📚 Year 10 Edexcel Computer Science: Quick Reference Formulae & Theorems | Edexcel Year 10 计算机：公式定理速查手册
This quick reference handbook collects the most important formulae, conversion rules and theorems you will encounter in the Year 10 Edexcel Computer Science course. From data storage calculations to Boolean algebra and search efficiency, these are the fundamentals you need to memorise and apply confidently in your written exam and practical tasks.
这份速查手册汇集了 Year 10 Edexcel 计算机科学课程中最重要的公式、换算规则和定理。从数据存储计算到布尔代数、再到搜索效率，这些都是你需要牢记并自信应用于笔试和实践任务的基础知识。
1. Data Storage Units & Conversions | 数据存储单位与换算
A bit is the smallest unit of data, storing a single binary value 0 or 1. All file sizes and memory capacities are built from bits using standard prefixes defined by the exam board, where each step is a factor of 1000.
比特（bit）是最小的数据单位，存储一个二进制的 0 或 1。所有文件大小和内存容量都由比特通过考试局定义的标准前缀构成，每一级换算的倍率为 1000。
caling: 1 KB = 1000 bytes, 1 MB = 1000 KB, 1 GB = 1000 MB, 1 TB = 1000 GB, 1 PB = 1000 TB.
4 个比特组成一个半位元组（nibble）；8 个比特组成一个字节（byte）。更大的单位遵循十进制换算：1 KB = 1000 字节，1 MB = 1000 KB，1 GB = 1000 MB，1 TB = 1000 GB，1 PB = 1000 TB。
| Unit | Equivalent in bytes |
|---|---|
| 1 bit | 1/8 byte |
| 1 nibble | 4 bits = 0.5 bytes |
| 1 byte (B) | 8 bits |
| 1 kilobyte (KB) | 1000 bytes |
| 1 megabyte (MB) | 1000 KB = 1×10⁶ bytes |
| 1 gigabyte (GB) | 1000 MB = 1×10⁹ bytes |
| 1 terabyte (TB) | 1000 GB = 1×10¹² bytes |
| 1 petabyte (PB) | 1000 TB = 1×10¹⁵ bytes |
To convert bits to bytes, divide by 8. To convert any larger unit to the next smaller unit, multiply by 1000; to convert to the next larger unit, divide by 1000.
将比特转换为字节，除以 8。将较大单位转换为相邻较小单位时乘以 1000；转换为相邻较大单位时除以 1000。
2. Image File Size Formula | 图像文件大小公式
The uncompressed file size of a bitmap image depends on the pixel dimensions and the colour depth. Each pixel's colour is stored using a fixed number of bits, and the total number of bits grows rapidly as resolution increases.
位图图像未经压缩的文件大小取决于像素尺寸和色彩深度。每个像素的颜色使用固定位数的比特存储，随着分辨率提高，总比特数迅速增长。
ge file size (bits) = width (px) × height (px) × colour depth (bits per pixel)
图像文件大小（比特） = 宽度（像素） × 高度（像素） × 色彩深度（每像素比特数）
To express the size in bytes, divide the result by 8. For kilobytes, divide bytes by 1000, and so on. The colour depth also determines the maximum number of distinct colours: Number of colours = 2colour depth. For example, a colour depth of 24 bits yields 2²⁴ ≈ 16.8×10⁶ colours.
要将结果表示为字节，需除以 8。转换为千字节则再将字节数除以 1000，以此类推。色彩深度同时决定了可表示的最大颜色数量：颜色数量 = 2色彩深度。例如，24 位的色彩深度可产生 2²⁴ ≈ 1.68×10⁷ 种颜色。
imes; 24 = 288×10⁶ bits, which is 36×10⁶ bytes (36 MB) before any compression.
示例：一张 4000 × 3000 像素、24 位色彩深度的照片需要 4000 × 3000 × 24 = 288×10⁶ 比特，即 36×10⁶ 字节（36 MB），不含压缩。
3. Sound File Size Formula | 声音文件大小公式
Digital audio is captured by taking regular samples of the sound wave. The file size is a product of how often you sample, how detailed each sample is, the number of audio channels and the duration of the recording.
数字音频通过定期对声波采样而获取。文件大小取决于采样频率、每次采样的详细程度、声道数量以及录音时长。
bits) = sample rate (Hz) × bit depth × number of channels × duration (s)
声音文件大小（比特） = 采样率（Hz） × 位深 × 声道数 × 时长（秒）
Convert bits to bytes by dividing by 8, then to larger units by dividing by 1000 as needed. Standard CD-quality audio uses a sample rate of 44 100 Hz, 16-bit depth and 2 stereo channels, resulting in roughly 10 MB per minute.
将比特转换为字节需除以 8，再根据需要除以 1000 转为更大的单位。标准 CD 音质使用 44 100 Hz 的采样率、16 位的位深和 2 个立体声声道，每分钟大约生成 10 MB 的数据。
s; 2 × 180 = 254×016×000 bits = 31×752×000 bytes ≈ 31.75 MB.
示例：一曲 3 分钟的立体声（44.1 kHz，16 位）产生：44×100 × 16 × 2 × 180 = 254×016×000 比特 = 31×752×000 字节 ≈ 31.75 MB。
4. Data Transfer Time | 数据传输时间
When moving data across a network or between storage media, the time taken grows with the file size and shrinks as the transfer rate increases. You must ensure both size and rate use the same unit of bits or bytes.
通过网络或在不同存储介质之间传输数据时，所花时间随文件大小增大而增加，随传输速率提高而缩短。你必须确保文件大小和传输速率使用相同的单位（同为比特或同为字节）。
Transfer time (s) = File size (bits) ÷ Transfer rate (bits per second)
传输时间（秒） = 文件大小（比特） ÷ 传输速率（比特每秒）
0⁶ to get bits per second. For example, downloading a 200 MB file over a 40 Mbps connection: file size = 200 × 8 × 10⁶ = 1600×10⁶ bits; rate = 40×10⁶ bps; time = 1600×10⁶ / 40×10⁶ = 40 seconds.
如果速率以字节每秒给出，需相应转换文件大小。常见的速率有 Mbps（兆比特每秒），乘以 1×10⁶ 得到比特每秒。例如，通过 40 Mbps 连接下载 200 MB 文件：文件大小 = 200 × 8 × 10⁶ = 1600×10⁶ 比特；速率 = 40×10⁶ bps；时间 = 1600×10⁶ / 40×10⁶ = 40 秒。
5. Basic Boolean Algebra Laws | 布尔代数基本定律
Boolean algebra simplifies logic circuits and expressions using a set of fundamental laws. These laws work with variables that can only be 0 or 1, and the operators AND (.) and OR (+).
布尔代数通过一组基本定律简化逻辑电路和表达式。这些定律作用于只能取 0 或 1 的变量，以及运算符 AND（.）和 OR（+）。
Below is a summary of the key identities for a single variable A:
以下是针对单个变量 A 的核心恒等式总结：
| Law | AND form | OR form |
|---|---|---|
| Identity | A . 1 = A | A + 0 = A |
| Null | A . 0 = 0 | A + 1 = 1 |
| Idempotent | A . A = A | A + A = A |
| Complement | A . A' = 0 | A + A' = 1 |
| Double negation | (A')' = A | |
B = B+A; A.(B.C) = (A.B).C; A+(B+C) = (A+B)+C; A.(B+C) = A.B + A.C; A+(B.C) = (A+B).(A+C).
交换律、结合律和分配律允许重新排列和组合运算：A.B = B.A；A+B = B+A；A.(B.C) = (A.B).C；A+(B+C) = (A+B)+C；A.(B+C) = A.B + A.C；A+(B.C) = (A+B).(A+C)。
6. De Morgan's Laws | 德摩根定律
De Morgan's laws are essential for converting between AND and OR expressions when negation is applied to a whole group. They are used extensively in circuit design and logic simplification.
德摩根定律对于在否定作用于整个组合时实现 AND 与 OR 表达式之间的转换至关重要，广泛用于电路设计和逻辑化简。
(A . B)' = A' + B'
(A . B)' = A' + B'
(A + B)' = A' . B'
(A + B)' = A' . B'
f a product becomes the sum of the negated variables, and vice versa. For example, (X.Y.Z)' = X' + Y' + Z'.
这些规则可扩展至任意多个变量：乘积的否定变为各变量否定之和，反之亦然。例如，(X.Y.Z)' = X' + Y' + Z'。
7. Logic Gate Truth Tables & Expressions | 逻辑门真值表与表达式
Each basic logic gate produces a specific output for every combination of binary inputs. The truth table summarises all possibilities and forms the basis for building and testing circuits.
每种基本逻辑门对每一组二进制输入产生特定的输出。真值表概括了所有可能，并构成搭建与测试电路的基础。
Below are the truth tables for the gates required in the Edexcel specification. A and B are inputs, Q is the output.
以下是 Edexcel 规范中要求掌握的逻辑门真值表，A 和 B 为输入，Q 为输出。
| Gate | A | B | Q |
|---|---|---|---|
| AND (Q = A.B) | 0 | 0 | 0 |
| 0 | 1 | 0 | |
| 1 | 0 | 0 | |
| 1 | 1 | 1 | |
| OR (Q = A+B) | 0 | 0 | 0 |
| 0 | 1 | 1 | |
| 1 | 0 | 1 | |
| 1 | 1 | 1 | |
| NOT (Q = A') | 0 | — | 1 |
| NOT | 1 | — | 0 |
| NAND (Q = (A.B)') | 0 | 0 | 1 |
| 0 | 1 | 1 | |
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