【Hacker News搬运】康威生命梯度

[hackernews]

Title: Conway's Gradient of Life

康威生命梯度

Text:

hn link

Url: https://hardmath123.github.io/conways-gradient.html

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Post by: networked

Comments:

versteegen: The objective function here defines a Markov random field (MRF) with boolean random variables and certain local statistics of nearest neighbours, either uniform if the target is a white image, or varying with location to produce an image. MRFs define Gibbs probability distributions, which you can sample from (which will already produce a good image here) or perform gradient ascent on to reach a local maxima. The negative log-likelihood of the MRF distribution is equal to the loss function of the original optimisation problem, so the maximum likelihood estimate (MLE) (there will often be multiple due to symmetry) of the MRF is the optimal solution(s) to the original problem. (But in general the MLE can look completely different to a sample.)<p>The statistics are 9th-order (of 3x3 blocks of pixels) but of a simple form which are hardly more expressive than 2nd-order nearest neighbour statistics (in terms of the different textures that they can reproduce) which are well known. In the approximate case where you only care about the average value of each pixel I think it would collapse to 2nd-order. Texture synthesis with MRFs with local statistics is discretized (in space) Turing reaction-diffusion. I did my PhD on this topic.<p>Probably the most influential early paper on this kind of simple texture model, where you will see similar patterns, is:<p>Cross & Jain, 1983, PAMI, Markov Random Field Texture Models

versteegen: 这里的目标函数定义了一个马尔可夫随机场（MRF），其中包含布尔随机变量和最近邻的某些局部统计数据，如果目标是白色图像，则这些数据是统一的，或者随着位置的变化而变化以生成图像。MRF定义了吉布斯概率分布，您可以从中采样（这在这里已经产生了一个好的图像），也可以对其进行梯度上升以达到局部最大值。MRF分布的负对数似然等于原始优化问题的损失函数，因此MRF的最大似然估计（MLE）（由于对称性，通常会有多个）是原始问题的最优解。（但一般来说，MLE看起来与样本完全不同。）<p>统计数据是9阶（3x3像素块），但形式简单，几乎不比众所周知的2阶最近邻统计数据（就它们可以再现的不同纹理而言）更具表现力。在你只关心每个像素的平均值的近似情况下，我认为它会崩溃到二阶。使用具有局部统计的MRF进行纹理合成是（在空间中）图灵反应扩散的离散化。我在这个课题上读了博士学位<p> 关于这种简单的纹理模型，最有影响力的早期论文可能是：<p>Cross&amp；Jain，1983，PAMI，马尔可夫随机场纹理模型

drofnarc: I'm so glad to see others working on this. I've attempted it too, but not with any measure of success.<p>Trying to trade space for time, I used a model that gives every cell a set of all 512 of the possible 3x3 neighborhoods that could have caused that cell's present state ("alibis"). It then goes to each cell, comparing its alibis to those of neighboring cells and eliminating mutually impossible ones from either set. This step has to be repeated until no more alibis are shed in a pass.<p>When it finally stabilizes, the model is a solution kernel that can then field further manual guesses against it. If a cell's alibis all agree it was dead in the "before", there's no need to guess, but if they're not unanimous, what if we hazard a guess one way or the other for a bit? How does that ripple through the rest of the board? If any of the cells ran completely out of alibis given a certain guess, that guess was clearly not a proper solution, and it's time to back out and try a different one. If there's no solution at all, that's a Garden of Eden.<p>Ultimately I wanted to generate not just one solution, but all the solutions for a given board. I got stumped because I wasn't convinced I wasn't still working in 2**(n*m) time or worse trying guesses against the kernel.<p>It's a really fascinating problem, so much so that I even made a pico8 game about it years ago! Even the 6x6 grids are really tough!

drofnarc: 我；我很高兴看到其他人在做这件事。我；我也尝试过，但没有取得任何成功<p> 为了用空间换取时间，我使用了一个模型，该模型为每个单元格提供了一组可能导致该单元格的所有512个3x3邻域；目前的状况（“不在场证明”）。然后，它进入每个单元格，将其不在场证明与相邻单元格的不在场证明进行比较，并从任何一组中消除相互不可能的不在场证据。必须重复此步骤，直到没有更多的不在场证明。<p>当它最终稳定下来时，模型是一个解决方案内核，然后可以对其进行进一步的手动猜测；所有的不在场证明都认为它已经死了；在“之前”；，那里；无需猜测，但如果他们；我们不是一致同意的，如果我们冒险猜测一下呢？这对董事会其他成员有何影响？如果任何细胞在给定一定猜测的情况下完全失去了不在场证明，那么该猜测显然不是一个合适的解决方案；是时候退一步，尝试另一种了。如果存在；这根本不是解决方案，那；这是一个伊甸园<p> 最终，我想为给定的电路板生成的不仅仅是一个解决方案，而是所有解决方案。我被难住了，因为我不是；我不相信自己是；t在2**（n*m）时间或更糟糕的情况下仍在尝试对内核进行猜测<p> 它；这是一个非常有趣的问题，以至于几年前我甚至制作了一个关于它的pico8游戏！即使是6x6的网格也非常坚固！

dvh: Few days ago I posted this: <a href="https://news.ycombinator.com/item?id=41743887">https://news.ycombinator.com/item?id=41743887</a>

dvh: 几天前，我发布了以下内容：<a href=“https:&#x2F；news.ycombinator.com&#x2M；item？id=41743887”>https:&#x2F；news.ecombinator.com；项目？id=41743887</a>

mbauman: Atavising was new to me. From <a href="https://nbickford.wordpress.com/2012/04/15/reversing-the-game-of-life-for-fun-and-profit/" rel="nofollow">https://nbickford.wordpress.com/2012/04/15/reversing-the-gam...</a> :<p>> First of all, while I said “Predecessorifier” in the talk, “Ataviser” seems to be the accepted word, coming from “Atavism”, which the online Merriam-Webster dictionary defines as “recurrence of or reversion to a past style, manner, outlook, approach, or activity”.

mbauman: Atavising对我来说是全新的。来自<a href=“https:nbickford.wordpress.com”2012年4月15日“为了乐趣和利润而逆转生活游戏”rel=“nofollow”>https:&#x2F；nbickford.wordpress.com；2012年；04；15°F；逆转游戏</a> ：<p>&gt；首先，当我在演讲中说“前置词”时，“Ataviser”似乎是公认的词，来自“Atavism”，在线Merriam-Webster词典将其定义为“过去风格、方式、观点、方法或活动的再现或回归”。

rustybolt: Feels to me like there is no need for backpropagation. I think you can just iteratively grab a random pixel and flip it of that would bring you closer to the target after one step.<p>It would probably work even better if you tweak the loss function with some kind of averaging/blurring filter.

rustybolt: 在我看来，没有必要进行反向传播。我认为你可以迭代地抓取一个随机像素并翻转它，这会让你在一步后更接近目标<p> 如果你用某种平均值来调整损失函数，它可能会更好地工作；模糊滤镜。