java,recursion,coding-style,drawrectangle,spiral

public class DrawIt { public static void main(String[] args) { JFrame frame = new JFrame(); final int width = 400; final int height = 400; frame.setSize(width, height); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); JComponent component = new JComponent() { public void paintComponent(Graphics graph) { draw(graph); } }; frame.add(component); frame.setVisible(true); } public static void drawpuzzle(Graphics g,...

php,sorting,coordinates,spiral

Algorithm design First, free your mind and don't think of a spiral! :-) Then, let's formulate the algorithms constraints (let's use the salesman's perspective): I am currently in a city and am looking where to go next. I'll have to find a city: where I have not been before that...

python,function,recursion,turtle-graphics,spiral

Depending on the values of initialLength and multiplier, it is very possible that your function will never be exactly 1. You check for this right here: if initialLength == 1 or newLength ==1: up() If it never reaches exactly one, the turtle will never stop drawing. Try changing it to:...

If you are entering your angles in degress, the proper conversion to radians is: float radians = degrees * M_PI / 180.0f; Your input to the function appears to be wrong, too. float aroundStep = coils/sides;// 0 to 1 based. This comment suggests that coils < sides. You have entered...

You could make the size of each div a function of the radius. One-seventh seems to work well: size = radius/7; $("#terrain") .append("<div class='drag' style='top:"+ y +"px;left:"+x+ "px;width:"+size+"px;height:"+size+"px;'></div>" ); Working Fiddle Edit Based on our comments, I now understand the problem better. You could declare a distance variable, which is...

c,algorithm,recursion,matrix,spiral

#include <stdio.h> void build_matrix(int msize, int a[msize][msize], int size, int value){ int i, row, col; if(size < 1) return; row = col = (msize - size) / 2; if(size==1){ a[row][col] = value; return; } for(i=0;i<size-1;++i) a[row][col++] = value++;//RIGHT for(i=0;i<size-1;++i) a[row++][col] = value++;//DOWN for(i=0;i<size-1;++i) a[row][col--] = value++;//LEFT for(i=0;i<size-1;++i) a[row--][col] = value++;//UP...

I was able to get an affect you are looking for by doing node.radius = a * Math.exp(b * offset)/6; 6 is an arbitrary number to adjust the size of the circle....

Without loss of generality, let's write the array as: arr = [ [ 1, 2, 3, 4,], [12,13,14, 5,], [11,16,15, 6,], [10, 9, 8, 7,] ] The desired result is: [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] I will use a helper: def rotate(arr) arr.map(&:reverse).transpose end For example: rotate(arr) #=> [[4, 5, 6, 7], [3, 14,...