📚 Year 8 CCEA Computer Science: Formula & Theorem Quick Reference Handbook | 八年级CCEA计算机科学：公式定理速查手册
Welcome to your essential quick reference guide for Year 8 CCEA Computer Science. Mastering fundamental formulas and theorems is crucial for understanding data representation, Boolean logic, and digital media calculations. This handbook provides clear explanations and examples to help you succeed in your studies.
欢迎使用八年级CCEA计算机科学必备速查手册。掌握基本公式和定理对于理解数据表示、布尔逻辑和数字媒体计算至关重要。本手册提供清晰的解释和示例，助力你的学习。
1. Binary to Denary Conversion | 二进制转十进制
In binary, each digit represents a power of 2, from right to left starting at 2⁰. To convert a binary number to denary, multiply each binary digit by its place value and sum the results.
在二进制中，每个数字代表2的幂，从右向左起始幂次为2⁰。将二进制数转换为十进制，需将每位数字乘以其位值并求和。
Example: Convert 1101₂ to denary.
示例：将1101₂转换为十进制。
1 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰ = 8 + 4 + 0 + 1 = 13₁₀
1 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰ = 8 + 4 + 0 + 1 = 13₁₀
Another example: 10110₂ → 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22₁₀.
另一示例：10110₂ → 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22₁₀。
2. Denary to Binary Conversion | 十进制转二进制
The most reliable method is repeated division by 2. Divide the denary number by 2, record the remainder (0 or 1), then divide the quotient by 2 again. Repeat until the quotient is 0. Read the remainders in reverse order to obtain the binary equivalent.
最可靠的方法是重复除以2。将十进制数除以2，记录余数（0或1），再将商除以2。重复直到商为0。逆序读取余数即得二进制数。
1 r 0; 11 ÷ 2 = 5 r 1; 5 ÷ 2 = 2 r 1; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Read upwards: 101101₂.
示例：将45₁₀转为二进制。45 ÷ 2 = 22 余 1; 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：101101₂。
You can also use a place-value table. Find the largest power of 2 not exceeding the number, write a 1, subtract, and continue with the remainder.
也可使用位值表。找出不超过该数的最大2的幂，写1，减去，再对余数继续该过程。
3. Binary Addition Rules | 二进制加法规则
Binary addition follows four simple rules. The last rule generates a carry to the next higher bit.
二进制加法遵循四条简单规则。最后一条会产生向高位的进位。
| 0 + 0 = 0 | No carry |
| 0 + 1 = 1 | No carry |
| 1 + 0 = 1 | No carry |
| 1 + 1 = 0, carry 1 | Carry 1 to next column |
When adding multiple bits, a carry can cascade. For example, 111₂ + 001₂ = 1000₂.
多个位相加时，进位可级联。例如111₂ + 001₂ = 1000₂。
Always align binary numbers to the right and add column by column, just like denary addition.
总是右对齐二进制数，逐列相加，与十进制加法类似。
4. Hexadecimal Number System | 十六进制数系统
Hexadecimal uses base‑16. Digits 0–9 represent the same values, while letters A (10), B (11), C (12), D (13), E (14), F (15) are used.
十六进制使用基数为16。数字0–9表示相同值，字母A（10）、B（11）、C（12）、D（13）、E（14）、F（15）被使用。
to nibbles (4 bits) from the right, then convert each nibble. Example: 1010 1111₂ → A F₁₆.
将二进制转换为十六进制，从右开始将二进制位每4位一组分组，然后转换每个四位组。示例：1010 1111₂ → A F₁₆。
To convert denary to hex, repeatedly divide by 16 and read remainders in reverse. Remainders 10–15 become A–F.
将十进制转换为十六进制，重复除以16并逆序读取余数。余数10–15变为A–F。
| Denary | Binary (4‑bit) | Hex |
|---|---|---|
| 0 | 0000 | 0 |
| 9 | 1001 | 9 |
| 10 | 1010 | A |
| 15 | 1111 | F |
5. Data Storage Unit Conversions | 数据存储单位换算
s follow binary multiples: 1 byte = 8 bits. Larger units use 2¹⁰ = 1024 as the multiplier.
计算机存储量基于位。单位遵循二进制倍数：1字节=8位。更大单位使用2¹⁰=1024作为乘数。
1 Kilobyte (KB) = 1024 bytes
1 Megabyte (MB) = 1024 KB
1 Gigabyte (GB) = 1024 MB
1 Terabyte (TB) = 1024 GB
1千字节（KB）= 1024字节
1兆字节（MB）= 1024 KB
1吉字节（GB）= 1024 MB
1太字节（TB）= 1024 GB
a smaller unit, divide when moving to a larger one. Example: 5 MB = 5 × 1024 KB = 5120 KB.
在单位之间转换时，向较小单位转换用乘法，向较大单位转换用除法。示例：5 MB = 5 × 1024 KB = 5120 KB。
Sometimes manufacturers use 1000 for simplicity, but in computer science exams, the binary definition (1024) is standard.
有时制造商为简便使用1000，但在计算机科学考试中，二进制定义（1024）是标准。
6. Boolean Logic Gates | 布尔逻辑门
A logic gate is a physical device that implements a Boolean function. The three basic gates are AND, OR, and NOT.
逻辑门是实现布尔函数的物理器件。三种基本门是与门、或门和非门。
The AND gate outputs 1 only if all inputs are 1. Symbolically: Output = A · B or A AND B.
与门仅当所有输入为1时输出1。符号表示为：输出 = A · B 或 A AND B。
The OR gate outputs 1 if at least one input is 1. Output = A + B or A OR B.
或门至少有一个输入为1时输出1。输出 = A + B 或 A OR B。
The NOT gate inverts the input: Output = NOT A or A bar (Ā). If A=0, output=1; if A=1, output=0.
非门反转输入：输出 = NOT A 或 A 上划线（Ā）。若A=0，输出1；若A=1，输出0。
Combining gates gives rise to NAND, NOR, XOR and XNOR. For Year 8, focus on AND, OR, NOT and simple combinations.
组合门可产生与非、或非、异或和同或。八年级重点在于与、或、非及简单组合。
7. Truth Tables for Basic Gates | 基本门真值表
A truth table lists all possible input combinations and the corresponding output for a logic circuit.
真值表列出了逻辑电路所有可能的输入组合及其对应的输出。
| A | B | Output |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
| A | B | Output |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
| A | Output |
|---|---|
| 0 | 1 |
| 1 | 0 |
You can build truth tables for combined circuits by working column by column through intermediate gate outputs.
你可以通过逐列计算中间门输出来构建组合电路的真值表。
8. Boolean Algebra Identities | 布尔代数恒等式
Boolean algebra helps simplify logic expressions. These identities use AND (·), OR (+), and NOT (overbar).
布尔代数有助于简化逻辑表达式。这些恒等式使用与（·）、或（+）和非（上划线）。
| Identity Law | A · 1 = A | A + 0 = A |
| Null Law | A · 0 = 0 | A + 1 = 1 |
| Idempotent Law | A · A = A | A + A = A |
| Inverse Law | A · Ā = 0 | A + Ā = 1 |
| Double Negation | (Ā)⁼⁼ = A | |
A and A + B = B + A. Associative: (A·B)·C = A·(B·C). Distributive: A·(B+C) = (A·B)+(A·C).
交换律：A·B = B·A 及 A+B = B+A。结合律：(A·B)·C = A·(B·C)。分配律：A·(B+C) = (A·B)+(A·C)。
De Morgan’s Laws: NOT (A AND B) = NOT A OR NOT B; NOT (A OR B) = NOT A AND NOT B.
德摩根定律：NOT (A AND B) = NOT A OR NOT B；NOT (A OR B) = NOT A AND NOT B。
These identities are useful for reducing the number of gates in a circuit.
这些恒等式对于减少电路中的门数量很有用。
9. Calculating Image File Size | 计算图像文件大小
An image is made of pixels. The colour depth is the number of bits used to store the colour of each pixel. The basic formula is:
图像由像素组成。颜色深度是用于存储每个像素颜色的位数。基本公式为：
Image file size (bits) = Width (pixels) × Height (pixels) × Colour Depth (bits)
图像文件大小（位）= 宽度（像素）× 高度（像素）× 颜色深度（位）
o bytes, divide by 8. For KB, divide by (8 × 1024). Example: A 200×300 pixel image with 24‑bit colour depth: 200 × 300 × 24 = 1,440,000 bits = 180,000 bytes ≈ 175.8 KB.
要转换为字节，除以8。转为KB则除以（8×1024）。示例：200×300像素、24位颜色深度的图像：200×300×24 = 1,440,000位 = 180,000字节 ≈ 175.8 KB。
Remember that metadata (file header) slightly increases the actual file size, but the exam formula ignores it.
记得元数据（文件头）会略微增加实际文件大小，但考试公式忽略它。
10. Calculating Sound File Size | 计算声音文件大小
Digitised sound is represented by samples taken at a fixed rate. The key factors are sample rate (Hz), bit depth, duration (seconds), and number of channels (1 for mono, 2 for stereo).
数字化声音由固定速率采集的样本表示。关键因素是采样率（Hz）、位深度、时长（秒）和声道数（单声道1，立体声2）。
Sound file size (bits) = Sample Rate (Hz) × Duration (s) × Bit Depth × Channels
声音文件大小（位）= 采样率（Hz）× 时长（秒）× 位深度 × 声道数
Example: 30 seconds of stereo audio at 44,100 Hz, 16‑bit depth: 44,100 × 30 × 16 × 2 = 42,336,000 bits ≈ 5.04 MB.
示例：30秒立体声音频，44,100 Hz，16‑位深度：44,100 × 30 × 16 × 2 = 42,336,000位 ≈ 5.04 MB。
Often you need to give the answer in appropriate units; use the storage conversion table from Section 5.
通常需要以合适单位给出答案；使用第5节的存储换算表。
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