I am considering a society where there are an arbitrary number of people. Each person has just two choices. Either he or she stays with her current choice or she switches. In the code that I want to write, the probability that the person switches is inputted by the user.
To make clear what I am trying to do, suppose that the user tells the computer that there are 3 people in the society where the probabilities that each person chooses to switch is given by (p1,p2,p3). Consider person 1. He has probability of p1 of switching. Using him as a base for our calculation, the probability given person 1 as a base, that exactly no one in the society chooses to switch is given by
P_{1}(0)=(1-p2)*(1-p3)
and the probability using person 1 as a base, that exactly one person in the society chooses to switch is given by
P_{1}(1)=p2*(1-p3)+(1-p2)*p3.
I can't figure out how to write this probability function in C++ without writing out every term in the sum. I considered using the binomial coefficient but I can't figure out a closed form expression for the sum since depending on user input, there are arbitrarily many probabilities that need to be considered.
I have attached what I have. The probability function is only a part of what I am trying to do but it is also the hardest part. I named the probability function probab and what I have in the for loop within the function is obviously wrong.
EDIT: Basically I want to calculate the probability of choosing a subset where each element in that subset has a different probability of being chosen.
I would appreciate any tips on how to go about this. Note that I am a beginner at C++ so any tips on improving my programming skills is also appreciated.
#include <iostream>
#include <vector>
using namespace std;
unsigned int factorial(unsigned int n);
unsigned int binomial(unsigned int bin, unsigned int cho);
double probab(int numOfPeople, vector<double> probs, int p, int num);
int main() {
char correctness;
int numOfPeople = 0;
cout << "Enter the # of people: ";
cin >> numOfPeople;
vector<double> probs(numOfPeople); // Create a vector of size numOfPeople;
for (int i = 1; i < numOfPeople+1; i++) {
cout << "Enter the probability of person "<< i << " will accept change: ";
cin >> probs[i-1];
}
cout << "You have entered the following probabilities of accepting change: (";
for (int i = 1; i < numOfPeople+1; i++) {
cout << probs[i-1];
if (i == numOfPeople) {
cout << ")";
}
else {
cout << ",";
}
}
cout << endl;
cout << "Is this correct? (Enter y for yes, n for no): ";
cin >> correctness;
if (correctness == 'n') {
return 0;
}
return 0;
}
unsigned int factorial(unsigned int n){ // Factorial function
unsigned int ret = 1;
for(unsigned int i = 1; i <= n; ++i) {
ret *= i;
}
return ret;
}
unsigned int binomial(unsigned int totl, unsigned int choose) { // Binomial function
unsigned int bin = 0;
bin = factorial(totl)/(factorial(choose)*factorial(totl-choose));
return bin;
}
double probab(int numOfPeople, vector<double> probs, int p, int num) { // Probability function
double prob = 0;
for (int i = 1; i < numOfPeople; i++) {
prob += binomial(numOfPeople, i-1)/probs[p]*probs[i-1];
}
return prob;
}
For future reference, for anybody attempting to do this, the probability function will look something like:
double probability (vector<double> &yesprobabilities, unsigned int numOfPeople, unsigned int yesNumber, unsigned int startIndex) {
double kprobability = 0;
// Not enough people!
if (numOfPeople-1 < yesNumber) {
kprobability = 0;
}
// n == k, the only way k people will say yes is if all the remaining people say yes.
else if (numOfPeople-1 == yesNumber) {
kprobability = 1;
for (int i = startIndex; i < numOfPeople-1; ++i) {
kprobability = kprobability * yesprobabilities[i];
}
}
else if (yesprobabilities[startIndex] == 1) {
kprobability += probability(yesprobabilities,numOfPeople-1,yesNumber-1,startIndex+1);
}
else {
// The first person says yes, k - 1 of the other persons have to say yes.
kprobability += yesprobabilities[startIndex] * probability(yesprobabilities,numOfPeople-1,yesNumber-1,startIndex+1);
// The first person says no, k of the other persons have to say yes.
kprobability += (1 - yesprobabilities[startIndex]) * probability(yesprobabilities,numOfPeople-1,yesNumber,startIndex+1);
}
return probability;
}
Something called a recursive function is used here. This is completely new to me and very illuminating. I credit this to Calle from Math stack exchange. I modified his version slightly to take vectors instead of arrays with some help.
Related
I am trying to use pointers whenever possible in the following code and am having difficulty figuring out how, exactly, to institute the pointers and how to return a pointer value at the end of my first function. I have done some research on the subject but none of the methods I found have been helpful so far, so I was hoping you may have some specialized tips.
Note: I am a beginner.
#include <iostream>
using namespace std;
int mode(int *pies[], int size) {
int count = 1;
int max = 0;
int *mode=pies[0];
for (int i=0; i<size-1; i++)
{
if (pies[i] == pies[i+1])
{
count++;
if (count>max)
{
max = count;
mode = pies[i];
}
}
else
count = 1;
}
return *mode;
}
int main() {
int n;
cout<<"Input the number of people: "<<endl;
cin>>n;
int survey[n];
cout << "Enter the amount of pie eaten by each person:" << endl;
for(int i = 0; i < n; i++) {
cout <<"Person "<<(i + 1)<< ": "<<endl;
cin>>survey[i];
}
cout<<"Mode: "<<mode(survey, n)<< endl;
return 0;
}
Here is an attempt to answer.
In your main(), you call the mode() function with mode(survey, n) while int survey[n]; is an array of int, so you may use int mode(int *pies, int size) instead of int mode(int *pies[], int size) (as the array int survey[n] can be implicitly converted into pointer).
However, you need to modify two more things in your function:
int *mode=pies[0]; is wrong as pies[0] is the first element of an array of int, thus is an int, while int* mode is a pointer on an int which is incompatible. mode should be an int to receive pies[0]. The correct code is then int mode = pies[0].
Your function signature is int mode(int *pies, int size), thus, again, you should return an int. You should then just return mode;
These are only hints on how to make the code compile.
Your next step is to formalize what you would like it to do and then modify the code accordingly
NB: The correct practice is to think about what you would like to achieve first and then code afterwards (but let us say that this is for the sake of helping each other)
To get started using pointers, you may look at some simple tutorials at first:
http://www.cplusplus.com/doc/tutorial/arrays/
https://www.programiz.com/c-programming/c-pointers
https://www.programiz.com/c-programming/c-pointers-arrays
https://www.geeksforgeeks.org/pointer-array-array-pointer/
https://www.geeksforgeeks.org/how-to-return-a-pointer-from-a-function-in-c/
https://www.tutorialspoint.com/cprogramming/c_return_pointer_from_functions.htm
Here is the modified code with the stated modifications above (it compiles):
#include <iostream>
using namespace std;
int mode(int *pies, int size) {
int count = 1;
int max = 0;
int mode=pies[0];
for (int i=0; i<size-1; i++)
{
if (pies[i] == pies[i+1])
{
count++;
if (count>max)
{
max = count;
mode = pies[i];
}
}
else
count = 1;
}
return mode;
}
int main() {
int n;
cout<<"Input the number of people: "<<endl;
cin>>n;
int survey[n];
cout << "Enter the amount of pie eaten by each person:" << endl;
for(int i = 0; i < n; i++) {
cout <<"Person "<<(i + 1)<< ": "<<endl;
cin>>survey[i];
}
cout<<"Mode: "<<mode(survey, n)<< endl;
return 0;
}
Program in C++ that takes 3 numbers and send them to a function and then calculate the average function of these 3 numbers.
I know how to do that without using a function ,for example for any n numbers I have the following program:
#include<stdio.h>
int main()
{
int n, i;
float sum = 0, x;
printf("Enter number of elements: ");
scanf("%d", &n);
printf("\n\n\nEnter %d elements\n\n", n);
for(i = 0; i < n; i++)
{
scanf("%f", &x);
sum += x;
}
printf("\n\n\nAverage of the entered numbers is = %f", (sum/n));
return 0;
}
Or this one which do that using arrays:
#include <iostream>
using namespace std;
int main()
{
int n, i;
float num[100], sum=0.0, average;
cout << "Enter the numbers of data: ";
cin >> n;
while (n > 100 || n <= 0)
{
cout << "Error! number should in range of (1 to 100)." << endl;
cout << "Enter the number again: ";
cin >> n;
}
for(i = 0; i < n; ++i)
{
cout << i + 1 << ". Enter number: ";
cin >> num[i];
sum += num[i];
}
average = sum / n;
cout << "Average = " << average;
return 0;
}
But is it possible to use functions?if yes then how? thank you so much for helping.
As an alternative to using fundamental types to store your values C++ provides std::vector to handle numeric storage (with automatic memory management) instead of plain old arrays, and it provides many tools, like std::accumulate. Using what C++ provides can substantially reduce your function to:
double avg (std::vector<int>& i)
{
/* return sum of elements divided by the number of elements */
return std::accumulate (i.begin(), i.end(), 0) / static_cast<double>(i.size());
}
In fact a complete example can require only a dozen or so additional lines, e.g.
#include <iostream>
#include <vector>
#include <numeric>
double avg (std::vector<int>& i)
{
/* return sum of elements divided by the number of elements */
return std::accumulate (i.begin(), i.end(), 0) / static_cast<double>(i.size());
}
int main (void) {
int n; /* temporary integer */
std::vector<int> v {}; /* vector of int */
while (std::cin >> n) /* while good integer read */
v.push_back(n); /* add to vector */
std::cout << "\naverage: " << avg(v) << '\n'; /* output result */
}
Above, input is taken from stdin and it will handle as many integers as you would like to enter (or redirect from a file as input). The std::accumulate simply sums the stored integers in the vector and then to complete the average, you simply divide by the number of elements (with a cast to double to prevent integer-division).
Example Use/Output
$ ./bin/accumulate_vect
10
20
34
done
average: 21.3333
(note: you can enter any non-integer (or manual EOF) to end input of values, "done" was simply used above, but it could just as well be 'q' or "gorilla" -- any non-integer)
It is good to work both with plain-old array (because there is a lot of legacy code out there that uses them), but equally good to know that new code written can take advantage of the nice containers and numeric routines C++ now provides (and has for a decade or so).
So, I created two options for you, one use vector and that's really comfortable because you can find out the size with a function-member and the other with array
#include <iostream>
#include <vector>
float average(std::vector<int> vec)
{
float sum = 0;
for (int i = 0; i < vec.size(); ++i)
{
sum += vec[i];
}
sum /= vec.size();
return sum;
}
float average(int arr[],const int n)
{
float sum = 0;
for (int i = 0; i < n; ++i)
{
sum += arr[i];
}
sum /= n;
return sum;
}
int main() {
std::vector<int> vec = { 1,2,3,4,5,6,99};
int arr[7] = { 1,2,3,4,5,6,99 };
std::cout << average(vec) << " " << average(arr, 7);
}
This is an example meant to give you an idea about what needs to be done. You can do this the following way:
// we pass an array "a" that has N elements
double average(int a[], const int N)
{
int sum = 0;
// we go through each element and we sum them up
for(int i = 0; i < N; ++i)
{
sum+=a[i];
}
// we divide the sum by the number of elements
// but we first have to multiply the number of elements by 1.0
// in order to prevent integer division from happening
return sum/(N*1.0);
}
int main()
{
const int N = 3;
int a[N];
cin >> a[0] >> a[1] >> a[2];
cout << average(a, N) << endl;
return 0;
}
how to do that without using a function
Quite simple. Just put your code in a function, let's call it calculateAverage and return the average value from it. What should this function take as input?
The list of numbers (array of numbers)
Total numbers (n)
So let's first get the input from the user and put it into the array, you have already done it:
for(int i = 0; i < n; ++i)
{
cout << i + 1 << ". Enter number: ";
cin >> num[i];
}
Now, lets make a small function i.e., calculateAverage():
int calculateAverage(int numbers[], int total)
{
int sum = 0; // always initialize your variables
for(int i = 0; i < total; ++i)
{
sum += numbers[i];
}
const int average = sum / total; // it is constant and should never change
// so we qualify it as 'const'
//return this value
return average
}
There are a few important points to note here.
When you pass an array into a function, you will loose size information i.e, how many elements it contains or it can contain. This is because it decays into a pointer. So how do we fix this? There are a couple of ways,
pass the size information in the function, like we passed total
Use an std::vector (when you don't know how many elements the user will enter). std::vector is a dynamic array, it will grow as required. If you know the number of elements beforehand, you can use std::array
A few problems with your code:
using namespace std;
Don't do this. Instead if you want something out of std, for e.g., cout you can do:
using std::cout
using std::cin
...
or you can just write std::cout everytime.
int n, i;
float num[100], sum=0.0, average;
Always initialize your variables before you use them. If you don't know the value they should be initialized to, just default initialize using {};
int n{}, i{};
float num[100]{}, sum=0.0, average{};
It is not mandatory, but good practice to declare variables on separate lines. This makes your code more readable.
So, here is my homework problem. It states "Enter five numbers five times. Every time, the program chooses the biggest number you enter, and it returns the arithmetic mean of those five largest numbers." Now, what I've done is use max function, however, I learned that it isn't useable in this way. Here is what I've tried:
#include <iostream>
using namespace std;
int main() {
int zbir = 0;
for (int i = 1; i < 6; i++) {
int a, b, c, d, e;
for (int j = 1; j < 6; j++) {
cin >> a >> b >> c >> d >> e;
}
int maks = max(a, b, c, d, e);
zbir = zbir + maks;
}
cout << zbir / 5;
}
There are two versions of max: the two-argument version, and the initializer list version. In this case, you have five arguments, so use the initializer list version:
std::max({a, b, c, d, e})
(You need to #include <algorithm> to use std::max.)
Within your code you do not need to have an inner for-loop because you are already collecting all 5 of the user numbers with the cin >> a >> b >> c >> d >> e; statement. Having a second for loop around it will cause you to collect 5 numbers from the user 25 times total.
This is an example of the alternative to using the max function in which case a single number is collected 5 times making use of an inner for-loop:
int main()
{
int sumOfMaxNums = 0;
int userNum = 0;
int maxNum = 0;
// it is a good practice to have a const that will bound the loop
// this way you can change it from here
// in other cases you can have two different const bounds, one for the inner loop and
// one for the outer loop because they may differ
const int NUM_ITERS = 5;
// This will handle asking the user to enter the numbers 5 times
for(int i = 0; i < NUM_ITERS; ++i)
{
// this loop will asks the user to enter a number 5 times and keep track of the max
for(int j = 0; j < NUM_ITERS; ++j)
{
cin >> userNum;
// here we want to set the maxNum to be the very first number that the user enters
// or we could set it to smallest negative number
if(j == 0)
{
maxNum = userNum;
}
else if(userNum > maxNum)
{
maxNum = userNum;
}
} // end of inner for-loop
sumOfMaxNums += maxNum;
} // end of outer for-loop
cout << sumOfMaxNums / static_cast<float>(NUM_ITERS) << "\n";
} // end of main
You probably want to calculate the mean as a float rather than an int, otherwise the program will round the final answer down to the nearest whole number. Also, you really don't need to use five variables to store each cycle's inputs, since you can ignore any inputs that are less than the running maximum for that cycle. This means you don't need to use std::max at all.
#include<iostream>
int main()
{
float running_total = 0;
for (int cycle = 1; cycle < 6; ++cycle)
{
float cycle_max;
for (int entry = 1; entry < 6; ++entry)
{
float input = 0;
std::cin >> input;
if (entry == 1 || input > cycle_max) cycle_max = input;
}
running_total += cycle_max;
}
std::cout << running_total / 5 << std::endl;
}
I filled an array with 30 random numbers and calculated average. I want to display how many numbers are higher than the average. I tried making a function "aboveAverage" and check if the numbers are higher than the average and than just increase the count "num_over_average++". The problem is I don't know how to pass a value "avg" from function to another function.
#include <iostream>
#include <ctime>
using namespace std;
const int n = 30;
void fillArray(int age[], int n) {
srand(time(NULL));
for (int index = 0; index < n; index++) {
age[index] = (rand() % 81) + 8;
}
}
void printArray(int age[], int n) {
for (int index = 0; index < n; index++) {
cout << age[index] << endl;
}
}
double printAverage(int age[], int n) {
double sum;
double avg = 0.0;
for (int i = 0; i < n; i++) {
sum = sum + age[i];
}
avg = ((double) sum) / n;
cout << avg << endl;
return avg;
}
void aboveAverage(int age[], int n) {
double avg;
int num_over_average = 0;
for(int i = 0; i < n; i++){
if(age[i] > avg) {
num_over_average++;
}
}
cout<<num_over_average;
}
int main(int argc, char *argv[]) {
int age[n];
fillArray(age, n);
cout << "array: " << endl;
printArray(age, n);
cout << endl;
aboveAverage(age, n);
//example: Days above average: 16
}
This should be a comment, but I don't have enough reps :(
Change aboveAverage to void aboveAverage(int age[], int n, double avg)
Return avg from printAverage function
Change the last part of your main code to
double avg;
avg = printAverage(age, n);
aboveAverage(age, n, avg);
Hope this helps!
You have two solutions using your code:
Either you call printAverage() to initialise avg in aboveAverage() :
void aboveAverage(int age[], int n) {
double avg = printAverage();
...
}
Or you pass the average at parameter of aboveAverage() after having computed it with printAverage() :
void aboveAverage(int age[], int n, double avg) {
...
}
If you use the standard library you can do that with two lines of code:
double average = std::accumulate(std::begin(age), std::end(age), 0.0) / std::size(age);
int above_average = std::count_if(std::begin(age), std::end(age),
[average](double value) { return average < value; });
Okay, you might count that as three lines.
One major advantage of this approach over the code in the question is that you can change the container type to, say, vector<double> without having to change any of this code.
Well is pretty simple but dependent on your situation, I'll elaborate.
I'm the case when it's part of a bigger function (do-somthing())
You could calculate the average value like so and pass it to your "aboveAverage" function and print it:
double n_average = printAverage(nArr_ages, n_agesArraySize);
aboveAverage(nArr_ages, n_agesArraySize, n_averag);
Myself would probably rewrite the printAverage function as two functions, one that returns the average value based on the array and another that prints it not both at once because it violates the SOLID principals of a single responsibility and that a function name should reflect exactly what it does, in this case maybe calculateAverage or getAverageAge or any other appropriate name will do (try and name your functions like the english language so your code will be read like a song.
For example:
const size_t n = 30;
double calculateAverage(int nArr_ages[], int n_agesArraySize) {
double sum = 0.0;
double avg = 0.0;
for (int indexInArray = 0; indexInArray < n_agesArraySize; indexInArray++) {
sum = sum + age[indexInArray];
}
average = ((double) sum) / n_agesArraySize;
return average;
}
int aboveAverageCells(int ages[], int n_agesArraySize ) {
double average = calculateAverage(ages, n);
int num_over_average = 0;
for(int indexInArray = 0; indexInArray < n_agesArraySize; indexInArray++) {
if(ages[indexInArray] > avg) {
num_over_average++;
}
}
return num_over_average;
}
Now just call them in order, save the returned values to local variables in the main function and print using cout also locally in main.
As a side note next time maybe choose different names for the const and the local functions variable for the array size.
So im trying to write a program that will calculate the sum of terms but each term is 3 times the previous term minus the second previous term so it looks like this 0, 1, 3, 8, 21, 55 and so on. For example if the user wants 4 terms then the program should output 21. The part im having problems with is setting up the variables to store the previous number and the second previous number. This is what i have so far.
#include <iostream>
using namespace std;
int main(){
int num;
int last;
int last2;
int current;
cout << "Number of terms to be shown: ";
cin >> num;
for(int i = 0; i < num; i++){
for(int term; term <= i; i++){
//THIS IS WHERE IM STUCK
}
}
}
The way i see it is the first for loop will tell the nested for loop how many times to run. In the nested for loop i think is where the math should go (current = (last * 3) - last2) while updating the last and last2 variables to keep the term list going. And then outside the loop i would cout << current so it would display the term. Like always, any help is appreciated!
There is an Undefined Behavior in your code in:
for(int i = 0; i < num; i++){
for(int term; term <= i; i++){ // term not initiaized. and the loop is infinte
//THIS IS WHERE IM STUCK
}
}
You are using term without being intialized. Also you are stuck in the inner loop because you should increment term not i in the inner loop.
So you can make it this way:
for(int i = 0; i < num; ++i){
for(int term = 0; term <= i; ++term){
// now rock here
}
}
You would typically remember the last two values, and simply go on computing the next one:
#include <iostream>
using namespace std;
int main()
{
int num;
cout << "Number of terms to be shown: ";
cin >> num;
int p1 = 1;
int p2 = 0;
cout << p2 << " " << p1 << " ";
num -= 2;
while (num > 0)
{
int current = 3 * p1 - p2;
cout << current << " ";
p2 = p1;
p1 = current;
num--;
}
}
This is the algorithm as I see it in my head when I read your question:
unsigned term(unsigned num) {
// the previous term*3 minus the second previous term
if(num > 1) return term(num - 1) * 3 - term(num - 2);
return num; // 0 or 1
}
It uses recursion to call itself, which is a great way to visualize what needs to be done for problems like this. The function works as-is but only for small nums, or else you'll get a stack overflow. It's also rather time consuming since it's doing function calls and calculates all terms many, many times.