数学英语翻译有人会吗?

ON THE NUMBER OF CONGRUENCE CLASSES OF PATHS
ZHICONG LIN AND JIANG ZENG
Abstract. Let Pn denote the undirected path of length n − 1. The cardinality of the set of congruence classes induced by the graph homomorphisms from Pn onto Pk is determined. This settles an open problem of Michels and Knauer (Disc. Math., 309 (2009) 5352-5359). Our result is based on a new proven formula of the number of homomorphisms between paths.
Keywords: Graph, graph endomorphisms, graph homomorphisms, paths, lattice paths
1. Introduction
We use standard notations and terminology of graph theory in [3] or [6, Appendix]. The graphs considered here are finite and undirected without multiple edges and loops. Given a graph G, we write V (G) for the vertex set and E(G) for the edge set. A homomorphism from a graph G to a graph H is a mapping f : V (G) → V (H) such that the images of adjacent vertices are adjacent. An endomorphism of a graph is a homomorphism from the graph to itself. Denote by Hom(G,H) the set of homomorphisms from G to H and by End(G) the set of endomorphisms of a graph G. For any finite set X we denote by |X| the cardinality of X. A path with n vertices is a graph whose vertices can be labeled v1, ..., vn so that vi and vj are adjacent if and only if |i − j| = 1; let Pn denote such a graph with vi = i for 1 ≤ i ≤ n. Every endomorphism f on G induces a partition ρ of V (G), also called the congruence classes induced by f, with vertices in the same block if they have the same image.
Let C (Pn) denote the set of endomorphism-induced partitions of V (Pn), and let |ρ| denote the number of blocks in a partition ρ. For example, if f ∈ End(P4) is defined by f(1) = 3, f(2) = 2, f(3) = 1, f(4) = 2, then the induced partition ρ is {{1}, {2, 4}, {3}} and |ρ| = 3.
The problem of counting the homomorphisms from G to H is difficult in general. How- ever, some algorithms and formulas for computing the number of homomorphisms of paths have been published recently (see [1, 2, 5]). In particular, Michels and Knauer [5] give an algorithm based on the epispectrum Epi(Pn) of a path Pn. They define Epi(Pn) = (l1(n), ..., ln−1(n)), where
lk(n) = |{ρ ∈ C (Pn) : |ρ| = n − k + 1}|. (1.1)
Here a misprint in the definition of lk(n) in [5] is corrected.
In [5], based on the first values of lk(n), Michels and Knauer speculated the following conjecture.
同余类的路径
ZHICONG林,江曾
摘要的数量。令Pn表示无向路径长度为n - 1。确定从的Pn到PK的图形同态扰洞诱导的同余类的集合的基数。这解决的一个公开问题的的米歇尔斯和克瑙尔(Disc.数学系,309(2009)5352-5359)。我们的结果是基于一个新的行之有效的路径之间的数目同态。
关键词:图表,曲线图的自同态,图同态,路径,晶格路径
1。介绍
我们使用标准的符号和术语,图论[3]或[6,附录]。这里考虑的是有限的图表和指导,没有多条边和循环。我们写的给定图G,V(G)的顶点集和边集E(G)。从曲线图G甲同态到一个图H是一个映射f:V(G)→V(H),使得相邻顶点的图像相邻。自同态映射的曲线图,从图中本身是一个同态。记磡(缓态枯G,H)的设置从G到H的同态和结束(G)图G的自同态的集合,我们表示对于任何有限集X | X |新路径的基数n个顶点vi和vj是一个图,它的顶点可以被标记为V1,...,VN,使相邻当且仅当| I - J | = 1;令Pn表示这样的图,用vi = I 1≤I ≤N。每一个自同态f的ĝ诱导
C(PN)表示集合的自同态引起的分区分区ρ为V(G),也被称为由f诱导的同余类,如果它们具有相同的图像的顶点在同一闭亩个街区。的V(Pn)的,并让|ρ|表示的块的数目在一个分区ρ。例如,如果f∈结束(P4),被定义为F(1)= 3,F(2)= 2,F(3)= 1,F(4)= 2,然后诱导分区ρ是{{1 },{2,4},{3}}和|ρ| = 3。
计数的同态从G到H的问题是难以普遍。然而,一些算法和公式计算同态的路径最近发表的(见[1,2,5])。特别是米歇尔斯和克瑙尔的[5]给出了一种算法的基础上的epispectrum长盈集团(PN)的路Pn。他们长盈集团(PN)=(L1(N),...,LN-1(N)),其中
LK(N)= | {ρ∈C(PN):|ρ| = - K + 1} |。(1.1)
LK(N)的定义[5]一个印刷错误被纠正。
[5]力劲(n)的,米歇尔斯和克瑙尔的第一个值的基础上,推测如下猜想。
的余类路径
张志聪林江曾

总结的数量。 PN代表无向路径长度为N - 1。从Pn决定PK图形同态引起的配型同余类的集合的基数。这解决了一个公开问题米歇尔斯和柯纳尔(Disc.数学系,309(2009)5352-5359)。我们的研究结果是基于同态之间的有效数量的一个新的路径。

关键词:图表,图自同态,同态,路径,格子路径

1。引进

我们使用标准的符号和术语,图形理论[3]或[6附录]。这里考虑的是有限的图表和指导,无毛边和回收。我们写信给定图G,顶点集V(G)和边集E(G)。从图G中,前一个图H是一个同态映射f:V(G)→V(H),使得相邻的顶点相邻的图像。从图同态,从图本身是一个同态。记住从G到相同的状态,并最终在H(G)图G的同态收集磡(G,H),我们表示任何有限集基地X | X |新路径n个顶点vi和vj为图它的顶点可以被标记为V1,...,VN相邻当且仅当| I - J | = 1;图PN代表,VI = I 1≤I≤N。每个自同态F?诱导

(PN),收集来自同一个国家的地区教育培行猜ρ引起的V(G),也被称为由f诱导的同余类,如果它们具有相同的顶点在同一个街区的图形。 V(Pn)的,并让|对|表示在一个分区中ρ块的数目。例如,如果f∈端的(P4)被定义为F(1)= 3,F(2)= 2,F(3)= 1,F(4)= 2,则诱导分区ρ{{1 } {2,4},{3}和|ρ| = 3。

计数从G到H的同态,问题是很难常见。然而,最近发表的算法和公式同态的路径([1,2,5])。米歇尔斯和克瑙尔[5]显示上:EPI epispectrum(PN)PN算法的方式。他们长带衡盈集团(PN)=(L1(N),...,LN-1(N)),其中

LK(N)= | {P∈C(PN):|ρ| = - K + 1} |。 (1.1)

的LK(N)的定义[5]打印错误得到纠正。
[5] LK(N),米歇尔斯和柯纳尔值的基础上的猜测如下猜想。
目测是数学相关的论文 还有摘要 简介 其他的看不太懂。太专业。