组合数学中的代数方法

注：本文为 “组合数学代数解法” 相关合辑。
英文引文，机翻未校。
如有内容异常，请看原文。

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Algebraic Methods in Combinatorics

组合数学中的代数方法

Po-Shen Loh
June 2011

1 Linear Algebra review

1. 线性代数复习

1.1 Matrix multiplication, and why we care

1.1 矩阵乘法及其重要性

Suppose we have a system of equations over some field $F$, e.g.
假设我们有一个在某个域 $F$ 上的方程组，例如：

$$\begin{gathered} 3x_1+x_2-8x_3=1 \\ 9x_1-x_2-x_3=2 \end{gathered}$$

The set of ordered triples $(x_1,x_2,x_3)$ that solve the system is precisely the set of 3-element vectors $x\in F^3$ that solve the matrix equation
解这个方程组的有序三元组 $(x_1,x_2,x_3)$ 正好是解矩阵方程的 3 元素向量 $x\in F^3$：

$$Ax=\left(\begin{array}{c}1\\ 2\end{array}\right)$$
where $A=\left(\begin{array}{ccc}3& 1& -8\\ 9& -1& -1\end{array}\right)$。

Suppose that $A$ is a square matrix. The equation $Ax=0$ always has a solution (all zeros). An interesting question is to study when there are no more solutions.
假设 $A$ 是一个方阵。方程 $Ax=0$ 总是有解的（全零解）。一个有趣的问题是研究何时没有其他解。

Definition 1 A square matrix $A$ is nonsingular if $Ax=0$ has only one solution: the all-zeros vector.
定义 1 方阵 $A$ 是非奇异的，如果 $Ax=0$ 只有一个解：全零向量。

Often, nonsingular matrices are also called invertible, for the following reason.
通常，非奇异矩阵也被称为可逆矩阵，原因如下。

Theorem 1 The following are equivalent:
定理 1 下列条件等价：

(i) The square matrix $A$ is nonsingular.
(i) 方阵 $A$ 是非奇异的。

(ii) There exists another matrix, denoted $A^{-1}$ such that $A^{-1}A=I=A{A}^{-1}$.
(ii) 存在另一个矩阵，记为 $A^{-1}$，使得 $A^{-1}A=I=A{A}^{-1}$。

Solution:
解：

• (i) to (ii) row-reduction, find that $A$ always has RREF = identity.
  (i) 到 (ii)：通过行简化，发现 $A$ 总是有简化行阶梯形式（RREF）等于单位矩阵。

• then can always solve $Ax=$ (anything).
  然后总是可以解 $Ax=$（任何值）。

• so can for example solve $Ax=I$, by doing it column by column.
  例如，可以逐列解 $Ax=I$。

• now we know there is some $B$ such that $AB=I$.
  现在我们知道存在某个 $B$，使得 $AB=I$。

• also, the row operations themselves were some left multiplication of matrices.
  同时，行操作本身是一些矩阵的左乘。

• so we also have some $C$ so that $CA=I$.
  因此，我们也有一些 $C$，使得 $CA=I$。

• Then $B=IB=CAB=CI=C$, so they are the same.
  那么 $B=IB=CAB=CI=C$，因此它们是相同的。

• (ii) to (i): just multiply by inverse.
  (ii) 到 (i)：只需乘以逆矩阵。

For non-square matrices, the most important fact is the following:
对于非方阵，最重要的事实如下：

Theorem 2 If $A$ has more columns than rows, then the equation $Ax=0$ has more solutions than just the all-zeros vector.
定理 2 如果 $A$ 的列数多于行数，那么方程 $Ax=0$ 的解不止全零向量。

Solution: RREF, run out of rows before columns, so we don’t have the sentinel 1 in every column. This enables us to choose arbitrary nonzero values for all non-sentinel columns, and then still read off a valid solution by backtracking the sentinel columns.
解： 通过简化行阶梯形式（RREF），列数多于行数，因此并非每列都有主元 1。这使我们可以为所有非主元列选择任意非零值，然后通过回溯主元列得到一个有效的解。

1.2 Vectors

1.2 向量

A vector is typically represented as an arrow, and this notion is widely used in Physics, e.g., for force, velocity, acceleration, etc. In Mathematics, vectors are treated more abstractly. In this lecture, we will mostly use concrete vectors, which one can think of as columns of numbers (the coordinates). The fundamental operation in Linear Algebra is the linear combination. It is called “linear” because vectors are not multiplied against each other.
向量通常被表示为箭头，这一概念在物理学中被广泛使用，例如用于力、速度、加速度等。在数学中，向量被更抽象地处理。在本讲中，我们将主要使用具体的向量，可以将其视为数字列（坐标）。线性代数中的基本运算是线性组合。之所以称为“线性”，是因为向量之间不会相互乘积。

Definition 2 Given vectors $v_1,\dots ,v_k$, and real coefficients $c_1,\dots ,c_k$, the sum $c_1{v}_1+\cdots +c_k{v}_k$ is called a linear combination.
定义 2 给定向量 $v_1,\dots ,v_k$，以及实系数 $c_1,\dots ,c_k$，和 $c_1{v}_1+\cdots +c_k{v}_k$ 被称为线性组合。

It’s intuitive that two vectors “usually” do not point in the same direction, and that three vectors “usually” do not all lie in the same plane. One can express both of the previous situations as follows:
直觉上，两个向量“通常”不会指向同一方向，三个向量“通常”不会全部位于同一平面。可以用以下方式表达这两种情况：

Definition 3 Let { v 1 , … , v k } \{v_1, \ldots, v_k\} {v1​,…,vk​} be a collection of vectors. One says that they are linearly dependent if:
定义 3 设 { v 1 , … , v k } \{v_1, \ldots, v_k\} {v1​,…,vk​} 是一组向量。如果：

• One of the vectors can be expressed as a linear combination of the others.
  其中一个向量可以表示为其他向量的线性组合。

• There is a solution to $c_1{v}_1+\cdots +c_k{v}_k=0$, using real numbers $c_1,\dots ,c_k$, where not all of the $c_i$’s are zero.
  存在解 $c_1{v}_1+\cdots +c_k{v}_k=0$，其中实数 $c_1,\dots ,c_k$ 不全为零。

Solution:
解：

• (i) to (ii): swap signs on the linear combination.
  (i) 到 (ii)：在线性组合中交换符号。

• (ii) to (i): pick a nonzero, then shift all others to other side, and divide by the nonzero coefficient.
  (ii) 到 (i)：选择一个非零值，然后将其余的移到另一边，并除以非零系数。

A fundamental theorem in Linear Algebra establishes that it is impossible to have “too many” linearly independent vectors. This is at the core of many Combinatorial applications.
线性代数的一个基本定理表明，不可能有“太多”线性无关的向量。这是许多组合应用的核心。

Theorem 3 The maximum possible number of linearly independent vectors in ${\mathbb{R}}^n$ is $n$.
定理 3 在 ${\mathbb{R}}^n$ 中，线性无关向量的最大数量是 $n$。

Solution: Suppose we have more than $n$, say $v_1,\dots ,v_{n+1}$. Try to solve for the coefficients. This sets up a matrix equation with more columns than rows, which by previous we know to have a nontrivial solution.
解： 假设我们有超过 $n$ 个，比如 $v_1,\dots ,v_{n+1}$。尝试解系数。这建立了一个列数多于行数的矩阵方程，根据前面的内容，我们知道它有非平凡解。

2 Combinatorics of sets

2. 集合的组合学

We begin with a technical lemma.
我们先从一个技术性引理开始。

Lemma 1 Let $A$ be a square matrix over $\mathbb{R}$, for which all non-diagonal entries are all equal to some $t\ge 0$, and all diagonal entries are strictly greater than $t$. Then $A$ is nonsingular.
引理 1 设 $A$ 是一个实数域上的方阵，所有非对角线元素都等于某个 $t\ge 0$，所有对角线元素都严格大于 $t$。那么 $A$ 是非奇异的。

Proof. If $t=0$, this is trivial. Now suppose $t>0$. Let $J$ be the all-ones square matrix, and let $D=A-tJ$. Note that $D$ is nonzero only on the diagonal, and in fact strictly positive there. We would like to solve $(tJ+D)x=0$, which is equivalent to $Dx=-tJx$. Let $s$ be the sum of all elements in $x$, and let the diagonal entries of $D$ be $d_1,\dots ,d_n$, in order. Then, we have $d_i{x}_i=-ts\Rightarrow x_i=-\frac{t}{d_i}s$. But since $t$ and $d_i$ are both strictly positive, this forces every $x_i$ to have the opposite sign from $s$, which is impossible unless all $x_i=0$. Therefore, $A$ is nonsingular.
证明： 如果 $t=0$，这是显然的。现在假设 $t>0$。设 $J$ 是全 1 的方阵，设 $D=A-tJ$。注意 $D$ 只在对角线上有非零值，且实际上在对角线上是严格正的。我们希望解 $(tJ+D)x=0$，这等价于 $Dx=-tJx$。设 $s$ 是 $x$ 中所有元素的和，设 $D$ 的对角线元素依次为 $d_1,\dots ,d_n$。那么，我们有 $d_i{x}_i=-ts\Rightarrow x_i=-\frac{t}{d_i}s$。但由于 $t$ 和 $d_i$ 都是严格正的，这迫使每个 $x_i$ 与 $s$ 符号相反，除非所有 $x_i=0$，否则这是不可能的。因此，$A$ 是非奇异的。

Solution: ALTERNATE: Let $J$ be the all-ones square matrix, and let $D=A-tJ$. Note that $D$ is nonzero only on the diagonal, and in fact strictly positive there, so it is a positive definite matrix. Also, $J$ is well-known to be positive semidefinite (easy to verify by hand), so $A$ is positive definite. In particular, this means that $x^{T}Ax=0$ only if $x=0$, implying that $Ax=0$ only for $x=0$. This is equivalent to $A$ being nonsingular. $■$
解： 另一种方法：设 $J$ 是全 1 的方阵，设 $D=A-tJ$。注意 $D$ 只在对角线上有非零值，且实际上在对角线上是严格正的，因此它是一个正定矩阵。此外，众所周知 $J$ 是半正定的（手工验证很容易），因此 $A$ 是正定的。特别是，这意味着只有当 $x=0$ 时，$x^{T}Ax=0$，这意味着只有当 $x=0$ 时，$Ax=0$。这等价于 $A$ 是非奇异的。 $■$

Now try the following problems. The last two come from 102 Combinatorial Problems, by T. Andreescu and Z. Feng.
现在尝试以下问题。最后两个问题来自 T. Andreescu 和 Z. Feng 的《102 个组合问题》。

2.1 Problems

2.1 问题

1. (A result of Bourbaki on finite geometries; also appeared in St. Petersburg Olympiad.) Let $X$ be a finite set, and let $F$ be a family of distinct proper subsets of $X$. Suppose that for every pair of distinct elements in $X$, there is a unique member of $F$ which contains both elements. Prove that $|F|\ge |X|$.
   1.（布尔巴基关于有限几何的结果；也出现在圣彼得堡奥林匹克竞赛中。）设 $X$ 是一个有限集合，$F$ 是 $X$ 的不同真子集的一个族。假设对于 $X$ 中的每一对不同元素，都有 $F$ 中的唯一一个成员包含这两个元素。证明 $|F|\ge |X|$。

   Solution: Let $X=[n]$ and F = { A 1 , … , A m } F = \{A_1, \ldots, A_m\} F={A1​,…,Am​}. We need to show that $n\le m$. Define the $m\times n$ incidence matrix $A$ over $\mathbb{R}$ by putting 1 in the $i$-th row and $j$-th column if $j\in A_i$. Consider the product $A^{T}A$, which is an $n\times n$ matrix. For $i\ne j$, its entry at $(i,j)$ is precisely 1. Also, the diagonal entries are strictly larger than 1, because if some element $j\in X$ belongs to only one set $A_k\in F$, then the condition implies that every element $i\in X$ is also in $A_k$, contradicting the requirement that $A_k$ be proper. Therefore, $A^{T}A$ is nonsingular by Lemma 1. But if $A$ has more rows than columns, then it would have some $x\ne 0$ such that $Ax=0$, hence $A^{T}Ax=0$. Therefore, $A$ doesn’t have more rows than columns, i.e., $n\le m$.
   解： 设 $X=[n]$， F = { A 1 , … , A m } F = \{A_1, \ldots, A_m\} F={A1​,…,Am​}。我们需要证明 $n\le m$。定义 $m\times n$ 的关联矩阵 $A$，如果 $j\in A_i$，则在第 $i$ 行第 $j$ 列放置 1。考虑乘积 $A^{T}A$，这是一个 $n\times n$ 的矩阵。对于 $i\ne j$，其 $(i,j)$ 位置的元素正好是 1。此外，对角线元素严格大于 1，因为如果某个元素 $j\in X$ 只属于一个集合 $A_k\in F$，那么条件意味着 $X$ 中的每个元素 $i$ 也在 $A_k$ 中，这与 $A_k$ 是真子集的要求矛盾。因此，根据引理 1，$A^{T}A$ 是非奇异的。但如果 $A$ 的行数多于列数，那么它会有一些 $x\ne 0$，使得 $Ax=0$，因此 $A^{T}Ax=0$。因此，$A$ 的行数不会多于列数，即 $n\le m$。

2. (Fisher’s inequality) Let C = { A 1 , … , A r } C = \{A_1, \ldots, A_r\} C={A1​,…,Ar​} be a collection of distinct subsets of { 1 , … , n } \{1, \ldots, n\} {1,…,n} such that every pairwise intersection $A_i\cap A_j$ $(i\ne j)$ has size $t$, where $t$ is some fixed integer between 1 and $n$ inclusive. Prove that $|C|\le n$.
   2.（费舍不等式）设 C = { A 1 , … , A r } C = \{A_1, \ldots, A_r\} C={A1​,…,Ar​} 是 { 1 , … , n } \{1, \ldots, n\} {1,…,n} 的不同子集的一个集合，使得每对交集 $A_i\cap A_j$ $(i\ne j)$ 的大小为 $t$，其中 $t$ 是介于 1 和 $n$ 之间的某个固定整数。证明 $|C|\le n$。

   Solution: Consider the $n\times r$ matrix $A$, where the $i$-th column of $A$ is the characteristic vector of $A_i$. Then, $A^{T}A$ is an $r\times r$ matrix, all of whose off-diagonal entries are $t$. We claim that the diagonal entries are all $>t$. Indeed, if there were some $|A_i|$ which were exactly $t$, then the structure of $C$ must look like a “flower,” with one set $A_j$ of size $t$, and all other sets fully containing $A_j$ and disjointly partitioning the elements of $[n]\setminus A_j$ among them. Any such construction has size at most $1+(n-t)\le n$, so we would already be done. Therefore, $A^{T}A$ is nonsingular by Lemma 1, and the previous argument again gives $r\le n$.
   解： 考虑 $n\times r$ 的矩阵 $A$，其中 $A$ 的第 $i$ 列是 $A_i$ 的特征向量。那么，$A^{T}A$ 是一个 $r\times r$ 的矩阵，其所有非对角线元素都是 $t$。我们声称对角线元素都 $>t$。确实，如果某个 $|A_i|$ 恰好是 $t$，那么 $C$ 的结构看起来像一个“花”，有一个大小为 $t$ 的集合 $A_j$，其他所有集合都完全包含 $A_j$，并且在它们之间不相交地划分 $[n]\setminus A_j$ 的元素。任何这样的构造的大小最多为 $1+(n-t)\le n$，所以我们的证明已经完成。因此，根据引理 1，$A^{T}A$ 是非奇异的，前面的论证再次给出 $r\le n$。

3. Let $A_1,\dots ,A_r$ be a collection of distinct subsets of { 1 , … , n } \{1, \ldots, n\} {1,…,n} such that all $|A_i|$ are even, and also all $|A_i\cap A_j|$ are even for $i\ne j$. How big can $r$ be, in terms of $n$?
   3. 设 $A_1,\dots ,A_r$ 是 { 1 , … , n } \{1, \ldots, n\} {1,…,n} 的不同子集的一个集合，使得所有 $|A_i|$ 都是偶数，且所有 $|A_i\cap A_j|$ 对于 $i\ne j$ 也是偶数。用 $n$ 来表示，$r$ 能有多大？

   Solution: Arbitrarily cluster $[n]$ into pairs, possibly with one element left over. Then take all possible subsets where we never separate the pairs; this gives $r$ up to $2^{\lfloor n/2\rfloor }$. But this is also best possible. Suppose that $S$ is the set of characteristic vectors of the sets in the extremal example. The condition translates into $S$ being self-orthogonal. But $S\perp S\Rightarrow ⟨S⟩\perp ⟨S⟩$，so extremality implies that $S$ is in fact an entire linear subspace, which is self-orthogonal (i.e., $S\subset S^{\perp }$). We have the general fact that for any linear subspace, $\dim S^{\perp }=n-\dim S$. This is because if $d=\dim S$，we can pick a basis $v_1,\dots ,v_d$ of $S$，and write them as the rows of a matrix $A$。Then, the kernel of $A$ is precisely $S^{\perp }$，but any kernel has dimension equal to $n$ minus the dimension of the row space ($d$)。Therefore, $S\subset S^{\perp }$ implies that $\dim S\le \dim S^{\perp }=n-\dim S$，which forces $\dim S\le \lfloor n/2\rfloor$，so we are done.
   解： 随意将 $[n]$ 分成对，可能剩下一个元素。然后取所有可能的子集，我们从不分开这些对；这给出 $r$ 最多 $2^{\lfloor n/2\rfloor }$。但这也是最好的可能。假设 $S$ 是极值例子中集合的特征向量的集合。条件转化为 $S$ 是自正交的。但 $S\perp S\Rightarrow ⟨S⟩\perp ⟨S⟩$，因此极值性意味着 $S$ 实际上是一个完整的线性子空间，它是自正交的（即 $S\subset S^{\perp }$）。我们有一个一般的事实，对于任何线性子空间，$\dim S^{\perp }=n-\dim S$。这是因为如果 $d=\dim S$，我们可以选择 $S$ 的一个基 $v_1,\dots ,v_d$，并将它们写成矩阵 $A$ 的行。那么，$A$ 的核正是 $S^{\perp }$，但任何核的维度等于 $n$ 减去行空间的维度（$d$）。因此，$S\subset S^{\perp }$ 意味着 $\dim S\le \dim S^{\perp }=n-\dim S$，这迫使 $\dim S\le \lfloor n/2\rfloor$，因此我们完成了证明。

4. What happens in the previous problem if we instead require that all $|A_i|$ are odd? We still maintain that all $|A_i\cap A_j|$ are even for $i\ne j$.
   4. 如果我们在前面的问题中要求所有 $|A_i|$ 都是奇数，会发生什么？我们仍然保持所有 $|A_i\cap A_j|$ 对于 $i\ne j$ 是偶数。

   Solution: Answer: $r\le n$. Work over ${\mathbb{F}}_2$. The characteristic vectors $v_i$ of the $A_i$ are orthonormal, so they are linearly independent: given any dependence relation of the form $\sum c_i{v}_i=0$，we can dot product both sides with $v_k$ and conclude that $c_k=0$。Thus, there can only be $\le n$ of them.
   解： 答案：$r\le n$。在 ${\mathbb{F}}_2$ 上工作。$A_i$ 的特征向量 $v_i$ 是正交归一的，因此它们是线性无关的：给定任何形式的依赖关系 $\sum c_i{v}_i=0$，我们可以对两边取 $v_k$ 的点积，并得出 $c_k=0$。因此，它们最多有 $n$ 个。

   ALTERNATE: Let $A$ be the $n\times r$ matrix where the columns are the characteristic vectors of the $A_i$。Then $A^{T}A$ equals the $r\times r$ identity matrix, which is of course of full rank $r$。Thus KaTeX parse error: Undefined control sequence: \rank at position 5: r = \̲r̲a̲n̲k̲(A^T A) \leq \r…。
   另一种方法： 设 $A$ 是 $n\times r$ 的矩阵，其列是 $A_i$ 的特征向量。那么 $A^{T}A$ 等于 $r\times r$ 的单位矩阵，其秩当然是 $r$。因此 KaTeX parse error: Undefined control sequence: \rank at position 5: r = \̲r̲a̲n̲k̲(A^T A) \leq \r…。

5. Prove that if all codegrees in a graph on $n$ vertices are odd, then $n$ is also odd.
   5. 证明如果一个有 $n$ 个顶点的图中所有共度数都是奇数，那么 $n$ 也是奇数。

   Solution: First we show that all degrees are even. Let $v$ be an arbitrary vertex. All vertices $w\in N(v)$ have odd codegree with $v$，which means they all have odd degree in the graph induced by $N(v)$。Since the number of odd-degree vertices in any graph must always be even, we immediately find that $|N(v)|$ is even, as desired. Let $A$ be the adjacency matrix. Then $A^{T}A=J-I$。But consider right-multiplying by $1$。$A1=0\Rightarrow A^{T}A1=0$ and $I1=1$，so we need to have $J1=1$，which implies that $n$ is odd.
   解： 首先我们证明所有度数都是偶数。设 $v$ 是任意一个顶点。所有顶点 $w\in N(v)$ 与 $v$ 的共度数都是奇数，这意味着它们在 $N(v)$ 诱导的图中的度数都是奇数。由于任何图中奇度数顶点的数量总是偶数，我们立即发现 $|N(v)|$ 是偶数，正如我们所期望的。设 $A$ 是邻接矩阵。那么 $A^{T}A=J-I$。但是考虑右乘以 $1$。$A1=0\Rightarrow A^{T}A1=0$ 而 $I1=1$，所以我们需要有 $J1=1$，这表明 $n$ 是奇数。

   ALTERNATE ENDING: Now, let S = { 1 , v 1 , … , v n } S = \{ \mathbf{1}, v_1, \ldots, v_n \} S={1,v1​,…,vn​} be the set of $n+1$ vectors in ${\mathbb{F}}_2^n$ where $1$ is the all-ones vector and $v_i$ is the characteristic vector of the neighborhood of the $i$-th vertex. There must be some nontrivial linear dependence $b1+\sum a_i{v}_i=0$。But note that if we take the inner product of this equation with $v_k$，we obtain ${\sum }_{i\ne k}{a}_i=0$ because $1\cdot v_k=0=v_k\cdot v_k$ and $v_i\cdot v_k=1$ for $i\ne k$。Hence all the $a_i$ are equal. Yet if they are all zero, then $b$ is also forced to be zero, contradicting the nontriviality of this linear combination. Therefore, all $a_i$ are 1, and the equation ${\sum }_{i\ne k}{a}_i=0$ forces $n-1$ to be even, and $n$ to be odd.
   另一种结尾： 现在，设 S = { 1 , v 1 , … , v n } S = \{ \mathbf{1}, v_1, \ldots, v_n \} S={1,v1​,…,vn​} 是 ${\mathbb{F}}_2^n$ 中的 $n+1$ 个向量的集合，其中 $1$ 是全 1 向量，$v_i$ 是第 $i$ 个顶点的邻域的特征向量。必须存在某个非平凡的线性依赖 $b1+\sum a_i{v}_i=0$。但请注意，如果我们取这个方程与 $v_k$ 的内积，我们得到 ${\sum }_{i\ne k}{a}_i=0$，因为 $1\cdot v_k=0=v_k\cdot v_k$ 而 $v_i\cdot v_k=1$ 对于 $i\ne k$。因此所有 $a_i$ 都相等。但如果它们都为零，那么 $b$ 也被迫为零，这与这个线性组合的非平凡性相矛盾。因此，所有 $a_i$ 都是 1，方程 ${\sum }_{i\ne k}{a}_i=0$ 迫使 $n-1$ 为偶数，而 $n$ 为奇数。

6. (Introductory Problem 38) There are $2n$ people at a party. Each person has an even number of friends at the party. (Here, friendship is a mutual relationship.) Prove that there are two people who have an even number of common friends at the party.
   6.（入门问题 38）在一个派对上有 $2n$ 个人。每个人在派对上有偶数个朋友。（这里，友谊是一种相互关系。）证明有两个人在派对上有偶数个共同朋友。

   Solution: Let $A$ be adjacency matrix. Suppose for contradiction that every pair of people has an odd number of common friends. Then over ${\mathbb{F}}_2$，we have $A^{T}A=J-I$，where $J$ is the all-ones matrix and $I$ is the identity. Since all degrees even, $A1=0$。Hence $A^{T}A1=0$。But $J1=0$ because $J$ is a $2n\times 2n$ matrix, and $I1=1$。Thus we have $0=A^{T}A1=(J-I)1=1$，contradiction.
   解： 设 $A$ 是邻接矩阵。假设矛盾，即每对人都有奇数个共同朋友。那么在 ${\mathbb{F}}_2$ 上，我们有 $A^{T}A=J-I$，其中 $J$ 是全 1 矩阵，$I$ 是单位矩阵。由于所有度数都是偶数，$A1=0$。因此 $A^{T}A1=0$。但 $J1=0$，因为 $J$ 是一个 $2n\times 2n$ 矩阵，而 $I1=1$。因此我们有 $0=A^{T}A1=(J-I)1=1$，矛盾。

7. (Advanced Problem 49) A set $T$ is called even if it has an even number of elements. Let $n$ be a positive even integer, and let $S_1,\dots ,S_n$ be even subsets of the set { 1 , … , n } \{1, \ldots, n\} {1,…,n}。Prove that there exist some $i\ne j$ such that $S_i\cap S_j$ is even.
   7.（高级问题 49）如果一个集合 $T$ 有偶数个元素，则称其为偶集合。设 $n$ 是一个正偶数，$S_1,\dots ,S_n$ 是集合 { 1 , … , n } \{1, \ldots, n\} {1,…,n} 的偶子集。证明存在某个 $i\ne j$，使得 $S_i\cap S_j$ 是偶数。

   Solution: Let $A$ be $n\times n$ matrix over ${\mathbb{F}}_2$ with columns that are the characteristic vectors of the $S_i$。Then $A^{T}A=J-I$，but $A$ is singular because $A^{T}1=0$。Square the equation. We have $(J-I)(J-I)=J^2-2J+I$ since $I$，$J$ commute. But $n$ is even, and we are in ${\mathbb{F}}_2$，so it is just $I$，and we get $A^{T}A{A}^{T}A=I$。Contradicts singularity of $A$。
   解： 设 $A$ 是 $n\times n$ 的矩阵，其列是 $S_i$ 的特征向量。那么 $A^{T}A=J-I$，但 $A$ 是奇异的，因为 $A^{T}1=0$。对这个方程平方。我们有 $(J-I)(J-I)=J^2-2J+I$，因为 $I$，$J$ 可交换。但 $n$ 是偶数，我们在 ${\mathbb{F}}_2$ 中，所以它只是 $I$，我们得到 $A^{T}A{A}^{T}A=I$。这与 $A$ 的奇异性相矛盾。

   (Uses $A$ nonsingular implies $A^T$ nonsingular. Indeed, we need $(AB{)}^T=B^T{A}^T$。So, in particular, if $A$ had inverse $B$，then we have a matrix $B^T$ such that it is left inverse of $A^T$。In particular, whenever we go to solve $A^{T}x=0$，we can left-multiply by $B^T$，and get $x=B^{T}0=0$，so no nontrivial solutions.)
   （使用 $A$ 非奇异意味着 $A^T$ 非奇异。确实，我们需要 $(AB{)}^T=B^T{A}^T$。所以，特别是，如果 $A$ 有逆矩阵 $B$，那么我们有一个矩阵 $B^T$，它是 $A^T$ 的左逆。特别是，当我们解 $A^{T}x=0$ 时，我们可以左乘以 $B^T$，得到 $x=B^{T}0=0$，所以没有非平凡解。）

   ALTERNATE: Singularity implies $\det A^{T}A=(\det A{)}^2=0$。However, $\det (J-I)$ is precisely the parity of $D_n$，the number of derangements of $[n]$。It remains to prove that for even $n$，$D_n$ is odd. But this follows from the well-known recursion $D_n=(n-1)(D_{n-1}+D_{n-2})$，which can be verified by looking at where the element $n$ is permuted to.
   另一种方法： 奇异性意味着 $\det A^{T}A=(\det A{)}^2=0$。然而，$\det (J-I)$ 正是 $D_n$ 的奇偶性，即 $[n]$ 的错排数。剩下要证明的是，对于偶数 $n$，$D_n$ 是奇数。但这遵循众所周知的递推关系 $D_n=(n-1)(D_{n-1}+D_{n-2})$，可以通过查看元素 $n$ 被置换到哪里来验证。

3 Bonus problems (not all linear algebra)

3. 加分题（并非全部涉及线性代数）

1. (Caratheodory.) A convex combination of points $x_i$ is defined as a linear combination of the form $\sum {\alpha }_i{x}_i$，where the ${\alpha }_i$ are non-negative coefficients which sum to 1. Let $X$ be a finite set of points in ${\mathbb{R}}^d$，and let KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(X) denote the set of points in the convex hull of $X$，i.e., all points expressible as convex combinations of the $x_i\in X$。Show that each point KaTeX parse error: Undefined control sequence: \conv at position 7: x \in \̲c̲o̲n̲v̲(X) can in fact be expressed as a convex combination of only $d+1$ points of $X$.
   1.（卡拉西奥多里定理。）点 $x_i$ 的凸组合定义为形式为 $\sum {\alpha }_i{x}_i$ 的线性组合，其中 ${\alpha }_i$ 是非负系数，它们的和为 1。设 $X$ 是 ${\mathbb{R}}^d$ 中的一个有限点集，设 KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(X) 表示 $X$ 的凸包中的点集，即所有可以表示为 $x_i\in X$ 的凸组合的点。证明 KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(X) 中的每个点 $x$ 实际上都可以表示为 $X$ 中仅 $d+1$ 个点的凸组合。

   Solution: Given a convex combination with $d+2$ or more nonzero coefficients, find a new vector with which to perturb the nonzero coefficients. Specifically, seek $\sum {\beta }_i{x}_i=0$ and $\sum {\beta }_i=0$，which is $d+1$ equations, but with $d+2$ variables ${\beta }_i$。So there is a non-trivial solution, and we can use it to reduce another ${\alpha }_i$ coefficient to zero.
   解： 给定一个有 $d+2$ 个或更多非零系数的凸组合，找到一个新的向量来扰动这些非零系数。具体来说，寻找 $\sum {\beta }_i{x}_i=0$ 和 $\sum {\beta }_i=0$，这是 $d+1$ 个方程，但有 $d+2$ 个变量 ${\beta }_i$。因此存在一个非平凡解，我们可以用它将另一个 ${\alpha }_i$ 系数减少到零。

2. (Radon.) Let $A$ be a set of at least $d+2$ points in ${\mathbb{R}}^d$。Show that $A$ can be split into two disjoint sets $A_1\cup A_2$ such that KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(A_1) and KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(A_2) intersect.
   2.（拉东定理。）设 $A$ 是 ${\mathbb{R}}^d$ 中至少有 $d+2$ 个点的集合。证明 $A$ 可以分成两个不相交的集合 $A_1\cup A_2$，使得 KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(A_1) 和 KaTeX parse error: Undefined control sequence: \conv at position 1: \̲c̲o̲n̲v̲(A_2) 相交。

   Solution: For each point, create an ${\mathbb{R}}^{d+1}$-vector $v_i$ by adding a “1” as the last coordinate. We have a non-trivial dependence because we have at least $d+2$ vectors in ${\mathbb{R}}^{d+1}$，say $\sum {\alpha }_i{v}_i=0$。Split $A=A_1\cup A_2$ by taking $A_1$ to be the set of indices $i$ with ${\alpha }_i\ge 0$，and $A_2$ to be the rest. By the last coordinate, we have ${\sum }_{i\in A_1}{\alpha }_i={\sum }_{i\in A_2}(-{\alpha }_i)$。Let $Z$ be that sum. Then if we use ${\alpha }_i/Z$ as the coefficients, we get a convex combination from $A_1$ via the first $d$ coordinates, which equals the convex combination from $A_2$ we get by using $(-{\alpha }_i)/Z$ as the coefficients.
   解： 对于每个点，通过在最后添加一个“1”作为坐标，创建一个 ${\mathbb{R}}^{d+1}$ 向量 $v_i$。我们有一个非平凡的依赖，因为在 ${\mathbb{R}}^{d+1}$ 中我们至少有 $d+2$ 个向量，比如说 $\sum {\alpha }_i{v}_i=0$。将 $A=A_1\cup A_2$ 分成两部分，取 $A_1$ 为 ${\alpha }_i\ge 0$ 的索引集合，其余的为 $A_2$。通过最后一个坐标，我们有 ${\sum }_{i\in A_1}{\alpha }_i={\sum }_{i\in A_2}(-{\alpha }_i)$。设 $Z$ 为这个和。那么如果我们使用 ${\alpha }_i/Z$ 作为系数，我们通过前 $d$ 个坐标从 $A_1$ 得到一个凸组合，这等于我们通过使用 $(-{\alpha }_i)/Z$ 作为系数从 $A_2$ 得到的凸组合。

3. (Helly.) Let $C_1,C_2,\dots ,C_n$ be convex sets of points in ${\mathbb{R}}^d$，with $n\ge d+1$。Suppose that every $d+1$ of the sets have a non-empty intersection. Show that all $n$ of the sets have a non-empty intersection.
   3.（赫利定理。）设 $C_1,C_2,\dots ,C_n$ 是 ${\mathbb{R}}^d$ 中的凸点集，且 $n\ge d+1$。假设每 $d+1$ 个集合都有非空交集。证明所有 $n$ 个集合都有非空交集。

   Solution: Induction on $n$。Clearly true for $n=d+1$，so now consider $n\ge d+2$，and assume true for $n-1$。Then by induction, we can define points $a_i$ to be in the intersection of all $C_j$，$j\ne i$。Apply Radon’s Lemma to these $a_i$，to get a split of indices $A\cup B$。Crucially, note that for each $i\in A$ and $j\in B$，the point $a_i$ is in $C_j$。So, each $i\in A$ gives $a_i\in {\bigcap }_{j\in B}{C}_j$，and hence the convex hull of points in $A$ is entirely contained in all $C_j$，$j\in B$。Similarly, the convex hull of points in $B$ is entirely contained in all $C_j$，$j\in A$。Yet Radon’s Lemma gave intersecting convex hulls, so there is a point in both hulls, i.e., in all $C_j$，$j\in A\cup B=[n]$。
   解： 对 $n$ 进行归纳。对于 $n=d+1$ 显然成立，所以现在考虑 $n\ge d+2$，并假设对于 $n-1$ 成立。那么根据归纳，我们可以定义点 $a_i$ 在所有 $C_j$ 的交集中，$j\ne i$。将拉东引理应用于这些 $a_i$，得到索引的划分 $A\cup B$。关键要注意，对于每个 $i\in A$ 和 $j\in B$，点 $a_i$ 在 $C_j$ 中。所以，每个 $i\in A$ 给出 $a_i\in {\bigcap }_{j\in B}{C}_j$，因此 $A$ 中点的凸包完全包含在所有 $C_j$ 中，$j\in B$。同样，$B$ 中点的凸包完全包含在所有 $C_j$ 中，$j\in A$。然而，拉东引理给出了相交的凸包，所以有一个点在两个包中，即在所有 $C_j$ 中，$j\in A\cup B=[n]$。

4. (From Peter Winkler.) The 60 MOPpers were divided into 8 teams for Team Contest 1. They were then divided into 7 teams for Team Contest 2. Prove that there must be a MOPper for whom the size of her team in Contest 2 was strictly larger than the size of her team in Contest 1.
   4.（来自彼得·温克勒。）60 名数学奥林匹克选手被分成 8 个队参加团队赛 1。然后他们被分成 7 个队参加团队赛 2。证明必须有一个选手，她在团队赛 2 中的团队规模严格大于她在团队赛 1 中的团队规模。

   Solution: In Contest 1, suppose the team breakdown was $s_1+\cdots +s_8=60$。Then in the $i$-th team, with $s_i$ people, say that each person did $1/s_i$ of the work. Similarly, in Contest 2, account equally for the work within each team, giving scores of $1/s_i^{\prime }$。However, the total amount of work done by all people in Contest 1 was then exactly 8, and the total amount of work done by all people in Contest 2 was exactly 7. So somebody must have done strictly less work in Contest 2. That person saw $1/s_i^{\prime }<1/s_i$，i.e., the size of that person’s team on Contest 2 was strictly larger than her team size on Contest 1.
   解： 在团队赛 1 中，假设团队分配是 $s_1+\cdots +s_8=60$。那么在第 $i$ 个团队中，有 $s_i$ 人，假设每个人完成了 $1/s_i$ 的工作。同样，在团队赛 2 中，按照每个团队内部的工作量平均分配，给出 $1/s_i^{\prime }$ 的分数。然而，所有人在团队赛 1 中完成的总工作量正好是 8，所有人在团队赛 2 中完成的总工作量正好是 7。所以有人在团队赛 2 中完成的工作量严格少于团队赛 1。那个人看到 $1/s_i^{\prime }<1/s_i$，即她在团队赛 2 中的团队规模严格大于她在团队赛 1 中的团队规模。

5. (MOP 2007/4/K2.) Let $S$ be a set of $10^6$ points in 3-dimensional space. Show that at least 79 distinct distances are formed between pairs of points of $S$.
   5.（2007 年数学奥林匹克计划 4/K2。）设 $S$ 是 3 维空间中 $10^6$ 个点的集合。证明 $S$ 中点对之间至少形成 79 种不同的距离。

   Solution: Zarankiewicz counting for the excluded $K_{3,3}$ in the unit distance graph. This upper-bounds the number of edges in each constant-distance graph, and therefore lower-bounds the number of distinct distances.
   解： 对单位距离图中排除的 $K_{3,3}$ 进行 Zarankiewicz 计数。这为每个固定距离图中的边数提供了上限，因此为不同距离的数量提供了下限。

6. (MOP 2007/10/K4.) Let $S$ be a set of $2n$ points in space, such that no 4 lie in the same plane. Pick any $n^2+1$ segments determined by the points. Show that they form at least $n$ (possibly overlapping) triangles.
   6.（2007 年数学奥林匹克计划 10/K4。）设 $S$ 是空间中 $2n$ 个点的集合，其中没有 4 个点共面。选择由这些点确定的任意 $n^2+1$ 条线段。证明它们至少形成 $n$ 个（可能重叠的）三角形。

   Solution: In fact, every $2n$-vertex graph with at least $n^2+1$ edges already contains at least $n$ triangles. No geometry is needed.
   解： 事实上，每个有 $2n$ 个顶点且至少有 $n^2+1$ 条边的图已经至少包含 $n$ 个三角形。不需要几何知识。

7. (Sperner capacity of cyclic triangle, also Iran 2006.) Let $A$ be a collection of vectors of length $n$ from ${\mathbb{Z}}_3$ with the property that for any two distinct vectors $a,b\in A$ there is some coordinate $i$ such that $b_i=a_i+1$，where addition is defined modulo 3. Prove that $|A|\le 2^n$.
   7.（循环三角形的斯珀纳容量，也是 2006 年伊朗数学奥林匹克。）设 $A$ 是 ${\mathbb{Z}}_3$ 中长度为 $n$ 的向量的集合，具有这样的性质：对于 $A$ 中的任意两个不同的向量 $a,b$，存在某个坐标 $i$，使得 $b_i=a_i+1$，其中加法定义为模 3。证明 $|A|\le 2^n$。

   Solution: For each $a\in A$，define the ${\mathbb{Z}}_3$-polynomial $f_a(x):={\prod }_{i=1}^n(x_i-a_i-1)$。Observe that this is multilinear. Clearly, for all $a\ne b\in A$，$f_a(b)=0$，and $f_a(a)\ne 0$；therefore, the $f_a$ are linearly independent, and bounded in cardinality by the dimension of the space of multilinear polynomials in $n$ variables, which is $2^n$。
   解： 对于每个 $a\in A$，定义 ${\mathbb{Z}}_3$ 多项式 $f_a(x):={\prod }_{i=1}^n(x_i-a_i-1)$。观察到这是多线性的。显然，对于所有 $a\ne b\in A$，$f_a(b)=0$，而 $f_a(a)\ne 0$；因此，$f_a$ 是线性无关的，并且其数量受到 $n$ 个多线性多项式空间的维度的限制，即 $2^n$。

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How to Solve Complex Combinatorics Assignment Problems Using Algebraic Methods

如何利用代数方法解决复杂的组合数学问题

July 10, 2025
Dr. Alan Merrick
United Kingdom
Algebra

Dr. Alan Merrick has over 11 years of experience in teaching and researching combinatorics and algebra. He completed his Ph.D. in Mathematics from University of Central Lancashire, UK.
艾伦·梅里克博士拥有超过 11 年的组合数学和代数教学及研究经验。他在英国中央兰开夏大学完成了数学博士学位。

Algebra and combinatorics are two foundational branches of mathematics, each offering distinct tools for understanding patterns, structures, and logical relationships. While algebra focuses on equations, vector spaces, and transformations, combinatorics centers on counting, arrangements, and discrete structures. When these areas intersect, they unlock powerful techniques capable of solving even the most complex mathematical problems with precision and elegance.
代数和组合数学是数学的两个基础分支，分别提供了理解模式、结构和逻辑关系的不同工具。代数侧重于方程、向量空间和变换，而组合数学则侧重于计数、排列和离散结构。当这两个领域相交时，它们解锁了强大的技术，能够以精确和优雅的方式解决最复杂的数学问题。

One key advantage of combining algebra with combinatorics is the ability to translate challenging discrete problems into algebraic forms—using vectors, polynomials, or matrices—which can then be analyzed with linear algebra or modular arithmetic. This approach simplifies problems that might be overwhelming using purely combinatorial strategies. Concepts like characteristic vectors, the Frankl–Wilson inequality, and Lindström’s theorem all demonstrate how algebra adds depth and clarity to combinatorial reasoning.
将代数与组合数学相结合的一个关键优势是能够将具有挑战性的离散问题转化为代数形式——使用向量、多项式或矩阵——然后可以使用线性代数或模运算进行分析。这种方法简化了那些使用纯组合策略可能令人不知所措的问题。特征向量、弗兰克尔 - 威尔逊不等式和林德斯特罗姆定理等概念都展示了代数如何为组合推理增添深度和清晰度。

For students seeking help with combinatorics assignment tasks, especially those involving abstract or advanced topics, understanding these algebraic methods can be a game changer. They not only provide alternative ways to approach a problem but also improve overall mathematical maturity. Whether it’s optimizing a graph, solving intersection problems, or understanding chromatic numbers in geometry, algebraic techniques make the solutions more accessible and logically sound. And if you’re struggling with algebraic tools along the way, don’t hesitate to get assistance with algebra assignment using guided support—mastering both subjects together strengthens your mathematical foundation.
对于那些在组合数学作业任务中寻求帮助的学生，尤其是那些涉及抽象或高级主题的学生来说，理解这些代数方法可能会改变游戏规则。它们不仅为解决问题提供了替代方法，还提高了整体的数学成熟度。无论是优化图、解决交集问题，还是理解几何中的色数，代数技术使解决方案更易于获取且逻辑上更合理。如果你在途中对代数工具感到困惑，不要犹豫，使用指导性支持来获得代数作业的帮助——同时掌握这两个主题可以增强你的数学基础。

Where Algebra Meets Combinatorics

代数与组合数学的交汇点

Combinatorics is often about counting, arranging, and optimizing discrete objects. Algebra, on the other hand, deals with equations, functions, vector spaces, and structures that obey certain rules. At first glance, these topics might appear unrelated, but many combinatorial problems can be translated into algebraic language. Once this is done, powerful algebraic theorems and tools can be applied to derive results that would be difficult or impossible to obtain by pure counting methods.
组合数学通常涉及计数、排列和优化离散对象。另一方面，代数处理方程、函数、向量空间以及遵守特定规则的结构。乍一看，这些主题似乎毫不相关，但许多组合问题可以转化为代数语言。一旦完成这一步，强大的代数定理和工具就可以被应用，得出通过纯计数方法难以获得甚至无法获得的结果。

This translation often simplifies the problem and reveals hidden structures or patterns. It also leads to new theorems that connect different areas of mathematics.
这种转化通常简化了问题，并揭示了隐藏的结构或模式。它还导致了连接数学不同领域的新的定理。

Understanding Lindström’s Theorem

理解林德斯特罗姆定理

One of the simplest examples of algebra in action is Lindström’s Theorem. The idea is to represent sets as vectors—specifically, characteristic vectors where each element of a set corresponds to a 1 in the vector, and non-elements correspond to 0. These vectors live in a real or finite-dimensional vector space.
林德斯特罗姆定理是代数发挥作用的最简单例子之一。其思想是将集合表示为向量——具体来说，是特征向量，集合中的每个元素对应向量中的 1，非元素对应 0。这些向量存在于实数或有限维向量空间中。

Using the idea of linear dependence of vectors, the theorem shows that in any large enough collection of subsets, there must be at least two disjoint collections whose union is the same. While the algebraic proof is short and elegant, the combinatorial equivalent would require much more effort.
利用向量的线性相关性，该定理表明，在足够大的子集集合中，至少存在两个不相交的集合，它们的并集是相同的。虽然代数证明简洁优雅，但组合等价物将需要更多的努力。

This is a perfect example of how a basic concept from algebra—like vector spaces and dimensions—can lead to meaningful conclusions in combinatorics.
这是代数中的一个基本概念——如向量空间和维度——如何在组合数学中得出有意义的结论的完美例子。

Graphs and the Addressing Problem

图与地址问题

Consider a communication network where messages need to be routed from one node to another. The addressing problem asks how we can label nodes so that each one can forward a message to its destination using only local information. This question becomes interesting in graphs where not all nodes are directly connected.
考虑一个通信网络，其中消息需要从一个节点路由到另一个节点。地址问题询问我们如何标记节点，以便每个节点都可以仅使用本地信息将消息转发到其目的地。在并非所有节点都直接连接的图中，这个问题变得有趣。

In this context, algebra plays a role through labeling schemes using vectors and sets over special alphabets like {0, 1, *}. A key result here is the Graham–Pollak Theorem, which proves that a complete graph on $n$ vertices cannot have its edges partitioned into fewer than $n-1$ complete bipartite graphs. This proof uses the rank of adjacency matrices and properties of linear algebra—demonstrating again how algebraic methods can handle intricate combinatorial structures.
在这种情况下，代数通过使用向量和在特殊字母表（如 {0, 1, *}）上的集合的标记方案发挥作用。这里的一个关键结果是格雷厄姆 - 波拉克定理，它证明了在 $n$ 个顶点上的完全图不能将其边划分为少于 $n-1$ 个完全二分图。这个证明使用了邻接矩阵的秩和线性代数的性质——再次证明了代数方法可以处理复杂的组合结构。

The Club Rules Problem

俱乐部规则问题

Some of the most creative uses of algebra in combinatorics appear in puzzle-like problems. Imagine a university trying to regulate student clubs by introducing rules like: every club must have an even number of members, and every pair of clubs must share an even number of members. This setup leads to interesting mathematical questions about how many such clubs can exist.
组合数学中代数的一些最具创造性的应用出现在类似谜题的问题中。想象一所大学试图通过引入这样的规则来规范学生俱乐部：每个俱乐部必须有偶数个成员，每对俱乐部必须共享偶数个成员。这种设置引发了关于可以存在多少这样的俱乐部的有趣数学问题。

The solution uses linear algebra over finite fields—specifically over ${\mathbb{F}}_2$ (the field with two elements). By representing each club as a vector, and examining the dot products of these vectors, it becomes possible to prove strong upper bounds on the number of possible clubs. Even when the rules are changed—such as requiring clubs to have an odd number of members—the same methods apply.
解决方案使用有限域上的线性代数——具体来说是在 ${\mathbb{F}}_2$（有两个元素的域）上。通过将每个俱乐部表示为一个向量，并检查这些向量的点积，可以证明可能的俱乐部数量的强上界。即使规则发生变化——例如要求俱乐部有奇数个成员——同样的方法仍然适用。

This kind of problem shows how abstract algebraic reasoning can guide real-world policy questions in unexpected ways.
这类问题展示了抽象代数推理如何以意想不到的方式指导现实世界的政策问题。

Frankl–Wilson Inequality and Its Modular Version

弗兰克尔 - 威尔逊不等式及其模版本

A major part of advanced combinatorics is studying intersecting families of sets. The Frankl–Wilson Inequality provides a bound on the size of a collection of sets where intersections follow certain rules. Its modular version refines this by considering set sizes and intersections modulo a prime number.
高级组合数学的一个重要部分是研究集合的相交族。弗兰克尔 - 威尔逊不等式为遵循某些规则的集合交集的集合大小提供了界限。其模版本通过考虑集合大小和模素数的交集来细化这一点。

This approach uses multilinear polynomials, modular arithmetic, and a clever method known as the Diagonal Principle to prove that such families must be relatively small. These results are not only theoretically important but also useful in areas like coding theory, where such constraints naturally arise.
这种方法使用多线性多项式、模运算和一种称为对角线原理的巧妙方法来证明这样的族必须相对较小。这些结果不仅在理论上很重要，而且在编码理论等领域也很有用，因为这些约束自然会出现。

Chromatic Numbers in High Dimensions

高维中的色数

One of the most visually appealing combinatorics problems is coloring the Euclidean space ${\mathbb{R}}^n$. The chromatic number $\chi ({\mathbb{R}}^n)$ is the smallest number of colors needed to color every point in $n$-dimensional space such that no two points exactly 1 unit apart have the same color.
最具视觉吸引力的组合数学问题之一是对欧几里得空间 ${\mathbb{R}}^n$ 进行着色。色数 $\chi ({\mathbb{R}}^n)$ 是在 $n$ 维空间中对每个点进行着色所需的最小颜色数，使得任何两个恰好相距 1 个单位的点没有相同的颜色。

At low dimensions, this problem is very hard—it’s still not known if three colors are enough for ${\mathbb{R}}^2$. But at higher dimensions, algebraic methods help to establish lower and upper bounds. By constructing special point sets using characteristic vectors or other algebraic structures, mathematicians have shown that $\chi ({\mathbb{R}}^n)$ grows exponentially with $n$. These constructions rely on scalar products, norms, and intersection properties of sets, all translated into the language of vectors.
在低维度中，这个问题非常困难——仍然不知道三个颜色是否足以用于 ${\mathbb{R}}^2$。但在高维度中，代数方法有助于建立下界和上界。通过使用特征向量或其他代数结构构造特殊的点集，数学家已经证明 $\chi ({\mathbb{R}}^n)$ 随 $n$ 指数增长。这些构造依赖于标量积、范数和集合的交集性质，所有这些都转化为向量的语言。

Disproving Borsuk’s Conjecture

反驳博苏克猜想

For decades, Borsuk’s Conjecture stated that any bounded set in ${\mathbb{R}}^n$ can be split into $n+1$ parts of smaller diameter. This seemed to be true for small dimensions, but the story changed in higher ones.
数十年来，博苏克猜想声称，任何在 ${\mathbb{R}}^n$ 中的有界集合都可以被分成 $n+1$ 个直径较小的部分。这在低维度中似乎是正确的，但在高维度中情况发生了变化。

Using algebraic constructions and the Frankl–Wilson Inequality, researchers found counterexamples in higher dimensions. For instance, sets of binary vectors can be embedded in such a way that their geometry contradicts the conjecture. This disproved Borsuk’s claim and opened a new field of investigation into how geometry and combinatorics interact when guided by algebra.
利用代数构造和弗兰克尔 - 威尔逊不等式，研究人员在高维度中发现了反例。例如，二进制向量集合可以被嵌入，使得它们的几何形状与猜想相矛盾。这反驳了博苏克的说法，并开辟了一个新的研究领域，即在代数的引导下，几何和组合学如何相互作用。

Practical Implications

实际意义

While many of these problems seem theoretical, they have important real-world applications:
尽管许多这些问题看起来是理论性的，但它们在现实世界中有重要的应用：

• Network routing and error correction use addressing and intersection ideas.
  网络路由和错误校正使用地址和交集思想。

• Data clustering and resource allocation relate to chromatic numbers in high dimensions.
  数据聚类和资源分配与高维中的色数有关。

• Coding theory benefits from constraints on set intersections and algebraic bounds.
  编码理论从集合交集的约束和代数界限中受益。

• Cryptography often relies on modular arithmetic and set systems.
  密码学通常依赖于模运算和集合系统。

In academic contexts, understanding these methods gives students a huge advantage. They not only improve problem-solving but also develop a deeper appreciation for the structure of mathematics.
在学术环境中，理解这些方法给学生带来了巨大的优势。它们不仅提高了解决问题的能力，还培养了对数学结构的更深层次的欣赏。

Conclusion

结论

Algebraic methods bring a powerful blend of structure and logic to the field of combinatorics, making it easier to solve problems that at first may seem overwhelmingly complex. From identifying patterns in set systems to analyzing network structures and disproving long-held conjectures, the application of algebra in combinatorics has led to some of the most elegant solutions in modern mathematics. Students who once viewed these two areas as distinct quickly realize how effectively they complement each other when used together.
代数方法为组合数学领域带来了强大的结构和逻辑的结合，使得解决最初可能显得极其复杂的问题变得更加容易。从识别集合系统中的模式到分析网络结构以及反驳长期存在的猜想，代数在组合数学中的应用导致了现代数学中最优雅的一些解决方案。曾经将这两个领域视为不同的学生很快就会意识到，当它们一起使用时，它们如何有效地相互补充。

By translating combinatorial problems into algebraic language—using tools like vector spaces, polynomials, and modular arithmetic—learners gain a new perspective that often simplifies the problem-solving process. This integrated approach not only enhances understanding but also leads to stronger mathematical reasoning. For anyone seeking help with math assignment tasks involving combinatorics, using algebraic techniques can unlock new pathways to success. Whether it’s tackling advanced coursework or preparing for competitive exams, mastering this intersection is invaluable. In today’s academic landscape, recognizing the interconnectedness of algebra and combinatorics can make a significant difference in both comprehension and performance. Rather than approaching them in isolation, treating these subjects as part of the same problem-solving toolkit makes complex challenges more approachable and rewarding.
通过将组合问题转化为代数语言——使用向量空间、多项式和模运算等工具——学习者获得了一个新的视角，这通常简化了解题过程。这种综合方法不仅增强了理解，还导致了更强的数学推理。对于任何寻求涉及组合数学的数学作业帮助的人来说，使用代数技术可以开辟新的成功途径。无论是应对高级课程还是准备竞赛考试，掌握这一交叉领域都是无价的。在当今的学术环境中，认识到代数和组合数学的相互联系可以在理解和表现方面产生显著的差异。与其孤立地对待它们，不如将这些主题视为同一个解决问题工具包的一部分，这使得复杂的挑战更加易于接近和值得。

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via:

• Algebraic Methods in Combinatorics
  https://www.math.cmu.edu/~ploh/docs/math/mop2011/alg-comb.pdf
• Solving Combinatorics Problems using Algebraic Methods
  https://www.mathsassignmenthelp.com/blog/solve-combinatorics-problems-using-algebra/