📚 Year 8 Cambridge Computer Science Formulas & Theorems Quick Reference | 剑桥八年级计算机科学公式定理速查手册
This quick reference handbook compiles the essential formulas, conversion rules, and fundamental theorems required for the Year 8 Cambridge Computer Science curriculum. Use it to revise data representation, logic, Boolean algebra, file sizes, and transmission calculations.
这本速查手册汇编了剑桥八年级计算机科学课程所需的关键公式、转换规则和基础定理。可用于复习数据表示、逻辑、布尔代数、文件大小与传输计算等内容。
1. Data Units and Conversions | 数据单位与换算
Computers store information in binary digits (bits). A group of 8 bits is called a byte. Larger units are defined as powers of 2, not powers of 10.
计算机以二进制位（bit）存储信息。8 个位组成 1 个字节（byte）。更大的单位按照 2 的幂次定义，而不是 10 的幂次。
1 nibble = 4 bits
1 byte = 8 bits
1 KB (kibibyte) = 2¹⁰ bytes = 1024 bytes
1 MB (mebibyte) = 2²⁰ bytes = 1,048,576 bytes
1 GB = 2³⁰ bytes
When converting between units, multiply or divide by 1024. For example, to convert 2048 KB to MB, divide by 1024: 2048 / 1024 = 2 MB.
进行单位换算时，乘以或除以 1024。例如，把 2048 KB 转换为 MB，除以 1024：2048 / 1024 = 2 MB。
2. Binary to Denary Conversion | 二进制转十进制
Each binary digit (bit) has a place value based on powers of 2, starting from 2⁰ on the right. Sum all place values where the bit is 1.
每个二进制位根据 2 的幂次具有不同的位权，从最右侧的 2⁰ 开始。将所有出现 1 的位权相加即可得到十进制值。
Denary value = Σ (bitₙ × 2ⁿ)
ple: binary 10110₂ = 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22 in denary.
示例：二进制 10110₂ = 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22（十进制）。
3. Denary to Binary Conversion | 十进制转二进制
The most reliable method is repeated division by 2. Write down the remainder at each step; the binary number is the remainders read from bottom to top.
最可靠的方法是不断除以 2 取余数。每一步记录余数，从下往上读取余数即为二进制数。
Method: Divide the denary number by 2. Record the remainder (0 or 1). Continue dividing the quotient until it becomes 0. The binary result is the remainders read in reverse order.
方法：将十进制数除以 2，记下余数（0 或 1）。继续用商除以 2，直到商为 0。将余数倒序读出即得到二进制结果。
11 ÷ 2 = 5 rem 1; 5 ÷ 2 = 2 rem 1; 2 ÷ 2 = 1 rem 0; 1 ÷ 2 = 0 rem 1. Read upwards: 10110₂.
示例：将 22 转为二进制。22 ÷ 2 = 11 余 0；11 ÷ 2 = 5 余 1；5 ÷ 2 = 2 余 1；2 ÷ 2 = 1 余 0；1 ÷ 2 = 0 余 1。从下往上读：10110₂。
4. Hexadecimal Overview | 十六进制简介
Hexadecimal (base 16) is a shorthand for binary. Each hex digit represents exactly 4 bits (a nibble). Digits 10–15 are written as A, B, C, D, E, F.
十六进制（基数为 16）是二进制的一种缩写形式。每个十六进制数正好代表 4 个二进制位（一个半字节）。数字 10–15 用 A、B、C、D、E、F 表示。
Binary → Hex: Group bits into nibbles from the right, convert each nibble to a hex digit.
Hex → Binary: Replace each hex digit with its 4‑bit binary equivalent.
11010₂ → split into 1101 (D) and 1010 (A) → hex DA. Hex 2F → 0010 1111 → binary 00101111₂.
示例：二进制 11011010₂ → 拆分为 1101（D）和 1010（A）→ 十六进制 DA。十六进制 2F → 0010 1111 → 二进制 00101111₂。
5. Image File Size Calculation | 图像文件大小计算
The size of a bitmap image is determined by its resolution (pixel dimensions) and colour depth (bits used per pixel to represent colour).
位图图像的大小由分辨率（像素尺寸）和位深（每个像素表示颜色所用的位数）决定。
Image size (bits) = width (px) × height (px) × colour depth (bpp)
Image size (bytes) = (width × height × colour depth) / 8
with 24-bit true colour uses 200 × 300 × 24 = 1,440,000 bits, or 180,000 bytes (≈ 176 KB).
例如，一幅 200 × 300 像素、24 位真彩色图像需要 200 × 300 × 24 = 1,440,000 位，即 180,000 字节（约 176 KB）。
Common colour depths: 1 bit (black & white), 8 bits (256 colours), 24 bits (16.7 million colours). Metadata (headers) may add a few extra bytes.
常见位深：1 位（黑白）、8 位（256 色）、24 位（1670 万色）。元数据（文件头）可能会使文件再增加一些字节。
6. Sound File Size Calculation | 声音文件大小计算
Digital audio is captured by sampling the sound wave at regular intervals. The file size depends on sample rate, bit depth, channels, and duration.
数字音频通过对声波进行等间隔采样获得。文件大小取决于采样率、采样比特深度、声道数和时长。
e (bits) = sample rate (Hz) × sample resolution (bits) × number of channels × duration (s)
File size (bytes) = (sample rate × resolution × channels × duration) / 8
100 Hz with 16-bit resolution is 44,100 × 16 × 2 × 10 = 14,112,000 bits, or about 1.68 MB.
示例：一段 10 秒的立体声录音，采样率 44,100 Hz，16 位分辨率，文件大小为 44,100 × 16 × 2 × 10 = 14,112,000 位，约 1.68 MB。
Standard CD quality uses 44.1 kHz, 16 bits, 2 channels. Telephone quality often uses 8 kHz, 8 bits, 1 channel.
标准 CD 音质使用 44.1 kHz、16 位、2 声道。电话音质通常使用 8 kHz、8 位、单声道。
7. Transmission Time Calculation | 传输时间计算
When sending data over a network, the time taken equals the total amount of data divided by the transmission speed (bit rate). Always match units.
通过网络传输数据时，耗时等于数据总量除以传输速度（比特率）。务必统一单位。
Time (seconds) = File size (bits) / Bit rate (bits per second, bps)
: 10 × 1024 × 1024 × 8 = 83,886,080 bits. Then time = 83,886,080 / 2,000,000 ≈ 42 seconds.
如果通过 2 Mbps 连接传输一个 10 MB 的文件，先把 10 MB 转为位：10 × 1024 × 1024 × 8 = 83,886,080 位。传输时间 = 83,886,080 / 2,000,000 ≈ 42 秒。
8. Logic Gates and Truth Tables | 逻辑门与真值表
Logic gates are the building blocks of digital circuits. The Year 8 syllabus covers NOT, AND, OR, XOR, NAND, and NOR gates. Each gate’s behaviour is defined by its truth table.
逻辑门是数字电路的基础构件。八年级课程涵盖 NOT、AND、OR、XOR、NAND 和 NOR 门。每种门的行为都通过真值表来定义。
NOT gate (inverter): Output is the opposite of the input.
NOT 门（反相器）：输出与输入相反。
| A | Q = ¬A |
|---|---|
| 0 | 1 |
| 1 | 0 |
AND gate: Output is 1 only if all inputs are 1.
AND 门：仅当所有输入都为 1 时，输出才是 1。
| A | B | A·B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
OR gate: Output is 1 if at least one input is 1.
OR 门：只要至少有一个输入为 1，输出就是 1。
| A | B | A+B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
XOR gate (exclusive OR): Output is 1 when inputs are different.
XOR 门（异或）：当输入不相同时输出 1。
| A | B | A⊕B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
NAND gate: Inverse of AND; output 0 only when all inputs are 1.
NAND 门：是 AND 的非；仅当所有输入都是 1 时输出 0。
NOR gate: Inverse of OR; output 1 only when all inputs are 0.
NOR 门：是 OR 的非；仅当所有输入都是 0 时输出 1。
9. Boolean Algebra Laws | 布尔代数定律
Boolean algebra allows simplification of logic expressions. The key identities and laws are essential for circuit design and problem solving.
布尔代数可以用来简化逻辑表达式。核心恒等式和定律对电路设计和解题至关重要。
- 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)
- Identity Law / 同一律: A + 0 = A ; A · 1 = A
- Annulment Law / 零一律: A + 1 = 1 ; A · 0 = 0
- Idempotent Law / 幂等律: A + A = A ; A · A = A
- Complement Law / 互补律: A + ¬A = 1 ; A · ¬A = 0
- Double Negation / 双重否定: ¬(¬A) = A
- De Morgan's Theorems / 德摩根定理: ¬(A · B) = ¬A + ¬B ; ¬(A + B) = ¬A · ¬B
These laws are used to reduce the number of gates in a digital circuit, saving cost and power.
这些定律用于减少数字电路中的门数量，从而降低成本和功耗。
10. Data Compression Concepts | 数据压缩概念
Compression reduces file size. Two main types exist: lossless compression allows exact reconstruction of original data; lossy compression discards some information permanently to achieve greater reduction.
压缩可以减小文件大小。主要分为两类：无损压缩能精确还原原始数据；有损压缩会永久丢弃部分信息以获得更显著的压缩效果。
Lossless examples: Run-length encoding (RLE), where repeated values are stored as count and value (e.g., 'AAAABBB' → '4A3B'). Huffman coding and LZW are also used in ZIP files and GIF images.
无损压缩示例：行程长度编码（RLE），用计数值和字符代替重复值（如 'AAAABBB' → '4A3B'）。霍夫曼编码和 LZW 编码常用于 ZIP 文件和 GIF 图像。
Lossy examples: JPEG for photos (discards colour details not easily seen by human eye), MP3 for audio (removes frequencies outside normal hearing range).
有损压缩示例：JPEG 用于照片（丢弃人眼不易察觉的颜色细节），MP3 用于音频（移除可听范围以外的频率）。
l size / compressed size. A ratio of 5:1 means the compressed file is 1/5 of the original.
压缩比 = 原始大小 / 压缩后大小。比值为 5:1 表示压缩后文件仅为原始文件的五分之一。
11. Error Detection Methods | 错误检测方法
When data is transmitted or stored, errors can occur. Simple error detection methods help identify if data has been corrupted, though they cannot usually fix it.
数据在传输或存储时可能发生错误。简单的错误检测方法可以帮助识别数据是否损坏，但通常无法修复错误。
Parity bit: An extra bit is added to a binary string to make the number of 1s either even (even parity) or odd (odd parity). The receiver counts the 1s; if the parity does not match, an error is detected.
奇偶校验位：在二进制串中加入一个额外位，使得 1 的总数为偶数（偶校验）或奇数（奇校验）。接收方统计 1 的个数，如果校验结果不匹配，就检测到了错误。
transmitted 01011000. If a single bit flips, the number of 1s becomes odd, flagging an error.
偶校验示例：数据 1011000 有 4 个 1（偶数），校验位设为 0，发送 01011000。如果有一位翻转，1 的个数变为奇数，从而报错。
Checksum: All data bytes are summed using a chosen modulus (often 256). The low-order byte of the sum is appended. The receiver performs the same sum and compares.
校验和：将所有数据字节按所选模数（通常为 256）求和，取低字节附加在后面。接收方做同样的求和并比对。
Parity can detect any odd number of bit errors; checksums can detect many common error patterns. These concepts form the basis of more advanced error correction codes.
奇偶校验可以检测到任意奇数个位错误；校验和能检测许多常见的错误模式。这些概念为更高级的纠错码奠定了基础。
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