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ffmpeg -i /dev/video0 -filter_complex "[0:v]split[orig][copy];[copy]setpts=PTS+0.4/TB[delayed]; [orig][delayed]blend=all_expr='if(lt(mod(XX+YY\,2000)\,1000)\,A\,B)'" -f rawvideo -pix_fmt yuv420p - | ffplay -f rawvideo -video_size 640x480 -

sleeping

1,992 subscribers

44,427 次观看 • 11 个月前 •via X (Twitter)

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$ffmpeg -i "/dev/video" -filter_complex \ "[0:v]split=2[orig][copy]; \ [copy]setpts=PTS+0.5/TB[delayed]; \ [orig][delayed]blend=all_expr='if(gt(mod(sin((X*X - Y*Y)/3000 + T*2)*cos((X+Y)/2000)\,1)\,0.5)\,A\,B)', \ hue=h='360*t/10':s=2,format=yuv420p[v]" \ -map "[v]" -map 0:a -c:v libx264 -crf 18 -preset fast -c:a$

ffmpeg -i "/dev/video" -filter_complex \ "[0:v]split=2[orig][copy]; \ [copy]setpts=PTS+0.5/TB[delayed]; \ [orig][delayed]blend=all_expr='if(gt(mod(sin((XX - YY)/3000 + T2)cos((X+Y)/2000)\,1)\,0.5)\,A\,B)', \ hue=h='360*t/10':s=2,format=yuv420p[v]" \ -map "[v]" -map 0:a -c:v libx264 -crf 18 -preset fast -c:a

𝐬𝐨𝐯

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This implicit function fascinates me. f(x, y) = sin(a · (x² + y²)) − cos(b · x · y)

This implicit function fascinates me. f(x, y) = sin(a · (x² + y²)) − cos(b · x · y)

Joshua Guo

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Observer vec2 p=(FC.xy*2.-r)/r.y/.2,v;for(float i,l,f;i++<1e1;o+=.03/max(l=length(v)-i,-l*3.)*(cos(t-i*.4+.1/l+vec4(0,1,2,3))+1.1))for(v=p,f=0.;f++<9.;v+=sin(ceil(v*f+i*.9)-t/2.)/f);o=max(tanh(o+(o=texture(b,(FC.xy+r.y*.05*sin(FC.xy+FC.yx/.6))/r))*o),.0);

Observer vec2 p=(FC.xy2.-r)/r.y/.2,v;for(float i,l,f;i++<1e1;o+=.03/max(l=length(v)-i,-l3.)(cos(t-i.4+.1/l+vec4(0,1,2,3))+1.1))for(v=p,f=0.;f++<9.;v+=sin(ceil(vf+i.9)-t/2.)/f);o=max(tanh(o+(o=texture(b,(FC.xy+r.y.05sin(FC.xy+FC.yx/.6))/r))*o),.0);

Xor

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Still excited about these 3Blue1Brown-style videos generated by AI. Here’s a beautiful illustration of a classic analysis problem: Let f be convex, nonnegative on [0,∞), with f(0)>0 and f(x)→0 as x→∞. Place a light source at (0,b) with 0<b<f(0). Rays reflect off the graph of f and the x-axis following the usual laws of optics. Question: can any ray escape to +∞, or do all trajectories eventually come back? Prompt to video

Still excited about these 3Blue1Brown-style videos generated by AI. Here’s a beautiful illustration of a classic analysis problem: Let f be convex, nonnegative on [0,∞), with f(0)>0 and f(x)→0 as x→∞. Place a light source at (0,b) with 0<b<f(0). Rays reflect off the graph of f and the x-axis following the usual laws of optics. Question: can any ray escape to +∞, or do all trajectories eventually come back? Prompt to video

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Event 2 vec2 p=(FC.xy*2.-r)/r.y/.3,v;for(float i,l,f;i++<9.;o+=.1/abs(l=dot(p,p)-5.-2./v.y)*(cos(i/3.+.1/l+vec4(1,2,3,4))+1.))for(v=p,f=0.;f++<9.;v+=sin(ceil(v.yx*f+i*.3)+r-t/2.)/f);o=max(tanh(o+(o=texture(b,(FC.xy+r.y*.04*sin(FC.xy+FC.yx/.6))/r))*o),.0);

Event 2 vec2 p=(FC.xy2.-r)/r.y/.3,v;for(float i,l,f;i++<9.;o+=.1/abs(l=dot(p,p)-5.-2./v.y)(cos(i/3.+.1/l+vec4(1,2,3,4))+1.))for(v=p,f=0.;f++<9.;v+=sin(ceil(v.yxf+i.3)+r-t/2.)/f);o=max(tanh(o+(o=texture(b,(FC.xy+r.y.04sin(FC.xy+FC.yx/.6))/r))*o),.0);

Xor

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"Waveform" for(float i,d,z,f;i++<9e1;o+=(cos(z*.5+t+vec4(0,2,4,3))+1.3)/d/z){vec3 p=z*normalize(FC.rgb*2.-r.xyy)+1.;f=max(-p.y,0.);p.y+=f+f;for(d=1.;d<3e1;d+=d)p.y+=cos(p*d+2.*t*cos(d)+z).x/d;z+=d=(.1*f+abs(p.y-1.)/(++f*f)+max(d=p.z+3.,-d*.1))/8.;}o=tanh(o/9e2);

"Waveform" for(float i,d,z,f;i++<9e1;o+=(cos(z.5+t+vec4(0,2,4,3))+1.3)/d/z){vec3 p=znormalize(FC.rgb2.-r.xyy)+1.;f=max(-p.y,0.);p.y+=f+f;for(d=1.;d<3e1;d+=d)p.y+=cos(pd+2.tcos(d)+z).x/d;z+=d=(.1f+abs(p.y-1.)/(++ff)+max(d=p.z+3.,-d*.1))/8.;}o=tanh(o/9e2);

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