📚 Year 9 CAIE Computer Science: Formula & Theorem Quick Reference Handbook | Year 9 CAIE 计算机：公式定理速查手册
Welcome to your quick-reference guide for Year 9 CAIE Computer Science. This handbook compiles all the essential formulas, theorems, and conversion rules you will encounter throughout the course. Whether you are revising binary arithmetic, calculating file sizes, or simplifying logic circuits, having these formulas at your fingertips will save you time and boost your confidence in the exam. Each section pairs a concise explanation in English with a matching Chinese version, ensuring you grasp every concept thoroughly.
欢迎使用 Year 9 CAIE 计算机科学速查手册。本手册汇集了课程中所有关键的公式、定理和换算规则。无论你正在复习二进制算术、计算文件大小还是化简逻辑电路，随时查阅这些公式都能帮你节省时间并提升考试信心。每个小节都配有简洁的中英文对照解释，确保你能透彻掌握每一个概念。
1. Binary to Decimal Conversion | 二进制与十进制转换
To convert a binary number to decimal, multiply each bit by 2 raised to the power of its position (starting from 0 on the right). Sum all the products to obtain the decimal equivalent.
要将二进制数转换为十进制，将每一位数字乘以 2 的位权次方（从右侧的 0 次方开始）。将所有乘积相加即得到十进制等值。
Decimal = bₙ × 2ⁿ + ... + b₂ × 2² + b₁ × 2¹ + b₀ × 2⁰
Example: Convert 1011₂ to decimal.
示例：将 1011₂ 转换为十进制。
- 1 × 2³ = 8
- 0 × 2² = 0
- 1 × 2¹ = 2
- 1 × 2⁰ = 1
Sum = 8 + 0 + 2 + 1 = 11₁₀
总和 = 8 + 0 + 2 + 1 = 11₁₀
2. Decimal to Binary Conversion | 十进制转二进制
To convert a decimal number to binary, repeatedly divide the number by 2 and record the remainders. Read the remainders from bottom to top to form the binary equivalent.
要将十进制数转换为二进制，反复将该数除以 2 并记录余数。从下往上读取余数，即为二进制等值。
Example: Convert 25₁₀ to binary.
示例：将 25₁₀ 转换为二进制。
2 = 6 remainder 0 → 6 ÷ 2 = 3 remainder 0 → 3 ÷ 2 = 1 remainder 1 → 1 ÷ 2 = 0 remainder 1
Reading from bottom: 11001₂
从下往上读取：11001₂
3. Binary Addition Rules | 二进制加法法则
Binary addition follows four fundamental rules. When the sum of bits exceeds 1, a carry is generated to the next column on the left.
二进制加法遵循四条基本规则。当各位之和超过 1 时，会向左侧下一列产生进位。
- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 = 0 (carry 1 to the next column)
- 1 + 1 + 1 (with carry) = 1 (carry 1 to the next column)
Overflow: When adding two 8-bit numbers produces a result requiring 9 bits, an overflow error occurs because the register can only store 8 bits.
溢出：当两个 8 位二进制数相加产生需要 9 位的结果时，由于寄存器只能存储 8 位，就会发生溢出错误。
4. Logical Shift Operations | 逻辑移位操作
Logical shifts move all bits in a binary number to the left or right. Vacant positions are filled with zeros, and bits shifted beyond the register are discarded.
逻辑移位将二进制数中的所有位向左或向右移动。空出的位置用零填充，移出寄存器范围的位被丢弃。
Left shift (multiplication): Shifting left by n places multiplies the number by 2ⁿ.
左移（乘法）：左移 n 位相当于将数值乘以 2ⁿ。
Value after left shift by n = Original value × 2ⁿ
左移 n 位后的值 = 原值 × 2ⁿ
Right shift (division): Shifting right by n places divides the number by 2ⁿ (integer division, discarding the remainder).
右移（除法）：右移 n 位相当于将数值除以 2ⁿ（整数除法，舍弃余数）。
Value after right shift by n = Original value ÷ 2ⁿ
右移 n 位后的值 = 原值 ÷ 2ⁿ
5. Logic Gates and Truth Tables | 逻辑门与真值表
Logic gates are the building blocks of digital circuits. Each gate implements a Boolean function. Below are the six fundamental gates you must memorise for the CAIE syllabus, along with their truth tables and Boolean expressions.
逻辑门是数字电路的基本构件。每种门实现一个布尔函数。以下是 CAIE 考纲要求掌握的六个基本逻辑门及其真值表和布尔表达式。
| Gate | Symbol | Expression | Truth Table Summary |
| NOT | Triangle + bubble | Q = NOT A | A=0 → Q=1; A=1 → Q=0 |
| AND | D-shape | Q = A AND B | Q=1 only when A=1 and B=1 |
| OR | Curved shape | Q = A OR B | Q=0 only when A=0 and B=0 |
| NAND | AND + bubble | Q = NOT (A AND B) | Q=0 only when A=1 and B=1 |
| NOR | OR + bubble | Q = NOT (A OR B) | Q=1 only when A=0 and B=0 |
| XOR | OR with extra curve | Q = A XOR B | Q=1 when A and B differ |
6. Boolean Algebra Laws | 布尔代数定律
Boolean algebra provides rules for simplifying logic expressions. Mastering these laws helps you reduce the number of gates in a circuit, making it more efficient.
布尔代数提供化简逻辑表达式的规则。掌握这些定律有助于减少电路中的门数量，使其更加高效。
| Law | AND Form | OR Form |
| Identity | A AND 1 = A | A OR 0 = A |
| Null | A AND 0 = 0 | A OR 1 = 1 |
| Idempotent | A AND A = A | A OR A = A |
| Complement | A AND (NOT A) = 0 | A OR (NOT A) = 1 |
| Commutative | A AND B = B AND A | A OR B = B OR A |
| Associative | (A AND B) AND C = A AND (B AND C) | (A OR B) OR C = A OR (B OR C) |
| Distributive | A AND (B OR C) = (A AND B) OR (A AND C) | A OR (B AND C) = (A OR B) AND (A OR C) |
| Double Negation | NOT (NOT A) = A | |
De Morgan's Theorems are essential for circuit simplification:
德摩根定理对于电路化简至关重要：
NOT (A AND B) = (NOT A) OR (NOT B)
NOT (A OR B) = (NOT A) AND (NOT B)
7. Storage Units Conversion | 存储单位换算
Computer storage is measured in bytes. All units in the syllabus use powers of 2. Memorising these relationships is vital for answering data-storage questions accurately.
计算机存储以字节为单位。考纲中的所有单位均采用 2 的幂次。牢记这些换算关系对于准确回答数据存储类题目至关重要。
| Unit | Abbreviation | Equivalent in bytes |
| kilobyte | KB | 1024 bytes (2¹⁰) |
| megabyte | MB | 1024 KB = 1,048,576 bytes (2²⁰) |
| gigabyte | GB | 1024 MB = 1,073,741,824 bytes (2³⁰) |
| terabyte | TB | 1024 GB = 1,099,511,627,776 bytes (2⁴⁰) |
Bit to byte: = 8 bits. To convert bits to bytes, divide by 8; to convert bytes to bits, multiply by 8.
位与字节：1 字节 = 8 位。将位转换为字节需要除以 8；将字节转换为位需要乘以 8。
Bytes = Bits ÷ 8 | Bits = Bytes × 8
字节 = 位 ÷ 8 | 位 = 字节 × 8
8. Data Transfer Rate | 数据传输速率
The time required to transfer a file depends on the file size and the transfer rate (bandwidth). Use the formula below to calculate download time. Make sure your units are consistent — convert everything to bits before calculating.
传输文件所需的时间取决于文件大小和传输速率（带宽）。使用以下公式计算下载时间。请确保单位一致——计算前将所有数据换算为位。
Transfer Time (seconds) = File Size (bits) ÷ Transfer Rate (bits per second)
传输时间（秒）= 文件大小（位）÷ 传输速率（位/秒）
Example: ansfer rate = 2 × 1,000,000 = 2,000,000 bps. Time = 83,886,080 ÷ 2,000,000 ≈ 41.9 seconds.
示例：一个 10 MB 的文件以 2 Mbps 的速率传输。首先换算：10 MB = 10 × 1024 × 1024 × 8 = 83,886,080 位。传输速率 = 2 × 1,000,000 = 2,000,000 bps。时间 = 83,886,080 ÷ 2,000,000 ≈ 41.9 秒。
9. Image File Size Calculation | 图像文件大小计算
on the image resolution (width × height) and the colour depth (number of bits per pixel).
位图图像分别存储每个像素的颜色。未压缩位图文件的大小取决于图像分辨率（宽 × 高）和颜色深度（每像素位数）。
Image File Size (bits) = Width × Height × Colour Depth
图像文件大小（位）= 宽度 × 高度 × 颜色深度
bit = 2 colours, 2 bits = 4 colours, 4 bits = 16 colours, 8 bits = 256 colours, 24 bits = ~16.7 million colours (True Colour).
颜色深度决定可用的颜色数量：1 位 = 2 种颜色，2 位 = 4 种颜色，4 位 = 16 种颜色，8 位 = 256 种颜色，24 位 ≈ 1670 万种颜色（真彩色）。
Example: ith 24-bit colour depth: Size = 640 × 480 × 24 = 7,372,800 bits = 921,600 bytes ≈ 0.88 MB.
示例：一幅 640 × 480 的 24 位颜色深度图像：大小 = 640 × 480 × 24 = 7,372,800 位 = 921,600 字节 ≈ 0.88 MB。
10. Sound File Size Calculation | 音频文件大小计算
Digital audio is created by sampling an analogue sound wave at regular intervals. The file size of uncompressed audio depends on the sample rate, bit depth, duration, and number of channels.
数字音频通过定期采样模拟声波而创建。未压缩音频的文件大小取决于采样率、位深度、时长和声道数。
Sound File Size (bits) = Sample Rate × Bit Depth × Duration (seconds) × Channels
音频文件大小（位）= 采样率 × 位深度 × 时长（秒）× 声道数
- Sample rate: Number of samples taken per second, measured in Hz. Common rates: 44,100 Hz (CD quality).
- 采样率：每秒采集的样本数，以 Hz 为单位。常见采样率：44,100 Hz（CD 音质）。
- Bit depth: Number of bits used to store each sample. Common depths: 16 bits.
- 位深度：用于存储每个样本的位数。常见位深：16 位。
- Channels: Mono = 1 channel, Stereo = 2 channels.
- 声道：单声道 = 1 个声道，立体声 = 2 个声道。
Example: stereo audio at 44.1 kHz, 16-bit: Size = 44,100 × 16 × 30 × 2 = 42,336,000 bits ≈ 5.04 MB.
示例：30 秒立体声音频，44.1 kHz，16 位：大小 = 44,100 × 16 × 30 × 2 = 42,336,000 位 ≈ 5.04 MB。
11. Number Systems Overview | 数制系统概述
binary values. Hexadecimal uses digits 0-9 and letters A-F (where A=10, B=11, ..., F=15).
除了十进制和二进制，Year 9 学生还会学习十六进制（基数为 16），它是以更紧凑的方式表示二进制数值的方法。十六进制使用数字 0-9 和字母 A-F（其中 A=10, B=11, ..., F=15）。
| Denary (Decimal) | Binary (4-bit) | Hexadecimal |
| 0 | 0000 | 0 |
| 1 | 0001 | 1 |
| 10 | 1010 | A |
| 15 | 1111 | F |
| 16 | 0001 0000 | 10 |
| 255 | 1111 1111 | FF |
Each hex digit corresponds to exactly 4 binary bits (a nibble), making conversions straightforward. To convert binary to hex, group the binary digits into nibbles starting from the right, then convert each nibble to its hex equivalent.
每个十六进制数字正好对应 4 个二进制位（一个半字节），使得转换非常简单。要将二进制转换为十六进制，从右侧开始将二进制数字分成每 4 位一组，然后将每个半字节转换为对应的十六进制数字。
12. Basic Programming Constructs | 基本编程结构
All algorithms can be built using three fundamental programming constructs: Sequence, Selection, and Iteration. Understanding these is key to writing pseudocode and flowcharts in the exam.
所有算法都可以使用三种基本的编程结构来构建：顺序、选择和迭代。理解这些结构是在考试中编写伪代码和流程图的关键。
- Sequence: Instructions are executed one after another, in the order they are written. No steps are skipped.
- 顺序：指令按照编写顺序逐一执行，不跳过任何步骤。
- Selection: The program makes a decision based on a condition (e.g., IF...THEN...ELSE). Different paths execute depending on whether the condition evaluates to TRUE or FALSE.
- 选择：程序根据条件做出判断（例如 IF...THEN...ELSE）。根据条件判定为 TRUE 还是 FALSE，执行不同的路径。
- Iteration: A block of code repeats either a fixed number of times (FOR loop) or until a condition is met (WHILE loop / REPEAT...UNTIL).
- 迭代：代码块重复执行固定次数（FOR 循环）或直到满足某个条件为止（WHILE 循环 / REPEAT...UNTIL）。
Counter variable: In FOR loops, a counter variable tracks the current iteration. Formula for total iterations of a loop from start to end inclusive: Total iterations = end - start + 1 (if step = 1).
计数器变量：在 FOR 循环中，计数器变量跟踪当前迭代次数。包含起止值的循环总迭代次数公式：总迭代次数 = 结束值 - 起始值 + 1（步长 = 1 时）。
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