a program for the 3x+1 problem

rec3g · 03-29-2008, 04:17 PM

okay so I'm trying to write a program that will calculate the number of steps it takes to get to 1 in the 3x+1 problem. The rules for the 3x+1 problem are as follows:

If a number is even divide by 2
If a number is odd, triple it and add 1
Stop if you ever get to 1

I have a program written but it only gives me an answer of 4 everytime. I don't understand why though! Can anyone point out what my problem is???

Quote:

#include <cmath>
#include <iostream>

using namespace std;

//Function prototype.
int UlamSequenceLength(int& number);

int main()
{
int number;
long count;
long x;

cout << "Enter number: " << endl;
cin >> number;

while (number != 1)
{
UlamSequenceLength(number);
count++;
}

x = UlamSequenceLength(number);

cout << x << endl;

}

int UlamSequenceLength(int& number)
{
if(number%2 == 0)
number = number/2;
else
number = (number * 3) + 1;
return number;
}

Dan04 · 03-29-2008, 04:51 PM

By the time the statement

Code:

x = UlamSequenceLength(number);

is executed, number will always be 1. Thus, you'll always get the hailstones sequence 1, 4, 2, 1 when you call UlamSequenceLength a second time. Why are you doing that?

matthewg42 · 03-29-2008, 04:56 PM

Please post code in code tags, not quote tags. code tags will preserve indentation, making your posted code a lot easier to read (assuming you have sane indentation in th original).

It looks like homework, so I won't try to answer the question directly. I would suggest stepping through the program with a debugger to understand what it is doing.

osor · 03-29-2008, 08:35 PM

Also, don’t you want to print count (not x)?

rec3g · 03-30-2008, 10:19 AM

I'm so confused... if I assigned count to x...why would it matter which one I print? but I did remove the x and tried to just print count and it gave me some off the wall negative number. And Dan04..why am I doing what?

rec3g · 03-30-2008, 10:37 AM

Code:

#include <cmath>
#include <iostream>

using namespace std;

//Function prototype.
int UlamSequenceLength(int& number);
	

int main()
{
	int number;
	long count;

	
	cout << "Enter number: " << endl;
	cin >> number;
	
	while (number != 1)
	{
		UlamSequenceLength(number);
		count++;	
	}
	
	
	cout << count << endl;
	
}

int UlamSequenceLength(int& number)
{
	if(number%2 == 0)
		number = number/2;
	else
		number = (number * 3) + 1;
	return number;
}

So this is my latest program..still not working

..I stink at programming

Dan04 · 03-30-2008, 11:30 AM

You aren't initializing count.

rec3g · 03-30-2008, 12:42 PM

wow...I'm stupid haha

osor · 03-30-2008, 01:01 PM

Quote:

Originally Posted by rec3g

if I assigned count to x...why would it matter which one I print?

(In your first attempt) you never assigned count to x. You assigned to x the number in the sequence after 1 (which was invariably 4).

Quote:

Originally Posted by Dan04

You aren't initializing count.

Yes. Remember that local variables with automatic storage (those without the keyword static) are not initialized to anything (so the value they usually have is whatever data which was already on the stack).