组合零点定理的一个简短证明
A short proof of Combinatorial Nullstellensatz
摘要:
本文给出了著名的组合零点定理的一个简短而直接的证明。
英文摘要:
In this note we give a short, direct proof of the well known Combinatorial Nullstellensatz.
机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。
A short proof of Combinatorial Nullstellensatz
Mateusz Michalek
本文给出了著名的组合零点定理的一个简短而直接的证明。
In this note we give a short, direct proof of the well known Combinatorial Nullstellensatz.
机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。
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Computational Aspects of the Combinatorial Nullstellensatz Method
在这里,我们讨论一些计算方面的组合零stellenerance参数。我们的主要结果表明,数量级的对称群与置换的变量在代数约束,决定了算法的性能自然推导出从阿隆的组合Nullstellenerance参数。最后给出了一个证明组合问题解的存在性或不存在性的原始-对偶多项式构造。
A generalization of Combinatorial Nullstellensatz
本文给出了一个推广的组合零点定理,并对非零单项式作了较弱的假设。我们还提出了我们的结果的应用在原来的定理似乎不工作的情况下。
The Generalized Combinatorial Lason-Alon-Zippel-Schwartz Nullstellensatz Lemma
本文综述了Alon组合零点定理与Schwartz-Zippel引理的若干强化与推广,证明了当变量在足够大范围内独立取值时多元多项式非零值的存在性。
We initiate the study of applying the Combinatorial Nullstellensatz to the DP-coloring of graphs even though, as is well-known, the Alon-Tarsi theorem does not apply to DP-coloring. We define the notion of good covers of prime order which allows us to apply the Combinatorial Nullstellensatz to DP-coloring. We apply these tools to DP-coloring of the cones of certain bipartite graphs and uniquely 3-colorable graphs. We also extend a result of Akbari, Mirrokni, and Sadjad (2006) on unique list colorability to the context of DP-coloring. We establish a sufficient algebraic condition for a graph $G…
By a well-known theorem of Thomassen and a planar graph depicted by Voigt, we know that every planar graph is $5$-choosable, and the bound is tight. In 1999, Lam, Xu and Liu reduced $5$ to $4$ on $C_4$-free planar graphs. In the paper, by applying the famous Combinatorial Nullstellensatz, we design an effective algorithm to deal with list coloring problems. At the same time, we prove that a planar graph $G$ is $4$-choosable if any two $4$-cycles having distance at least $5$ in $G$, which extends the result of Lam et al.
The long-standing Erdős-Faber-Lovász conjecture states that every $n$-uniform linear hypergaph with $n$ edges has a proper vertex-coloring using $n$ colors. In this paper we propose an algebraic framework to the problem and formulate a corresponding stronger conjecture. Using the Combinatorial Nullstellensatz, we reduce the Erdős-Faber-Lovász conjecture to the existence of non-zero coefficients in certain polynomials. These coefficients are in turn related to the number of orientations with prescribed in-degree sequences of some auxiliary graphs. We prove the existence of certain orientations,…
Residues and the Combinatorial Nullstellensatz
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Combinatorial Nullstellensatz Techniques
本文介绍多种将组合零点定理用于有限域多项式的技术,推广了Noga Alon的相关定理并提出若干新定理。
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应用类似于组合零点的技术,我们证明了一个较低的估计$|f(A,B)|$有限子集$A$,$B$的一个字段,和多项式$f(x,y)$的形式$f(x,y)=g(x)+yh(x)$,其中程度的$g$大于程度的$h$。
Motivated exposition of combinatorial Nullstellensatz
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Expressing Combinatorial Optimization Problems by Systems of Polynomial Equations and the Nullstellensatz
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本文通过重复应用零和网格的组合零点定理,证明了 Martin Gardner 提出的最小无三连线问题在 $n \equiv 1 \pmod 4$ 时下界为 $n+1$ 的猜想。
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Low Degree Nullstellensatz Certificates for 3-Colorability
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Application of the Combinatorial Nullstellensatz to magic-type graph labelings
本文利用组合零点定理,对简单图\(G\),在给定整数\(k\geq 2\)下,研究边标记\(h\)诱导顶点标记的情况,分析\(G\)成为\(t -\)和\(\mathbb Z_k -\)魔术图以及\(\mathbb{Z}_{k}-\)反魔术图的条件。
Ramsey Numbers through the Lenses of Polynomial Ideals and Nullstellensätze
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Quantitative Combinatorial Nullstellensatz
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Combinatorial Alexander Duality -- a Short and Elementary Proof
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Algebraic combinatorial geometry: the polynomial method in arithmetic combinatorics, incidence combinatorics, and number theory
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A combinatorial determinant
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In this article, we present a short, non-exhaustive study of an important and well-known property of combinatorial sequences - unimodality. We shall have a look at a sample of classical results on unimodality and related properties, and then proceed to understand the unimodality of the Gaussian polynomial in more detail. We will look at an outline of O'Hara's proof of the unimodality of the Gaussian polynomial. In order to grasp the challenge of the problem of obtaining an injective proof of the unimodality of the Gaussian polynomial (which is still an open question), we make several attempts …
In this article, we consider a generalization of Young tableaux in which we allow some consecutive pairs of cells with decreasing labels. We show that this leads to a rich variety of combinatorial formulas, which suggest that these new objects could be related to deeper structures, similarly to the ubiquitous Young tableaux. Our methods rely on variants of hook-length type formulas, and also on a new efficient generic method (which we call the density method) which allows not only to generate constrained combinatorial objects, but also to enumerate them. We also investigate some repercussions …
A combinatorial proof of the extension property for partial isometries
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Let $Q_{n,d}$ denote the random combinatorial matrix whose rows are independent of one another and such that each row is sampled uniformly at random from the subset of vectors in $\{0,1\}^n$ having precisely $d$ entries equal to $1$. We present a short proof of the fact that $\Pr[\det(Q_{n,d})=0] = O\left(\frac{n^{1/2}\log^{3/2} n}{d}\right)=o(1)$, whenever $d=ω(n^{1/2}\log^{3/2} n)$. In particular, our proof accommodates sparse random combinatorial matrices in the sense that $d = o(n)$ is allowed. We also consider the singularity of deterministic integer matrices $A$ randomly perturbed by a s…
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Combinatorics of random tensor models
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In this short review we first recall combinatorial or ($0-$dimensional) quantum field theory (QFT). We then give the main idea of a standard QFT method, called the intermediate field method, and we review how to apply this method to a combinatorial QFT reformulation of the celebrated Jacobian conjecture on the invertibility of polynomial systems. This approach establishes a related theorem concerning partial elimination of variables that implies a reduction of the generic case to the quadratic one. Note that this does not imply solving the Jacobian conjecture, because one needs to introduce a …
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