Square Root of a number

This is a discussion on Square Root of a number within the C++ Programming forums, part of the General Programming Boards category; Doesnt work, what do I need to change? Code: #include <iostream> using namespace std; double U; double L = 0; ...

  1. #1
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    Square Root of a number

    Doesnt work, what do I need to change?

    Code:
    #include <iostream>
    using namespace std;
    
    double U;
    double L = 0;
    double guess;
    
    double abs(int a,int b)
    {
    	if ((a-b) < 0)
    	{
    		return (a-b)*-1;
    	}
    	else
    	{
    		return (a-b);
    	}
    }
    double sqrt(double x)
    
    {
    	if (x == 0 || x == 1)
    	{
    		return x;
    	}
    	else 
    	{
    		guess = x/2;
    	}
    		while (abs((guess*guess)-x) != .0000000000000000001)
    		if (guess*guess > x)
    		{		
    				U = guess;
    				guess = ((U - L)/2) + L;
    		}
    		else 
    		{
    				L = guess;
    				guess = ((U - L)/2) + L;
    		}		
    		cout<<guess;
    }

  2. #2
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    Can you describe HOW it doesn't work.

    --
    Mats
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    Please don't PM me for help - and no, I don't do help over instant messengers.

  3. #3
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    Doesnt print the "guess".

  4. #4
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    Is is possible that
    Code:
    while (abs((guess*guess)-x) != .0000000000000000001)
    never gets fulfilled?

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  5. #5
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    So whats wrong? is it the syntax or the algorithm?

  6. #6
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    It's the convergence criterion.

    The thing to remember is that floating point types cannot represent any arbitrary value: there is a smallest positive value that can be represented in a floating point variable. If your 0.000 (many zeros) 001 is less than that smallest representable value, then your algorithm may never satisfy your convergence criterion.

    Out of curiosity, why not use the sqrt() function from <cmath>?

    Incidentally, if you're not printing your guess, the function is never returning either ..... you have an infinite loop.

  7. #7
    Registered User Cpro's Avatar
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    Code:
    while (abs((guess*guess)-x) != .0000000000000000001)
    I don't think your abs function is being called (correctly anyways), because you are only passing one argument.
    I could be wrong, but in the test I ran:
    Code:
    int main()
    {
    	sqrt(9);
    
    	return 0;
    }
    it wasn't being called.
    However, I'm not sure why I wasn't receiving any errors when I compiled.
    IDE - Visual Studio 2005
    Windows XP Pro

  8. #8
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    Right!

  9. #9
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    Quote Originally Posted by freddyvorhees View Post
    Right!
    a) Those are not integers.
    b) Once they have turned into integers, how can the difference from subtracting a-b be a floating point value?

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  10. #10
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    noted,but still noting

  11. #11
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    This is another version, it also doesnt work,prints nothing

    Code:
    double abs(int a,int b)
    {
    	if ((a-b) < 0)
    	{
    		return (a-b)*-1;
    	}
    	else
    	{
    		return (a-b);
    	}
    }
    double sqrt(int x)
    {
    double U = x;
    double L = 0;
    double guess = 0;
    
    	if (x == 0 || x == 1)
    	{
    		return x;
    	}
    
    		while (abs((x/2*x/2),x) > .001)
    			guess = (x/2);	
    			if (guess*guess > x)
    			{
    				guess = (U - L)/2;
    			}
    			else
    			{
    				L = x;
    				guess = (U - L)/2;
    			}
    			return x/2;
    }

  12. #12
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    Code:
    while (abs((x/2*x/2),x) > .001)
        guess = (x/2);
    You don't change x in the loop, so the loop never ends.

  13. #13
    Malum in se abachler's Avatar
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    The problem is the above mentioned convergence problem. You need to increase 0.0000~001 or change the line

    Code:
    while (abs((guess*guess)-x) != .0000000000000000001)
    to

    Code:
    while (abs((guess*guess)-x) != 0.000001)
    First get it working, then worry abotu being acurate.

    Also, I noticed you dont have a main(), is what you posted all the code? There may be problems in how you are calling the function.

    On secodn thought just replace that whoel section with -

    Code:
    while (abs((guess*guess)-x) > .00000001) guess = (guess + x/guess)/2;
    Last edited by abachler; 07-30-2008 at 08:53 AM.
    Until you can build a working general purpose reprogrammable computer out of basic components from radio shack, you are not fit to call yourself a programmer in my presence. This is cwhizard, signing off.

  14. #14
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    you always check that your error is GREATER THAN OR EQUAL TO your error tolerance; not exactly equal to.

    the odds of it hitting that exact number are infinitesimal.

  15. #15
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    Working now but results are inaccurate:
    Code:
    #include <iostream>
    using namespace std;
    
    
    double abs(int a,int b)
    {
    	if ((a-b) < 0)
    	{
    		return (a-b)*-1;
    	}
    	else
    	{
    		return (a-b);
    	}
    }
    double sqrt(double x)
    
    
    
    {double U = x;
    double L = 0;
    double guess= x/2;
    	if (x == 0 || x == 1)
    	{
    		return x;
    	}
    	else 
    		while (abs((guess*guess),x) > .000000000001)
    		if (guess*guess > x)
    		{
    				U = guess;
    				guess = ((U - L)/2) + L;
    		}
    		else 
    		{
    				L = guess;
    				guess = ((U - L)/2) + L;
    		}		
    		return guess;
    }
    int main()
    {
    	cout<<"Root ";
    	cout<<sqrt(9);
    }
    prints 3.09375

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