Code Golf：绘制 ASCII 艺术明星

Question

由于本周没有人发布代码高尔夫挑战，我将尝试一下。我这样做是为了让你可以在漫长的编译周期中做一些除了玩剑之外的事情。

挑战：

绘制 ASCII 艺术星星，给定标准输入上的三个数字（尖峰数量、星星类型（星星是通过连接相隔 n 个顶点的顶点绘制的）以及星星的直径）。示例：

Input:                        Input:                       Input:

5 2 20                        7 2 20                       7 3 20

Output:                       Output:                      Output:

             x                        x                        x
            xx                        xx                       x
            xx                       x  xx   xx                xx      x
           x x                       x   xxxx x                xx     xx
  xx      x  x                      xxxxx  x  x                x x  xxx
   xxxx   x  x                   xxxx       x x                x x x  x
   x   xxx   x                 xx  x         xx         xxx    x  x  x
    x   x xxxx                 x   x          xx         x xxxxx xx  x
     x  x    xxx                x x           x x         xx   xxxxxx
     x x     x  xxx             x x           x  x          x xx   xxxxxx
      x      x     xx            x            x   x          x x   x     xxx
     x x     x  xxx              x            x  x          x xx   xxxxxx
     x x     xxx                x x           x x         xx   xxxxxx
    x   x xxxx                  x x           xx         x xxxxx xx  x
   x   xxx   x                 xx  x         xx         xxx    x  x  x
   xxxx   x  x                   xxxx       x x                x x x  x
  xx      x  x                      xxxxx  x  x                x x  xxx
           x x                      x    xxxx x                xx     xx
            xx                       x  xx   xx                xx      x
            xx                       x x                       x
             x                        x                        x

由于正确光栅化这些线条可能是代码高尔夫挑战的 PITA，因此我会留一些余地，但不会太多。更多示例：

足够好：

    x           x                   x       x
     xx       xx                    x       x
       x     x                       x     x
     x  xx xx  x                      x   x
          x                            x x
         x x                            x
      xxx   xxx                      x     x
      x       x               xxxxxxxxxxxxxxxxxxxxx
     x         x                xx    x   x    xx
   xx  x     x  xx                xx x     x xx
  x               x                 xxx   xxx
xxxxxxxxxxxxxxxxxxxxx                 xxxxx
        x   x                      x  xx xx  x
                                  x xx     xx x
                                 xxx         xxx
         x x                    xx     x x     xx


          x
                                        x

没问题：

            x                  xx               xx
            xx                   x             x
            xx                    x           x
           x x                     xx       xx
 xx       x  x                       x     x
  xxxx   x   x                     x  xx xx  x
  x   xxxx   x                          x
   x    xxxx x                         x x
    x  x    xxx                     xxx   xxx
     xx      x xxxxxx               x       x
     xx      x xxxxxx              x         x
     xx     xxx                  xx  x     x  xx
    x  x xxx x                  x               x
    x xxx    x                xxxxxxxxxxxxxxxxxxxxx
   xxx  x    x                        x   x
 xx      x   x
          x  x
          x  x                         x x
           x x
            xx
                                        x

 Lack of precission              Lack of clipping

玩得开心！

Answer 1

红宝石 — 323 276 304 297

a,b,c=gets.split.map &:to_i
o=(0..c).map{' '*c}
m=(1..a).map{|i|r=2*h=Math::PI*i/a
[(1+Math.sin(r))*c/2+0.5,(1+Math.cos(r))*c/2]}
a.times{|n|j,i,x,y=*m[n],*m[n-b]
j,x,i,y=i,y,j,x if k=((i-y)/(j-x))**2>1
s,z=[x,j].sort
s.to_i.upto(z){|m|t=i+(m-j)*(i-y)/(j-x)
k ?(o[t][m]='x'):(o[m][t]='x')}}
puts o

             x              xx                   x                      xx          
             x              x x                  x                      x x         
            xx              x  x    xx           xx       x            x   x        
           x x             x    xxxx x           xx      xx      xxxxxxxxxxxxxxxx   
 xxx      x  x             xxxxx xx  x           x x    xx       x   x       x  x   
  x xxx   x  x         xxxxx       x x    x      x x   x x       x  x         x x   
  x    xxx   x       xx   x         xx     xxxx  x  xxx x        x x           xx   
   x    x xxxx        x  x           x       x xxxxxx   x        xx             x   
    x  x     xxx      x  x           xxx      x  x xxxxx         x              xx  
     x x     x  xxx    xx            x  x      x xx  x xxxxx    xx              x x 
     xx      x     xx  xx            x   x      xx    x     xxxx x              x  x
     xx      x  xxx    xx            x  x      x xx   x xxxx    xx              x x 
     x x     xxx       xx            xxx      x  x xxxxx         x              xx  
    x  x  xxxx        x  x           x       x xxxxxxx x         xx             x   
   x   xxx   x       xx  x          xx     xxxx  x  xxx x        x x           xx   
  x xxx  x   x         xxxxx       x x    x      x  x  xx        x  x         x x   
 xxx      x  x            x xxxx xx  x           x x    xx       x   x       x  x   
          x  x             x    xxxx x           x x     x       xxxxxxxxxxxxxxxx   
           x x             x   x    xx           xx       x            x   x        
            xx              x x                  xx                     x x         
             x              xx                   x                       x          

Answer 2

C# : 555 428 个字符

using System.Drawing;class P{static void Main(string[]a){int p=int.Parse(a[0 ]),i=int.Parse(a[1]),l=int.Parse(a[2]),n;var o=System.Console.Out;var b=新位图(l*4/3, l*4/3);var g=Graphics.FromImage(b);g.TranslateTransform(l/8,l/3);for(n=0;n

0?"#":" ");}}

星星 5 2 20

            #
           # #
           # #
           # #
          #   #
          #   #
  #####################
   ##    #     #    ##
     #   #     #   #
      #  #     #  #
       ##       ##
        ##     ##
        # #   # #
       #   ###   #
       #   # #   #
       # ##   ## #
      # #       # #
      ##         ##
      #           #

Answer 3

OCaml，659 个字符 (wc -c)。

使用经典的 Bresenham 算法来绘制线条，为了好玩而递归。
另存为 stars.ml 并使用 echo 5 2 20 | 运行ocaml star.ml。

let pl b x y=b.(x).[y]<-'x'
let ln b x0 y0 x1 y1 =
let rec h p x0 y0 x1 y1 =
if x0>x1 then h p x1 y1 x0 y0
else(
let dx,dy,ys=x1-x0,abs(y1-y0),if y0<y1 then 1 else -1 in
let rec k x y e=
if x<=x1 then(
p y x;
if e<dy then k(x+1)(y+ys)(e-dy+dx)
else k(x+1)y(e-dy)
)in
k x0 y0(dx/2)
)in
if abs(y1-y0)>abs(x1-x0)then
h(fun x y->pl b y x)y0 x0 y1 x1
else h(pl b)x0 y0 x1 y1
let f=float
let g n s d=
let b=Array.init d(fun _->String.make d ' ')in
let r=f(d/2)-.0.1 in
let k h i=int_of_float(r+.r*.h(6.2831853*.(f(i mod n)/.(f n))))in
let x,y=k cos,k sin in
for i=0 to n do ln b(x i)(y i)(x(i+s))(y(i+s))done;Array.iter print_endline b
let()=Scanf.scanf"%d %d %d"g

输出

            x                  x                       x
           xx                  xxx                     xx
           xx                 x   x   xxx              xx       x
          x x                 x  xxxxx  x              x x     x
 xx      x  x              xxxxxx   xx  x              x x   xxx
  xxxxx x   x           xxx  x        x x       xxx    x  x x x
  x    xxxx x           x   x          xx        xxxxxxx  xx  x
   x   x   xxx           x  x           xx         x   xxxxx x
    x x     x xxxx       x x            x x         x  xx  xxxxxx
     x      x     xx      xx            x  x         xxx    x    xxx
     x      x  xxx        x             x x          xxx    x xxxx
    x x     xxx           xx            xx          x  xx xxxx
   x   x xxxx            x x           xx          x  xxxxxx x
  x   xxx   x            x  x         x x        xxxxx x  xx  x
  xxxx  x   x           xxx x        x  x       xx     x  x x x
 xx      x  x              xxxxxx   x   x              x x   xxx
          x x                x   xxxxx  x              x x     x
           xx                 x  xx   xxx              xx       x
           xx                 x x                      xx
            x                  x                       x

Answer 4

我第一次尝试编码高尔夫...

Python - 550 533 507 484 451 > 437 个字符

另存为 k.py。以 python k.py 5 2 20 运行

import sys,math as m
A=abs;R=range;X=round
p,t,w=map(int,sys.argv[1:])
G=[[' ']*-~w for i in R(w+1)]
def L(a,b,c,d):
 s=A(d-b)>A(c-a)
 if s:a,b,c,d=b,a,d,c
 if a>c:a,b,c,d=c,d,a,b
 f=c-a;e=f/2;y=int(X(b));x=int(X(a))
 while x<X(c):
  e-=A(d-b);p,q=((x,y),(y,x))[s];G[p][q]='x';x+=1
  if e<0:y+=(-1,1)[y<d];e+=f
r=w/2;k=m.pi*2/p
s=[(m.sin(k*i)*r+r,m.cos(k*i)*r+r)for i in R(p+t)]
for i in R(p):L(*(s[i]+s[i+t]))
for i in G:print''.join(i)

（通过用制表符替换第二级缩进，可以节省 2 个字符...）

此处提供扩展版本：http://gist.github.com/591485

示例输出：

             x                x                     x
            xx                xx                    x
            xx               x  xx    x             xx      x
           x x               x    xxxxx             xx     xx
  xx      x  x              x xxxx x  x             x x  xxx
   xxxx  x   x            xxxx      x x             x x x  x
   x   xxx   x         xxx x         xx      xxx    x  x  x
    x   x xxxx         x   x          xx      x xxxxx xx  x
     x x     xxx        x x           x x      xx   xxxxxx
      x      x  xxx     x x           x  x       x xx   xxxxx
      x      x     x     x            x           x x   xx   xxx
     x x     x  xxx      x            x  x       x xx   xxxxx
    x   x    xxx        x x           x x      xx   xxxx  x
    x    xxxxx          x x           xx      x xxxxx xx  x
   x   xxx   x         xxx x         xx      xxx    x xx   x
  x xxx   x  x            xxxx      x x             x x x  x
  xx       x x              x xxxx x  x             xx   xx x
           x x              x     xxxxx             xx     xx
            xx               x  xx                  x
             x               x x                    x
                              x

Answer 5

VBScript，584 511 482 字节

我还对线条使用了 Bresenham 算法。

g=Split(WScript.StdIn.ReadAll):t=g(1):d=g(2):ReDim r(d*d)
For i=0 To d*d:r(i)=" ":Next
c=1.57:u=(d-1)/2:b=c*4/g(0):For i=1 To g(0)
Z n:x=g:Z n+c:y=g:Z n-b*t:f=g:Z n-b*t+c:n=n+b
s=Abs(g-y)>Abs(f-x):If s Then W x,y:W f,g
If x>f Then W x,f:W y,g
p=f-x:q=Abs(g-y):h=p\2:v=(y>g)*2+1
For x=x To f:r((s+1)*(y*d+x)-s*(x*d+y))="x"
h=h-q:If h<0 Then y=y+v:h=h+p
Next:Next:For i=0 To d:WScript.Echo Mid(Join(r,""),i*d+1,d):Next
Sub W(a,b):e=a:a=b:b=e:End Sub
Sub Z(e):g=Int(u*Cos(e)+u):End Sub

输出：

            x               x                    x            
           xx               xx                   x            
           xx              x  xx  xxx            xx      x    
          x x              x  xxxx  x            xx     xx    
  xx     x  x            xxxxx   x  x            x x   xx     
   xxxx  x  x         xxx x       x x      xx    x x xx x     
   x   xxx  x         x  x         xx       xxxxxx  x  x      
    x  x  xxx          x x          xx       x   xxxx  x      
     x x    xxx        x x          x x       xx xx  xxx      
     xx     x  xxx      x           x  x        xx   xx xxxx  
     xx     x     xx    x           x   x       xx   xx    xxx
     x x    x xxxx     x x          x  x      xx xx  xxxxxx   
    x  x   xxx         x x          xxx      x   xxxxxxx      
   x    xxx x         x  x          x       x xxxx  x  x      
   xxxxx x  x         xxx x        xx      xxx   x x xx x     
  xx     x  x            xxxxx   xx x            x x   xx     
          x x              x  xxxx  x            xx     xx    
           xx              x   x  xxx            xx      x    
           xx               xxx                  x            
            x               x                    x            

扩展代码：

Dim PI, args, output, i
PI = 4 * Atn(1)

args = Split(WScript.StdIn.ReadAll, " ")
output = Join(Star(args(0), args(1), args(2)), vbNullString)

For i = 1 To Len(output) Step args(2)
    WScript.Echo Mid(output, i, args(2))
Next

Function Star(spikes, star_type, diameter)
    Dim result(), i, vertexes(), angle, radius, p1, p2
    ReDim result(diameter * diameter - 1)
    ReDim vertexes(spikes - 1)

    For i = 0 To UBound(result)
        result(i) = " "
    Next

    radius = (diameter - 1) / 2
    For i = 0 To UBound(vertexes)
        vertexes(i) = Array(CLng(radius * Cos(angle) + radius), _
            CLng(radius * Sin(angle) + radius))
        angle = angle - (2 * PI / spikes)
    Next

    For i = 0 To UBound(vertexes)
        p1 = vertexes(i)
        p2 = vertexes((i + star_type) Mod spikes)
        DrawLine p1(0), p2(0), p1(1), p2(1), result, diameter
    Next

    Star = result
End Function

Sub DrawLine(ByVal x0, ByVal x1, ByVal y0, ByVal y1, arr, diameter)
    Dim steep, deltax, deltay, error, ystep

    steep = Abs(y1 - y0) > Abs(x1 - x0)
    If steep Then
        Swap x0, y0
        Swap x1, y1
    End If
    If x0 > x1 Then
        Swap x0, x1
        Swap y0, y1
    End If

    deltax = x1 - x0
    deltay = Abs(y1 - y0)
    error = deltax \ 2
    If y0 < y1 Then ystep = 1 Else ystep = -1

    For x0 = x0 To x1
        If steep Then
            arr(x0 * diameter + y0) = "x"
        Else
            arr(y0 * diameter + x0) = "x"
        End If
        error = error - deltay
        If error < 0 Then
            y0 = y0 + ystep
            error = error + deltax
        End If
    Next
End Sub

Function Swap(a, b)
    Dim temp
    temp = a
    a = b
    b = temp
End Function

Answer 6

Javascript（基于Isc的python解决方案），598 591 586 559字节（与rhino shell一起使用：另存为'stars.js'，运行与“rhinostars.js 7 2 20”）：

for(var a=Math,b=parseInt,c=a.floor,d=a.abs,e=arguments,f=b(e[0]),g=b(e[1]),h=b(e[2]),i=[],j=0;j<(h+1)*h;j++)i.push((j+1)%(h+1)?" ":"\n");function k(w,x){i[x*(h+1)+w]="x"}var l=(h-1)/2,m=[];for(j=0;j<f;j++)m.push([c(l*a.cos(a.PI*2/f*j)+h/2),c(l*a.sin(a.PI*2/f*j)+h/2)]);
for(j=0;j<f;j++){var n=m[j][0],o=m[j][1],p=m[(j+g)%f][0],q=m[(j+g)%f][1],r=void 0,s=void 0;if(r=d(q-o)>d(p-n)){s=n;n=o;o=s;s=p;p=q;q=s}if(n>p){s=n;n=p;p=s;s=o;o=q;q=s}for(var t=p-n,u=0,v=d(q-o)/t,y=o<q?1:-1,z=o,A=n;A<=p;A++){r?k(z,A):k(A,z);u+=v;if(u>=0.5){z+=y;u-=1}}}print(i.join(""));

输出：

           x       
           xx       
           xx       
          x x       
  xx     x  x       
   xxxx  x  x       
   x   xxx  x       
    x  x  xxx       
     x x    xxx     
     xx     x  xxx  
     xx     x     xx
     x x    x xxxx  
    x  x   xxx      
   x    xxx x       
   xxxxx x  x       
  xx     x  x       
          x x       
           xx       
           xx       
            x       

       x            
       xx           
      x  x   xxx    
      x  xxxx  x    
    xxxxx   x  x    
 xxx x       x x    
 x  x         xx    
  x x          x    
  x x          xxx  
   x           x  x 
   x           x   x
  x x          x xx 
  x x          xx   
 x  x          x    
 xxx x       xxx    
    xxxxx   x  x                                                                                                                                              
      x  xxxx  x                                                                                                                                              
      x  xx  xxx                                                                                                                                              
       xx                                                                                                                                                     
       x            

       x            
       x            
       xx      x    
       xx     xx    
       x x   xx     
 xx    x x xx x     
  xxxxxx  x  x      
   x   xxxx  x      
    xx xx xxxx      
      xx   xx xxxx  
      xx   xx    xxx
    xx xx  xxxxxx   
   x   xxxxxxx      
  x xxxx  x  x      
 xxx   x x xx x     
       x x   xx     
       xx     xx    
       xx      x    
       x            
       x            

扩展代码：

var p = parseInt(arguments[0]),
    t = parseInt(arguments[1]),
    w = parseInt(arguments[2]);

var g = [];
for(var i = 0; i < (w + 1) * w; i++)
        g.push((i + 1) % (w + 1) ? ' ' : '\n');

function plot(x, y) { g[y * (w + 1) + x] = 'x';}
function line(x0, y0, x1, y1) {
        var steep = Math.abs(y1 - y0) > Math.abs(x1 - x0), temp;
        if(steep) {
                temp = x0; x0 = y0; y0 = temp;
                temp = x1; x1 = y1; y1 = temp;
        }
        if(x0 > x1) {
                temp = x0; x0 = x1; x1 = temp;
                temp = y0; y0 = y1; y1 = temp;
        }
        var deltax = x1 - x0;
        var deltay = Math.abs(y1 - y0);
        var error = 0;
        var deltaerr = deltay / deltax;
        var ystep
        var y = y0;
        if(y0 < y1) ystep = 1; else ystep = -1;
        for(var x = x0; x <= x1; x++) {
                if(steep) plot(y,x); else plot(x,y);
                error = error + deltaerr;
                if(error >= 0.5) {
                        y = y + ystep;
                        error = error - 1.0;
                }
        }
}

var r = (w-1)/2, points = [];
for(var i = 0; i < p; i++) {
    points.push([
        Math.floor(r * Math.cos(Math.PI * 2 / p * i) + w/2),
        Math.floor(r * Math.sin(Math.PI * 2 / p * i) + w/2)
    ]);
}

for(var i = 0; i < p; i++) {
    line(points[i][0], points[i][1], points[((i + t) % p)][0], points[((i + t) % p)][1]);
}

print(g.join(''));