📚 KS3 AQA Computer Science: Formula & Theorem Quick Reference Guide | KS3 AQA 计算机：公式定理速查手册
Mastering the key formulas and rules in KS3 AQA Computer Science will help you solve problems faster and understand how computers process data. This guide brings together all the essential calculations, conversions, and logical theorems you need in one place. Whether you are working with binary, measuring file sizes, or predicting network speeds, these reference notes will support your learning and revision.
掌握 KS3 AQA 计算机科学中的核心公式和规则，能够帮助你更快地解决问题并理解计算机如何处理数据。这份指南将你需要的所有关键计算、进制转换和逻辑定理汇集在一起。无论你是在处理二进制、计算文件大小还是预测网络速度，这些参考笔记都将为你的学习和复习提供支持。
1. Data Units and Conversion | 数据单位与换算
Data size is measured in bits and bytes. A single bit is the smallest unit, holding either a 0 or a 1. Eight bits make one byte. Larger units use binary prefixes where each step multiplies by 1024.
数据大小以位和字节衡量。单个位是最小单位，保存 0 或 1。八位组成一个字节。更大的单位使用二进制前缀，每一步乘以 1024。
| Unit | Size in bytes | 单位 | 字节大小 |
|---|---|---|---|
| 1 bit | 1/8 byte | 1 b | 1/8 B |
| 1 nibble | 4 bits = 0.5 B | 1 nibble | 4 b = 0.5 B |
| 1 byte (B) | 8 bits | 1 B | 8 b |
| 1 kilobyte (KB) | 1024 B | 1 KB | 1024 B |
| 1 megabyte (MB) | 1024 KB | 1 MB | 1024 KB |
| 1 gigabyte (GB) | 1024 MB | 1 GB | 1024 MB |
| 1 terabyte (TB) | 1024 GB | 1 TB | 1024 GB |
To convert between units, multiply when moving to a smaller unit and divide when moving to a larger unit. For example, to change 5 MB into KB, multiply by 1024: 5 × 1024 = 5120 KB.
在不同单位之间换算时，转换为更小单位用乘法，转换为更大单位用除法。例如，将 5 MB 转换为 KB，乘以 1024：5 × 1024 = 5120 KB。
2. Binary to Decimal Conversion | 二进制转十进制
Every binary digit (bit) has a place value that doubles from right to left, starting at 1. Add the place values where the bit is 1 to get the decimal number.
每个二进制位（bit）都有一个从右到左依次翻倍的位值，从 1 开始。将位为 1 的位值相加，就得到十进制数。
The formula for an 8-bit number: decimal = b₇×128 + b₆×64 + b₅×32 + b₄×16 + b₃×8 + b₂×4 + b₁×2 + b₀×1, where bₙ is 0 or 1.
对于一个 8 位数，公式为：十进制 = b₇×128 + b₆×64 + b₅×32 + b₄×16 + b₃×8 + b₂×4 + b₁×2 + b₀×1，其中 bₙ 是 0 或 1。
Example: 1101₂ = (1×8) + (1×4) + (0×2) + (1×1) = 8+4+0+1 = 13.
示例：1101₂ = (1×8) + (1×4) + (0×2) + (1×1) = 8+4+0+1 = 13。
3. Decimal to Binary Conversion | 十进制转二进制
Divide the decimal number repeatedly by 2, writing down the remainder each time. The binary result is the remainders read from bottom to top.
反复将十进制数除以 2，每次写出余数。将余数从下往上读，就是二进制结果。
| Divide by 2 | Remainder |
|---|---|
| 13 ÷ 2 = 6 | 1 |
| 6 ÷ 2 = 3 | 0 |
| 3 ÷ 2 = 1 | 1 |
| 1 ÷ 2 = 0 | 1 |
Read remainders from bottom to top: 1101₂. So 13 in decimal is 1101 in binary. For larger numbers, continue dividing until the quotient reaches 0.
从下往上读余数：1101₂。所以十进制 13 的二进制是 1101。对于更大的数，继续除直到商为 0。
4. Hexadecimal Basics | 十六进制基础
Hexadecimal (hex) uses sixteen symbols: 0–9 and A–F, where A=10, B=11, C=12, D=13, E=14, F=15. Each hex digit represents four bits (a nibble), making it a compact way to write binary.
十六进制使用十六个符号：0–9 和 A–F，其中 A=10, B=11, C=12, D=13, E=14, F=15。每个十六进制位代表四位二进制（一个 nibble），因此可以更紧凑地书写二进制数。
To convert binary to hex, split the binary number into groups of four from the right, then convert each group. Example: 1010 1111₂ becomes A (1010) and F (1111), giving AF₁₆.
要将二进制转换为十六进制，从右边开始将二进制数四位一组分开，然后转换每一组。示例：1010 1111₂ 转换为 A (1010) 和 F (1111)，得到 AF₁₆。
Decimal to hex uses repeated division by 16, recording remainders, and converting remainders above 9 into letters. Hex to decimal multiplies each digit by 16 to the power of its position.
十进制转十六进制则反复除以 16，记录余数，并将大于 9 的余数转换为字母。十六进制转十进制则将每位数字乘以 16 的位权次方。
5. Binary Addition | 二进制加法
Binary addition follows simple rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10 (write 0, carry 1). When adding 1+1+1, write 1 and carry 1.
二进制加法遵循简单规则：0+0=0，0+1=1，1+0=1，1+1=10（写 0，进 1）。当计算 1+1+1 时，写 1 并进 1。
| Input A | Input B | Sum | Carry |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 (with carry in) | 1 | 1 | 1 |
Overflow happens when an 8-bit result exceeds 11111111₂ (255). The CPU sets an overflow flag to indicate the error. This is important when programming or designing circuits.
当 8 位计算结果超过 11111111₂（255）时，就会发生溢出。CPU 会设置溢出标志来指示错误。这在编程或设计电路时非常重要。
6. Image File Size Calculation | 图像文件大小计算
An image is made of pixels. The file size in bits depends on the width in pixels, the height in pixels, and the colour depth (bits per pixel).
图像由像素组成。以位为单位的文件大小取决于图像的像素宽度、像素高度以及颜色深度（每个像素的位数）。
Image file size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
图像文件大小（位） = 宽度（像素） × 高度（像素） × 颜色深度（每像素位数）
To get the size in bytes, divide the result by 8. For kilobytes, divide by (8 × 1024). Example: a 200×100 image with 24-bit colour depth uses 200 × 100 × 24 = 480,000 bits = 60,000 bytes ≈ 58.6 KB.
要得到以字节为单位的大小，将结果除以 8。要得到 KB，再除以 1024。示例：一个 200×100 像素、24 位颜色深度的图像，需要 200 × 100 × 24 = 480,000 b = 60,000 B ≈ 58.6 KB。
7. Sound File Size Calculation | 声音文件大小计算
Sound is stored digitally by sampling the amplitude at regular intervals. The file size depends on sample rate, sample resolution (bit depth), length in seconds, and number of channels.
声音通过定期对振幅进行采样来以数字方式存储。文件大小取决于采样率、采样分辨率（位深度）、时长（秒）以及声道数。
Sound file size (bits) = sample rate (Hz) × sample resolution (bits) × duration (seconds) × number of channels
声音文件大小（位） = 采样率（Hz） × 采样分辨率（位） × 时长（秒） × 声道数
Convert bits to bytes by dividing by 8. CD-quality audio uses 44,100 Hz, 16-bit samples, and 2 channels. For a 10-second stereo recording: 44100 × 16 × 10 × 2 = 14,112,000 bits = 1,764,000 bytes ≈ 1.68 MB.
将位转换为字节要除以 8。CD 品质的音频使用 44,100 Hz、16 位采样和 2 声道。一段 10 秒的立体声录音：44100 × 16 × 10 × 2 = 14,112,000 b = 1,764,000 B ≈ 1.68 MB。
8. Boolean Logic: AND, OR, NOT | 布尔逻辑：与、或、非
Boolean algebra uses variables that are either TRUE (1) or FALSE (0). The three basic operations are AND, OR, and NOT. Truth tables define their behaviour.
布尔代数使用只有 TRUE (1) 或 FALSE (0) 两种值的变量。三种基本运算是 AND、OR 和 NOT。真值表定义了它们的行为。
AND gate | 与门
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
AND outputs 1 only when both inputs are 1. It is like multiplication: A · B = 1 only if A=1 and B=1.
AND 门仅当两个输入都为 1 时输出 1。这类似于乘法：A · B = 1 当且仅当 A=1 且 B=1。
OR gate | 或门
| A | B | A OR B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
OR outputs 1 if at least one input is 1. It behaves like addition limited to 1 (1+1=1).
OR 门只要至少一个输入为 1 就输出 1。它类似于以 1 为上限的加法（1+1=1）。
NOT gate | 非门
| A | NOT A |
|---|---|
| 0 | 1 |
| 1 | 0 |
NOT simply flips the input: 0 becomes 1, 1 becomes 0. In expressions we write ¬A or A’.
NOT 只是将输入翻转：0 变成 1，1 变成 0。在表达式中写作 ¬A 或 A’。
9. Network Transmission Time | 网络传输时间
The time needed to send a file over a network depends on the file size and the data transfer rate. Always ensure units match (bits with bits per second, or bytes with bytes per second).
通过网络发送文件所需的时间取决于文件大小和数据传输速率。请始终确保单位匹配（位对应 bps，字节对应 Bps）。
Transmission time (seconds) = file size ÷ transfer rate
传输时间（秒）= 文件大小 ÷ 传输速率
Example: a 20 MB file over a 50 Mbps (megabits per second) connection. Convert 20 MB to megabits: 20 × 8 = 160 megabits. Time = 160 ÷ 50 = 3.2 seconds. Alternatively, use bytes: 20 MB = 20,000,000 bytes, rate = 50 Mbps = 6.25 MBps, 20 ÷ 6.25 = 3.2 s.
示例：一个 20 MB 的文件通过 50 Mbps（兆比特每秒）的连接传输。将 20 MB 转换为兆比特：20 × 8 = 160 兆比特。时间 = 160 ÷ 50 = 3.2 秒。或者使用字节：20 MB = 20,000,000 B，速率 = 50 Mbps = 6.25 MBps，20 ÷ 6.25 = 3.2 s。
10. Sorting and Searching Algorithms Quick Reference | 排序与搜索算法速查
Understanding how algorithms work helps you choose the most efficient method. KS3 focuses on the step-by-step logic rather than big-O notation, but simple comparisons are useful.
理解算法的工作原理有助于你选择最有效的方法。KS3 关注分步逻辑而非大 O 表示法，但简单的比较仍然很有用。
- Linear Search / 线性搜索: Checks each item one by one. In the worst case it examines every element. If a list has n items, the maximum number of checks is n. 它逐个检查每一项。在最坏情况下会检查每个元素。如果列表有 n 项，最大检查次数为 n。
- Binary Search / 二分搜索: Works on sorted lists. Repeatedly divides the list in half, discarding the half that cannot contain the target. Maximum checks is roughly log₂(n). 二分搜索适用于有序列表。反复将列表分成两半，丢弃不可能包含目标的那一半。最大检查次数大约为 log₂(n)。
- Bubble Sort / 冒泡排序: Compares adjacent pairs and swaps them if they are in the wrong order. Repeats passes until no swaps are needed. Worst-case number of comparisons is about n²/2. 冒泡排序比较相邻元素并在顺序错误时交换它们。重复遍历直到无需交换。最坏情况下的比较次数约为 n²/2。
- Merge Sort / 归并排序: Splits the list into halves repeatedly down to single items, then merges them back in order. It is efficient but requires extra memory. 归并排序将列表反复分成两半直至只剩单个元素，然后按顺序合并回来。它效率高但需要额外内存。
These are core algorithm patterns. Practice tracing them on small data sets to see how the number of steps grows.
这些是核心算法模式。在小型数据集上追踪它们，以观察步骤数量是如何增长的。
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