int cellSize = 30;
int radius = cellSize/2;
int ellipseSize = cellSize;
int ellipseSizeMini = 5;
int gap = 0;
void setup()
{
size (300, 300);
background(255);
noStroke();
for (int j=0; j<height; j=j+cellSize)
for (int i=0; i<width; i=i+cellSize)
fill(0);
ellipse(i+radius, j+radius, ellipseSize, ellipseSize);
ellipseSizeMini = calculateSizeMini(i, j);
gap = radius-(ellipseSizeMini/2);// [10%] to write an equation which calculates the length of gap between the smaller white circle and the bigger black circle. Hard coding value does not receive any mark
float shiftX = random(-gap, gap);
float shiftY = random(-gap, gap);
fill(255);
ellipse(i+radius+shiftX,j+radius+shiftY,ellipseSizeMini,ellipseSizeMini); // [10%] fill in the four parameters (each takes 2.5% mark) for this ellipse() so that it draw a white circle on top of its corresponding black circle with size of ellipseSizeMini
}
}
int calculateSizeMini(int i, int j)
{
int sizeMax = (int)0.8*ellipseSize;// [5%] to write an equation which generates a value of 80% of ellipseSize to be the maximum possible size of the small white circle. Hard coding value does not receive any mark
int sizeMin = (int) 0.1*ellipseSize;// [5%] to write an equation which generates a value of 10% of ellipseSize to be the minimum possible size of the small white circle. Hard coding value does not receive any mark
int answer = ellipseSizeMini;
// [20%] to develop equation(s) and write lines of code here to calculate a value for ellipseSizeMini. This value changes according to i and j.
// The value is closed to sizeMax if i and j are closed to the center of the application window, and it closed to sizeMin if i and j are
// closed to the edge of the application window. This value will be assigned to the variable of answer as a returning value of this function.
// The simpler the approach, the higher is the mark.
return answer;
}
i finshed the previous part and the last part just make me crazy. i can’t work this out.
Can someone help me ? The closer a white dot locates toward the center, the bigger it is and vice versa. You should try to solve this problem by simply coding.