Algorithms in IGCSE Mathematics | IGCSE 数学：算法考点精讲

📚 Algorithms in IGCSE Mathematics | IGCSE 数学：算法考点精讲

In IGCSE Mathematics, the term “algorithm” refers to a clearly defined, step-by-step procedure used to carry out a calculation, solve a problem, or process data. Understanding algorithms helps students develop logical thinking and ensures accuracy when tackling multi-step questions in topics such as number operations, algebra, and sequences. Whether you are finding the highest common factor (HCF) of two large numbers or solving a linear equation, following a systematic algorithm will lead to the correct answer efficiently.

在 IGCSE 数学中，“算法” 是指用于进行计算、解决问题或处理数据的一系列明确定义的步骤。理解算法有助于学生培养逻辑思维，在处理数字运算、代数和数列等涉及多步骤的题目时确保准确性。无论是求两个大数的最大公因数 (HCF) 还是解一元一次方程，遵循系统化的算法都能高效地得出正确答案。

1. What is an Algorithm in Mathematics? | 什么是数学中的算法？

An algorithm in mathematics is a finite sequence of instructions that, when followed, produces a specific result. Just like following a recipe to bake a cake, a mathematical algorithm breaks down a complex problem into smaller, manageable steps. Common examples include the steps for long division, the order of operations (BODMAS/PEMDAS), and the Euclidean algorithm for HCF. In IGCSE exam questions, you may be asked to describe an algorithm, apply it to given numbers, or trace a flowchart that represents an algorithm.

数学中的算法是一系列有限的指令，按照这些指令执行可以产生特定的结果。就像按照食谱烤蛋糕一样，数学算法将复杂问题分解为更小、可管理的步骤。常见的例子包括长除法的步骤、运算顺序 (BODMAS/PEMDAS) 以及求 HCF 的欧几里得算法。在 IGCSE 考题中，你可能会被要求描述某个算法、将其应用于给定的数字，或追踪表示算法的流程图。

Recognising algorithmic patterns is a key skill. For instance, when evaluating an expression like 3 + 6 × (5 – 2)² ÷ 3, simply calculating from left to right gives a wrong answer. The correct algorithm — the order of operations — guarantees a unique, correct result every time.

识别算法模式是一项关键技能。例如，计算 3 + 6 × (5 – 2)² ÷ 3 这样的表达式时，简单地从左到右计算会得出错误答案。正确的算法 —— 即运算顺序 —— 保证每次都能得到唯一正确的结果。

2. The Order of Operations Algorithm (BODMAS/PEMDAS) | 运算顺序算法

The BODMAS (Brackets, Orders, Division/Multiplication, Addition/Subtraction) or PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction) rule is one of the most fundamental algorithms in mathematics. It defines the exact sequence in which operations must be performed:

BODMAS (括号、幂次、除/乘、加/减) 或 PEMDAS (圆括号、指数、乘/除、加/减) 规则是数学中最基本的算法之一。它定义了必须执行运算的精确顺序：

B/P: Perform calculations inside brackets/parentheses first.
O/E: Evaluate powers (exponents) and roots.
DM/MD: Perform division and multiplication as they appear from left to right.
AS: Perform addition and subtraction as they appear from left to right.

括号： 首先计算括号内的表达式。
幂次/指数： 计算幂和根。
除/乘： 按从左到右的顺序进行除法和乘法。
加/减： 按从左到右的顺序进行加法和减法。

Example: Evaluate 2 + 3 × (4² – 6) ÷ 5.

例题：计算 2 + 3 × (4² – 6) ÷ 5。

Step 1: Brackets → 4² – 6 = 16 – 6 = 10.
Step 2: Expression becomes 2 + 3 × 10 ÷ 5.
Step 3: Multiplication and division left to right → 3 × 10 = 30, then 30 ÷ 5 = 6.
Step 4: Addition → 2 + 6 = 8.

步骤 1：括号 → 4² – 6 = 16 – 6 = 10。
步骤 2：表达式变为 2 + 3 × 10 ÷ 5。
步骤 3：从左到右乘除 → 3 × 10 = 30，然后 30 ÷ 5 = 6。
步骤 4：加法 → 2 + 6 = 8。

3. Prime Factorisation Algorithms | 质因数分解算法

Prime factorisation is the process of breaking down a composite number into a product of its prime factors. IGCSE students are expected to apply two systematic algorithms: the factor tree method and the repeated division method. Both ensure that no factor is missed.

质因数分解是将一个合数分解为其质因子乘积的过程。IGCSE 学生需要掌握两种系统化算法：因子树法和重复除法。这两种方法都能确保不会遗漏任何因子。

Factor tree algorithm: Start with the number. Split it into any two factors. If a factor is prime, circle it; if not, continue splitting. Repeat until all branches end in prime numbers. Write the original number as the product of all circled primes.

因子树算法： 从该数开始。将其拆分为任意两个因子。如果因子是质数，圈起来；如果不是，继续分解。重复直到所有分支都以质数结尾。将原数写为所有圈出质数的乘积。

Example: Find prime factors of 72. 72 = 8 × 9 → 8 = 2 × 4, 4 = 2 × 2 → primes: 2,2,2; 9 = 3 × 3 → primes: 3,3. So 72 = 2³ × 3².

例题：求 72 的质因数。72 = 8 × 9 → 8 = 2 × 4，4 = 2 × 2 → 质因数：2,2,2；9 = 3 × 3 → 质因数：3,3。因此 72 = 2³ × 3²。

Repeated division algorithm: Divide the number by the smallest prime (2, 3, 5, …) that divides it exactly. Repeat with the quotient until the quotient is 1. The prime factors are the divisors used.

重复除法算法： 用能整除该数的最小质数 (2, 3, 5, …) 去除它。对商重复此过程，直到商为 1。所使用的除数即为该数的质因数。

4. Euclidean Algorithm for HCF | 求最大公因数的欧几里得算法

The Euclidean algorithm provides an efficient way to find the highest common factor (HCF, also called greatest common divisor, GCD) of two numbers without needing to list all factors. This algorithm is particularly useful for large numbers and is a classic example of an iterative procedure.

欧几里得算法提供了一种无需列出所有因子即可高效求出两个数最大公因数 (HCF, 也称最大公约数 GCD) 的方法。该算法对于大数尤其有用，是迭代过程的经典范例。

Algorithm steps:
1. Divide the larger number by the smaller number.
2. Find the remainder.
3. Replace the larger number with the smaller number, and the smaller number with the remainder.
4. Repeat steps 1-3 until the remainder is 0. The last non-zero remainder is the HCF.

算法步骤：
1. 用较大的数除以较小的数。
2. 求余数。
3. 将较大的数替换为较小的数，较小的数替换为余数。
4. 重复步骤 1-3 直到余数为 0。最后一个非零余数即为 HCF。

Example: Find HCF(252, 105).
252 ÷ 105 = 2 remainder 42
105 ÷ 42 = 2 remainder 21
42 ÷ 21 = 2 remainder 0
HCF = 21.

例题：求 HCF(252, 105)。
252 ÷ 105 = 2 余 42
105 ÷ 42 = 2 余 21
42 ÷ 21 = 2 余 0
HCF = 21。

In IGCSE exams, you may need to show these divisions clearly in a table or list each step. Always state the final answer explicitly.

在 IGCSE 考试中，你可能需要用表格清晰展示这些除法步骤，或逐一列出每一步。最后务必明确地写出答案。

5. Algorithm for LCM Using HCF | 通过 HCF 求 LCM 的算法

Once you have found the HCF of two numbers, a very simple algorithm gives their lowest common multiple (LCM). This method avoids lengthy listing of multiples and is mathematically elegant:

一旦你求出两个数的 HCF，一个非常简单的算法即可给出它们的最小公倍数 (LCM)。该方法避免了冗长的倍数罗列，在数学上十分简洁：

LCM(a, b) = (a × b) ÷ HCF(a, b)

Algorithm steps:
1. Use the Euclidean algorithm (or any method) to find HCF(a, b).
2. Multiply a by b.
3. Divide the product by the HCF. The result is the LCM.

算法步骤：
1. 使用欧几里得算法 (或任何方法) 求出 HCF(a, b)。
2. 将 a 与 b 相乘。
3. 将乘积除以 HCF，结果即为 LCM。

Example: Find LCM(252, 105). HCF = 21.
LCM = (252 × 105) ÷ 21 = 26460 ÷ 21 = 1260.

例题：求 LCM(252, 105)。HCF = 21。
LCM = (252 × 105) ÷ 21 = 26460 ÷ 21 = 1260。

This algorithm is extremely useful for both calculator and non-calculator papers. Remember that it only works for two numbers; for three or more numbers, use prime factorisation to find LCM directly.

该算法在可用和不可用计算器的试卷中都非常有用。请记住它仅适用于两个数；对于三个及以上的数，应直接通过质因数分解求 LCM。

6. Algorithmic Approach to Solving Linear Equations | 解一元一次方程的算法

Solving linear equations in IGCSE requires a structured, algorithmic approach to isolate the unknown variable. A reliable sequence of steps ensures you do not make sign errors or lose solutions:

解 IGCSE 中的一元一次方程需要一种结构化的算法方法来隔离未知变量。一个可靠的步骤序列能确保你不出现符号错误或丢失解：

Standard algorithm for ax + b = cx + d:
1. Expand any brackets on both sides.
2. Simplify each side by collecting like terms.
3. Move all terms containing the variable to one side (usually left) by adding/subtracting.
4. Move constant terms to the opposite side by adding/subtracting.
5. Divide both sides by the coefficient of the variable to get x = ?
6. Check your solution by substituting it back into the original equation.

形如 ax + b = cx + d 的方程标准算法：
1. 展开两边所有的括号。
2. 通过合并同类项化简每一边。
3. 通过加减将所有含变量的项移到一边 (通常是左边)。
4. 通过加减将常数项移到另一边。
5. 两边同时除以变量的系数，得到 x = ?
6. 将解答代入原方程检验。

Example: Solve 3(x + 2) – 4 = 2x + 10.
Step 1: Expand → 3x + 6 – 4 = 2x + 10.
Step 2: Simplify LHS → 3x + 2 = 2x + 10.
Step 3: Subtract 2x from both sides → x + 2 = 10.
Step 4: Subtract 2 from both sides → x = 8.
Check: 3(8+2)-4 = 3×10-4=26; 2×8+10=26. Correct.

例题：解 3(x + 2) – 4 = 2x + 10。
步骤 1：展开 → 3x + 6 – 4 = 2x + 10。
步骤 2：化简左边 → 3x + 2 = 2x + 10。
步骤 3：两边减 2x → x + 2 = 10。
步骤 4：两边减 2 → x = 8。
检验：3(8+2)-4 = 3×10-4=26；2×8+10=26，正确。

7. Trial and Improvement Algorithm for Equations | 方程求根的试位与改进算法

For equations that cannot be solved easily by algebraic rearrangement, IGCSE often tests the “trial and improvement” algorithm. This iterative method refines an estimate until a solution of a required accuracy is found. It is a practical application of an algorithm that uses substitution and comparison.

对于无法通过代数移项轻松求解的方程，IGCSE 常考查 “试位与改进” 算法。这种迭代方法不断细化估计值，直到找到所需精度的解。这是通过代入和比较来应用算法的一个实际例子。

Algorithm:
1. Rearrange the equation to the form f(x) = 0 or use the given expression.
2. Choose a starting value (may be given).
3. Substitute the value; observe if the result is too high or too low.
4. Try a next value that brings the result closer to zero.
5. Repeat until the difference between two successive trials is within the required accuracy.
6. State the solution to the specified decimal place.

算法：
1. 将方程整理为 f(x) = 0 的形式或使用给定表达式。
2. 选择一个起始值 (可能给出)。
3. 代入该值，观察结果是偏大还是偏小。
4. 尝试下一个能使结果更接近零的值。
5. 重复直到两次连续试值之差在要求精度范围内。
6. 给出精确到指定小数位的解。

Example: The equation x³ – 5x = 8 has a solution between 2 and 3. Use trial and improvement to find it to 1 d.p. Write trials in a table:

例题：方程 x³ – 5x = 8 在 2 和 3 之间有一个解。用试位与改进法求其至 1 位小数。将试值写入表格：

x	x³ – 5x	Comment
2.5	15.625 – 12.5 = 3.125	Too low (less than 8)
2.8	21.952 – 14 = 7.952	Too low
2.9	24.389 – 14.5 = 9.889	Too high (greater than 8)
2.85	23.149 – 14.25 = 8.899	Too high

Since 2.8 gives 7.952 (< 8) and 2.85 gives 8.899 (> 8), the solution lies between 2.8 and 2.85. To 1 decimal place, approximating gives x ≈ 2.8. (Further refinement would be needed for 2 d.p.).

因为 2.8 得到 7.952 (< 8) 而 2.85 得到 8.899 (> 8)，解位于 2.8 与 2.85 之间。精确到 1 位小数的近似解为 x ≈ 2.8 (如果要求更高精度需进一步细化)。

8. Generating Sequences via Position-to-Term Rule | 通过通项公式生成数列的算法

An algorithm for sequences is the step-by-step application of the position-to-term rule (nth term formula). This algorithm converts a given formula into the actual terms of a sequence by substituting successive integer values for n.

数列的生成算法是逐步应用通项公式 (第 n 项公式) 的过程。该算法通过将连续的整数值代入 n，将给定的公式转化为数列的具体项。

Algorithm:
1. Start with n = 1.
2. Substitute n into the given formula.
3. Calculate and write down the term.
4. Increase n by 1.
5. Repeat until the required number of terms is obtained.

算法：
1. 从 n = 1 开始。
2. 将 n 代入给定公式。
3. 计算并写出该项。
4. n 增加 1。
5. 重复直到获得所需数量的项。

Example: Find the first four terms of the sequence given by the nth term T(n) = n² + 2n – 3.
n=1 → 1²+2-3 = 0
n=2 → 4+4-3 = 5
n=3 → 9+6-3 = 12
n=4 → 16+8-3 = 21
Sequence: 0, 5, 12, 21.

例题：求由通项公式 T(n) = n² + 2n – 3 给出的数列的前四项。
n=1 → 1²+2-3 = 0
n=2 → 4+4-3 = 5
n=3 → 9+6-3 = 12
n=4 → 16+8-3 = 21
数列：0, 5, 12, 21。

IGCSE questions often reverse this algorithm: given a sequence, you identify the pattern and derive the nth term. Both directions rely on systematic numerical experimentation — a form of algorithmic thinking.

IGCSE 题目常反向运用该算法：给定一个数列，你识别模式并推导出第 n 项公式。这两个方向都依赖于系统化的数值试验 —— 一种算法思维形式。

9. Algorithm for Standard Form Calculations | 标准形式计算算法

Calculations with numbers in standard form (scientific notation) follow a clear algorithm to avoid errors in handling powers of ten. This algorithm is especially tested in calculator papers but the steps remain essential for estimation and non-calculator work.

用标准形式 (科学记数法) 表示的数字进行计算时，有一个清晰的算法可避免处理 10 的幂时出错。该算法在可用计算器的试卷中尤其常考，但步骤对于估算和非计算器题目同样至关重要。

Multiplication: (a × 10ᵐ) × (b × 10ⁿ) = (a × b) × 10ᵐ⁺ⁿ
Steps: 1. Multiply the coefficients a and b. 2. Add the exponents m and n. 3. If necessary, adjust the coefficient so it is between 1 and 10 by shifting the decimal point and modifying the exponent accordingly.

乘法：(a × 10ᵐ) × (b × 10ⁿ) = (a × b) × 10ᵐ⁺ⁿ
步骤：1. 系数 a 和 b 相乘。2. 指数 m 和 n 相加。3. 必要时，通过移动小数点并相应调整指数，使系数调整到 1 和 10 之间。

Division: (a × 10ᵐ) ÷ (b × 10ⁿ) = (a ÷ b) × 10ᵐ⁻ⁿ
Steps: 1. Divide the coefficients a by b. 2. Subtract the exponent n from m. 3. Adjust coefficient to standard form if needed.

除法：(a × 10ᵐ) ÷ (b × 10ⁿ) = (a ÷ b) × 10ᵐ⁻ⁿ
步骤：1. 系数 a 除以 b。2. 指数 m 减去 n。3. 若需要，调整系数至标准形式。

Example (multiplication): (3.2 × 10⁵) × (4 × 10³) = (3.2×4) × 10⁵⁺³ = 12.8 × 10⁸ = 1.28 × 10⁹ (after adjustment).

例题 (乘法)：(3.2 × 10⁵) × (4 × 10³) = (3.2×4) × 10⁵⁺³ = 12.8 × 10⁸ = 1.28 × 10⁹ (调整后)。

Always present the final answer in correct standard form: a × 10ⁿ where 1 ≤ a < 10 and n is an integer.

最终答案始终要以正确的标准形式呈现：a × 10ⁿ，其中 1 ≤ a < 10 且 n 为整数。

10. Exam Tips for Algorithmic Questions | 算法类考题应试技巧

When facing IGCSE questions that require an algorithmic approach, keep the following strategies in mind to maximise marks:

在面对需要算法应用的 IGCSE 题目时，牢记以下策略以最大化得分：

Show every step clearly. The algorithm is worthless if the examiner cannot see your logical flow. For procedures like Euclidean algorithm or trial and improvement, present your working in a well-organised table or numbered steps. This not only earns method marks but also helps you spot mistakes.

清晰地展示每一步。如果考官看不到你的逻辑流程，算法就毫无价值。对于欧几里得算法或试位改进法等过程，将你的解题过程整理成清晰的表格或编号步骤。这不仅能赚取方法分，还能帮助你发现错误。

Always check the final result. For solving equations, substitute your answer back; for HCF/LCM, test if the original numbers divide exactly by the HCF; for standard form, verify the coefficient range. These checks take seconds but can catch careless errors.

始终检查最终结果。对于解方程，将答案代回原式；对于 HCF/LCM，检验原数是否能被 HCF 整除；对于标准形式，验证系数范围。这些检查仅需数秒，却能发现粗心导致的错误。

Practise converting verbal descriptions into algorithmic steps and vice versa. Some IGCSE questions describe a procedure in words and ask you to carry it out; others give a flowchart and require interpretation. Being fluent in reading, writing, and executing algorithms is a high-level skill rewarded in the exam.

练习将文字描述转化为算法步骤，反之亦然。有些 IGCSE 题目用文字描述一个流程让你执行；另一些给出流程图要求解读。能熟练阅读、书写和执行算法是一项高级技能，在考试中会得到回报。

Finally, familiarise yourself with the precise language of algorithms — words like “input”, “output”, “while”, “do”, “repeat”, “if…then”. Although formal programming syntax is not required, understanding these can make description questions easier.

最后，熟悉算法的精确用语 —— 诸如 “input”、“output”、“while”、“do”、“repeat”、“if…then” 等。尽管不要求正式的编程语法，但理解这些词汇能让描述性题目变得更容易。

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Algorithms in IGCSE Mathematics | IGCSE 数学：算法考点精讲