📚 Formula & Theorem Quick Reference Guide for Year 10 Eduqas Computer Science | 公式定理速查手册
This quick reference handbook brings together the essential formulas, theorems, and conversion rules from the Year 10 Eduqas Computer Science specification. Use it to revise binary arithmetic, data representation, Boolean logic, compression, and basic algorithmic efficiency. Each topic is presented with a concise explanation followed by the key equation expressed in clear notation.
这本速查手册汇整了 Year 10 Eduqas 计算机科学课程的核心公式、定理和转换规则。你可以用它来复习二进制运算、数据表示、布尔逻辑、压缩以及基本算法效率。每个主题都配有简明解释，其后用清晰符号给出关键公式。
1. Binary to Decimal Conversion | 二进制转十进制
To convert a binary number into decimal, multiply each bit by its place‑value power of 2 and sum the results. The rightmost bit holds position 0, the next position 1, and so on.
将二进制数转换为十进制，将每一位乘以其位值的 2 次幂，然后求和。最右边的位为 0 位，向左依次为 1、2 等。
Decimal = dₙ₋₁ × 2ⁿ⁻¹ + … + d₁ × 2¹ + d₀ × 2⁰
For example, to convert the 8‑ ⁵) + (1×2⁴) + (0×2³) + (1×2²) + (1×2¹) + (0×2⁰) = 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182.
例如，转换 8 位二进制数 10110110₂： (1×2⁷) + (0×2⁶) + (1×2⁵) + (1×2⁴) + (0×2³) + (1×2²) + (1×2¹) + (0×2⁰) = 182。
2. Decimal to Binary Conversion | 十进制转二进制
Repeatedly divide the decimal number by 2, recording the remainder each time. Continue until the quotient is zero. The binary result is the sequence of remainders read from bottom (last remainder) to top (first remainder).
反复将十进制数除以 2，记录余数。持续执行直到商为零。二进制结果是从下往上读取的余数序列。
Binary = (remainderₖ … remainder₁ remainder₀)₂
1, 2 ÷ 2 = 1 remainder 0, 1 ÷ 2 = 0 remainder 1. Reading remainders upwards gives 101101₂.
将 45 转换为二进制：45 ÷ 2 商 22 余 1，继续得到余数 0、1、1、0、1。向上读取余数为 101101₂。
3. Binary to Hexadecimal Conversion | 二进制转十六进制
Group the binary digits into nibbles (groups of four) from right to left. Pad the leftmost group with leading zeros if necessary. Convert each nibble to its equivalent hexadecimal digit (0–9, A–F).
从右向左将二进制数分成四位一组的半字节。必要时在左侧补零。将每个半字节转换为对应的十六进制数字 (0–9, A–F)。
Hex digit = value of 4‑bit nibble (0–9, A–F)
Binary 11011110₂: groups are 1101 (D) and 1110 (E), giving DE₁₆. The whole‑byte conversion can be expressed as DE where each hex digit represents exactly four bits.
二进制 11011110₂：组为 1101 (D) 和 1110 (E)，得到 DE₁₆。每个十六进制数字恰好代表四位二进制位。
4. Data Unit Conversions | 数据单位转换
Understanding the relationships between bits, bytes, kilobytes, megabytes, and gigabytes is fundamental for storage and transmission problems. In the Eduqas specification, units are typically powers of two for internal memory and powers of ten for storage device capacities.
理解比特、字节、千字节、兆字节和吉字节之间的关系是解决存储与传输问题的基础。在 Eduqas 规范中，内存单位通常采用 2 的幂，而存储设备容量采用 10 的幂。
1 byte (B) = 8 bits (b)
1 字节 (B) = 8 比特 (b)
1 kibibyte (KiB) = 2¹⁰ B = 1024 B (binary)
1 千比字节 (KiB) = 1024 B
1 kilobyte (kB) = 10³ B = 1000 B (decimal)
1 千字节 (kB) = 1000 B
1 mebibyte (MiB) = 2²⁰ B ≈ 1 048 576 B
1 兆比字节 (MiB) ≈ 1,048,576 B
1 megabyte (MB) = 10⁶ B = 1 000 000 B
1 兆字节 (MB) = 1,000,000 B
Always read the context: operating systems often report sizes in binary multiples, whereas hard‑drive manufacturers use decimal multiples.
务必根据上下文判断：操作系统通常以二进制倍数显示大小，而硬盘制造商使用十进制倍数。
5. File Size – Image | 图像文件大小
The uncompressed size of a bitmap image depends on its width, height, and colour depth (bits per pixel). The formula multiplies the total number of pixels by the bits needed to represent each pixel and converts the result into bytes.
未压缩的位图图像大小取决于宽度、高度和颜色深度（每像素位数）。将像素总数乘以表示每像素所需的位数，然后转换为字节。
Image size (bytes) = (width × height × colour depth in bits) ÷ 8
For a 1024 × 768 image with 24‑ calculation is: (1024 × 768 × 24) ÷ 8 = 2 359 296 bytes, which is approximately 2.25 MiB.
对于 1024 × 768、24 位真彩色的图像：(1024 × 768 × 24) ÷ 8 = 2,359,296 字节，约为 2.25 MiB。
If the image uses indexed colour (e.g. 8‑bit palette), substitute the colour depth accordingly – 8 instead of 24.
若图像使用索引色（如 8 位调色板），则将颜色深度替换为 8。
6. File Size – Sound | 声音文件大小
Uncompressed sound file size is determined by the sample rate, bit depth, number of channels, and duration. Multiply these four factors together to obtain the total number of bits, then divide by 8 to get bytes.
未压缩的声音文件大小由采样率、位深度、声道数和时长决定。将这四个因素相乘得到总比特数，再除以 8 转换为字节。
Sound size (bytes) = (sample rate × bit depth × channels × duration in seconds) ÷ 8
Stereo CD‑quality audio (44 100 Hz, 16‑ ls) for 3 minutes (180 seconds): (44100 × 16 × 2 × 180) ÷ 8 = 31 752 000 bytes ≈ 30.3 MiB.
立体声 CD 音质（44,100 Hz、16 位、2 声道）3 分钟（180 秒）：(44100 × 16 × 2 × 180) ÷ 8 ≈ 30.3 MiB。
Mono halves the channel count, and reducing the sample rate or bit depth lowers the file size proportionally.
单声道将声道数减半，降低采样率或位深度也会成比例地减小文件大小。
7. Compression Ratio | 压缩比
Compression ratio quantifies how much a file has been reduced in size. It is the uncompressed size divided by the compressed size. A higher ratio indicates more aggressive compression.
压缩比用于量化文件大小减少的程度，它是未压缩大小除以压缩后大小。比值越高表示压缩越强。
Compression ratio = uncompressed size ÷ compressed size
imply 4). Space saving percentage can be calculated as (1 − compressed/uncompressed) × 100%.
如果一幅图像原始为 4 MB，经无损压缩后变为 1 MB，压缩比为 4:1（或直接写 4）。节省空间百分比 = (1 − 压缩后/压缩前) × 100%。
In the exam you may also be asked to describe the difference between lossy and lossless compression. Lossy removes some data permanently to achieve higher ratios; lossless allows perfect reconstruction.
考试中可能要求说明有损与无损压缩的区别。有损压缩会永久移除部分数据以获得更高压缩比；无损压缩可完美重建原始数据。
8. Boolean Algebra – Basic Laws | 布尔代数基本定律
Boolean algebra describes the behaviour of logic circuits using operators AND, OR, and NOT. The following laws are essential for simplifying logic expressions in the Eduqas specification.
布尔代数用 AND、OR、NOT 运算符描述逻辑电路的行为。以下定律是 Eduqas 规范中简化逻辑表达式的关键。
Identity laws: A AND 1 = A, A OR 0 = A
恒等律：A AND 1 = A，A OR 0 = A
Annulment (null) laws: A AND 0 = 0, A OR 1 = 1
零一律：A AND 0 = 0，A OR 1 = 1
Idempotent laws: A AND A = A, A OR A = A
幂等律：A AND A = A，A OR A = A
Complement laws: A AND NOT A = 0, A OR NOT A = 1
互补律：A AND NOT A = 0，A OR NOT A = 1
Double negation: NOT (NOT A) = A
双重否定：NOT (NOT A) = A
Commutative laws: A AND B = B AND A, A OR B = B OR A
交换律：A AND B = B AND A，A OR B = B OR A
Associative laws: (A AND B) AND C = A AND (B AND C)
结合律：(A AND B) AND C = A AND (B AND C)
Distributive laws: A AND (B OR C) = (A AND B) OR (A AND C)
分配律：A AND (B OR C) = (A AND B) OR (A AND C)
De Morgan's laws: NOT (A AND B) = (NOT A) OR (NOT B); NOT (A OR B) = (NOT A) AND (NOT B)
德摩根定律：NOT (A AND B) = (NOT A) OR (NOT B)；NOT (A OR B) = (NOT A) AND (NOT B)
These laws are used to minimise logic expressions and reduce the number of gates needed in a circuit.
这些定律用于简化逻辑表达式，并减少电路中所需门电路的数量。
9. Logic Gate Symbols & Truth Tables | 逻辑门符号与真值表
Eduqas candidates must recognise the standard symbols for NOT, AND, OR, XOR, NAND, and NOR gates and be able to complete a truth table for up to three inputs.
Eduqas 考生必须能识别 NOT、AND、OR、XOR、NAND、NOR 门的标准符号，并能填写最多三个输入的真值表。
| Gate | Boolean expression | Output description |
| NOT | Q = ¬A | Inverts the input |
| AND | Q = A ∧ B | True only if all inputs are 1 |
| OR | Q = A ∨ B | True if at least one input is 1 |
| XOR | Q = A ⊕ B | True when inputs differ |
| NAND | Q = ¬(A ∧ B) | Opposite of AND |
| NOR | Q = ¬(A ∨ B) | Opposite of OR |
For a two‑ gives 0 only for A=0, B=0; NAND gives 0 only for A=1, B=1; NOR gives 1 only for A=0, B=0.
对应两输入真值表：AND 只在 A=1, B=1 时输出 1；OR 只在 A=0, B=0 时输出 0；NAND 只在 A=1, B=1 时输出 0；NOR 只在 A=0, B=0 时输出 1。
10. Network Transmission Time | 网络传输时间
When a file is sent across a network, the transmission time depends on the file size and the network bit rate. Convert large file sizes to bits and divide by the bit rate.
文件通过网络传输时，传输时间取决于文件大小和网络比特率。将大文件大小转换为比特，再除以比特率。
Time (seconds) = file size (bits) ÷ bit rate (bps)
00 000 000 ≈ 8.39 seconds. Note the use of binary MiB and decimal Mbps as per common convention.
假设通过 100 Mbps 连接传输 100 MiB 文件，先将文件大小转换为位：100 × 1024 × 1024 × 8 = 838,860,800 比特。再除以速率：838,860,800 ÷ 100,000,000 ≈ 8.39 秒。注意 MiB 使用二进制、Mbps 使用十进制是常见惯例。
11. Algorithm Efficiency – Searching | 算法效率 – 搜索
Two fundamental search algorithms appear in the specification: linear search and binary search. Their worst‑case time complexity is expressed in Big‑O notation.
规范中包含两种基本搜索算法：线性搜索和二分搜索。它们的最坏情况时间复杂度用大 O 符号表示。
Linear search checks each element in turn: O(n).
线性搜索逐个检查每个元素：O(n)。
Binary search (requires sorted data) repeatedly halves the search space: O(log n).
二分搜索（要求数据已排序）反复将搜索区间减半：O(log n)。
1 000 comparisons; binary search requires at most about 10 comparisons because 2¹⁰ = 1024.
对于 1000 个元素的列表，线性搜索最多需要 1000 次比较；二分搜索最多需要约 10 次比较，因为 2¹⁰ = 1024。
The Eduqas course expects you to be able to explain the steps of each algorithm and compare their efficiency in different contexts, not to write complex complexity proofs.
Eduqas 课程要求你能够解释每种算法的步骤，并在不同情境中比较其效率，而无需写出复杂的时间复杂度证明。
12. Character Encoding – ASCII & Unicode | 字符编码 – ASCII 与 Unicode
Characters are stored as numerical codes. Extended ASCII uses 8 bits to represent up to 256 characters, while Unicode (typically UTF‑8 or UTF‑16) can encode over a million characters from all major writing systems. The key formula relates the number of bits to the number of representable symbols.
字符以数字编码存储。扩展 ASCII 用 8 位表示最多 256 个字符，而 Unicode（通常为 UTF‑8 或 UTF‑16）可编码超过一百万个字符，涵盖所有主要书写系统。关键公式将位数与可表示符号数联系起来。
Number of symbols = 2n (where n = number of bits)
With 7‑bit standard ASCII, 2⁷ = 128 characters; with 8‑bit extended ASCII, 2⁸ = 256 characters. Unicode’s greater bit depth (up to 32 bits in UTF‑ e in ASCII uses fewer bytes per character than one using Unicode with diacritics or emoji.
7 位标准 ASCII 有 2⁷ = 128 个字符；8 位扩展 ASCII 有 2⁸ = 256 个字符。Unicode 使用更大位深度（UTF‑32 可达 32 位）支持庞大的字符集。这解释了为什么纯 ASCII 文本文件每字符占用的字节数少于包含变音符号或表情符号的 Unicode 文本。
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