Code Golf: Triforce

问题

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This is inspired by/taken from this thread: http://www.allegro.cc/forums/thread/603383

The Problem

Assume the user gives you a numeric input ranging from 1 to 7. Input should be taken from the console, arguments are less desirable.

When the input is 1, print the following:

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

Values greater than one should generate multiples of the pattern, ending with the one above, but stacked symmetrically. For example, 3 should print the following:

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

Bonus points if you print the reverse as well.

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

Can we try and keep it to one answer per language, that we all improve on?

回答1:

Assembler, 165 bytes assembled

Build Instructions

1. Download A86 from here
2. Add a reference to the A86 executable into your DOS search path
3. Paste the code below into a text file (example: triforce.asm)
4. Invoke the assembler: a86 triforce.asm
5. This will create a .COM file called triforce.com
6. Type triforce to run

This was developed using the standard WinXP DOS box (Start->Programs->Accessories->Command Prompt). It should work with other DOS emulators.

Assemble using A86 and requires WinXP DOS box to run the .COM file it produces. Press 'q' to exit, keys 1-7 to draw the output.

  l20:mov ah,7
      int 21h
      cmp al,'q'
      je ret
      sub al,'0'
      cmp al,1
      jb l20
      cmp al,7
      ja l20
      mov [l0-1],al
      mov byte ptr [l7+2],6
      jmp $+2
      mov ah,2
      mov ch,0
      mov bh,3
   l0:mov bl,1
   l1:mov dh,0
   l3:cmp dh,ch
      je l2
      mov dl,32
      int 21h
      inc dh
      jmp l3
      ret
   l2:mov dh,bh
   l6:mov cl,12
   l5:mov dl,42
      cmp cl,bl
      ja l4
      mov dl,32
      cmp dh,1
      je l21
   l4:int 21h
      dec cl
      jnz l5
  l21:dec dh
      jnz l6
      mov dl,10
      int 21h
      mov dl,13
      int 21h
  l10:inc ch
   l9:add bl,2
   l7:cmp ch,6
      jne l1
  l13:add byte ptr [l7+2],6
  l11:dec bh
  l12:cmp bh,0
      jne l0
      xor byte ptr [l0+1],10
      xor byte ptr [l9+1],40
      xor byte ptr [l10+1],8
      xor byte ptr [l13+1],40
      sub byte ptr [l7+2],12
      mov dh,[l0-1]
      inc dh
      xor [l12+2],dh
      xor byte ptr [l11+1],8
      xor byte ptr [l1+1],1
      inc bh
      cmp byte ptr [l0+1],11
      je l0
      jmp l20

It uses lots of self-modifying code to do the triforce and its mirror, it even modifies the self-modifying code.

回答2:

GolfScript - 43 chars

~:!6*,{:^' '
 *'*'12*' '
  ^6%.+)*+
   -12>!^
    6/-*
     n}
     /

~:!6*,{:^' '*'*'12*' '^6%.+)*+-12>!^6/-*n}/

48 Chars for the bonus

~:!6*,.-1%+{
 :^' '*'*'12
  *' '^6%.+
   )*+-12>
    !^6/-
     *n}
      /

~:!6*,.-1%+{:^' '*'*'12*' '^6%.+)*+-12>!^6/-*n}/

回答3:

Python - 77 Chars

n=input()
for k in range(6*n):print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6)

n=input()
for k in range(6*n):j=1+k%6*2;print' '*k+('*'*(12-j)+' '*j)*(n-k/6)

89 Chars for the bonus

n=input();R=range(6*n)
for k in R+R[::-1]:print' '*k+('*'*11+' '*11)[k%6*2:][:12]*(n-k/6)

114 Chars Version just using string replacements

u,v=' *';s=(v*11+u)*input()
while s.strip():print s;s=u+s.replace(*((v*2+u,u*3),(v*1+u*10,v*11))[' * 'in s])[:-2]

---

Unk Chars all in one statement, should work w/ 2.x and 3.x. The enumerate() is to allow the single input() to work for both places you need to use it.

print ('\n'.join('\n'.join(((' '*(6*n))+' '.join(('%s%s%s'%(' '*(5-x),'*'*(2*x+1),' '*(5-x)) for m in range(i + 1)))) for x in range(5,-1,-1)) for n, i in enumerate(range(int(input())-1,-1,-1))))

Yet Another Method

def f(n): print '\n'.join(' '*6*(n-r)+(' '*(5-l)+'*'*(l*2+1)+' '*(5-l)+' ')*r for r in xrange(1, n+1) for l in xrange(6))
f(input())

回答4:

Ruby - 74 Chars

(6*n=gets.to_i).times{|k|puts' '*k+('*'*(11-(j=k%6*2))+' '*(j+1))*(n-k/6)}

回答5:

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.

回答6:

sed, 117 chars

s/$/76543210/
s/(.).*\1//
s/./*********** /gp
:
s/\*(\**)\*/ \1 /gp
t
:c
s/\* {11}\*/ ************/
tc
s/\*  /   /p
t

Usage: $ echo 7 | sed -rf this.sed

First attempt; improvements could probably be made...

回答7:

Ruby 1.9 - 84 characters :

v=gets.to_i
v.times{|x|6.times{|i|puts' '*6*x+(' '*i+'*'*(11-2*i)+' '*i+' ')*(v-x)}}

回答8:

Perl - 72 chars

die map$"x$_.("*"x(12-($l=1+$_%6*2)).$"x$l)x($n-int$_/6).$/,0..6*($n=<>)

78 chars

map{$l=$_%6*2;print$"x$_,("*"x(11-$l).$"x$l.$")x($n-int$_/6),$/}0..6*($n=<>)-1

87 chars

$n=<>;map{$i=int$_/6;$l=$_%6*2;print$"x$_,("*"x(11-$l).$"x$l.$")x($n-$i),$/}(0..6*$n-1)

97 chars

$n=<>;map{$i=int$_/6;$l=$_%6;print$"x(6*$i),($"x$l."*"x(11-2*$l).$"x$l.$")x($n-$i),$/}(0..6*$n-1)

108 chars

$n=<>;map{$i=int$_/6;$l=$_%6;print ""." "x(6*$i),(" "x$l."*"x(11-2*$l)." "x$l." ")x($n-$i),"\n";}(0..6*$n-1)

回答9:

Powershell, 78 characters

0..(6*($n=read-host)-1)|%{" "*$_+("*"*(12-($k=1+$_%6*2))+" "*$k)*(.4+$n-$_/6)}

Bonus, 92 characters

$a=0..(6*($n=read-host)-1)|%{" "*$_+("*"*(12-($k=1+$_%6*2))+" "*$k)*(.4+$n-$_/6)}
$a
$a|sort

The output is stored in an array of strings, $a, and the reverse is created by sorting the array. We could, of course, just reverse the array, but it would be more characters to type :)

回答10:

Haskell - 131 138 142 143 Chars

(⊗)=replicate
z o=[concat$(6*n+m)⊗' ':(o-n)⊗((11-m-m)⊗'*'++(1+m+m)⊗' ')|n<-[0..o-1],m<-[0..5]]
main=getLine>>=mapM_ putStrLn.z.read

---

This one is longer (146 148 chars) at present, but an interesting, alternate line of attack:

(⊗)=replicate
a↑b|a>b=' ';_↑_='*'
z o=[map(k↑)$concat$(6*n)⊗' ':(o-n)⊗"abcdefedcba "|n<-[0..o-1],k<-"abcdef"]
main=getLine>>=mapM_ putStrLn.z.read

回答11:

FORTRAN - 97 Chars

Got rid of the #define and saved 8 bytes thanks to implict loops!

$ f95 triforce.f95  -o triforce && echo 7 | ./triforce

READ*,N
DO K=0,N*6
M=2*MOD(K,6)
PRINT*,(' ',I=1,K),(('*',I=M,10),(' ',I=0,M),J=K/6+1,N)
ENDDO
END

125 bytes for the bonus

READ*,N
DO L=1,N*12
K=L+5
If(L>N*6)K=N*12-L+6
M=2*MOD(K,6)
PRINT"(99A)",(32,I=7,K),((42,I=M,10),(32,I=0,M),J=K/6,N)
ENDDO
END

FORTRAN - 108 Chars

#define R REPEAT
READ*,N
DO I=0,6*N
J=MOD(I,6)*2
PRINT*,R(' ',I)//R(R('*',11-J)//R(' ',J+1),N-I/6)
ENDDO
END

回答12:

JavaScript 1.8 - SpiderMonkey - 118 chars

N=readline()
function f(n,c)n>0?(c||' ')+f(n-1,c):''
for(i=0;i<N*6;i++)print(f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1)))

w/ bonus - 151 chars

N=readline()
function f(n,c)n>0?(c||' ')+f(n-1,c):''
function l(i)print(f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1)))
for(i=0;i<N*6;i++)l(i)
for(;i--;)l(i)

Usage: js thisfile.js

JavaScript - In Browser - 154 characters

N=prompt()
function f(n,c){return n>0?(c||' ')+f(n-1,c):''}
s='<pre>'
for(i=0;i<N*6;i++)s+=f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1))+'\n'
document.write(s)

The non-obfuscated version (before optimizations by gnarf):

var N = prompt();
var S = ' ';

function fill(c, n) {
    for (ret=''; n--;)
        ret += c;
    return ret;
}

var str = '<pre>';

for (i=0; i<N*6; i++) {
    str += fill(S, i);
    for (j=0; j<N-i/6; j++)
        str += fill('*', 11-i%6*2) + fill(S, i%6*2+1);
    str += '\n';
}

document.write(str);

Here's a different algorithm that uses replace() to go from one line to the next of each line of a triangle row:

161 characters

N=readline()
function f(n,c){return n>0?(c||' ')+f(n-1,c):''}l=0
for(i=N;i>0;){r=f(i--,f(11,'*')+' ');for(j=6;j--;){print(f(l++)+r)
r=r.replace(/\*\* /g,'   ')}}

回答13:

F#, 184 181 167 151 147 143 142 133 chars

let N,r=int(stdin.ReadLine()),String.replicate
for l in[0..N*6-1]do printfn"%s%s"(r l" ")(r(N-l/6)((r(11-l%6*2)"*")+(r(l%6*2+1)" ")))

Bonus, 215 212 198 166 162 158 157 148 chars

let N,r=int(stdin.ReadLine()),String.replicate
for l in[0..N*6-1]@[N*6-1..-1..0]do printfn"%s%s"(r l" ")(r(N-l/6)((r(11-l%6*2)"*")+(r(l%6*2+1)" ")))

回答14:

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))));}

回答15:

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!

回答16:

Written in C

Bonus points (492 chars):

p(char *t, int c, int s){int i=0;for(;i<s;i++)printf("      ");for(i=0;i<c;i++)printf("%s",t);printf("\n");}main(int a, char **v){int i=0;int k;int c=atoi(v[1]);for(;i<c;i++){p("*********** ",c-i,i);p(" *********  ",c-i,i);p("  *******   ",c-i,i);p("   *****    ",c-i,i);p("    ***     ",c-i,i);p("     *      ",c-i,i);}for(i=0;i<c;i++){k=c-i-1;p("     *      ",1+i,k);p("    ***     ",1+i,k);p("   *****    ",1+i,k);p("  *******   ",1+i,k);p(" *********  ",1+i,k);p("*********** ",i+1,k);}}

Without bonus points (322 chars):

p(char *t, int c, int s){int i=0;for(;i<s;i++)printf("      ");for(i=0;i<c;i++)printf("%s",t);printf("\n");}main(int a, char **v){int i=0;int k;int c=atoi(v[1]);for(;i<c;i++){p("*********** ",c-i,i);p(" *********  ",c-i,i);p("  *******   ",c-i,i);p("   *****    ",c-i,i);p("    ***     ",c-i,i);p("     *      ",c-i,i);}}

First time posting, too!

回答17:

Lua, 121 chars

R,N,S=string.rep,io.read'*n',' 'for i=0,N-1 do for j=0,5 do X=R(S,j)print(R(S,6*i)..R(X..R('*',11-2*j)..X..S,N-i))end end

123

R,N,S=string.rep,io.read'*n',' 'for i=0,N-1 do for j=0,5 do print(R(S,6*i)..R(R(S,j)..R('*',11-2*j)..R(S,j)..S,N-i))end end

回答18:

PHP, 153

<?php $i=fgets(STDIN);function r($n,$c=' '){return$n>0?$c.r($n-1,$c):'';}for($l=0;$l<$i*6;){$z=$l%6*2;echo r($l).r($i-$l++/6,r(11-$z,'*').r($z+1))."\n";}

with Bonus, 210

<?php $i=fgets(STDIN);function r($n,$c=' '){return$n>0?$c.r($n-1,$c):'';}$o=array();for($l=0;$l<$i*6;){$z=$l%6*2;$o[]=r($l).r($i-$l++/6,r(11-$z,'*').r($z+1));}print join("\n",array_merge($o,array_reverse($o)));

回答19:

dc 105 chars

123 129 132 139 141

[rdPr1-d0<P]sP?sn
0sk[1lk6%2*+sj32lkd0<Plnlk6/-si
[[*]12lj-d0<P32ljd0<Pli1-dsi0<I]dsIx
10Plk1+dskln6*>K]dsKx

回答20:

Mathematica, 46 characters

The answer prints sideways.

TableForm@{Table["*",{l,#},{l},{j,6},{2j-1}]}&

回答21:

HyperTalk - 272 chars

function triforce n
    put"******" into a
    put n*6 into h
    repeat with y=0 to h-1
        put"   " after s
        put char 1 to y of s after t
        repeat n-y div 6
            get y mod 6*2
            put char 1 to 11-it of (a&a)&&char 1 to it of s after t
        end repeat
        put return after t
    end repeat
    return t
end triforce

Indentation is neither needed nor counted (HyperCard automatically adds it).

Miscellanea:

Since there is no notion of console or way to access console arguments in HyperCard 2.2 (that I know of), a function is given instead. It can be invoked with:

on mouseUp
    ask "Triforce: "
    put triforce(it) into card field 1
end mouseUp

To use this, a card field would be created and set to a fixed-width font. Using HyperCard's answer command would display a dialog with the text, but it doesn't work because:

• The answer dialog font (Chicago) is not fixed-width.
• The answer command refuses to display long text (even triforce(2) is too long).

回答22:

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))))

回答23:

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

来源：https://stackoverflow.com/questions/2406780/code-golf-triforce