按照黄老师的计划，我们今天在 Rocky Mountains National Park。出门之前我看了一下温度，好家伙最高 35 度，那我立刻把长裤换成短裤，外套也懒得带了。
好了又到了选择司机的时间。这次我们有三个司机：
人挡杀人车挡超车高速上来去自如的黄老师
稳如老狗波澜不惊慢速平稳还省油的 zyw
奇思妙想妙语连珠多线程并行处理的 cxq
上午去爬了两个 trail：Bear Lake Loop 和 Emerald Lake Trail。第一个 trail 很简单，围着湖走一圈就完事了。 Bear Lake 其实没有什么特别让我惊艳的地方，反正也看不到熊 →_→ ……第二个 trail 就好一些，一路上依次经过 Nymph Lake、Dream Lake 最后到达 Emerald Lake。整个 trail 难度也不大，只要休息的足够多，总能上去的。尽管如此，我们还是不断损兵折将。在 Dream Lake 之前损失了爬山健将脸哥和董先生，Dream Lake 损失了 cls 和 jls，最后走走停停，七人成功抵达 Emerald Lake。由于这地方海拔已经不低了，有三千多米，所以有些人已经开始有一定的高反了。
Bear Lake
Dream Lake
Emerald Lake
爬完山后我们开始思考午饭问题。这次准备吃个简单的，中午在 Estes Park 吃了个 McDonald's。由于毕老师没有在美国长期呆过，他的手机号没有注册过 McDonald's 账号，这可馋坏了 cxq：新客户优惠不用白不用啊！所以尽管 cxq 已经吃的差不多了，可是他还是怂恿着毕老师整了一个账号，以超低价拿到了个啥汉堡，这才心满意足的吃饱了。
吃完饭后我们准备去搞个 road trip，从 Estes Park 由 Fall River Entrance Station 进入公园。出发的时候有几个跟着黄老师走，有几个人跟着 zyw 走。可是开着开着就有人在群里喊：有人知道 cxq 在哪里吗？我们顿感不太妙，一顿沟通后才发现 cls 和 jls 成为了街友。本来三辆车十一个人应该是一个 4+4+3 的配置，但是 cxq 出于某种原因独自走了，留下 cls 和 jls 在刚刚的 McDonald's 徘徊。更加蛋疼的是，cxq 车上没人，所以都不好联系到他。万幸的是，cxq 的车就在我们后面，所以等我们开了十几分钟到公园门口的时候，我们车先检票进公园，然后在路边停着等着 cxq。cxq 进来后和我们一 bibi 才意识到出问题了。原来当我们离开的时候，我们挥手致意，而 cxq 将其理解成了车上有五个人，于是 cxq 就 assume 了一个 5+5+1 的配置……于是 cxq 马上调转车头前往 Estes Park，捡上 cls、jls 这街友二人组……
之后我们就去 Denver downtown 上闲逛。首先去了 Colorado Convention Center，主要是去参观它的 Big Blue Bear，甚至网站的 favicon 都是大蓝熊。
Big Blue Bear
之后就走马观花的看了一下，主要想沿着 16th Street 走一走，没想到脸哥走到一半又走不动了，主要也是比较热……于是脸哥就找了个星巴克开始原地修整。说实话我也感觉意思不大，之后我们就坐了免费的 the 16th Street Free Ride 去了 Union Station。Union Station 比较有趣的一点是，它刚好在西经 105° 上（105°00′00″W），门口还有一根线标着西经 105°。
Denver Union Station
离开 Denver 前最后去逛的地方是 Rocky Mountain Arsenal National Wildlife Refuge，本来听说里面有野牛，但是我们在里面逛了半天，啥也没见到，最后不死心，终于在路边找到了几只土拨鼠……想想也 make sense，这么热的天，野牛当然会躲在阴凉处啦，为何要出来闲逛……
路上好像没啥可说的，直奔宾馆。宾馆我们就订的门口的 Yosemite View Lodge，优点就是离门口特别近，比里面的 The Ahwahnee 便宜好订。吐槽一下，这里的网络真烂。
Glacier Point
我们一开始计划是到得早就去 Taft Point 看一看，到得晚就直奔 Glacier Point，中间可以在 Tunnel View 那里看看风景，因为这三个地方在一块，和其他的地方不挨着。虽然地图上看 Glacier Point 离 Yosemite Valley 挺近的，但是除非你走 Four Mile Trail，不然只能从 Tunnel View 那里开车过去，路上要一个小时。如果你真的想走 Four Mile Trail 从山脚走上来，单程爬升将近 1km，那我还是服气的……
然而我们将近七点才到宾馆，这个时候还逛什么 Taft Point 呀，能赶上 Glacier Point 的日落就不错了。Tunnel View 那里路过都没停，怕晚了赶不上 Glacier Point 的日落。尽管如此，等我们到 Glacier Point 的时候还是有点晚，太阳已经下山了，但是没有完全落下，还有 blue hour。从这个角度看过去，Half Dome 的那个面真是被削得好平……
Blue hour 时刻的 Glacier Point 景色。左边三座山峰的样子很像 emoji
Half Dome 来个大特写
Glacier Point 上还有座小房子，于是试图把它和 Half Dome 拍在一起
月光，篝火
等我们下山到宾馆，已经将近十点了，随便吃了点小零食，洗洗睡了……
周六
大早起！七点钟就起来了！今天是爬山的一天！在我们宏伟的计划中，上午拳打 Upper Yosemite Fall Trail，下午脚踢 Mist Trail！起码是这么计划的……
Upper Yosemite Fall Trail 地图。我们从 1 出发，2 是 Columbia Rock，3 是 Oh-My-Gosh Point，5 是终点。地图上 1-2 这一段全是 zigzag，于是从此我患上了 zigzag PTSD，看到长这样的路逃跑也是可以的呢……过了 4 后还可以走另外一条路去 Yosemite Point
Columbia Rock，拍 Yosemite Valley 的好地方
Oh-My-Gosh Point 拍的 Upper Yosemite Fall，看着挺秀气的，但是感觉比 Lower Yosemite Fall 要大
Lower Yosemite Fall, Yosemite Valley
决定了在 Oh-My-Gosh Point 调头后，今天的 hiking 之旅就简单很多了。我们下山的速度明显比上山快，很省力。这个时候登山杖就发挥了它最大的作用。用之前我还觉得，它只有那一个尖尖，能戳的稳吗，用了之后发现比我想象中的稳得多，即使有些细沙也不会滑。
从 Upper Yosemite Fall Trail 上下来后我们就走路去了 Lower Yosemite Fall。这路上就不难了，都是平地，主要困难在于天气很热，我当时有点缺水。我就带了两瓶水，已经喝完了，真是个大失误。我甚至觉得脑袋有点晕晕的，可能有点小中暑或者脱水。
Lower Yosemite Fall 稍微逛逛就差不多了，但是这地方确实比较凉快，所以我们在这里又休息了一会。接下来就讨论该去哪里了。wzc 建议我们可以在中间的大草坪走一圈，然后去 Exploration Center 那里休息一下，然后可以走一下昨天没有去的 Taft Point。中间这个大草坪草挺深的，往远处看可以看到 Half Dome 的正面，非常壮观。
到了 Exploration Center，本来我们以为快一点了，过了饭点了，是不是人会少一些，但是里面还是人山人海，我们运气特别好，上来就有个位置空了，我们就在那里坐着。我实在是困了，就趴下来睡了一会。等我醒来后，wzc 也困了，于是他也趴在桌子上睡了一会，就这样，两人在 Exploration Center 轮流睡觉白嫖了两个小时座位。到接近三点的时候，我们就可以出发了。Exploration Center 附近有个 gift shop 可以买冰箱贴、明信片、拼图。
对于 Taft Point，我们计划走的 trail 是这个，估计两个小时左右。这个 trail 还是非常简单的，主要是没啥海拔爬升。我们先去了 Taft Point，这里的特色是陡峭的悬崖，我们 literally 站在悬崖边上，一失足成千古恨的那种。Taft Point 最高处，在悬崖峭壁上，有一个栏杆，可以让游客靠着来一个自拍，栏杆都被压弯了……在去 Sentinel Dome 的路上没多久就感觉有点小雨，一想起码还要一个小时才能返回，我们就半路折返了。另外需要注意的是这 trail 上有一堆蚊子，真的是拼命往身上扑，由于 trail 难度不大，我们就没怎么休息，但是蚊子仍然可以一路追过来。我有时候感觉脖子后似乎有点小动静，伸手一拍，再拿过来一看，好大一只蚊子，就这样拍死了三四只……
Taft Point 离 Glacier Point 不远，所以我们又去了一趟 Glacier Point。景色和昨天差不多，但是今天晚上就有很多云，所以甚至不如昨天……我们没待多久就准备跑路了，一来昨天已经看得差不多了，二来我们还想着能不能去酒店那里吃点饭。下山路上倒是看到了好看的日落，可惜不好拍。如果这个时候还在 Glacier Point 应该风景不错吧。
我觉得这个安排很合理：下午到达 Yosemite，可以去 Tunnel View 拍照，然后去 Taft Point 搞个两三个小时的 trail，最后去 Glacier Point 拍日落，整个流程可能三四个小时，然后返回酒店休息。
Tunnel View，左边是 El Capitan 酋长岩，中间隐约能看到 Half Dome，右边是 Bridalveil Fall
Taft Point. 放大后可以看到，最高处被游客压弯的栏杆。过曝了救不回来了……
阴天的 Glacier Point
加油
从 Glacier Point 跑路后，我们突然发现了一个严重的问题：车，快，没，油，了……我们从湾区出发的时候加了次油，到 Yosemite 的时候差不多还有 180miles，感觉在园区内是够用的。但是我们开了两次 Glacier Point，一次单程 34miles，中间可能还有一些杂七杂八的……等我们意识到这个问题的时候，我们赶紧从 sport mode 换成了 eco mode，看看能不能再挤出点 mileage。最后在 wzc 的精妙操作下，我们从 Glacier Point 回到宾馆的时候，车里还剩下 26miles。
开车去一家加油站。我们找到了 Mariposa 的一家加油站，再一看距离，离宾馆所在的 El Portal 刚好 27miles，嗯？你是在逗我吗？理论上这是不可到达的，但是如果我们 eco mode 一下，说不定能行？？？但是万一不行，车在路上抛锚了，在这荒郊野外的，电话都不好使，只能寄希望于巡逻的警察或者路过的车辆能帮我们一把……其实更近一点的还有一家，甚至就在 El Portal，但是 Google Maps 显示它周日周一不开，这是什么诡异的营业时间……
但是，中文字体部署在网页上都有一个相同的问题：字体文件太大了……光是一个 Regular 型号就有 11MB，一来下载慢，用户需要等一阵才能看到，二来也浪费带宽。很多常用的英文字体没这烦恼，本来字体就好，还可以直接 Google Fonts 就好。霞骛文楷虽然有 Google Fonts，但是只有繁体中文版，而且我需要的还是 LxgwBright……
]]><div class="note default"><p>一个集合 <span class="math inline">\(S\)</span> 被称为 sum-free set 当且仅当 <span class="math inline">\(\forall a, b \in Ten Days in the Zed Editorhttps://www.roosephu.net/2025/05/18/Ten-Days-in-Zed-Editor/2025-05-18T22:45:39.000Z2025-05-18T22:45:39.000ZRecently I'm playing with a new code editor, Zed. I used it for about ten days and here write a review to share my experiences with it. This review is based on the latest version (v0.186.9).
TL;DR
My feelings about Zed are mixed: as an editor, it's pretty good; as a (Python) code editor, it's pretty bad; as an AI code editor, it's reasonable. As of its current state (v0.186.9), it's hard for me to recommend it to colleagues due to its poor Python experience, but I think I'll keep an eye on it.
Update (@2025-05-19): After tuning LSP configurations, the Python experience is much better. See below.
Metadata
Zed is very actively maintained. Their commit activity shows that there are more than 200 commits almost every week. Their releases are also very frequent: one release every one or two business days. It's still pre-1.0, so they have the chance to make breaking changes, and the so-called "Stable Release" is almost nightly. Take a random example: this PR was submitted on May 13th, was merged on May 14th, and then a new release, v0.186.8, was published on the same day. It does sound a bit aggressive to me... Maybe after 1.0, the Stable version can be less aggressive?
They released their roadmap.
Zed as a Text Editor
It runs pretty smoothly, which is not surprising. It's one of the main selling points, so they have to do it well.
I really like their text-based configuration method.
The auto-complete works amazingly well. Zed also checks the schema and quickly prompts.
It also formats the configuration file automatically, so the configuration always looks clean. However, I wish it didn't remove the trailing comma.
One missing point is that it doesn't sync settings. This has been discussed for a long time, but there's still no official solution.
It has built-in git support (announcement). Works as expected.
I have an annoying issue: after it restarts, the git gutter doesn't work for certain files. I have to close the file and reopen it. I'm not sure if it's related to my specific setup.
Currently, there is no conflict resolver. I really like the one in VSCode.
I'm also using jj recently. There is no jj integration, but it's totally understandable—even VSCode doesn't have it.
It supports remote development, which is essential for work. The overall experience is pretty good.
Read the section on settings before you proceed. I spent a lot of time figuring out why Ruff doesn't work, why I can't hide certain files from the project panel, etc., and it turns out the reason is that I should configure them in the server setting instead of the local setting.
One issue is that I don't know how to open a file in the current window.
Another issue is that I sometimes add multiple folders to the same workspace, but after it restarts, the extra folders are gone.
The key binding setting is very flexible. The format is similar to VSCode.
Rant: project panel is cmd-shift-e, outline panel cmd-shift-b, but why is the git panel ctrl-shift-g?
A note is that to use cmd-b in the terminal, you should send the escape string instead of cmd-b.
It has multibuffers, where multiple files are shown in the same buffer. Some usages include (a) git diff, (b) multi-cursor over multiple files (demo), (c) changes from an AI agent, (d) all calls to the function.
It has built-in vim mode, and it's very well integrated with the editor.
It also integrates a few popular vim plugins, e.g., vim-surround, vim-textobj-entire. I still miss flash.nvim. Zed has vim-sneak emulated (hidden deep in the docs), but vim-easymotion, leap.nvim, or my (current) favorite, flash.nvim (tracking issue) are missing.
The search experience is quite similar to vim: /xxx<enter>. Now you need to use n/N to find the next/previous occurrence. This is exactly how vim works, although I'm pretty used to using <enter> to go to the next occurrence (which doesn't work anymore).
Off-topic: Among all editors with vim integration, VSCode has the best experience so far due to vscode-noevim. It runs a neovim instance so all neovim's plugins can run—no emulation needed! But recently I've seen a lot of this issue, which is very annoying... Though, Zed has a blog post on why it doesn't embed neovim.
It also has a nice write-up on its plan to improve vim mode.
Some mini-features that I miss:
Sticky scroll in the editor (here, here), notebook, and file tree (tracking issue).
Minimap, which was released in v10.187.0-pre and will be in the stable release soon.
Zed as a (Python) Code Editor (IDE)
My main programming language for work is Python (I'm doing Machine Learning research), so here I focus on support for Python. But the result is disappointing...
The biggest issue is that the Python LSP is extremely slow (>5 seconds to see the auto-complete popup, tracking issue), making auto-complete unusable. I can't believe the speed of auto-complete for Python and Zed configuration is from the same editor...
Changing basedpyright.analysis.diagnosticMode from workspace to openFilesOnly mitigates it a little bit, but it's still too slow.
I tried to disable PyLSP, but it's still too slow.
My hypothesis is that I have two (or more) LSP requests: one for the last edit and one for autocomplete. One of Zed or LSP is handling them sequentially—either the LSP can only give the autocomplete list after reanalysis is done, or Zed waits for the LSP's response for the edit before sending the autocomplete request. But I could be totally wrong—I haven't read a single line of Zed's code.
Update (@2025-05-19): It turns out that the reason why Python LSP is extremely slow is due to basedpyright: after turning off the auto import feature (basedpyright.analysis.autoImportCompletions=false), the performance is much better and is now acceptable. I strongly suggest the extension to turn it off by default, as in Pylance.
Other features (inlay hints, type checking, go to definition) work pretty well.
To make inlay hints work, use basedpyright as the Python LSP. I tried very hard to make it work with pyright, until I realized that pyright doesn't come with inlay hints. The one that works in VSCode is actually pylance, which is closed-source.
To use basedpyright in Zed, I followed the guide here, but basically you just need to install the Zed extension, install the basedpyright binaries, and configure the path.
There is also a discussion on the Python experience, and people are complaining about it.
I really wish astral-sh/ty can finish soon. Meta also released Pyrefly a few days ago, which is also written in Rust and is IDE first. The community acts so fast that we now have an Zed extension, but I haven't tried it.
One missing feature is to attach icons for each auto-completion item (tracking issue).
An annoying bug is that basedpyright complains about receiving "redundant open text document command" during restart ( tracking issue). I suspect it's related to the Git gutter issue that I've mentioned above. For more context, I'm using remote development but I'm not sure if it's related.
Syntax highlighting works out of the box (thanks to tree-sitter), but semantic highlighting is not ready (tracking issue).
Sometimes syntax highlighting is incorrect...
The PR for semantic highlighting hasn't been merged, but I do see some sort of semantic highlighting (i.e., f"{x}" and "{x}" are different).
One missing feature is notebook support, which is on their roadmap but has low priority. However, this is important to me as I always have a notebook open for interactive data analysis and experiment management.
Zed as an AI Code Editor
It has Edit Prediction (I like the short demo), which is essentially the Cursor Tab feature.
An issue that occurs for me is the conflict between tab for indentation and Edit Prediction. The document says that "Zed requires a modifier key to accept predictions when your cursor isn't at the right indentation level." However, my experience is different: I just want to add one level of indentation, while <tab> accepts all predictions...
To use the free 2000 Edit Predictions, you need to click to enroll in the Free plan.
I'm super interested in this blog post—I also work for an AI company, but not a generative AI company.
The integration with model providers is ok, but it can still be improved.
Integration includes internal model/completion providers, external completion providers (GitHub Copilot, Supermaven), local model providers (Ollama, LM Studio), and external model providers (OpenAI, Anthropic, Google, etc.) through BYOK.
I tried the Ollama integration and found it a little disappointing. It doesn't stream text, so I have to wait until all output is generated.
I tried local Qwen2.5-Coder-7B, but the result was miserable. The generated code is incomplete: it ends with ... (other code remains the same). The output path is wrong, so Zed can't diff it and I have to manually copy over the generated code.
I also tried local Qwen3-30B-A3B, but it's very slow on my MBP.
Agentic editing and tool use are also supported.
One issue is that the <thinking> token is not hidden (tracking issue).
I like that the user prompts are editable in Agent Panel. My suggestion is that we use a button to enter the edit mode instead of clicking the text, because sometimes I just want to copy it.
Another issue is that the Agent Panel doesn't render MathJax, but I agree that's such a niche feature request...
I do like the feature of using LLMs to generate commit messages. I just don't like writing commit messages...
One improvement would be to allow custom prompts (tracking issue).
As for data privacy, to my understanding, if we don't opt in, Zed does not retain any data ("Zed does not persistently store user content or use user content to evaluate and/or improve our AI features") for both Agent Panel and Edit Prediction, but third-party services for Agent Panel might do so (inferred from "Anthropic may not train models on Customer Content from paid Services.").
I do appreciate that the default is opt-out.
For comparison, both Cursor (in Privacy Mode) and Windsurf (in zero-data retention mode) have a zero-data retention policy that prevents third-party services from retaining data, if I understand it correctly. I wish Zed could also have a zero-data retention policy for the Agent Panel.
For Edit Prediction, Zed itself doesn't retain any data (when not opted in), but Zed uses Baseten to serve the model, and it's not clear whether Baseten retains any data.
Another improvement is to have an option to disable opt-in in project settings in case the user enables it by accident.
In terms of the quality of generated code, it's hard for me to evaluate and compare as I haven't used many AI code editors before, and during my evaluation, I was using GitHub Copilot as the service provider instead of Zed's own services... I only found GitHub Copilot mediocre. I also used Cursor for a little bit and was impressed by the Tab completion feature. Honestly, I feel like Zed's Edit Prediction is slightly slower and worse than Cursor Tab, but I say that with pretty low confidence.
Random Thoughts
As a programmer, I'm a heavy user of code editors. I've used (Neo)Vim, Emacs, Atom, VSCode, etc. When I first heard of Zed, I wasn't that interested. I kind of forgot what exactly I read, but all I remember is that it's an editor with Rope and it's written in Rust, both of which are buzzwords. I later found their blog post, How CRDTs make multiplayer text editing part of Zed's DNA, but given the prior art, the Rope science series, Zed's blog post didn't really impress me (but the animation is pretty cool!). When I first heard "Rust" and "Rope", I was thinking, isn't the Xi-editor doing the same thing?
I tried Zed on and off for a few sessions, each of which took only half an hour or so. The reason I didn't invest my time in it is that I have concerns about its future: What if the product goes unmaintained? It's hard to monetize a text editor. The market is competitive and there are many tough-to-beat free options, like the OG Vim/Emacs and modern options like VSCode. Moreover, it also competes with full IDEs like Visual Studio and JetBrains products. So before they joined the AI Code Editor army, I didn't see a good business model (B2B is just hard). Maybe they can commercialize the product with certain features (e.g., collaborative editing), but I'm not convinced. Until recently, they announced being the fastest AI code editor. I finally pulled the trigger, as this might be a viable business model (B2C). AI code editor is also a very competitive market (just to name a few: Cursor, Windsurf, OpenAI Codex, Claude Code, Augment Code, Fitten Code, Tabby, GitHub Copilot, Devin), but now Zed has its own strength: they understand the editor down to every line of code and every feature, and they have full control.
Regarding the future of AI code editors, I do have some thoughts, but they're pretty immature and I haven't investigated enough. It's possible that the problems have been studied intensively in the industry while I live in too small a bubble to see it. Perhaps I should revisit this later after using more AI code editors for longer to think more about the features, integrations (of AI/LSP/Editor), business model, research, etc.
I also read two more posts on others' experiences with Zed 12 after writing this post, so the opinions are my own.
]]><p>Recently I'm playing with a new code editor, <span class="exturl" data-url="aHR0cHM6Ly96ZWQuZGV2">Zed</span>. I used it for about ten days and here write a review to share my experiences with it. This review is based on the latest version (v0.186.9).</p>
<h1 id="tldr">TL;DR</h1>
<p>My feelings about Zed are mixed: as an <em>editor</em>, it's pretty good; as a <em>(Python) code editor</em>, it's pretty bad; as an <em>AI code editor</em>, it's reasonable. As of its current state (v0.186.9), it's hard for me to recommend it to colleagues due to its poor Python experience, but I think I'll keep an eye on it.</p>
<p>Update (<span class="citation" data-cites="2025-05-19">@2025-05-19</span>): After tuning LSP configurations, the Python experience is much better. <a href="/2025/05/18/Ten-Days-in-Zed-Editor/#zed-as-a-python-code-editor-ide">See below</a>.</p>Integer and Rational Solutions of a Binary Quadratic Equation (0): General binary quadratic equationhttps://www.roosephu.net/2025/03/23/Integer-and-Rational-Solutions-of-a-Binary-Quadratic-Equation-0-General-Binary-Quadratic-Equation/2025-03-24T02:16:59.000Z2025-03-24T02:16:59.000ZIn this post, we are interested in finding all integer and rational solutions for a general binary quadratic equation:\[\begin{equation}ax^2 + bxy + cy^2 + dx + ey + f = 0,\label{eq:general}\end{equation}\]for any integer coefficients \(a, b, c, d, e, f\). The equation might degenerate; we only point it out when it does but do not solve it, as it's typically easier to solve a degenerate equation.
This post mainly serves as an entry point for a series of posts and also contains a little bit of the story behind the posts. That is why this post has index 0 in the title.
The story
As readers might know, I have been practicing Project Euler for years. Project Euler contains many problems related to number theory, some of which involve solving binary Diophantine equations. One of the most famous examples of quadratic Diophantine equations is the Pythagorean triples (\(x^2 + y^2 = z^2\)).
One day, I suddenly had an idea: What's the general method for solving quadratic equations? How can we enumerate the solutions? Given that there might be infinite solutions, probably we want to enumerate bounded solutions efficiently (mostly for solving Project Euler problems). So I spent time researching, reading numerous papers, notes, books and blog posts... Until I thought I was done and ready to summarize what I had learned into posts.
While finishing my writing, I found Quadratic diophantine equations BCMATH programs, and immediately realized that my literature review was so incomplete. The website covers most related topics and is maintained by Keith Matthews, who seems to be a domain expert, with regular updates. Reading his papers made me question the necessity of my posts...
Nevertheless, I decided to proceed with my writing. Perhaps it's for my own understanding, or perhaps I simply don't want to admit I haven't published anything on my blog for two months (so I even break the topic down into many small parts)...
What's next? I still have two incomplete things, but it depends on whether my interests fade.
Implement the algorithms, which is a perfect chance to learn a new programming language, just like what I did for my previous side project (Analytic prime counting, Part 1 and Part 2).
More efficient enumeration of primitive solutions of the Legendre equation. Although an upper bound of the GCD exists, I still want to make it more efficient by removing GCD. The requirement of enumerating the primitive input pair \((u, v)\) probably can't be removed, though.
Common techniques
Suppose we want to solve \(f(\mathbf{x}) = n\).
(Homogenization) If \(f\) is not homogeneous, it may be beneficial to homogenize it;
(Diagonalization) If \(f(\mathbf{x}) = x^\T Q \mathbf{x}\) is quadratic but \(Q\) is not diagonal, diagonalize it;
(Primitive solutions) Only consider the primitive solutions. For most cases, an algorithm for primitive solutions can be easily used to find non-primitive solutions (Turing reduction);
(Partition of solution space) Consider the equation \(f(\mathbf{x}) \equiv 0 \pmod{n}\). The solution gives us strong constraints on \(\mathbf{x}\), which is enough to partition the solution space.
Integer solutions
Let \(\Delta := b^2 - 4ac\) be the discriminant of the equation.
(Case 1) If \(a = c = 0\):
(Case 1.1) If \(b = 0\): it simplifies to a linear case \(dx + ey + f = 0\);
(Case 1.2) If \(b \neq 0\): This results in a degenerate case: \((bx + e)(by + d) = de - bf\). To solve, factor \(de - bf\) to express it in terms of \(bx + e\) and \(by + d\).
(Case 2) If \(\Delta = 0\) while \(a \neq 0\) or \(c \neq 0\): Without loss of generality (W.L.O.G.), we assume \(a \neq 0\). Note\[ 4a(a x^2 + bxy + cy^2 + dx + ey + f) = (2ax + by + d)^2 + (4ae - 2bd)y - (d^2 - 4af). \]Thus:
(Alternative Case 3) If \(\Delta \neq 0\), note\[ \Delta^2 (ax^2 + bxy + cy^2 + dx + ey + f) = ax'^2 + bx'y' + cy'^2 + \Delta n, \]where\[ \begin{aligned} x' &= \Delta x - 2cd + be, \\ y' &= \Delta y - 2ae + bd, \\ n &= ae^2 - bed + cd^2 + \Delta f. \end{aligned} \]Then we can use the methods about solving binary quadratic forms to find all integer solutions to \(ax'^2 + bx'y' + cy'^2 = n\). The advantage of the method is that we don't need to diagonalize the equation, although many methods implicitly include this step.
Note that many changes of variable are not unimodular, so we need to verify whether the result is an integer solution.
Rational solutions
We write the equation in the matrix form:\[\mathbf{x}^\T Q \mathbf{x} = 0, \quadQ := \begin{pmatrix}2a & b & d \\b & 2c & e \\d & e & 2f \\\end{pmatrix}, \quad\mathbf{x} := \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}.\]
(Diagonalization) The matrix \(Q\) may not be diagonal. If so, we can diagonalize it by finding an invertible matrix \(P \in \QQ^{3 \times 3}\) and a diagonal matrix \(D \in \ZZ^{3 \times 3}\) such that\[Q = P^\T D P.\]It follows that\[\mathbf{x}^\T Q \mathbf{x} = 0 \quad \Longleftrightarrow \quad (P \mathbf{x})^\T D (P \mathbf{x}) = 0.\]One possible approach is to use the LDL decomposition. However, the matrix \(Q\) may not have a LDL decomposition if, at some step (e.g., \((x+y)^2 + yz\)), the diagonal element becomes zero. If this occurs, we can conclude that the equation at that step is actually linear in the variable (e.g., \(y\) in the example). Thus we can always treat other variables (\(x' = x+y, z\)) as free variables and compute \(y\) in the end.
(Legendre equation) Let \(Q = \diag(a', b', c')\), we consider the primitive solutions to the Legendre equation \(a'x'^2 + b' y'^2 + c'z'^2 = 0\). For any primitive solution \(\mathbf{x}' = (x', y', z')\) to the Legendre equation, write \((x, y, z) = P \mathbf{x'}\). If \(z \neq 0\), then \((x/z, y/z)\) is a rational solution to the original equation.
(Solving Legendre equation) For the Legendre equation \(a'x'^2 + b' y'^2 + c'z'^2 = 0\), we assume \(|a'| \geq |b'| \geq |c'|\) W.L.O.G. We analyze the number of zero coefficients among \(a', b', c'\).
There are 3 zeros. Any \((x', y', z')\) for any \(x', y', z'\) is a solution;
There are 2 zeros. Any \((0, y', z')\) for any \(y', z'\) is a solution;
There is 1 zero. Any \((ty', y', z')\) for any \(y', z'\), and any rational solution \(t\) to \(a' t^2 + b' = 0\), is a solution.
Quadratic diophantine equations BCMATH programs by Keith Matthews.
Also check out the history of his algorithm for his line of work.
Methods to solve quadratic Diophantine equations by Dario Alejandro Alpern;
The validity of the following statement is uncertain:
If \(4F^2<B^2-4AC\), the solutions of the equation will be amongst the convergents of the continued fraction of the roots of the equation \(At^2+ Bt + C = 0\).
Sander Verdonschot's repo, which implements many Keith Matthews' papers in Java;
Thilina Rathnayake's blog when developing the Diophantine module of SciPy;
Diophantine Equation--2nd Powers from Wolfram MathWorld;
SciPy's doc on Diophantine equations;
Writings in Keith Conrad's websites. It may be elementary, but it is sufficient for amateurs like me.
Textbooks and notes:
[Dickson19] : Dickson, L.E., 1919. History of the Theory of Numbers (No. 256). Carnegie Institution of Washington.
Matthews15: Matthews, K., 2015. Solving the Diophantine Equation \(ax^2 + bxy + cy^2 + dx + ey + f = 0\).
]]><p>In this post, we are interested in finding all <strong><em>integer</em></strong> and <strong><em>rational</em></strong> solutions for a general binary quadratic equation:
<span class="math display">\[
\begin{equation}
ax^2 + bxy + cy^2 + dx + ey + f = 0,
\label{eq:general}
\end{equation}
\]</span>
for any integer coefficients <span class="math inline">\(a, b, c, d, e, f\)</span>. The equation might degenerate; we only point it out when it does but do not solve it, as it's typically easier to solve a degenerate equation.</p>
<p>This post mainly serves as an entry point for a series of posts and also contains a little bit of the story behind the posts. That is why this post has index 0 in the title.</p>Integer and Rational Solutions of a Binary Quadratic Equation (5): Binary quadratic formhttps://www.roosephu.net/2025/03/22/Integer-and-Rational-Solutions-of-a-Binary-Quadratic-Equation-5-Binary-Quadratic-Form/2025-03-22T11:12:20.000Z2025-03-22T11:12:20.000ZIn this post, we are interested in finding all integer solutions of the following equation:\[\begin{equation}ax^2 + bxy + cy^2 = n.%\label{eq:ibqf}\end{equation}\]where \(acn \neq 0\), \(b^2 - 4ac \neq 0\). This equation can certainly be solved using methods from the generalized Pell equation, but here, we explore other methods.
The function \(ax^2 + bxy + cy^2\) is also called the binary quadratic form (BQF), which we denote as \(\left< a,b,c \right>\) for brevity. The discriminant of \(\left< a,b,c \right>\) is defined as \(\Delta := b^2 - 4ac\).
Theory of binary quadratic forms
Binary quadratic forms have been studied for centuries and many textbooks discuss them. The goal of this section is to provide a brief introduction. Advanced topics (e.g., composition) will not be discussed.
(Equivalence) We define an equivalence relation between BQFs: \(\left< a,b,c \right> \sim \left< a',b',c' \right>\) if and only if there exists an \(M \in \ZZ^{2\times 2}\) such that \(\det M = 1\) and\[\begin{equation}M^\T \begin{pmatrix}a & b/2 \\b / 2 & c \\\end{pmatrix} M = \begin{pmatrix}a' & b' / 2 \\b' / 2 & c' \\\end{pmatrix},%\label{eq:equivalence}\end{equation}\]The matrix determinants of both sides show that \(\left< a,b,c \right>\) and \(\left< a',b',c' \right>\) have the same discriminant.
(Representing numbers) One way to understand this equivalence is by examining the set of numbers it represents. We say \(\left< a,b,c \right>\) represents \(n\) if the equation \(ax^2 + bxy + cy^2 = n\) for some \(x, y \in \ZZ\), and it properly represents \(n\) if \(\gcd(x, y) = 1\). If \(\left< a,b,c \right> \sim \left< a',b',c' \right>\) and \(\left< a',b',c' \right>\) represents \(n\) by \([x, y]\), then \(\left< a,b,c \right>\) represents \(n\) through \(M [x, y]\), which is a simple change of variables (under the constraint that \(\det M = 1\)).
(Positive definite) A BQF \(\left< a,b,c \right>\) is positive definite if for any \((x, y) \neq 0\), \(ax^2 + bxy + cy^2 > 0\). An alternative definition states that \(a > 0, c > 0\) and \(\Delta < 0\).
(Lagrange-reduced form) A positive definite BQF \(\left< a,b,c \right>\) is (Lagrange-)reduced if it meets both of the following conditions: (1) \(|b| \leq a \leq c\), (2) if \(|b| = a\) or \(a = c\), then \(b \geq 0\). Every positive definite BQF is equivalent to a unique reduced BQF.
(Reduction) We have two rules to reduce a positive definite BQF, similar to the Euclidean algorithm:\[\begin{equation}\left< a,b,c \right> \sim \left< c,-b,a \right>, \quad M = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}.\label{R.1}\tag{R.1}\end{equation}\]and for any \(t \in \ZZ\),\[\begin{equation}\left< a,b,c \right> \sim \left<a, b - 2at, at^2 - bt + c\right>, \quad M = \begin{pmatrix} 1 & -t \\ 0 & 1 \end{pmatrix},\label{R.2}\tag{R.2}\end{equation}\]A special case of \(\eqref{R.2}\) is \(\left< a,a,c \right> \sim \left< a,-a,c \right>\) for \(t = 1\).
We can always apply \(\eqref{R.1}\) to ensure that \(a \leq c\), and \(\eqref{R.2}\) to ensure \(|b| \leq a\) by setting \(t = \left\lfloor \frac{b}{2a} \right\rceil\). Both \(\eqref{R.1}\) and \(\eqref{R.2}\) reduces either \(a\) or \(|b|\) while keeping the other the same, so the process must terminate with a BQF \(\left< a,b,c \right>\) where \(|b| \leq a \leq c\). Finally, if \(a = c\) or \(a = |b|\), we choose the one with \(b \geq 0\).
See William Stein's lecture notes and [Manguba-Glover18, Chapter 2] for more details.
Normalization
(Normalization) The equation is first normalized as follows:
Assume \(\gcd(a, b, c) = 1\). Otherwise, we divide both sides by \(\gcd(a, b, c)\).
Assume \(\gcd(a, n) = 1\). Otherwise, we can always find \(u, s, v, r \in \ZZ\) such that \(us - vr = 1\) and \(\gcd(au^2 + buv + cv^2, n) = 1\), and then apply \(\begin{pmatrix} u & r \\ v & s\end{pmatrix}\) to \(\left< a,b,c \right>\). See here for how to find \((u, v)\). The high level idea is that for any prime \(p\), it's enough to try \((1, 0), (0, 1), (1, 1)\), and we use CRT to combine the results.
(Primitive solutions) We only care about the primitive solutions \((x, y)\) where \(g := \gcd(x, y) = 1\). If \(g \neq 1\), we find the primitive solutions of \(ax^2 + bxy + cy^2 = n/g^2\). It's also clear that \(\gcd(y, n) = 1\) for primitive solutions \((x, y)\).
Ellipse case (\(\Delta < 0\))
(This section is mostly taken from [Matthews15].)
We assume the BQF \(\left< a,b,c \right>\) is positive definite. Otherwise, we negate \(a, b, c\) and \(n\).
For a primitive solution \((x, y)\) to \(ax^2 + bxy + cy^2 = n\), define \(\lambda := x y^{-1} \bmod{|n|}\), which certainly satisfies \(a\lambda^2 + b\lambda + c \equiv 0 \pmod{|n|}\). We write \(x = x'n - \lambda y\), hence\[ax^2 + bxy + cy^2 = n, \quad \Longleftrightarrow \quad an x'^2 + (b-2\lambda a) x'y + \frac{a\lambda^2 + b\lambda + c}{n}y^2 = 1.\]Note that these two BQFs have the same discriminant.
The theorem below determines the minimum positive value of a reduced positive definite BQF:
([Matthews15, Lemma 0.1], [Manguba-Glover18, Lemma 1]) For a reduced positive definite BQF \(\left< a, b, c \right>\), the minimum positive value of \(ax^2 + bxy + cy^2\) for \(x, y \in \ZZ\) is \(a\), and is achieved when \((x, y)\) is:
\((\pm 1, 0)\) if \(a < c\);
\((\pm 1, 0), (0, \pm 1)\) if \(b < a = c\);
\((\pm1, \mp1)\) if \(b = a = c\).
After reducing \(\left<an, b-2 \lambda a, \frac{a\lambda^2+b\lambda+c}{n} \right>\) to \(\left< a', b', c' \right>\) using some \(M_{\lambda}\), we can use the theorem above to determine the solutions of \(a'x^2 + b'xy + c'y^2 = 1\).
Here is an algorithm to solve the case:
defsolve_bqf_neg_disc_primitive(a, b, c, n): """find primitive solutions to ax^2 + bxy + cy^2 = n. Assume gcd(a, b, c) = 1 and gcd(c, n) = 1, b^2 - 4ac < 0. """
Δ = b*b - 4*a*c assert Δ < 0
for λ in sqrtmod_all(Δ, n): λ = invmod(2*a, n) * (λ - b) % n
(a2, b2, c2), M = bqf_neg_disc_reduce((a*n, b - 2*a*λ, (a*λ*λ + b*λ + c) // n)) if a2 != 1: continue
# transform back to the original equation M = [[n, -λ], [0, 1]] @ M
(This section is mostly taken from [Matthews15] and [MRS17].)
Structure and transformation
(Equivalence relation) [Matthews23, Section 2] Let \((x^*, y^*)\) be the fundamental solution to \(x^2 - \Delta y^2 = 4\), and \((x, y)\) be a solution to \(ax^2 + bxy + cy^2 = n\), then any \((x', y')\) such that\[\begin{pmatrix}x' \\y' \\\end{pmatrix}= \pm U^k \begin{pmatrix}x \\y\end{pmatrix}, \quad U := \begin{pmatrix}\frac{x^* - by^*}{2} & -cy^* \\ay^* & \frac{x + by^*}{2}\end{pmatrix}.\]for \(k \in \ZZ\) is also a solution to \(ax^2 + bxy + cy^2 = n\). Hence we can use the associated equivalence relation from the group action of a group \(\left\{ \pm U^k: k \in \ZZ \right\}\) on the set of all solutions, which is the Stolt equivalence.
(Partition of solution space) Similarly as the ellipse case, we define \(\lambda := x y^{-1} \bmod{|n|}\) for a primitive solution \((x, y)\) and do the same transformation: write \(x = x'n - \lambda y\), hence\[ax^2 + bxy + cy^2 = n, \quad \Longleftrightarrow \quad an x'^2 + (b-2\lambda a) x'y + \frac{a\lambda^2 + b\lambda + c}{n}y^2 = 1.\]One can also show ([Stolt57, Theorem 5]) that two primitive solutions are Stolt equivalent if and only if they have the same \(\lambda\), implying that the number of Stolt equivalent classes is the finite.
Find Stolt fundamental solutions
Our goal in this subsection is to find the Stolt fundamental solution for a specific \(\lambda\), which solves the equation\[an x'^2 + (b-2\lambda a) x'y + \frac{a\lambda^2 + b\lambda + c}{n}y^2 = 1,\]with minimum non-negative \(y\) (and maximum \(x\) for tie-breaking).
With abuse of notation, we study the equation \(ax^2 + bxy + cy^2 = n\) where \(\Delta > 0\) is not a square, \(a > 0\) and \(n = \pm 1\). Additionally, we require \(y > 0\). Since \(\Delta \equiv 1 \pmod 4\) and \(\Delta\) is not a square, we have \(\Delta \geq 5\).
We first present a very interesting result in [Serret66]:
([Serret66]) Let \(0 < n < \sqrt{ \Delta } / 2\), and let \((x, y)\) be a primitive solution to \(ax^2 + bxy + cy^2 = n\) with \(y > 0\) where \(a > 0\). In addition, let \(\omega, \omega' = (-b \pm \sqrt{ \Delta }) / 2a\) be the roots of \(ar^2 + br + c = 0\) where \(\omega < \omega'\). Then \(p/q\) is a convergent to either \(\omega\) or \(\omega'\).
[MRS17] gives a simple proof: Note \(a(x / y - \omega)(x / y - \omega') = n / y^2\). If \(x / y > \omega'\),\[\frac{x}{y} - \omega' = \frac{n}{a (x / y - \omega) y^2} < \frac{\sqrt{ \Delta } / 2}{a(\omega' - \omega )y^2} = \frac{1}{2y^2},\]then by Dirichlet's approximation theorem, we conclude that \(x / y\) is a convergent to \(\omega'\). Similarly, if \(x / y < \omega\), it's a convergent to \(\omega\).
However, the theorem is no longer true for negative \(n\). [Pavone86] generalized the theorem for\[|n| = \mu :=\min\limits_{\substack{(x, y) \in \ZZ^2 \\ (x, y) \neq (0, 0)}} |ax^2 + bxy + cy^2|,\]where he proved that Serret's theorem still holds unless \(\Delta = 5\), \(n < 0\), in which case \(x / y\) can also be a special semiconvergent of both \(\omega\) and \(\omega'\). [MRS17] further generalizes [Pavone86]'s result, allowing \(|n| < \sqrt{ \Delta } / 2\).
Anyway, [Pavone86]'s result is enough for us since we only care about the case \(n = \pm 1\). One way to find the fundamental solution is to examine the candidates from the following categories to see if they solve the equation, and pick the most fundamental one:
Check the smallest convergent of \(\omega\) before the first period ends;
Check the smallest convergent of \(\omega'\) before the first period ends;
If \(\Delta = 5\) and \(n < 0\), also check the special semi-convergent.
[Matthews02] also designed an algorithm based on [Pavone86], but added many optimizations.
References
General:
Robertson09: Robertson, J., 2009. Computing in quadratic orders. at jpr2718. org.
Ellipse case (\(\Delta < 0\)):
Matthews15: Matthews, K.R., 2015. On a transformation of Lagrange [online]
Matthews14: Matthews, K.R., 2014. Lagrange's Algorithm Revisited: Solving \(at^2+ btu+ cu^2= n\) in the case of negative discriminant. Journal of Integer Sequences, 17(11).
It uses continued fraction and is more complicated. Also need to handle a few exceptional cases.
Manguba-Glover18: Manguba-Glover, K., 2018. Reduction Theory of Binary Quadratic Forms (Doctoral dissertation, University of Hawai ‘i at Manoa).
Parabola case (\(\Delta > 0\)):
Matthews23: Matthews, K., 2023. Finding the Fundamental Solutions of \(ax^2+ bxy+ cy^2= n\).
Notes on Stolt equivalence.
MR21: Matthews, K.R. and Robertson, J.P., 2021. On solving a binary quadratic Diophantine equation. Rocky Mountain Journal of Mathematics, 51(4), pp.1369-1385.
Unfortunately it's paywalled so I haven't read it yet. My guess is that it's pretty similar to [Matthews23].
Matthews02: Matthews, K., 2002. The diophantine equation \(ax^ 2+ bxy+ cy^ 2= N\), \(D= b^ 2-4ac>0\). Journal de théorie des nombres de Bordeaux, 14(1), pp.257-270.
Continued fraction-based algorithm.
MRS15: Matthews, K.R., Robertson, J.P. and Srinivasan, A., 2015. On fundamental solutions of binary quadratic form equations.
This bounds the scale of the Stolt fundamental solutions.
MRS17: Matthews, K.R., Robertson, J.P. and Srinivasan, A., 2017. Extending Theorems of Serret and Pavone. J. Integer Seq., 20(10), pp.17-10.
Pavone86: Pavone, M., 1986. A Remark on a Theorem of Serret. Journal of Number Theory, 23(2), pp.268-278.
Stolt57: Stolt, B., 1957. On a Diophantine equation of the second degree. Arkiv för Matematik, 3(4), pp.381-390.
]]><p>In this post, we are interested in finding all <em><strong>integer</strong></em> solutions of the following equation:
<span class="math display">\[
\begin{equation}
ax^2 + bxy + cy^2 = n.
%\label{eq:ibqf}
\end{equation}
\]</span>
where <span class="math inline">\(acn \neq 0\)</span>, <span class="math inline">\(b^2 - 4ac \neq 0\)</span>. This equation can certainly be solved using methods from the <a href="/2025/03/15/Integer-and-Rational-Solutions-of-a-Binary-Quadratic-Equation-4-Pell-equation-and-generalized-Pell-equation/#generalized-pell-equation-x2--d-y2-n">generalized Pell equation</a>, but here, we explore other methods.</p>Integer and Rational Solutions of a Binary Quadratic Equation (4): Pell equation and generalized Pell equationhttps://www.roosephu.net/2025/03/15/Integer-and-Rational-Solutions-of-a-Binary-Quadratic-Equation-4-Pell-equation-and-generalized-Pell-equation/2025-03-15T12:03:16.000Z2025-03-15T12:03:16.000ZIn this post, we are interested in finding all integer solutions of Pell equation\[\begin{equation}x^2 - d y^2 = 1, \label{eq:pell} % \tag{Pell Equation}\end{equation}\]and generalized Pell equation\[\begin{equation} x^2 - d y^2 = n, \label{eq:generalized-pell} % \tag{Generalized Pell Equation}\end{equation}\]for a non-square integer \(d > 0\) and \(n \neq 0\).
Connection to \(\ZZ[\sqrt{ d }]\)
For any element \(s = x + y \sqrt{ d }\) in \(\ZZ[\sqrt{ d }]\), we define \(\bar{s} := x - y \sqrt{ d }\). It is easy to show that \(\bar{s} \bar{t} = \overline{st}\) for any \(s, t \in \ZZ[\sqrt{d}]\).
Note that for \(s = x + y \sqrt{ d }\), \(x^2 - dy^2 = s \bar{s}.\) Now we define\[S_{n} := \{ s \in \ZZ[\sqrt{ d }]: s \bar{s} = n \}.\]so solving Pell equations is equivalent to calculating \(S_1\) and solving generalized Pell equations is equivalent to calculating \(S_n\).
Most references I found online use the \((x, y)\) notation, but in certain cases, it might be cleaner to present the results using operations in \(\ZZ[\sqrt{ d }]\), especially when presenting the structure of solutions.
Primitive solutions
(Primitive solutions) A solution \((x, y)\) is primitive if \(\gcd(x, y) = 1\). If \((x, y)\) is primitive, \(\gcd(x, n) = \gcd(y, n) = 1\).
Similar to solving \(ax^2 + by^2 = n\), once we have an algorithm to find all primitive solutions, we can derive all solutions. Let \(S'_n\) be the set of all primitive solutions of \(x^2 - d y^2 = n\):\[S'_{n}:= \left\{ s\in \ZZ[\sqrt{ d }]: s \bar{s} = n, \quad s = x + y \sqrt{ d }, \quad \gcd(x, y) = 1 \right\},\]then it is straightforward to see that\[S_{n} = \left\{ gs: g \in \NN, \quad g^2 | n, \quad s \in S'_{n / g^2} \right\} .\]Clearly, \(S_1 = S'_1\).
Pell equation \(x^2 - d y^2 = 1\)
The Pell equation has been studied extensively over time. See [Lenstra02] for a comprehensive overview. There are so many tutorials (e.g. [Conrad22A] and [Conrad22B]) on solving Pell equations, so here I only list a few key conclusions and the algorithm.
(Structure of solutions) It follows from Dirichlet's unit theorem that \(S_{1}\) has the structure\[S_{1} = \{ \pm s^{*k}: k \in \ZZ \},\]for a unique fundamental solution\(s^* := x^* + y^* \sqrt{d}\) where \(x^* > 0, y^* > 0\). Note \(n\) can be negative, as \(\bar{s} = s^{*-1}\). See [Conrad22A, Theorem 5.3] for proof.
(Fundamental solution). It is also shown that \(x^*/y^*\) must be a convergent of \(\sqrt{d}\). We can use the PQa algorithm below to find all convergents. See [Conrad22B, Theorem 5.1] for proof.
(PQa algorithm) Given \(p_0, q_0\) and \(d\) such that \(p_0^2 \equiv d \pmod {q_{0}}\), the PQa algorithm computes the continued fraction \([ a_0, a_1, a_2, \dots]\) of \((p_0 + \sqrt{d}) / q_0\). It works as follows:\[\begin{aligned}a_i & = \floor{(p_i + \sqrt{d}) / q_i}, \\p_{i+1} & = a_i q_i - p_i, \\q_{i+1} & = (d -p_{i+1}^2)/q_i. \\\end{aligned}\]Note that the division for calculating \(q_{i+1}\) is always exact.
Below is an implementation of the PQa algorithm optimized for solving the Pell equation. See [Robertson04] for more details.
defpell(d): """find the fundamental solution of x^2 - dy^2 = 1 and -1""" p, q = 0, 1
(The idea in this section is taken from [Matthews00].)
(Nagell equivalence) For any \(s \in S'_{n}\) and \(t \in S'_1\), we observe\[(s t) \overline{(st)} = (s \overline{s})(t \overline{t}) = n, \quad (st) \bar{t} = s \in S'_{n},\]implying that \(s t \in S'_{n}\). This motivates us to define equivalence relation, the Nagell equivalence, for elements in \(S'_{n}\): Two solutions \(s \sim s'\) if and only if there exists an element \(t \in S'_{1}\), such that \(st = s'\).
(Partition of solution space) Let \(s = x + y \sqrt{ d }\) and \(s' = x' + y' \sqrt{ d }\) be two primitive solutions, and define \(\lambda(s) := -x y^{-1} \bmod |n|\). Since \(\lambda(s)^2 \equiv d \pmod {|n|}\), it follows that\[\begin{aligned}s \sim s' \quad & \Longleftrightarrow \quad s s'^{-1} \in \ZZ[\sqrt{ d }] \\& \Longleftrightarrow \quad x x' \equiv yy' d \pmod{|n|}, \quad xy' \equiv x' y \pmod {|n|} \\& \Longleftrightarrow \quad \tau(s) = \tau(s').\end{aligned}\]So an alternative definition of Nagall equivalence for primitive solutions is that \(s \sim s'\) if and only if \(\lambda(s) = \lambda(s')\).
(Structure of solutions) The alternative definition of Nagall equivalence directly implies the number of equivalent classes in \(S'_n\) is finite, that is, there exists a finite set \(F_{n} \subset \ZZ[\sqrt{ d }]\) such that\[S'_{n} = \left\{f t: f \in F_{n}, t \in S'_{1} \right\} = \left\{ \pm f s^{*n}: F \in F_{n, }n \in \ZZ \right\} ,\]where \(s^*\) is the fundamental solution of \(x^2 - dy^2 = 1\).
(Finding a minimum positive solution) Now we focus on one equivalence class, represented by \(\lambda \equiv -x y^{-1} \pmod{|n|}\) (with abuse of notation) where \(-|n| / 2 < \lambda \leq |n| / 2\). If we write \(x = x' |n| - \lambda y\), it is proved in [Matthews00, Theorem 1] that \(x' / y\) is a convergent of \((\lambda + \sqrt{ d }) / |n|\). Thus, we can find a minimum positive solution\(s = x + y \sqrt{ d }\) with smallest positive \(x\) and \(y\) using the PQa algorithm.
(Finding a fundamental solution) It might be more convenient to find a fundamental solution for an equivalence class, with smallest positive \(y\). If multiple solutions have the same smallest positive \(y\), choose the one with positive \(x\). A fundamental solution can be found as follows:
If \(\lambda \in \{ 0, n / 2\}\), the minimum positive solution is the fundamental solution;
Otherwise, let \((x, y)\) and \((x', y')\) be the minimum positive solutions of \(\lambda\) and \(-\lambda\). If \(y < y'\), the fundamental solutions are \((x, y)\) and \((-x, y)\); otherwise they're \((-x', y')\) and \((x', y')\).
The algorithm is almost the same as [Mollin01, Algorithm 4.1]. The Lagrange-Matthews-Mollin (LMM) algorithm, proposed in [Matthews00, Section 5], improves the process by considering properties of the continued fraction of \((\lambda + \sqrt{ d }) / |n|\).
Here is the pseudocode:
defsolve_generalized_pell_primitive(d, n): """find all fundamental solutions of x^2 - d y^2 = n"""
for λ in sqrtmod_all(d, n): if (s := PQa(d, λ, n)) isNone: continue
if λ == 0or λ * 2 == n: yield s elif λ * 2 < n: p1, q1 = s p2, q2 = PQa(d, -λ, n)
A special generalized Pell equation \(x^2 - dy^2 = 4\)
We also discuss special generalized Pell equation \(x^2 - dy^2 = 4\), as it also shares many properties as the Pell equation \(x^2 - dy^2 = 1\).
As we have mentioned above, the set \(S_1\) has a special structure \(S_1 = \{s_{1}^{*n}: n \in \ZZ\}\) for a fundamental solution \(s_{1}^*\). Similarly, the solutions for \(x^2 - dy^2 = 4\) have a similar structure\[T_{4} = \left\{ \pm t_{4}^{*k}: k \in \ZZ \right\}, \quad T_{4} := \left\{ \frac{x+y\sqrt{ d }}{2} : x^2 - dy^2 = 4 \right\}.\]for a fundamental solution \(t_4^* = \frac{x^* + y^* \sqrt{ d }}{2}\) where \((x^*, y^*)\) is minimum positive solution to \(x^2 - dy^2 = 4\) [Stolt57, p.4]. There is actually a connection between \(s_{1}^{*}\) and \(t_4^{*}\): \(s_1^{*} = t_{4}^{*k}\) for \(k \in \left\{ 1,2,3 \right\}\) ([Stolt57, p.4], [Matthews23, Theorem 1.1]).
(Equivalence relation) Observing that Negall equivalence is the associated equivalence relation of the group action of group \(\{\pm s_1^{*k}: k \in \ZZ\}\) on the set \(S'_n\). As \(T_4\) is also a group, we can define a similar equivalence relation, the Stolt equivalence. In some sense, the Stolt equivalence is more fundamental than the Negall equivalence, in the sense that \(s_1^*\) can be generated by \(t_4^*\) but \(t_4^*\) might not be generated by \(s_1^*\). As a consequence, the Stolt equivalence might have fewer equivalence classes than the Negall equivalence.
To find the fundamental solution \(t_4^*\), check out [Johnson04] for an efficient algorithm implemented below:
defpell4(d): """find the mininum positive solution to x^2 - d y^2 = 4 or -4""" if d % 4 == 1: p, q = 1, 2
It gives a complete description of the LMM algorithm.
Stolt57: Stolt, B., 1957. On a Diophantine equation of the second degree. Arkiv för Matematik, 3(4), pp.381-390.
Update @2025-03-20: Added the discussion of \(x^2 - dy^2 = 4\) and the optimized algorithm for the Pell equation.
]]><p>In this post, we are interested in finding all <em><strong>integer</strong></em> solutions of <em>Pell equation</em>
<span class="math display">\[
\begin{equation}
x^2 - d y^2 = 1, \label{eq:pell} % \tag{Pell Equation}
\end{equation}
\]</span>
and <em>generalized Pell equation</em>
<span class="math display">\[
\begin{equation} x^2 - d y^2 = n, \label{eq:generalized-pell} % \tag{Generalized Pell Equation}
\end{equation}
\]</span>
for a non-square integer <span class="math inline">\(d > 0\)</span> and <span class="math inline">\(n \neq 0\)</span>.</p>Integer and Rational Solutions of a Binary Quadratic Equation (3): $ax^2+by^2=n$https://www.roosephu.net/2025/03/11/Integer-and-Rational-Solutions-of-a-Binary-Quadratic-Equation-3-ax-2-by-2-c/2025-03-11T09:45:32.000Z2025-03-11T09:45:32.000ZIn this post, we're interested in finding all integer solutions of a binary quadratic equation:\[\begin{equation}ax^2 + by^2 = n,\label{eq:ellipse}\end{equation}\]for \(a, b, n > 0\).
Normalization
(Pairwise coprime) Equation \(\eqref{eq:ellipse}\) is reduced if \(a, b, n\) are pairwise coprime. We can reduce it as follows:
If \(p | a\) and \(p | b\), then \(p\) must divide \(n\) (otherwise there is no solution), so we can reduce it to \((a / p, b / p, n / p)\).
If \(p | a\) and \(p | n\) but \(p \nmid b\), then \(p | y\), so we can reduce it to \((a / p, yp, n / p)\).
Similarly for \(p | b\) and \(p | n\) but \(p \nmid a\).
Every time we make a reduction, \(|abn|\) decreases so it must terminate with a reduced equation.
(Primitive solutions) A solution \((x, y)\) is primitive if and only if \(\gcd(x, y) = 1\). It's trivial to use an algorithm which only finds all primitive solutions to find all solutions: If \(g := \gcd(x, y) > 1\), then \((x / g, y / g)\) is a primitive solution to the equation \(ax^2 + by^2 = n/g^2\).
From this point forward, we assume that Equation \(\eqref{eq:ellipse}\) is reduced unless stated otherwise, and we focus on the primitive solutions:
For pairwise coprime integers \(a, b, n > 0\), find all primitive solutions \((x, y)\) such that
\[ax^2 + by^2 = n,\]
Method 1: Lattice
(Partition of solution space) We take the idea from a method on solving Legendre equations. Let \(\mathcal{S}\) be a set of integer points in \(\ZZ^2\):\[S = \{(x, y): a x^2 + by^2 \equiv 0 \pmod n, \quad \gcd(x, y) = 1\}.\]Obviously, \(S\) contains all solutions to Equation \(\eqref{eq:ellipse}\). We may partition \(S\) into subsets: Let \(\Lambda_{g} = \{\lambda: \lambda^2 \equiv -ab \pmod{n}\}\), then\[S = \bigcup_{\lambda \in \Lambda} S_{\lambda}, \quad S_{\lambda} := \{(x, y): a x \equiv \lambda y \pmod{n}, \quad \gcd(x, y) = 1\}.\]Next, we need to determine the solutions in each individual \(S_{\lambda}\).
(Lattice) To find the solutions in \(S_{\lambda}\), we do it in a superset of \(S_{\lambda}\), denoted as \(L_{\lambda}\), and check if the solution is in \(L_{\lambda}\). The superset \(L_{\lambda}\) is defined as:\[L_{\lambda} := \{(x, y): a x \equiv \lambda y \pmod{n}\}.\]It turns out that \(L_{\lambda}\) is a lattice with basis \(\mathbf{u} = (n, 0)\), \(\mathbf{v} = (\lambda a^{-1} \bmod{n}, 1)\). It's also easy to check that every point \((x, y)\) in \(L_{\lambda}\) satisfies \(ax^2 + by^2 \equiv 0 \pmod n\), thus, checking all minimizers of \(ax^2 + by^2\) in \(L_{\lambda}\) suffices to find all solutions to Equation \(\eqref{eq:ellipse}\).
(Shortest vector) As in solving Legendre equation, we can use the same lemma for 3D lattice to efficiently find the shortest vector after reducing the basis by either Lenstra–Lenstra–Lovász algorithm or Lagrange-Gauss algorithm.
Let \(\mathcal{L} = \{\mathbf{w} \mathbf{B}: \mathbf{w} \in \ZZ^2\}\) be a 2D lattice where the (row-based) basis \(\mathbf{B}\) is LLL-reduced. Then all shortest vectors in \(\mathcal{L}\) have form of \(\mathbf{w} \mathbf{B}\) where \(\mathbf{w} \in \{-1, 0, 1\}^2\).
The following is the pseudocode for the complete algorithm. We need an efficient algorithm to solve \(\lambda^2 \equiv -ab \pmod n\), which was covered before. It's also interesting to explore whether it is possible to avoid repeatly solving \(\lambda\)'s but that is left for future work. (Well, probably there won't be any future work because the other method is much easier...)
defsolve_primitive(a, b, n): """find all *primitive* solutions of (x, y) such that a x^2 + b y^2 = n""" assert gcd(a, b) == 1and gcd(b, n) == 1and gcd(a, n) == 1 assert1 <= a and1 <= b and1 <= n
dot = lambda u, v: sum(u * v * [a, b]) for λ in sqrtmod_all(-a * b, n): B = np.array([ [n, 0], [λ * invmod(a, n) % n, 1], ]) B = LLL(B, dot=dot) for w in product([-1, 0, 1], [-1, 0, 1]): if w != (0, 0) and gcd(x := w @ B) == 1and dot(x, x) == n: yield x
Method 2: The Hardy-Muskat-Williams algorithm
The most relevant paper is [HMW90], which introduces the Hardy-Muskat-Williams (HMW) algorithm. For pairwise copime integers \(a, b, n > 0\) such that \(n > a + b\), it finds all primitive solutions \((x, y)\) to \(ax^2 + by^2 = n\) with \(x \geq 1, y \geq 1\).
(Parition of solutions) Similarly, the HMW algorithm partitions the solutions using a linear congruence: For any \(0 < \lambda < n / 2\) such that \(a \lambda^2 + b \equiv 0 \pmod n\), define\[S_\lambda = \{(x, y): a x^2 + b y^2 = n, \quad \gcd(x, y) = 1, \quad x \equiv \pm \lambda y \pmod n, \quad x \geq 1, \quad y \geq 1 \},\]where pairs are unordered when \(ab = 1\), that is, \((x, y)\) and \((y, x)\) is considered the same solution when \(ab = 1\). It can be shown that \(|S_\lambda| \leq 1\).
(Structure of \(S_\lambda\)) The HMW algorithm is based on the following elegant theorem:
(Theorem 1, [HMW90]) Run the Euclidean algorithm on \(n\) and \(\lambda\) and let \(r_i\) be all the reminders:
i.e., \(r_{i+1} = r_{i-1} \bmod r_{i}\). Let \(r_k\) be the first remainder such that \(a r_k^2 < n\). If \(S_\lambda\) is not empty, then
\[S_\lambda = \{ (x, y ) \}, \quad x := r_k, \quad y:= \sqrt{(n - a r_k^2) / b}.\]
If \(ab = 1\), then \(y = r_{k+1}\).
The implementation is surprisingly easy:
defsolve_primitive(a, b, n): """find all positive primitive (unordered if a = b = 1) solutions such that ax^2 + by^2 = n""" for t in sqrtmod_all(-b * invmod(a, n) % n): if t >= n // 2: continue p, x = n, t while a * x * x >= n: p, x = x, p % x y = int(sqrt((n - a * x * x) // b)) if a * x * x + b * y * y == n: yield x, y
A special case of the HMW algorithm is Cornacchia's algorithm where \(a = 1\). [Basilla04] proved the correctness of Cornacchia's algorithm using same lattice as ours.
A failed attempt
The Brahmagupta–Fibonacci identity states that\[(x_1 + d y_1^2)(x_2 + d y_2^2) = (x_1 x_2 - n y_1 y_2)^2 + n (x_1 y_2 + x_2 y_1)^2,\]so the attempt is to find all prime factors of \(n\) and solve subproblems for each prime factors. However, not all solutions can be generated this way. A simple counterexample is \(1^2 + 5 \times 1^2 = 6\), while \(x^2 + 5 y^2 = 2\) and \(x^2 + 5 y^2 = 3\) have no solutions.
It's also related to the fact that \(\ZZ[\sqrt{ -D }]\) is not a UFD for \(D \geq 3\).
Reference
Due to an incomplete literature review, I initially overlooked a significant portion of previous work before arriving with the lattice algorithm. Later I came across Cornacchia's algorithm algorithm, which can be further generalized to the HMW algorithm.
Rathnayake13: Rathnayake T., 2013. Improving the solving process using Cornacchia [online]
Basilla04: Basilla, J.M., 2004. On the solution of \(x^2+dy^2=m\).
Deng16: Deng, X., 2016. An introduction to lenstra-lenstra-lovasz lattice basis reduction algorithm. Massachusetts Institute of Technology (MIT).
Nguyen09: Nguyen, P.Q., 2009. Hermite’s constant and lattice algorithms. In The LLL Algorithm: Survey and Applications (pp. 19-69). Berlin, Heidelberg: Springer Berlin Heidelberg.
HMW90: Hardy, K., Muskat, J.B. and Williams, K.S., 1990. A deterministic algorithm for solving \(n = fu^2 + gv^2\) in coprime integers 𝑢 and 𝑣. Mathematics of Computation, 55(191), pp.327-343.
Update @2025-03-13: Added the Hardy-Muskat-Williams algorithm.
]]><p>In this post, we're interested in finding all <em><strong>integer</strong></em> solutions of a binary quadratic equation:
<span class="math display">\[
\begin{equation}
ax^2 + by^2 = n,
\label{eq:ellipse}
\end{equation}
\]</span>
for <span class="math inline">\(a, b, n > 0\)</span>.</p>Integer and Rational Solutions of a Binary Quadratic Equation (2): Quadratic Residuehttps://www.roosephu.net/2025/03/10/Integer-Solutions-of-a-Binary-Quadratic-Equation-Quadratic-Residue/2025-03-10T07:21:17.000Z2025-03-10T07:21:17.000ZIn this post, we're interested in finding any or all integer solutions of a binary quadratic equation:\[x^2 - ny - a = 0.\]We assume that the equation is non-degenerate in the sense that \(n \neq 0\).
Normalization
We first normalize the problem to simplify its properties.
Assume \(n > 0\). Otherwise, reduce it to \(x^2 - (-n)(-y) - a = 0\).
Assume \(n = p^e\) is a prime power. Otherwise, write \(n = \prod\limits_{i=1}^k p_i^{e_i}\) and solve \(x^2 \equiv a \pmod {p_i^{e_i}}\) for each \(p_i^{e_i}\) and use CRT to combine all results.
Assume \(a \neq 0\). If \(a = 0\), all solutions are \(\{p^i \bmod p^e: e \leq 2i \leq 2 e\}\).
Assume \(p\) does not divide \(a\). Otherwise, write \(a = a' p^{e'}\) where \(p\) does not divide \(a'\).
If \(e'\) is odd, there exists no solution.
If \(e'\) is even, \(x\) must be a multiple of \(p^{e'/2}\), so reduce it to \((x/p^{e'/2})^2 \equiv a' \pmod{p^{e-e'}}\).
After normalization, the problem that we're interested in is:
For a prime \(p\), an integer \(a \neq 0\) such that \(p\) does not divide \(a\), find one or all integer \(x\) such that
\[\begin{equation}x^2 \equiv a \pmod{p^e}.\label{eq:quad-residue}\end{equation}\]
We say that \(x\) is a quadratic residue modulo \(n\) if \(x^2 \equiv a \pmod{n}\) has a solution, and a quadratic non-residue modulo \(n\) otherwise.
The case of \(p = 2\)
(Solubility) For \(p = 2\), \(a\) is a quadratic residue modulo \(p^e\) if and only if \(a \equiv 1 \pmod 8\).
To find one or all solutions, we first consider the two easy cases:
If \(e = 1\), certainly the case is \(a \equiv 1 \pmod 2\) and all solutions are \(x \equiv 1 \pmod 2\).
If \(e = 2\):
If \(a \equiv 1 \pmod{4}\), all solutions are \(x \equiv 1 \pmod 2\).
Otherwise, there is no solution.
Now we assume \(e > 2\).
(Any solution) Observe that:\[\left(x + \frac{x^2 - a}{2} \right)^2 - a = (x+1)(x^2 - a) + \left( \frac{x^2 - a}{2} \right)^2.\]So if \(x_{e-1}\) satisfies \(x^2 \equiv a \pmod{2^{e-1}}\), then \(x_e = x_{e-1} + \frac{x_{e-1}^2 - a}{2}\) satisfies \(x^2 \equiv a \pmod{2^e}\). Thus, we can begin with \(x_2 = 1\).
(All solutions) Let \(x^*\) be any solution to \(x^2 \equiv a \pmod{2^e}\). It is clear that \(x^*\) must be odd. Consider \(\frac{x^* - x}{2}\) and \(\frac{x^* + x}{2}\). The sum is \(x^*\), so only one of them is even. The product is \(\frac{x^{*2} - x^2}{4} \equiv 0 \pmod{2^{e-2}}\), thus one of them is 0 modulo \(2^{e-2}\). Thus, the set of all solutions is:\[x \in \{\pm x^*\} \pmod{2^{e-1}}.\]
The case of an odd prime \(p\)
(Solubility) The solubility is often determined by the Jacobi symbol, which is the generalization of the Legendre symbol: For an odd prime \(p\), \(a\) is a quadratic residue modulo \(p^e\) if and only if\[\left( \frac{a}{p^e} \right) = 1, \quad \left( \frac{a}{p^e} \right) := \left( \frac{a}{p} \right)^e,\]where \(\left( \frac{a}{p} \right)\) is the Legendre symbol.
Note that this does not apply to \(p = 2\) (e.g., \(x^2 \equiv 3 \pmod 4\)) or general composite numbers (e.g., \(x^2 \equiv 8 \pmod {15}\)).
(Modulo a prime (\(e=1\))) There exist well-known algorithms for finding a solution to \(x^2 \equiv a \pmod p\), e.g., the Tonelli-Shanks algorithm and Cipolla's algorithm.
(Modulo a prime power (\(e > 1\))) Let \(x'\) be any solution to \(x^2 \equiv a \pmod{p}\). Tonelli shows that\[x^* \equiv x'^{p^{e - 1}} \cdot a^{(p^e - 2 p^{e-1} + 1) / 2} \pmod{p^e},\]is a solution to Equation \(\eqref{eq:quad-residue}\).
Alternatively, Cipolla's algorithm can be extended to find a solution to Equation \(\eqref{eq:quad-residue}\) directly: Let \(u\) be any integer such that \(u^2 - a\) is a quadratic non-residue modulo \(p\), then\[x^* \equiv \frac{1}{2} a^{(p^e - 2 p^{e-1} + 1) / 2} \left(\left(u + \sqrt{u^2 - a}\right)^s + \left(u - \sqrt{u^2 - a}\right)^s \right). \pmod{p^e}\]where \(s := p^{e-1} (p+1)/2\).
Moreover, it's also possible to use the identity\[\left( \frac{x^2 + a}{2x} \right)^2 - a = \frac{(x^2 - a)^2}{4x^2}\]to generate a solution from \(x'\) by iterating it for \(\log e\) times, but a division modulo \(p^e\) takes \(O(\log p^e)\) time so it won't be faster.
(All solutions) Similarly, let \(x^*\) be any solution to \(x^2 \equiv a \pmod{p^e}\). Since \((x^* - x)(x^* +x) \equiv 0 \pmod {p^e}\), and at most one of \(x^* - x\) and \(x^* + x\) is a multiple of \(p\), we conclude that all solutions are\[x \in \{\pm x^*\} \pmod{p^e}.\]
References
Tonelli-Shanks algorithm
Cipolla's algorithm
[Dickson]: Dickson, L.E., 1919. History of the Theory of Numbers (No. 256). Carnegie Institution of Washington.
]]><p>In this post, we're interested in finding any or all <strong><em>integer</em></strong> solutions of a binary quadratic equation:
<span class="math display">\[
x^2 - ny - a = 0.
\]</span>
We assume that the equation is non-degenerate in the sense that <span class="math inline">\(n \neq 0\)</span>.</p>Integer and Rational Solutions to a Binary Quadratic Equation (1): Legendre equationhttps://www.roosephu.net/2025/03/08/Rational-Solutions-to-a-Binary-Quadratic-Equation/2025-03-08T12:38:17.000Z2025-03-08T12:38:17.000ZIn this post, we're interested in finding any or all rational solutions of the following equation:\[ax^2 + by^2 + c = 0,\]or alternatively, all integer solutions to the Legendre equation:\[\begin{equation}ax^2 + by^2 + cz^2 = 0,\label{eq:legendre}% \tag{Legendre Equation}\end{equation}\]where \(abc \neq 0\).
Legendre equation and normal form
(Normalization) We say the Legendre equation is in normal form if \(abc\) is square-free. The following transformation by Gauss reduces a Legendre equation to normal form: Write \(bc = \alpha^2 s, ca=\beta^2 t, ab=\gamma^2 u\) for square-free integers \(s, t, u\) and integers \(\alpha, \beta, \gamma\), then\[\frac{a \alpha}{\beta \gamma} (x/\alpha)^2 + \frac{b \beta}{\gamma \alpha} (y/\beta)^2 + \frac{c \gamma}{\alpha \beta} (z/\gamma)^2 = 0,\]and \(\frac{a \alpha}{\beta \gamma}, \frac{b \beta}{\gamma \alpha}, \frac{c\gamma}{\alpha \beta}\) are square-free and pair-wise coprime.
(Solubility) Legendre gave an elegant theorem to determine the solubility of an equation in normal form:
(Legendre) Equation \(\eqref{eq:legendre}\) in normal form has a non-trivial integer solution iff. \(a, b, c\) have different sign, and \(-bc, -ca, -ab\) are quadratic residues modulo \(a, b, c\) respectively.
The if direction follows by considering modulo \(a, b, c\) individually.
Find a single non-trivial integer solution
Brute-force
Holtz proved that there exists a small non-trivial solution:
(Small solution, Holzer) If Equation \(\eqref{eq:legendre}\) in normal form has a non-trivial solution, then there exists a non-trivial solution \((\hat x, \hat y, \hat z)\) such that:
Additionally, \(|bc|\) is a square if and only if \(|bc| = 1\), since \(bc\) is square-free. The same holds for \(|ca|\) and \(|ab|\).
See [Mordell, Chapter 7, Theorem 5], [CR03] and [CM97] for more details.
Legendre's descent
(This subsection is mostly taken from [Aitken, Section 5].)
Apparently, \(ax^2 + by^2 + cz^2 = 0\) is equivalent to \((-ac) x^2 + (-bc) y^2 = (cz)^2\). Legendre's descent is based on the following observation:\[ax^2 + by^2 = z^2 \quad \Longleftrightarrow \quad a ( ux + z)^2 + b' (b y)^2 = (uz + ax)^2.\]for \(u^2 = a + b'b\). It allows us to reduce the problem \(ax^2 + by^2 = z^2\) to another problem \(ax^2 + b' y^2 = z^2\). When \(|b| \geq \max(2, |a|)\), the other problem is smaller as we can always choose \(|u| \leq |b|/2\) so \(|b'| \leq (u^2 + |a|) / |b| \leq |b|/4+1 < |b|\).
The algorithm runs as follows:
# given a, b, find (x, y, z) such that a x^2 + b y^2 = z^2 defsolve(a, b): assert is_squarefree(a) and is_squarefree(b) assert a >= 0or b >= 0
ifabs(a) > abs(b): y, x, z = solve(b, a) return (x, y, z)
if a == 0: return (1, 0, 0) if b == 1: return (0, 1, 1) if a == 1: return (1, 0, 1)
# find u such that u^2 == a (mod b) and -b / 2 <= u < b / 2 # WHAT IF u DOES NOT EXIST? u = sqrtmod(a, b) u = min(u, b - u) b2 = (u * u - a) // b
# b2 = sqf_b2 * sq_b2^2 where sqf_b2 is square-free sqf_b2, sq_b2 = squarefree(b2)
x2, y2, z2 = solve(a, sqf_b2) return ( u * x2 + z2, sqf_b2 * sq_b2 * y2, u * z2 + a * x2, )
The key point of the algorithm is that \(u\) must exist during all recursions. It's proved that if \(ax^2 + by^2 + cz^2 = 0\) is in normal form, \(u\) always exists when calling \(\operatorname*{solve}(-ac, -bc)\).
See [CR03, Section 2.3], [Aitken] and [Hillgarter] for more details. [CR03] also optimizes the method but it's beyond the scope of the post.
Lattice-based algorithm
(This section is taken from [CR03, Section 2.6].)
Another idea arises when going deeper in Legendre's theorem.
We first discuss the case where any of \(|ab|, |bc|, |ca|\) is 1. W.L.O.G. we assume \(|ab| = 1\).
If \(a \neq b\), then \((x, y, z) = (1, 1, 0)\) is a non-trivial solution.
If \(a = b\), we can use Cornacchia's algorithm to solve \(x^2 + y^2 = -c / a\), and \((x, y, 1)\) is a non-trivial solution.
Now we consider the general case. We define a set of points \(\mathcal{L} \subset \ZZ^3\):\[\begin{aligned}\mathcal{L} = \{ (x, y, z): \ & by \equiv \lambda_a z \pmod{|a|}, \\& cz \equiv \lambda_b x \pmod{|b|}, \\& ax \equiv \lambda_c y \pmod{|c|},\\& a x^2 + b y^2 + c z^2 \equiv 0 \pmod{|2abc|}\},\end{aligned}\]
where \(\lambda_a\) is an arbitrary solution to \(\lambda^2 \equiv -bc \pmod{|a|}\), similarly for \(\lambda_b\) and \(\lambda_c\). Let \((\bar x, \bar y, \bar z)\) be the non-zero point in \(\mathcal{L}\) minimizing \(|a| x^2 + |b| y^2 + |c| z^2\). It can be proved that there is a small solution \((\hat x, \hat y, \hat z)\) in \(\mathcal{L}\), which can be used to bound \(|a\bar x^2 + b\bar y^2 + c \bar z^2|\):\[\left |a \bar x^2 + b \bar y^2 + c \bar z^2 \right| \leq |a| \hat x^2 + |b| \hat y^2 + |c| \hat z^2 < 2|abc|.\]so we conclude that \(a \bar x^2 + b \bar y^2 + c \bar z^2 = 0\) and the minimizer itself satisfies Holzer's condition.
The final step is to determine the minimizer. [CR03, Section 2.6, p.17] shows that \(\mathcal{L}\) is a lattice, that is, \(\mathcal{L} = \{\mathbf{w} \mathbf{B}: \mathbf{w} \in \ZZ^3 \}\) for a (row-based) basis \(\mathbf{B} \in \ZZ^{3 \times 3}\). It also gives an explicit expression for a basis \(\mathbf{B}\). Moreover, the shortest vector in a 3D lattice can be found efficiently using LLL, as long as we have an arbitrary basis \(\mathbf{B}\) for \(\mathcal{L}\):
([CR03, Lemma 2.7, p.16]) Let \(\mathcal{L} = \{\mathbf{w} \mathbf{B}: \mathbf{w} \in \ZZ^3\}\) be a 3D lattice where \(\mathbf{B}\) is LLL-reduced. Then the shortest vector in \(\mathcal{L}\) is in the form of \(\mathbf{w}\mathbf{B}\) where \(\mathbf{w} \in \{-1, 0, 1\}^3\).
defget_B(): λ_a = sqrtmod(-b * c, a) λ_b = sqrtmod(-a * b, c) λ_c = sqrtmod(-a * c, b)
u, a_2 = invmod(b, c), invmod(a, b * c) v, b_2 = (1 - u * b) // c, (1 - a * a_2) // (b * c)
α = b_2 * c * λ_a % a β = u * a_2 * b * λ_b % (b * c) γ = v * a_2 * c * λ_c % (b * c)
# the basis of {(x, y, z): a x^2 + b y^2 + c z^2 = 0 mod abc} B1 = np.array([ [b * c, 0, 0], [a * β, a, 0], [α * β + γ, α, 1], ])
# the basis of {(x, y, z): a x^2 + b y^2 + c z^2 = 0 mod 2abc} eps = lambda v: f(v) // (a * b * c) % 2 r = next(r for r in B1 if eps(r) != 0) return np.array([v if eps(v) == 0else v + r for v in B])
B = LLL(get_B(), dot=lambda u, v: sum(u * v * np.abs([a, b, c])))
for w in product([-1, 0, 1], [-1, 0, 1], [-1, 0, 1]): if w != (0, 0, 0) and f(v := w @ B) == 0: return v
# unreachable!
Check out [CR03, Section 2.6] and [CM97] for more details. [CM97] uses a different lattice and [CR03] improves upon it.
A single non-trivial rational solution to \(\mathbf{x}^\top Q \mathbf{x} = 0\) with \(Q \in \ZZ^{d \times d}\)
Assume \(\det Q \neq 0\) and \(Q\) is symmetric. We here only list a few references below:
[Simon05] removes the need of diagonalization, which often bloats the size of coefficients. It gives an algorithm where \(d = 3\). It works by reducing \(Q\) repeatedly until \(\det Q = \pm 1\), and then solving the case where \(\det Q = \pm 1\).
[Simon06] extends [Simon05] and gives an algorithm for \(d > 3\). However, I have not read the paper.
All non-trivial solutions from a single non-trivial solution
Given a single non-trivial solution \(\mathbf{\hat x} \in \QQ^d\) to a quadratic equation \(F(\mathbf{x}) = 0\), we parametrize \(\mathbf{x} \in \QQ^d\) by \(\mathbf{x} = \mathbf{\hat x} - t\mathbf{u}\), where \(\mathbf{u} \in \ZZ^d, t \in \QQ\) and \(\mathbf{u}\) is primitive (i.e., \(\gcd(\mathbf{u}) = 1\)).
Since \(F(\mathbf{\hat x}) = 0\), we solve \(t\) in \(F(\mathbf{\hat x} - t\mathbf{u}) = 0\) using the second-order Taylor expansion of \(F\) at \(\mathbf{\hat x}\):\[F(\mathbf{\hat x}) - t \mathbf{u}^\T \nabla F(\mathbf{\hat x}) + \frac{t^2}{2} \mathbf{u}^\T \nabla^2 F(\mathbf{\hat x}) \mathbf{u} = 0 \quad \Longrightarrow \quad t = \frac{\mathbf{u}^\T \nabla F(\mathbf{\hat x})}{\frac{1}{2} \mathbf{u}^\T \nabla^2 F(\mathbf{\hat x}) \mathbf{u}}.\]Since \(\mathbf{u}\) and \(-\mathbf{u}\) produce the same solution, we require that \(\mathbf{u}\) is lexicographically larger than \(-\mathbf{u}\). Note this parametrization also generates the initial solution \(\mathbf{\hat x}\) when \(\mathbf{u}^\top \nabla F(\mathbf{\hat x}) = 0\), i.e., \(\mathbf{u}\) is tangent to the curve.
Connection to integer solutions of a homogenous ternary quadratic equation
A homogeneous polynomial \(F(\mathbf{x})\) of degree \(d\) satisfies \(F(\lambda \mathbf{x}) = \lambda^d F(\mathbf{x})\), so every rational solution can be normalized to a primitive integer solution. Furthermore, any primitive integer solution \(\mathbf{x}\) with \(\mathbf{x}_1 \neq 0\) is equivalent to a rational solution of a heterogeneous equation \(F([1, \mathbf{x'}]) = 0\) for \(\mathbf{x}' \in \QQ^{d-1}\).
When the goal is to enumerate integral solutions, a good strategy is to first find a non-trivial rational solution to \(F(\mathbf{x}) = 0\) and then apply the parametrization trick to a heterogeneous equation to avoid duplicates.
Consider the following example. We consider the homogenous equation \(ax^2+bxy+cy^2=d z^2\) in \(\ZZ^3\) with a non-trivial primitive solution \((x_0, y_0, z_0)\) where \(z_0 \neq 0\), which is equivalent rational solutions to the heterogenous equation \(a x^2 + b xy + c y^2 - d = 0\) with a non-trivial solution \(\mathbf{\hat x} = (x_0 / z_0, y_0 / z_0)\). Write \(\mathbf{u} = [u, v]\) and apply the results above:\[\begin{aligned}\mathbf{u}^\top \nabla F(\mathbf{\hat x}) & = z_0^{-1}(u (2ax_0 + b y_0) + v(b x_0 + 2c y_0) ), \\\frac{1}{2} \mathbf{u}^\T \nabla^2 F(\mathbf{\hat x}) \mathbf{u} & = a u^2 + b uv + c v^2. \\\end{aligned}\]After simplification, we have\[\begin{pmatrix}x \\ y \\ z\end{pmatrix} =A \mathbf{r}, \quadA := \begin{pmatrix}ax_0 + by_0 & 2cy_0 & -cx_0 \\-ay_0 & 2ax_0 & c y_0 + b x_0 \\a z_0 & b z_0 & c z_0 \\\end{pmatrix}, \quad \mathbf{r} := \begin{pmatrix}u^2 \\ uv \\ v^2\end{pmatrix},\]which are exactly the Equations (8-10) in Wolfram MathWorld. The equations might not generate primitive solutions directly, but they do produce all primitive tuples after normalization.
To use the equations to generate bounded primitive solutions, the common factor \(g := \gcd(A \mathbf{r})\) needs to be determined. It can be shown that \(g\) divides \(\det A\) by writing \(\mathbf{r}\) using \(A \mathbf{r}\):\[\mathbf{r} = A^{-1} A \mathbf{r} = \frac{\adj A}{\det A} A \mathbf{r},\]and observing that if \(g\) doesn't divide \(\det A\), \(\gcd(\mathbf{r})\) can't be 1.
In our case, \(\det A = -\Delta d z_0^3\) where \(\Delta = b^2 - 4ac\). The bound can be slightly improved to \(\Delta d z_0^2\) by finding the LCM of denominators in \(A^{-1}\):\[A^{-1} = \frac{1}{\Delta d z_0^2}\begin{pmatrix}\Delta x_0 + cT & cS & -2cdz_0 \\-aS & -cT & b d z_0 \\aT & \Delta y_0 + a S & -2adz_0\end{pmatrix},\quad\begin{cases}S := 2cy_0+bx_0, \\T := 2ax_0+by_0.\end{cases}\]If \(b = 0\), the denominator can be further reduced to \(2acdz_0^2\) instead of \(4acdz_0^2\).
References
Rathnayake13: Rathnayake T., 2013. Solving Ternary Quadratic forms by descent method [online]
CR03: Cremona, J. and Rusin, D., 2003. Efficient solution of rational conics. Mathematics of Computation, 72(243), pp.1417-1441.
Simon05: Simon, D., 2005. Solving quadratic equations using reduced unimodular quadratic forms. Mathematics of Computation, 74(251), pp.1531-1543.
Simon06: Simon, D., Quadratic equations in dimensions 4, 5 and more. Preprint, 2005 [online]
CM97: Cochrane, T. and Mitchell, P., 1998. Small solutions of the Legendre equation. Journal of Number Theory, 70(1), pp.62-66.
Aitken: Wayne Aitken, Legrendre's Theorem, Legrange's Descent
Hillgarter: Hillgarter, E., 1996. Rational points on conics. na.
Holzer: Holzer, L., 1950. Minimal solutions of diophantine equations. Canadian Journal of Mathematics, 2, pp.238-244.
It's weird that I can't find the first name of Holzer...
]]><p>In this post, we're interested in finding any or all <strong><em>rational</em></strong> solutions of the following equation:
<span class="math display">\[
ax^2 + by^2 + c = 0,
\]</span>
or alternatively, all <strong><em>integer</em></strong> solutions to the Legendre equation:
<span class="math display">\[
\begin{equation}
ax^2 + by^2 + cz^2 = 0,
\label{eq:legendre}
% \tag{Legendre Equation}
\end{equation}
\]</span>
where <span class="math inline">\(abc \neq 0\)</span>.</p>六个解谜游戏https://www.roosephu.net/2025/01/05/6-puzzle-games/2025-01-06T05:15:44.000Z2025-01-06T05:15:44.000Z转眼间 2024 年快已经过去了……隔了好久没写娱乐相关的文章了，随便写几个作为今去年的年终总结得了……（然后震惊地发现我上次写游戏已经是四年前了……）
Return of the Obra Dinn
The Case of the Golden Idol 和 The Rise of the Golden Idol
Patrick's Parabox
COCOON
ANIMAL WELL
（可能有略微剧透，手动狗头）
Return of the Obra Dinn
早就听说过这款游戏的大名。作者 Lucas Pope 之前开发的 Papers, Please 让他一举成名顺便赚了不少钱可以没有后顾之忧的做新的游戏。
在这个游戏里，你扮演一个侦探，任务在 Obra Dinn 这艘船上寻找线索，判断出每个人各是谁，死因是什么。这种侦探游戏的难点是如何设计线索。我之前玩过 The Room 系列或者 Gorogoa，玩法是一个线性叙事，不停和物品交互获取新的线索，揭开一个谜题后开始下一个，点点点式的交互才是玩法核心，没啥推理成分。Return of the Obra Dinn 则不同，它给你的主要线索是若干个快照，一个快照就是一个静态场景，你可以仔细看，但是没法交互，所以不存在那种哪个地方没点到就解锁不了下一关的痛苦。而且 Return of the Obra Dinn 是一个非线性叙事，你不需要任何推理结果就找到所有线索，剩下的就是推理了，推理才是游戏的玩法核心。
The Case of the Golden Idol 和 The Rise of the Golden Idol
这俩游戏是同一个系列的两款作品，The Case of the Golden Idol 是一代，The Rise of the Golden Idol 是二代，玩法一样，但是二代优化了用户体验。
这俩游戏玩法和 Return of the Obra Dinn 类似，但是谜题的颗粒度不同。Gorogoa 非常 fine-grained，一个小谜题就一分钟；Return of the Obra Dinn 非常 coarse-grained，整个游戏就一个大谜题；the Golden-Idol 系列则取了一个中间，10-20min 一个。对于一个谜题，the Golden Idol 系列和 Return of the Obra Dinn 类似，你需要找到线索，直到游戏告诉你找到了所有线索，剩下的就是推理了。当然，推理的过程中你也需要不停回去看各个场景。另一个不同点是解谜要求不同：Return of the Obra Dinn 要求填补 这张脸 的名字是 谁，死于 死因，不要求你补完故事，而 the Golden Idol 系列需要填补剧情，这张脸 是 谁谁谁，谁谁谁 做了 什么事 ，所以 谁谁谁 做了 什么事，从而 什么东西怎么怎么 了。官方还有 网页试玩版 让你体验具体玩法。要比较 the Golden Idol 系列和 Return of the Obra Dinn 的话，我还是更喜欢 the Golden Idol 系列一些，可能是我玩 Return of the Obra Dinn 时解谜不顺吧，解完了感觉缺了一堆细节不清楚……但是我还是得承认像 Return of the Obra Dinn 一样设计一个特别大的谜题是一件特别难的事。
]]><p>转眼间 2024 年<del>快</del>已经过去了……隔了好久没写娱乐相关的文章了，随便写几个作为<del>今</del>去年的年终总结得了……（然后震惊地发现我上次写游戏已经是四年前了……）</p>
<ul>
<li><span class="exturl" data-url="aHR0cHM6Ly9vYnJhZGlubi5jb20v">Return of the Obra Dinn</span></li>
<li><span class="exturl" data-url="aHR0cHM6Ly93d3cudGhlZ29sZGVuaWRvbC5jb20v">The Case of the Golden Idol</span> 和 <span class="exturl" data-url="aHR0cHM6Ly93d3cucmlzZW9mdGhlZ29sZGVuaWRvbC5jb20v">The Rise of the Golden Idol</span></li>
<li><span class="exturl" data-url="aHR0cHM6Ly93d3cucGF0cmlja3NwYXJhYm94LmNvbS8=">Patrick's Parabox</span></li>
<li><span class="exturl" data-url="aHR0cHM6Ly93d3cuY29jb29uZ2FtZS5jb20v">COCOON</span></li>
<li><span class="exturl" data-url="aHR0cHM6Ly93d3cuYW5pbWFsd2VsbC5uZXQv">ANIMAL WELL</span></li>
</ul>
<p>（可能有略微剧透，手动狗头）</p>论 First Principlehttps://www.roosephu.net/2024/12/30/on-First-Principles/2024-12-30T09:10:47.000Z2024-12-30T09:10:47.000Z前几天和 hls bibi 了一下。hls 推荐了 这个视频 给我。我是以一个批判的角度看视频的，就像我之前 review paper 一样，先把屁股坐住了再来找论据，所以大部分观点都是 biased 的，可能就是在 nitpicking。
先抛出我对 First Principle 的理解：在需要 innovate 的场合，挑选合适的 assumption 来定义问题，然后 end-to-end 来解决问题。
如何定义问题？（assumption 的合理性）
First Principle 说，对于解决一个问题，我们应该从基本公理出发来思考，而不是从类比、经验来思考：
[…] You kind of boil things down to the most fundamental truths, and say ok what do we sure is true or as sure as possible is true, and reason up from there. […]
[…] people take too many things as they assume too many things to be true without a sufficient basis in that belief. It's very important that people closely analyze what is supposed to be true […]
[…] the second thing is try to delete the whatever the step is, the part or the process step […]
就是我们有了一个 solution 后，试图去减少其中的步骤来去除冗余。
视频里 Elon explicitly 提到了一个 anti-pattern:
[…] through most of our life, we get through life by reasoning by analogy, which essentially means kind of copying what other people do with slight variations. […]
[…] don't just follow the trend […]
这句话可以 apply 到一件具体的事上。那如果我们直接把它 apply 到 First Principle 自己呢？如果我们 get through our life by applying First Principle with slight variations 呢？例如，大部分人思考问题都是 hierarchical thinking，Elon 他思考方式会有什么本质不同的 variation 吗？不同粒度的 hierarchical thinking 算不算 slight variation？Elon 会直接 end-to-end thinking，跑一个全局最短路吗？
[…] When you want to do something new, you have to apply the physical approach. The physics is about how to discover new things that are counterintuitive. […]
[…] if you're trying to do something new, it's the best way to think. […]
只是，我们需要在所有地方 innovate 吗？
视频中提到：
[…] they'll say we do that because it's always done that way, or they'll not do it because nobody's ever done that way so it must not be good […]
it's always done that way 或者 nobody's ever done that way 这两句话确实不太行，但是为此否定 prior experience 就太草率了。我更加关心的问题是：没人做过，是因为有 strong reason 大家不做，还是没人想到在这里 innovate，还是有人做了我们不知道？别人做出过哪些尝试，有哪些经验教训可以汲取？如果无法回答这些问题，盲目地 assume 我们不应该 follow 别人的做法，在不懂的地方强行 innovate，难免会重复别人踩过的坑。难道真的是人类从历史中学到的唯一教训就是人类不能从历史中学到任何教训？
In conclusion，我不认为 First Principle is nonsense。我觉得 First Principle 有一点是有道理的：不要被 convention 束缚住，因为一件事是 convention 就认为它好。但是只想着 First Principle 就是 nonsense 了。而且，即使 First Principle 了，或许还可以更加 First Principle。用 First Principle 来分析是需要 mental energy，需要时间去研究，这都是有限的资源，所以我们需要把 First Principle 留在关键的地方，ROI 大的地方。
Bonus：请用 First Principle 来分析 First Principle 的合理性。
]]><p>前几天和 hls bibi 了一下。hls 推荐了 <span class="exturl" data-url="aHR0cHM6Ly93d3cueW91dHViZS5jb20vd2F0Y2g/dj1GNVdaQ3BUTkVDOA==">这个视频</span> 给我。我是以一个批判的角度看视频的，就像我之前 review paper 一样，先把屁股坐住了再来找论据，所以大部分观点都是 biased 的，可能就是在 nitpicking。</p>The Optimal Order to Purchase Vampire Survivors Power Upshttps://www.roosephu.net/2024/10/20/optimal-order-to-purchase-Vampire-Survivors-Power-Ups/2024-10-21T02:52:13.000Z2024-10-21T02:52:13.000Z
The cost for Power Ups has changed since version 0.7, so this post is no longer applicable.
\(\newcommand{\eps}{\varepsilon} \newcommand{\R}{\mathbb{R}}\)Vampire Survivors is a game released in early access near the end of 2021. It received 110k recommendations in one month, which is pretty amazing given its minimal aesthetics and seemingly low production costs. I also played it a lot back then, but I won't talk about its gameplay here. Instead, this post discusses the (old) Power Up mechanism and some of the math behind it.
The Power Up mechanism in Vampire Survivors is similar to Kingdom Rush's Upgrades, where you spend the gold earned during gameplay to make permanent gains (e.g., increase move speed by x%). As per the notes in Fandom, the cost for a Power Up is defined as:\[Price = \text{InitialPrice} \cdot (1 + \text{Bought}) \cdot \left(1 + \frac{\text{TotalBought}}{10} \right),\]where \(\text{InitialPrice}\) is the initial price of the Power Up, \(\text{Bought}\) is the number of purchased ranks for this Power Up, and \(\text{TotalBought}\) is the total number of purchased ranks among all Power Ups. What's interesting is that the order in which you buy Power Ups actually matters!
Intuitively, we should purchase expensive Power Ups first. For example, IGN recommends maxing out Power Ups with the highest initial costs first. However, this is not optimal. Suppose we have two Power Ups, A and B. A has 5 ranks with an initial price of 1, while B has only 1 rank with an initial price of 2. If we max out A first, the total price is:\[1 \times 1 + 1.1 \times 2 + 1.2 \times 3 + 1.3 \times 4 + 1.4 \times 5 + 1.5 \times 2 = 22,\]and if we max out B first, the total price is:\[1 \times 2 + 1.1 \times 1 + 1.2 \times 2 + 1.3 \times 3 + 1.4 \times 4 + 1.5 \times 5 = 22.5.\]
So, here is our central topic today:
What is the optimal order for buying all Power Ups?
A General Formulation and Its Hardness
The first attempt is to formulate the problem as a directed acyclic graph (DAG). More specifically, each rank \(j\) of a certain Power Up \(i\) is a node \(v_{i, j}\), and it has a weight defined as\[w(v_{i, j}) = \frac{1}{10} \text{InitialPrice}_i \cdot (1 + j),\]with edges \(v_{i, j-1} \to v_{i, j}\), representing that \(v_{i, j}\) depends on \(v_{i, j - 1}\). A valid order of purchasing Power Ups is a topological sort of all nodes, and our goal is to find the optimal topological sort. We may notice that the graph has a special property (i.e., it only contains chains), but for now, let's focus on a general graph.
Weighted Topological Sort
Given a directed acyclic graph \(G = (V, E)\) and a node weight function \(w: V \to \mathbb{N}\), find a topological ordering \(v_{i_1}, \dots, v_{i_n}\) to minimize \(\sum\limits_{i=1}^n i w(v_{a_i}).\)
An equivalent objective can also be defined in terms of completion time\(c(v)\). The completion time of a specific node \(v'\) in an ordering \(v_{a_1}, \dots, v_{a_n}\) is the \(i\) such that \(v' = v_{a_i}\), so the objective can also be written as\[\sum_{i} i w(v_{a_i}) = \sum_{v \in V} c(v) w(v).\]
One might try to solve the general problem but may struggle to make significant progress. The reason is simple: it's NP-hard! We present a proof, which is essentially a combination of Theorem 4.15 and Exercise 4.19 from Elements of Scheduling with (very) minor modifications.
Proof of NP-hardness of Weighted Topological Sort
We start with Linear Arrangement, an NP-hard problem, which will be defined below and used for reduction.
Linear Arrangement
Given an undirected graph \(G = (V, E)\), find a one-to-one mapping \(f: V \to [n]\) to minimize \(\sum_{(u, v) \in E} |f(u) - f(v)|\).
One can easily show that Linear Arrangement is equivalent to minimizing \(\sum_{(u, v) \in E} \max(f(u), f(v))\), as \(|f(u) - f(v)| = 2 \max(f(u), f(v)) - f(u) - f(v)\). We now reduce the Linear Arrangement instance to a Weighted Topological Sort instance.
Let \(K = |E|^2\). Construct a graph \(G' = (V', E')\) from \(G = (V, E)\) as follows:
For each vertex \(v_i \in V\):
Add \(K\) nodes \(V_i'\) to \(V'\): \(v_i^{(j)}\) for \(j \in \{1, \dots, K\}\);
Add \(K - 1\) dependencies: \(v_i^{(j)} \to v_i^{(j + 1)}\) for \(j \in \{1, \dots, K - 1\}\).
For each edge \(e = (v_i, v_j) \in E\):
Add \(e\) to \(V'\);
Add two dependencies: \(v^{(K)}_i \to e\) and \(v^{(K)}_j \to e\).We also define the vertex weight as \(w(v_{i}^{(j)}) = 0\) and \(w(e) = 1\) for each edge \(e\).
Let \(OPT\) be the optimal \(\sum_{(u, v) \in E} \max(f(u), f(v))\) in the Linear Arrangement instance, and let \(f^*\) be the corresponding mapping. Similarly, let \(OPT'\) be the minimum total cost in the constructed instance, and let \(v'^*\) be the corresponding solution.
We first show that \(OPT' < K \cdot OPT + K\). We determine the order of \(V'_i\) as \[\left[V'_{f^{*-1}(1)}, V'_{f^{*-1}(2)}, \dots, V'_{f^{*-1}(n)} \right]\] and then insert all edges \(e \in E\) as early as possible. Since at most \(|E|\) edges are inserted, the completion time for an edge \(e\) is bounded by \(c(e) < K \max(f^*(u), f^*(v)) + |E|\), which gives a total cost of\[OPT' < \sum_{e = (u, v) \in E} \left(K \max(f^*(u), f^*(v)) + |E| \right) = K \cdot OPT + K.\]
Now, we construct an optimal solution \(f\) from \(v'^*\). We consider the subsequence \(\left\{v_{a_i}^{(K)} \right\}_i\) in \(v'^*\) and define \(f\left(v_{a_i} \right) = i\). The completion time \(c(e)\) for an edge \(e = (u, v)\) has a lower bound of \(c_e > K \max(f(u), f(v))\), which in turn provides an upper bound on \(\sum_e \max(f(u), f(v))\):\[OPT \leq \sum_{(u, v) \in E} \max(f(u), f(v)) < \frac{1}{K} \sum_{e \in E} c_e = \frac{1}{K} OPT' < OPT + 1.\]Note that \(f\) maps nodes to integers, so the first inequality must be an equality, which means \(f\) is an optimal mapping for the Linear Arrangement instance.
The Optimal Order for Buying Power Ups?
The result above is definitely frustrating. However, we might want to take a step back. Remember that our goal is only to find the optimal order to buy Power Ups? It's sufficient to consider only chains instead of a general graph.
Weighted Topological Sort for Chains
Given a directed acyclic graph \(G = (V, E)\) and a node weight function \(w: V \to \mathbb{N}\), find a topological ordering \(v_{i_1}, \dots, v_{i_n}\) to minimize \(\sum\limits_{i=1}^n i w(v_{a_i}).\)The graph \(G\) contains only \(k\) chains:\[\begin{cases}V = \{v _{i, j}\} _{i \in \{1, \dots, k\}, j \in \{1, \dots, n_i\}}, \\E = \{v _{i, j} \to v _{i, j + 1}\} _{i \in \{1, \dots, k\}, j \in \{1, \dots, n_i - 1\}}.\end{cases}\]
It turns out that there is an efficient (polynomial time) algorithm!
A polynomial algorithm for Weighted Topological Sort for Chains
Let \(v^*\) be the optimal ordering. Let a segment be an interval of nodes from the same chain. Since \(v^*\) is optimal, swapping adjacent segments that come from different chains won't improve the objective. This can be formulated as: let \(v_{i_1, \{l_1, \dots, r_1\}}\) and \(v_{i_2, \{l_2, \dots, r_2\}}\) be adjacent segments where \(i_1 \neq i_2\), then:\[\sum_{j=l_1}^{r_1} (j - l_1) w(v_{i_1, j}) + \sum_{j=l_2}^{r_2} (j - l_2 + r_1 - l_1 + 1) w(v_{i_2, j}) \leq \sum_{j=l_2}^{r_2} (j - l_2) w(v_{i_2, a}) + \sum_{j=l_1}^{r_1} (j - l_1 + r_2 - l_2 + 1) w(v_{i_1, j}),\]After simplification, it becomes:\[\frac{\sum\limits_{a=l_2}^{r_2} w(v_{i_2, a})}{r_2 - l_2 + 1} \leq \frac{\sum\limits_{a=l_1}^{r_1} w(v_{i_1, a})}{r_1 - l_1 + 1}.\]
We define the weight of a segment \(w(v_{i, \{l, \dots, r\}}) := \frac{1}{r - l + 1} \sum\limits_{j=l}^r w(v_{i, j})\), so the inequality above simply means that \(w(v_{i_1, \{l_1, \dots, r_1\}}) \geq w(v_{i_2, \{l_2, \dots, r_2\}})\), and all maximal segments in \(v^*\) are sorted in descending order.
Now, let us consider the chain \(V_i\). We start with \(n_i\) chains, each containing only one node. Now we're trying to merge adjacent nodes: If there exist two adjacent segments \(v_{i, \{a, \dots, b\}}\) and \(v_{i, \{b + 1, \dots, c\}}\) such that $w(v_{i, {a, , b}}) < $ \(w(v_{i, \{b + 1, \dots, c\}})\), we can merge them into one segment \(v_{i, \{a, \dots, c\}}\). The reason is: If there is another segment \(v_{i', \{l', \dots, r'\}}\), one of the following must be true:\[w(v_{i', \{l', \dots, r'\}}) > w(v_{i, \{a, \dots, b\}}), \quad \text{or} \quad w(v_{i', \{l', \dots, r'\}}) < w(v_{i, \{b + 1, \dots, c\}}),\]which contradicts our previous observation. The process of merging can be summarized in the following pseudocode:
>defmerge_segments(chain): segments = [] for v in chain: cur_seg = [v] while segments and w(segments[-1]) < w(cur_seg): cur_seg = segments.pop() + cur_seg segments.append(cur_seg) return segments
After merging segments, all segments in chain \(i\) are sorted in descending order of \(w\). This naturally leads to a greedy algorithm: The best candidate in each chain will be the current first segment, so we don't have any dependency issues!
>defoptimal_order(chains): segments = [seg for chain in chains for seg in merge_segments(chain)] segments = sorted(segment, key=w) return [v for seg in segments for v in seg]
The code above is inefficient in its representation of segments and can be easily optimized to \(O(n \log n)\), where \(n = |V|\).
How About a General Completion Time Weight Function?
We have considered the objective in the form of \(\sum_{v} c(v) w(v) = \sum_{v} \mathnormal{Id}(c(v)) w(v)\), where \(Id\) is the identity function. Is it possible to generalize this to a more complex completion time weight function \(f: \mathbb{N} \to \R\) instead of the identity function?
Unfortunately, this is also NP-hard, even if the graph contains only chains. We will reduce the 3-Partition problem to it.
3-Partition
Let \(S\) be a multiset containing \(3n\) positive integers. Can we partition \(S\) into \(n\) multisets, each containing 3 elements with a sum \(s = \frac{1}{n} \sum\limits_{e \in S} e\)?
Note that 3-Partition is strongly NP-complete: it's NP-complete even when all numbers are bounded by a polynomial of \(n\).
Proof of NP-hardness of General Completion Time Weighted Topological Sort
Given an instance of the 3-Partition problem (where we assume every element in \(S\) is strictly greater than 1), we construct the following problem:
Instance of General Completion Time Weighted Topological Sort
The graph \(G\) contains \(3n\) chains, where the \(i\)-th chain \(V_i\) has \(S_i\) nodes.
The weight function \(w\) and completion time weight function \(f\) are:\[w(v_{i, j}) = \begin{cases} 1 & j = 1 \\ 0 & j = S_i \\ 2 & \text{otherwise} \end{cases}, \quad f(t) = \begin{cases} 1 & t \bmod s \in \{1, 2, 3\} \\ 2 & t \bmod s \in \{0, s - 1, s - 2\} \\ 0 & \text{otherwise} \end{cases}\]
From the Rearrangement Inequality, \(3n\) is a lower bound of the objective, and it is only attained if \(c(v_{i, 1}) \bmod s \in \{1, 2, 3\}\) and \(c(v_{i, S_i}) \bmod s \in \{0, s - 1, s - 2\}\) for all chains \(i\). In this case, the optimal ordering can be split into \(n\) intervals evenly, each consisting of 3 complete chains, which means the 3-Partition instance is satisfiable.
Are Chains and Identity Completion Time Weight Function the Limit?
We found an efficient algorithm for Weighted Topological Sort for Chains, but it failed for two extensions:
A general graph;
A general completion time weight function,
both of which have been proven NP-hard. So what is the right problem to focus on for a successful generalization?
Section 4.3 of Elements of Scheduling shows that the following extensions still have polynomial time algorithms:
A series-parallel graph;
A cost function with a preference order on sequences.
We refer readers to the book for more details.
Notes
In fact, this post has been in draft for more than a year. The reason I didn’t post it earlier is that… I realized I was completely wrong when it was almost finished. I thought it should align with a classic problem:
A Classic Problem
Given a DAG, the task is to sort all vertices topologically, such that vertex 1 has the smallest index. Break ties by minimizing the index of vertex 2, 3, and so on, until \(n\).
which can be seen as a special case (\(w(i)= \eps^i\)) of the Weighted Topological Sort problem. The analysis, however, is entirely different. When I realized this, the post was nearly done, except for the (wrong) proof. It was awkward to admit I couldn’t solve it anymore! I then modeled the problem as the Weighted Topological Sort problem and spent quite some time trying to solve it. Unfortunately, I made no progress in designing an efficient algorithm. Recently, I tried again and, after changing some keywords, I found a relevant paper. After checking out a few more papers and reading their Introduction and Related Work sections, I discovered the right keywords, which led me to Elements of Scheduling and a satisfying solution.
As for writing, I feel it’s better to present and solve the Weighted Topological Sort problem first and then attempt to generalize it. However, I’m too lazy to reorder the sections, as it already took me a lot of effort just to finish the post…
References
Scheduling (single machine)
Elements of Scheduling
]]><div class="note warning">
<p><em>The cost for Power Ups has changed since version 0.7, so this post is no longer applicable.</em></p>
</div>
<p><span class="math inline">\(\newcommand{\eps}{\varepsilon} \newcommand{\R}{\mathbb{R}}\)</span> <span class="exturl" data-url="aHR0cHM6Ly9zdG9yZS5zdGVhbXBvd2VyZWQuY29tL2FwcC8xNzk0NjgwL1ZhbXBpcmVfU3Vydml2b3JzLw==">Vampire Survivors</span> is a game released in early access near the end of 2021. It received 110k recommendations in one month, which is pretty amazing given its minimal aesthetics and seemingly low production costs. I also played it a lot back then, but I won't talk about its gameplay here. Instead, this post discusses the (old) Power Up mechanism and some of the math behind it.</p>
<p>The Power Up mechanism in Vampire Survivors is similar to Kingdom Rush's Upgrades, where you spend the gold earned during gameplay to make permanent gains (e.g., increase move speed by x%). As per the notes in <span class="exturl" data-url="aHR0cHM6Ly92YW1waXJlLXN1cnZpdm9ycy5mYW5kb20uY29tL3dpa2kvUG93ZXJVcHM/b2xkaWQ9MzU2NQ==">Fandom</span>, the cost for a Power Up is defined as:
<span class="math display">\[
Price = \text{InitialPrice} \cdot (1 + \text{Bought}) \cdot \left(1 + \frac{\text{TotalBought}}{10} \right),
\]</span>
where <span class="math inline">\(\text{InitialPrice}\)</span> is the initial price of the Power Up, <span class="math inline">\(\text{Bought}\)</span> is the number of purchased ranks for this Power Up, and <span class="math inline">\(\text{TotalBought}\)</span> is the total number of purchased ranks among all Power Ups. What's interesting is that the order in which you buy Power Ups actually matters!</p>
<p>Intuitively, we should purchase expensive Power Ups first. For example, <span class="exturl" data-url="aHR0cHM6Ly93d3cuaWduLmNvbS93aWtpcy92YW1waXJlLXN1cnZpdm9ycy9Qb3dlclVwX09yZGVy">IGN recommends maxing out Power Ups with the highest initial costs first.</span> However, this is not optimal. Suppose we have two Power Ups, A and B. A has 5 ranks with an initial price of 1, while B has only 1 rank with an initial price of 2. If we max out A first, the total price is:
<span class="math display">\[
1 \times 1 + 1.1 \times 2 + 1.2 \times 3 + 1.3 \times 4 + 1.4 \times 5 + 1.5 \times 2 = 22,
\]</span>
and if we max out B first, the total price is:
<span class="math display">\[
1 \times 2 + 1.1 \times 1 + 1.2 \times 2 + 1.3 \times 3 + 1.4 \times 4 + 1.5 \times 5 = 22.5.
\]</span></p>
<p>So, here is our central topic today:</p>
<div class="note info">
<p>What is the optimal order for buying all Power Ups?</p>
</div>Lake Tahoe 游记https://www.roosephu.net/2024/10/12/trip-to-Lake-Tahoe/2024-10-12T23:20:42.000Z2024-10-12T23:20:42.000Z几个星期前，我和朋友们去了趟湾区的后花园 Lake Tahoe，这里就随便写点东西记录一下。由于时间比较短，我们只过了一个周末，所以显得这篇文章尤为流水帐 23333
划了 kayak 后，我们就去吃饭了。由于我们也不期望 Incline Village 这边有啥好吃的餐厅，我们就在 Google Map 上随便看看了。环顾一圈，我们找到了一家 炸鸡店，评分高达 4.9，堪称是 Incline Village 评分最高的餐馆。于是我们就一致决定是它了。没想到的是，这家炸鸡店做的特别慢，而我们还准备去 Cave Rock 去看日落，就只好在车上吃了。
Cave Rock 那边不是很好停车，我们几个人先下车之后有人去停了车。Cave Rock 那里一堆大石头，坡度不小，手脚并用才好上去，幸亏这里石头很实，踩上去不会晃悠。由于就在湖边上，拍的日落还挺好看的。刚日落的时候云是灿烂的金黄色，太阳落下去后云变成了赤陶色，甚至有种不真实的模糊感。lzy 和 tjc 来的比较晚，想爬上去的时候已经比较黑了，担心安全就没上去了，特别是下山比上山还危险。
爬 Cave Rock
Cave Rock 上的晚霞
爬完 Cave Rock 后我们就去买了点东西，给明天的活动准备一下，然后租的 AirBnB 休息了。说起来这一段时间刚好 Incline Village 有山火，好像就在我们租的 AirBnB 旁边。经过 lzy 缜密分析查资料，最后决定风向应该是远离我们的方向。要是突然风向变了，睡到一半被烟薰起来就尴尬了。
今天的第一站是 Ed Z'berg Sugar Pine Point State Park。公园地上一堆松果，你说的我都懂，可是为啥松果这么大？胡老师大为震撼，决定捡一个松果给汤姐做礼物。胡老师又注意到这松果上有些粘液，还有种奇特的芬芳，然后意识到这就是传说中的松脂吗……而且这个公园叫做 Sugar Pine Point State Park，那想必这里的松树都是 Sugar Pine 咯，有没有勇士来尝尝是不是真的是甜的 2333
第二站是 Emerald Bay State Park。今天的风明显比昨天大，我在 Vikingsholm Trail Head 被风吹的不行了，但是这里的风景确实不错，从这里能看到 Fannette Island。理论上附近还有一个 Eagle Fall，到了之后实在让我大跌眼镜，就这？？？别说是 Fall 了，你说这是个 Creek 我都将信将疑……
FACE => face FOOD => food 1DEA => idea 1CE => ice BEEF => beef
形容词
A11 => all 01D => old BAD => bad C001 => cool DEAD => dead
如果只看 wordfreq small list 里的，那更少，只有 160 个。我自己稍微组了一些词：
FEE1BAAD => feel bad 这个其实不太可行，因为我用的 BAAD，也可以考虑 FEE15BAD A10EFOOD => aloe food DEB81DEA => debate idea BADCAB1E => bad cable ADD1ABE1 => add label DEADF001 => dead fool 差一点变成了 deadpool 1EAFFA11 => leaf fall 让我想起了 e.e. cummings 的诗 l(a
如果我们不允许任何 alias，只用 A-F 来组词，也是能组出一些的：
DEADBEEF => dead beef 经典 FEEDBABE => feed babe 张嘴，啊～ FADEDDAD => faded dad 我去买个橘子？
我挑了一些词：
部分只包含 A-F 的词
BAD => bad FACE => face DEAD => dead ADDED => added ADD => add BED => bed DAD => dad FEED => feed DECADE => decade FED => fed FEE => fee FACED => faced BEEF => beef ACE => ace BABE => babe BEE => bee DEAF => deaf CAB => cab FADE => fade DEED => deed FADED => faded BEAD => bead FACADE => facade DAB => dab FAD => fad BEADED => beaded EBB => ebb
可以看到长度为 5 的词很少，所以 3+5 和 5+3 这种搭配不好组词，只能 4+4 的组。
代码不长我就直接放这里了：
代码
import wordfreq from pathlib import Path
deftransform(x): result = x.upper() for c in x: ifnot c.isalpha(): returnNone if"I"in result and"L"in result: returnNone
# if result.endswith("ATE"): # result = result[:-3] + "8" # result = result.replace("L", "1") # result = result.replace("O", "0") # result = result.replace("I", "1") # result = result.replace("S", "5")
for c in result: ifnot c.isdigit() and c > "F": returnNone
words = {} for w in wordfreq.iter_wordlist("en", wordlist="small"): if w notin dictionary: continue w2 = transform(w) if w2 and w2 notin words and3 <= len(w2) <= 8: words[w2] = w
for w2, w in words.items(): print(f"{w:12} => {w2:12}")
出门就是 Waikiki Beach，但是里面人太多了（而且我也不会游泳）。拍了几张游客照我就想跑路了。接下来我就面临人生两大问题之一：中午吃什么？这个时候自然要问问万能的 xhs 啦～xhs 告诉我，这里可以吃日料，但是一个个都需要预订，很显然，我完全没有做准备。不要慌，我还有 wp 的 O'ahu 旅游宝典。wp 推荐的第一个餐馆叫做 Gina's BBQ。可以啊，让我体验一下什么才是正宗的 Hawaiian BBQ！我准备一路走过去，只要走四五十分钟就到了哦～于是打定主意，我就走了过去。这一路上都没啥树，在大太阳下走着晒死我了，路过了几家日料店，没有位置，路过了一个 Taco Bell，有啥好吃的，路过了一个 Mango Mango，都到夏威夷了怎么能吃这种大路货！最后终于走到了 Gina's BBQ，等等，为啥店名下面一个大大的 Korean BBQ？？？我再看了一眼菜单，没错，就是 Korean BBQ……不知道点啥就点了个 Gina's special，里面肉挺多的，但是问题是，我去哪里吃……这个店就一个门店，没有吃饭的地方啊……不慌，我看夏威夷这么多的公园，找个公园里的桌子不成问题！于是我又原路走回，这一路上都没啥树，在大太阳下走着晒死我了，路过了几家日料店，没有位置，路过了一个 Mango Mango，都到夏威夷了怎么能吃这种……啊真香！热的时候吃冷饮太香了！我路上试图招一个空地上的长桌来吃饭，但是问题是这些桌子全在草坪里，蚂蚁苍蝇来开会，我实在吃不下去，不能再现 SIGYAO30 '22 @ LA 的三和大神荣光了，直到——我有一次路过 Taco Bell。虽然我没买东西，但是 Taco Bell 有室外的水泥餐桌凳子，看起来不赶人，重点是，还有遮阳伞！我没有丝毫犹豫，直接坐在了 Taco Bell 的凳子上开始吃饭，吃饱喝足后旁边还有个垃圾桶，一气呵成！
之后我准备往 Diamond Head 那边走走看，拍几张游客照。走到一半收到短信说房间好了可以去 check in 了。那我自然应该往酒店走了。到酒店取出之前的行李后，ydl 就过来了，打了声招呼，我就去 check in 了，zw 到了之后一起去房间。
稍微修整了一下后我们就去下面玩，因为 zw 来的时候堵车了，原因就是有一个 festival 封路。既然来都来了，不如我们就去逛逛吧。这个 festival 叫做 Kanikapila Waikiki Festival，就在我们酒店旁边，但是去了之后有点大路货……所谓的 festival 就是一堆小贩在街边支个摊子，然后卖一些有的没的的东西。有些东西还稍微有点特色，例如自称 Hawaii 本地蜂蜜，但是有些东西就毫无特色了，例如烤鱿鱼，写着名字的竹子……
晚上到了宾馆之后我和 zw 再来规划明天的行程。看了一下，周围似乎没啥好玩的了，有个 aquarium 但是似乎评价不是很高。红警玩家 zw 推荐了珍珠港，因为 Arizona Memorial 在红警里出现过，那我们就去那里了。这里有个问题是 Arizona Memorial 的摆渡船需要预约，而我们没有预约上，但是 anyway 我们还是决定去那里看一下拍个队，随缘吧。说起来这边的 researvation 就很迷，只能预约要么明天的，要么两个星期之后的。预约明天的基本上都被黄牛抢了……
周一
我们本可以选择去大大大早起去 Pearl Harbor 排 Arizona Memorial 的队，但是经过前两天的特种兵旅游，我们决定躺了，吃完饭再去。卡着点 checkout 后，我们去 Sato's Seafood 吃了一个 Poke Bowl。我点了一个 spicy salmon 感觉酱调的非常不错，就是鱼有点少。这家店店内没有桌椅，就在店门口有一两个小桌子，随意配上三四把椅子，我又回忆起了前几天吃 Gina's BBQ 的场景……
吃完饭后打个车去了 Pearl Harbor。门口存了个包后我们就进去排队，幸亏这里的排队是签到制，登记了名字之后不需要排在那里，不过我对排上队不报啥希望了，一来我们吃了饭才来，里面好多人，二来我要赶飞机，如果一个小时内排不上队的话我就看不了了。之后就去里面的珍珠港事件纪念馆随便看看。正当我觉得马上到时间了看不上 Arizona Memorial 的时候，zw 突然发现排到了，哟太凑巧了。
Arizona Memorial 建立在珍珠港事件中被摧毁的 USS Arizona 残骸上，在纪念馆还能看到一些残骸。它不大，里面没有多少东西可以参观。我有时候看到水面上还有一些漂浮的五光十色的东西，zw 告诉我那是不停冒出来的机油，漂浮在水面上，不同地方油层厚度不同，在阳光下反射的光的颜色也不同，这东西还有一个名字，叫做 Tears of the Arizona。Arizona Memorial 附近还有另外一艘战舰，是退役的 USS Missouri，就是在这艘船上，日本正式投降了。将 USS Missouri 停在珍珠港内，算是给故事添上一个句号吧。
Arizona Memorial
Tears of the Arizona
]]><p>今年上半年，ydl 找到我，说他和 maggie 订婚了，诚邀我去 Hawaii 参加婚礼。这我怎么能不参加呢？况且我还没去过 Hawaii，自然应该趁这个机会小小旅游一下。这里就随手记一下流水帐。这次旅游之后，终于知道 Hawaii, O'ahu, Waikiki, Honolulu 之间的关系了……</p>
<p>后来发现一个有趣的 trivia：夏威夷的 state fish 叫做 Humuhumunukunukuapua'a，虽然看起来很长，但是读起来非常有节奏感 23333</p>IMO 2024 P1 & P5https://www.roosephu.net/2024/07/23/IMO-2024-P1-P5/2024-07-23T08:36:35.000Z2024-07-23T08:36:35.000Z听说今年这两道题比较简单，我就来蹭蹭热度了。 P1 是一个不太难的题，但是我很可能把它想复杂了；P5 是一个非常巧妙的 brainteaser 。
"/> 
A 点左上方的点没有坏人是因为 A 点是我们从上往下扫找到的第一个坏人。
