[RESOLVED] Hailstone Sequence Problem
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Thread: [RESOLVED] Hailstone Sequence Problem

  1. #1
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    [RESOLVED] Hailstone Sequence Problem

    Error
    Last edited by Squall_890; November 3rd, 2010 at 08:04 PM. Reason: Violation

  2. #2
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    Re: Hailstone Sequence Problem

    Please use code tags when posting code, [code]your code here[/code]. Press EDIT and add them and some formatting.

    Most of the code look alright but a few observations:

    - You assign both the lower and the upper value to the same variable (num)
    - In Hailstone you don't assign anything for an even number
    - You need a loop in main that couts every returned value from the calls to Hailstone (similar to the code in lengthHail)
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  3. #3
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    Re: Hailstone Sequence Problem

    Code:
    int Hailstone(int n)
    {
        if(n % 2 == 0)
        {
            n/2; <- This doesn't do anything Perhaps you want to assign the result to n?
        }
        else
        {
            n=3*n+1;    //number is odd
        }
        return n;        //returns the value 
    }
    
    //Calculates the length of the hailstone sequence
    int lengthHail(int num)
    {
        int count=1;
        while(num != 1)
        {
            num = Hailstone(num);
            count++;        //adds one to count everytime a calculation is done 
        }
        return count;
    }
    
    //Ask for user input and reads in two integers that represent the limit . 
    int main()
    {    
        int num;;
        cout << "Enter Lower Limit.\t";
        cin >> num;
        cout << "Enter Upper Limit.\t";
        cin >> num;
        cout << "Hailstone Sequence starting at " << num << " is " << Hailstone(num) << endl; //<- This doesn't print a sequence
        Hailstone(num);
        cout << "\nLength of sequence is" << lengthHail(num) << ".\n" << endl;
    }
    Looks OK. I have n idea what a lower/upper limit in a hailstone sequence is supposed to represent. All I know is that it has a starting sequence.

    Your hailstone function works at calculating a single iteration. You use it correctly in lengthHail with a while loop. However, in you main, you are not.

    If you want to store and see the entire sequence, you have to start either calling Hailstone in a loop, or recursivelly. Then, you can either print the result as you go, or store the result in a container.
    Is your question related to IO?
    Read this C++ FAQ LITE article at parashift by Marshall Cline. In particular points 1-6.
    It will explain how to correctly deal with IO, how to validate input, and why you shouldn't count on "while(!in.eof())". And it always makes for excellent reading.

  4. #4
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    Re: Hailstone Sequence Problem

    Quote Originally Posted by Squall_890 View Post
    So like this? or do I have to make one for upper as well?
    Did you try running the code? It looks fine to me, and since you don't really have a question, there's not much to add.
    Is your question related to IO?
    Read this C++ FAQ LITE article at parashift by Marshall Cline. In particular points 1-6.
    It will explain how to correctly deal with IO, how to validate input, and why you shouldn't count on "while(!in.eof())". And it always makes for excellent reading.

  5. #5
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    Re: Hailstone Sequence Problem

    So I was able to run the code but it seems that no matter what the length really is I keep getting the same length. So I'm guessing I need to do a for loop for upper as well?

    Code:
    #include <iostream>		//I/O Library C++
    using namespace std;	     // defining reserved names	
    
    //Hailstone sequence takes any number as an input and divides it by 2 if the number is even otherwise multiply by 3 and add 1 if the number is odd.
    //The idea is that no matter what the user inputs you will eventually come to the answer of 1 at the end of the sequence. 		
    
    //Hailstone function takes the input of the current number then calculates the next number in the hailstone sequence 
    //Replaced f with Hailstone because couldn't get past error message otherwise
    int Hailstone(int n)
    {
    	if(n % 2 == 0)
    	{
    		n /= 2;		//number is even 
    	}
    	else
    	{
    		n = 3 * n + 1;	//number is odd
    	}
    	return n;		//returns the value 
    }
    
    //Calculates the length of the hailstone sequence
    int lengthHail(int num)
    {
    	int count = 1;
    	while(num != 1)
    	{
    		num = Hailstone(num);
    		count++;		//adds one to count everytime a calculation is done 
    	}
    	return count;
    }
    
    //Ask for user input and reads in two integers that represent the limit . 
    int main()
    {	
    	int lower, upper;
    	int count = 0;	
    	cout << "Enter Lower Limit.\t";
    	cin >> lower;
    	cout << "Enter Upper Limit.\t";
    	cin >> upper;
    	for(int i = lower;i <= upper; i++)
    	{
    		lower = i;
    		while(lower != 1)
    		{
    			cout << " " << Hailstone(lower) << " " << endl;
    			lower = Hailstone(lower);
    			count++;
    		}
    		cout << "\nLength of sequence is " << lengthHail(upper) << ".\n" << endl;
    	}
    	return 0;
    }

  6. #6
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    Re: Hailstone Sequence Problem

    Quote Originally Posted by Squall_890 View Post
    So I was able to run the code but it seems that no matter what the length really is I keep getting the same length. So I'm guessing I need to do a for loop for upper as well?

    Code:
        ...
        for(int i = lower;i <= upper; i++)
        {
            lower = i;
            while(lower != 1)
            {
                cout << " " << Hailstone(lower) << " " << endl;
                lower = Hailstone(lower);
                count++;
            }
            cout << "\nLength of sequence is " << lengthHail(upper) << ".\n" << endl;
        }
        return 0;
    }
    You seem to be mixing and confusin the lower bound, the upper bound, the current index, and your current hailstone value.
    Is your question related to IO?
    Read this C++ FAQ LITE article at parashift by Marshall Cline. In particular points 1-6.
    It will explain how to correctly deal with IO, how to validate input, and why you shouldn't count on "while(!in.eof())". And it always makes for excellent reading.

  7. #7
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    Re: Hailstone Sequence Problem

    I think I may have the off by one bug but I can't seem to figure out how to resolve the issue.

  8. #8
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    Re: Hailstone Sequence Problem

    Quote Originally Posted by Squall_890 View Post
    For instance, if I were to input 7 for lower limit and 20 for upper limit I get the sequence correct but the length comes out to be 8 when the correct answer is 18.
    You get 8 because you don't really track the maximum in your program at all. You are just looking at the lenght of the last one of the checked sequences which is correctly 8. But the correct maximum for these limits isn't 18 either, it's 21 (length of the sequence starting with 19).

    I checked this with a program that is the result of applying my own corrections to your original code. And I'm pretty sure it works...

    But then in another case, if I make the lower 9 and the upper 28 it produces both the correct sequence and length.
    I can't see how you get to that result with the code you posted. I get a maximum legth of 112 for these limits, which is the length of the sequence starting at 27, not the last one (28).

  9. #9
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    Smile Re: Hailstone Sequence Problem

    Yes, you're correct sorry must of typed the wrong number was studying for an exam. I understand that I don't really follow the upper but I already tried using a loop to follow it but when I tried running the program it kept printing out, "Length of Sequence is 8" so I'm not really sure how exactly to go about following upper correctly. If you could give me a hint, I'd greatly appreciate it.

  10. #10
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    Re: Hailstone Sequence Problem

    You don't need an extra loop to track the maximum; you can just do that "on the fly" in the outermost loop (for) in main().

    A short breakdown on how to do it: Declare a variable for the maximum before entering the loop, and initialize it to an impossibly low maximum value. 0 would be a natural choice for that. During each of the iterations, check whether the current sequence length is greater than the maximum you have so far, and if yes, adjust the maximum accordingly. Once you're done with the loop, you have your maximum.

    The C++ code required to do this, BTW, would be much less characters than my description above.

    HTH

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