Year 10 CAIE Computer Science: Quick Reference Formulas and Theorems | Year 10 CAIE 计算机：公式定理速查手册

📚 Year 10 CAIE Computer Science: Quick Reference Formulas and Theorems | Year 10 CAIE 计算机：公式定理速查手册

This handbook provides a concise collection of essential formulas and theorems for the Year 10 CAIE Computer Science syllabus. Use it as a quick revision tool to master calculations in data representation, logic gates, and more.

本手册为 Year 10 CAIE 计算机科学课程整理了所有核心公式与定理，可作为考前速查工具，涵盖数据表示、逻辑门等计算重点。

---

1. Data Storage Units | 数据存储单位

The fundamental units of digital storage are the bit and the byte. All larger units are derived as powers of 2, and it is critical to remember the exact byte equivalents for each prefix.

数字存储的基本单位是比特和字节。所有更大的单位都以 2 的幂次方衍生，必须准确记住每个前缀对应的字节数。

Unit	Abbreviation	Equivalent in bytes
bit	b	1/8 byte
nibble	-	4 bits = 1/2 byte
byte	B	8 bits
kibibyte / kilobyte	KiB / kB	2¹⁰ bytes = 1 024 bytes
mebibyte / megabyte	MiB / MB	2²⁰ bytes = 1 048 576 bytes
gibibyte / gigabyte	GiB / GB	2³⁰ bytes = 1 073 741 824 bytes
tebibyte / terabyte	TiB / TB	2⁴⁰ bytes = 1 099 511 627 776 bytes

To convert a value from a larger unit to bytes, multiply by the appropriate power of 2. To go from bytes to a larger unit, divide by that same power.

将较大单位转换为字节时，乘以相应的 2 的幂；从字节转换为较大单位时，除以相同的 2 的幂。

Number of bytes = Value × 2ⁿ (where n = 10 for KiB, 20 for MiB, 30 for GiB, etc.)

---

2. Text File Size Calculation | 文本文件大小计算

The storage size of a plain text file depends on the number of characters and the number of bits used to encode each character. Common encoding standards include ASCII (7-bit or 8-bit) and Unicode (UTF-16 with 16 bits, UTF-32 with 32 bits).

纯文本文件的存储大小取决于字符数量以及编码每个字符所用的比特数。常见编码标准包括 ASCII（7 位或 8 位）和 Unicode（UTF-16 为 16 位，UTF-32 为 32 位）。

File size (bits) = Number of characters × Bits per character

File size (bytes) = (Number of characters × Bits per character) ÷ 8

For example, a file containing 500 characters encoded in standard 8‑bit ASCII requires 500 × 8 = 4 000 bits, or 500 bytes. The same text in UTF‑16 would need 500 × 16 = 8 000 bits (1 000 bytes).

例如，一个包含 500 个字符、使用标准 8 位 ASCII 编码的文件需要 500 × 8 = 4 000 bit，即 500 字节。同样的文本若使用 UTF‑16 编码，则需要 500 × 16 = 8 000 bit（1 000 字节）。

---

3. Image File Size Calculation | 图像文件大小计算

Bitmap images are stored as a grid of pixels. The file size (before compression) is determined by the image resolution and the colour depth (bit depth). Colour depth specifies how many bits are used to represent each pixel’s colour.

位图图像以像素网格形式存储。文件大小（压缩前）由图像分辨率和颜色深度（位深度）决定。颜色深度规定了用多少比特表示每个像素的颜色。

File size (bits) = Width (pixels) × Height (pixels) × Colour depth (bits per pixel)

File size (bytes) = (Width × Height × Colour depth) ÷ 8

An image of dimensions 800 × 600 with 24‑ 11 520 000 bits, which is 1 440 000 bytes, or approximately 1.37 MiB (dividing by 1024²).

一幅 800 × 600 像素、24 位真彩色的图像占用 800 × 600 × 24 = 11 520 000 bit，即 1 440 000 字节，约合 1.37 MiB（除以 1024²）。

---

4. Sound File Size Calculation | 声音文件大小计算

Digital audio storage depends on the sample rate, bit depth, duration, and the number of audio channels. The sample rate is the number of samples taken per second (Hz); bit depth is the number of bits per sample.

数字音频的存储取决于采样率、位深度、时长和声道数。采样率指每秒采集的样本数（Hz），位深度是每个样本所用的比特数。

File size (bits) = Sample rate (Hz) × Bit depth × Duration (seconds) × Number of channels

File size (bytes) = (Sample rate × Bit depth × Duration × Channels) ÷ 8

For a 3‑minute (180 s) stereo recording at 44 100 Hz with 16‑ s 44 100 × 16 × 180 × 2 = 254 016 000 bits, which equals 31 752 000 bytes (about 30.3 MB).

一段 3 分钟（180 秒）的立体声录音，采用 44 100 Hz 采样率和 16 位精度，其原始文件大小为 44 100 × 16 × 180 × 2 = 254 016 000 bit，即 31 752 000 字节（约 30.3 MB）。

---

5. Data Transfer Rate Calculation | 数据传输速率计算

Data transfer rate describes how fast data can be transmitted over a network or between components. The time required to transfer a file is calculated by dividing the file size by the transfer rate, ensuring that units match.

数据传输速率描述数据在网络或组件之间传输的快慢。传输文件所需的时间通过文件大小除以传输速率来计算，并需确保单位一致。

Transfer time (seconds) = File size (bits) ÷ Transfer rate (bits per second)

× 1024 × 8 = 419 430 400 bits, then divide by 100 000 000 bps, giving about 4.19 seconds.

若要通过 100 Mbps 连接发送一个 50 MiB 的文件，首先将文件转换为兆比特：50 × 1024 × 1024 × 8 = 419 430 400 bit，然后除以 100 000 000 bps，得到约 4.19 秒。

---

6. Binary and Denary Conversions | 二进制与十进制转换

Denary (decimal) numbers are base‑10; binary numbers are base‑2. To convert a binary integer to denary, sum the products of each bit multiplied by its corresponding power of 2, starting from the rightmost position (2⁰).

十进制数是以 10 为基数的数，二进制数是以 2 为基数的数。要将二进制整数转换为十进制，从最右侧的位（2⁰）开始，将每一位乘以对应的 2 的幂次方并求和。

Example: Binary 1011₂ = (1 × 2³) + (0 × 2²) + (1 × 2¹) + (1 × 2⁰) = 8 + 0 + 2 + 1 = 11₁₀.

示例：二进制 1011₂ = (1×2³)+(0×2²)+(1×2¹)+(1×2⁰) = 8+0+2+1 = 11₁₀。

To convert denary to binary, repeatedly divide the number by 2 and record the remainders. The binary equivalent is the sequence of remainders read from the last remainder upwards.

要将十进制转换为二进制，反复将数除以 2 并记录余数。二进制数即为从最后一个余数开始向上读取的余数序列。

---

7. Binary and Hexadecimal Conversions | 二进制与十六进制转换

Hexadecimal is base‑16, using digits 0–9 and letters A–F. Each hex digit directly represents a group of four binary digits (a nibble), making conversion straightforward.

十六进制以 16 为基数，使用数字 0–9 和字母 A–F。每个十六进制数字直接代表一组四位二进制数字（一个 nibble），转换非常简便。

To convert binary to hex, split the binary number into groups of four bits from the right, then replace each group with its hex equivalent. To convert hex to binary, replace each hex digit with its 4‑bit binary representation.

二进制转十六进制：从右向左每四位一组划分，不足左边补零，然后把每组转换成对应的十六进制数字。十六进制转二进制：把每个十六进制数字替换成其四位二进制表示。

Binary 11010111₂ split as 1101 0111 → D7₁₆. Hex 3F₁₆ → 0011 1111₂.

二进制 11010111₂ 分组为 1101 0111 → D7₁₆。十六进制 3F₁₆ → 0011 1111₂。

---

广告 / Sponsored

8. Binary Addition and Overflow | 二进制加法与溢出

Binary addition follows simple rules similar to denary addition. When the sum of a column exceeds 1, a carry is generated to the next significant column.

二进制加法遵循与十进制加法类似的简单规则。当一列的和超过 1 时，会产生一个进位到下一个更重要的列。

The basic addition rules:

基本加法规则：

• 0 + 0 = 0

  0 + 0 = 0

• 1 + 0 = 1

  1 + 0 = 1

• 1 + 1 = 0, carry 1

  1 + 1 = 0，进位 1

• 1 + 1 + 1 = 1, carry 1

  1 + 1 + 1 = 1，进位 1

Overflow occurs when the result of an addition exceeds the maximum value that can be stored in the allocated number of bits. For an 8‑bit unsigned integer, adding 11111111₂ (255) and 00000001₂ (1) gives 1 00000000₂, but the 9th bit is lost, causing overflow and an incorrect result.

当加法结果超出了分配的位数能存储的最大值时，就会发生溢出。对于 8 位无符号整数，11111111₂（255）加 00000001₂（1）得到 1 00000000₂，但第 9 位会丢失，导致溢出和错误结果。

---

9. Logic Gate Truth Tables | 逻辑门真值表

Logic gates are the fundamental building blocks of digital circuits. You must memorise the truth table and symbol for each gate.

逻辑门是数字电路的基本构建模块，必须熟记每种门的真值表和符号。

NOT gate (inverter)

非门（反相器）

Input A	Output Q
0	1
1	0

AND gate

与门

A	B	Q = A AND B
0	0	0
0	1	0
1	0	0
1	1	1

OR gate

或门

A	B	Q = A OR B
0	0	0
0	1	1
1	0	1
1	1	1

NAND gate

与非门

A	B	Q = A NAND B
0	0	1
0	1	1
1	0	1
1	1	0

NOR gate

或非门

A	B	Q = A NOR B
0	0	1
0	1	0
1	0	0
1	1	0

XOR gate

异或门

A	B	Q = A XOR B
0	0	0
0	1	1
1	0	1
1	1	0

---

10. Boolean Algebra Laws | 布尔代数定律

Boolean algebra is used to simplify logic expressions. The following laws are essential for reducing circuits and proving equivalence.

布尔代数用于化简逻辑表达式。以下定律对于简化电路和证明等价性至关重要。

Commutative Law (交换律): A + B = B + A; A • B = B • A

Associative Law (结合律): (A + B) + C = A + (B + C); (A • B) • C = A • (B • C)

Distributive Law (分配律): A • (B + C) = (A • B) + (A • C); A + (B • C) = (A + B) • (A + C)

Identity Law (同一律): A + 0 = A; A • 1 = A

Published by TutorHao | Year 10 Computer Science Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）