We introduce a class of n×n matrices of 0's and l's which can be regarded as generalizations of the classical Ferrets boards of rook theory. Assuming the matrices are fully indecomposable, we determine th...We introduce a class of n×n matrices of 0's and l's which can be regarded as generalizations of the classical Ferrets boards of rook theory. Assuming the matrices are fully indecomposable, we determine the minimum permanent and the minimum numberof l's as a function of n. We also characterize these matrices in terms of weighted,top-rooted trees.展开更多
In this paper,we show that there exist precisely W(A) Ferrers matrices F(C1,C2,…,cm)such that the rook polynomials is equal to the rook polynomial of Ferrers matrix F(b1,b2,…,bm), where A={b1,b2-1,…,bm-m+1} is a re...In this paper,we show that there exist precisely W(A) Ferrers matrices F(C1,C2,…,cm)such that the rook polynomials is equal to the rook polynomial of Ferrers matrix F(b1,b2,…,bm), where A={b1,b2-1,…,bm-m+1} is a repeated set,W(A) is weight of A.展开更多
整数分拆是指将正整数n表示成一些正整数的无序和。周长为n的整数边不全等三角形个数问题是整数分拆里的一个特殊情况。目前对于整数边三角形问题的研究已有许多结果。本文将采用两种方法证明整数边三角形个数的表达式。方法一采用组合...整数分拆是指将正整数n表示成一些正整数的无序和。周长为n的整数边不全等三角形个数问题是整数分拆里的一个特殊情况。目前对于整数边三角形问题的研究已有许多结果。本文将采用两种方法证明整数边三角形个数的表达式。方法一采用组合学上整数分拆的方法,方法二是运用空间格点方法证明。在方法一中介绍了整数分拆理论求解的常规方法,利用Ferrers图把整数边三角形个数问题求解转化为4x1+2x2+3x3=n-3的非负整数解个数求解,继而可采用生成函数法求解(x1,x2,x3)的个数,也即原问题中周长为的不全等整数边三角形的个数。在方法二中,借助于几何方法,把原问题中三角形三边x、y、z所需满足的条件:x+y+z=n且x,y,z≤n/2转化为三维坐标轴中对应的平面图,因为x、y、z为整数,所以实则对应于一网格点图,通过研究网格点的性质可求出整数边三角形的个数表达式。此外,本文还进一步研究了三角形的各类型个数与其间关系。例如,其中包含的等腰、等边三角形的个数表达式。对于直角、锐角、钝角三角形个数问题,目前只得出相关性质的一些结论和猜想。Integer partitions refer to a representation of the positive integer n as a sum of integers. We do not consider the order of terms of the sum. The problem of counting non-congruent triangles with integer sides is just a case of partition of integers. Now, there have been many results about the study of triangles with integer sides problem. In this article, we will solve the problem in two ways. Firstly, we take the common version using the theory of integer partitions to give a proof. Here, we will require generating functions. By using Ferrers diagram, the integer triangles problem will cross to the solution with integers xi≥0, i=1,2,3 of 4x1+2x2+3x3=n-3, while the sum of (x1,x2,x3) is equal to the solution of triangles with integer sides problem using the method of generating function. Secondly, we give a geometric approach using triangular coordinates which is easier to understand. Since x+y+z=n, we can view (x,y,z) as a point in the space x+y+z=n, in the triangle cutting off by the planes x=0, y=0, z=0. Then, the sum of the integral values of x,y,z corresponds to the number of non-congruent triangles with integer sides. Also, we bring out several further properties, including the number of non-congruent triangles types, such as Isosceles triangles and Equilateral triangles. At the end, we study more about right triangles, acute triangles and obtuse triangles in the non-congruent triangles. But we can just get some relevant properties and conjectures now.展开更多
文摘We introduce a class of n×n matrices of 0's and l's which can be regarded as generalizations of the classical Ferrets boards of rook theory. Assuming the matrices are fully indecomposable, we determine the minimum permanent and the minimum numberof l's as a function of n. We also characterize these matrices in terms of weighted,top-rooted trees.
文摘In this paper,we show that there exist precisely W(A) Ferrers matrices F(C1,C2,…,cm)such that the rook polynomials is equal to the rook polynomial of Ferrers matrix F(b1,b2,…,bm), where A={b1,b2-1,…,bm-m+1} is a repeated set,W(A) is weight of A.
文摘整数分拆是指将正整数n表示成一些正整数的无序和。周长为n的整数边不全等三角形个数问题是整数分拆里的一个特殊情况。目前对于整数边三角形问题的研究已有许多结果。本文将采用两种方法证明整数边三角形个数的表达式。方法一采用组合学上整数分拆的方法,方法二是运用空间格点方法证明。在方法一中介绍了整数分拆理论求解的常规方法,利用Ferrers图把整数边三角形个数问题求解转化为4x1+2x2+3x3=n-3的非负整数解个数求解,继而可采用生成函数法求解(x1,x2,x3)的个数,也即原问题中周长为的不全等整数边三角形的个数。在方法二中,借助于几何方法,把原问题中三角形三边x、y、z所需满足的条件:x+y+z=n且x,y,z≤n/2转化为三维坐标轴中对应的平面图,因为x、y、z为整数,所以实则对应于一网格点图,通过研究网格点的性质可求出整数边三角形的个数表达式。此外,本文还进一步研究了三角形的各类型个数与其间关系。例如,其中包含的等腰、等边三角形的个数表达式。对于直角、锐角、钝角三角形个数问题,目前只得出相关性质的一些结论和猜想。Integer partitions refer to a representation of the positive integer n as a sum of integers. We do not consider the order of terms of the sum. The problem of counting non-congruent triangles with integer sides is just a case of partition of integers. Now, there have been many results about the study of triangles with integer sides problem. In this article, we will solve the problem in two ways. Firstly, we take the common version using the theory of integer partitions to give a proof. Here, we will require generating functions. By using Ferrers diagram, the integer triangles problem will cross to the solution with integers xi≥0, i=1,2,3 of 4x1+2x2+3x3=n-3, while the sum of (x1,x2,x3) is equal to the solution of triangles with integer sides problem using the method of generating function. Secondly, we give a geometric approach using triangular coordinates which is easier to understand. Since x+y+z=n, we can view (x,y,z) as a point in the space x+y+z=n, in the triangle cutting off by the planes x=0, y=0, z=0. Then, the sum of the integral values of x,y,z corresponds to the number of non-congruent triangles with integer sides. Also, we bring out several further properties, including the number of non-congruent triangles types, such as Isosceles triangles and Equilateral triangles. At the end, we study more about right triangles, acute triangles and obtuse triangles in the non-congruent triangles. But we can just get some relevant properties and conjectures now.
