Homework #4

Completing this assignment

For the each of the problems below, write a separate C++ program named in the problem problemN.cpp, where N is the problem number.

Important

Make sure your that g++ can compile your code. For some of the problems here, you will need to use -std=c++20.

Upload your C++ source files (not the executables produced by g++) to Brightspace.

Important

All work must be your own.

You may not use generative AI / large-language models for any part of this assignment.

1. In our vector example, Averaging, we computed the average of a vector of data.

   Starting with this example, extend the code to compute the standard deviation of the vector, defined as:

   \[\sigma = \sqrt{ \frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2 }\]

   where \(\mu\) is the average.

   Have your code output the average and standard deviation.

   solution

   Listing 129 hw4_p1_stddev.cpp

   #include <cmath>
   #include <format>
   #include <iostream>
   #include <vector>
   
   int main() {
   
       std::vector<double> vec{1.0, 2.2, 10.5, 21.3, 25.4, 6.6, 4.2};
   
       double sum{};
       for (auto e : vec) {
           sum += e;
       }
   
       double avg = sum / vec.size();
   
       double sigma{};
   
       for (auto e : vec) {
           sigma += std::pow(e - avg, 2.0);
       }
       sigma = std::sqrt(sigma / vec.size());
   
       std::cout << std::format("The average / std dev is: {:.3f} +/- {:.3f}\n",
                                avg, sigma);
   
   }
   

2. Consider a struct that defines a point in the 2D plane:

   struct Point {
       double x;
       double y;
   };
   

   You want to compute the distance from the origin for a collection of Point objects. For this problem, we will work with the following points \((x, y)\):

   \[(1.5, 2.3), (3.2, 1.6), (5.3, 0.1), (8.6, 2.5)\]

   Your task:

   • Create a std::vector that holds Point data and initialize it with the data given above.

   • Loop over the points in your vector and compute the distance from the origin, \(d = \sqrt{x^2 + y^2}\), for each point

   • Within the loop body, output the distance to the screen using std::format to give a line for each point of the form:

     Point ( 1.50, 2.30) is  2.746 from the origin.
     

   solution

   Listing 130 hw4_p2_points.cpp

   #include <cmath>
   #include <format>
   #include <iostream>
   #include <vector>
   
   struct Point {
       double x;
       double y;
   };
   
   int main() {
   
       std::vector<Point> points{Point{.x=1.5, .y=2.3},
                                 Point{.x=3.2, .y=1.6},
                                 Point{.x=5.3, .y=0.1},
                                 Point{.x=8.6, .y=2.5}};
   
       for (auto p : points) {
           double dist = std::sqrt(std::pow(p.x, 2.0) + std::pow(p.y, 2.0));
           std::cout << std::format("Point ({:5.2f}, {:5.2f}) is {:6.3f} from the origin.\n",
                                    p.x, p.y, dist);
   
       }
   
   }
   

   Tip

   Alternately, we could use std::hypot to compute the distance. This will do the calculation just like we did here, but adds additional logic for floating point overflows.

3. Let’s find the maximum element in a vector. The algorithm library provides std::max(a, b) which will return whichever of a or b is the largest.

   Starting with the vector:

   std::vector<double> vec{-100.3, 5.64, -25.3, 10.9, 50.3, -2.4, 13.9};
   

   With a program that uses a ranged-for loop (what we’ve been using in class) and std::max to find the maximum value in the vector.

   Tip

   You’ll want to start with the maximum value initialized to a very small value so the first element you check in the vector will become the maximum. Using std::numeric_limits can help here.

   Note

   You should add:

   #include <algorithm>
   

   to your code to get access to std::max.

   solution

   Listing 131 hw4_p3_maximum.cpp

   #include <algorithm>
   #include <iostream>
   #include <limits>
   #include <vector>
   
   int main() {
   
       std::vector<double> vec{-100.3, 5.64, -25.3, 10.9, 50.3, -2.4, 13.9};
   
       double max{std::numeric_limits<double>::lowest()};
   
       for (auto e : vec) {
           max = std::max(e, max);
       }
   
       std::cout << "the maximum element is " << max << std::endl;
   }
   

   Tip

   Later we’ll see that can use the std::max_element function for this. But we need to learn about pointers first.