Fibonacci Spiral Help?

I'm trying to write a Java program that will display the Fibonacci Spiral as shown here . For right now, I just have it printing the starting coordinates of the rectangle that's being drawn, along with its length. There's nothing wrong with my program, except for a little array problem that I can fix, but I think there is a fundamental flaw in my thinking for this program. Can someone take a look at my code below and see why it does not work?
*Note: This is a school project, so I don't want too much help telling me what I should do, I just need help telling me what I shouldn't have done and why.
Thanks!

Code :

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public class FibSpiralPanel extends JPanel{
 
	public void paintComponent(Graphics g){
		super.paintComponent(g);
 
		int xCenter = getWidth()/2;
		int yCenter = getHeight()/2;
		int x=0, number, nextNumber = 0, count;
		int[] xStart = {0,0,0,0,0,0,0,0}, yStart={0,0,0,0,0,0,0,0};
 
		xStart[0] = xCenter;
		yStart[0] = yCenter;
 
		for(x=1; x<5; x++){
			number = (int)Fibonacci.getFibonacci(x);  //Get the length of the square.
 
 
 
			if(x%4 == 0){
				xStart[x] = xStart[x-1] - number;
				yStart[x] = yStart[x-1];
			}
 
			else if(x%3==0){
				xStart[x] = xStart[x-3];
				yStart[x] = yStart[x-3] - number;
			}
 
			else if(x%2 ==0){
				xStart[x] = xStart[x-2] + number;
				yStart[x] = yStart[x-2];
			}
 
			else{
				xStart[x] = xStart[x-1];
				yStart[x] = yStart[x-1] + number;
 
			}
			System.out.println("Fibonacci "+x+" is "+number);
			System.out.println("xStart: "+xStart[x]+"\nyStart: "+yStart[x]);
		}				
	}
}

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Inspect the way the code determines the location of a Fibonacci square (eg the uses of modulus). My understanding is that the squares are repetitive in fours, and so checking the remainder from dividing by 4 may be more appropriate. Also, I assume the getFibonacci method returns the value at position x in the series? Because that code isn't listed I'll just comment that this is in the loop of a drawing routine so it should be fast (precalculated)

Calculating Fibonacci can be fast... O(n) with O(1) memory space :)

It might be easier to work with the corners rather than the centers because they are always aligned by at least one axis (either x or y), so there's no need to calculate offsets between different centers. Maybe stick to the top-left corner? Also, in your modulus section, you don't want to modulate with different numbers, but rather with 4 all the time and check to see what the remainder value is.

Code :

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for(int x = 1; x < 5; x++)
{
     if (x % 4 == 0)
     { }
     else if (x % 4 == 1)
     { }
     else if (x % 4 == 2)
     { }
     else if (x % 4 == 3)
     { }
}

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Quote:

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Originally Posted by helloworld922

Calculating Fibonacci can be fast... O(n) with O(1) memory space

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I dislike big O notation, mainly because I stink at calculating it #-o...but do that in a loop it becomes something like O(n*n) (I know I know, we're talking a loop of five here though ;) )

I'll second Helloworld's advice on working with corners, especially if you plan to do the drawing using g.drawArc(...)

Code :

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public static int fib(int n)
{
     if (n < 3)
     {
          return 1;
     }
     int num1 = 1, num2 = 1;
     for (int i = 3; i < n; i++)
     {
          int temp = num1 + num2;
          num1 = num2;
          num2 = temp;
     }
     return num2;
}

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In a loop, still O(n) time. True dat it's only a loop of 5, but it's still nice to be efficient :)

Quote:

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Originally Posted by helloworld922

In a loop, still O(n) time. True dat it's only a loop of 5, but it's still nice to be efficient :)

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Sorry, I meant to calculate the series over and over again in a loop. For example (using your fib example function)

Code :

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int n = 5;
for ( int i = 0; i < n; i++ ){
    int fib = fib(i);
}

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As the loop goes through, it recalculates the previously calculated fibonacci values, doing so in a drawing routine which may be called often. Of course, for 5 its not a big deal, but larger values the computation time starts to pile up

This is as opposed to previously computing the series up to n, then accessing these as needed

Code :

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/**
*Returns an array containing the Fibonacci sequence up to n in the series
*/
public static int[] fib(int n)
{
     int fib[] = new int[n];
     if (n < 3)
     {
 
          fib[0] = 1;
          if ( n == 2 ) 
             fib[1] = 1;
     }
     for (int i = 3; i < n; i++)
     {
          fib[i] = fib[i-1] + fib[i-2];
     }
     return fib;
}
 
int fib[] = fib(n);//even better to do this upon construction or whenever the value of n is set
for ( int i = 0; i < n; i++ ){
    int num = fib[i];
     ///do the calculations and drawing
}

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