Stack Practice questions
Duration: 7 min
This video lesson is available to enrolled students.
AI Summary
An AI-generated summary of this video lecture.
The video presents a computer science problem asking to identify the functionality of a specific C function named `fun`. The instructor systematically analyzes the code by tracing its execution with a concrete example, `n = 81`. He explains the recursive logic where the function divides the input by 3 repeatedly. The core concept is determining whether a number is a power of 3 based on its divisibility properties. The instructor contrasts a valid case (81) with an invalid case (21) to verify the function's behavior against the provided multiple-choice options. The lesson emphasizes understanding base cases and recursive steps in algorithm analysis.
Chapters
0:00 – 2:00 00:00-02:00
The session begins with the instructor writing `n = 81` on the whiteboard to serve as a test input. He notes that 81 is a multiple of 3 ($3 imes 27$). He starts the trace for `fun(81)`. The first condition `if (n == 0 || n == 1)` evaluates to false. The second condition `if (n % 3 != 0)` is also false because 81 is perfectly divisible by 3. Consequently, the code executes `return fun(n/3)`, which translates to a recursive call `fun(27)`. He writes `fun(81) -> fun(27)` on the board to visualize the call stack. He highlights that the function keeps dividing by 3 as long as the remainder is zero.
2:00 – 5:00 02:00-05:00
The instructor continues the trace for `fun(27)`. He writes `27 % 3 = 0` and `27/3 = 9`, leading to `fun(9)`. The process repeats: `fun(9)` calls `fun(3)`, and `fun(3)` calls `fun(1)`. Upon reaching `n = 1`, the base case `if (n == 0 || n == 1)` becomes true, and the function returns `1`. To ensure this isn't a coincidence, he tests `n = 21`. `21 % 3 = 0` leads to `fun(7)`. Since `7 % 3` is not 0, the function returns `0`. This confirms the function returns 1 for powers of 3 and 0 otherwise. He writes `21/3 = 7` and `7 % 3 = 1` to show the failure point where the condition `if (n % 3 != 0)` becomes true.
5:00 – 7:20 05:00-07:20
The instructor summarizes the findings by drawing a visual chain: $81 ightarrow 27 ightarrow 9 ightarrow 3 ightarrow 1$. He explicitly writes $3^4 = 81$ to reinforce that 81 is a power of 3. He reviews the options: (A) claims it returns 1 for multiples of 3, which is incorrect because 21 is a multiple but returns 0. (C) and (D) are also incorrect based on the return values. He selects option (B): "It returns 1 when n is a power of 3, otherwise returns 0". He crosses out the incorrect options to finalize the answer. He emphasizes that the function essentially checks if the number can be reduced to 1 by repeated division by 3, which is the definition of a power of 3.
The lecture provides a clear methodology for debugging recursive functions. By manually stepping through the recursion stack with specific inputs, the instructor deduces the mathematical property the code implements. The distinction between "multiple of 3" and "power of 3" is the key takeaway, demonstrated through the contrasting traces of 81 and 21. The visual aids, such as the call chain and the modulo calculations, are crucial for understanding the flow. The final conclusion solidifies the relationship between the code logic and the mathematical concept of powers.