A New Solution to the Red‑Blue Eye Paradox!
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2026年2月9日 23:36 #3400
Many years ago, Terence Tao, the Chinese-Australian mathematical prodigy, posted a problem online called The Blue-Eyed Islanders Puzzle for public discussion. Online debates have varied widely, and no definitive conclusion has been reached to this day.
On a remote, isolated island live 100 islanders, whose eyes are only of two colors: 5 have red eyes, and 95 have blue eyes. All islanders follow the same religion and possess the following traits:
Perfect rationality: Everyone has flawless logical reasoning ability and can draw every possible logical conclusion.
Devout obedience: They strictly follow all religious rules without exception.
No communication: Any form of discussion or hint about eye color is forbidden.
No self-observation: There are no mirrors, reflective water surfaces, or any reflective objects on the island, so no one can directly see their own eye color.
Daily assembly: Every noon, all islanders gather in the square and can clearly see everyone else’s eye color.
Core Religious Rules
No one may learn their own eye color by any means (including direct viewing, indirect asking, hints from others, etc.).
No one may tell another person their eye color.
Fatal rule: Anyone who logically deduces that they have red eyes must commit suicide at midnight that same day.
Key Event and Question
The islanders lived peacefully until one day, an unaware foreign visitor arrived on the island. At the assembly with all islanders present, the visitor announced publicly and loudly:
“There are people with red eyes here.”
After speaking, the foreigner left and had no further influence.
Question: Assuming the islanders are perfectly intelligent, their reasoning is unassailable, and they strictly obey the rules, what will happen on the island, and why?
The “accepted” answer to this problem is derived using mathematical induction, as shown below.
Proof by Induction
1. Base Case
Case
k=1
(1 red-eyed, 99 blue-eyed):
The only red-eyed islander sees everyone else has blue eyes.
Given that “there is at least one red-eyed person”, they immediately conclude: I must be the red-eyed one.
Conclusion: Suicide at midnight on Day 1.
The base case holds.
2. Inductive Hypothesis
Assume that if there are
k=m
red-eyed islanders,
all
m
red-eyed islanders will commit suicide together at midnight on Day
m
(where
m
is any positive integer).
3. Inductive Step
We prove the statement holds for
k=m+1
.
Suppose there are
m+1
red-eyed islanders.
For any red-eyed islander:
They see exactly
m
red-eyed islanders.
They reason:
If I do not have red eyes, then there really are only
m
red-eyed islanders.
By the inductive hypothesis, these
m
red-eyed islanders should all commit suicide at midnight on Day
m
.
When Day
m
ends and no one dies,
they immediately reject their premise: I must also have red eyes (total is
m+1
).
All
m+1
red-eyed islanders complete this reasoning simultaneously.
Conclusion: All
m+1
red-eyed islanders commit suicide together at midnight on Day
m+1
.
The inductive step holds.
Actually, the Answer Is Wrong — No One Will Commit Suicide
The explanation is as follows:
Definitions
One red-eyed person forms a red-eye reasoning unit (E).
Two red-eyed people form a red-eye reasoning module (M).
Three red-eyed people form a basic red-eye reasoning system (S).
Four red-eyed people form an advanced red-eye reasoning system (SA).
The immediately preceding structure is called the antecedent (A) of the next.
Definition of Reasoning Process
An observer participates in the reasoning, becoming both a disturber to the antecedent’s reasoning and a member of the new combined structure.
After the foreigner’s announcement, the observer reasons based on the antecedent’s behavior.
The time at which the antecedent independently confirms its conclusion is called the antecedent confirmation time (TA).
The time at which the new system (including the observer) independently confirms its conclusion is called the system confirmation time (TF).
(1) Reasoning in Unit E
A single red-eyed islander sees no other red eyes, infers they are the only one, and chooses to sacrifice themselves (denoted KK).
(2) Feedback Reasoning in Module M
If I have blue eyes, the red-eyed person I see will act at TA (following the reasoning of antecedent E).
If no KK occurs at TA, I infer that I am red-eyed and have disturbed E.
Other observers reason identically.
Result: Collective KK at TF.
(3) Feedback Reasoning in System S
If I have blue eyes, the red-eyed people I see will act at TA (following the reasoning of antecedent M).
If no KK occurs at TA, I infer that I am red-eyed and have disturbed M.
Others reason identically.
Result: Collective KK at TF.
(4) Feedback Reasoning in Advanced System SA
If I have blue eyes, the red-eyed people I see will act at TA (following the reasoning of antecedent S).
If no KK occurs at TA, I infer that I am red-eyed and have disturbed S.
Others reason identically.
Result: Collective KK at TF.
The “reference” to antecedent reasoning in M and S is valid, but the reference to S in SA is invalid.
System S can reach a conclusion only because the foreigner’s announcement applies directly to S. In reality, the foreigner made only one announcement, which applies to SA as a whole, not separately to its embedded S subsystem.
Why, then, is the “reference” valid in M and S?
Because M and S do not truly rely on the antecedent’s conclusion — their reasoning structure merely matches the antecedent’s form.
In M:
If I am blue-eyed, the red-eyed person I observe must act immediately (KK). If no KK occurs, I am red-eyed. This is a simple deductive inference, with no need to reference E’s reasoning or conclusion.
In S:
If I am blue-eyed, another red-eyed person may assume they are blue-eyed. That person would then infer that the single red-eyed person they see must act immediately (KK). If no KK occurs, that person realizes they are red-eyed and will act on Day 2. If no KK occurs on Day 2, I — the observer — must be red-eyed.
This is iterative elimination, more complex, but still forms a necessary chain without referencing M’s conclusion.
In SA:
If I am blue-eyed, and another red-eyed person assumes they are blue-eyed, that person cannot validly infer that the two red-eyed people they observe would act on Day 2 under their own blue-eye assumption.
One might ask: Why can they not infer this? Isn’t that person just the observer in S? If the observer in S can deduce it, so can the person in SA.
This is incorrect.
The observer in S can reach a conclusion about M because, when assuming they are blue-eyed, they can confirm that the two red-eyed people in M each see exactly one red-eyed person, creating a unique reasoning path.
When the system expands from S to SA, and the observer in S becomes the self-assumed blue-eyed person in SA, they can no longer confirm — under their own blue-eye assumption — that the two red-eyed people in M each see exactly one red-eyed person. Thus, no unique reasoning path exists.
One might argue: We can confirm this within the first-level assumption (the SA observer assuming they are blue-eyed), which includes a second-level assumption (the implied assumption about the self-assumed blue-eyed person).
This confuses concepts. Reasoning requires assumptions, but the possibility of an assumption is entirely different from the validity of a deduction. Confirmation within an assumption cannot provide genuine confirmation for actual reasoning.
Specifically, let the four red-eyed islanders be A, B, C, D.
A can assume that, in B’s mind, B assumes B is blue-eyed.
But A cannot expect C, D, or anyone else to treat B — who is actually red-eyed — as blue-eyed in any real reasoning.
While assuming A is blue-eyed, A may further assume that B assumes B is blue-eyed, and that B can confirm C and D each see only one red-eyed person.
But A cannot expect all four islanders (including A) to ever accept that there are at most two red-eyed people among them in any actual deduction.
Conclusion
Since already with 4 red-eyed islanders, no one can logically deduce their own eye color, no suicides will occur at all on the island with 5 red-eyed islanders.
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