📚 Year 11 CAIE Computer Science: Formula and Theorem Quick Reference | 公式定理速查手册
This quick reference handbook brings together all the essential formulas, theorems and conversion rules you need for the CAIE IGCSE Computer Science (0478) examination. Every topic from data representation and logic to file sizes and error detection is covered, with clear English explanations followed by their Chinese equivalents. Use this guide to sharpen your problem-solving skills and boost your exam confidence.
这本速查手册汇总了 CAIE IGCSE 计算机科学（0478）考试所需的全部核心公式、定理与转换规则。涵盖数据表示、逻辑电路、文件大小、错误检测等各个主题，每条要点均以英文说明搭配中文解释。借助本手册，你可以强化解题能力，提升应试信心。
1. Data Storage Units | 数据存储单位
Digital storage is measured in bits and bytes. A bit is a single binary digit (0 or 1). Eight bits make one byte. Because computers work in binary, larger units are based on powers of 2, not powers of 10.
数字存储以位（bit）和字节（byte）为单位。一位是一个二进制数字（0 或 1）。8 位组成一个字节。由于计算机采用二进制，更大的单位都是基于 2 的幂次，而不是 10 的幂次。
1 byte = 8 bits
1 KB = 2¹⁰ bytes = 1024 bytes
1 MB = 2²⁰ bytes = 1024 KB
1 GB = 2³⁰ bytes = 1024 MB
1 TB = 2⁴⁰ bytes = 1024 GB
Always use 1024 (2¹⁰) as the multiplier when converting between kilobytes, megabytes and gigabytes, unless the question specifically states that 1 KB = 1000 bytes. In typical CAIE tasks, 1024 is the standard.
在千字节、兆字节和吉字节之间换算时，始终使用 1024（2¹⁰）作为乘数，除非题目明确说明 1 KB = 1000 字节。CAIE 常规题目中，1024 是标准。
| Unit | Power of 2 | Exact bytes |
|---|---|---|
| 1 KB | 2¹⁰ | 1,024 |
| 1 MB | 2²⁰ | 1,048,576 |
| 1 GB | 2³⁰ | 1,073,741,824 |
2. Image File Size Calculation | 图像文件大小计算
The size of an uncompressed bitmap image is determined by its resolution and colour depth. Resolution is the number of pixels across the width and height, while colour depth is the number of bits used to represent the colour of a single pixel.
未压缩位图图像的大小由其分辨率和色彩深度决定。分辨率是宽度和高度上的像素数量，色彩深度是用于表示单个像素颜色的位数。
Image file size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
To obtain the size in bytes, divide the result by 8. For larger units, divide further by 1024 as needed. Metadata such as header information is typically ignored in exam calculations.
要得到以字节为单位的大小，可将结果除以 8。如需更大单位，可继续除以 1024。考试计算中通常忽略文件头等元数据信息。
Example: A 200 × 300 pixel image with 16-bit colour depth requires 200 × 300 × 16 = 960,000 bits = 120,000 bytes ≈ 117.2 KB.
示例：200 × 300 像素、16 位色彩深度的图像需要 200 × 300 × 16 = 960,000 位 = 120,000 字节 ≈ 117.2 KB。
3. Sound File Size Calculation | 声音文件大小计算
Uncompressed sound files are sampled at a regular rate. Each sample has a bit depth (sample resolution), and recordings can be mono or stereo. The total size depends on these factors and the duration of the audio.
未压缩的声音文件以固定速率进行采样。每个样本具有位深（采样分辨率），录音可以是单声道或立体声。总大小取决于这些因素以及音频时长。
Sound file size (bits) = sampling rate (Hz) × sample resolution (bits) × duration (seconds) × number of channels
Sampling rate is measured in hertz (Hz), e.g. 44,100 Hz for CD quality. Common sample resolutions are 8-bit or 16-bit. For stereo, number of channels = 2; for mono, it is 1. After calculating the size in bits, convert to bytes by dividing by 8.
采样率以赫兹（Hz）为单位，例如 CD 音质的 44,100 Hz。常见样本分辨率为 8 位或 16 位。立体声声道数为 2；单声道则为 1。计算出位大小后，除以 8 转换为字节。
Example: 30 seconds of stereo audio at 44.1 kHz with 16-bit resolution: 44,100 × 16 × 30 × 2 = 42,336,000 bits = 5,292,000 bytes ≈ 5.05 MB.
示例：30 秒立体声音频，采样率 44.1 kHz、16 位分辨率：44,100 × 16 × 30 × 2 = 42,336,000 位 = 5,292,000 字节 ≈ 5.05 MB。
4. Nyquist Theorem | 奈奎斯特采样定理
The Nyquist theorem states that to accurately reconstruct an analogue signal from its digital samples, the sampling rate must be at least twice the highest frequency present in the original signal. This avoids aliasing.
奈奎斯特定理指出，要从数字采样中准确重建模拟信号，采样率必须至少为原始信号中最高频率的两倍。这可以避免混叠现象。
Sampling rate ≥ 2 × fmax
where fmax is the maximum frequency in the analogue signal (measured in Hz). For example, human speech contains frequencies up to about 4 kHz, so a minimum sampling rate of 8 kHz is required. CD-quality audio targets a max audible frequency of around 20 kHz, hence the common 44.1 kHz rate.
其中 fmax 是模拟信号中的最高频率（单位 Hz）。例如人声频率最高约 4 kHz，因此最低需要 8 kHz 的采样率。CD 音质的目标可听最高频率约为 20 kHz，因此常采用 44.1 kHz 的采样率。
5. Logic Gates and Truth Tables | 逻辑门与真值表
Logic gates are the building blocks of digital circuits. Each gate performs a Boolean function on one or more inputs. The table below summarises the six fundamental gates, their expressions and truth tables.
逻辑门是数字电路的基本构件。每种门对一或多个输入执行布尔函数。下表汇总了六种基本门、它们的表达式和真值表。
| Gate | Symbol (Boolean) | Truth Table (A B → Output) |
|---|---|---|
| AND | X = A ∧ B | 0 0 → 0; 0 1 → 0; 1 0 → 0; 1 1 → 1 |
| OR | X = A ∨ B | 0 0 → 0; 0 1 → 1; 1 0 → 1; 1 1 → 1 |
| NOT | X = ¬A | 0 → 1; 1 → 0 |
| NAND | X = ¬(A ∧ B) | 0 0 → 1; 0 1 → 1; 1 0 → 1; 1 1 → 0 |
| NOR | X = ¬(A ∨ B) | 0 0 → 1; 0 1 → 0; 1 0 → 0; 1 1 → 0 |
| XOR | X = A ⊕ B | 0 0 → 0; 0 1 → 1; 1 0 → 1; 1 1 → 0 |
Memorising these truth tables is essential for analyzing and designing logic circuits. NAND and NOR gates are known as universal gates because any other gate can be built using only NAND or only NOR gates.
熟记这些真值表对于分析与设计逻辑电路至关重要。NAND 和 NOR 门被称为通用门，因为仅用 NAND 或仅用 NOR 门就可以构造出任何其他门。
6. Boolean Algebra Laws | 布尔代数定律
Boolean algebra is used to simplify logic expressions. The following laws and theorems allow you to reduce complex circuits and expressions.
布尔代数用于简化逻辑表达式。下列定律和定理可以帮助你化简复杂的电路和表达式。
- Commutative Law | 交换律: A ∧ B = B ∧ A; A ∨ B = B ∨ A
- Associative Law | 结合律: (A ∧ B) ∧ C = A ∧ (B ∧ C); (A ∨ B) ∨ C = A ∨ (B ∨ C)
- Distributive Law | 分配律: A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C); A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C)
- Identity Law | 同一律: A ∧ 1 = A; A ∨ 0 = A
- Null Law | 零律: A ∧ 0 = 0; A ∨ 1 = 1
- Idempotent Law | 幂等律: A ∧ A = A; A ∨ A = A
- Inverse Law | 互补律: A ∧ ¬A = 0; A ∨ ¬A = 1
- Absorption Law | 吸收律: A ∧ (A ∨ B) = A; A ∨ (A ∧ B) = A
- De Morgan’s Laws | 德摩根定律:
¬(A ∧ B) = ¬A ∨ ¬B
¬(A ∨ B) = ¬A ∧ ¬B - Double Negation | 双重否定: ¬(¬A) = A
These identities are extremely useful when simplifying Boolean circuits or writing efficient program conditions. De Morgan’s laws, in particular, are frequently tested.
这些恒等式在化简布尔电路或编写高效程序条件时非常有用。德摩根定律尤其常见于考题中。
7. Binary, Decimal and Hexadecimal Conversions | 二进制、十进制与十六进制转换
Understanding number systems is fundamental to computer science. The most common conversions tested include binary to decimal, decimal to binary, hexadecimal to binary and hexadecimal to decimal.
理解数制是计算机科学的基础。考试中最常见的转换包括二进制转十进制、十进制转二进制、十六进制转二进制和十六进制转十进制。
Binary to decimal: sum of (digit × 2position) — start from 0 on the right.
Decimal to binary: repeatedly divide by 2 and record remainders.
Hex to binary: replace each hex digit with its 4-bit binary equivalent.
Hex to decimal: sum of (digit × 16position).
二进制转十进制：每位数字乘以2的位置次幂后求和，位置从右起0开始。十进制转二进制：反复除以2并记录余数。十六进制转二进制：将每位十六进制数字替换为对应的4位二进制。十六进制转十进制：每位数字乘以16的位置次幂后求和。
| Hex | Binary | Decimal |
|---|---|---|
| 0 | 0000 | 0 |
| 1 | 0001 | 1 |
| 2 | 0010 | 2 |
| 3 | 0011 | 3 |
| 4 | 0100 | 4 |
| 5 | 0101 | 5 |
| 6 | 0110 | 6 |
| 7 | 更多咨询请联系16621398022（同微信）
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