📚 Year 8 AQA Computer Science: Formula & Theorem Quick Reference Guide | Year 8 AQA 计算机科学：公式定理速查手册
This quick reference handbook brings together the essential formulae, theorems and conversion rules needed for the Year 8 AQA Computer Science syllabus. From binary arithmetic and logic gates to data units and file size calculations, every key idea is presented in clear, bilingual pairs so you can revise with confidence.
这本速查手册汇集了 Year 8 AQA 计算机科学所需的核心公式、定理及转换法则。从二进制算术与逻辑门，到数据单位与文件大小计算，每一个关键知识点都以清晰的中英对照形式呈现，帮助你自信复习。
1. Decimal and Binary Conversion | 十进制与二进制转换
To convert a decimal number to binary, repeatedly divide the number by 2 and record the remainders from bottom to top.
将十进制数转换为二进制，反复除以 2 并从下往上记录余数。
Decimal ÷ 2 → remainder 0 or 1 → read remainders bottom-up
十进制 ÷ 2 → 余数 0 或 1 → 从下往上读取余数
To convert a binary number to decimal, multiply each bit by 2 raised to the power of its place value (starting at 0 from the right).
将二进制转换为十进制，将每一位乘以其位值对应的 2 的幂，然后求和（最右边位值为 0）。
Value = dₙ₋₁×2ⁿ⁻¹ + … + d₁×2¹ + d₀×2⁰
Example / 示例: 1101₂ = 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀
2. Binary Addition and Overflow | 二进制加法与溢出
Binary addition follows the same principles as decimal addition, carrying any value of 2 or greater to the next column.
二进制加法与十进制加法原理相同，任何结果 ≥ 2 的列都会产生进位。
0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0 carry 1, 1 + 1 + 1 = 1 carry 1
When the result of an addition exceeds the maximum number that can be stored in the available bits, an overflow error occurs.
当加法结果超出可用位数所能表示的最大值时，会发生溢出错误。
Overflow = carry out of the most significant bit in a fixed‑width register
Example / 示例: Adding 1111₂ + 0001₂ in a 4‑bit register gives 0000₂ with a carry out → overflow.
示例: 在 4 位寄存器中计算 1111₂ + 0001₂ 得到 0000₂ 并产生进位 → 溢出。
3. Logic Gate Symbols and Boolean Expressions | 逻辑门符号与布尔表达式
A logic gate is an electronic circuit that performs a Boolean function. The three basic gates are AND, OR and NOT.
逻辑门是执行布尔函数的电子电路。三种基本门是 AND、OR 和 NOT。
- AND gate / 与门: Output is 1 only if all inputs are 1.
仅当所有输入为 1 时输出为 1。
Expression / 表达式: Q = A · B
- OR gate / 或门: Output is 1 if at least one input is 1.
至少有一个输入为 1 时输出即为 1。
Expression / 表达式: Q = A + B
- NOT gate / 非门: Output is the inverse of the single input.
输出为单一输入的反相。
Expression / 表达式: Q = Ā (or A’)
Combining gates creates circuits that can perform complex logical operations, and their behaviour is summarised by truth tables.
组合多个门可创建执行复杂逻辑运算的电路，其行为由真值表总结。
4. Truth Tables for Basic Logic Gates | 基本逻辑门真值表
A truth table lists every possible combination of inputs and the corresponding output of a logic gate or circuit.
真值表列出所有可能的输入组合及其对应输出，用于描述逻辑门或电路的行为。
AND gate truth table (A · B) / 与门真值表:
| A | B | Q |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
OR gate truth table (A + B) / 或门真值表:
| A | B | Q |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
NOT gate truth table (Ā) / 非门真值表:
| A | Q |
| 0 | 1 |
| 1 | 0 |
Truth tables are used to verify circuit designs and simplify expressions before building real hardware.
真值表用于在实际搭建硬件前验证电路设计并化简表达式。
5. Boolean Algebra Laws | 布尔代数定律
Boolean algebra provides a set of laws that allow us to manipulate and simplify logic expressions. These are fundamental for reducing the number of gates in a circuit.
布尔代数提供了一组可用于操作和化简逻辑表达式的定律，是减少电路门数的基本工具。
Identity Law / 恒等律: A + 0 = A, A · 1 = A
Null Law / 零律: A + 1 = 1, A · 0 = 0
Idempotent Law / 幂等律: A + A = A, A · A = A
Complement Law / 互补律: A + Ā = 1, A · Ā = 0
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)
Students can apply these laws step by step to prove equivalence between different logic expressions.
学生可以逐步应用这些定律来证明不同逻辑表达式之间的相等性。
6. Simplification Using Boolean Algebra | 使用布尔代数化简
Simplifying a Boolean expression reduces the number of logic gates required, making circuits faster and cheaper.
化简布尔表达式可减少所需逻辑门的数量，使电路更快、成本更低。
Common simplification techniques include factoring, applying laws, and eliminating redundant terms.
常用化简技巧包括提取公因子、应用定律以及消去冗余项。
Example / 示例: Simplify / 化简 Q = A·B + A·B’
Factor out A: Q = A·(B + B’)
By the complement law, B + B’ = 1, so Q = A·1 = A
Thus the circuit reduces to just a wire (or buffer).
因此电路简化为一条直通线（或缓冲器）。
Always double‑check your simplified expression with a truth table to ensure no mistakes were introduced.
始终用真值表再次检查化简后的表达式，确保没有引入错误。
7. Data Units and Conversions | 数据单位与换算
All digital data is stored as bits. A group of 8 bits is called a byte. Larger units are defined as powers of 2 (binary) or powers of 10 (decimal, often used by manufacturers).
所有数字数据均以位（bit）存储。8 个位组成一个字节（Byte）。更大的单位按 2 的幂（二进制）或 10 的幂（十进制，制造商常用）定义。
| Unit / 单位 | Symbol / 符号 | Binary size / 二进制大小 |
| kilobyte | KB | 2¹⁰ bytes = 1024 B |
| megabyte | MB | 2²⁰ bytes = 1,048,576 B |
| gigabyte | GB | 2³⁰ bytes = 1,073,741,824 B |
| terabyte | TB | 2⁴⁰ bytes |
When calculating file sizes, always pay attention to whether the problem expects binary (e.g., 1 KB = 1024 B) or simplified decimal (1 KB = 1000 B) conversions; in AQA KS3, binary units are standard.
在计算文件大小时，务必注意题目要求是按二进制（如 1 KB = 1024 B）还是简化的十进制（1 KB = 1000 B）进行换算；AQA KS3 课程中二进制单位是标准。
8. Calculating Image File Size | 图像文件大小计算
The size of a bitmap image file depends on its width in pixels, height in pixels, and colour depth (bits per pixel).
位图图像文件的大小取决于其像素宽度、像素高度以及颜色深度（每像素位数）。
Image file size = Width × Height × Colour depth
图像文件大小 = 宽度 × 高度 × 颜色深度
Colour depth is the number of bits used to represent the colour of a single pixel. For example, 1 bit gives 2 colours (black and white), 8 bits give 256 colours, and 24 bits give over 16 million colours.
颜色深度是用于表示单个像素颜色的位数。例如，1 位可表示 2 种颜色（黑白），8 位可表示 256 色，24 位可表示超过 1600 万种颜色。
Example / 示例: A 200×300 pixel image with a colour depth of 8 bits:
Size = 200 × 300 × 8 = 480,000 bits. Convert to bytes: 480,000 ÷ 8 = 60,000 B ≈ 58.6 KB (÷ 1024).
大小 = 200 × 300 × 8 = 480,000 位。转换为字节：480,000 ÷ 8 = 60,000 B ≈ 58.6 KB（除以 1024）。
9. Calculating Sound File Size | 声音文件大小计算
Digital sound is created by sampling an analogue wave. The file size is determined by the sample rate, bit depth, length in seconds, and number of channels.
数字声音通过对模拟波形采样而生成。文件大小由采样率、位深度、时长（秒）及声道数决定。
Sound file size = Sample rate × Bit depth × Time (sec) × Channels
声音文件大小 = 采样率 × 位深度 × 时长（秒） × 声道数
The sample rate is measured in hertz (Hz) and indicates how many times per second the sound is measured. Bit depth is the number of bits used to store each sample. Stereo sound uses 2 channels, while mono uses 1.
采样率以赫兹（Hz）为单位，表示每秒对声音的测量次数。位深度是存储每个采样所使用的位数。立体声使用 2 个声道，单声道使用 1 个声道。
Example / 示例: 10 seconds of stereo audio at 44,100 Hz with a 16‑bit depth:
Size = 44,100 × 16 × 10 × 2 = 14,112,000 bits = 1,764,000 bytes ≈ 1.68 MB.
10. Hexadecimal Representation and Conversion | 十六进制表示与转换
Hexadecimal (base‑16) is a compact way to represent large binary numbers. It uses digits 0–9 and letters A–F, where A = 10, B = 11, …, F = 15.
十六进制（基数为 16）是一种紧凑表示大二进制数的方式，使用数字 0–9 和字母 A–F，其中 A=10，B=11，…，F=15。
Each hex digit represents exactly 4 bits (a nibble). To convert binary to hex, group bits into sets of four from the right, then replace each group with the corresponding hex symbol.
每个十六进制数字代表 4 位（一个半字节）。将二进制转换为十六进制时，从右向左每 4 位分为一组，然后用相应的十六进制符号替换。
1101 1010₂ → D A₁₆
To convert hex to decimal, multiply each digit by 16 raised to the appropriate power, then sum the results.
将十六进制转换为十进制，将每位数字乘以相应位置的 16 的幂，然后求和。
2F₁₆ = 2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀
Hexadecimal is widely used to represent memory addresses, colour codes (e.g., #FF00FF), and error codes because it is easier to read than long binary strings.
十六进制广泛用于表示内存地址、颜色代码（如 #FF00FF）以及错误码，因为它比长串二进制更易读。
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