Code Golf: Triforce

Answer 1

77 char alternative python solution based on gnibbler's:

n=input()
k=0
exec"print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6);k+=1;"*6*n

Amazingly the bonus came out exactly the same also (101 chars, oh well)

n=input()
l=1
k=0
s="print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6);k+=l;"*6*n
exec s+'l=-1;k-=1;'+s

Answer 2

Common Lisp, 150 characters:

(defun f(n o)(unless(= n 0)(dotimes(x 6)(format t"~v@{~a~:*~}~-1:*~v@{~?~2:*~}~%"
 o" "n"~11@: "(list(- 11(* 2 x))#\*)))(f(1- n)(+ 6 o))))

Answer 3

COBOL - 385 Chars

$ cobc -free -x triforce.cob && echo 7| ./triforce

PROGRAM-ID.P.DATA DIVISION.WORKING-STORAGE SECTION.
1 N PIC 9.
1 M PIC 99.
1 value '0100***********'.
2 I PIC 99.
2 K PIC 99.
2 V PIC X(22). 
2 W PIC X(99).
PROCEDURE DIVISION.ACCEPT N
COMPUTE M=N*6
PERFORM M TIMES
DISPLAY W(1:K)NO ADVANCING
PERFORM N TIMES
DISPLAY V(I:12)NO ADVANCING
END-PERFORM
DISPLAY ''
ADD 2 TO I
IF I = 13 MOVE 1 TO I ADD -1 TO N END-IF
ADD 1 TO K
END-PERFORM.

K could be returned to outside the group level. An initial value of zero for a numeric with no VALUE clause is compiler-implementation dependent, as is an initial value of space for an alpha-numeric field (W has been cured of this, at no extra character cost). Moving K back would save two characters. -free is compiler-dependant as well, so I'm probably being over-picky.

Answer 4

C - 120 Chars

main(w,i,x,y){w=getchar()%8*12;for(i=0;i<w*w/2;)y=i/w,x=i++%w,putchar(x>w-2?10:x<y|w-x-1<y|(x-y)%12>=11-2*(y%6)?32:42);}

Note that this solution prints some trailing spaces (which is okay, right?). It also relies on relational operators having higher precedence than bitwise OR, saving two characters.

124 Chars

main(n,i,k){n=getchar()&7;for(k=0;k<6*n;k++,putchar(10))for(i=-k-1;++i<12*n-2*k-1;putchar(32+10*(i>=0&&(11-i%12>2*k%12))));}

Answer 5

C - 177 183 Chars

#define P(I,C)for(m=0;m<I;m++)putchar(C)
main(t,c,r,o,m){scanf("%d",&t);for(c=t;c>0;c--)for(r=6;r>0;r--){P((t-c)*6+6-r,32);for(o=0;o<c;o++){P(r*2-1,42);P(13-r*2,32);}puts("");}}

C - 222 243 Chars (With Bonus Points)

#define P(I,C)for(m=0;m<I;m++)putchar(C)
main(t,c,r,o,m){scanf("%d",&t);for(c=t-1;-c<2+t;c-=1+!c)for(r=c<0?1:6;c<0?r<7:r>0;r+=c<0?1:-1){P((t-abs(c+1))*6+6-r,32);for(o=0;o<abs(c+1);o++){P(r*2-1,42);P(13-r*2,32);}puts("");}}

This is my first Code Golf submission as well!