Stack Practice questions
Duration: 10 min
This video lesson is available to enrolled students.
AI Summary
An AI-generated summary of this video lecture.
The video lecture analyzes two C functions, `fun` and `fun2`, to determine the mathematical operation performed by `fun2`. The instructor first identifies that `fun(x, y)` implements multiplication (`x * y`). He then traces the execution of `fun2(a, b)` using specific inputs (a=2, b=3) to find the output. By expanding the recursive calls, he calculates the result as 8. Finally, he generalizes the recursive structure of `fun2` to show that it computes `a` raised to the power of `b` (exponentiation), confirming option (C) as the correct answer.
Chapters
0:00 – 2:00 00:00-02:00
The instructor introduces the problem by displaying two C functions on the screen: `fun(int x, int y)` and `fun2(int a, int b)`. He begins by analyzing `fun`, noting that it recursively adds `x` to itself `y` times, effectively implementing multiplication (`x * y`). He then sets up a trace for `fun2` using the specific input values `a=2` and `b=3`. He writes down the initial recursive expansion on the whiteboard: `fun2(2, 3)` calls `fun(2, fun2(2, 2))`. He continues to expand the inner call, writing `fun(2, fun(2, fun2(2, 1)))` and then `fun(2, fun(2, fun(2, fun2(2, 0))))`. This establishes the recursive structure needed to evaluate the function.
2:00 – 5:00 02:00-05:00
The instructor proceeds to evaluate the nested function calls starting from the base case. He identifies that `fun2(2, 0)` returns `1` based on the base condition `if (b == 0) return 1;`. Substituting this back, the expression becomes `fun(2, fun(2, fun(2, 1)))`. He evaluates the innermost `fun(2, 1)` as `2 * 1 = 2`. The expression simplifies to `fun(2, fun(2, 2))`. Next, he evaluates `fun(2, 2)` as `2 * 2 = 4`. The expression further simplifies to `fun(2, 4)`. Finally, he calculates `fun(2, 4)` as `2 * 4 = 8`. He writes `2^3 = 8` on the board, noting that the result matches the power operation.
5:00 – 9:52 05:00-09:52
To confirm the general behavior, the instructor analyzes the recursive step of `fun2` algebraically. He observes that `fun2(a, b)` returns `fun(a, fun2(a, b-1))`. Since `fun(x, y)` is multiplication, this translates to `a * fun2(a, b-1)`. This recurrence relation, `f(a, b) = a * f(a, b-1)` with base case `f(a, 0) = 1`, is the standard definition of exponentiation `a^b`. He compares this with the given options: (A) `a*b`, (B) `a+a*b`, (C) `a^b`, and (D) `b^a`. Although option (B) also yields 8 for inputs (2,3), the recursive structure `a * fun2(...)` clearly matches exponentiation, not addition. He circles option (C) `a^b` as the correct answer, concluding that `fun2` computes `a` raised to the power of `b`.
The lecture demonstrates a method for analyzing recursive C functions by first simplifying helper functions and then tracing specific examples to find a pattern. By identifying `fun` as multiplication and tracing `fun2(2, 3)` to get 8, the instructor establishes a numerical pattern. He then generalizes the recursive step `fun(a, fun2(a, b-1))` into the algebraic form `a * fun2(a, b-1)`, which definitively proves that `fun2` implements exponentiation `a^b`.