📚 Pre-U OCR Computer Science: Interdisciplinary Integrated Question Practice | Pre-U OCR 计算机：跨学科综合题型训练
Interdisciplinary questions in Pre-U OCR Computer Science require students to synthesise knowledge from multiple domains, such as mathematics, physics, logic, and even economics. This article presents a series of practice problems designed to strengthen your ability to tackle such cross-topic challenges effectively.
Pre-U OCR 计算机科学中的跨学科问题要求学生综合运用来自数学、物理、逻辑甚至经济学等多个领域的知识。本文提供一系列练习题，旨在强化您有效应对此类跨主题挑战的能力。
1. Boolean Algebra and Logic Circuits | 布尔代数与逻辑电路
This section explores how algebraic manipulation of Boolean expressions directly translates into digital circuit design, blending mathematical logic with computer hardware.
本节探讨如何将布尔表达式的代数操作直接转化为数字电路设计，融合数学逻辑与计算机硬件。
Problem: Simplify the Boolean function F(A,B,C)=A'BC+AB'C+ABC'+ABC and implement it using only NAND gates.
问题：化简布尔函数 F(A,B,C)=A'BC+AB'C+ABC'+ABC，并仅用与非门实现。
Identify missing minterms: The function includes minterms 3,5,6,7 (binary 011,101,110,111). This is the majority function.
识别缺少的最小项：该函数包含最小项3、5、6、7（二进制011,101,110,111）。这是多数函数。
Algebraic simplification: F = BC(A'+A) + AC(B'+B) + AB(C'+C) = BC + AC + AB.
代数化简：F = BC(A'+A) + AC(B'+B) + AB(C'+C) = BC + AC + AB。
NAND realisation: Apply De Morgan: AB+BC+AC = ((AB)'(BC)'(AC)')'. The circuit uses three NAND gates for the products and one NAND for the final inversion.
与非门实现：应用德摩根定律：AB+BC+AC = ((AB)'(BC)'(AC)')'。电路使用三个与非门产生乘积项，一个与非门完成最终反相。
This problem demonstrates the synergy between algebraic reasoning and physical gate implementation.
此题演示了代数推理与物理门实现之间的协同作用。
2. Data Representation and Numerical Analysis | 数据表示与数值分析
Floating-point arithmetic introduces round-off errors that can be analysed using numerical methods. Here we examine error propagation in a simple calculation.
浮点运算会引入舍入误差，可使用数值方法进行分析。这里我们检查一个简单计算中的误差传播。
Problem: c and a+(b+c) using IEEE single precision for a=1.0, b=1×10⁻⁸, c=-1.0. Explain the result.
问题：使用IEEE单精度计算 (a+b)+c 与 a+(b+c) 的差，其中 a=1.0, b=1×10⁻⁸, c=-1.0。解释结果。
, 1.0 + 1×10⁻⁸ attempts to increment the mantissa, but 10⁻⁸ is below the machine epsilon (~1.19×10⁻⁷) for numbers around 1.0. Thus rounding causes the sum to remain 1.0 exactly.
在单精度中，1.0 + 1×10⁻⁸ 试图增加尾数，但10⁻⁸小于机器精度（约1.19×10⁻⁷），因此舍入导致和仍恰好为1.0。
Therefore (a+b)+c = (1.0+1×10⁻⁸) -1.0 = 1.0 -1.0 = 0.0.
因此 (a+b)+c = (1.0+1×10⁻⁸) -1.0 = 1.0 -1.0 = 0.0。
.0 = -0.99999999, which is representable. Then 1.0 + (-0.99999999) = 1×10⁻⁸ approximately.
对于 a+(b+c)：首先 b+c = 1×10⁻⁸ - 1.0 = -0.99999999，可精确表示。然后 1.0 + (-0.99999999) ≈ 1×10⁻⁸。
The associative law fails due to loss of significance. This connects computer arithmetic with numerical analysis concepts like condition number and catastrophic cancellation.
由于有效数字丢失，结合律不再成立。这将计算机算术与条件数和灾难性抵消等数值分析概念联系起来。
3. Recursion and Mathematical Induction | 递归与数学归纳法
Recursive algorithms often require mathematical induction to prove their correctness and complexity. Here we link a divide-and-conquer recurrence to asymptotic analysis.
递归算法常需用数学归纳法证明其正确性和复杂度。这里我们将分治递推与渐近分析联系起来。
Problem: rrence T(n) = 2T(n/2) + n for n>1, with T(1)=1. Prove by induction that T(n) = O(n log n).
问题：考虑递推式 T(n) = 2T(n/2) + n (n>1), T(1)=1。用归纳法证明 T(n) = O(n log n)。
Inductive hypothesis: Assume for all k < n, T(k) ≤ c·k log₂ k for some constant c>0.
归纳假设：假设对所有 k < n，存在常数 c>0 使得 T(k) ≤ c·k log₂ k。
recurrence: T(n) ≤ 2·c·(n/2) log₂(n/2) + n = c·n (log₂ n - 1) + n = c·n log₂ n - c·n + n.
代入递推式：T(n) ≤ 2·c·(n/2) log₂(n/2) + n = c·n (log₂ n - 1) + n = c·n log₂ n - c·n + n。
g T(n) ≤ c·n log₂ n. The base case holds for sufficiently large c. Thus T(n) = O(n log n).
选择 c ≥ 1 使额外项非正，得到 T(n) ≤ c·n log₂ n。基础情况对足够大的c成立。因此 T(n) = O(n log n)。
This exercise merges algorithmic design with discrete mathematics, reinforcing the role of proof techniques in computer science.
此练习将算法设计与离散数学融合，强化了证明技术在计算机科学中的作用。
4. Graph Theory and Network Routing | 图论与网络路由
Network routing protocols rely on graph algorithms to determine optimal paths. This problem asks you to apply Dijkstra's algorithm and reflect on its mathematical underpinnings.
网络路由协议依赖图算法来确定最优路径。本题要求应用Dijkstra算法并反思其数学基础。
Problem: ith vertices A-E and edge weights: AB=4, AC=2, BC=1, BD=5, CD=8, CE=10, DE=2. Run Dijkstra' and explain why the algorithm relies on the triangle inequality for non-negative weights.
问题：给定带权图，顶点A-E，边权：AB=4, AC=2, BC=1, BD=5, CD=8, CE=10, DE=2。从A运行Dijkstra算法，并解释为何该算法依赖非负权重的三角不等式。
Initialise distances: d(A)=0, others = ∞. Extract A, relax neighbours: d(C)=2, d(B)=4.
初始化距离：d(A)=0，其余为∞。提取A，松弛邻居：d(C)=2, d(B)=4。
allest tentative). Relax through C: d(B) becomes min(4, 2+1)=3; d(D)=2+8=10; d(E)=2+10=12.
提取C（最小暂定值）。通过C松弛：d(B)变为 min(4, 2+1)=3；d(D)=2+8=10；d(E)=2+10=12。
elax E: d(E) becomes min(12, 8+2)=10. Extract E. Shortest paths: A to C=2, B=3, D=8, E=10.
提取B，松弛：d(D)变为 min(10, 3+5)=8。提取D，松弛E：d(E)变为 min(12, 8+2)=10。提取E。最短路径：A到C=2, B=3, D=8, E=10。
rtex with ≥ current distance, which cannot yield a smaller total due to non-negative edges (a form of triangle inequality).
正确性依赖于以下事实：当一个顶点被提取时，其距离最小，因为任何替代路径都会经过另一个距离≥当前提取距离的顶点，由于边权非负（三角不等式的一种形式），不可能得到更小的总距离。
5. Relational Databases and Set Theory | 关系数据库与集合论
Database query languages like SQL are founded on relational algebra, which is deeply rooted in set theory. This problem bridges the two.
像SQL这样的数据库查询语言建立在关系代数之上，而关系代数深深植根于集合论。本题连接二者。
Problem: Two relations: R(SID, Name) and S(SID, Course). Write an SQL query to find names of students taking both 'Maths' and 'CS'. Then express the same query using relational algebra operations and explain the connection to set intersection.
问题：两个关系：R(SID, Name) 和 S(SID, Course)。编写SQL查询找出同时选修'Maths'和'CS'的学生姓名。然后用关系代数表达相同查询，并解释与集合交集的联系。
SQL: SELECT DISTINCT Name FROM R WHERE SID IN (SELECT SID FROM S WHERE Course='Maths') AND SID IN (SELECT SID FROM S WHERE Course='CS');
SQL：SELECT DISTINCT Name FROM R WHERE SID IN (SELECT SID FROM S WHERE Course='Maths') AND SID IN (SELECT SID FROM S WHERE Course='CS');
Relational algebra: πName(R ⨝ (πSID(σCourse='Maths'(S)) ∩ πSID(σCourse='CS'(S)))). The intersection of SID sets directly applies set theory.
关系代数：πName(R ⨝ (πSID(σCourse='Maths'(S)) ∩ πSID(σCourse='CS'(S))))。SID集合的交集直接应用了集合论。
The query can also be done with a natural join and division, illustrating how database operations mirror mathematical set manipulations.
该查询也可用自然连接和除法完成，说明数据库操作如何反映数学集合处理。
6. Algorithmic Complexity and Summation Techniques | 算法复杂度与求和技巧
Analysing nested loops often requires evaluating arithmetic series. This section ties programming constructs to discrete summation identities.
分析嵌套循环常常需要计算等差级数。本节将编程结构和高散求和恒等式联系起来。
Problem: Determine the time complexity of the following code: for i=1 to n: for j=i to n: print('*'). Use summation to derive the exact number of iterations.
问题：确定以下代码的时间复杂度：for i=1 to n: for j=i to n: print('*')。使用求和推导精确的迭代次数。
For a fixed i, the inner loop runs (n - i + 1) times. Total operations = Σi=1n (n - i + 1).
对于固定的i，内层循环运行 (n - i + 1) 次。总操作数 = Σi=1n (n - i + 1)。
Let k = n - i + 1. As i goes from 1 to n, k goes from n down to 1. The sum becomes Σk=1n k = n(n+1)/2.
令 k = n - i + 1。当 i 从1变到n时，k从n降到1。总和变为 Σk=1n k = n(n+1)/2。
Hence the time complexity is O(n²), and the exact count is n(n+1)/2. This connects basic summation formulas to the analysis of iterative algorithms.
因此时间复杂度为O(n²)，精确次数为n(n+1)/2。这连接了基本求和公式和迭代算法分析。
7. Computer Graphics and Linear Transformations | 计算机图形学与线性变换
Transformations of 2D objects are expressed using matrices, a central topic in linear algebra. This problem asks you to combine multiple transformations and understand the order of operations.
二维物体的变换用矩阵表示，这是线性代数的核心主题。本题要求合并多个变换并理解操作顺序。
Problem: A point P(2,3) undergoes a rotation of 90° anticlockwise about the origin, followed by a translation by (1, -2). Compute the final coordinates using homogeneous coordinates. Show the composite transformation matrix.
问题：点P(2,3)先绕原点逆时针旋转90°，再平移(1, -2)。使用齐次坐标计算最终坐标，并展示复合变换矩阵。
= [[0, -1, 0], [1, 0, 0], [0, 0, 1]]. Translation matrix T = [[1,0,1], [0,1,-2], [0,0,1]].
90°旋转矩阵：R = [[0, -1, 0], [1, 0, 0
Published by TutorHao | Pre-U Computer Science Revision Series | aleveler.com
广告 / Sponsored现货 Cambridge International AS & ALevel IT Coursebook剑桥国际IGC…计算机 备考教辅 · 正版现货 · 与本文内容相关查看详情 / View on Taobao参考价 ¥467.0更多咨询请联系16621398022（同微信）
备考资讯每日更新，关注不错过考纲变化 · 科目分析 · 学习方法
在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图公众号：tutorhao