📚 Year 10 WJEC Computer Science Formula & Theorem Quick Reference | WJEC 十年级计算机科学公式定理速查手册
This is your go-to revision guide for all the essential formulas, theorems, and quick rules you need to master in Year 10 WJEC Computer Science. Use the paired English–Chinese explanations to deepen your understanding and boost your exam confidence.
这是你备考 WJEC 十年级计算机科学时必不可少的速查手册，涵盖了所有核心公式、定理与快速法则。中英文对照讲解将帮助你加深理解，提升应试信心。
1. Number Systems: Binary, Denary, Hexadecimal | 数制：二进制、十进制与十六进制
Binary values are base‑2 numbers using only 0 and 1. Each place represents a power of 2, starting from 2⁰ on the right. To convert a binary number to denary, add the place values where a 1 appears.
二进制是基数为 2 的数制，只使用 0 和 1。每一位代表 2 的幂，最右侧为 2⁰。将二进制数转换为十进制时，只需将所有出现 1 的位值相加。
Example: 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀.
示例：1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀。
Hexadecimal (base 16) uses digits 0–9 and letters A–F, where A=10, B=11, … bits), so conversion between binary and hex is straightforward by grouping bits in fours.
十六进制（基数为 16）使用数字 0–9 以及字母 A–F（A=10, B=11, …, F=15）。每个十六进制位代表半个字节（4 位），因此通过将二进制每四位一组进行分组，即可方便地在二进制与十六进制之间转换。
| Decimal | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| Hex | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | A | B | C | D | E | F |
2. Binary Arithmetic: Addition and Shifts | 二进制算术：加法与移位
ork from right to left, carrying into the next column when the sum equals 2 (10₂) or 3 (11₂).
二进制加法遵循四条简单规则：0+0=0，0+1=1，1+0=1，1+1=0 进 1（1+1+进位=1 进 1）。始终从右向左进行计算，每当和等于 2（10₂）或 3（11₂）时，就向更高位列进位。
A left shift by n places multiplies the binary number by 2ⁿ. A right logical shift divides the unsigned binary number by 2ⁿ and discards the remainder.
左移 n 位相当于二进制数乘以 2ⁿ。逻辑右移 n 位相当于无符号二进制数除以 2ⁿ，并舍弃余数。
Example: 00001100 (12) shifted left twice = 00110000 (48). t shift twice = 00000011 (3). Be careful with overflow if a shifted 1 leaves the register.
示例：00001100（12）左移两次 = 00110000（48）。右移两次 = 00000011（3）。注意，如果移出的 1 超出寄存器，就会发生溢出。
3. Units of Data Storage | 数据存储单位
A bit is the smallest unit of data, storing a 0 or 1. A nibble is 4 bits, and a byte is 8 bits. All larger units are based on powers of 2, but exam questions often use the simplified 10³ rule unless stated otherwise.
比特（位）是数据的最小单位，存储 0 或 1。一个 nibble（半字节）为 4 位，一个字节为 8 位。所有更大的单位都基于 2 的幂，但除非另有说明，考试题目通常采用简化的 10³ 规则。
megabyte (MB) = 1000 KB (10⁶ bytes). 1 gigabyte (GB) = 1000 MB. 1 terabyte (TB) = 1000 GB.
1 千字节（KB）= 1000 字节（常见）或 1024（2¹⁰）字节。1 兆字节（MB）= 1000 KB（10⁶ 字节）。1 吉字节（GB）= 1000 MB。1 太字节（TB）= 1000 GB。
Petabyte (PB), exabyte (EB) are larger units. To convert between units, multiply or divide by 1000 (or 1024, where binary precision matters). Always check the context of the question.
拍字节（PB）、艾字节（EB）是更大的单位。进行单位换算时，乘以或除以 1000（或 1024，在二进制精度较重要的场合）。做题时务必结合题目背景。
4. Image File Size Calculation | 图像文件大小计算
. To convert the result into bytes, divide by 8; into kilobytes, divide by 8000 (or 8192).
图像文件大小（以位为单位）= 像素宽度 × 像素高度 × 颜色深度。将结果转换为字节，则除以 8；转换为千字节，则除以 8000（或 8192）。
Colour depth is the number of bits used to represent the colour of each pixel. For example, a 1‑bit image can display only 2 colours (black and white), while 24‑bit colour (True Colour) can display about 16.7 million different colours.
颜色深度是为每个像素分配的色彩位数。例如，1 位图像只能显示 2 种颜色（黑白），而 24 位色彩（真彩色）可显示约 1670 万种颜色。
Formula for the number of possible colours = 2colour depth. For 8‑ 2⁸ = 256 colours. Always express final file size answers in appropriate units (B, KB, MB).
可显示的颜色数公式 = 2颜色深度。对于 8 位色彩：2⁸ = 256 种颜色。最终答案应使用合适的单位（B、KB 或 MB）。
Example: 640×480 image, 24‑ its = 640×480×24 = 7,372,800 bits; in bytes = 7,372,800 ÷ 8 = 921,600 B; in KB ≈ 921.6 KB.
示例：640×480 图像，24 位色彩 → 位大小 = 640×480×24 = 7,372,800 位；以字节计 = 7,372,800 ÷ 8 = 921,600 B；以 KB 计 ≈ 921.6 KB。
5. Sound File Size Calculation | 声音文件大小计算
pth × duration (seconds) × number of channels. For mono, use 1 channel; for stereo, use 2.
声音文件大小（位）= 采样率（Hz）× 位深度 × 时长（秒）× 声道数。单声道时声道数为 1，立体声时为 2。
Sample rate is the number of samples taken per second. Standard rates are 44.1 kHz (CD quality) and 48 kHz. Bit depth is the number of bits per sample, often 16‑bit or 24‑bit.
采样率是每秒采集的样本数。标准采样率为 44.1 kHz（CD 品质）和 48 kHz。位深度是每个样本的位数，通常为 16 位或 24 位。
t depth × channels. You can use bit rate to estimate file size by multiplying by duration.
比特率（每秒位数）= 采样率 × 位深度 × 声道数。你可利用比特率乘以时长来估算文件大小。
Example: 10 seconds of stereo audio at 44.1 kHz, 16‑bit → size = 44,100 × 16 × 10 × 2 = 14,112,000 bits = 1,764,000 bytes ≈ 1.76 MB.
示例：10 秒立体声，44.1 kHz、16 位 → 大小 = 44,100 × 16 × 10 × 2 = 14,112,000 位 = 1,764,000 字节 ≈ 1.76 MB。
6. Text Encoding: ASCII and Unicode | 文本编码：ASCII 与 Unicode
ASCII (American Standard Code for Information Interchange) uses 7 bits to represent 128 characters, including English letters, digits, punctuation, and control codes. Extended ASCII uses 8 bits for 256 characters, adding accented letters and graphical symbols.
ASCII（美国信息交换标准代码）使用 7 位表示 128 个字符，包括英文字母、数字、标点符号及控制码。扩展 ASCII 使用 8 位表示 256 个字符，增加了重音字母和图形符号。
Unicode was introduced to support global languages and emojis. The most common form, UTF‑8, uses 8‑bit blocks and can encode over 1 million characters, remaining backward‑compatible with ASCII.
Unicode 旨在支持全球语言和表情符号。最常见的 UTF‑8 使用 8 位块，可对超过一百万个字符进行编码，同时保持与 ASCII 的向后兼容。
Plain ASCII text uses 7 or 8 bits per character; Unicode files may use 8, 16, or 32 bits per character depending on encoding.
估算文本文件大小：字符数 × 每个字符的位数。普通 ASCII 文本每个字符占用 7 或 8 位；Unicode 文件根据编码方式可能每个字符占用 8、16 或 32 位。
7. Compression: Lossy, Lossless and Ratios | 压缩：有损、无损与压缩比
Compression reduces file size to save storage space and speed up transmission. Lossless compression allows the original data to be reconstructed perfectly; lossy compression permanently removes some data, reducing quality.
压缩通过减小文件大小来节省存储空间并加快传输速度。无损压缩可完全还原原始数据；有损压缩会永久性移除部分数据，导致质量下降。
/ (compressed size). A higher ratio indicates stronger compression. For example, a file reduced from 800 KB to 200 KB has a ratio of 4:1.
压缩比 =（原始大小）/（压缩后大小）。比值越高，表示压缩程度越强。例如，一个文件从 800 KB 缩小至 200 KB，其压缩比为 4:1。
Saving percentage = ((original − compressed) / original) × 100%. In the example above, saving = (600/800) × 100% = 75%.
节省百分比 =（（原始大小 − 压缩大小）/ 原始大小）× 100%。上述例子中，节省 =（600/800）× 100% = 75%。
Run‑length encoding (RLE) is a simple lossless method: consecutive repeated symbols are replaced by a count and the symbol. ‘AAABBB’ becomes ‘3A3B’.
游程编码（RLE）是一种简单的无损方法：连续的重复符号被替换为计数与该符号。“AAABBB”变为“3A3B”。
8. Network Performance: Latency and Bandwidth | 网络性能：延迟与带宽
Bandwidth is the maximum amount of data that can be transmitted over a network in a given time, measured in bits per second (bps). Higher bandwidth allows more data to flow simultaneously.
带宽是在给定时间内可通过网络传输的最大数据量，以每秒比特数（bps）衡量。带宽越高，可同时传输的数据就越多。
le size (bits) / bandwidth (bps). Always convert file size to bits if bandwidth is in bps.
传输时间（秒）= 文件大小（位）/ 带宽（bps）。如果带宽以 bps 为单位，务必将文件大小换算为位。
aused by propagation time and processing delays. Total time = transmission time + latency.
延迟是传输开始前的等待时间，通常由传播时间和处理延迟引起。总时间 = 传输时间 + 延迟。
5×8 = 40 megabits. Time = 40 / 100 = 0.4 seconds (ignoring latency and protocol overhead).
示例：在 100 Mbps 连接上传输 5 MB 文件。5 MB = 5×8 = 40 兆位。时间 = 40 / 100 = 0.4 秒（忽略延迟和协议开销）。
9. Boolean Algebra Laws | 布尔代数定律
Boolean algebra deals with variables that can be TRUE (1) or FALSE (0). The three basic operations are AND (conjunction), OR (disjunction), and NOT (negation).
布尔代数处理取值为 TRUE（1）或 FALSE（0）的变量。三种基本运算为 AND（合取）、OR（析取）和 NOT（否定）。
dempotent laws: A AND A = A; A OR A = A. Complement laws: A AND NOT A = 0; A OR NOT A = 1.
同一律：A AND 1 = A；A OR 0 = A。零一律：A AND 0 = 0；A OR 1 = 1。幂等律：A AND A = A；A OR A = A。互补律：A AND NOT A = 0；A OR NOT A = 1。
De Morgan' OT (A OR B) = (NOT A) AND (NOT B). Use them to simplify expressions or convert gate types.
德摩根定理至关重要：NOT (A AND B) = (NOT A) OR (NOT B)；NOT (A OR B) = (NOT A) AND (NOT B)。可用于化简表达式或转换门类型。
tive law: A AND (B OR C) = (A AND B) OR (A AND C); A OR (B AND C) = (A OR B) AND (A OR C).
分配律：A AND (B OR C) = (A AND B) OR (A AND C)；A OR (B AND C) = (A OR B) AND (A OR C)。
10. Logic Gates and Truth Tables | 逻辑门与真值表
A logic gate is a physical device implementing a Boolean function. Key gates: NOT (inverter), AND, OR, XOR (exclusive OR), NAND, and NOR.
逻辑门是实现布尔函数的物理器件。主要门电路：NOT（反相器）、AND、OR、XOR（异或）、NAND 和 NOR。
AND gate: output is 1 only when all inputs are 1. OR gate: output is 1 when at least one input is 1. NOT gate: produces the inverse of its single input.
AND 门：仅当所有输入均为 1 时，输出才为 1。OR 门：只要至少一个输入为 1，输出即为 1。NOT 门：对其单一输入取反。
XOR gate: output is 1 when inputs are different. NAND and NOR are the opposite of AND and OR; they are known as universal gates because any Boolean function can be implemented using only NAND or only NOR gates.
XOR 门：当输入不同时输出为 1。NAND 和 NOR 分别是 AND 和 OR 的反相；它们被称为通用门，因为任何布尔函数均可仅用 NAND 门或仅用 NOR 门来实现。
| A | B | A AND B | A OR B | A XOR B | NAND | NOR |
| 0 | 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 | 0 | 0 |
11. Sorting and Searching Algorithm Efficiency | 排序与查找算法效率
Linear search examines each element in a list until the target is found. In the worst case it checks n elements; time complexity is O(n). It works on unsorted data.
线性查找逐个检查列表中元素，直到找到目标值。最坏情况下需检查 n 个元素；时间复杂度为 O(n)。可用于无排序数据。
Binary search requires a sorted list. It repeatedly divides the list in half, comparing the middle element with the target. Each step halves the search space, leading to a worst‑case time complexity of O(log n).
二分查找要求列表已排序。它重复将列表一分为二，将中间元素与目标值比较。每一步将搜索空间减半，最坏情况时间复杂度为 O(log n)。
Bubble sort repeatedly steps through the list, swapping adjacent out‑of‑order elements. It requires up to n(n‑1)/2 comparisons, so complexity is O(n²). Insertion sort builds a sorted list one item at a time, also O(n²) on average. Merge sort uses a divide‑and‑conquer approach and guarantees O(n log n).
冒泡排序反复遍历列表，交换相邻的顺序错误元素。最多需要 n(n‑1)/2 次比较，因此复杂度为 O(n²)。插入排序一次构建一个有序项，平均也为 O(n²)。归并排序采用分治法，复杂度保证为 O(n log n)。
Students need to be able to trace these algorithms and compare their performance on small data sets.
学生需要能够追踪这些算法，并比较它们在小型数据集上的表现。
12. Error Detection and Parity Bits | 错误检测与奇偶校验位
Errors can occur during data transmission or storage. A parity bit is a simple error‑detection method. For even parity, the number of 1‑bits in the data plus the parity bit must be even. For odd parity, that total must be odd.
数据在传输或存储过程中可能发生错误。奇偶校验位是一种简单的错误检测方法。对于偶校验，数据中 1 的个数加上校验位后必须为偶数。对于奇校验，该总和必须为奇数。
The sender calculates the parity bit and appends it to the data. The receiver recalculates the parity and compares: if they differ, an error has occurred. Parity can only detect an odd number of bit errors.
发送方计算校验位并将其附加到数据中。接收方重新计算奇偶性并比较：若不一致，则表示发生了错误。奇偶校验只能检测奇数个比特错误。
Checksums use an arithmetic sum of data bytes, often with wrapping, to create a check value. They can detect more complex errors but still are not foolproof. Check digits (e.g., MOD 11) are used in barcodes and ISBN numbers.
校验和利用数据字节的算术和（通常采用回绕）生成校验值。它能检测更复杂的错误，但仍非万无一失。校验数位（如模 11）用于条形码和 ISBN 号。
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