📚 Year 7 Cambridge Computer Science: Formula & Theorem Quick Reference | Year 7 剑桥计算机：公式定理速查手册
This quick reference handbook gathers the essential formulas, conversion rules, and logical theorems that every Year 7 Cambridge Computer Science student needs. Mastering these building blocks will strengthen your understanding of how computers represent data, perform calculations, and make simple decisions.
本速查手册汇集了 Year 7 剑桥计算机课程中每一位学生都必须掌握的基本公式、转换规则和逻辑定理。熟练掌握这些基础知识将帮助你更深刻地理解计算机如何表示数据、执行计算以及做出简单决策。
1. Binary to Decimal Conversion | 二进制转十进制
To convert a binary number to denary (decimal), assign place values as powers of 2, starting from 20 on the right. Multiply each binary digit (0 or 1) by its place value and sum the results.
要将二进制数转换为十进制，请从右向左依次为每一位分配 2 的幂次作为位值，最右侧为 20。将每个二进制位（0 或 1）乘以其位值，然后求和。
Value = dn-1 × 2n-1 + ... + d1 × 21 + d0 × 20
Example: 10112 = 1×23 + 0×22 + 1×21 + 1×20 = 8 + 0 + 2 + 1 = 11 in denary.
示例：10112 = 1×23 + 0×22 + 1×21 + 1×20 = 8 + 0 + 2 + 1 = 11（十进制）。
A shortcut for 8-bit values is to remember the place values: 128, 64, 32, 16, 8, 4, 2, 1. Add up the values wherever a 1 appears.
对于 8 位二进制数，一个速记方法是记住位值：128, 64, 32, 16, 8, 4, 2, 1。但凡出现 1 的位值就相加。
2. Decimal to Binary Conversion | 十进制转二进制
Use the 'divide by 2' method: repeatedly divide the denary number by 2, keep the remainder (0 or 1), and read the remainders from bottom to top. Alternatively, use the subtraction method with binary place values.
使用“除以 2”法：将十进制数反复除以 2，保留余数（0 或 1），然后从下向上读取所有余数。也可以使用二进制位值表进行减法转换。
umber ÷ 2 → Quotient + Remainder, repeat until quotient = 0; then read remainders upwards.
Example: Convert 2510 2 = 3 r 0; 3 ÷ 2 = 1 r 1; 1 ÷ 2 = 0 r 1. Reading remainders from last to first gives 110012.
示例：将 2510 转换为二进制。25 ÷ 2 = 12 余 1；12 ÷ 2 = 6 余 0；6 ÷ 2 = 3 余 0；3 ÷ 2 = 1 余 1；1 ÷ 2 = 0 余 1。从下向上读取余数得到 110012。
Always check your answer by converting back to denary.
务必通过反向转换回十进制来检查答案。
3. Binary Addition Rules | 二进制加法规则
ws four simple rules. Remember that 1 + 1 = 0 with a carry of 1 to the next higher column.
二进制加法遵循四条简单规则。请记住 1 + 1 = 0，同时向高位进1。
| 0 + 0 = 0 | 0 + 1 = 1 |
| 1 + 0 = 1 | 1 + 1 = 0, carry 1 |
When adding multiple columns, handle carries just as in denary addition. If a column adds up to 2 (102), write down 0 and carry 1; if it adds up to 3 (112), write down 1 and carry 1.
当进行多列加法时，像十进制加法一样处理进位。如果某一列相加结果为 2（102），则写 0 进 1；如果结果为 3（112），则写 1 进 1。
Example: 1011 + 0110 = 10001. Check: 11 + 6 = 17 in denary.
示例：1011 + 0110 = 10001。验算：11 + 6 = 17（十进制）。
4. Logic Gates & Symbols | 逻辑门与符号
Logic gates are the fundamental building blocks of digital circuits. The three basic gates are AND, OR and NOT. Their symbols and shapes are standardised in circuit diagrams.
逻辑门是数字电路的基本构建块。三种基本门是与门、或门和非门。在电路图中，它们的符号与形状都已标准化。
AND gate: inputs are 1. Symbol: a D-shaped block with a flat back. Boolean expression: Q = A AND B.
与门：仅当所有输入均为 1 时输出才为 1。符号：半圆形，背面是平的。布尔表达式：Q = A AND B。
OR gate: Symbol: a pointed shape like a shield, curving to a point. Boolean expression: Q = A OR B.
或门：只要至少一个输入为 1，输出就是 1。符号：形如盾牌，后端弯曲成尖角。布尔表达式：Q = A OR B。
NOT gate: ut. : Output is the opposite of the input. Symbol: a triangle followed by a small circle (bubble). Boolean expression: Q = NOT A.
非门：输出与输入相反。符号：一个三角形后跟一个小圆圈。布尔表达式：Q = NOT A。
Combinations of these gates form more complex circuits such as NAND, NOR and XOR.
这些门的组合可以构成更复杂的电路，如与非门、或非门和异或门。
5. Truth Tables for Basic Gates | 基本逻辑门真值表
A truth table lists every possible combination of inputs and the corresponding output for a logic gate. Learning to construct and read truth tables is an essential skill.
真值表列出了逻辑门所有可能的输入组合及其对应的输出。学会构建和阅读真值表是一项基本技能。
AND gate truth table
与门真值表
| A | B | Q = A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
OR gate truth table
或门真值表
| A | B | Q = A OR B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
NOT gate truth table
非门真值表
| A | Q = NOT A |
|---|---|
| 0 | 1 |
| 1 | 0 |
For two-input gates there are 22 4 possible input combinations. Always list them in binary counting order: 00, 01, 10, 11.
对于双输入门，共有 22 = 4 种输入组合。务必按二进制计数的顺序列出：00, 01, 10, 11。
6. Data Storage Units | 数据存储单位
Computers store data in binary. The smallest unit is a bit (binary digit). All larger units are based on powers of 2. Use these conversion formulas when moving between units.
计算机以二进制形式存储数据。最小的单位是比特（二进制位）。所有更大的单位都基于 2 的幂。在不同单位之间转换时，请使用下列换算公式。
1 nibble = 4 bits; 1 byte = 8 bits; 1 kilobyte (KB) = 1024 bytes = 210 bytes.
1 nibble = 4 比特；1 字节 = 8 比特；1 千字节 (KB) = 1024 字节 = 210 字节。
1 megabyte (MB) = 1024 KB = 220 bytes; 1 gigabyte (GB) = 1024 MB = 230 bytes; 1 terabyte (TB) = 1024 GB = 240 bytes.
1 兆字节 (MB) = 1024 KB = 220 字节；1 吉字节 (GB) = 1024 MB = 230 字节；1 太字节 (TB) = 1024 GB = 240 字节。
go smaller to larger, divide. For example, 5 MB = 5 × 1024 × 1024 bytes = 5,242,880 bytes.
将较大单位转换为较小单位时用乘法；将较小单位转换为较大单位时用除法。例如，5 MB = 5 × 1024 × 1024 字节 = 5,242,880 字节。
Remember that hard drive manufacturers sometimes use powers of 1000, but in Computer Science we follow the binary standard.
请记住，硬盘制造商有时使用 1000 的幂次，但在计算机科学中我们遵循二进制标准。
7. Image Representation Formulas | 图像表示公式
Bitmap images are made of pixels. Two key factors control file size and quality: resolution and colour depth. The formula for the file size of an uncompressed bitmap image is straightforward.
位图图像由像素组成。两个关键因素控制着文件大小和质量：分辨率和颜色深度。未压缩位图文件大小的计算公式非常直观。
mage file size (bits) = Width in pixels × Height in pixels × Colour depth (bits per pixel)
图像文件大小（比特）= 宽（像素）× 高（像素）× 颜色深度（每像素比特数）
lour depth determines how many different colours can be represented. Number of colours = 2colour depth. For example, a colour depth of 4 bits gives 24 = 16 colours; 8 bits gives 256 colours; 24 bits gives about 16.7 million colours.
颜色深度决定了可以表示多少种不同颜色。颜色数 = 2颜色深度。例如，4 位颜色深度可得出 24 = 16 种颜色；8 位为 256 种颜色；24 位约 1670 万种颜色。
To convert the result into bytes, simply divide by 8. Always state whether your answer is in bits or bytes.
要将结果转换为字节，只需除以 8。务必注明你的答案单位是比特还是字节。
8. Sound Representation Formulas | 声音表示公式
Sound is sampled at regular intervals. The quality and file size depend on the sample rate and bit depth. The formula for the size of a mono audio clip is:
声音以固定的间隔进行采样。质量和文件大小取决于采样率和位深度。单声道音频片段大小的计算公式为：
Sound file size (bits) = Sample rate (Hz) × Bit depth (bits) × Duration (seconds)
音频文件大小（比特）= 采样率 (Hz) × 位深度（比特） × 时长（秒）
For stereo sound, multiply the result by 2 because there are two channels. A common sample rate is 44.1 kHz (44,100 Hz), and a standard bit depth is 16 bits.
对于立体声，需将结果乘以 2，因为有两个声道。常见的采样率为 44.1 kHz（44,100 Hz），标准位深度为 16 位。
Higher sample rates capture higher frequencies more accurately, and greater bit depths allow a wider dynamic range. The Nyquist theorem states that the sample rate must be at least twice the highest frequency to be reproduced; this is why CD audio uses around 44.1 kHz.
采样率越高，越能准确地捕捉高频声音；位深度越大，动态范围越广。奈奎斯特定理指出，采样率必须至少是所要重现的最高频率的两倍，这也是 CD 音频使用约 44.1 kHz 的原因。
9. Hexadecimal Introduction | 十六进制入门
Hexadecimal (base-16) is a shorthand for binary. Each hex digit represents exactly 4 bits (a nibble). It uses digits 0–9 and letters A–F for values 10–15.
十六进制（以 16 为基数）是二进制的简写形式。每一位十六进制数恰好代表 4 个比特（一个 nibble）。它使用 0–9 的数字以及字母 A–F 表示 10–15 的数值。
To convert binary to hex, split the binary number into groups of 4 from the right, then convert each group. Example: 101110102 → 1011 (B) and 1010 (A) → BA16.
要将二进制转换为十六进制，从右开始将二进制数分成四个一组的片段，然后转换每个组。示例：101110102 → 1011 (B) 和 1010 (A) → BA16。
The formula for hexadecimal to denary is: Value = dn-1 × 16n-1 + ... + d0 × 160. BA16 = 11 × 161 + 10 × 160 = 176 + 10 = 18610.
十六进制转换为十进制的公式为：数值 = dn-1 × 16n-1 + ... + d0 × 160。BA16 = 11 × 161 + 10 × 160 = 176 + 10 = 18610。
Hexadecimal is widely used to represent memory addresses and colour codes in web design (e.g., #FF0088).
十六进制广泛用于表示内存地址和网页设计中的颜色代码（例如 #FF0088）。
10. Parity & Error Detection | 奇偶校验与错误检测
Parity checking is a simple method to detect single-bit errors in transmitted data. A parity bit is added to a binary string so that the total number of 1s is either even (even parity) or odd (odd parity).
奇偶校验是一种用于检测传输数据中单个比特错误的简单方法。将一个奇偶校验位添加到二进制串中，使 1 的总数为偶数（偶校验）或奇数（奇校验）。
Even parity rule: Total number of 1s (including the parity bit) must be even. If the original data has an odd number of 1s, set the parity bit to 1; otherwise set it to 0.
偶校验规则：1 的总数（包括奇偶校验位）必须为偶数。若原始数据中 1 的个数为奇数，则将奇偶校验位设为 1；否则设为 0。
Odd parity rule: Total number of 1s must be odd. If the original data has an even number of 1s, set the parity bit to 1; otherwise set it to 0.
奇校验规则：1 的总数必须为奇数。若原始数据中 1 的个数为偶数，则将奇偶校验位设为 1；否则设为 0。
If the receiver counts the 1s and the parity does not match the agreed scheme, an error has occurred. Parity cannot correct errors, only detect an odd number of bit flips.
如果接收端统计 1 的个数，发现奇偶性与约定的方案不符，则说明发生了错误。奇偶校验不能纠正错误，只能检测奇数个比特翻转的情况。
More advanced codes, such as checksums and cyclic redundancy checks, build on these ideas but are covered in later years.
更高级的编码，如校验和与循环冗余校验，建立在这些思想之上，但会在更高年级学习。
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