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Picture a butterfly with no memory of its last wingbeat.

演示视频

Picture a butterfly with no memory of its last wingbeat. — 1

提示词

Picture a butterfly with no memory of its last wingbeat.
Ask it for frame 720 and you get frame 0, pixel for pixel.

720 frames. 12 seconds. 6 counters. All at 0.

Opus 5.5 can't hand over a video file, one builder says. He made that gap the method.

> STAGE - one fixed 1000x1080 canvas, fonts baked in
> RENDER - a windowless engine calls draw(f) once per frame number
> ENCODE - ffmpeg writes H.264, pixel aspect ratio declared
> CHECK - three passes after every build

By his account the model writes only the first, a drawing program. The other three get reused.

Each picture depends on its number alone: no clock, no timer, no dice that land differently each run.

So any frame renders the same every time. A fix is a small edit and a re-render.

The sample loop is 720 pictures in 12 seconds. A hand animator at 12 drawings a second would need 144.

The smoke test plays one loop unseen and counts six kinds of defect, from script errors to text on text.
Six yes/no gates before a human looks.

The seam check puts frame 0 against frame 720. One changed pixel means some motion isn't periodic.

He reports a last render of 1,820 frames with six zeros. At the sample's 60 fps that would be about half a minute.

Let one frame read a clock and the same number stops giving the same picture.

We drew the rule as an X-ray butterfly, its wingbeat computed from the frame number ↓

中文译文

想象一只不记得自己上一次振翅的蝴蝶。
向它要第 720 帧,你会得到第 0 帧,一个像素都不差。

720 帧。12 秒。6 个计数器。全部为 0。

一位开发者说,Opus 5.5 交不出视频文件。他把这个缺口变成了方法。

> STAGE - 一块固定的 1000x1080 canvas,字体预先烘焙进去
> RENDER - 一个无窗口的引擎按帧号调用一次 draw(f)
> ENCODE - ffmpeg 写出 H.264,声明像素宽高比
> CHECK - 每次构建后跑三遍检查

据他所说,模型只写第一个,也就是一个绘图程序。另外三个会被复用。

每一幅画面只依赖它的编号:没有时钟,没有定时器,没有每次都会掷出不同点数的骰子。

所以任何一帧每次都渲染出同样的结果。修一个地方,就是小改一下再重新渲染。

这段示例循环是 12 秒里的 720 幅画面。手绘动画师按每秒 12 张来算,需要 144 张。

冒烟测试会在无人观看的情况下播放一轮,并统计六类缺陷,从脚本错误到文字叠文字。
在人看之前,先过六道是/否关卡。

接缝检查把第 0 帧和第 720 帧放在一起比对。只要有一个像素不同,就说明某处运动不是周期性的。

他报告的最后一次渲染是 1,820 帧,六项计数全为零。按示例的 60 fps 计算,大约半分钟。

只要让某一帧去读时钟,同一个编号就不再给出同一幅画面。

我们把这条规则画成一只 X 光蝴蝶,它的振翅由帧号计算得出 ↓

溯源

原作者
0xKarmi
发布日期
2026-10-09(由推文 ID 反解)
原始出处
查看原帖 ↗
收录来源
yihui-dev/awesome-opus5-5-videos · 0xkarmi-129839 · MIT