Recursion Practice questions
Duration: 3 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 for the output of a specific C function `fun(int n)` when called with `n = 25`. The instructor displays the function code, which includes a base case check `if (n == 0) return;`, a print statement `printf("%d", n%2);`, and a recursive call `fun(n/2);`. He explains that the function executes the print statement before the recursive call, crucial for determining output order. He underlines key lines, pointing out `n%2` and `n/2` operations. The background features a watermark for KnowledgeGate.
Chapters
0:00 – 2:00 00:00-02:00
In the first segment, the instructor introduces the problem and begins tracing `fun(25)`. He notes that since 25 is not 0, the function prints `25 % 2` (1) and calls `fun(12)`. He continues, showing `fun(12)` prints 0 and calls `fun(6)`, and `fun(6)` prints 0 and calls `fun(3)`. The instructor draws a tree diagram on the right side of the whiteboard to visualize this recursive structure, writing `fun(25)` at the top and branching down to `fun(12)`, `fun(6)`, `fun(3)`, and `fun(1)`. He uses a digital pen to write on the screen, clearly marking the flow of execution.
2:00 – 3:00 02:00-03:00
In the final segment, the instructor completes the trace by showing that `fun(1)` prints 1 and calls `fun(0)`. He demonstrates that `fun(0)` hits the base case condition `if (n == 0)` and returns immediately without printing anything. He then aggregates the printed values in the order they occurred: 1, 0, 0, 1, 1. He writes the final sequence "10011" in a green box on the screen. Comparing this result with the given multiple-choice options—(A) 11001, (B) 10011, (C) 11111, (D) 00000—he identifies option (B) as the correct answer and places a green checkmark next to it to signify the correct choice.
The lesson demonstrates tracing recursive functions by following the order of operations: checking the base case, executing the print statement, and then making the recursive call. The key takeaway is that the function prints remainders of division by 2 as encountered in recursion, corresponding to binary representation but in reverse order (Least Significant Bit first). By visualizing the call stack as a tree, students understand the flow of control and the state of variable `n` at each step. This method is essential for debugging complex recursive algorithms in computer science, ensuring output matches expected binary conversion logic.