idk, idk, idk, ik - wtfff

28 Mar, 2025

My favorite logic puzzle is the Blue eyes prison island one. If you've never heard of it, you should check it out! I recently came across an easier version of this puzzle and I want to go over this version.

The Question

There's three people in a circle. Each person has a positive integer floating above their heads, such that each person can see the other two numbers but not their own. The sum of two of the numbers is equal to the third. The first person is asked for his number, and he says that he doesn't know. The second person is asked for his number, and he says that he doesn't know. The third person is asked for his number, and he says that he doesn't know. Then, the first person is asked for his number again, and he says: 65. What is the product of the three numbers?

Important Assumptions

There are some implicit assumptions here that aren't obvious to many people:

Everyone is honest and truthful. And everyone knows everyone is honest. And everyone knows everyone...
No one is guessing anything; everything said is true and follows from pure logic on the speaker's end
Everyone involved is a perfect logician - if something can be inferred from logic, it will be inferred

You should really spend some time at least thinking about this puzzle. It's not easy, feel free to grab a pen/paper. Who knows, maybe this puzzle could keep you amused for a few hours or more.

Note: It's not quite the same as the blue eyes puzzle, and is less absurd as well. Particularly because in the original no new information seems to be revealed. It is completely reliant on common knowledge.

Solution

spoiler

This explanation is inspired by u/Leet_Noob.

Let me first define common knowledge for the rest of the post. A statement $S$ is common knowledge if everyone involved knows that $S$ is true and everyone knows that everyone else also knows that $S$ is true and everyone knows that everyone knows... and so on.

In our example, this means Person 1 knows $S$ , Person 1 knows that Person 2 knows $S$ , Person 1 knows that Person 2 knows that Person 3 knows $S$ . And all other permutations of persons.

Another assumption is that the assumptions listed above are all common knowledge.

Let's denote the three people as A, B, C. Let the numbers be denoted as $[x_{1}, x_{2}, x_{3}]$ .

Constraints:

$x_{i} \in ℤ^{+} ⟹ x_{i} > 0$ - all numbers are positive
$x_{i} \in {| x_{j} - x_{k} |, x_{j} + x_{k}}$ - for any person, their number is one of two possible from observing the other two numbers

Let's go through what each person now says, and what that conveys to everyone (becomes common knowledge):

Event 1: A says IDK

If $x_{2} = x_{3}$ , then $x_{1}$ would have to be $x_{2} + x_{3}$ because $x_{2} - x_{3} = 0$ is not possible.

Common Knowledge: Everyone knows $[2 x, x, x]$ is not possible.

Event 2: B says IDK

We now have to figure out the scenarios which, if true, would allow B to know the answer. This way we can eliminate more possibilities.

Since $x_{2} \neq 0$ , we know $x_{1} \neq x_{3}$ . Otherwise for $x_{2}$ we would have possibilities ${x_{1} + x_{3}, 0}$ . This implies $[x, 2 x, x]$ is also not possible.

Since $[2 x, x, x]$ is not possible, we also cannot have $x_{1} = 2 x_{3}$ . This is the crucial logic to understand. If $x_{1} = 2 x_{3}$ was actually true, then B would see two possibilities for his own number - ${x_{3}, 3 x_{3}}$ . And since $x_{2}$ cannot be equal to $x_{3}$ (or A wouldn't have said IDK in that case), B can conclude $x_{2} = 3 x_{3}$ . And since this does not happen, everyone figures out that we don't live in the $[2 x, 3 x, x]$ world.

In short, if B saw $x_{1} = 2 x_{3}$ he would immediately conclude $x_{2} = 3 x_{3}$ since $x 2 = x 3$ is not possible.

Common Knowledge of impossible worlds:

$[2 x, x, x]$
$[x, 2 x, x]$
$[2 x, 3 x, x]$

Event 3: C says IDK

Following similar logic - we list more scenarios in which C wouldn't say IDK, and thus can be eliminated:

Since $x_{3} \neq 0$ , we know $x_{1} \neq x_{2}$ . Thus $[x, x, 2 x]$ is eliminated.
Since [2x,x,x] is not possible, x1≠2x2. Eliminate [2x,x,3x].
- Otherwise, looking at $x_{1} = 2 x$ and $x_{2} = x$ , C would conclude $x_{3} = 3 x$ (since $x_{3} \neq x_{2}$ or A wouldn't have said IDK).
Since [x,2x,x] is not possible, 2x1≠x2. Eliminate [x,2x,3x].
- Otherwise, seeing $x_{1} = x$ and $x_{2} = 2 x$ , C would conclude $x_{3} = 3 x$ (since $x_{3} \neq x_{1}$ or B wouldn't have said IDK).
Since [2x,3x,x] is not possible, 3x1≠2x2. Eliminate [2x,3x,5x].
- Otherwise, seeing $x_{1} = 2 x$ and $x_{2} = 3 x$ , C would conclude $x_{3} = 5 x$ (since $[2 x, 3 x, x]$ or B wouldn't have said IDK).

Common knowledge of impossible worlds:

$[2 x, x, x]$
$[x, 2 x, x]$
$[2 x, 3 x, x]$
$[x, x, 2 x]$
$[2 x, x, 3 x]$
$[x, 2 x, 3 x]$
$[2 x, 3 x, 5 x]$

Event 4: A says IK

And you're like WTFF???

Until now we've figured out the scenarios in which the speaker would know their number to eliminate those worlds. Now we do the same thing, but to figure out which world we live in.

From A's point of view, there are always two possible worlds - ${| x_{2} - x_{3} |, x_{2} + x_{3}}$ . Since we have a list of eliminated worlds (as does A), we can list down worlds in which A would be able to eliminate one of his two options:

This is calculated by looking at the eliminated worlds and figuring out the complementary world from A's pov. Complementary in ${| x_{2} - x_{3} |, x_{2} + x_{3}}$ , so if the eliminated world is $| x_{2} - x_{3} |$ its complement is $x_{2} + x_{3}$ and vice versa.

Eliminated world $\to$ Only possible world in given scenario

$[2 x, x, x] \to [0, x, x]$
$[x, 2 x, x] \to [3 x, 2 x, x]$
$[2 x, 3 x, x] \to [4 x, 3 x, x]$
$[x, x, 2 x] \to [3 x, x, 2 x]$
$[2 x, x, 3 x] \to [4 x, x, 3 x]$
$[x, 2 x, 3 x] \to [5 x, 2 x, 3 x]$
$[2 x, 3 x, 5 x] \to [8 x, 3 x, 5 x]$

Of these, the possible worlds are all but $[0, x, x]$ . Since A says 65, and $65 = 5 \times 13$ , we know that the only possible world is $[65, 26, 39]$ where $x = 13$ and we have $[5 x, 2 x, 3 x]$ .

So the product, $x_{1} \times x_{2} \times x_{3} = 65 \times 26 \times 39 = 65, 910$ .

The original is a great follow-up to this, and likely significantly more difficult and fun.

#fun #im-1 #mine #puzzle