size of expected intersection should be $O(n)$.
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@@ -237,19 +248,22 @@ We randomly pick one set in the span-partition $\mathcal S_F$ and merge it into
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For graphic matroid, the desired bound is equivalent to the following conjecture.
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\begin{conjecture}
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Let $G=(V,E)$ be a connected graph with $n$ vertices. Contract edges uniformly random (ignore parallel edges) and remove loops until the remaining graph $H$ has 2 vertices.
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Then for any spanning tree $T$ of $G$, the expected number of edges in $H\cap T$ is at most 2.
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Let $G=(V,E)$ be a connected graph with $n$ vertices. Contract edges uniformly random (ignore parallel edges) and remove loops until the remaining graph $H$ has 2 vertices.
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Then for any spanning tree $T$ of $G$, the expected number of edges in $H\cap T$ is at most 2.
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\end{conjecture}
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However, this is not the case.
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Consider an edge $(u,v)$ and one round.
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\[
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\Pr[\text{$(u,v)$ is not contracted}]\leq 1-\frac{1}{|E|} \leq
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\frac{n(n-1)-2}{n(n-1)}=\frac{(n+1)(n-2)}{n(n-1)}
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\Pr[\text{$(u,v)$ is not contracted}]\leq 1-\frac{1}{|E|} \leq
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\frac{n(n-1)-2}{n(n-1)}=\frac{(n+1)(n-2)}{n(n-1)}
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\]
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Then the probability that edge $(u,v)$ survives in the end is at most $\prod_{k=3}^n \frac{(n+1)(n-2)}{n(n-1)}=\frac{n+1}{3(n-1)}$.
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Then the number of remaining edges in any spanning tree is at most $(n+1)/3$.
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\subsection{Hypergraphic matroid}
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Hypergraphic matroids are not closed under contraction (cf. Tamás Király's thesis).
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\section{Greedy base packing}
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\section{Principal sequence + KT contraction}
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