Question

5.1 Collatz's hypothesis- 6 points In 1937, a mathematician named Lothar Collatz e formulated an intriguing hypothesis (aka. Collatz conjecture e ) that remains unsolved to this day (perhaps this would be a good challenge for you?) which can be described in the following way: 1. take any non-negative and non-zero integer number and name it cO; 2. if it's even, evaluate a new cO as cO/2 3. otherwise, if it's odd, evaluate a new cO as 3 cO+1 4. if cO 1, skip to point 2 The hypothesis says that, regardless of the initial value of cO, it will always (always!) go to 1 Of course, it's an extremely complex task to use a computer in order to prove the hypothesis for any natural number (it may in fact need artificial intelligence), but you can use C++ to check some individual numbers. Maybe you can find the one that disproves the hypothesis and become a famous mathematician Given the above scenario, write a program which reads one natural number and executes the above steps as long as co remains different from 1 Moreover, we'll give you another task - we want you to count the steps needed to achieve the goal. Your code should output all intermediate values of cO, too - it'll be very illustrative, won't it? Hint: the most important part of the problem is how to transform Collatz's idea into a "while" loop- this is the key to success Input validation: Do not accept a zero or negative value for the initial input. Include appropriate volidation routines If this happens, include a validation routine that clears the error state and allows the user to input a positive value. number Sample screen output 1: Enter a non negative number greater than 0: 15 46 70 35 106 53 160 e0 40 20

Answer 1

If you have any doubts, please give me comment...

#include<iostream>

using namespace std;

int main(){

int c0;

while(1){

cout<<"Enter a non negative number greater than 0: ";

cin>>c0;

if(c0>0)

break;

cout<<"Invalid value input"<<endl;

}

int steps = 0;

while(c0!=1){

if(c0%2==0)

c0 = c0/2;

else

c0 = 3*c0 +1;

steps++;

cout<<c0<<endl;

}

cout<<"steps = "<<steps<<endl<<endl;

return 0;

}