If you are submitting this as homework there are some elements of style that may help you. What are the roles of 200, 100 and 300? These are "magic constants" which should be avoided. Are they related or is it just chance that they have these values? Suggest you use symbols such as:
int NPOINTS = 200;
or
double radius = 100.0
That would reveal whether the 300 was actually the value you want. I haven't checked.
Personally I wouldn't write
y*-1
but
-y
as it's too easy to mistype the former.
I would also print out the 200 points as floats and see if you can tell by eye where the error is. It's highly likely that the spurious lines are either drawn at the start or end of the calculation - it's easy to make "end-effect" errors where exactly one point is omitted or calculated twice.
Also it's cheap to experiment. Try iterating c from 0 to 100. or 0 to 10, or 0 to 198 or 1 to 200. Does your spurious line/triangle always occur?
UPDATE Here is what I think is wrong and how to tackle it. You have made a very natural graphics error and a fence-post error (http://en.wikipedia.org/wiki/Off-by-one%5Ferror) and it's hard to detect what is wrong because your variable names are poorly chosen.
What is mypoints? I believe it is the bottom half of the squircle - if you had called it bottomHalf then those replying woulod have spotted the problem quicker :-).
Your graphics problem is that you are drawing TWO HALF-squircles. Your are drawing CLOSED curves - when you get to the last point (c==199) the polygon is closed by drawing back to c==0. That makes a D-shape. You have TWO D-shapes, one with the bulge UP and one DOWN. Each has a horizontal line closing the polygon.
Your fence-post error is that you are drawing points from 0 to 199. For the half-squircle you want to draw from 0 to 200. That's 201 points! The loss of one point means that you have a very slightly sloping line. The bottom lines slopes in tghe opposite direction from the top. That gives you a very then wedge shape, which you refer to as a triangle. I'm guessing that your triangle is not actually closed but like a slice from a pie but very then/sharp.
(The code below could be prettier and more compact. However it is often useful to break symmetrical problems into quadrants or octants. It would also be interesting to use an anngle to sweep out the polygon).
You actually want ONE polygon. The code should be something like:
int NQUADRANT = 100;
int NPOINTS = 4*NQUADRANT ; // closed polygon
double[] xpoints = new double[NPOINTS];
double[] ypoints = new double[NPOINTS];
Your squircle is at 100, 100 with radius 100. I have chosen different values here
to emphasize they aren't related. By using symbolic names you can easily vary them.
double xcenter = 500.0;
double ycentre = 200.0;
double radius = 100.;
double deltax = radius/(double) NQUADRANT;
// let's assume squircle is centered on 0,0 and add offsets later
// this code is NOT complete or correct but should show the way
// I might have time later
for (int i = 0; i < NPOINTS; i++) {
if (i < NQUADRANT) {
double x0 = -radius + i* deltax;
double y0 = fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}else if (i < 2*NQUADRANT) {
double x0 = (i-NQUADRANT)* deltax;
double y0 = fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}else if (i < 3*NQUADRANT) {
double x0 = (i-2*NQUADRANT)* deltax;
double y0 = -fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}else {
double x0 = -radius + (i-3*NQUADRANT)* deltax;
double y0 = -fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}
}
// draw single polygon
private double fourthRoot(double radius, double x) {
return Math.sqrt(Math.sqrt(radius*radius*radius*radius - x*x*x*x));
}