📚 Year 10 CIE Computer Science: Formulae & Theorems Quick Reference | Year 10 CIE 计算机：公式定理速查手册
This handbook provides a quick reference to essential formulae and theorems covered in the CIE IGCSE Computer Science (0478) syllabus. Each section is presented with concise explanations in both English and Chinese to support bilingual learners.
本手册为 CIE IGCSE 计算机科学（0478）课程中涉及的核心公式与定理提供快速查阅。每个小节都配有简明中英文解释，帮助双语学习者掌握要点。
1. Data Units and Conversion | 数据单位与换算
ll larger units are based on powers of 2, where 1 KB = 1024 bytes (2¹⁰ B), not 1000 bytes.
数据的最小单位是比特（bit，二进制位），取值为 0 或 1。8 个比特组成 1 个字节（Byte）。所有更大单位都以 2 的幂为基础，1 KB = 1024 字节（2¹⁰ B），而不是 1000 字节。
| 1 Byte (B) | 8 bits | 2³ bits |
| 1 Kilobyte (KB) | 1024 B | 2¹⁰ B |
| 1 Megabyte (MB) | 1024 KB | 2²⁰ B |
| 1 Gigabyte (GB) | 1024 MB | 2³⁰ B |
| 1 Terabyte (TB) | 1024 GB | 2⁴⁰ B |
divide by 1024 at each step. For example, 5 MB = 5 × 1024 × 1024 bytes = 5,242,880 bytes.
单位换算时，每上升一级乘以 1024，每下降一级除以 1024。例如，5 MB = 5 × 1024 × 1024 字节 = 5,242,880 字节。
2. Image File Size Calculation | 图像文件大小计算
s resolution (width × height in pixels) and colour depth (bits per pixel). The formula is:
位图图像的大小取决于其分辨率（宽×高，单位像素）和颜色深度（每像素位数）。公式为：
Image size (bytes) = (Width × Height × Colour depth) ÷ 8
Colour depth determines the number of bits used to represent the colour of each pixel. For instance, 24-bit colour uses 24 bits per pixel, allowing over 16 million colours.
颜色深度决定了每个像素颜色所用的位数。例如，24 位色彩每像素用 24 位，可表示超过 1600 万种颜色。
00 bytes ≈ 6.22 MB. Note that file headers and metadata slightly increase the actual size.
一张 1920×1080、24 位深度的图像，原始文件大小为 (1920 × 1080 × 24) / 8 = 6,220,800 字节 ≈ 6.22 MB。注意，文件头和元数据会使实际大小略增。
3. Audio File Size Calculation | 音频文件大小计算
Uncompressed audio file size is determined by four factors: sample rate, bit depth, duration, and number of channels. The formula is:
未压缩音频文件大小由四个因素决定：采样率、位深度、时长和声道数。公式为：
Audio size (bits) = Sample rate (Hz) × Bit depth × Duration (s) × Channels
16-bit) has a size of 44,100 × 16 × 60 × 2 = 84,672,000 bits = 10,584,000 bytes ≈ 10.1 MB.
将位数除以 8 转换为字节。例如，1 分钟 CD 音质立体声录音（44,100 Hz、16 位）大小为 44,100 × 16 × 60 × 2 = 84,672,000 比特 = 10,584,000 字节 ≈ 10.1 MB。
Often, you need to give the answer in appropriate units like KB or MB by dividing further by 1024.
通常需要将答案转换为合适单位，如 KB 或 MB，再除以 1024。
4. Text File Size and Character Encoding | 文本文件大小与字符编码
The size of a plain text file is the product of the number of characters and the number of bytes per character in the encoding scheme used. Standard ASCII uses 7 bits per character (stored in 1 byte), Extended ASCII uses 1 byte, and UTF-8 can use 1 to 4 bytes per character.
纯文本文件的大小等于字符数乘以所用编码方案中每个字符的字节数。标准 ASCII 每字符 7 位（存为 1 字节），扩展 ASCII 每字符 1 字节，UTF-8 每字符 1 到 4 字节。
Text size (bytes) = Number of characters × Bytes per character
hly 5 characters long, would be about 25,000 characters, occupying 25,000 bytes ≈ 24.4 KB in ASCII.
如果一个文档含 2000 个 ASCII 字符，文件大小约为 2000 字节（另加文件头）。一篇 5000 词的英语短文，平均每词 5 字符，约 25,000 字符，用 ASCII 存储约 25,000 字节 ≈ 24.4 KB。
5. Data Transmission Time | 数据传输时间
Transmission time over a network is calculated by dividing the total data size by the transfer rate (bandwidth). Be careful to match units: if bandwidth is in bits per second (bps), data size must be in bits.
网络传输时间等于数据总量除以传输速率（带宽）。注意单位匹配：若带宽用每秒比特（bps），数据大小必须转换为比特。
Time (s) = Data size (bits) ÷ Bandwidth (bps)
≈ 41.94 seconds. Real-world factors like latency and protocol overhead may increase this.
通过 10 Mbps 连接下载 50 MB 文件，先将 50 MB 转换为比特：50 × 1024 × 1024 × 8 = 419,430,400 比特。时间 = 419,430,400 / 10,000,000 ≈ 41.94 秒。实际延迟和协议开销会增加时间。
6. Compression Ratio | 压缩比
Compression ratio measures the effectiveness of file compression. A higher ratio indicates more space saved. It is defined as:
压缩比衡量文件压缩的有效性。比值越高，节省的空间越多。定义如下：
Compression ratio = Uncompressed size ÷ Compressed size
s means the compressed file is one-quarter the original size. You may also express it as a percentage saving.
如果一个 20 MB 文件压缩到 5 MB，压缩比为 20 / 5 = 4:1，即压缩后是原来的四分之一。亦可用节省百分比表示。
7. Boolean Algebra Laws | 布尔代数定律
Boolean algebra uses variables that have only two possible values: 0 (False) and 1 (True). Key laws simplify logic circuits. Here, '+' denotes OR, '·' denotes AND, and ' (prime) denotes NOT.
布尔代数中的变量只有 0（假）和 1（真）两种取值。主要定律用于化简逻辑电路。此处 '+' 表示或，'·' 表示与，' ' 表示非。
Commutative Law: A + B = B + A; A · B = B · A
交换律：A + B = B + A; A · B = B · A
Associative Law: (A + B) + C = A + (B + C); (A · B) · C = A · (B · C)
结合律：(A + B) + C = A + (B + C); (A · B) · C = A · (B · C)
Distributive Law: A · (B + C) = (A · B) + (A · C); A + (B · C) = (A + B) · (A + C)
分配律：A · (B + C) = (A · B) + (A · C); A + (B · C) = (A + B) · (A + C)
Absorption Law: A + (A · B) = A; A · (A + B) = A
吸收律：A + (A · B) = A; A · (A + B) = A
These laws are used extensively in simplifying logic expressions and reducing the number of logic gates in a circuit.
这些定律广泛用于化简逻辑表达式和减少电路中逻辑门的数量。
8. De Morgan's Theorems | 德摩根定理
De Morgan's theorems are essential for converting between AND and OR gates, particularly for NAND and NOR gate analysis. The two theorems are:
德摩根定理对于在 AND 和 OR 门之间进行转换至关重要，尤其是分析 NAND 和 NOR 门时。两个定理如下：
(A · B)' = A' + B'
(A + B)' = A' · B'
In words, the complement of (A AND B) is equal to (NOT A) OR (NOT B). The complement of (A OR B) is equal to (NOT A) AND (NOT B). These theorems are used in digital circuit design and Boolean simplification.
换句话说，(A 与 B) 的非等于 (非 A) 或 (非 B)；(A 或 B) 的非等于 (非 A) 与 (非 B)。这两个定理用于数字电路设计和布尔化简。
9. Logic Gates and Truth Tables | 逻辑门与真值表
A truth table lists all possible input combinations and the corresponding output for a logic gate. Below are the standard gates and their Boolean expressions.
真值表列出了逻辑门所有可能的输入组合及对应输出。以下是标准门及其布尔表达式。
AND gate (A · B): Output is 1 only if both inputs are 1.
与门 (A · B)： 仅当两个输入均为 1 时输出 1。
AND: Q = A · B
OR gate (A + B): Output is 1 if at least one input is 1.
或门 (A + B)： 至少一个输入为 1 时输出 1。
OR: Q = A + B
NOT gate (A'): Output is the inverse of the input.
非门 (A')： 输出为输入的相反值。
NAND gate: Q = (A · B)'
与非门： Q = (A · B)'
NOR gate: Q = (A + B)'
或非门： Q = (A + B)'
XOR gate: Q = A' · B + A · B' (true when inputs are different).
异或门： Q = A' · B + A · B'（输入不同时为真）。
The truth tables can be constructed by evaluating these expressions. Knowing them is crucial for understanding how computer circuits work.
通过计算这些表达式可以构建真值表。理解它们对掌握计算机电路工作原理至关重要。
10. Checksum and Parity Check | 校验和与奇偶校验
Checksum is an error-detection technique where data bytes are summed, and a check digit (often the sum modulo 256) is transmitted with the data. The receiver recomputes the sum and checks for a match.
校验和是一种错误检测技术：将所有数据字节求和，传输数据时附带一个校验位（常为和对 256 取模）。接收方重新计算并比对。
Checksum = (Sum of data bytes) mod 256
Parity checking uses a single bit (parity bit) to make the total number of 1s in a byte either even (even parity) or odd (odd parity). Even parity bit is computed as the XOR of all data bits.
奇偶校验用一个位（奇偶位）使一个字节中 1 的总数为偶数（偶校验）或奇数（奇校验）。偶校验位等于所有数据位的异或（XOR）。
Even parity bit = D₁ XOR D₂ XOR ... XOR Dₙ
For example, the byte 1101001 has five 1s (odd). To achieve even parity, the parity bit must be 1, making six 1s (even).
例如字节 1101001 有五个 1（奇数），要成为偶校验，校验位需为 1，使总数为六个 1（偶数）。
11. Encryption Formulae (Caesar Shift) | 加密公式（凯撒移位）
. For a shift key K, encryption and decryption (assuming letters A=0, B=1, ..., Z=25) are:
凯撒密码将明文中每个字母按固定位数（密钥）移位。设密钥为 K，加密与解密（设字母 A=0, B=1, ..., Z=25）公式为：
Encryption: C = (P + K) mod 26
Decryption: P = (C − K) mod 26
Where P is the plaintext letter's position, C is the ciphertext letter's position. If the shift goes past 'Z', it wraps around to 'A' using modulo 26.
其中 P 为明文字母位置，C 为密文字母位置。若移位超过 'Z'，则通过模 26 从 'A' 重新循环。
For a shift of 3, 'A' (0) becomes D (3), 'X' omes A ((23+3) mod 26 = 0). This is the coding principle behind the classic Caesar cipher.
若移位为 3，'A' (0) 变成 D (3)，'X' (23) 变成 A ((23+3) mod 26 = 0)。这就是经典凯撒密码的编码原理。
12. Number of Colours and Bit Depth | 颜色数与位深度
The number of distinct colours a pixel can represent is determined by its bit depth. If a pixel uses n bits, the number of possible colours is 2ⁿ.
一个像素能表示的不同颜色数由其位深度决定。若像素使用 n 位，则可表示的颜色数为 2ⁿ。
Number of colours = 2ⁿ
Thus, 1
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