Ship of Theseus: after every plank is replaced, which ship is the original?

01预测Predict

02修船Repair the ship

每次修理换掉几块最朽的木板，新板照着被换下的那块仿制。先修几次，再打开收藏家，一直修到船 A 上一块原木板都不剩。边修边看图下方的四条判定标准。

Each repair swaps out the most rotten planks; each new plank is copied from the one it replaces. Run a few repairs, switch on the collector, and keep going until no original plank is left on Ship A. Watch the four criteria under the figure as you go.

修理次数（每年一次）Repairs (one per year)

收藏家保留旧木板，拼成船 BA collector keeps the old planks and rebuilds them as Ship B

03解释：标准分道扬镳Explain: the criteria come apart

“同一艘船”平时像是一个判断，其实是四个判断碰巧一致：木料、经历、设计、名字都指向同一个东西。修船本身没让问题变难。收藏家把这四样东西分到了两艘船上，标准们才第一次给出不同答案。

“The same ship” feels like one judgment. It is really four that usually agree: the timber, the history, the design and the name all point at one object. The repairs never made the question harder. The collector split those four things across two ships, and only then did the criteria give different answers.

四条标准，以及它们在你这次运行中的判定Four criteria, and what each says in your run
标准Criterion	它追踪什么What it tracks	判定Verdict
材料Matter	哪艘船拥有过半原木料Which ship holds most of the original timber
时空连续Continuity	能否用一连串小改动连回最初那艘Whether a chain of small changes leads back to the first ship
形式（设计）Form (design)	哪艘还保有原设计（≥ ⅔ 的设计信息）Which ship still carries the original design (≥ ⅔ of its information)
名称（登记）Name (registry)	城邦把哪艘叫作忒修斯之船、让它继续服役Which ship the city calls Theseus’s and keeps in service

先看船 A：它一直在服役、一直挂着名牌，所以时空连续和名称都跟着它。Start with Ship A: it never left service and kept its name plate, so continuity and the name follow it.
再看船 B：原木板全在这里，所以材料标准跟着它走。Then Ship B: every original plank ended up here, so the matter criterion follows it.
再看设计信息：新板照旧板仿制、没有出错，所以形式标准认出两艘都是。Now the design information: the copies were exact, so the form criterion names both ships.
最后看判定板：平时一致的四条标准，被分到了两艘船上。Finally the verdict board: four criteria that normally agree are now split across two ships.

一个奇怪的后果

如果按材料标准，船 A 是不是忒修斯之船，取决于城另一头有没有一个收藏家。可船 A 本身什么都没变。很多哲学家认为同一性应该只由两个对象自身之间的关系决定（“只看 x 和 y”原则），材料标准在这里违反了它。

“两艘都是”为什么也不行

同一性是一对一的：如果 A 和 B 都等于原来那艘船，那 A 就等于 B，但它们明明是两艘船，可以开往不同方向。所以“两艘都是”会在逻辑上出问题。四维论者给出另一种说法：从一开始就有两艘重叠的船，它们早年共用同一段历史，后来才分开。

这个问题的来历

普鲁塔克在《忒修斯传》里记载，雅典人一直保存这艘三十桨船，直到德米特里乌斯·法勒鲁斯的时代，旧木板一坏就换新的，哲学家们拿它当“生长之物”问题的范例。约两千年后，霍布斯在《论物体》（1655）里加了收藏家这一步，并回答：答案取决于你用哪个名称来问，问的是这堆材料、这个形式，还是这整个东西。

A strange consequence

On the matter criterion, whether Ship A is Theseus’s depends on whether a collector exists across town. Nothing about Ship A itself changed. Many philosophers hold that identity should turn only on the two things being compared (the “only x and y” principle). The matter criterion breaks it here.

Why “both” fails too

Identity is one-to-one. If A and B are each identical to the original, then A is identical to B, yet they are two ships that can sail in different directions. A four-dimensionalist offers another description: there were two overlapping ships all along, sharing their early history and parting later.

Where the puzzle comes from

Plutarch’s Life of Theseus reports that the Athenians preserved the thirty-oared ship down to the time of Demetrius of Phalerum, replacing decayed timbers as they went, and that philosophers used it as their standard example for “things that grow.” About two thousand years later Hobbes, in De Corpore (1655), added the collector and answered that it depends on the name under which you ask: the matter, the form, or the whole.

用你自己的话说。Put it in your own words.

“平时这四条标准一致，是因为 ___；收藏家让 ___ 和 ___ 分开了。”“The criteria normally agree because ___; the collector separated ___ from ___.”

04多慢才算“逐渐”？How gradual is gradual?

连续性标准要有个门槛：一次换得太多，就不算“修”，而是“重造”。这个模型规定，单次修理换掉的木板超过容忍值，连续链就断了。试试一次换 24 块：一夜之间整艘船换新。

The continuity criterion needs a threshold. Replace too much at once and it is a rebuild, not a repair. In this model the chain breaks when one repair replaces more planks than the tolerance. Try replacing all 24 overnight.

每次修理换几块Planks per repair

连续性容忍值Continuity tolerance

现在试着找出那条线：把每次换的块数调到 7，链断了；再把容忍值调到 7，链又接上了。换 7 块算修理，换 8 块就算重造吗？没有哪一块木板让差别突然变得明显。这就是连锁悖论（堆悖论）：每一步都“差不多一样”，但一步步累积起来就完全不同。门槛总得有人去定。

Now look for the line. Set planks per repair to 7 and the chain breaks; raise the tolerance to 7 and it holds again. Is swapping 7 a repair but 8 a rebuild? No single plank makes the difference obvious. This is the sorites paradox (the paradox of the heap): every step is “nearly the same,” yet the steps add up to something entirely different. Someone has to choose the threshold.

05复印件的复印件A copy of a copy

目前每块新板都和旧板一模一样，所以设计标准说“两艘都是”。真实的工匠会出错。木板上的小点是设计（每块 8 个比特）。设一个抄错率，新板照着它替换的那块来做，错误就会一代代传下去。

So far every new plank matched its predecessor exactly, which is why the form criterion says “both.” Real carpenters make mistakes. The dots on each plank are its design, 8 bits per plank. Set a copy error, and since each new plank copies the plank it replaces, errors pass down the generations.

记下预测后：把抄错率调到 5%，再让每块板都被仿制 3 代，看图下方的“船 A 的设计信息”。

Then set copy error to 5% and keep repairing until every plank has been copied three times. Read “Design information in A” under the figure.

抄错率（每比特）Copy error per bit

保留原始图纸：每块新板都照图纸做Keep the original blueprint and copy every new plank from it

每仿制一次，就是让设计通过一条有噪声的信道：每个比特有 ε 的概率翻转。复印件再复印，就是把信道串起来。数据处理不等式说，串联只会让结果包含的关于原件的信息不增：

Each copy sends the design through a noisy channel: every bit flips with probability ε. Copying a copy chains the channels. The data processing inequality says a chain can never increase what the result tells you about the original:

I(X; Y₃) ≤ I(X; Y₂) ≤ I(X; Y₁)
ε_g = (1 − (1 − 2ε)^g) / 2
I = 1 − H(ε_g) bits per design bit

这里 g 是仿制的代数，H 是二元熵。5% 的错误抄 3 代，翻转率累积到 13.6%，留下的信息只有 43%。

为什么“相同比特”还有 86%，信息却只剩 43%？因为瞎猜也能对一半：随机比特和原件有 50% 相同，但信息量是零。信息衡量的是超过瞎猜的那部分，而且按对数计算。

原始图纸让每块板都只离原件一步，错误不再累积，信息稳定在 1 − H(ε)。母带、计量基准、权威版本的文本都靠这一招。代价是：设计标准又重新认出“两艘都是”。

Here g is the number of copy generations and H is binary entropy. At 5% per copy, three generations compound to a 13.6% flip rate, leaving only 43% of the information.

Why do 86% of bits still match while only 43% of the information remains? Guessing gets half right. Random bits match the original 50% of the time and carry zero information. Information counts only agreement beyond chance, measured on a log scale.

The blueprint keeps every plank one step from the source. Errors stop compounding and information holds at 1 − H(ε). Master recordings, measurement standards and authoritative texts all rely on this. The cost: the form criterion goes back to naming both ships.

06迁移：传送机Transfer: the teleporter

这个模型的适用范围Where this model applies

船是示意：24 块木板、4 列船板，没有龙骨和肋骨；修理顺序固定，用来代表“最朽的先换”。每块板的设计简化成 8 个比特。

三个门槛是模型的选择，不是事实：材料按过半计，连续性按容忍值计，设计按 ⅔ 的信息计。改变门槛就会改变判定，这本身就是本课要说的一点。信息量的计算假设原设计的每个比特等可能为 0 或 1，抄错彼此独立（二元对称信道）；“相同比特”用的是固定随机种子下的一次抽样。

四条标准并不完整。四维论、“没有船”的取消论、帕菲特“同一性并不重要”的观点，都给出不同的处理方式。本页不判定哪种是对的。

The ship is schematic: 24 planks in 4 strakes, with no keel or frames. Repairs follow a fixed order standing in for “most rotten first.” Each plank’s design is reduced to 8 bits.

The three thresholds are model choices, not facts: matter uses a majority, continuity a tolerance, form two thirds of the information. Moving a threshold changes a verdict, which is part of the lesson. The information measure assumes each design bit is equally likely 0 or 1 and copy errors are independent (a binary symmetric channel). “Bits matching” is one sample from a fixed random seed.

These four criteria are not the whole field. Four-dimensionalism, eliminativism about ships, and Parfit’s view that identity is not what matters each handle the case differently. This page does not decide between them.

离线可用。用 Tab 切换控件，方向键调节滑块。没有快捷键会干扰你的写作。Works offline. Use Tab to reach controls and arrow keys on sliders. No keyboard shortcuts override your writing.

换掉每一块木板之后，哪艘才是忒修斯之船？Once every plank is replaced, which ship is Theseus’s?