function sierpinski(A, B, C, n)
if n == 0
    fill ([A(1), B(1), C(1)], [A(2), B(2), C(2)], [0.0 0.0 0.0]);
    hold on
else
   sierpinski(A, (A + B)/2, (A + C)/2, n-1)
   sierpinski(B, (B + A)/2, (B + C)/2, n-1)
   sierpinski(C, (C + A)/2, (C + B)/2, n-1)
end
%sierpinski([0 0], [1 0], [.5 .8], 8)

sierpinski[{a_, b_, c_}] := With[{ab = (a + b)/2, bc = (b + c)/2, ca = (a + c)/2},
  {{a, ab, ca}, {ab, b, bc}, {ca, bc, c}}];
pts = {{0, 0}, {1, 0}, {.5, .8}};
d = Nest[Join @@ sierpinski /@ # &, {pts}, 8];
Graphics[{EdgeForm@Black, Polygon@d}]

3楼: Originally posted by Simewe at 2013-09-11 18:47:10
这样是快了点，但我想知道怎么把递归改成循环呢？matlab里的递归据说比python更慢，
正好我也有python版的代码（使用循环）

from matplotlib import collections as coll, pyplot as plt
import numpy as np
...

tic;
% 原始三角形顶点,保存在一个n*3*2的数组,第三维共2层,第一层是x,第二层是y
pts(:,:,1) = [0,1,0.5];
pts(:,:,2) = [0,0,0.8];
% 层数
layer = 8;
% 当前层的三角形顶点
cur_pts = pts;
while layer>0
    next_pts = []; % 利用当前层顶点计算出来下一层的三角形顶点
    for num=1:size(cur_pts,1)
        % 利用三顶点计算分形三角形顶点坐标
        a = cur_pts(num,1,:);
        b = cur_pts(num,2,:);
        c = cur_pts(num,3,:);
        next_pts(end+1:end+3,:,:) = [a,(a+c)/2,(a+b)/2;
            (a+b)/2,b,(b+c)/2;
            (a+c)/2,(b+c)/2,c];
    end
    % 迭代当前层顶点
    cur_pts = next_pts;
    % 将顶点加入结果数组
    pts(end+1:end+size(cur_pts,1),:,:) = cur_pts;
    % 层数递减
    layer = layer-1;
end
% 作图
patch(pts(:,:,1)',pts(:,:,2)','w');
toc;

function sierpinski(A, B, C, n)
if n == 0
    fill ([A(1), B(1), C(1)], [A(2), B(2), C(2)], [0.0 0.0 0.0]);
else
   sierpinski(A, (A + B)/2, (A + C)/2, n-1)
   sierpinski(B, (B + A)/2, (B + C)/2, n-1)
   sierpinski(C, (C + A)/2, (C + B)/2, n-1)
end

%figure;hold on;
%sierpinski([0 0], [1 0], [.5 .8], 8)

2楼: Originally posted by libralibra at 2013-09-11 17:18:13
递归可能就这样,你把hold on放在递归函数外面的话,时间会减少一半左右.
原代码需要3.7s在我的机器上运行,hold on提出来后1.8s左右,这个递归的方法挺好的了.三角形都是完整的,如果用chaos game的办法,画出来只是远看 ...

from matplotlib import collections as coll, pyplot as plt
import numpy as np
from time import time
st=time()

def sierpinski(a,b,c):
    return [[a, (a + c)/2, (a + b)/2], [(a + b)/2, b, (b + c)/2], [(a + c)/2, (b + c)/2, c]]

pts=np.array([[[0.,0.], [1.,0.], [.5,.8]]])
for n in range(8):
    pts=[y for x in [sierpinski(*i) for i in pts] for y in x]

print np.shape(pts)
plt.gca().add_collection(coll.PolyCollection( pts ))
plt.axis('scaled')
print time()-st
plt.show()

4楼: Originally posted by libralibra at 2013-09-11 23:28:12
matlab也用循环的话,快的可怕,0.12s左右
tic;
% 原始三角形顶点,保存在一个n*3*2的数组,第三维共2层,第一层是x,第二层是y
pts(:,:,1) = ;
pts(:,:,2) = ;
% 层数
layer = 8;
% 当前层的三角形顶点
cur_pts ...

function r = sierpinski(a, b, c, n)
    if n == 0
        r = [a; b; c];
    else
        r=[sierpinski(a, (a+b)/2, (a+c)/2, n-1);
           sierpinski((b+a)/2, b, (b+c)/2, n-1);
           sierpinski((c+a)/2, (c+b)/2, c, n-1)];
    end
end

%{
tic;
n=8;
V = sierpinski([0,0],[1,0],[0.5,0.8],n);
F = reshape(1:length(V),3,[])';
patch('Vertices',V,'Faces', F);
toc;
%}