How to create as many nested loops as you want? - c++

Question

I am trying to create a program that can take many numbers as I want in C++ language.
Then it find what operators can make the equation true and show all correct possible operation.
Example: If I put 3 5 15
Then it output 3x5 = 15
If I put 1 2 3 4 4
Then it outputs 1+2-3+4 =4
The following code is my written program:
The problem about it is that when I want to reduce the number of input or increase the number of input I need to add/reduce nested loops EVERYTIME. I want to know what is a more effective way of a more flexible nested loops or recursion method.
#include <iostream>
#include <cmath>
using namespace std;
char getOperator(int);
double operate(int, double, double);
int main() {
double a, b, c, d, e, result;
short noOfAnswers = 0;
cout << "Input first 5 numbers to make it equal to another 1 number.\n" <<
"I'll find what are the operators needed to make 2 sides of the equation equal.\n";
cin >> a >> b >> c >> d >> e >> result;
int noOfOperators = 5;
for (int i = 0; i <= noOfOperators; i++) {
double firstTwo = operate(i, a, b);
for (int j = 0; j <= noOfOperators; j++) {
double firstThree = operate(j, firstTwo, c);
for (int k = 0; k <= noOfOperators; k++) {
double firstFour = operate(k, firstThree, d);
for (int l = 0; l <= noOfOperators; l++) {
double firstFive = operate(l, firstFour, e);
if (firstFive == result) {
cout << ++noOfAnswers << ')' << a << getOperator(i) << b << getOperator(j) << c
<< getOperator(k) << d << getOperator(l) << e << '=' << result << endl;
}
}
}
}
}
if (noOfAnswers) cout << "I have found " << noOfAnswers << " solutions for this extremely hard problem for humanity \nin less than a second." << endl;
else cout << "I cannot find any solutions to this problem.\n"
<<"They're just a bunch of random numbers & That is UNSOLVABLE!" << endl;
cout << "Do not doubt my judgment. I am always right!" << endl << "(Please note that all calculations are done from the left side first.)" << endl;
return 0;
}
double operate(int iteration, double num1, double num2) {
switch (iteration) {
case 0: return num1+num2;
case 1: return num1-num2;
case 2: return num1*num2;
case 3: return num1/num2;
case 4: return pow(num1, num2);
case 5: return fmod(num1, num2);
}
return 0;
}
char getOperator(int pos) {
switch (pos) {
case 0: return '+';
case 1: return '-';
case 2: return 'x';
case 3: return '/';
case 4: return '^';
case 5: return '%';
}
return ' ';
}

Answer 1

You might want to use while() loops, cause you dont know when the loop terminates.
int main() {
double numbers[] = {3,5,15};//consider storing the number as an array
//the last element is the result
double result;
int arr_len = sizeof(numbers)/sizeof(double);
int i,j;
while(1)
{
j = 0;
while(j++ < 5)//over all operators
{i = 0;
result = numbers[0];//start with first element
while(i < arrlen - 2)//over all numbers, exclude the result
{
result = operate(j, result, numbers[++i]);
//something like this...this does not work correctly
//it might give you a hint in the right direction
if(result == numbers[arr_len - 1])//compare to last element
return 0;
}
}
}
return 0;
}

Answer 2

Following may help:
// increment v where each value is a digit with maximal value maxSize
// so {0,1,2}, 3 lead to {0,2,0}
// return false on overflow.
bool increment(std::vector<int>& v, int maxSize)
{
for (auto it = v.rbegin(); it != v.rend(); ++it) {
++*it;
if (*it != maxSize) {
return true;
}
*it = 0;
}
return false;
}
// display something like 1 + 2 * 3 = 9 // with the following meaning ((1 + 2) * 3) = 9
void display(const std::vector<double>& operands, const std::vector<int>& operators, double total)
{
const char operators_string[] = "+-*/^%";
std::cout << operands[0];
for (std::size_t i = 0; i != operators.size(); ++i) {
std::cout << " " << operators_string[operators[i]] << " " << operands[i + 1];
}
std::cout << " = " << total << std::endl;
}
// Compute something like {1 2 3 4}, {+ * /} as (((1 + 2) * 3) / 4)
double compute(const std::vector<double>& operands, const std::vector<int>& operators)
{
std::function<double(double, double)> fs[] = {
[](double a, double b) { return a + b; },
[](double a, double b) { return a - b; },
[](double a, double b) { return a * b; },
[](double a, double b) { return a / b; },
[](double a, double b) { return pow(a, b); },
[](double a, double b) { return fmod(a, b); },
};
double res = operands[0];
for (std::size_t i = 0; i != operators.size(); ++i) {
res = fs[operators[i]](res, operands[i + 1]);
}
return res;
}
void display_combinaison(const std::vector<double>& operands, double total)
{
std::vector<int> operators(operands.size() - 1);
do {
if (compute(operands, operators) == total) {
display(operands, operators, total);
}
} while (increment(operators, 6));
}
Live example