The Erds–Ko–Rado Theorem for Integer Sequences
On a coloring problem for the integer grid

On a coloring problem for the integer gridAdrian Dumitrescu Radoˇs Radoiˇc i´cAbstractJ´a nos Pach asked whether for any-coloring of the points of the integer grid in the plane, there exist arbitrary long sequences of consecutive collinear points of the same color.We answerthis question in the negative for the plane,and more generally for any number of dimensions.More precisely,we show that for any,there exists a maximum positive integersuch that for any-coloring of,there exist consecutive collinear points of the same color.We also give upper and lower bounds on for all.1IntroductionConsider the set of points having integer coordinates,where,which we refer to as the integer grid in dimensions(or-grid).Van der Waerden’s theorem states that if the set of integers is partitioned into two classes then at least one of the classes must contain arbitrary long arithmetic progressions(see e.g.[GRS90]),a result which holds for anyfixed number of classes.Gallai’s theorem provides the following generalization of van der Waerden’s theorem in two dimensions:Let.If isfinitely colored,there exist so that all points of the form,are of the same color(see[GRS90]).The following statement,due to E.Witt,is also known:If isfinitely colored,and ifis afinite subset,then at least one of the color classes contains a subset that is affinely equivalent,or even more generally,homothetic to(cf.[HD64],page31).Consider again a-coloring of.If one tries instead tofind long arithmetic progressions con-sisting of consecutive integers in the same class,the maximum length over all partitions is one(as determined by the partition into even and odd numbers).What happens in higher dimensions?J´a nos Pach asked whether for any-coloring of,there exist arbitrary long sequences of consecutive collinear points of the same color.He also noticed that one can alwaysfind three such points,a result which we show is best possible(see Theorem2below),and thus answer his question in the negative.We further show that the answer is negative for any number of dimensions,by exhibiting a-coloring of the grid obtained by tiling.The coloring of the tile itself is found using the prob-abilistic method.From the opposite direction,we prove that for any-coloring of the-grid there exist consecutive collinear points of the same color,where is bounded from below by a function growing slowly to.This result is a simple consequence of the Hales-Jewett theorem [HJ63,GRS90].A combinatorial line in,the-cube over elements, is a sequence of points,,so that in each coordinate,, either or,for,the latter occurring for at least one.Theorem1(Hales-Jewett Theorem)For all there exists so that for,the following holds:If the vertices of are-colored there exists a monochromatic combinatorial line.A sequence of consecutive collinear points(on some line in)is said to form a segment in. Note that a combinatorial line in is a segment of points in(and thus in),although the opposite does not hold.The best upper bound on[GRS90]can be obtained from Shelah’s proof of the Hales-Jewett theorem[Sh88],in terms of a rapidly growing function from the Ackerman hierarchy.The Ackerman hierarchy is a sequence of functions,having the set of positive integers as domain and range[GRS90],where can be defined inductively byLet be the functional inverse of defined by.Clearly ,although its growth rate is very slow.Theorem2For any,there exists a maximum positive integer such that for any-coloring of,there exist consecutive collinear points of the same color.Furthermore,(i)and.(ii)For,.2Proof of Theorem2(i)We start with the lower bound,and refer to Fig.1(the case,is immediate from the earlier comment).Consider a-coloring of the grid points in the plane by white or black.If for each point,all its four neighbors in the grid(in the N,S,E,W directions)have a color different from that of,any diagonal is monochromatic,thus one can select three consecutive collinear points of the same color.Assume therefore that there exist two adjacent points on a horizontal line of the same color,say and in Fig.1(left)are white.If point is white,the following colors are forced when one has to avoid a monochromatic segment of three points:black,black,white,black. No matter what the color of is,three points(or)as desired are found.Finally assume that no triplet of points(as above)forms a monochromatic shape,as in Fig.1(right).By symmetry,if and are white,and are black,and are black,and the coloring is completely determined as a domino tiling:white,black,etc.Then form a monochromatic (black)segment of three points.This concludes the proof of the lower bound in theplane.Figure1:Proof of the lower bound.Before showing that,we introduce some notation regarding periodic colorings and monochromatic segments in.For a positive integer,consider the square tile in:,and some-coloring of.Then extend the coloring of to the whole grid,so that for each grid point,the grid point of has the same color.We call the trace of.For any segment of points in the grid,let,and call it the trace of. Since the grid coloring is a tile coloring,is bichromatic if and only if the trace of is bichromatic.A segment of points in the-grid is of the form,for some.We also say that is defined by the pair.Since the grid coloring is a tile coloring,we can replace and by their traces and.Denote by the greatest common divisor of and,and in general,denote by the greatest common divisor of. Since consists of consecutive points on some line in,we must have,where .We next show that.For this purpose,consider the coloring of the grid obtained by extending the coloring of a tile,as shown in Fig.2.Figure2:Proof of the upper bound:a portion of the tiling;tile is at the bottom left corner.We have to prove that any segment of four points is bichromatic.The conditionimplies that the trace of any segment of four points in the grid is one of the following:It is easy to check that all the traces above are bichromatic.Therefore,any segment of four points in the grid is bichromatic as well.(ii)Wefirst show the upper bound on.We describe a-coloring of the grid points,so that there are no more than consecutive collinear points of the same color.Let be a positive integer(which will be selected later large enough),and a random-coloring of a square tile in,:each integer point in thetile is colored white or black,independently with probability.Then extend the coloring of to the whole grid.Now use the Lov´a sz Local Lemma[EL75]stated below.Let be events in a probability space.A graph on is said to be a dependency graph of if,for all,the event is mutually independent of. Theorem3(LLL)Let be events with for all and with a dependency graph of maximum degree at most,that is,for all,If,then,there exists a-coloring of,whose extension to the entire grid does not allow for a monochromatic segment of points.This proves that exists.Furthermore,a straightforward calculation shows that the above inequality holds for.Next we show the lower bound on.Let be the number obtained from Shelah’s proof of the Hales-Jewett theorem[Sh88].Conform with[GRS90],for, .This gives a lower bound on in terms of the functional inverse of.For, ,thus by the definition of,for we havefrom which the lower bound follows.3Concluding remarksSince our construction is probabilistic,an explicit constructive proof of the upper bound would be more enlightening,and perhaps also yield a better bound.For,thefirst case where we do nothave an exact value for,,where the lower bound follows from the monotonicity of,and the upper bound was computer generated using a program we have written.The question of obtaining a better lower bound on remains an interesting question to be answered.In particular,what is the minimum for which?It may be that such an improved bound could lead to further improved upper bounds on van der Waerden numbers.The goal of obtaining many consecutive points of the same color in a line is precisely the goal in(the many variants of)the two-player game of tic-tac-toe.In the version of the positional game tic-tac-toe,the“board”is the-cube over elements[HJ63,JT95].In this game,two players alternately select a previously un-selected vertex(point)of,and the goal of each player is to collect all vertices of on some line in(i.e.not only combinatorial lines form winning configurations).[BCG82]presents other variants and interesting results regarding tic-tac-toe.This game has been the object of research(and play!)for many years,and its study has lead to significant advances in the theory of positional games on hypergraphs[HJ63,ES73,Pa78,Be81]. References[AS00]N.Alon and J.Spencer,The Probabilistic Method,second edition,Wiley,New York,2000. [Be81]J.Beck,On positional games,Journal of Combinatorial Theory,Series A,30,117–133, 1981.[BCG82]E.R.Berlekamp,J.H.Conway and R.K.Guy,Winning Ways,Vol.1and2,Academic Press,1982.[EL75]P.Erd˝o s and L.Lov´a sz,Problems and results on-chromatic hypergraphs and some related problems,in:Infinite and Finite Sets,Colloquia Mathematica Societatis J´a nos Bolyai,vol.10(A.Hajnal et.al.,eds.),609–617,North-Holland,Amsterdam,1975.[ES73]P.Erd˝o s and J.L.Selfridge,On a combinatorial game,Journal of Combinatorial Theory, Series B,14,298–301,1981.[GRS90]R.L.Graham,B.L.Rothschield and J.H.Spencer,Ramsey Theory,second edition,John Wiley,New York,1990.[HD64]H.Hadwiger,H.Debrunner and V.Klee,Combinatorial Geometry in the Plane,Holt, Rinehart and Winston,New York,1964.[HJ63]A.W.Hales and R.I.Jewett,Regularity and positional games,Transactions of American Mathematical Society,106,222–229,1963.[JT95]T.J.Jensen and B.Toft,Graph Coloring Problems,John Wiley,New York,1995.[Le77]W.J.LeVeque,Fundamentals of Number Theory,Dover,New York,1977.[Pa78]J.L.Paul,Tic-tac-toe in dimensions,Mathematical Magazine,51,45–49,1978.[Sh88]S.Shelah,Primitive recursive bounds for van der Waerden numbers,Journal of American Mathematical Society,1,683–697,1988.。
rado定理

rado定理在数学领域中,常常会有一些重要的定理被提出,这些定理对于问题的解决起到了关键的作用。
本文将重点介绍一条名为"rado定理"的定理,该定理对于组合数学中的一个重要问题提供了一个非常有用的工具。
Rado定理是由匈牙利数学家Rado于1933年提出的,该定理的核心内容是关于超图的一个性质。
超图是一种表示多个点和边之间关系的图结构,它与普通的图不同之处在于,超图中的边可以连接多个点,而不仅仅是两个点。
Rado定理在多项式时间内判断一个给定的超图是否存在一种特殊的分解。
具体来说,Rado定理的内容是:对于给定的一个超图,如果该超图满足某些特定的条件,那么我们可以将超图分解为几个子超图的并集,其中每个子超图都满足一个特定的性质。
这个性质是与超图的边的组成有关的,即每个子超图中的边的集合是超图中边的某个子集。
为了更好地理解Rado定理,让我们来举一个生活中的例子。
假设有一家咖啡店,他们提供了多种不同口味的咖啡和多种不同类型的饼干。
超图可以被看作是一家咖啡店中所有咖啡和饼干的组合方式。
而Rado定理就可以通过判断这个超图是否存在某种分解,来说明是否存在一种特殊的组合方式,使得每个子超图中的咖啡和饼干的组合方式满足特定条件。
根据Rado定理,我们可以通过以下步骤来判断一个给定的超图是否存在Rado分解:1. 判断超图中是否存在两个子超图,使得它们的边集合分别为E1和E2,且E1为E2的子集。
这意味着E1中的边集合是E2中边集合的一部分。
2. 对于每个子超图,继续判断是否存在另外两个子超图,使得它们的边集合分别为E3和E4,且E3为E4的子集。
同样地,E3中的边集合是E4中边集合的一部分。
3. 不断重复上述步骤,直到无法找到满足条件的子超图为止。
如果找到了满足条件的子超图,那么就可以将超图分解为多个子超图的并集。
通过Rado定理,我们可以得出一个结论:如果一个超图满足所有的判断条件,那么它就可以被Rado分解。
large_integer

large_integerLarge IntegerIntroduction:In the field of mathematics and computer science, the concept of integers plays a crucial role. An integer is a whole number that can be positive, negative, or zero. While integers can cover a wide range of values, there are cases where extremely large numbers need to be handled. In this document, we will explore the concept of large integers, their significance, and different approaches to handle them efficiently.What are Large Integers?Large integers, as the name suggests, refer to numbers that exceed the range of the standard integer data type in a programming language. In most programming languages, an integer is typically stored using a fixed number of bits, such as 32 or 64 bits, which limits the maximum and minimum values that can be represented. However, in certain applications, such as cryptography, number theory, and scientific computing, there is a need to handle numbers that are much larger than what can be accommodated by the standard integer data type.Why are Large Integers Important?Large integers find their significance in various fields. Let's explore a few areas where large integers are crucial:1. Cryptography: In modern cryptography, large integers are used for secure communication and encryption algorithms. Public key algorithms, such as RSA, rely on the factorization of large integers for security. Handling large integers efficiently is essential for secure communication and data protection.2. Number Theory: Large integers are extensively studied in number theory, which is a branch of mathematics that deals with properties and relationships of numbers. Many number theory problems require the manipulation and computation of large integers for their solutions.3. Scientific Computing: In certain scientific computations, such as simulations and modeling, large integers are used to represent physical quantities or to simulate complex systems. For example, in astronomical calculations, large integers are used to represent distances, masses, and time scales.Approaches to Handle Large Integers:There are various approaches to handle large integers efficiently. Let's delve into a few commonly used techniques:1. Arbitrary-Precision Arithmetic: Arbitrary-precision arithmetic libraries provide a way to handle large integers by dynamically allocating memory as needed to store the digits of the number. These libraries offer operations such as addition, subtraction, multiplication, and division on large integers, allowing efficient computation with numbers of any size. Examples of popular arbitrary-precision arithmetic libraries include GNU MP (GMP) and BigInteger in Java.2. Modular Arithmetic: Modular arithmetic involves performing arithmetic operations on the remainder of the division of integers by a fixed modulus. This approach is often employed in cryptography and number theory to handle large integers efficiently. Modular arithmetic allows computations to be performed on smaller numbers, reducing the complexity of the calculations.3. Karatsuba Multiplication: Karatsuba multiplication is a fast multiplication algorithm that can significantly reduce the number of multiplications needed to multiply two large integers. It utilizes the divide-and-conquer approach torecursively break down the multiplication into smaller subproblems, reducing the complexity of the overall computation. This algorithm is widely used in large integer multiplication implementations.4. Big Integer Libraries: Many programming languages provide built-in or external libraries specifically designed to handle large integers efficiently. These libraries often implement optimized algorithms and data structures to perform operations on large integers. Using such libraries can simplify the handling of large integers and improve the performance of computations.Conclusion:Large integers are essential in various fields, ranging from cryptography to scientific computing. By employing techniques like arbitrary-precision arithmetic, modular arithmetic, Karatsuba multiplication, and big integer libraries, efficient handling of large integers becomes achievable. The ability to manipulate and perform calculations with large integers opens up possibilities for solving complex problems and enables secure communication and data protection. As technology advances, the need for handling large integers will continue to grow, and researchers and developers will continue to explore new algorithms and approaches to efficiently manipulate these essential numbers.。
On the Traces of Complete Hypergraphs

2
ex(n, Hp (F )), where Hp (F ) is the family of all p-graphs H with |F | edges such that H → F . p q Let Hq,t be the member of Hp (Kt ) with the maximum number of vertices. In other words, p q t Hq,t is the p-graph obtained from Kt by enlarging each of its q edges with a (different) p p set of p − q new vertices. Trivially Hp,t = Kt .
q p Proposition 1.2. Fix 2 ≤ q < p < t. Then Lp (n, Kt ) = ex(n, Hq,t ) + o(np ). q Our main result determines Lp (n, Kt ), p ≥ t in many cases.
Theorem 1.3. Fix 2 ≤ q < t ≤ p. When either q = t − 1 or p = t, we have
q q Lp (n, Kt ) = Lt−1 (n, Kt ) + o(nt−1 ).
(1)
When p ≤ t or q = t − 1, Proposition 1.2 and Theorem 1.3 together imply that
q r Lp (n, Kt ) = ex(n, Hq,t ) + o(nr ).
On the Traces of Complete Hypergraphs
python 德摩根定律

python 德摩根定律德摩根定律是逻辑学中的一条基本定律,也是计算机科学中常用的逻辑原则之一。
在Python编程语言中,德摩根定律可以帮助我们更好地理解与使用逻辑运算符,进而提高代码的可读性和效率。
本文将介绍德摩根定律的概念、适用场景以及具体应用。
德摩根定律是由英国逻辑学家奥古斯塔斯·德·摩根于19世纪提出的,它描述了逻辑运算符not、and和or之间的关系。
首先,我们来看德摩根定律的原始形式:德摩根定律一:非(A and B)等于非A或非B。
德摩根定律二:非(A or B)等于非A且非B。
在Python中,我们可以利用德摩根定律对表达式进行优化和简化。
考虑以下示例:```x = 5y = 10z = 15if not (x < y and y < z):print("条件不满足")```根据德摩根定律一,我们可以将表达式`not (x < y and y < z)`转化为`not x < y or not y < z`。
因此,可以重写为以下形式:```if not x < y or not y < z:print("条件不满足")```同理,根据德摩根定律二,我们可以将表达式`not (x < y or y < z)`转化为`not x < y and not y < z`。
因此,可以重写为以下形式:```if not x < y and not y < z:print("条件不满足")```德摩根定律的应用不仅仅局限于逻辑运算符的组合,还可以在条件语句、循环结构以及布尔表达式等方面发挥作用。
通过合理地运用德摩根定律,我们可以简化代码、提高可读性,并且减少逻辑错误的出现。
在实际应用中,德摩根定律经常与其他逻辑原则和技巧一起使用。
例如,结合短路求值的特性,可以进一步简化复杂的布尔表达式。
r语言决条件推断树

r语言决条件推断树1.引言1.1 概述本文主要介绍了R语言中的决策条件推断树算法。
决策树算法是一种常用的机器学习方法,通过构建一个树形结构,将数据集划分为多个子集,从而实现对数据的分类或回归。
决策条件推断树是决策树算法的一种变体,它在构建决策树的过程中,根据不同的条件对数据进行推断。
R语言是一种功能强大的数据分析和统计建模工具,它提供了丰富的函数和包,用于实现各种机器学习算法。
本文将首先介绍R语言的基本特点和应用领域,然后详细讲解决策树算法的原理和实现方法,最后评述决策条件推断树在分类和回归问题上的优势,并展望其在未来的应用前景。
通过本文的学习,读者将能够全面了解R语言中的决策条件推断树算法,并能够灵活运用该算法解决实际问题。
1.2文章结构文章结构部分的内容可以包括以下内容:文章结构部分主要围绕整篇文章的组织架构进行说明,以帮助读者了解整个文章的脉络和主要内容安排。
在这部分可以介绍文章的章节划分和各个章节的内容概要。
以下是可能的内容:文章结构部分可以简要介绍整篇长文的章节划分和各个章节的主要内容。
具体而言,文章结构部分可以包括以下段落:1. 引言:简要介绍文章的开篇部分,包括引言的主题、目的和意义,以及文章的整体结构。
2. 正文部分:详细介绍每个章节的主要内容。
可以提及每个章节的标题、主题和重点,以及各个章节之间的逻辑关系。
3. 结论部分:简要总结文章的主要内容,并展望下一步的研究方向或者讨论可能的应用前景。
通过这样的文章结构部分的介绍,读者可以更好地了解整篇长文的组织结构和主要内容安排,从而更好地理解和阅读后续章节的内容。
1.3 目的本文的目的是介绍和探讨R语言中的决策条件推断树。
通过对决策树算法概述和R语言介绍的讨论,我们将深入了解决策条件推断树算法的原理和应用。
同时,我们还将探索决策条件推断树在实际问题中的优势和应用前景。
首先,我们将介绍R语言的基本概念和特点,为读者提供相关的背景知识。
了解R语言将为我们进一步理解决策条件推断树算法的实现和操作提供必要的前提。
约翰·戈尔德斯坦问题解决的算法
约翰·戈尔德斯坦问题解决的算法约翰·戈尔德斯坦是20世纪著名的数学家和计算机科学家,他在计算机领域的贡献被广泛认可。
其中,他提出的问题解决算法在计算机科学领域得到了广泛的应用。
本文将介绍约翰·戈尔德斯坦问题解决算法的原理和应用。
一、算法的基本原理约翰·戈尔德斯坦问题解决的基本原理是将问题分解成小的子问题,然后通过递归的方式解决这些子问题,最终将结果合并成原始问题的解。
这种分而治之的策略能够有效地解决各种复杂的问题,尤其在计算机科学领域有着重要的应用。
二、算法的具体步骤约翰·戈尔德斯坦问题解决算法的具体步骤如下:1. 将原始问题分解成小的子问题;2. 递归地解决这些子问题;3. 将子问题的解合并成原始问题的解。
以排序算法为例,可以将一个大的数组分解成多个小的子数组,然后分别对这些子数组进行排序,最后将排序好的子数组合并成一个有序的数组。
这种分而治之的策略能够大大提高问题的解决效率和性能。
三、算法的应用领域约翰·戈尔德斯坦问题解决算法在计算机科学领域有着广泛的应用,其中包括但不限于以下几个方面:1. 排序算法:如归并排序、快速排序等;2. 查找算法:如二分查找、哈希查找等;3. 图算法:如最短路径算法、最小生成树算法等;4. 动态规划算法:如背包问题、最长公共子序列等。
这些应用领域中的问题通常都可以通过约翰·戈尔德斯坦问题解决算法得到高效的解决方案。
四、算法的优势和局限性约翰·戈尔德斯坦问题解决算法具有以下几个优势:1. 高效性:能够有效地解决各种复杂的问题;2. 可复用性:可以用于解决多种不同类型的问题;3. 可扩展性:能够适应不同规模和复杂度的问题。
然而,约翰·戈尔德斯坦问题解决算法也存在一些局限性,例如在处理规模较大的问题时可能会出现性能问题,需要合理地选择递归的终止条件以避免出现无限递归等。
五、算法的发展和未来展望约翰·戈尔德斯坦问题解决算法作为经典的计算机科学算法,随着计算机科学技术的不断发展,其应用领域和性能也在不断扩展和提高。
01背包问题的数学逻辑
01背包问题的数学逻辑1.引言1.1 概述01背包问题是一类经典的组合优化问题,它是数学逻辑中的一个重要问题之一。
在实际生活中,我们经常会面对资源有限的情况,而如何在有限的资源下做出最佳决策,已经成为一个重要的研究领域。
01背包问题就是在给定总容量和一组物品的情况下,选取其中的一些物品放入背包中,使得背包中物品的总价值最大化,而不超过背包的总容量。
这个问题由G. Dantzig在1957年首次提出,并且成为组合优化中的一个经典问题。
它的名字来源于背包只能放入0或1个同样特性的物品。
虽然问题看似简单,但由于问题的解空间庞大,是一个NP完全问题,因此求解过程通常使用一些近似算法。
1.2 目的本文的目的是探究01背包问题的数学逻辑,并介绍一些常用的求解方法。
通过深入研究01背包问题,我们可以更好地理解其数学模型,在实际应用中解决类似的优化问题。
具体目标包括:1. 分析01背包问题的数学模型,并介绍相关的定义和术语;2. 探讨01背包问题的求解方法,包括动态规划、贪心算法和近似算法等;3. 介绍优化问题的评价指标,包括背包的总价值、总重量和可行性等;4. 分析不同情况下的算法复杂性,讨论解决01背包问题的时间和空间复杂性;5. 举例说明01背包问题在实际生活中的应用,如旅行行李、采购决策等。
通过对01背包问题的研究,我们能够更好地理解和应用数学逻辑,提高问题求解的能力。
了解背包问题的求解方法和评价指标,对我们在实际生活中面对资源有限的情况下做出最佳决策具有重要意义。
无论是在物流管理、金融投资还是其他领域,都可以通过对01背包问题的研究,提高决策的效率和准确度。
在接下来的文章中,将会详细介绍01背包问题的数学逻辑,分析不同求解方法的优劣,并给出实际应用的例子,以便读者更好地理解和应用该问题。
2.正文2.1 01背包问题的定义和背景介绍01背包问题是运筹学中的一个经典问题,在算法和动态规划中有重要的应用。
该问题的核心是在给定的背包容量和一组物品的情况下,如何选择物品放入背包中,使得背包中的物品总价值最大化。
多项式的正交性与其零点有界性的一个注记
其中 Φ0 ( z) = 1 . j = 1, 2, …, n 代入式( 1 ) 当中, 证明: 把 z n, j , j = 1, 2, …, n 是 Φ n 的零点, 由假设: z n, 不难推 j , 假设式 ( 2 ) 成立, 让 εn 代 出式( 2 ) 成立. 反过来, zΦ n -1 ( z) + ε n Φ * 表式( 2 ) 当中的比值. 那么, n -1 ( z) 是首 项 系 数 为 1 , 次 数 为 n 的 多 项 式, 记为 j = 1, 2, …, n 是其所有 Ψ n ( z) . 由式 ( 2 ) 知 z n, j , 零点, 这表明 Ψ n ( z) 和 Φ n ( z) 的零点完全一样, 又两 者 都 是 首 项 系 数 为 1 的 多 项 式, 从而必有 Ψ n ( z) = Φ n ( z) . 所以式 ( 2 ) 对 任 意 的 n ≥ 1 , n ∈N 成立. 证完. 接下来, 我们给出本文的中心结果. 定理 2 设 { Φ n ( z) }" 是次数为 n 的首一复 n =0 n∈N , 多项式序列, 若对每一个 n ≥ 1 , 式( 2 ) 成 * 立, 同时 Φ n ( z) 与 Φ n ( z) 的零点互不相同. 则下 面三个条件相互等价: ( i) 序列 { Φ n ( z) }" 关于唯一的有限的正波 n =0 且测度的支集是无限的. 雷尔测度正交, ( ii) | ε n | < 1 , 其中 - ε n 是式 ( 2 ) 当中的 比值. ( iii ) Φn ( z) 的 零 点 都 在 单 位 圆 内, 即 | zn, j = 1, 2, …, n. j | < 1, 证明: 由( i) 到( ii) 是平凡的, 而由( ii) 到 ( iii ) 是正交多项式的经典结果. 反过来, 由 ( ii ) 到 ( i ) 5]中定理 来自于单位圆上 Favard 定理, 见文献[ 8. 1 , p. 156 和文献[ 6]中定理 8. 3 , p. 140. 因此, 我们只需证明: 如果( iii) 成立, 那么( ii) 成立. j = 1, 2, …, n 代表 Φ n ( z) 的所有零 设 z n, j , 点, 他们当中的某些零点可以相等. 所以若( iii ) 成 j = 1, 2, …, n . 把 z n, j = 1, 立, 则有 | z n, j | < 1 , j , 2, …, n 代入式( 1 ) , 可得 | ε n | = | z n, j | Φ n - 1 ( z n, j) * Φ n - 1 ( z n, j)
Mathematisches Forschungsinstitut Oberwolfach Report No. 482006 Combinatorics, Probability
Combinatorics, Probability and Computing
2877
Workshop: Combinatorics, Probability and Computing Table of Contents
J´ ozsef Balogh (joint with B´ ela Bollob´ as and Robert Morris) Majority bootstrap percolation on the hypercube . . . . . . . . . . . . . . . . . . . . . . 2879 Christian Borgs (joint with J.T. Chayes, L. Lov´ asz, V.T. S´ os and K. Vesztergombi) Convergent sequences of dense graphs II. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2882 Peter Bro Miltersen The two probe conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2887 Jennifer Chayes (joint with C. Borgs, L. Lov´ asz, V.T. S´ os and K. Vesztergombi) Convergent sequences of dense graphs I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2888 Amin Coja-Oghlan (joint with Michael Behrisch and Mihyun Kang) Local limit theorems for the giant component . . . . . . . . . . . . . . . . . . . . . . . . 2892 Benjamin Doerr (joint with Tobias Friedrich) Deterministic random walks on the two-dimensional grid . . . . . . . . . . . . . 2896 Uriel Feige (joint with Jeong Han Kim and Eran Ofek) Witnesses for non-satisfiability of dense random 3CNF formulas . . . . . . 2897 Ehud Friedgut (joint with Irit Dinur) Intersecting families are essentially contained in juntas . . . . . . . . . . . . . . . 2898 Alan Frieze (joint with Jon Kleinberg, R. Ravi and Warren Debany) Line-of-sight networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2902 Stefanie Gerke (joint with Paul Balister and B´ ela Bollob´ as) Connectivity of random addable graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2906 Svante Janson The probability that a random multigraph is simple . . . . . . . . . . . . . . . . . . 2907 Peter Keevash (joint with Dhruv Mubayi, Benny Sudakov and Jacques Verstra¨ ete) Rainbow Tur´ an problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2910 Yoshiharu Kohayakawa (joint with Vojtˇ ech R¨ odl and Eduardo Tengan) The size-Ramsey number of short subdivisions . . . . . . . . . . . . . . . . . . . . . . . 2913 Michael Krivelevich (joint with Dan Hefetz and Tibor Szab´ o) Hamilton cycles in expanding and highly connected graphs . . . . . . . . . . . . 2916 Nati Linial (joint with Adi Shraibman) Complexity measures of sign matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2917
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= q n−t−2r
i=0
t + 2r (q we prove the case q ≥ t + 1 (and t ≥ 1) directly. Independently, Ahlswede and Khachatrian [2] obtained Theorem 2 as a diametric theorem in Hamming spaces. They used a different method. See [6] or [2] for the history of the problem.
1
Introduction
Let n, q, t be positive integers with q ≥ 2, n ≥ t, and let [q ] := {1, 2, . . . , q }. Then H ⊂ [q ]n is a set of integer sequences (a1 , . . . , an ), 1 ≤ ai ≤ q . We say that H is t-intersecting if any two sequences intersects in at least t positions, more precisely, |{i : ai = a i }| ≥ t holds for all (a1 , . . . , an ), (a1 , . . . , an ) ∈ H. In this paper, we determine the exact value of the following function. f (n, q, t) := max{|H| : H ⊂ [q ]n , H is t-intersecting}. A family A ⊂ 2[n] is called t-intersecting if |A ∩ A | ≥ t holds for all A, A ∈ A. Define a weighted size of A by w(A) := A∈A (q − 1)n−|A| . Using a shifting technique, it is not difficult to check the following: Lemma 1 (Proposition 2 in [5]) f (n, q, t) = maxA w(A), where A ⊂ 2[n] runs over all t-intersecting families. If q = 2 then w(A) = |A|. Thus, f (n, 2, t) is simply the maximal size of t-intersecting family A ⊂ 2[n] , which is given by the Katona Theorem. This case was solved by Kleitman [7]. Let us define a t-intersecting family Ar ⊂ 2[n] by Ar := {A ⊂ [n] : |A ∩ [t + 2r]| ≥ t + r}. 1
Hence we have
r
g (q, t) := nlim f (n, q, t)q −n ≤ q −t−2r →∞ On the other hand, (1) implies
r
i=0
t + 2r (q − 1)i . i
(4)
g (q, t) ≥ q −t−2r
i=0
t + 2r (q − 1)i . i
Case I 0 < δ < 1. Note that δ depends only on t and q . Lemma 3 For k ∈ I and sufficiently large n, (k − t + 1)(2 + Proof (2) is equivalent to (2 + (t − 1)/r)−1 n + t − 1 ≤ k ≤ (2 + (t − 1)/(r + 1))−1 n + t − 1 Let us show the right half. Since k < (1 + )(n/q ), it is sufficient to show (1 + )(n/q ) ≤ (2 + (t − 1)/(r + 1))−1 n + t − 1 or (1 + )(2 + (t − 1)/(r + 1)) < q. This follows from q = 2 + (t − 1)/(r + δ ) > 2 + (t − 1)/(r + 1) and < can prove the left half of (3) similarly.
0 (q, t).
t−1 t−1 ) ≤ n ≤ (k − t + 1)(2 + ) r+1 r
(2)
(3)
One
Thus, by the Ahlswede–Khachatrian theorem we have |B (k )| ≤ AK(n, k, t, r). Therefore, f (n, q, t)q −n < q −n
k ∈I
w(B (k )) + AK(n, k, t, r)(q − 1)n−k +
k ∈I t+2r
≤ q −n = q
−n
k∈I j =t+r t+2r
t + 2r j
n − t − 2r (q − 1)n−k + k−j n − t − 2r (q − 1)n−k + k−j
< q −n
j =t+r t+2r
2
Proof of the theorem
r := t−1 t−1 = − δ. q−2 q−2
Throughout this section, we fix q and t and set
Let us recall the following easy probabilistic result. Lemma 2 (Proposition 3 in [5]) For every > 0 the number of sequences (a1 , . . . , an ) ∈ [q ]n which contain more than (1 + )(n/q ) 1’s or less than (1 − )(n/q ) 1’s is less than q n for n > n0 ( ). Choose any sufficiently small positive , i.e., 0 < < 0 (q, t), and set an open interval I := ((1 − )(n/q ), (1 + )(n/q )). In view of Lemma 1, f (n, q, t)q −n = w(A)q −n for some t-intersecting family A. Moreover Lemma 2 gives that f (n, q, t)q −n < w(B )q −n + where B := {B ∈ A : |B | ∈ I }. Set B (k ) := {B ∈ B : |B | = k }. 2
n−t−2r t+2r t−1 q −2
. Then f (n, q, t) = w(Ar ) for n ≥ t + 2r.
w(Ar ) =
j =0 n−t−2r i=t+r
t + 2r i
n − t − 2r (q − 1)n−i−j j
=
j =0
t+2r t + 2r n − t − 2r (q − 1)n−t−2r−j (q − 1)t+2r−i j i i=t+r r
[n] k
Theorem 1 ([1]) Let 1 ≤ t ≤ k ≤ n and B ⊂ (k − t + 1)(2 +
be t-intersecting. If
t−1 t−1 ) ≤ n ≤ (k − t + 1)(2 + ) r+1 r for some r ∈ N, then |B| ≤ AK(n, k, t, r). Using the above result, we prove the following in section 2. Theorem 2 Let q ≥ 3 and set r := Note that
In [5], Frankl and F¨ uredi conjectured f (n, q, t) = maxr≥0 w(Ar ). If q ≥ t + 1 then the conjecture claims f (n, q, t) = q n−t . They showed that this is true if t ≥ 15. Now we introduce the full Erd˝ os–Ko–Rado theorem, which was conjectured by Frankl in [4], and proved by Ahlswede and Khachatrian in [1]. Set AK(n, k, t, r) := |{B ∈ [n] : |B ∩ [t + 2r]| ≥ t + r}|. k
The Erd˝ os–Ko–Rado Theorem for Integer Sequences
Peter Frankl CNRS, ER 175 Combinatoire, 54 Bd Raspail, 75006 Paris, France Norihide Tokushige College of Education, Ryukyu University, Nishihara, Okinawa, 903-0213, Japan hide@edu.u-ryukyu.ac.jp