📚 Year 11 AQA Computer Science: Key Formulas & Theorems Quick Reference | Year 11 AQA 计算机：公式定理速查手册
Mastering the key formulas and theorems is essential for success in the Year 11 AQA Computer Science exam. This guide consolidates all the critical calculations, logic rules, and algorithmic concepts you need to memorize for Paper 1 and Paper 2, presented clearly with concise explanations and paired English-Chinese text to support EAL learners.
掌握关键的公式和定理对于在 Year 11 AQA 计算机科学考试中取得成功至关重要。本指南整合了您在 Paper 1 和 Paper 2 中需要记忆的所有关键计算、逻辑规则和算法概念，并附有清晰简要的英中配对解释，以支持 EAL学习者。
1. Data Storage Units & Conversions | 数据存储单位与转换
Digital data is measured in bits (b) and bytes (B), where 1 byte is exactly 8 bits. Both decimal (base-10) and binary (base-2) prefix systems are used to describe larger volumes of data, and it is vital to distinguish between them for accurate storage calculations.
数字数据以位 (bit, b) 和字节 (byte, B) 为单位，其中 1 字节恰好是 8 位。十进制（基数为 10）和二进制（基数为 2）两种前缀系统都用于描述更大的数据量，为了准确进行存储计算，区分它们至关重要。
| Unit | 单位 | Decimal Value (Standard) | 十进制值（标准） | Binary Value (IEC) | 二进制值（IEC） |
|---|---|---|
| 1 Kilobyte (KB) | 10³ bytes = 1,000 B | – |
| 1 Kibibyte (KiB) | – | 2¹⁰ bytes = 1,024 B |
| 1 Megabyte (MB) | 10⁶ bytes = 1,000,000 B | – |
| 1 Mebibyte (MiB) | – | 2²⁰ bytes ≈ 1.05 MB |
| 1 Gigabyte (GB) | 10⁹ bytes = 1,000,000,000 B | – |
| 1 Gibibyte (GiB) | – | 2³⁰ bytes ≈ 1.07 GB |
The fundamental conversion to remember when shifting between the smallest units is: number of bits = number of bytes × 8.
在最小单位之间转换时要记住的基本换算：比特数 = 字节数 × 8。
2. Binary & Decimal Conversions | 二进制与十进制转换
Binary numbers use a base-2 system, utilizing only the digits 0 and 1. Each position in a binary number represents an increasing power of 2, starting from 2⁰ on the far right. To convert a binary number to its decimal equivalent, sum the place values (weights) wherever a 1 appears.
二进制数字使用基数为 2 的系统，仅使用数字 0 和 1。二进制数中的每个位置代表一个递增的 2 的幂，从最右侧的 2⁰ 开始。要将二进制数转换为其十进制等效值，请将出现 1 的所有位置的位值（权值）相加。
| Binary Place Value | 二进制位值 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|
| Example: 10010110 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
Calculation: 128 + 16 + 4 + 2 = 150 in decimal. To convert a decimal number to binary, you can repeatedly divide the decimal number by 2 and record the remainder (which will always be 0 or 1), reading from bottom to top; alternatively, subtract the largest possible power of 2 sequentially.
计算：128 + 16 + 4 + 2 = 十进制 150。要将十进制数转换为二进制，您可以将十进制数反复除以 2 并记录余数（始终是 0 或 1），从下往上读取；或者，依次减去可能的最大的 2 的幂。
3. Hexadecimal Conversions | 十六进制转换
Hexadecimal is a base-16 numbering system that uses the digits 0-9 and the letters A-F to represent values 10 to 15 respectively. It serves as a compact and human-readable shorthand for binary, because every single hexadecimal digit corresponds to a group of exactly four binary digits (a nibble).
十六进制是一种基数为 16 的计数系统，使用数字 0-9 和字母 A-F 分别表示值 10 到 15。它作为二进制的一种紧凑且易于人阅读的速记法，因为每个十六进制数字恰好对应一组四位二进制数字（一个半字节）。
To convert a binary number to hexadecimal, start from the right-hand side and split the binary string into groups of four bits. If the leftmost group has fewer than four bits, pad it with leading zeros. Then, convert each 4-bit group into its single hex equivalent using a lookup table.
要将二进制数转换为十六进制，请从右侧开始，将二进制串分成四位一组。如果最左侧的组少于四位，请在其前面补零。然后，使用查找表将每个四位组转换为其等效的单个十六进制数字。
Example 1: Binary 11011110 → Split into 1101 1110 → Hex D E → Result: 0xDE.
Example 2: Binary 101 → Pad to 0101 → Hex 5.
示例 1：二进制 11011110 → 分成 1101 1110 → 十六进制 D E → 结果：0xDE。
示例 2：二进制 101 → 补零为 0101 → 十六进制 5。
4. Image File Size Calculation | 图像文件大小计算
The storage size of a bitmap image is directly determined by its spatial resolution and colour depth. Resolution is the total number of pixels in the image, calculated by multiplying the width in pixels by the height in pixels. Colour depth (or bit depth) defines how many bits are used to store the colour of each individual pixel.
位图图像的存储大小直接由其空间分辨率和颜色深度决定。分辨率是图像中的像素总数，通过将像素宽度乘以像素高度来计算。颜色深度（或位深度）定义用于存储每个单独像素颜色的位数。
The core formula to memorize is: Image File Size (bits) = Width (px) × Height (px) × Colour Depth (bits per pixel, bpp).
需要记住的核心公式是：图像文件大小（比特）= 宽度（像素）× 高度（像素）× 颜色深度（每像素比特数，bpp）。
To express the calculated bit-size in bytes, divide by 8. To convert further into standard decimal megabytes (MB) or binary mebibytes (MiB), divide by 1,000,000 or 1,048,576 respectively, depending on the context of the exam question.
要将计算出的比特大小以字节表示，请除以 8。要根据考试题目的上下文，进一步转换为标准十进制兆字节 (MB) 或二进制兆比字节 (MiB)，请分别除以 1,000,000 或 1,048,576。
Worked Example: An image with dimensions 2048 × 1024 pixels and a 24-bit colour depth.
Size in bits = 2048 × 1024 × 24 = 50,331,648 bits.
Size in bytes = 50,331,648 / 8 = 6,291,456 B ≈ 6.29 MB.
示例：一张尺寸为 2048 × 1024 像素、24 位颜色深度的图像。
比特大小 = 2048 × 1024 × 24 = 50,331,648 比特。
字节大小 = 50,331,648 / 8 = 6,291,456 B ≈ 6.29 MB。
5. Sound File Size Calculation | 声音文件大小计算
Digital audio quality and the resultant file size are governed by the sample rate, bit depth, and the number of channels. The sample rate, measured in Hertz (Hz), specifies how many samples of the analog sound wave are taken per second; a higher sample rate captures higher frequencies more accurately.
数字音频的质量和生成的文件大小由采样率、位深度和声道数决定。采样率以赫兹 (Hz) 为单位，指定每秒从模拟声波中采样的次数；更高的采样率能更准确地捕获更高的频率。
The fundamental formula for calculating uncompressed audio size is: File Size (bits) = Sample Rate (Hz) × Bit Depth (bits) × Number of Channels × Duration (seconds).
计算未压缩音频大小的基本公式是：文件大小（比特）= 采样率（赫兹）× 位深度（比特）× 声道数 × 持续时间（秒）。
Always check the units: sample rate is given in Hz (which equates to samples per second), bit depth is the bits per sample, and duration in seconds. A standard stereo track has 2 channels. Follow the same byte/megabyte conversion rules as for image files.
务必检查单位：采样率以 Hz 为单位（等同于每秒采样数），位深度是每样本的比特数，持续时间以秒为单位。标准立体声轨道有 2 个声道。遵循与图像文件相同的字节/兆字节转换规则。
Worked Example: A 3-minute stereo audio clip recorded at 44.1 kHz with 16-bit depth.
File size (bits) = 44,100 × 16 × 2 × 180 = 254,016,000 bits.
File size (MB) = 254,016,000 / (8 × 1,000,000) ≈ 31.75 MB.
示例：一段 3 分钟的立体声音频剪辑，以 44.1 kHz 录制，16 位深度。
文件大小（比特）= 44,100 × 16 × 2 × 180 = 254,016,000 比特。
文件大小（MB）= 254,016,000 / (8 × 1,000,000) ≈ 31.75 MB。
6. Text Encoding & Character Requirements | 文本编码与字符需求
The capability of a character set to represent distinct characters is entirely determined by the number of bits assigned to encode each character. Increasing the bit length exponentially increases the theoretical number of unique symbols available, allowing the character set to cover more languages and symbols.
字符集表示不同字符的能力完全取决于分配给编码每个字符的位数。增加位长度会按指数级增加可用的理论唯一符号数量，使字符集能够覆盖更多的语言和符号。
The governing theorem is the combinations formula: Total Number of Characters = 2ⁿ, where n is the number of bits per character code.
其支配定理是组合公式：字符总数 = 2ⁿ，其中 n 是每个字符代码的位数。
Standard ASCII uses a 7-bit code, which defines 2⁷ = 128 distinct characters, covering the English alphabet, digits, and basic control codes. Extended ASCII uses an 8-bit code capable of 2⁸ = 256 characters, adding accents and graphical symbols. Modern Unicode encoding
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