📚 Year 10 Eduqas Computer Science: Formula & Theorem Quick Reference Handbook | 10年级Eduqas计算机：公式定理速查手册
This handbook brings together the essential formulas, laws, and theorems you need to master for the Year 10 Eduqas GCSE Computer Science course. From binary conversions and file size calculations to Boolean algebra and data transmission, every topic is distilled into clear reference material with worked examples. Use it alongside your revision notes to test yourself and build confidence for the exam.
本手册汇集了10年级Eduqas GCSE计算机科学课程必须掌握的核心公式、定律和定理。从二进制转换、文件大小计算到布尔代数与数据传输，每个主题都提炼为清晰的参考资料并配有示例。请与复习笔记配合使用，进行自测并提升应考信心。
1. Number Systems & Conversions | 数制与转换
l digit using the mapping 10=A, 11=B, 12=C, 13=D, 14=E, 15=F. Converting from hexadecimal to binary reverses the process by writing each hex digit as a 4-bit binary number.
二进制转十六进制时，从右起每四位一组划分二进制数，将每个半字节替换为对应的十六进制数字（10=A, 11=B, 12=C, 13=D, 14=E, 15=F）。十六进制转二进制则反向操作，将每个十六进制数字写作四位二进制数。
1 = 13₁₀. Denary to binary uses repeated division by 2, reading the remainders backwards.
二进制转十进制时，将出现1的位权值相加。例：1101₂ = 8 + 4 + 0 + 1 = 13₁₀。十进制转二进制则反复除以2，余数逆序读取。
2. Units of Storage | 存储单位
Storage capacities are measured in bytes. Common units and their conversions are:
存储容量以字节为单位。常见单位及其换算如下：
| Unit (单位) | Equivalent (等值) | Power of 2 (2的幂) |
|---|---|---|
| 1 kilobyte (KB) | 1,024 bytes | 2¹⁰ |
| 1 megabyte (MB) | 1,024 KB | 2²⁰ |
| 1 gigabyte (GB) | 1,024 MB | 2³⁰ |
| 1 terabyte (TB) | 1,024 GB | 2⁴⁰ |
These values are based on binary multiples (IEC standard). Some exam questions may treat 1 KB as 1,000 bytes – always read the question carefully to see which definition is used.
这些值基于二进制倍数（IEC标准）。部分考题可能将1 KB视为1,000字节——请仔细读题，确认使用了哪种定义。
3. Image File Size Calculation | 图像文件大小计算
An uncompressed bitmap image's size depends on its resolution and colour depth. The formula is:
未压缩的位图图像大小取决于分辨率和色深。公式为：
Image size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
图像大小 (位) = 宽度 (像素) × 高度 (像素) × 色深 (每像素位数)
el image with a 24-bit colour depth requires 100 × 100 × 24 = 240,000 bits = 30,000 bytes.
转换为字节需除以8。例如，100×100像素、24位色深的图像需要100 × 100 × 24 = 240,000 位 = 30,000 字节。
4. Sound File Size Calculation | 声音文件大小计算
The size of a digital audio file is determined by the sampling rate, resolution, duration, and number of channels. The basic formula for a mono file is:
数字音频文件的大小由采样率、分辨率、时长和声道数决定。单声道文件的基本公式为：
File size (bits) = sample rate (Hz) × sample resolution (bits) × duration (seconds)
文件大小 (位) = 采样率 (赫兹) × 采样分辨率 (位) × 时长 (秒)
io sampled at 44,100 Hz with 16-bit resolution produces 44,100 × 16 × 10 = 7,056,000 bits.
立体声文件则将结果乘以2。例：10秒音频，采样率44,100 Hz，16位分辨率，则大小为44,100 × 16 × 10 = 7,056,000 位。
5. Compression Ratio | 压缩比
Compression ratio measures how much file size is reduced. It compares the original size to the compressed size:
压缩比衡量文件大小减少的程度，比较原始大小与压缩后的大小：
Compression ratio = Uncompressed size ÷ Compressed size
压缩比 = 未压缩大小 ÷ 压缩后大小
of its original size. Space saving can be calculated as (1 − 1 / compression ratio) × 100%.
压缩比为4:1表示文件体积是原来的四分之一。空间节约率可计算为 (1 − 1 / 压缩比) × 100% 。
6. Data Transmission & Transfer Time | 数据传输与传输时间
When transferring data across a network, the time needed depends on the amount of data and the transfer speed. Units must be matched (bits or bytes):
通过网络传输数据时，所需时间取决于数据量和传输速率。单位必须统一（位或字节）：
Time (seconds) = Data size ÷ Transfer speed
时间 (秒) = 数据大小 ÷ 传输速率
check whether the speed is in bits per second (bps, Kbps, Mbps) or bytes per second (Bps).
例如，以50 Mbps下载200 MB文件：200 MB = 1,600 Mb，时间 = 1,600 Mb / 50 Mbps = 32秒。务必检查速率是每秒位 (bps, Kbps, Mbps) 还是每秒字节 (Bps)。
7. Logic Gates – Truth Tables & Boolean Expressions | 逻辑门 – 真值表与布尔表达式
Logic gates perform Boolean operations on one or more inputs to produce a single output. The most common gates are AND, OR, NOT, NAND, NOR, and XOR. Their truth tables and Boolean notations are summarised below.
逻辑门对一个或多个输入执行布尔运算，产生单一输出。最常见的门有AND、OR、NOT、NAND、NOR和XOR。其真值表和布尔符号总结如下。
| Gate (门) | Boolean Expression (布尔表达式) | Truth Table (真值表) |
|---|---|---|
| AND | Q = A · B | 0·0=0, 0·1=0, 1·0=0, 1·1=1 |
| OR | Q = A + B | 0+0=0, 0+1=1, 1+0=1, 1+1=1 |
| NOT | Q = A' | A=0 → Q=1; A=1 → Q=0 |
| NAND | Q = (A · B)' | opposite of AND; 1·1=0, others=1 |
| NOR | Q = (A + B)' | opposite of OR; 0+0=1, others=0 |
| XOR | Q = A ⊕ B | 0⊕0=0, 0⊕1=1, 1⊕0=1, 1⊕1=0 |
The prime symbol (') denotes NOT operation. In diagrams, small circles at inputs or outputs represent inversion.
撇号 (') 表示NOT运算。在电路图中，输入端或输出端的小圆圈代表取反。
8. Boolean Algebra Laws & Theorems | 布尔代数定律与定理
Boolean algebra provides a set of rules to simplify logic expressions. These identities are crucial for reducing the number of gates in a circuit.
布尔代数提供了一套简化逻辑表达式的规则。这些恒等式对于减少电路中的门数量至关重要。
Commutative Law: A + B = B + A, A · B = B · A
交换律：A + B = B + A, A · B = B · A
Associative Law: (A + B) + C = A + (B + C), (A · B) · C = A · (B · C)
结合律：(A + B) + C = A + (B + C), (A · B) · C = A · (B · C)
Distributive Law: A · (B + C) = A · B + A · C, A + (B · C) = (A + B) · (A + C)
分配律：A · (B + C) = A · B + A · C, A + (B · C) = (A + B) · (A + C)
De Morgan's Theorems: (A · B)' = A' + B', (A + B)' = A' · B'
德摩根定理：(A · B)' = A' + B', (A + B)' = A' · B'
= A, A · 1 = A), Idempotent Law (A + A = A, A · A = A), and Absorption Law (A + A·B = A).
其他有用恒等式包括Annulment律 (A + 1 = 1, A · 0 = 0)、Identity律 (A + 0 = A, A · 1 = A)、Idempotent律 (A + A = A, A · A = A) 和Absorption律 (A + A·B = A)。
9. Binary Arithmetic – Addition & Overflow | 二进制算术 – 加法与溢出
write 0, carry 1). When adding numbers of a fixed bit-width, an overflow occurs if the result requires more bits than are available. Overflow is detected when the carry into the most significant bit (MSB) differs from the carry out of the MSB.
二进制加法遵循四条简单规则：0+0=0, 0+1=1, 1+0=1, 1+1=10 (写0进1)。当对固定位宽的数进行加法时，若结果需要的位数超过可用位数，则发生溢出。当最高有效位(MSB)的进位输入与进位输出不同时，即可检测出溢出。
For example, in 4-bit addition, 0111₂ (7) + 0001₂ (1) = 1000₂ (−8 in two' . In unsigned arithmetic, overflow occurs simply when a carry out of the MSB is generated.
例如，在4位加法中，0111₂ (7) + 0001₂ (1) = 1000₂ (二进制补码中为−8)，即为溢出。在无符号算术中，仅当MSB产生进位时就发生溢出。
10. Error Detection – Parity & Checksums | 检错 – 奇偶校验与校验和
Parity checking adds a single bit to a byte to make the total number of 1s either even (even parity) or odd (odd parity). The receiver recalculates parity; a mismatch indicates a possible error. While simple, it cannot detect an even number of bit-flips.
奇偶校验为每个字节添加一个比特，使得比特1的总数为偶数（偶校验）或奇数（奇校验）。接收方重新计算奇偶位，不匹配则表明可能出错。这种方法简单，但无法检测偶数个比特翻转的错误。
A checksum involves sending a block of data along with a calculated sum value. The simplest checksum is the arithmetic sum of all bytes in the data block, modulo 256. The receiver computes the same sum and compares it with the transmitted checksum.
校验和是在发送数据块的同时发送一个计算出的总和值。最简单的校验和是数据块中所有字节的算术和，模256。接收方计算相同的和并与传输过来的校验和进行比较。
11. Encryption – Caesar Cipher | 加密 – 凯撒密码
The Caesar cipher is a substitution cipher that shifts each letter in the plaintext by a fixed number of positions. If shift is n, encryption can be expressed as:
凯撒密码是一种替换密码，将明文中的每个字母按固定位置数进行移位。若移位为n，加密可表示为：
Ciphertext letter = (Plaintext letter position + n) mod 26
密文字母 = (明文字母位置 + n) mod 26
Decryption uses the inverse: (Cipher letter position − n) mod 26. For example, with a shift of 3, 'A' becomes 'D', 'B' to 'E', …, 'Z' wraps to 'C'. The key is the shift value n.
解密使用逆运算：(密文字母位置 − n) mod 26。例如，移位3时，'A'变为'D'，'B'变为'E'，…… 'Z'回绕为'C'。密钥即为移位值n。
12. Binary Shifts | 二进制移位
Shifting a binary number left by one place multiplies it by 2; shifting right by one place divides by 2 (integer division discarding remainders). For logical shifts, the empty positions are filled with zeros.
二进制数左移一位相当于乘以2；右移一位相当于除以2（整数除法，舍去余数）。对于逻辑移位，空位用零填充。
- Left shift: 0010₂ (2) → 0100₂ (4)
- Right shift: 1000₂ (8) → 0100₂ (4)
- 左移： 0010₂ (2) → 0100₂ (4)
- 右移： 1000₂ (8) → 0100₂ (4)
Left shifts can be used for fast multiplication by powers of 2, and right shifts for division. Programmers must be aware of overflow when using left shifts with limited bit-width registers.
左移可用来快速实现2的幂次乘法，右移用于除法。程序员在有限位宽寄存器中使用左移时需注意溢出。
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