Suppose that you have the following recursive function written in C++. I've intentionally written the function so that it doesn't use descriptive names for anything, though that's obviously not a good design practice, but I aim not to bias your understanding of the techniques you'll be using by assuming anything about what the function's goal is; to stay "on task" here, you should resist the temptation to try to figure that out first. void kaboom( const std::vector& v, std::vector& w, unsigned int i) { if (i < w.size()) { int q = 0; for (int j : v) { 9 += j; w.at(i) = q; kaboom(v, w, i + 1); } 1. Using the techniques from our discussion of the Asymptotic Analysis of Recursion, write a recurrence that describes the time required to run the following call to kaboom for two vector v and w whose sizes are the same. Use the variable s in your recurrence to denote that size. kaboom(v, w, 0) 2. Using the repeated substitution technique, reduce the recurrence to a closed form, then give an asymptotic notation that best describes it.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
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Suppose that you have the following recursive function written in C++. I've intentionally written the function so that it doesn't use descriptive names for anything, though that's obviously not a good design
practice, but I aim not to bias your understanding of the techniques you'll be using by assuming anything about what the function's goal is; to stay "on task" here, you should resist the temptation to try to
figure that out first.
void kaboom(
const std::vector<int>& v, std::vector<int>& w,
unsigned int i)
{
if (i < w.size())
{
int q =
0;
for (int j : v)
{
q += j;
}
w.at(i) -
kaboom(v, w, i + 1);
}
= q;
}
1. Using the techniques from our discussion of the Asymptotic Analysis of Recursion, write a recurrence that describes the time required to run the following call to kaboom for two vector v and w whose
sizes are the same. Use the variable s in your recurrence to denote that size.
kaboom(v, w, 0)
2. Using the repeated substitution technique, reduce the recurrence to a closed form, then give an asymptotic notation that best describes it.
Transcribed Image Text:Suppose that you have the following recursive function written in C++. I've intentionally written the function so that it doesn't use descriptive names for anything, though that's obviously not a good design practice, but I aim not to bias your understanding of the techniques you'll be using by assuming anything about what the function's goal is; to stay "on task" here, you should resist the temptation to try to figure that out first. void kaboom( const std::vector<int>& v, std::vector<int>& w, unsigned int i) { if (i < w.size()) { int q = 0; for (int j : v) { q += j; } w.at(i) - kaboom(v, w, i + 1); } = q; } 1. Using the techniques from our discussion of the Asymptotic Analysis of Recursion, write a recurrence that describes the time required to run the following call to kaboom for two vector v and w whose sizes are the same. Use the variable s in your recurrence to denote that size. kaboom(v, w, 0) 2. Using the repeated substitution technique, reduce the recurrence to a closed form, then give an asymptotic notation that best describes it.
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