📚 Year 10 OCR Computer Science: Formula & Theorem Quick Reference | Year 10 OCR 计算机：公式定理速查手册
This quick reference guide brings together all the essential formulas and theorems you need to memorise for Year 10 OCR Computer Science. From calculating image file sizes to applying Boolean algebra laws, every key relationship is explained with clear examples and ready-to-use equations. Keep this handbook close for revision, homework, and exam preparation – it will save you time and help you avoid simple calculation errors.
这本速查手册汇总了 Year 10 OCR 计算机科学需要牢记的所有基本公式与定理。从计算图像文件大小到应用布尔代数定律，每一个关键关系都用清晰的示例和可直接使用的公式加以说明。请将本手册放在手边用于复习、作业和备考——它会帮你节省时间，避免简单的计算错误。
1. Units of Digital Storage | 数字存储单位
exam questions often use powers of 10 (1 kB = 1000 bytes). Always check whether the question specifies binary (kibibyte) or decimal (kilobyte) units; the OCR course primarily uses decimal prefixes.
数据的最小单位是比特 (b)，可以是 0 或 1。8 个比特组成一个字节 (B)。更大的单位基于 2 的幂，但为简单起见，考题常使用 10 的幂（1 kB = 1000 字节）。务必检查题目是指定二进制单位（kibibyte）还是十进制单位（kilobyte）；OCR 课程主要使用十进制前缀。
| Unit | Abbreviation | Value (decimal) |
|---|---|---|
| bit | b | 1 bit |
| byte | B | 8 bits |
| kilobyte | kB | 1000 bytes |
| megabyte | MB | 1000 kB |
| gigabyte | GB | 1000 MB |
| terabyte | TB | 1000 GB |
2. Image File Size Formula | 图像文件大小公式
our of each pixel. The basic formula is: File Size (bits) = Image Width (pixels) × Image Height (pixels) × Colour Depth (bpp). To convert to bytes, divide by 8.
位图图像文件的大小（以比特为单位）取决于图像分辨率和色深。色深是表示每个像素颜色所用的比特数。基本公式为：文件大小（比特）= 图像宽度（像素）× 图像高度（像素）× 色深（bpp）。要转换为字节，除以 8。
File Size (bits) = Width × Height × Colour Depth
File Size (bytes) = (Width × Height × Colour Depth) / 8
Example: An image with resolution 1024 × 768 and 24-bit colour depth gives:
4 × 768 × 24 = 18,874,368 bits = 2,359,296 bytes ≈ 2.36 MB (using 1 MB = 1,000,000 bytes).
示例：一幅分辨率为 1024 × 768、24 位色深的图像：
1024 × 768 × 24 = 18,874,368 比特 = 2,359,296 字节 ≈ 2.36 MB （使用 1 MB = 1,000,000 字节）。
3. Sound File Size Formula | 声音文件大小公式
per sample), and duration (seconds). For stereo tracks, multiply by 2. The formula is: File Size (bits) = Sample Rate × Sample Resolution × Duration × Number of Channels.
未压缩的声音文件会存储从模拟信号中采集的每个样本的数据。三个因素为：采样率 (Hz)、采样分辨率（每样本比特数）和时长（秒）。立体声需乘以 2。公式为：文件大小（比特）= 采样率 × 采样分辨率 × 时长 × 声道数。
File Size (bits) = Sample Rate (Hz) × Bit Depth × Duration (s) × Channels
File Size (bytes) = (Sample Rate × Bit Depth × Duration × Channels) / 8
Example: 30 seconds of CD-quality audio (44,100 Hz, 16-bit, stereo):
44,100 × 16 × 30 × 2 = 42,336,000 bits = 5,292,000 bytes ≈ 5.29 MB.
示例：30 秒 CD 音质音频（44100 Hz、16 位、立体声）：
44100 × 16 × 30 × 2 = 42,336,000 比特 = 5,292,000 字节 ≈ 5.29 MB。
4. Text File Size Calculation | 文本文件大小计算
its (usually 1 byte). With Unicode, the number of bytes per character varies (UTF-8 uses 1– UTF-16. The simple formula is: Size (bytes) = Number of Characters × Bytes per Character.
纯文本文件使用字符编码（如 ASCII 或 Unicode）存储每个字符。在标准 ASCII 中，每个字符使用 7 或 8 比特（通常为 1 字节）。对于 Unicode，每个字符的字节数可变（UTF-8 为 1–4 字节）。考试计算中，除非另有说明，ASCII 假定为每字符 1 字节，UTF-16 假定为每字符 2 字节。简单公式为：大小（字节）= 字符数 × 每字符字节数。
Text Size (bytes) = Number of Characters × Bytes per Character
Example: A tweet of 280 characters stored as ASCII uses 280 bytes. Stored as UTF-16, it would use 560 bytes.
示例：一条 280 个字符的推文，用 ASCII 存储需要 280 字节。用 UTF-16 存储则需要 560 字节。
5. Data Transfer Time | 数据传输时间
) = File Size (bits) / Transfer Rate (bits per second). You must ensure the units match. Often you will convert file size from bytes to bits by multiplying by 8, and convert transfer rate from Mbps to bps by multiplying by 1,000,000.
通过网络发送文件时，传输时间取决于文件大小和网速（带宽）。基本公式为：时间（秒）= 文件大小（比特）/ 传输速率（比特每秒）。必须保证单位一致。通常需要将文件大小从字节乘以 8 转换为比特，并将传输速率从 Mbps 乘以 1,000,000 转换为 bps。
Transfer Time (s) = File Size (bits) / Bandwidth (bps)
Example: A 100 MB file transferred over a 50 Mbps connection:
File size = 100 × 1,000,000 × 8 = 800,000,000 bits.
Bandwidth = 50 × 1,000,000 = 50,000,000 bps.
Time = 800,000,000 / 50,000,000 = 16 seconds.
示例：通过 50 Mbps 连接传输 100 MB 文件：
文件大小 = 100 × 1,000,000 × 8 = 800,000,000 比特。
带宽 = 50 × 1,000,000 = 50,000,000 bps。
时间 = 800,000,000 / 50,000,000 = 16 秒。
6. Boolean Algebra Laws – Commutative, Associative, Distributive | 布尔代数定律 – 交换、结合、分配
rs (AND, OR, NOT). The commutative law states that order does not matter for AND and OR: A ∧ B = B ∧ A; A ∨ B = B ∨ A. The associative law allows regrouping: (A ∧ B) ∧ C = A ∧ (B ∧ C); same for OR. The distributive law connects AND and OR: A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C); A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C).
布尔代数使用二进制变量和逻辑运算符（与、或、非）。交换律指出 AND 和 OR 的次序无关紧要：A ∧ B = B ∧ A；A ∨ B = B ∨ A。结合律允许重新分组：(A ∧ B) ∧ C = A ∧ (B ∧ C)；OR 同理。分配律连接 AND 和 OR：A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C)；A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C)。
| Law | AND Form | OR Form |
|---|---|---|
| Commutative | A · B = B · A | A + B = B + A |
| Associative | (A·B)·C = A·(B·C) | (A+B)+C = A+(B+C) |
| Distributive | A·(B+C) = (A·B)+(A·C) | A+(B·C) = (A+B)·(A+C) |
7. Boolean Algebra Laws – Identity, Null, Idempotent, Inverse | 布尔代数定律 – 同一、零、幂等、互补
Identity laws: A ∧ 1 = A, A ∨ 0 = A. Null laws: A ∧ 0 = 0, A ∨ 1 = 1. Idempotent laws: A ∧ A = A, A ∨ A = A. Inverse (complement) laws: A ∧ ¬A = 0, A ∨ ¬ 1. These simple rules help reduce complex logic expressions during circuit simplification.
同一律：A ∧ 1 = A, A ∨ 0 = A。零律：A ∧ 0 = 0, A ∨ 1 = 1。幂等律：A ∧ A = A, A ∨ A = A。互补律：A ∧ ¬A = 0, A ∨ ¬A = 1。这些简单规则有助于在简化电路时化简复杂的逻辑表达式。
| Law | AND Form | OR Form |
|---|---|---|
| Identity | A · 1 = A | A + 0 = A |
| Null | A · 0 = 0 | A + 1 = 1 |
| Idempotent | A · A = A | A + A = A |
| Inverse | A · ¬A = 0 | A + ¬A = 1 |
8. De Morgan's Theorems | 德摩根定理
De Morgan' verting OR operations into AND operations with inverted inputs, and vice versa. Theorem 1: ¬(A ∧ B) = ¬A ∨ ¬B. Theorem 2: ¬(A ∨ B) = ¬A ∧ ¬ rcuits using only NAND or NOR gates, and in simplifying logical conditions in programming.
德摩根定理对于将 OR 操作转换为带反相输入的 AND 操作至关重要，反之亦然。定理 1：¬(A ∧ B) = ¬A ∨ ¬B。定理 2：¬(A ∨ B) = ¬A ∧ ¬B。这些定理广泛用于仅使用 NAND 或 NOR 门设计逻辑电路，以及在编程中简化逻辑条件。
¬(A · B) = ¬A + ¬B
¬(A + B) = ¬A · ¬B
9. Binary Addition and Subtraction Rules | 二进制加法与减法规则
dding two 8-bit unsigned integers, a carry out of the most significant bit indicates an overflow if the result does not fit in 8 bits. Subtraction can be performed by adding the two's complement of the minuend.
二进制加法遵循以下逐位规则：0 + 0 = 0，0 + 1 = 1，1 + 0 = 1，1 + 1 = 0 进 1，1 + 1 + 进位 1 = 1 进 1。当两个 8 位无符号整数相加时，若最高位产生进位且结果超出 8 位范围，则发生溢出。减法可通过加上减数的补码来实现。
Two's complement subtraction: A - B = A + (two's complement of B). To find two's complement of a binary number, invert all bits and add 1.
二进制补码减法：A - B = A + (B 的补码)。求二进制数的补码方法：按位取反，再加 1。
10. Overflow and Binary Shifts | 溢出与二进制移位
Overflow occurs when the result of an arithmetic operation exceeds the maximum value an allocated number of bits can hold. In unsigned binary, after 11111111 + 00000001 the 8-bit result is 00000000 with a carry out – indicating overflow. In signed two's complement, adding two positive numbers and getting a negative result also signals overflow. Binary left shift multiplies a number by 2 for each shift position; right shift divides by 2 (integer division).
当算术运算的结果超出指定位数所能表示的最大值时，就会发生溢出。在无符号二进制中，11111111 + 00000001 后，8 位结果为 00000000 并产生进位——表示溢出。在有符号补码中，两个正数相加得到负数也表明溢出。二进制左移每移一位相当于乘以 2；右移相当于除以 2（整数除法）。
Left Shift n places = Multiply by 2ⁿ
Right Shift n places = Integer Division by 2ⁿ
Example: 00001100 (12) left shift 2 gives 00110000 (48). Right shift 2 gives 00000011 (3).
示例：00001100 (12) 左移 2 位得 00110000 (48)。右移 2 位得 00000011 (3)。
11. Compression Ratio | 压缩比
s: Compression Ratio = Original File Size / Compressed File Size. A higher ratio indicates more aggressive compression. For example, a 10:1 ratio means the compressed file is one tenth of the original.
压缩可以减小文件大小。压缩比告诉你压缩后的文件比原始文件小了多少。其计算公式为：压缩比 = 原始文件大小 / 压缩后文件大小。比值越大表示压缩越强。例如，10:1 表示压缩后的文件是原来的十分之一。
Compression Ratio = Original Size / Compressed Size
This is useful when comparing lossless and lossy compression efficiency – but remember, lossy compression can achieve much higher ratios.
这在比较无损和有损压缩效率时很有用——但请记住，有损压缩可以实现高得多的压缩比。
12. Number Base Conversions – Key Patterns | 数制转换 – 关键模式
and hexadecimal are often done with tables or by manual division, there are two important 'shortcut' eplace each group with the hex equivalent. For denary → binary, repeated division by 2 gives remainder bits in reverse order – this is an algorithm, not a formula, but a vital theorem of representation.
虽然十进制、二进制和十六进制之间的转换通常借助表格或手动除法完成，但有两个重要的“捷径”定理需要牢记：每个十六进制数字恰好对应一个 4 位二进制半字节。因此，将二进制转十六进制时，从右往左每四位一组，将每组替换为对应的十六进制数字。十进制转二进制时，重复除以 2 得到的余数位倒序排列——这是一种算法，而非公式，但却是表示法的一条重要定理。
0, 5=0101, 6=0110, 7=0111, 8=1000, 9=1001, A=1010, B=1011, C=1100, D=1101, E=1110, F=1111.
十六进制表格： 00, 5=0101, 6=0110, 7=0111, 8=1000, 9=1001, A=1010, B=1011, C=1100, D=1101, E=1110, F=1111。
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