#include <iostream>

// This program prints the sequence of values in the
// '3n + 1' problem, also known as the Collatz problem.
//
// Starting with a positive integer 'n', after printing
// the current value, the next value is
//
//     n/2       if 'n' is even, and
//     3*n + 1   if 'n' is odd.
//
// For example, starting with 6 produces
//
//     6, 3, 10, 5, 16, 8, 4, 2, 1.
//
// It is conjectured that every positive starting
// integer will eventually reach 1. Once 1 is reached,
// continuing the process produces the cycle
//
//     1, 4, 2, 1, 4, 2, 1, ...
//
// Despite its simple statement, no proof is known
// that every positive integer eventually reaches 1.
//
// Exercises:
//
// 1. Modify the program so that the user must enter
//    an integer greater than or equal to 1. If the
//    user enters a smaller value, repeatedly query
//    the user until a valid value is entered.
//
// 2. Add a counter that records the number of
//    iterations required to reach 1.
//
//    For example, if the user enters 42, the output
//    should be
//
//        42, 21, 64, 32, 16, 8, 4, 2, 1
//        Iterations: 8
//
//    Notice that the starting value is not itself
//    an iteration. Each change from one value to the
//    next counts as one iteration.
//
// 3. Modify the output so that there is no trailing
//    comma after the final value 1.

// Function declarations
int main();

// Function definitions
int main() {
    int n{};

    std::cout << "Enter a positive integer: ";
    std::cin  >> n;

    while ( n != 1 ) {
        std::cout << n << ", ";

        if ( n%2 == 0 ) {
            n /= 2;
        } else {
            n = 3*n + 1;
        }
    }

    std::cout << 1 << std::endl;

    return 0;
}



























// Hint for Exercise 1:
//
// Use a loop that repeatedly queries the user until
// the condition
//
//     n >= 1
//
// is satisfied.



























// Hint for Exercise 2:
//
// Each time the program changes 'n' to the next
// value in the sequence, one iteration has occurred.
//
// Create a counter before the loop and increment it
// once during each iteration of the loop.



























// Hint for Exercise 3:
//
// The existing program already avoids a trailing
// comma by printing each value followed by ", "
// only while n != 1.
//
// The final value 1 is then printed separately.



























// Here is one solution:
/*
int main() {
    int n{};

    while ( true ) {
        std::cout << "Enter a positive integer: ";
        std::cin  >> n;

        if ( n >= 1 ) {
            break;
        }
    }

    int iterations{ 0 };

    while ( n != 1 ) {
        std::cout << n << ", ";

        if ( n%2 == 0 ) {
            n /= 2;
        } else {
            n = 3*n + 1;
        }

        ++iterations;
    }

    std::cout << 1 << std::endl;

    std::cout << "Iterations: "
              << iterations << std::endl;

    return 0;
}
*/