📚 Year 11 Eduqas Computer Science: Quick-Reference Formulas & Theorems | 公式定理速查手册
This concise handbook brings together every essential formula, conversion rule, Boolean law and algorithmic theorem specified in the Year 11 Eduqas Computer Science syllabus. Use it to revise storage calculations, data representation, logic gates, compression ratios and search performance boundaries — all presented in a clear, exam‑ready structure with paired English‑Chinese explanations.
这本精炼的手册汇集了 Year 11 Eduqas 计算机科学大纲中的每一个必备公式、转换规则、布尔定律和算法定理。你可以用它来复习存储计算、数据表示、逻辑门、压缩比率和搜索性能边界，所有内容均以清晰的结构呈现，并配有中英双语解释，便于应考。
1. Storage Units & Conversions | 存储单位与转换
In computer science, storage capacity is measured in bits and bytes. One byte is exactly 8 bits. To convert between larger units, powers of 2 are applied — 1 kilobyte (KB) is 2¹⁰ bytes, not 1000 bytes. For Eduqas examinations, always use binary prefixes unless a question explicitly states otherwise.
在计算机科学中，存储容量以位和字节为单位。1 字节严格等于 8 位。在较大单位之间转换时使用 2 的幂——1 千字节 (KB) 是 2¹⁰ 字节，而不是 1000 字节。在 Eduqas 考试中，除非题目明确说明，否则始终使用二进制前缀。
- 1 byte = 8 bits | 1 字节 = 8 位
- 1 KB = 1024 bytes (2¹⁰) | 1 KB = 1024 字节
- 1 MB = 1024 KB (2²⁰ bytes) | 1 MB = 1024 KB
- 1 GB = 1024 MB (2³⁰ bytes) | 1 GB = 1024 MB
- 1 TB = 1024 GB (2⁴⁰ bytes) | 1 TB = 1024 GB
number of bits = number of bytes × 8
位 数 = 字节数 × 8
2. Image File Size | 图像文件大小
The file size of a bitmap image depends on its pixel dimensions and colour depth. Colour depth is the number of bits used to represent the colour of each pixel. The raw size in bits is found by multiplying width, height and colour depth.
位图图像的文件大小取决于像素尺寸和色彩深度。色彩深度是用来表示每个像素颜色的位数。以位为单位的原始大小由宽度、高度和色彩深度相乘得到。
image size (bits) = width (pixels) × height (pixels) × colour depth (bpp)
图像大小 (位) = 宽度 (像素) × 高度 (像素) × 色彩深度 (位/像素)
To express the size in bytes, divide by 8. For example, a 1920×1080 image with 24‑bit colour requires 1920 × 1080 × 24 = 49,766,400 bits ≈ 6,220,800 bytes ≈ 5.93 MB.
要以字节为单位表示大小，则除以 8。例如，一幅 1920×1080 的 24 位彩色图像需要 1920 × 1080 × 24 = 49 766 400 位 ≈ 6 220 800 字节 ≈ 5.93 MB。
3. Sound File Size | 声音文件大小
Uncompressed audio file size is calculated from the sample rate, bit depth, duration and number of channels. Sample rate is measured in hertz (Hz) and bit depth in bits per sample.
未压缩音频文件大小由采样率、位深度、时长和声道数计算得出。采样率以赫兹 (Hz) 为单位，位深度以位/样本为单位。
audio size (bits) = sample rate (Hz) × bit depth × duration (s) × channels
音频大小 (位) = 采样率 (Hz) × 位深度 × 时长 (秒) × 声道数
CD‑ channels) for one minute uses 44,100 × 16 × 60 × 2 = 84,672,000 bits, i.e. about 10.09 MB.
一分钟 CD 品质立体声音频 (44 100 Hz、16 位、2 声道) 使用 44 100 × 16 × 60 × 2 = 84 672 000 位，约 10.09 MB。
4. Video File Size | 视频文件大小
A video is effectively a sequence of still images displayed at a certain frame rate, often accompanied by audio. The uncompressed size of the video stream alone can be calculated by extending the image size formula.
视频实际上是以特定帧率显示的一系列静止图像，通常还伴有音频。仅视频流的未压缩大小可以通过扩展图像大小公式来计算。
video size (bits) = width × height × colour depth × frame rate × duration (s)
视频大小 (位) = 宽度 × 高度 × 色彩深度 × 帧率 × 时长 (秒)
and 30 fps lasting 10 seconds has 1280 × 720 × 24 × 30 × 10 = 6,635,520,000 bits ≈ 791 MB.
如果包含音频，则加上音频大小。例如，一段 720p 视频，24 bpp、30 fps，时长 10 秒，大小为 1280 × 720 × 24 × 30 × 10 = 6 635 520 000 位 ≈ 791 MB。
5. Data Transfer Time | 数据传输时间
The time needed to send or download a file is determined by its size and the network transfer rate. It is vital to ensure both quantities share the same unit — normally bits.
发送或下载文件所需的时间由文件大小和网络传输速率决定。确保两者使用相同的单位（通常为位）至关重要。
transfer time (s) = data size (bits) / transfer rate (bps)
传输时间 (秒) = 数据大小 (位) / 传输速率 (bps)
100 × 10⁶) ≈ 4.19 seconds (using binary MB with 1 Mbps = 1,000,000 bps as per convention).
当网络速度以 Mbps 给定时，务必将兆字节转换为位 (× 8 × 1024²)。一个 50 MB 的文件通过 100 Mbps 连接大约需要 (50 × 8 × 1024²) / (100 × 10⁶) ≈ 4.19 秒（使用二进制 MB，并按照惯例 1 Mbps = 1 000 000 bps）。
6. Binary Arithmetic & Negative Numbers | 二进制算术与负数表示
Computers represent signed integers using two's complement. This method allows the same binary addition circuitry to handle subtraction. The range of an n‑bit two's complement number is from −2ⁿ⁻¹ to 2ⁿ⁻¹ − 1.
计算机使用二进制补码表示有符号整数。这种方法使相同的二进制加法电路也能处理减法。n 位二进制补码数的范围是从 −2ⁿ⁻¹ 到 2ⁿ⁻¹ − 1。
Two's complement conversion rule: To negate a number, flip all bits and add 1. The most significant bit (MSB) indicates the sign — 1 for negative, 0 for non‑negative.
二进制补码转换规则： 对一个数取反（求其负数），将所有位取反然后加 1。最高有效位 (MSB) 表示符号——1 为负，0 为非负。
For an 8‑bit register, −1 is represented as 11111111₂ and +127 as 01111111₂. The sum of a number and its two's complement always yields zero (ignoring overflow).
对于 8 位寄存器，−1 表示为 11111111₂，+127 表示为 01111111₂。一个数与其补码之和总是得到零（忽略溢出）。
7. Hexadecimal & Binary Conversions | 十六进制与二进制转换
Hexadecimal (base‑16) provides a compact way to write binary strings. Each hex digit corresponds to exactly four bits, making conversion straightforward and error‑free.
十六进制 (基数 16) 提供了一种书写二进制串的简洁方式。每个十六进制数字精确对应四位二进制，使得转换直接且不易出错。
| Hex | 十六进制 | Binary | 二进制 | Decimal | 十进制 |
|---|---|---|
| 0 | 0000 | 0 |
| 1 | 0001 | 1 |
| A | 1010 | 10 |
| F | 1111 | 15 |
Conversion method: group binary digits into nibbles (4 bits) starting from the right, then replace each nibble with its hex equivalent.
转换方法：从右向左将二进制位每 4 位分为一组（半字节），然后将每个半字节替换为对应的十六进制数字。
8. Boolean Algebra Laws & De Morgan’s Theorems | 布尔代数定律与德摩根定理
Boolean algebra simplifies logic circuits and expressions. Mastery of its laws is essential for reducing gate count and evaluating logical equivalence in the Eduqas exam.
布尔代数可以简化逻辑电路和表达式。掌握其定律对于在 Eduqas 考试中减少门用量和评估逻辑等价性至关重要。
- Commutative Law: A + B = B + A; A · B = B · A | 交换律
- Associative Law: A + (B + C) = (A + B) + C; A · (B · C) = (A · B) · C | 结合律
- Distributive Law: A · (B + C) = A · B + A · C; A + (B · C) = (A + B) · (A + C) | 分配律
- Identity Law: A + 0 = A; A · 1 = A | 同一律
- Complement Law: A + ¬A = 1; A · ¬A = 0 | 互补律
- Idempotent Law: A + A = A; A · A = A | 幂等律
- Absorption Law: A + (A · B) = A; A · (A + B) = A | 吸收律
De Morgan’s Theorem 1: ¬(A · B) = ¬A + ¬B
德摩根定理 1: 非 (A 与 B) = 非 A 或 非 B
De Morgan’s Theorem 2: ¬(A + B) = ¬A · ¬B
德摩根定理 2: 非 (A 或 B) = 非 A 与 非 B
These theorems are invaluable for transforming NAND‑only or NOR‑only circuits and for simplifying conditions in programming.
这些定理在转换为纯与非门电路或纯或非门电路以及简化编程条件时都非常宝贵。
9. Logic Gates & Truth Tables | 逻辑门与真值表
Logic gates are the building blocks of digital circuits. The six fundamental gates each have a unique truth table and Boolean expression. Drawing and interpreting them is a core skill.
逻辑门是数字电路的基本组件。六种基本门各有唯一的真值表和布尔表达式。绘制和解读它们是一项核心技能。
| Gate | 门 | Boolean Expression | 布尔表达式 | Symbol (shape) | 符号 (形状) | Truth Table (A,B → Q) | 真值表 (A,B → Q) |
|---|---|---|---|
| NOT | Q = ¬A | Triangle + bubble | 0 → 1; 1 → 0 |
| AND | Q = A · B | D‑shaped | 00→0, 01→0, 10→0, 11→1 |
| OR | Q = A + B | Pointed shield | 00→0, 01→1, 10→1, 11→1 |
| NAND | Q = ¬(A · B) | AND + bubble | 00→1, 01→1, 10→1, 11→0 |
| NOR | Q = ¬(A + B) | OR + bubble | 00→1, 01→0, 10→0, 11→0 |
| XOR | Q = A ⊕ B | OR with extra curve | 00→0, 01→1, 10→1, 11→0 |
NAND and NOR gates are functionally complete, meaning any Boolean function can be implemented using only NAND or only NOR gates — a fact frequently tested in logic circuit design questions.
与非门和或非门具有功能完备性，这意味着任何布尔函数都可以仅用与非门或仅用或非门实现——这一事实在逻辑电路设计题中经常考查。
10. Compression Ratios | 压缩比率
Compression reduces file size by encoding data more efficiently. The compression ratio compares original and compressed sizes and is a key metric when evaluating lossless and lossy techniques.
压缩通过更高效地编码数据来减小文件大小。压缩比用于比较原始大小和压缩后大小，是评估无损和有损技术时的一个关键指标。
compression ratio = uncompressed size / compressed size
压缩比 = 未压缩大小 / 压缩后大小
Alternatively, the space saving percentage can be calculated:
此外，还可以计算空间节省百分比：
space saving (%) = [(uncompressed − compressed) / uncompressed] × 100
空间节省 (%) = [(未压缩 − 压缩后) / 未压缩] × 100
For instance, a 40 MB image compressed to 10 MB yields a compression ratio of 4:1 and a space saving of 75%.
例如，一张 40 MB 的图像压缩到 10 MB，压缩比为 4:1，空间节省率为 75%。
11. Search Algorithms – Maximum Comparisons | 搜索算法 – 最大比较次数
When analysing algorithm efficiency, the worst‑case number of comparisons is a useful metric. For an ordered list of size n, the linear search and binary search behave very differently as n grows.
在分析算法效率时，最坏情况下的比较次数是一个有用的度量。对于大小为 n 的有序列表，线性搜索和二分搜索的性能随着 n 的增长而差异显著。
Linear Search: In the worst case, the target is at the end or not present, requiring exactly n comparisons.
线性搜索：在最坏情况下，目标在末尾或不存在，需要精确比较 n 次。
Binary Search: The maximal number of comparisons is proportional to log₂(n). The exact worst‑case count for a list of n elements is:
二分搜索：最大比较次数与 log₂(n) 成正比。对于 n 个元素的列表，确切的最坏情况次数为：
max comparisons = ⌈log₂(n+1)⌉
最大比较次数 = ⌈log₂(n+1)⌉
ch needs up to 1000 comparisons, while binary search needs at most 10 comparisons, illustrating the dramatic speed‑up on sorted data.
另一种常用公式为 floor(log₂(n)) + 1。当 n = 1000 时，线性搜索最多需要 1000 次比较，而二分搜索最多只需要 10 次，这展示了在已排序数据上速度的显著提升。
Published by TutorHao | Computer Science Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图