Code Golf: Ulam Spiral

Question

The Challenge

The shortest code by character count to output Ulam's spiral with a spiral size given by user input.

Ulam's spiral is one method to map prime numbers. The spiral starts from the number 1 being in the center (1 is not a prime) and generating a spiral around it, marking all prime numbers as the character '*'. A non prime will be printed as a space ''.

alt text http://liranuna.com/junk/ulam.gif

Test cases

Input:
    2
Output:
    * *
      *
    *  

Input:
    3
Output:
    *   *
     * * 
    *  **
     *   
      *  

Input:
    5
Output:
        * *  
     *     * 
    * *   *  
       * * * 
      *  ** *
     * *     
    *   *    
     *   *   
    *     *  

Code count includes input/output (i.e full program).

Answer 1

C, 208 206 201 200 199 196 194 193 194 193 188 185 183 180 176 Bytes

(if newlines are removed):

main(int u,char**b){
for(int v,x,y,S=v=**++b-48;--v>-S;putchar(10))
for(u=-S;++u<S;){
x=u;y=v;v>-u^v<u?:(x=v,y=u);
x=4*y*y-x-y+1+2*(v<u)*(x-y);
for(y=1;x%++y;);
putchar(y^x?32:42);}}

Compiled with

> gcc -std=c99 -o ulam ulam.c

Warning. This program is slow, because is does a trial division up to 2^31. But is does produce the required output:

    * *
 *     *
* *   *
   * * *
  *  ** *
 * *
*   *
 *   *
*     *

In nicely formatted C and with redundant #includes:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char** argv) {

    int u,v,x,y,d,S = atoi(argv[1]);

    /* v is the y coordinate of grid */
    for (v=S; v>=-S; --v)

        /* u is the x coordinate. The second operand (!putchar...) of the boolean or
         * is only ececuted a a end of a x line and it prints a newline (10) */
        for (u=-S; u<=S || !putchar(10); ++u) {

            /* x,y are u,v after "normalizing" the coordintes to quadrant 0
               normalizing is done with the two comparisions, swapping and and
               an additional term later */
            d = v<u;
            x=u;
            y=v;

            if (v<=-u ^ d) {
                x=v;
                y=u;
            }

            /* reuse x, x is now the number at grid (u,v) */
            x = 4*y*y -x-y+1 +2*d*(x-y);   

           /* primality test, y resused as loop variable, won't win a speed contest */
            for (y=2; y<x && x%y; ++y)
                 ;

            putchar(y!=x?' ':'*');
        }
}

It works by transforming the coordinates of the grid to the appropriate number and then performing the primality test, intead of drawing in a snake-like manner. The different equations for the four "quadrants" can be collapsed into one with swapping x and y and an additional term for "backward counting".

Answer 2

MATLAB: 182 167 156 characters

Script ulam.m:

A=1;b=ones(1,4);for i=0:(input('')-2),c=b(4);b=b+i*8+(2:2:8);A=[b(2):-1:b(1);(b(2)+1:b(3)-1)' A (b(1)-1:-1:c+1)';b(3):b(4)];end;disp(char(isprime(A)*10+32))

And formatted a little nicer:

A = 1;
b = ones(1,4);
for i = 0:(input('')-2),
  c = b(4);
  b = b+i*8+(2:2:8);
  A = [b(2):-1:b(1); (b(2)+1:b(3)-1)' A (b(1)-1:-1:c+1)'; b(3):b(4)];
end;
disp(char(isprime(A)*10+32))

Test cases:

>> ulam
2
* *
  *
*  
>> ulam
3
*   *
 * * 
*  **
 *   
  *  
>> ulam
5
    * *  
 *     * 
* *   *  
   * * * 
  *  ** *
 * *     
*   *    
 *   *   
*     *  

Answer 3

J solution: 197 173 165 161 bytes (so far)
this does not use the method mentioned in the comments to the OP

p=:j./<.-:$g=:1$~(,])n=:<:+:".1!:1]3
d=:j.r=:1
(m=:3 :'if.y<*:n do.if.0=t=:<:t do.d=:d*0j1[t=:<.r=:r+0.5 end.m>:y[g=:y(<+.p=:p+d)}g end.')t=:2
1!:2&2(1 p:g){' *'

Answer 4

Python 2.x, 220C 213C 207C 204C 201C 198C 196C 188C

Special thanks to gnibbler for some hints in #stackoverflow on Freenode. Output includes a leading and trailing newline.

import math
v=input()*2
w=v-1
a=['\n']*w*v
p=w*v/2
for c in range(1,w*w):a[p]=' *'[(c>1)*all(c%d for d in range(2,c))];x=int(math.sqrt(c-1));p+=(-1)**x*((x*x<c<=x*x+x)*w+1)
print''.join(a)

(Python 3 compatibility would require extra chars; this uses input, the print statement and / for integer division.)

Answer 5

Lua, 302 characters

s=...t={" "}n=2 function p()for j=2,n-1 do if n%j==0 then n=n+1 return" "end
end n=n+1 return"*"end for i=2,s do for k=#t,1,-1 do t[k+1]=t[k]..p()end
t[1]=""for k=1,i*2-1 do t[1]=p()..t[1]end for k=2,#t do t[k]=p()..t[k]end
t[#t+1]=""for k=1,i*2-1 do t[#t]=t[#t]..p()end end print(table.concat(t,"\n"))

Output from lua ulam.lua 6:

*   *      
     * *   
* *     * *
 * *   *   
    * * *  
   *  ** * 
* * *      
 *   *     
* *   *   *
 *     *   
  *        

Answer 6

Python - 203 Characters

  _________________________________________________________
 /x=input();y=x-1;w=x+y;A=[];R=range;k,j,s,t=R(4)          \
| for i in R(2,w*w):                                        |
|  A+=[(x,y)]*all(i%d for d in R(2,i))                      |
|  if i==s:j,k,s,t=k,-j,s+t/2,t+1                           |
|  x+=j;y+=k                                                | 
| for y in R(w):print"".join(" *"[(x,y)in A]for x in R(w))  |
 \_________________________________________________________/
        \   ^__^
         \  (oo)\_______
            (__)\       )\/\
                ||----w |
                ||     ||


x=input();y=x-1;w=x+y
A=[];R=range;k,j,s,t=R(4)
for i in R(2,w*w): 
 A+=[(x,y)]*all(i%d for d in R(2,i))
 if i==s:j,k=k,-j;s,t=s+t/2,t+1
 x+=j;y+=k
for y in R(w):print"".join(" *"[(x,y)in A]for x in R(w))

How it works
The idea is to fill A with x,y coords that need to be printed as '*'
The algorithm starts at the cell corresponding to 2, so the special case of testing 1 for primality is avoided.
x,y is the cell of interest
j,k keep track of whether we need to inc or dec x or y to get to the next cell
s is the value of i at the next corner
t keeps track of the increment to s

all(i%d for d in R(2,i)) does the primality check

The last line is rather clumsy. It iterates over all the cells and decides whether to place a space or an asterisk