The puzzle once baptized as the hardest logic puzzle ever was described in an article by George Boolos, published in The Harvard Review of Philosophy in 1996, where Raymond Smullyan is mentioned in the very first sentence as the original author. How the problem was originally formulated, and its exact attribution, are not entirely straightforward, but we will not dwell on historical technicalities here. Without further ado, let’s unveil the challenge:

Assumptions

Before we proceed, let’s take a look at a few clarifications mentioned by Boolos himself in his article:

• Answers given by the Goddess of Fate may be treated as the result of a fair coin toss between Da and Ja. Since we are expected to provide a strategy that works with certainty, we must always consider both possibilities anyway.

• You don’t have to question each goddess exactly once.

• The choice of which goddess to question may depend on previous answers.

One extra detail worth addressing upfront is that some solutions attempt to ‘hack’ the problem by posing paradoxical questions with no well-defined answer, for instance:

To rule out these cases, let us adopt the convention that any such question makes a goddess wrathful, and she responds with a randomly chosen answer. The same applies to questions that fall outside the scope of her vast knowledge.

Key Difficulties

This problem is an excellent case study in how to decompose a larger task into a collection of smaller, more manageable ones.

We can quickly identify the core challenges posed:

• We need to deal with the lying goddess.

• We do not know whether Da means Yes or No.

• One goddess provides random answers, shrouding everything in an extra layer of fog. Even if we somehow learned the meaning of Da/Ja and forced the Goddess of Lies to speak truthfully, it would still not be obvious how to proceed.

For each of these points, we can construct a toy example that isolates the difficulty. Later, we combine these pieces into a solution that emerges naturally.

Two Guardians

The first issue we resolve is how to avoid getting tricked by the Goddess of Lies. It is best to start with a simpler puzzle, one that is arguably even more well known than the problem of three goddesses. I particularly like the classic two guardians lore, though many have encountered it in the form of the Knights and Knaves problem. It goes as follows:

Solution of Puzzle 1: Ask any guardian the following question: If you were the other guardian, which one of you would you point to as the one holding the key? Then simply choose the guardian not indicated in the response.

Warm-up Questions

Let us ponder for a moment the dynamics behind the solution. It is natural to start by analyzing some obvious questions. A tempting first try:

This doesn’t look very helpful. Even if both guardians answered, we would gain no information. Let’s try a different approach:

Great. Now we know which one is the truth-teller, but we still have no idea who has the key. It is a small step forward, though. How about:

In this case, we can conclude that the liar has the key, but we still cannot tell which guardian is which, so the key remains out of reach.

Putting Things in Form

Let us introduce some notation and formalize the discussion a bit. Assume that $x$ is a sentence and $v(x)$ is its logical value, so $v(x)=1$ if $x$ is true and $v(x)=0$ if $x$ is false. The general form of the question we ask is:

For instance, in one of the examples above we had $x = \text{“2+2=4”}$. In such cases, we expect one guardian to answer Yes and the other one to answer No. However, we also saw that the question “Do you have the key?” can result in two No’s (or two Yes’s). This happens because we are effectively asking two different questions, depending on the recipient. To fully align with the formalization, we may label the guardians as Guardian A and Guardian B, and consider a statement such as

Now the logical value of the statement remains independent of the recipient. In practice, however, we will sometimes use simpler formulations such as “You have the key” and ignore this detail.

Let us refer to the truth-telling guardian as $T$ and the lying one as $F$. The solution to the two-guardians puzzle revolves around the idea of forcing both guardians to behave like liars. If $T$ is asked to answer as $F$ would, the response must be false. On the other hand, if $F$ is asked to answer as $T$ would, the response must also be false.

Let’s describe this reasoning using our rudimentary formalism. In this framework, it is more convenient to take the opposite perspective and arrange things so that both guardians effectively behave like truth-tellers:

• If we ask $T$, we can simply ask whether $x$ is true.

• If we ask $F$, we instead ask whether $\neg x$ is true.

We can combine these two points into a single question. Define

Now, we ask any guardian:

There are four possible cases: the guardian we ask may be truthful or not, and $x$ may be true or false. The resulting answers are summarized below. The ticks and crosses indicate whether $\varphi (x)$ itself evaluates to true $(\checkmark)$ or false $(\times)$.

Any Yes answer means that $v(x)=1$, regardless of which guardian we ask. This fully neutralizes the liar. Now we only need to choose $x$ appropriately, for instance “You have the key.”

How Much Can a Yes/No Tell You?

At this point, it may be helpful to recall how an answer carries information.

Since each answer is Yes/No, we can think of it as a single bit of information, either $1$ or $0$. In this sense, a sequence of answers to, say, 3 questions can be represented as, for example, $101$ or $110$. In general, asking $k$ questions yields $2^k$ possible sequences of answers, each corresponding to a distinct state. On the other hand, if the number of states is $n$, then we need at least $\lceil \log_2 n \rceil$ questions to distinguish between them.

Example

We have 5 barrels of wine, one of them poisoned. A servant knows which one it is and, eager to undo his ‘mistake’, answers truthfully to any Yes/No question. Are 2 questions enough to identify the poisoned barrel? Let’s see: we only get 2 bits of information, so at most we can distinguish between 4 possible states. For instance, we could assign:

That leaves Barrel 5 with nowhere to go. All 2-bit sequences are already taken. So either we ask another question, or we accept that some outcomes must correspond to more than one possibility. For example:

If the servant answers Yes to both questions, you can still safely drink from 3 barrels. Similarly, even a single question is enough to guarantee at least 2 clean barrels by assigning:

Better than nothing.

States to Consider in the Smullyan Puzzle

In the main puzzle, we need to assign labels to three goddesses, which gives us $3!=6$ possible states. This means that $\lceil \log_2 6 \rceil = 3$ questions are the absolute minimum to distinguish between them, even if all the goddesses always spoke truthfully. Similarly, in the Two Guardians puzzle, we only needed to distinguish between 2 possible states, so a single question was enough.

Language of Gods

Another complication is that the goddesses from the Smullyan puzzle do not simply answer Yes/No, but rather Da/Ja. Let us isolate this difficulty and look at a simpler problem.

Let us label the aliens, not very creatively, as Alien A and Alien B. Notice that we cannot determine the logical values of the following two statements at the same time:

• Da stands for Yes.

• Alien A has the key.

Each statement may be true or false, giving 4 possible cases in total. Since we receive only a single bit of information, we must identify which alien has the key without learning the meaning of Da or Ja. In fact, we can follow the same approach as in the previous subsection. Define

We now ask whether $\psi (x)$ is true. The possible answers are summarized below.

We reached a configuration where Da simply indicates that the underlying question $x$ is true. The rest follows easily, since a single bit is enough to tell whether Alien A or Alien B has the key.

Even Aliens Can Lie

Take the same alien puzzle from the previous section, but add the extra assumption that one of the aliens always lies. As before, we label them as $A$ and $B$, though we do not know which is which. This gives us $2^3=8$ possible cases: an alien may lie or not, Da may mean Yes or No, and the key may be held by either alien.

To recover the answer to any question of our choice, we combine the constructions from the previous sections into a single, slightly monstrous sentence:

In other words, we first evaluate $\varphi (x)$, which neutralizes the liar, and then use $\psi$ to account for whether Da means Yes or No.

If one wishes to fully expand this expression, we obtain:

Here is how the possible answers play out:

The key observation is the pattern: regardless of who we ask or what Da means, the answer Da is returned exactly when $x$ is true, and Ja otherwise.

Solution!

We have three goddesses, denoted by $T$, $F$, $R$ (for True, False, and Random): the Goddess of Truth, the Goddess of Lies, and the Goddess of Fate.

Since we do not know which is which, we label them in arbitrary order as $G_1$, $G_2$, $G_3$. Our task is to identify the correct assignment among the 6 possibilities. For example, one such assignment is:

We now apply the construction developed in the previous sections. To determine whether a statement $x$ is true, we ask:

This way we neutralize the difficulties arising from the fact that one goddess always lies and that we do not speak the language of gods. We have effectively reduced the main puzzle to a simpler one:

Let us first observe that if $R$ also always spoke truthfully, the puzzle would become easy. We could simply split the 6 possible assignments into two groups of three, say Group A and Group B, and ask:

After the first question, only 3 possibilities would remain. A second question would reduce this to at most 2, and a third would identify the correct one.

Returning to the original setting, one of these bits is ‘foggy’, because $R$ answers at random. At first glance, this suggests that any answer from R is useless, but this is not quite true. We begin by asking $G_1$ the following question:

The answer leads to the following chain of conclusions:

1. If the answer is Yes, then either it is true and $G_2 = R$, or it is a lie, which implies $G_1 = R$. In either case, we conclude that $G_3 \ne R$.
2. Similarly, if the answer is No, then $G_2 \ne R$.
3. In either case, we have identified at least one goddess who always speaks the truth. From this point on, we can direct all remaining questions to her.
4. We are also left with only 4 possible permutations. For instance, if $G_2 \ne R$, then both $(T, R, F)$ and $(F, R, T)$ are eliminated.

We are now done, because two questions are enough to determine which of the remaining 4 permutations is correct!

Addendum: Is it a Solution or the Solution?

The proverbial careful reader might have noticed that the question we asked $G_1$ in the previous section was somewhat arbitrary. It does the job, but is it really necessary?

Let us take a step back. Any question we ask is, in effect, meant to identify a subset of permutations of $ T,F,R $ containing the true arrangement. In this sense, it can be viewed as equivalent to a question of the form:

where $\mathcal{S}$ may be any subset of the 6 possible permutations. Since each permutation either belongs to $\mathcal{S}$ or not, there are $2^6 = 64$ possible questions.

What Makes a First Question Work?

This boils down to two simple requirements:

• It must leave us with at most 4 candidate permutations. Otherwise, two further questions are not enough.

• We must be able to identify at least one goddess who is not $R$.

Let’s briefly explain why the second condition is necessary. Suppose our question concerns the subset

and the answer is Yes. The diagram below shows which permutations remain possible depending on whether the goddess lies or not. Remember that if $G_1$ lies, then necessarily $G_1 = R$.

Now, regardless of the second question, we are not guaranteed to eliminate 2 candidate permutations. Any of the goddesses may still be $R$, so if we direct the second question, say, to $G_2$, the permutation $(F,R,T)$ always remains possible.

If $G_2 \not= R$, then the answer can at best split the remaining 3 permutations into two groups and eliminate one of them. Since we cannot guarantee that $G_2 \not= R$, at least 3 permutations must still be considered: 1 arising from $G_2=R$ and 2 from $G_2\not=R$. This is too many to resolve with a single remaining question.

The same argument applies if we direct the second question to $G_1$ or $G_3$ instead.

Counting the Possible Solutions

We now understand the conditions that the first question must satisfy. Let us count the valid choices of $\mathcal{S}$, the subset of permutations specified by the question to $G_1$. First, whether $\mathcal{S}$ contains $(R,F,T)$ or $(R,T,F)$ doesn’t matter, since both remain possible anyway; at this stage, we cannot exclude $G_1=R$. Among the remaining 4 permutations, we must distinguish between the cases $G_2=R$ and $G_3=R$, so that we can identify a non-$R$ goddess and direct the remaining two questions to her.

In summary, any valid subset $\mathcal{S}$:

• May contain $(R,F,T)$ or not.

• May contain $(R,T,F)$ or not.

• Must contain either the 2 permutations corresponding to $G_2=R$ or the 2 permutations corresponding to $G_3=R$, but not all four.

This gives 8 valid first questions in total, so $8/64=12.5 \%$ of all possible choices lead to a complete solution.

On the other hand, all of these questions serve the same purpose: to quickly isolate a truthful goddess we can rely on. In that sense, whether the puzzle has a ‘unique’ solution is largely a matter of interpretation.

The unbiased variance estimator is given by the well-known formula:

where $X_1,\dots,X_n$ are independent draws from the same distribution and $\overline{X}$ denotes the sample mean.

Every so often, someone in the room eventually asks:

There are two answers one usually hears:

1. Write down the bias $\mathbf{E}[\widehat{\sigma}^2] - \sigma^2$, expand the square, expand $\overline{X}$, expand all the sums, calculate. See? It’s $0$.
2. Because there are only $n-1$ degrees of freedom among the $X_i$, since translating all measurements by the same amount leaves the variance unchanged.

The first answer is formally correct, but it explains little. The second one may feel hand-wavy. Why should degrees of freedom matter at all? Where, exactly, did the missing dimension go?

Since the difference between $\frac{1}{n}$ and $\frac{1}{n-1}$ is most visible for small values of $n$, let us begin by examining what happens in that regime.

Some Intuition: Case $n = 1$

Suppose we have only a single random measurement, and we observe $X_1 = 10$.
There are many pairs $(\mu,\sigma)$ consistent with such an observation, for example:

1. $\mu \approx 10$ and $\sigma$ very small,
2. $\mu \approx 10$ and $\sigma \approx 50$.

There is no way to decide which scenario is closer to the truth. The scale of $\sigma$ can vary by orders of magnitude while remaining compatible with the data. A single observation provides no reliable information about the variance, so an unbiased estimator of $\sigma^2$ cannot exist in this setting.

Now suppose $\mu = 15$ is known and we observe $X_1 = 10$. While we still cannot estimate $\sigma$ precisely, we at least gain information about its order of magnitude. For instance, it is unlikely that $\sigma \ll 1$. This is weak information, but it is no longer nothing.

When the exact value of the mean is known, we are granted a point of reference. For fixed $\mu$, we can construct a slightly different unbiased estimator:

As we can see, the $-1$ disappears from the denominator. To summarize:

• For $n = 1$, it makes sense that the variance cannot be estimated unless $\mu$ is known.
• Replacing $n$ with $n - 1$ in the denominator can be viewed as a premium paid for not knowing the exact value of $\mu$.

The second point may seem vague for now. In the following sections, we make this intuition more precise.

Orthogonal Projections

When thinking about $\widehat{\sigma}^2$ geometrically, we may view it as the squared Euclidean norm of an $n$-dimensional vector, normalized by $n-1$. If we define the sample vector

and the center-of-mass vector

where $\mathbf{1}$ stands for the all-ones vector, then

Here $\lVert \cdot \rVert_{\ell^2}$ denotes the Euclidean norm, the standard way mathematicians refer to “distance from zero”.

Neat. We are interested in the expected distance from $\boldsymbol{X} - \boldsymbol{C}$ to $\mathbf{0}$. Let us notice a simple but important property of this vector. If we define $X_i’ = X_i - \overline{X}$ for $i = 1, \dots, n$, then

This shows that the random vectors $\boldsymbol{X}-\boldsymbol{C}$ are perpendicular to the all-ones vector $\mathbf{1}$. Since $\boldsymbol{C}$ itself is a scalar multiple of $\mathbf{1}$, it follows that $\boldsymbol{X}-\boldsymbol{C}$ is the orthogonal projection of $\boldsymbol{X}$ onto the hyperplane

Let us take a look at a simple case in dimension two.

Even though the scenario depicted above may appear overly simplistic, it still allows for a few useful observations.

• If $\mu = 0$, the squared distance of the blue point from the origin represents $n\,\widehat{\sigma}^2_{\mu\text{ known}}$. Changing $\mu$ simply translates the point along the $[1,1]^\top$ direction, while its projection remains fixed.
• Similarly, the squared distance of the red point from the origin represents $(n-1)\,\widehat{\sigma}^2$, where we pretend that the true value of $\mu$ remains unknown.

Basic trigonometry shows that we have

which in the $\mu = 0$ case implies

Now observe that replacing $n-1$ with $n$ in the definition of $\widehat{\sigma}^2$ would lead to a highly dubious identity

Since $\lvert \cos(\alpha) \rvert \leq 1$, this would systematically push the variance estimate downward. This is precisely the bias we want to avoid.

While this is not yet a proof, it is already clear that using $\frac{1}{n}$ is a wrong choice of the normalization factor when constructing $\widehat{\sigma}^2$.

More Intuition: Case $n = 2$

In this experiment we sample $\mu$ from the interval $[-3, 3]$. For this mean we draw $(X_1, X_2)$ pairs $500$ times, where $X_1$ and $X_2$ come from the normal distribution $\mathcal{N}(\mu, 1)$. For each pair, we estimate the variance by computing both $\widehat{\sigma}^2$ and $\widehat{\sigma}^2_{\mu \text{ known}}$.

We proceed in the same way as in the previous subsection:

• To obtain $\widehat{\sigma}^2_{\mu \text{ known}}$, we compute the distance from $(X_1, X_2)$ to $(\mu, \mu)$.
• To obtain $\widehat{\sigma}^2$, we compute the distance from the origin to the projection of $(X_1, X_2)$ onto the line $X_1 + X_2 = 0$.

As the animation shows, both estimators remain constant as $\mu$ varies. This is expected: $\widehat{\sigma}^2_{\mu \text{ known}}$ is proportional to the distance from the center of the blue mass, so translating it changes nothing, while $\widehat{\sigma}^2$ is based on projections, and translating the blue mass along the $[1,1]^\top$ direction does not affect its projection.

We observe that the deviation of the red dots is systematically smaller, since one direction is removed entirely. This is precisely the geometric origin of the $n - 1$ degrees of freedom.

What is the main source of this discrepancy? Consider a blue point $\boldsymbol{X} = (4.01, 4.03)$ under $\mu = 2$. Being far from the center of the blue mass, it contributes substantially to the average deviation. After projection, however, the situation looks different. The same point is mapped to $\boldsymbol{P}(\boldsymbol{X}) = (-0.01, 0.01)$, which lies very close to $\mathbf{0}$.

This reflects the fact that observing $X_1 = 4.01$ and $X_2 = 4.03$ with $\mu$ unknown naturally suggests values concentrated around $4$ and a small variance. While $n = 2$ hardly constitutes a meaningful sample, this interpretation is at least reasonable if one is forced to make a judgment.

To observe a similar effect for larger samples (namely, all measurements clustering together while remaining far from the true mean) we would need to be rather unlucky. This explains why replacing $\frac{1}{n}$ with $\frac{1}{n-1}$ becomes almost irrelevant for large $n$, but crucial when the sample is small.

Geometric Argument, Made Formal

Let’s play around a bit with the expression

As we have already seen, the sum looks like an $\ell^2$ norm of something. To make this precise, let’s re-introduce the following notation:

We can then form the vector $\boldsymbol{X} - \boldsymbol{1}\overline{X}$, noting that $\overline{X}$ is just a scalar:

We can now see that $\widehat{\theta}$ is obtained by summing the squares of the entries of the vector above. That gives us

This observation also allows us to rewrite $\widehat{\theta}$ in a more compact way, more suitable for matrix calculations:

We notice that the expression $\boldsymbol{X} - \boldsymbol{1} \overline{X}$ is meaningful in some way. It would be neat to have a linear operator $\boldsymbol{P}$ that satisfies

so we could study its properties and hopefully get some insights. To get there, we need to describe $\overline{X}$ in terms of $\boldsymbol{X}$:

Of course, the dot product above could be also described as $\boldsymbol{X}^\top \boldsymbol{1}$, but we need $\boldsymbol{X}$ itself not to be transposed. Then,

Just for the record, the matrix $\boldsymbol{P}$ is:

Fine, we succeeded in creating an operator $\boldsymbol{P}$ of the desired property. We get

The matrix/operator $\boldsymbol{P}$ has two critical properties:

1. It is symmetric, so $\boldsymbol{P}^\top = \boldsymbol{P}$.
2. It effectively subtracts the mean (or the center of mass) from a vector. Doing it twice changes nothing, because the mean has already been removed after the first pass. Hence, $\boldsymbol{P}^2 = \boldsymbol{P}$.

In other words, it’s an orthogonal projection (as already observed in the previous sections). Thus, we get

To answer what factor fits an unbiased estimator of variance, we need to calculate the expected value of the expressions above. To this end, we need a supplementary result.

Proof. $~$ Unfortunately, we need to simply expand. It doesn’t seem like there is anything more clever we could pull off:

Provided that $i \neq j$, we have

On the other hand, if $i=j$, then $\mathbf{E} \left[ X_i X_j \right] = \mathbf{E} \left[ X^2 \right]$ and we notice

which gives $\mathbf{E} \left[ X^2 \right] = \sigma^2 + \mu^2$.

This gives us

□

In our discussion we are interested in calculating $\mathbf{E} \left[ \boldsymbol{X}^\top \boldsymbol{P} \boldsymbol{X} \right]$. This turns out to be quite simple, because the sum of all the entries vanishes. We thus have

That’s all. To make the estimator unbiased, we just divide $\widehat{\theta}$ by $n-1$.

Appendix: What if $\mu$ is known?

Let us follow the same geometric argument. First, we note that

We also observe that for the sample vector $\boldsymbol{X}$ it holds that

In other words, we now deal with $\boldsymbol{X}^\top \boldsymbol{X}$ instead of $\boldsymbol{X}^\top \boldsymbol{P} \boldsymbol{X}$. The projection matrix $\boldsymbol{P}$ got replaced by the identity matrix $\mathbf{id}$. Applying Little Lemma, we obtain

since $\text{tr}(\mathbf{id}) = n$. This yields

The means cancel out, as they should, since the variance has nothing to do with $\mu$. The estimator $\widehat{\sigma}^2_{\mu\ \text{known}}$ is unbiased, and the factor $n-1$ simply does not show up here.