MATLAB | 艺术就是画圈圈

什么是艺术？？艺术就是画圈圈！！

这两天刷到了Hamid Naderi Yeganeh大佬的一系列线条艺术，感觉非常惊艳，顺手就用MATLAB实现了一下。大佬的大部分作品都在以下网站，大家有兴趣可以去瞅瞅去试试实现其他的作品：

https://www.ams.org/publicoutreach/math-imagery/yeganeh

飞鸟一

This image shows 9,830 circles. For $k=1,2,3,\dots ,9830$ , the center of the circle is $(X(k),Y(k))$ and the radius of the k-th circle is $R(k)$ , where:
$$\begin{array}{rl}& X(k)={\left(\sin \left(\frac{\pi k}{20000}\right)\right)}^{12}\left(\frac{1}{2}{\left(\cos \left(\frac{31\pi k}{1000}\right)\right)}^{16}\sin \left(\frac{6\pi k}{10000}\right)+\frac{1}{6}{\left(\sin \left(\frac{31\pi k}{1000}\right)\right)}^{20}\right)+\frac{3k}{20000}+{\left(\cos \left(\frac{31\pi k}{10000}\right)\right)}^6\sin \left(\frac{\pi }{2}{\left(\frac{k-10000}{10000}\right)}^7-\frac{\pi }{5}\right)\\ & Y(k)=\frac{-9}{4}{\left(\cos \left(\frac{31\pi k}{10000}\right)\right)}^6\cos \left(\frac{\pi }{2}{\left(\frac{k-10000}{10000}\right)}^7-\frac{\pi }{5}\right)\left(\frac{2}{3}+{\left(\sin \left(\frac{\pi k}{20000}\right)\sin \left(\frac{3\pi k}{2000}\right)\right)}^6\right)+\frac{3}{4}{\left(\cos \left(3\pi \frac{k-10000}{10000}\right)\right)}^{10}{\left(\cos \left(9\pi \frac{k-10000}{10000}\right)\right)}^{10}{\left(\cos \left(36\pi \frac{k-10000}{100000}\right)\right)}^{14}+\frac{7}{10}{\left(\frac{k-10000}{10000}\right)}^2\\ & R(k)={\left(\sin \left(\frac{\pi k}{20000}\right)\right)}^{10}\left(\frac{1}{4}{\left(\cos \left(\frac{31\pi k}{10000}+\frac{25\pi }{32}\right)\right)}^{20}+\frac{1}{20}{\left(\cos \left(\frac{31\pi k}{10000}\right)\right)}^2\right)+\frac{1}{30}\left(\frac{3}{2}-{\left(\cos \left(\frac{62\pi k}{10000}\right)\right)}^2\right)\end{array}$$

完整代码

% bird 1
K=1:9830;
t=linspace(0,2*pi,200);

X=@(k) (sin(pi.*k./2e4)).^12.*(cos(31.*pi.*k./1e4).^16.*sin(6.*pi.*k./1e4)./2+sin(31.*pi.*k./1e4).^20./6)...
        +3.*k./2e4+cos(31.*pi.*k./1e4).^6.*sin(pi./2.*(k./1e4-1).^7-pi./5);
Y=@(k) -9./4.*cos(31.*pi.*k./1e4).^6.*cos(pi./2.*(k./1e4-1).^7-pi./5).*(2./3+(sin(pi.*k./2e4).*sin(3.*pi.*k./2e4)).^6)...
        +3./4.*cos(3.*pi.*(k-1e4)./1e5).^10.*cos(9.*pi.*(k-1e4)./1e5).^10.*cos(36.*pi.*(k-1e4)./1e5).^14+7./10.*((k-1e4)./1e4).^2;
R=@(k) sin(pi.*k./2e4).^10.*(1./4.*cos(31.*pi.*k./1e4+25.*pi./32).^20+1./20.*cos(31.*pi.*k./1e4).^2)+1./30.*(3./2-cos(62.*pi.*k./1e4).^2);

CX=[X(K')+cos(t).*R(K'),K'.*nan]';
CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
plot(CX(:),CY(:),'Color',[0,0,0,.2]);
set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');

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飞鸟二

This image shows all circles of form $(x-A(k){)}^2+(y-B(k){)}^2=R(k{)}^2$ for $k=-20000,-19999,\dots ,20000$, where:

$$\begin{array}{rl}& A(k)=\frac{3k}{45000}+\sin \left(\frac{17\pi }{20}{\left(\frac{k}{20000}\right)}^5\right){\cos }^6\left(\frac{41\pi k}{20000}\right)+\left(\frac{1}{3}{\cos }^{16}\left(\frac{41\pi k}{20000}\right)+\frac{1}{3}{\cos }^{80}\left(\frac{41\pi k}{20000}\right)\right){\cos }^{12}\left(\frac{\pi k}{40000}\right)\sin \left(\frac{6\pi k}{20000}\right)\\ & B(k)=\frac{15}{30}{\left(\frac{k}{20000}\right)}^4-\cos \left(\frac{17\pi }{20}{\left(\frac{k}{20000}\right)}^5\right)\left(\frac{11}{10}+\frac{45}{20}{\cos }^8\left(\frac{\pi k}{40000}\right){\cos }^6\left(\frac{3\pi k}{40000}\right)\right){\cos }^6\left(\frac{41\pi k}{20000}\right)+\frac{12}{20}{\cos }^{10}\left(\frac{3\pi k}{200000}\right){\cos }^{10}\left(\frac{9\pi k}{200000}\right){\cos }^{10}\left(\frac{18\pi k}{200000}\right)\\ & R(k)=\frac{1}{50}+\frac{1}{40}{\sin }^2\left(\frac{41\pi k}{20000}\right){\sin }^2\left(\frac{9\pi k}{200000}\right)+\left(\frac{1}{17}\right){\cos }^2\left(\frac{41\pi k}{20000}\right){\cos }^{10}\left(\frac{\pi k}{40000}\right)\end{array}$$

完整代码

% bird 2
K=-2e4:2e4;
t=linspace(0,2*pi,200);

X=@(k) k./15e3+sin(17.*pi./20.*(k./2e4).^5).*cos(41.*pi.*k./2e4).^6+...
        (1./3.*cos(41.*pi.*k./2e4).^16+1./3.*cos(41.*pi.*k./2e4).^80).*cos(pi.*k./4e4).^12.*sin(6.*pi.*k./2e4);
Y=@(k) 1./2.*(k./2e4).^4-cos(17.*pi./20.*(k./2e4).^5).*(11./10+45./20.*cos(pi.*k./4e4).^8.*cos(3.*pi.*k./4e4).^6).*cos(41.*pi.*k./2e4).^6+...
        12./20.*cos(3.*pi.*k./2e5).^10.*cos(9.*pi.*k./2e5).^10.*cos(8.*pi.*k./2e5).^10;
R=@(k) 1./50+1./40.*sin(41.*pi.*k./2e4).^2.*sin(9.*pi.*k./2e5).^2+1./17.*cos(41.*pi.*k./2e4).^2.*cos(pi.*k./4e4).^10;

CX=[X(K')+cos(t).*R(K'),K'.*nan]';
CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
plot(CX(:),CY(:),'Color',[0,0,0,.2]);
set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');

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蝴蝶一

This image shows 40000 circles. For $k=1,2,3,\dots ,40000$ , the center of the circle is $(X(k),Y(k))$ and the radius of the k-th circle is $R(k)$ , where:

$$\begin{array}{rl}X(k)=& (3/2)\left((\cos (141\pi k/40,000){)}^9\right)\ast \\ & (1-(1/2)\sin (\pi k/40,000))\ast \\ & (1-(1/4)\left((\cos (2\pi k/40,000){)}^{30}\right)\ast \\ & \left(1+(\cos (32\pi k/40,000){)}^{20}\right))\ast \\ & (1-(1/2)\left((\sin (2\pi k/40,000){)}^{30}\right)\ast \\ & \left((\sin (6\pi k/40,000){)}^{10}\right)\ast \\ & \left((1/2)+(1/2)(\sin (18\pi k/40,000){)}^{20}\right))\\ Y(k)=& \cos (2\pi k/40,000)\ast \\ & \left((\cos (141\pi k/40,000){)}^2\right)\ast \\ & (1+(1/4)\left((\cos (\pi k/40,000){)}^{24}\right)\ast \\ & \left((\cos (3\pi k/40,000){)}^{24}\right)\ast \\ & (\cos (21\pi k/40,000){)}^{24})\\ R(k)=& (1/100)+(1/40)\left(\left((\cos (141\pi k/40,000){)}^{14}\right)+(\sin (141\pi k/40,000){)}^6\right)\ast \\ & (1-\left((\cos (\pi k/40,000){)}^{16}\right)\left((\cos (3\pi k/40,000){)}^{16}\right)\ast \\ & (\cos (12\pi k/40,000){)}^{16}).\end{array}$$

完整代码

% butterfly 1
K=1:4e4;
t=linspace(0,2*pi,200);

X=@(k) 3./2.*cos(141.*pi.*k./4e4).^9.*(1-1./2.*sin(pi.*k./4e4)).*(1-1./4.*cos(2.*pi.*k./4e4).^30.*(1+cos(32.*pi.*k./4e4).^20)).*...
        (1-1./2.*sin(2.*pi.*k./4e4).^30.*sin(6.*pi.*k./4e4).^10.*(1./2+1./2.*sin(18.*pi.*k./4e4).^20));
Y=@(k) cos(2.*pi.*k./4e4).*cos(141.*pi.*k./4e4).^2.*(1+1./4.*cos(pi.*k./4e4).^24.*cos(3.*pi.*k./4e4).^24.*cos(21.*pi.*k./4e4).^24);
R=@(k) 1./100+1./40.*(cos(141.*pi.*k./4e4).^14+sin(141.*pi.*k./4e4).^6).*(1-cos(pi.*k./4e4).^16.*cos(3.*pi.*k./4e4).^16.*cos(12.*pi.*k./4e4).^16);

CX=[X(K')+cos(t).*R(K'),K'.*nan]';
CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
plot(CX(:),CY(:),'Color',[0,0,0,.2]);
set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');

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蝴蝶二

This image shows 40000 circles. For $k=1,2,3,\dots ,40000$ , the center of the circle is $(X(k),Y(k))$ and the radius of the k-th circle is $R(k)$ , where:

$$\begin{array}{rl}X(k)=& (6/5)\left((\cos (141\pi k/40,000){)}^9\right)\left(1-(1/2)(\sin (\pi k/40,000){)}^3\right)\ast \\ & (1-(1/4)\left((\cos (2\pi k/40,000){)}^{30}\right)\left(1+(2/3)(\cos (30\pi k/40,000){)}^{20}\right)-\\ & \left((\sin (2\pi k/40,000){)}^{10}\right)\left((\sin (6\pi k/40,000){)}^{10}\right)\ast \\ & \left((1/5)+(4/5)(\cos (24\pi k/40,000){)}^{20}\right))\\ Y(k)=& \cos (2\pi k/40,000)\left((\cos (141\pi k/40,000){)}^2\right)(1+(1/4)\left((\cos (\pi k/40,000){)}^{24}\right)\ast \\ & \left((\cos (3\pi k/40,000){)}^{24}\right)(\cos (19\pi k/40,000){)}^{24})\\ R(k)=& (1/100)+(1/40)(\left((\cos (2820\pi k/40,000){)}^6\right)+\\ & (\sin (141\pi k/40,000){)}^2)(1-\left((\cos (\pi k/40,000){)}^{16}\right)\ast \\ & \left((\cos (3\pi k/40,000){)}^{16}\right)(\cos (12\pi k/40,000){)}^{16}).\end{array}$$

完整代码

% butterfly 2
K=1:4e4;
t=linspace(0,2*pi,200);

X=@(k) 6./5.*cos(141.*pi.*k./4e4).^9.*(1-1./2.*sin(pi.*k./4e4).^3).*(1-1./4.*cos(2.*pi.*k./4e4).^30.*(1+2./3.*cos(30.*pi.*k./4e4).^20)-...
        sin(2.*pi.*k./4e4).^10.*sin(6.*pi.*k./4e4).^10.*(1./5+4./5.*cos(24.*pi.*k./4e4).^20));
Y=@(k) cos(2.*pi.*k./4e4).*cos(141.*pi.*k./4e4).^2.*(1+1./4.*cos(pi.*k./4e4).^24.*cos(3.*pi.*k./4e4).^24.*cos(19.*pi.*k./4e4).^24);
R=@(k) 1./100+1./40.*(cos(2820.*pi.*k./4e4).^6+sin(141.*pi.*k./4e4).^2).*(1-cos(pi.*k./4e4).^16.*cos(3.*pi.*k./4e4).^16.*cos(12.*pi.*k./4e4).^16);

CX=[X(K')+cos(t).*R(K'),K'.*nan]';
CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
plot(CX(:),CY(:),'Color',[0,0,0,.2]);
set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');

---

橄榄枝

This image shows 4000 circles. For $k=1,2,3,\dots ,4000$ , the center of the circle is $(X(k),Y(k))$ and the radius of the k-th circle is $R(k)$ , where:

$$\begin{array}{rl}X(k)=& (2k/4000)+(1/28)\sin (42\pi k/4000)\\ & +(1/9)\left((\sin (21\pi k/4000){)}^8\right)\\ & +(1/4)\left((\sin (21\pi k/4000){)}^6\right)\ast \\ & \sin \left((2\pi /5)(k/4000{)}^{12}\right)\\ Y(k)=& (1/4)(k/4000{)}^2\\ & +(1/4)(\left((\sin (21\pi k/4000){)}^5\right)\\ & +(1/28)\sin (42\pi k/4000))\ast \\ & \left(\cos \left((\pi /2)(k/4000{)}^{12}\right)\right)\\ R(k)=& (1/170)+(1/67)\left((\sin (42\pi k/4000){)}^2\right)\ast \\ & \left(1-\left((\cos (21\pi k/4000){)}^4\right)\right).\end{array}$$

完整代码

% olive
K=1:4e3;
t=linspace(0,2*pi,200);

X=@(k) 2.*k./4e3+1./28.*sin(42.*pi.*k./4e3)+1./9.*sin(21.*pi.*k./4e3).^8+1./4.*sin(21.*pi.*k./4e3).^6.*sin(2.*pi./5.*(k./4e3).^12);
Y=@(k) 1./4.*(k./4e3).^2+1./4.*(sin(21.*pi.*k./4e3).^5+1./28.*sin(41.*pi.*k./4e3)).*cos(pi./2.*(k./4e3).^12);
R=@(k) 1./170+1./67.*sin(42.*pi.*k./4e3).^2.*(1-cos(21.*pi.*k./4e3).^4);

CX=[X(K')+cos(t).*R(K'),K'.*nan]';
CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
plot(CX(:),CY(:),'Color',[0,0,0,.2]);
set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');

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当然除了画圈圈，该网站上还有一些其他有趣的图形，可以自行去查看，最后，给大家比个心~

心形

This image shows the following curve:

$$\begin{array}{rl}& X(t)=\frac{4}{9}\sin (2t)+\frac{1}{3}(\sin (t){)}^8\cos (3t)+\frac{1}{8}\sin (2t)(\cos (247t){)}^4\\ & Y(t)=\sin (t)+\frac{1}{3}(\sin (t){)}^8\sin (3t)+\frac{1}{8}\sin (2t)(\sin (247t){)}^4,0\le t\le \pi \end{array}$$

完整代码

% heart
t=linspace(0,pi,5e3);
X=@(t) 4./9.*sin(2.*t)+1./3.*sin(t).^8.*cos(3.*t)+1./8.*sin(2.*t).*cos(247.*t).^4;
Y=@(t) sin(t)+1./3.*sin(t).^8.*sin(3.*t)+1./8.*sin(2.*t).*sin(247.*t).^4;

hLine=plot(X(t),Y(t),'-','LineWidth',1);
colorNum=length(hLine.XData);
colorData=uint8([(pink(colorNum).*255)';255+t.*0]);
pause(1e-16) 
set(hLine.Edge,'ColorBinding','interpolated', 'ColorData',colorData)
set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');

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转载：https://blog.csdn.net/slandarer/article/details/127702433