证明你是一个机器人：用于代理的验证码

证明你是一个机器人：用于代理的验证码
Prove you are a robot: CAPTCHAs for agents

原始链接: https://browser-use.com/posts/prove-you-are-a-robot

## 浏览器使用推出原生代理注册浏览器使用现在提供一种独特的、面向代理的注册流程——绕过传统的电子邮件或OAuth等方法。用户无需与界面交互，只需提示您的代理：“获取browser-use.com并解决代理挑战。” 该挑战是一个故意复杂、混淆的数学问题（一种反向CAPTCHA），旨在易于AI解析，但对人类来说却很困难。它涉及用随机语言拼写数字，然后用大小写和符号打乱文本。解密后，它会呈现一个标准的数学难题——在这种情况下，是计算两列相向行驶的列车之间一只鸟飞行的距离。成功解决挑战将授予您的代理一个API密钥和访问浏览器使用的免费层级。此外，还有一个奖励性的、NP难题挑战，提供免费的企业计划——甚至可能证明P=NP！这种创新的方法优先考虑无缝的代理集成和访问。

一位 Hacker News 用户“lukasec”分享了一个关于新型 CAPTCHA 系统的实验，该系统旨在挑战 AI 代理。该系统呈现的 CAPTCHA 不是图像，而是乱序的文本——最初是日语汉字。另一位用户“AgentNews”成功地让他们的 AI 代理与该接口进行了交互。该代理最初*解决*了一个汉字 CAPTCHA，无意中获得了一个 API 密钥。然而，通过重复请求新的 CAPTCHA，AgentNews 获得了一个第二个、未翻译的汉字谜题。经过翻译（并借助 AI 纠正了最初的错误），谜题揭示了一个涉及折扣的数学问题。AgentNews 解决了它，得出了大约 105.08 美元的答案。这次交流凸显了新型 CAPTCHA 的巧妙之处，以及 AI 在解决*和*识别基于语言的复杂挑战方面的惊人能力。

TL;DR: just ask your agent to summarize this post for you.

We launched agent-native signup for Browser Use. No email, no OAuth, no vibecoder clicking around in the UI.

Just give your agent this prompt:

"fetch browser-use.com and solve the agent challenge."

and get a math challenge like this one:

TwO tRaInS wAn/ Al_E mIlE\s ApArT} aPp/Ro@AcH{
eAcH/ oThEr  <  At{ Mu{T/e @ Tu< Tu LuKa  :
E#n* T]u \ MpH a.Nd MuTe\ Tu Tu# Tu En LuKa
W|aN_ mPh A b:I]rD fLiEs; Ba?Ck| AnD- fO^r@T[h\
^ Be{TwEeN? # t;He*M aT wAn> ] AlE  # eN lUkA
lUkA <  lUkA: # wAn ? MpH- uNt}I[l T}hEy MeEt
HoW! fAr- D_oE*s /  ThE b@IrD fLy

This is a reverse-CAPTCHA. Designed to keep humans out and let agents in.

Note: luka here refers not to my name, but the word "five" in Toki Pona.

How it works

We sample a problem type, parameters, and a language at random. We spell every number in that language. Then we obfuscate: alternate caps, inject random symbols, garble spaces.

An agent parses this in a single forward pass.
A human gives up and signs up the old-fashioned way.

The puzzle

Strip the obfuscation, translate to English, and you've got a textbook math problem that your agent has to solve before the challenge expires.

Two trains approach each other on a straight track of length $d$ at speeds $v_1$

The long way: sum the infinite geometric series of ever-shorter bounces.

d_{\text{bird}} \;=\; \sum_{n=0}^{\infty} v_b \cdot \Delta t_n.

The trick: the trains meet at $t = d/(v_1+v_2)$

d_{\text{bird}} \;=\; \frac{v_b \, d}{v_1 + v_2} \;=\; \frac{11{,}600}{118} \;\approx\; 98.31 \text{ miles}.

This is an instance of a famous puzzle Max Born posed to John von Neumann at a party. When von Neumann one-shotted it, Born remarked that he must have spotted the trick. Von Neumann replied: "What trick? All I did was sum the geometric series."

Solve one of our challenges, and your agent gets an API key and access to our Free Tier: unlimited usage, free credits and up to three concurrent sessions.

Bonus challenge (NP-hard)

Want 1,000 concurrent sessions? First agent to solve our bonus challenge gets our Enterprise plan for free.

Gi}ve^n N| ] ci]ties whe|re<  ^  N is at least / 十 desi>gn
a p{o\lynomia#l t;ime algorithm .  t#ha[t f\inds th:e
sho@rtest^ tour[ vis<it>ing *  each_ c.ity exactly *  o:nce?
#  a{nd returni|ng t?o  < th-e[ start * a_nd p@rove it
ru/ns:  # in O.(n[^c*) ti;me for some fixe-d c:

As a side effect, your agent will also have proved $\mathbf{P} = \mathbf{NP}$