📚 Year 9 Edexcel Computer Science: Formula & Theorem Quick Reference Handbook | Year 9 Edexcel 计算机：公式定理速查手册
This quick reference handbook collects all the essential formulas, conversions and logical theorems needed for the Year 9 Edexcel Computer Science course. From binary arithmetic to encryption keys, you will find clear statements and practical examples that make revision quick and efficient. Every topic is presented with a concise English explanation immediately followed by its Chinese equivalent, so you can study in either language or both.
本速查手册汇集了 Year 9 Edexcel 计算机科学课程所需的所有关键公式、转换方法和逻辑定理。从二进制运算到加密密钥，清晰的理论陈述和实用示例将帮助你高效复习。每个主题都配有简洁的英文解释和对应的中文说明，方便你单语或双语学习。
1. Binary to Decimal Conversion | 二进制转十进制
To convert a binary number to decimal, multiply each binary digit (bit) by 2 raised to the power of its position, counting from 0 on the far right. Add all the resulting products.
将二进制数转换为十进制时，将每一位数字（比特）乘以 2 的位权次幂，位权从最右侧的 0 开始计数，最后将所有乘积相加。
Decimal = bₙ₋₁×2ⁿ⁻¹ + bₙ₋₂×2ⁿ⁻² + … + b₁×2¹ + b₀×2⁰
Example: 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀.
示例：1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀。
2. Decimal to Binary Conversion | 十进制转二进制
Repeatedly divide the decimal number by 2, recording the remainder each time. Read the remainders from bottom to top to form the binary equivalent.
将十进制数反复除以 2，每次记录余数。从下往上读取余数，就得到对应的二进制数。
Decimal / 2 → quotient, remainder → continue until quotient = 0
Example: 13₁₀: 13÷2=6 r1, 6÷2=3 r0, 3÷2=1 r1, 1÷2=0 r1 → read upwards: 1101₂.
示例：13₁₀：13÷2=6 余 1，6÷2=3 余 0，3÷2=1 余 1，1÷2=0 余 1 → 从下往上读：1101₂。
3. Hexadecimal Conversions | 十六进制转换
Hexadecimal (base‑16) uses digits 0-9 and letters A-F (A=10, B=11, …, F=15). To convert hex to decimal, multiply each hex digit by 16 raised to its position power, then sum. To convert binary to hex, group bits in sets of four from the right; replace each group with its hex equivalent.
十六进制（基数为 16）使用数字 0-9 和字母 A-F（A=10，B=11，……，F=15）。将十六进制转为十进制时，将每个十六进制位乘以 16 的位权次幂，然后求和。二进制转十六进制时，从右侧开始每 4 个比特一组，用对应的十六进制符号替换每一组。
Hex to Dec: hₙ₋₁×16ⁿ⁻¹ + hₙ₋₂×16ⁿ⁻² + … + h₀×16⁰
Binary to Hex: Group 4 bits → 0…F
Example: 2F₁₆ = 2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀. Binary 11011101₂ → 1101 1101 → D D → DD₁₆.
示例：2F₁₆ = 2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀。二进制 11011101₂ → 1101 1101 → D D → DD₁₆。
4. Data Storage Units | 数据存储单位
Digital data is measured in bits and bytes. A byte is 8 bits. Larger units follow powers of 2 (commonly approximated for exam purposes as 1 KB = 1000 bytes, but true binary units use powers of 1024).
数字数据以比特和字节为单位。一个字节（Byte）等于 8 个比特。更大的单位按照 2 的幂次定义（考试中通常近似为 1 KB = 1000 字节，但真正的二进制单位以 1024 为倍数）。
1 byte = 8 bits
| 1 Kilobyte (KB) | ≈ 1000 bytes (or 1024 bytes) |
| 1 Megabyte (MB) | ≈ 1000 KB |
| 1 Gigabyte (GB) | ≈ 1000 MB |
| 1 Terabyte (TB) | ≈ 1000 GB |
To convert any size, multiply or divide by the appropriate power of 1000 (or 1024 where specified).
转换任意大小时，乘以或除以相应的 1000（或根据题目要求使用 1024）的幂次。
5. Image File Size | 图像文件大小
The file size of a bitmap image depends on its resolution (width × height in pixels) and colour depth (bits per pixel).
位图图像的文件大小取决于分辨率（宽度 × 高度，单位像素）和色深（每像素比特数）。
Image file size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
Image file size (bytes) = (width × height × colour depth) ÷ 8
Example: An 800×600 image with 24-bit colour depth requires 800×600×24 = 11,520,000 bits, which is 11,520,000 ÷ 8 = 1,440,000 bytes (≈1.44 MB).
示例：一张 800×600 像素、24 位色深的图像需要 800×600×24 = 11,520,000 比特，即 11,520,000 ÷ 8 = 1,440,000 字节（约 1.44 MB）。
6. Sound File Size | 音频文件大小
Uncompressed sound file size is determined by sample rate, sample resolution (bit depth), duration in seconds, and number of channels (mono = 1, stereo = 2).
未压缩音频文件的大小取决于采样率、采样分辨率（位深）、时长（秒）以及声道数（单声道 = 1，立体声 = 2）。
Sound file size (bits) = sample rate (Hz) × bit depth × duration (seconds) × number of channels
Sound file size (bytes) = (sample rate × bit depth × duration × channels) ÷ 8
Example: 30 seconds of CD-quality stereo audio (44,100 Hz, 16-bit, 2 channels) = 44,100 × 16 × 30 × 2 = 42,336,000 bits ÷ 8 = 5,292,000 bytes (≈5.29 MB).
示例：30 秒 CD 质量的立体声音频（44,100 Hz、16 位、2 声道）= 44,100 × 16 × 30 × 2 = 42,336,000 比特，除以 8 后为 5,292,000 字节（约 5.29 MB）。
7. Data Transfer Time | 数据传输时间
When downloading or uploading data across a network, the time required is the total amount of data divided by the transfer rate. Ensure units match: convert everything to bits or bytes before dividing.
通过网络下载或上传数据时，所需时间为数据总量除以传输速率。务必统一单位：在相除之前将所有值都转换为比特或字节。
Time (seconds) = Data size (bits) ÷ Transfer rate (bits per second)
Example: A 100 MB file transferred at 50 Mbps: 100 MB = 100 × 8 = 800 megabits. Time = 800 Mb ÷ 50 Mbps = 16 seconds.
示例：传输一个 100 MB 的文件，传输速率为 50 Mbps：100 MB = 100 × 8 = 800 兆比特。时间 = 800 Mb ÷ 50 Mbps = 16 秒。
8. Boolean Algebra Theorems | 布尔代数定理
Boolean algebra is used to simplify logic expressions in digital circuits. The following theorems are fundamental for Year 9 Edexcel: Identity, Annulment, Idempotent, Complement, and Commutative laws.
布尔代数用于简化数字电路中的逻辑表达式。以下定理是 Year 9 Edexcel 课程的基础：同一律、零一律、幂等律、互补律和交换律。
Identity: A + 0 = A, A · 1 = A
同一律： A + 0 = A，A · 1 = A
Annulment: A + 1 = 1, A · 0 = 0
零一律： A + 1 = 1，A · 0 = 0
Idempotent: A + A = A, A · A = A
幂等律： A + A = A，A · A = A
Complement: A + ¬A = 1, A · ¬A = 0
互补律： A + ¬A = 1，A · ¬A = 0
Commutative: A + B = B + A, A · B = B · A
交换律： A + B = B + A，A · B = B · A
De Morgan’s laws are also introduced at this level:
本阶段也引入了德摩根定律：
¬(A · B) = ¬A + ¬B
¬(A + B) = ¬A · ¬B
9. Logic Gate Truth Tables | 逻辑门真值表
Every logic gate has a fixed truth table that defines its output for all possible inputs. The three basic gates are AND, OR and NOT.
每个逻辑门都有固定的真值表，规定了在所有可能输入下的输出。三种基本逻辑门是与门（AND）、或门（OR）和非门（NOT）。
AND gate (A · B): Output is 1 only when both inputs are 1.
与门（A · B）： 仅当两个输入都为 1 时输出为 1。
| A | B | Output |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
OR gate (A + B): Output is 1 if at least one input is 1.
或门（A + B）： 只要有任意一个输入为 1，输出即为 1。
| A | B | Output |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
NOT gate (¬A): Output is the inverse of the input.
非门（¬A）： 输出是输入的反相。
| A | Output |
| 0 | 1 |
| 1 | 0 |
These truth tables are the building blocks for all digital logic design.
这些真值表是所有数字逻辑设计的构建模块。
10. Search Algorithms | 搜索算法
Two searching algorithms are studied at this level: linear search and binary search. Linear search examines each element one by one, taking at most n comparisons for a list of n items. Binary search works on a sorted list, repeatedly dividing the search interval in half.
本阶段学习两种搜索算法：线性搜索和二分搜索。线性搜索逐个检查元素，对一个包含 n 个元素的列表最多进行 n 次比较。二分搜索在有序列表上进行，反复将搜索区间一分为二。
Linear search maximum comparisons = n
Binary search maximum comparisons = ⌈log₂(n + 1)⌉
For example, searching through 1000 sorted items with binary search takes at most 10 comparisons because 2¹⁰ = 1024; with linear search it could take 1000.
例如，在 1000 个有序项中使用二分搜索最多需要 10 次比较（因为 2¹⁰ = 1024），而使用线性搜索可能需要 1000 次比较。
11. Sorting Algorithms | 排序算法
Bubble sort is a simple algorithm that repeatedly steps through the list, compares adjacent items and swaps them if they are in the wrong order. In the worst case, the number of comparisons is given by the formula below.
冒泡排序是一种简单的算法，它反复遍历列表，比较相邻项并在顺序错误时交换它们。在最坏情况下，比较次数由以下公式给出。
Worst-case comparisons = n(n – 1) ÷ 2
If a list has 5 elements, the maximum number of comparisons is (5×4)÷2 = 10. Each pass reduces the number of unsorted elements by one.
如果列表有 5 个元素，最大比较次数为 (5×4)÷2 = 10。每一趟扫描都会使未排序元素的数量减少一个。
12. Caesar Cipher Encryption | 凯撒加密法
The Caesar cipher shifts each letter in the plaintext by a fixed number of positions along the alphabet. The shift key k is an integer between 1 and 25. Decryption shifts in the opposite direction.
凯撒密码将明文中的每个字母沿字母表移动固定的位数。移位密钥 k 是 1 到 25 之间的整数。解密则向相反方向移动。
Encryption: E(x) = (x + k) mod 26
Decryption: D(y) = (y − k) mod 26
Here x is the position of the plaintext letter (A=0, B=1, …, Z=25), and mod 26 ensures the result wraps around the alphabet. Example with k=3: ‘A’ becomes ‘D’, ‘X’ becomes ‘A’.
其中 x 是明文字母的位置（A=0，B=1，……，Z=25），mod 26 确保结果在字母表中循环。以 k=3 为例：’A’ 变为 ‘D’，’X’ 变为 ‘A’。
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