📚 Year 11 Edexcel Computer Science: Quick Reference Handbook of Formulas & Theorems | Year 11 Edexcel 计算机：公式定理速查手册
This quick reference handbook collects the essential formulas, theorems, and key numerical relationships required for the Edexcel GCSE Computer Science (9-1) specification. Use it alongside your revision to sharpen calculations, logic, and algorithmic reasoning.
本速查手册汇集了 Edexcel GCSE 计算机科学（9-1）考纲中必备的公式、定理和关键数量关系。在复习时配合使用，可以强化你的计算、逻辑和算法推理能力。
1. Data Units and Conversion | 数据单位与转换
All data is represented in bits (b). A bit is the smallest unit of data, taking the value 0 or 1. Larger units follow powers of 2.
所有数据都以比特（b）表示。比特是数据的最小单位，取值为 0 或 1。更大的单位遵循 2 的幂次关系。
- Bit (b): 0 or 1
比特 (b)：0 或 1 - Nibble: 4 bits
半字节 (Nibble)：4 比特 - Byte (B): 8 bits
字节 (B)：8 比特 - Kilobyte (KB): 10³ bytes = 1,000 bytes (often 10³ for storage, but note 1 KiB = 1024 bytes)
千字节 (KB)：10³ 字节 = 1,000 字节（存储常用 10³，但注意 1 KiB = 1024 字节） - Megabyte (MB): 10⁶ bytes
兆字节 (MB)：10⁶ 字节 - Gigabyte (GB): 10⁹ bytes
吉字节 (GB)：10⁹ 字节 - Terabyte (TB): 10¹² bytes
太字节 (TB)：10¹² 字节
se binary multiples: 1 KiB = 2¹⁰ bytes = 1024 bytes; 1 MiB = 2²⁰ bytes; 1 GiB = 2³⁰ bytes.
精确的二进制倍数：1 KiB = 2¹⁰ 字节 = 1024 字节；1 MiB = 2²⁰ 字节；1 GiB = 2³⁰ 字节。
2. Binary Arithmetic and Logical Shifts | 二进制算术与逻辑移位
Binary addition follows four basic rules. When adding 1 + 1, the result is 0 with a carry of 1 to the next column. Overflow occurs when the result of an addition exceeds the available number of bits (e.g. > 255 for 8-bit unsigned).
二进制加法遵循四条基本规则。1 + 1 时结果为 0，并向高位进 1。当加法结果超出可用位数（例如 8 位无符号数超过 255）时，发生溢出。
0 + 1 = 1 1 + 0 = 1 1 + 1 = 0, carry 1 1 + 1 + 1 = 1, carry 1
Logical left shift: each bit moves left by n places; vacant LSBs are filled with 0. This multiplies the number by 2ⁿ (integer effect).
逻辑左移：每位向左移动 n 位；空出的最低有效位 (LSB) 填入 0。相当于将数值乘以 2ⁿ（整数效果）。
Logical right shift: each bit moves right by n places; vacant MSBs are filled with 0. This divides the unsigned integer by 2ⁿ, discarding any remainder.
逻辑右移：每位向右移动 n 位；空出的最高有效位 (MSB) 填入 0。相当于将无符号整数除以 2ⁿ 并丢弃余数。
Value after left shift by n = original value × 2ⁿ
Value after right shift by n = ⌊original value ÷ 2ⁿ⌋
3. Hexadecimal System | 十六进制系统
Hexadecimal (base‑16) uses digits 0–9 and letters A– ). It provides a compact way to represent binary: one hex digit corresponds to exactly 4 bits.
十六进制（基数为 16）采用数字 0–9 和字母 A–F（A=10, B=11, C=12, D=13, E=14, F=15）。它能紧凑地表示二进制：一个十六进制位对应 4 个比特。
| Denary | Hex | Binary |
|---|---|---|
| 0 | 0 | 0000 |
| 1 | 1 | 0001 |
| 10 | A | 1010 |
| 15 | F | 1111 |
t), then replace each group with the corresponding hex digit. For example, 1101 0110 → D6.
将二进制转换为十六进制时，从右侧开始将二进制数划分为 4 位一组，然后将每组替换为对应的十六进制位。例如，1101 0110 → D6。
Denary to hex: repeatedly divide by 16 and read remainders in reverse.
十进制转十六进制：反复除以 16，然后逆序读取余数。
4. Boolean Logic and Truth Tables | 布尔逻辑与真值表
Boolean logic operates with TRUE (1) and FALSE (0). The fundamental gates AND, OR, and NOT form the building blocks of digital circuits.
布尔逻辑用 TRUE (1) 和 FALSE (0) 进行运算。基础门电路 AND、OR 和 NOT 是数字电路的构建模块。
AND gate: Output is 1 only if both inputs are 1.
AND 门：仅当两个输入都为 1 时输出才为 1。
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
OR gate: Output is 1 if at least one input is 1.
OR 门：只要至少一个输入为 1，输出即为 1。
| A | B | A OR B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
NOT gate: Outputs the inverse of the input.
NOT 门：输出与输入相反。
| A | NOT A |
|---|---|
| 0 | 1 |
| 1 | 0 |
5. Boolean Algebra and De Morgan's Laws | 布尔代数与德摩根定律
Boolean algebra supplies rules for simplifying logic expressions. De Morgan's laws are vital for circuit design and can be proved with truth tables.
布尔代数提供了化简逻辑表达式的规则。德摩根定律对电路设计至关重要，并可用真值表证明。
De Morgan's Law 1: ¬(A ∧ B) = (¬A) ∨ (¬B)
德摩根定律 1: NOT (A AND B) = (NOT A) OR (NOT B)
De Morgan's Law 2: ¬(A ∨ B) = (¬A) ∧ (¬B)
德摩根定律 2: NOT (A OR B) = (NOT A) AND (NOT B)
Other useful identities include Commutative (A ∧ B = B ∧ A), Associative (A ∧ (B ∧ C) = (A ∧ B) ∧ C), Distributive (A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C)), and Absorption laws.
其他有用的恒等式包括交换律 (A ∧ B = B ∧ A)、结合律 (A ∧ (B ∧ C) = (A ∧ B) ∧ C)、分配律 (A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C)) 以及吸收律。
6. Image Representation and File Size | 图像表示与文件大小
f pixels. The colour depth (bits per pixel) determines the number of available colours = 2colour depth.
位图图像由像素构成。颜色深度（每像素比特数）决定了可用颜色数 = 2颜色深度。
Number of colours = 2colour depth in bits
颜色数 = 2颜色深度，以比特计
The file size of an uncompressed bitmap (in bits) is calculated as:
未压缩位图的文件大小（以比特计）计算如下：
Image file size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
图像文件大小 (比特) = 宽度 (像素) × 高度 (像素) × 颜色深度 (每像素比特数)
btain size in bytes, divide the result by 8. For kB, divide by 8000 (or 8 × 1024 for KiB).
以字节为单位时，将结果除以 8。以千字节 (KB) 为单位时，除以 8000（若以 KiB 计则除以 8×1024）。
7. Sound Representation and File Size | 声音表示与文件大小
ach sample to a fixed bit depth. More channels (e.g. stereo = 2) increase the data volume.
声音通过按固定时间间隔（采样率）测量振幅并将每个样本量化为固定比特深度来采样。更多声道（如立体声 = 2）会增加数据量。
file size (bits) = sample rate (Hz) × bit depth × duration (seconds) × number of channels
声音文件大小 (比特) = 采样率 (Hz) × 比特深度 × 时长 (秒) × 声道数
Convert to bytes: divide by 8.
转换为字节：除以 8。
Typical sample rates: 44.1 kHz (CD quality), 48 kHz. Common bit depths: 16 bits, 24 bits.
典型采样率：44.1 kHz（CD 音质）、48 kHz。常见比特深度：16 比特、24 比特。
8. Data Transmission Time and Compression Ratios | 数据传输时间与压缩比
The time needed to transfer a file depends on its size and the available transfer rate (bandwidth). Always match units – convert file size to bits if the rate is in bps.
传输文件所需的时间取决于文件大小和可用的传输速率（带宽）。请务必统一单位——如果速率以 bps 表示，则将文件大小转换为比特。
Transfer time (seconds) = file size (bits) ÷ transfer rate (bps)
传输时间 (秒) = 文件大小 (比特) ÷ 传输速率 (bps)
Compression reduces storage and transmission needs. The compression ratio compares the original size to the compressed size.
压缩能降低存储和传输需求。压缩比用于比较原始大小与压缩后的大小。
Compression ratio = original size ÷ compressed size
压缩比 = 原始大小 ÷ 压缩后大小
Lossless compression allows perfect reconstruction (e.g. run‑length encoding, Huffman coding). Lossy compression sacrifices some detail to achieve smaller files (e.g. JPEG, MP3).
无损压缩允许完全重建（如游程编码、哈夫曼编码）。有损压缩牺牲部分细节以获得更小的文件（如 JPEG、MP3）。
9. Algorithm Analysis: Linear vs Binary Search | 算法分析：线性搜索与二分搜索
Linear search checks each element in turn. It works on unsorted data. The worst‑case number of comparisons is n, where n is the number of items.
线性搜索依次检查每个元素。它可用于未排序的数据。最坏情况下的比较次数为 n，其中 n 为元素个数。
Linear search worst‑case: n comparisons
线性搜索最坏情况：n 次比较
Average linear search: (n+1)/2 comparisons
平均线性搜索：(n+1)/2 次比较
Binary search repeatedly divides a sorted list in half. It eliminates half the remaining items with each check. The maximum number of comparisons is approximately log₂(n).
二分搜索重复地将已排序的列表对半分。每次检查能排除剩余元素的一半。最大比较次数约为 log₂(n)。
Binary search worst‑case: ⌊log₂(n)⌋ + 1 comparisons
二分搜索最坏情况：⌊log₂(n)⌋ + 1 次比较
For n = 1000, linear search may need 1000 checks; binary search needs at most 10 checks.
对于 n = 1000，线性搜索可能需要 1000 次检查；二分搜索最多只需 10 次检查。
10. Sorting Algorithms Overview | 排序算法概览
Sorting algorithms arrange data into order. Bubble sort repeatedly steps through the list, swapping adjacent elements if they are in the wrong order. The largest unsorted element "bubbles" to the end in each pass.
排序算法将数据按序排列。冒泡排序反复遍历列表，若相邻元素顺序错误则交换。每趟遍历将未排序的最大元素“冒泡”至末尾。
Bubble sort worst‑case comparisons:
冒泡排序最坏情况比较次数：
n(n − 1) ÷ 2
Merge sort uses a divide‑and‑conquer approach. It splits the list into halves recursively until each sub‑list contains one item, then merges them back in sorted order. It guarantees O(n log₂ n) comparisons.
归并排序采用分治法。它递归地将列表对半分，直至每个子列表只含一个元素，然后按序合并。它能保证 O(n log₂ n) 次比较。
Insertion sort builds a sorted portion by taking one element at a time and inserting it into its correct position. It is efficient for small or partially sorted data.
插入排序通过逐个取出元素并将其插入到已排序部分的正确位置来构建有序序列。它对于小规模或部分有序的数据很高效。
11. CPU Clock Speed and Cycle Time | CPU 时钟速度与周期时间
The CPU operates on a clock signal. Clock speed (frequency) measures cycles per second, expressed in hertz (Hz). A higher clock speed typically means more instructions can be executed per second.
CPU 依赖时钟信号运行。时钟速度（频率）衡量每秒的周期数，以赫兹 (Hz) 表示。更高的时钟速度通常意味着每秒能执行更多的指令。
Clock speed (Hz) = 1 ÷ cycle time (s)
时钟速度 (Hz) = 1 ÷ 周期时间 (秒)
one clock cycle. For a 3 GHz CPU, cycle time = 1 / (3 × 10⁹) ≈ 3.33 × 10⁻¹⁰ s (0.333 ns).
周期时间是一个时钟周期的时长。对于 3 GHz 的 CPU，周期时间 = 1 / (3×10⁹) ≈ 3.33×10⁻¹⁰
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