Simple while loop doesn't end..

[DrJones]

why does this not end...
It should end when either one of the values becomes 19 or above..

typedef std::pair<int,int> coordinate;
coordinate endpoints(double angle, coordinate start)
{
    double radians = (M_PI/180)*angle;
    double x2 = 0;
    double y2= 0;
    int lenght = 0;
    
    while (x2 < 19 ||  y2 < 19)
    {
        x2 = start.first + (lenght * cos(radians));
        y2 = start.second + (lenght * sin(radians));
        std::cout << x2 << y2 << std::endl;
        lenght++;
    }

    
    return std::make_pair(round(x2),round(y2));
}

[helios]

You probably meant to use &&, not ||. As it is, the loop will only end when both x2 and y2 become greater than 19 (x2 < 19 || y2 < 19 is equivalent to !(x2 >= 19 && y2 >= 19)). If angle == 0 then x2 will always equal start.first.

[DrJones]

i only have to end if one of being 19.. could i ask a related question to this ?

[Chervil]

What happens when x2 or y2 are negative, and become more negative?

[DrJones]

well.. That the case..


I have a problem with the negative value..
The idea with this code was to find an endpoint inside a matrix sixes 19x19.

from the center of this matrix (10,10) will a line with length 9 (hence x2 <19 or y2 < 19) try to find a endpoint at that given angle with the given length/displacement.

So sometimes i get negative values.. but at those cases it should output a proper value which fits the matrix, and is angled with the given angle from (10,10)

[DrJones]

in am not sure i get it..?

[JayhawkZombie]

Just ignore him. dhs2's trolling.

[helios]

You don't need a loop to calculate that.

Let L be a straight line that forms an angle theta with the horizontal axis and passes through (10, 10), then its equation is y = cos(theta) * (x - 10) + 10 if theta in (-pi / 2; pi / 2). This line crosses these lines at the following points:
x = 0 at (0, cos(theta) * (-10) + 10) = P1
x = 19 at (19, cos(theta) * 9 + 10) = P2
y = 0 at (-10 / cos(theta) + 10, 0) = P3
y = 19 at (9 / cos(theta) + 10, 19) = P4
If P3.x > 19 || P3.x < 0 then the line is approximately horizontal, and will leave the box at P1 and P2.
If P3.x < 19 && P3.x > 0 then the line is approximately vertical, and will leave the box at P3 and P4.
If P3.x == 19 || P3.x == 0 the line is exactly diagonal and either pair of points may be used.

[Chervil]

I'm not sure whether you want the maximum length which will fit within those boundaries. That would make a diagonal longer than a horizontal or vertical. Ot is it like the radius of a circle, a fixed length.

If you start from (10,10), 19 would be an upper limit. 1 would be the lower limit. You'd need to check both.
maybe?

    do {
        x2 = start.first + (length * cos(radians));
        y2 = start.second + (length * sin(radians));
        std::cout << std::setw(8) << x2 << std::setw(8) <<  y2 << std::endl;        
        length++;
    } while (x2<19 &&  y2<19 && x2>1 && y2>1);

[DrJones]

The idea is has to be computed from 0 - 2pi..

in between is computed using the equation.. and x = 9..

what about -pi/4 and pi/4..


what about X in between?

[DrJones]

@Chervill


19 is the fixed length (9 steps steps along either x or y ) the lower limit is 0.. array implementation..

[Chervil]

How do you handle negative values?

I don't.

Perhaps I misunderstood. I used a starting point of 10, 10
I think this way the value can never become negative.

I'm not sure about the change from while to do-while. it was just an idea I had to make sure that it started off from x2 = start.first and y2 = start.second.

[Chervil]

You don't need a loop to calculate that.

Yes, I agree.

[helios]

The idea is has to be computed from 0 - 2pi..

The range [pi; 2*pi] is equivalent to the range [0; pi], only the order of the solutions is inverted. I.e. if f(theta) = sequence[a, b], then f(theta + pi) = sequence[b, a]. If order doesn't matter then f(theta) = f(theta + pi).

in between is computed using the equation.. and x = 9..

I don't understand this sentence.

what about -pi/4 and pi/4..

Those values are explicitly included in my method. Do you mean (+/-)pi/2? For (+/-)pi/2 the solutions are known: (10, 0) and (10, 19).

[DrJones]

I am at the moment trying to iterate a 2d array [square matrix] using a line which scans the matrix with a given radius, and angle..

In this case the radius is 9 (9 being 9 different array position), and the angle being from 0 - 2pi.

The endpoint value should be a value within the 2d array, which was my first problem, and then the negative values occurred.

your solution seem to output values which is displaced less than 9 array positions array from the center...


0-90-180-270-360 seem to be correct.. There is some roundoff errors...

since you ignore negative values, you don't assure that the end positions is placed 9 array positions away from the start..

[helios]

If you just want the matrix cell that is at a certain polar coordinate from the center, then that's simply
(round(r * cos(theta) + centerx), round(r * sin(theta) + centery)
After rounding, coordinates outside the matrix should be ignored.