📚 Year 7 OCR Computer Science: Quick Reference Formulae & Theorems | Year 7 OCR 计算机：公式定理速查手册
This handbook brings together the essential formulae, conversion rules, boolean laws and size calculation theorems every Year 7 OCR Computer Science student needs at their fingertips. Use it to tackle binary challenges, simplify logic expressions and work out file sizes with confidence.
本手册汇集了所有 Year 7 OCR 计算机科学学生必须掌握的公式、转换规则、布尔定律和大小计算定理，方便随时查阅。用它来攻克二进制难题、化简逻辑表达式并自信地计算文件大小。
1. Bits and Bytes | 位与字节
A bit is the smallest unit of data in a computer, holding either a 0 or a 1. Eight bits together form one byte, which can represent 2⁸ = 256 different values.
位（bit）是计算机中最小的数据单位，只能存储 0 或 1。8 个位组成一个字节（byte），可以表示 2⁸ = 256 种不同的值。
When we group bits, each extra bit doubles the number of possible combinations. The formula for the number of values with n bits is 2ⁿ.
当我们把位组合起来时，每增加一位，可能的组合数就翻倍。用 n 位可以表示的数值数量公式是 2ⁿ。
2. Binary Number Basics | 二进制基础
Binary is a base‑2 number system using only the digits 0 and 1. Each column in a binary number represents a power of 2, starting from 2⁰ on the right.
二进制是一种以 2 为基数的数字系统，只使用 0 和 1 两个数字。二进制数中的每一列都代表 2 的幂，从最右边的 2⁰ 开始。
For an 8‑bit binary number, the place values are: 2⁷ (128), 2⁶ (64), 2⁵ (32), 2⁴ (16), 2³ (8), 2² (4), 2¹ (2), 2⁰ (1). We place 1s and 0s in these columns to build any number from 0 to 255.
对于 8 位二进制数，位值依次为：2⁷ (128)、2⁶ (64)、2⁵ (32)、2⁴ (16)、2³ (8)、2² (4)、2¹ (2)、2⁰ (1)。我们在这些列中填入 1 或 0，就可以构建从 0 到 255 的任意数字。
3. Decimal to Binary Conversion Formula | 十进制转二进制公式
To convert a decimal number to binary, repeatedly divide the number by 2 and record the remainder at each step. Read the remainders from bottom to top to obtain the binary equivalent.
要将十进制数转换为二进制，需要反复将该数除以 2 并记录每一步的余数。从下往上读取余数，就得到了二进制表示。
Formally, keep dividing until the quotient becomes 0: Remainder₀ (least significant bit), Remainder₁, … , Remainderₙ₋₁ (most significant bit).
正式的做法是持续除以 2 直到商为 0：余数₀（最低位）、余数₁、……、余数ₙ₋₁（最高位）。
Example: Convert 13 to binary. 13 ÷ 2 = 6 r 1; 6 ÷ 2 = 3 r 0; 3 ÷ 2 = 1 r 1; 1 ÷ 2 = 0 r 1. Reading upwards gives 1101₂.
示例：将 13 转换为二进制。13 ÷ 2 = 6 余 1；6 ÷ 2 = 3 余 0；3 ÷ 2 = 1 余 1；1 ÷ 2 = 0 余 1。自下而上读得 1101₂。
4. Binary to Decimal Conversion Formula | 二进制转十进制公式
The value of a binary number is found by summing each digit × 2 to the power of its position. For an n‑bit number bₙ₋₁ bₙ₋₂ … b₀, the decimal value = bₙ₋₁×2ⁿ⁻¹ + bₙ₋₂×2ⁿ⁻² + … + b₀×2⁰.
二进制数的值等于每一位数字乘以 2 的位权次幂再求和。对于 n 位二进制数 bₙ₋₁ bₙ₋₂ … b₀，十进制值 = bₙ₋₁×2ⁿ⁻¹ + bₙ₋₂×2ⁿ⁻² + … + b₀×2⁰。
Example: 10110₂ = 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22.
示例：10110₂ = 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22。
5. Binary Addition Rules | 二进制加法规则
Binary addition follows four simple rules: 0 + 0 = 0; 0 + 1 = 1; 1 + 0 = 1; 1 + 1 = 0, carry 1 to the next higher column.
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，并向高位进 1。
When adding two 1s and a carried 1, the sum is 1 with a carry of 1 (1 + 1 + 1 = 11₂). Always work from right to left, adding the carry into each column.
当两个 1 再加上一个进位 1 时，和为 1 并向高位进 1（1 + 1 + 1 = 11₂）。始终从右向左逐位相加，并将进位带入下一列。
6. Overflow in Binary Addition | 二进制加法中的溢出
Overflow occurs when the result of a binary addition exceeds the maximum value that can be stored in the fixed number of bits. For example, in 8‑bit unsigned binary, the largest number is 255 (11111111₂). Adding two 8‑bit numbers whose sum is 256 or more causes an overflow, and an extra bit would be needed to represent the result correctly.
当二进制加法的结果超出了固定位数所能存储的最大值时，就会发生溢出。例如，在 8 位无符号二进制中，最大值为 255（11111111₂）。两个 8 位数相加，若结果大于或等于 256，就会产生溢出，此时需要额外的位才能正确表示结果。
Overflow is usually signalled by a carry out of the most significant bit. In programming, it can lead to incorrect calculations, which is why we often use larger bit widths or detect overflow flags.
溢出通常表现为最高位产生进位。在编程中，溢出可能导致计算错误，因此我们常常使用更宽的位宽或检测溢出标志。
7. Storage Unit Conversion Formulae | 存储单位换算公式
Computer storage is measured in bytes, with larger units formed by multiplying repeatedly by 1024 (2¹⁰). The standard binary prefixes are:
计算机存储以字节为单位，更大的单位通过反复乘以 1024 (2¹⁰) 得到。标准的二进制前缀如下：
1 KiB (kibibyte) = 1024 bytes; 1 MiB = 1024 KiB = 1024² bytes; 1 GiB = 1024 MiB = 1024³ bytes; 1 TiB = 1024 GiB = 1024⁴ bytes.
1 KiB（千二进制字节）= 1024 字节；1 MiB = 1024 KiB = 1024² 字节；1 GiB = 1024 MiB = 1024³ 字节；1 TiB = 1024 GiB = 1024⁴ 字节。
Sometimes manufacturers use decimal prefixes (1 KB = 1000 bytes), but in OCR Computer Science we always use base‑2 conversions when discussing memory and file sizes.
有时制造商会使用十进制前缀（1 KB = 1000 字节），但在 OCR 计算机科学中，讨论内存和文件大小时我们始终使用以 2 为基数的换算。
8. Boolean Logic Truth Tables | 布尔逻辑真值表
The three fundamental logic gates produce predictable outputs based on their inputs. The truth table for the NOT gate contains one input and one output:
三种基本逻辑门根据输入产生可预测的输出。非门（NOT）的真值表有一个输入和一个输出：
| Input A | Output NOT A |
| 0 | 1 |
| 1 | 0 |
The AND gate gives an output of 1 only when all inputs are 1. Its truth table for two inputs A and B is:
与门（AND）仅当所有输入都为 1 时才输出 1。两输入 A、B 的真值表如下：
| A | B | A AND B |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
The OR gate gives an output of 1 if at least one input is 1. The two‑input truth table is:
或门（OR）只要至少有一个输入为 1 就输出 1。两输入真值表如下：
| A | B | A OR B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
9. Boolean Laws and Simplification | 布尔逻辑定律与化简
Boolean algebra allows us to simplify logic expressions using a set of laws. The commutative law states that the order of inputs does not matter: A AND B = B AND A and A OR B = B OR A.
布尔代数允许我们利用一套定律来化简逻辑表达式。交换律指出输入的顺序无关紧要：A AND B = B AND A 且 A OR B = B OR A。
The associative law lets us group inputs differently: (A AND B) AND C = A AND (B AND C); similarly for OR.
结合律允许我们以不同方式对输入进行分组：(A AND B) AND C = A AND (B AND C)；OR 运算类似。
Identity laws are useful: A AND 1 = A, A OR 0 = A. Domination laws: A AND 0 = 0, A OR 1 = 1. Idempotent laws: A AND A = A, A OR A = A.
恒等律十分有用：A AND 1 = A，A OR 0 = A。支配律：A AND 0 = 0，A OR 1 = 1。幂等律：A AND A = A，A OR A = A。
The complement law: A AND (NOT A) = 0, A OR (NOT A) = 1. The double negation law: NOT (NOT A) = A.
互补律：A AND (NOT A) = 0，A OR (NOT A) = 1。双重否定律：NOT (NOT A) = A。
These laws help reduce the number of gates needed in a circuit, making the design more efficient.
这些定律有助于减少电路中所需逻辑门的数量，从而使设计更加高效。
10. Image File Size Formula | 图像文件大小公式
The file size of a bitmap image is determined by its resolution and colour depth. The basic formula is: Image file size (bits) = Image width (pixels) × Image height (pixels) × Colour depth (bits per pixel).
位图图像的文件大小由其分辨率和色彩深度决定。基本公式为：图像文件大小（位）= 图像宽度（像素）× 图像高度（像素）× 色彩深度（每像素位数）。
Colour depth tells you how many bits are used to store the colour of each pixel. A 1‑bit depth gives 2 colours, 8‑bit depth gives 2⁸ = 256 colours, and 24‑bit depth yields 16.7 million colours.
色彩深度表示用于存储每个像素颜色的位数。1 位深度可产生 2 种颜色，8 位深度产生 2⁸ = 256 种颜色，24 位深度可产生约 1670 万种颜色。
To express the final size in bytes, divide the result by 8. For example, a 100 × 100 pixel image with 24‑bit colour: 100 × 100 × 24 = 240,000 bits ÷ 8 = 30,000 bytes.
若要以字节表示最终大小，只需将结果除以 8。例如，一幅 100 × 100 像素、24 位色彩的图像：100 × 100 × 24 = 240,000 位 ÷ 8 = 30,000 字节。
11. Sound File Size Formula | 音频文件大小公式
A digital audio file records sound by sampling the amplitude thousands of times per second. The size formula is: Sound file size (bits) = Sample rate (Hz) × Bit depth × Duration (seconds) × Number of channels.
数字音频文件通过每秒数千次对振幅进行采样来录制声音。大小公式为：音频文件大小（位）= 采样率（Hz）× 位深度 × 时长（秒）× 声道数。
Typical sample rates are 44.1 kHz for CD‑quality audio. Bit depth is often 16 bits. Stereo sound uses 2 channels; mono uses 1.
CD 品质音频的典型采样率为 44.1 kHz。位深度通常为 16 位。立体声使用 2 个声道，单声道使用 1 个声道。
Example: 30 seconds of stereo CD‑quality audio: 44,100 × 16 × 30 × 2 = 42,336,000 bits. Divide by 8 to get bytes (5,292,000 bytes), then by 1,024 twice to reach approximately 5.04 MiB.
示例：30 秒立体声 CD 品质音频：44,100 × 16 × 30 × 2 = 42,336,000 位。除以 8 得到字节数 (5,292,000 字节)，再除以两次 1024 则得到约 5.04 MiB。
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