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

Python, 299 characters:

from sys import *
def t(n):
 if n==1:return ' '
 for i in range(2,n):
  if n%i==0:return ' '
 return '*'
i=int(stdin.readline())
s=i*2-1
o=['\n']*(s+1)*s
c=1
g=2
d=0
p=(s+2)*(i-1)
for n in range(s**2):
 o[p]=t(n+1);p+=[1,-s-1,-1,s+1][d%4];g-=1
 if g==c:d+=1
 if g==0:d+=1;c+=1;g=2*c
print ''.join(o)

Answer 2

Mathematica 243

l = Length; t = Table; f = Flatten;
h@m_ := With[{x = l@m[[1]], y = l@m}, f[{{Reverse@t[w + y + (x y), {w, x + 2}]}, 
  t[f[{(x y) + x + y + 2 + w, m[[w]], (x y) + y - w + 1}], {w, y}], 
  {t[2 + y + x + y + w + (x y), {w, x + 2}]}}, 1]];
m_~g~z_ := Nest[h, m, z] /. {_?PrimeQ -> "\[Bullet]", _Integer -> ""};
  Grid[{{1}}~g~#, Frame -> All] &

Usage

13 windings:

Grid[{{1}}~g~#, Frame -> All] &[13]

Answer 3

MATLAB, 56 characters

based on @gnovice solution, improved by using MATLAB's spiral function :)

disp(char(isprime(flipud(spiral(2*input('')-1)))*10+32))

Test Cases:

>> disp(char(isprime(flipud(spiral(2*input('')-1)))*10+32))
2
* *
  *
*  
>> disp(char(isprime(flipud(spiral(2*input('')-1)))*10+32))
3
*   *
 * * 
*  **
 *   
  *  
>> disp(char(isprime(flipud(spiral(2*input('')-1)))*10+32))
5
    * *  
 *     * 
* *   *  
   * * * 
  *  ** *
 * *     
*   *    
 *   *   
*     *  

Answer 4

My first code golf!

Ruby, 309 301 283 271 265 characters

s=gets.to_i;d=s*2-1;a=Array.new(d){' '*d}
e=d**2;p='*'*e;2.upto(e){|i|2.upto(e/i){|j|p[i*j-1]=' '}};p[0]=' '
s.times{|i|k=s-i-1;l=2*i;m=l+1;o=l-1
m.times{|j|n=j+k;a[k][n]=p[l**2-j];a[n][k]=p[l**2+j];a[k+l][n]=p[m**2-m+j]}
l.times{|j|a[j+k][k+l]=p[o**2+o-j]}}
puts a

Answer 5

Haskell - 224 characters

(%)=zipWith(++)
r x=[x]
l 1=r[1]
l n=r[a,a-1..b]++(m r[a+1..]%l s%m r[b-1,b-2..])++r[d-s*2..d]where{d=(n*2-1)^2;b=d-s*6;a=d-s*4;s=n-1}
p[_]='*'
p _=' '
i n=p[x|x<-[2..n],n`mod`x==0]
m=map
main=interact$unlines.m(m i).l.read

i'm not the best at haskell so there is probably some more shrinkage that can occur here

output from echo 6 | runghc ulam.hs

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

---

this is a different algorithm (similar to @drhirsch's) unfortunately i cannot seem to get it below 239 characters

p[_]='*'
p _=' '
main=interact$unlines.u.read
i n=p[x|x<-[2..n],n`mod`x==0]
u(n+1)=map(map(i.f.o).zip[-n..n].replicate((n+1)*2-1))[n,n-1..(-n)]
f(x,y,z)=4*y*y-x-y+1+if z then 2*(x-y)else 0
o(u,v)=if(v> -u)==(v<u)then(v,u,v<u)else(u,v,v<u)

Answer 6

Ruby 1.8.7, 194 chars

n=2*gets.to_i-1
r=n**2
l,c=[nil]*r,r/2
r.times{|i|l[c]=i+1;c=i==0||l[c-n]&&!l[c+1]?c+1:l[c-1]&&!l[c-n]?c-n:l[c+n]?c-1:c+n}
r.times{|i|print"1"*l[i]!~/^1?$|^(11+?)\1+$/?'*':' ',i%n==n-1?"\n":''}

For some reason, ruby1.9 wants another space on line 4:

r.times{|i|l[c]=i+1;c=i==0||l[c-n]&&!l[c+1]?c+1:l[c-1]&&!l[c-n]?c-n :l[c+n]?c-1:c+n}