📚 Electoral Reform: A Mathematical Perspective | 选举改革：数学视角分析
Electoral reform is often framed as a political debate, but beneath the competing claims lies a rich set of mathematical ideas: ratios, quotas, inequalities and paradoxes. This article explores how A-level Mathematics, especially statistics and modelling, can be used to compare voting systems and to quantify fairness.
选举改革常常被看作一场政治辩论，但在这类争论背后，隐藏着丰富的数学思想：比率、配额、不等式以及悖论。本文探讨如何运用 A-level 数学（尤其是统计与建模）来比较投票制度，并量化公平性。
Although electoral reform is not a standard core topic in Edexcel A-level Mathematics, it provides an excellent applied context for revising proportional reasoning, integer division, summation and inequality. The ideas below link closely to the mathematical modelling and statistics strands.
尽管选举改革并非爱德思 A-level 数学的标准核心考点，但它为复习比例推理、整数除法、求和与不等式提供了极好的应用背景。以下内容与数学建模和统计模块密切相关。
1. Why Mathematics Matters in Electoral Reform | 为什么数学在选举改革中重要
Any electoral system converts a set of votes into a set of seats. That conversion is a function: the input is a vote distribution, and the output is a seat distribution. By writing that function explicitly, we can test whether a reform makes the system more proportional or more stable.
任何选举制度都是把一组选票转换为一组席位。这种转换本质上是一个函数：输入是选票分布，输出是席位分布。通过明确写出这个函数，我们可以检验某项改革是让制度更具比例性，还是更稳定。
For example, a constituency result with 20,000 votes for Party A and 15,000 for Party B may give 1 seat and 0 seats under first-past-the-post. The mathematical relationship between 20,000 and 15,000 is not reflected in the seat ratio 1:0. This mismatch is the core object of electoral reform.
例如，一个选区中 A 党获得 20,000 票、B 党获得 15,000 票，在简单多数制下可能产生 1 席和 0 席。20,000 与 15,000 的数学关系并没有体现在席位比 1:0 中。这种不匹配正是选举改革的核心研究对象。
2. Electoral Systems and Their Parameters | 选举制度及其参数
Common systems include plurality (first-past-the-post), majoritarian two-round, and proportional representation with party lists. Mathematically, the key variables are V (total valid votes), S (total seats), vᵢ (votes for party i), and sᵢ (seats won by party i).
常见制度包括简单多数制（领先者当选）、两轮多数制以及政党名单比例代表制。从数学角度看，关键变量有 V（总有效票数）、S（总席位数）、vᵢ（第 i 党的得票数）和 sᵢ（第 i 党赢得的席位数）。
- Plurality: each constituency elects one member; the candidate with the largest v in that constituency wins.
- PR list: seats are allocated so that sᵢ / S is as close as possible to vᵢ / V.
中文对应：
- 简单多数制：每个选区选出一名议员；该选区得票最多者获胜。
- 名单比例代表制：席位分配使 sᵢ / S 尽可能接近 vᵢ / V。
ᵢ / V for every party, but because seats are integers, exact equality is often impossible.
这些定义让我们能够把改革目标表示为不等式。例如，完全比例的制度应当对每个政党满足 sᵢ = S × vᵢ / V，但由于席位必须是整数，精确相等往往不可能。
3. Quotas: Hare and Droop | 配额：黑尔与德鲁普
A quota is the number of votes needed to win one seat. The Hare quota is defined as V / S, while the Droop quota is defined as ⌊V / (S + 1)⌋ + 1. The floor function ensures the quota is an integer.
配额是赢得一个席位所需的票数。黑尔配额定义为 V / S，而德鲁普配额定义为 ⌊V / (S + 1)⌋ + 1。向下取整函数确保配额为整数。
Hare quota Q = V / S
Droop quota Q = ⌊V / (S + 1)⌋ + 1
Suppose V = 100,000 and S = 4. Then the Hare quota is 25,000, while the Droop quota is ⌊100,000 / 5⌋ = 20,001. Droop is smaller, so it tends to favour larger parties and reduce fragmentation.
假设 V = 100,000，S = 4。那么黑尔配额为 25,000，而德鲁普配额为 ⌊100,000 / 5⌋ + 1 = 20,001。德鲁普配额更小，因此它往往有利于大党并减少政党碎片化。
4. Seat Allocation Algorithms | 席位分配算法
Once quotas are known, a reform must decide how to allocate remainders. The d'Hondt method uses divisors 1, 2, 3, ... to allocate seats sequentially. The Sainte-Laguë method uses divisors 1, 3, 5, ..., which treats small and large parties more evenly.
一旦确定了配额，改革方案就必须决定如何分配余数。d'Hondt 方法使用除数 1, 2, 3, ... 依次分配席位。Sainte-Laguë 方法使用
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