Cubic equation program not giving me right answers

This is a discussion on Cubic equation program not giving me right answers within the C++ Programming forums, part of the General Programming Boards category; Im getting the wrong roots when running the program. The maths the program is based on is in the attatched ...

  1. #1
    Refugee face_master's Avatar
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    Cubic equation program not giving me right answers

    Im getting the wrong roots when running the program. The maths the program is based on is in the attatched image.
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    
    #define pi 3.14159265
    
    int sign(float s);
    
    int main()
    {
    	float a1, a2, a3, x1, x2, x3, s, q, th, s_sq, q_cub, a1_sq, a1_cub, s_abs;	
    
    
    	printf("The cubic equation has the form x^3 + a1x^2 + a2x + a3 = 0\n");
    	
    	printf("Enter a1: ");
    	scanf("%f", &a1);
    	
    	printf("\nEnter a2: ");
    	scanf("%f", &a2);
    
    	printf("\nEnter a3: ");
    	scanf("%f", &a3);
    
    	a1_cub = pow(a1, 3);
    	a1_sq = pow(a1, 2);
    
    	/* calculate q and s */
    
    	q = ( a1_sq - (3 * a2) ) / 9;
    	
    	s = ( (2 * a1_cub) - (9 * (a1*a2) ) + (27 * a3) ) / 54;
    
    	/* get square / cube of s and q  */
    
    	s_sq = pow(s, 2);
    	q_cub = pow(q, 3);
    
    	s_abs = fabs(s);
    
    
    	/* ****** */
    
    	th = acos( (s/(sqrt(q_cub)) )); /* calculate theta */
    
    	if( (q_cub - s_sq) >= 0)	/* cubic equation has 3 roots */
    	{
    		x1 = ((-2)*(sqrt(q))*(th/3)) - (a1/3);
    		
    		x2 = ((-2)*(sqrt(q))*((th + (2*pi))/3)) - (a1/3);
    
    		x3 = ((-2)*(sqrt(q))*((th + (4*pi))/3)) - (a1/3);
    
    		printf("\nx1 = %f\nx2 = %f\nx3 = %f\n", x1, x2, x3);
    
    	}
    	else if( (s_sq - q_cub) > 0 ) /* cubic equation has 1 root */
    	{
    		
    		float part1 = sqrt(s_sq - q_cub) + s_abs;
    
    		x1 = -(sign(s)) * ( (pow(part1, (1/3))) + (q/(pow(part1, (1/3)))) ) - (a3/3);
    		                                                                                                                                           
    		printf("\nx1 = %f", x1);
    		
    	}
    	else
    	{
    		printf("something crap happened\n");
    	}
    
    
    	system("pause");
    	
    	return 0;		
    }
    
    int sign(float s)
    {
    	if(s > 0)
    	{
    		return 1;
    	}
    	else if(s < 0)
    	{
    		return -1;
    	}
    	
    }
    Attached Images Attached Images  

  2. #2
    C++ Witch laserlight's Avatar
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    Any example input, output, and expected output values? One potential problem for the single root solution is that 1/3 will be evaluated as 0.
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  3. #3
    Refugee face_master's Avatar
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    im testing my results with the results from this website

    http://www.agolda.com/CubicEquation.html

    which seem to be accurate

    I changed the (1/3)'s to (0.334)'s and the results are still wrong

  4. #4
    and the hat of wrongness Salem's Avatar
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    > (1/3)
    This sub-expression will be evaluated as integers, giving 0

    > int sign(float s)
    What does this return when s == 0 ?

    > x1 = ((-2)*(sqrt(q))*(th/3)) - (a1/3);
    Your graphic contains a cos() as well
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  5. #5
    Refugee face_master's Avatar
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    > x1 = ((-2)*(sqrt(q))*(th/3)) - (a1/3);
    Your graphic contains a cos() as well

    ARGH! Thanks...that was a really stupid mistake, a pretty much rewrote the code like 3 times trying to fix it

  6. #6
    C++ Witch laserlight's Avatar
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    Did that fix it?
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  7. #7
    Refugee face_master's Avatar
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    it worked for a few but when I tried

    a1 = -4
    a2 = -1
    a3 = -20

    I get an incorrect result

  8. #8
    The superhaterodyne twomers's Avatar
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    x1 = -(whatever) ... a1 => a1 = -(x1 whatever...)

    are they the right numbers, but with the wrong signs?

  9. #9
    Refugee face_master's Avatar
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    I assume theyre right...i just followed the information i was given (that image i attacthed)

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