Anyone on Veeky Forums smart enough to solve this one?

Anyone on Veeky Forums smart enough to solve this one?

Hint: use the axiom of choice

>shape rooms and boxes into spheres
>break into pieces, reassemble twice as many equal rooms and boxes
>now twice as many rooms, each mathematician gets 2 rooms
>they go into second room an open the box they left closed in the first room

I can't imagine how this is possible. Got a link to the solution?

If they can't communicate with each other after they enter the room then there is no difference in strategy between this game and a game where there is only one mathematician and one room. And for one mathematician in one room, there are an infinite number of possible reals in each box, so it's impossible to guess with any kind of certainty what it could be no matter how many boxes are opened. This game has no solution.

It's possible. The boxes leave a literal paper trail. Alternate question is the 100 prisoners question.

I think it's different because no particular mathematician has to guarantee that he's correct, but it should be equivalent to two rooms with one being correct.

Have each mathematician walk into the room next to theirs, open a box, and then 'guess' the corresponding box in their room

Oh I think you're right. I think the key is in the phrase
>Then each mathematician must guess the real number that is contained in a particular box of his choosing
The mathematicians should choose a box that they themselves have opened but another mathematician hasn't opened. For example, one solution is for the mathematician in room 1 to open box 2 and guess the number he sees for the unopened box 2 in room 2. And equally the mathematician in room n should open box n+1 and guess that number for the unopened box n+1 in room n+1. The mathematician in room 100 should open box 1 and guess for the unopened box 1 in room 1.

>what is the winning strategy
Don't play

the only strategy I can see that all mathematicians agree to pick the same box and guess the same real number
its unclear how they would deduce whats in the boxes so why even open them
I dislike how the puzzle never says that they only get one guess, so you could just guess until you die. You could order them by least to greatest, assign each person a box, then have them guess infinitely.

ignore the first bit obviously, I did think of another strategy

Define an equivalence relation on sequences of real numbers that 2 sequences are equivalent of they differ in only finitely many places. During planning The mathematicians choose a representative from each equivalence class and number themselves off 1-100

When mathematician k gets to his room he opens all of the boxes except those that are k mod 100. Then for each number j between 1 and 100 except k he looks at the subsequence of boxes that have numbers equal to j mod 100. He finds what equivalence class this sequence belongs to then he finds the greatest value where this subsequence differs from the representative of its equivalence class.

Now he turns his attention to the sequence of boxes with labels k mod 100. He chooses a box in this sequence that is farther along in the subsequence than any of the differences in the other 99 subsequences. He leaves this box unopened then opens all the other boxes in the subsequence. This lets him know what equivalence class this subsequence belongs to. Then he guesses that the number in the unopened box is the same as the corresponding number in the representative of that equivalence class.


The only possible way that mathematician k could guess wrong is if the subsequence of boxes with numbers k mod 100 differs from the representative of its equivalence class at a place farther along than the subsequence of boxes j mod 100 for every other j in the range 1-100.

This can happen for at most 1 k in the range 1-100 so only one of the mathematicians could possibly guess wrong. So at least 99 will guess right and we've found a winning strategy.

Is the key to this that a strategy that fails in a countable number of configurations guarantees success?

I wish I was intelligent enough to come up with this

These sorts of proofs always bother me because I can understand them (with a small bit of work) but I can't for the life of me see how anyone would ever make it through all the steps of coming up with the whole thing by themselves

I wonder how many people actually do compared to learning the solution

>t. brainlet

>When mathematician k gets to his room he opens all of the boxes except those that are k mod 100.
But there are only 100 rooms, so wouldn't k mod 100 always be k? Same with j mod 100. Why are you making this more complicated than it needs to be, it makes me think your solution is complete bullshit.

"all the boxes that are k mod 100" is boxes k, k+100,k+200,...

I'll try to give a run down of my though process

First theres a simpler version of this problem that goes like this. Infinitely many people stand in a line. They each have a real number floating above their heads. They can see the numbers above each person in front of them. Each page erson guesses the number above their own head. Find a strategy so all but finitely many will guess correctly.

The solution to this problem is to define the same equivalence relation. And again pick a representative for each class. Then each person just guesses the corresponding number from the representative. This guarantees at most finitely many guess wrong.

Now since this problem is kinda similar we know that starting with that equivalence relation and a representative for each class could be helpful. But in this problem we need more than just all but a finite number we need all but one. So we know we're gonna need another trick.

Te mathematicians get to pick which box they guess. The goal is to pick a box that's far enough along in the sequence that it's past all the places where the sequence differs from the representative. So we want our mathematicians to choose high numbers to guess on. More precisely we know that we need the box they guess on to be unboundedly big because if it was bounded we would get screwed when we had a sequence that differed past the bound.

They only way to get the guesses unboundedly big is If they're not guessing the same box each time. So we know that the strategy has to look something like "open up some boxes, choose a box, open up some more boxes, guess". Now when we do our first opening step we have to leave infinitely many boxes closed in order to leave the possibility of our guess being unboundedly large.

But now we have an issue. If we leave infinitely many boxes closed we don't know which equivalence class our sequence belongs to. But we still want to use our representatives somehow so what can we do? The answer is we must be looking at the equivalence class of a subsequence instead of the whole thing.

What subsequences should we look at? Well there's 100 mathematicians so let's try breaking it into 100 subsequences using mod 100.

From this point the rest of the solution kinda just falls out

>just know about a virtually identical case that uses the exact same equivalence class+AoC representative argument with a small modification in the remainder of the argument
well yeah, in that case it's not too hard to work out

every mathematician is assigned a number 1-100
every mathematician strategizes a way to index the boxes (ie, when they're in their rooms all the boxes have the same number system)
when they get to the room, each math guy starts by opening the box corresponding to his number
then he opens the box that was inside the box he previously opened, and repeat until he gets his number
not foolproof, but better than anything the other brainlets in the thread came up with

That's the solution to a completely different brain teaser. Are you trolling or did you not read

I wish Veeky Forums had more of these kinds of posts.

Wait, so, if 100 of 100 guess correctly then they (collectively) lose the game?

That's how Math works. You build up from problems you've already solved. But if that's not satisfying for you heres my thought process for the other problem.

First we notice that it is very unintuitive that this should have a solution. It send that the probability of any person guesding right should be 0 but at the end almost all of them guys right. That's weird.

An unintuitive result on a problem involving infinity hints that we should be using the axiom of choice.

We know from vitali's construction of a non measurable set that one way to get weird results with the axiom of choice is to pick a representative from each of infinitely many different equivalence classes. Let's see if we can apply that to this problem.

What should our equivalence relation be? Well we have sequence of real numbers so maybe we should look at relations over sequences of real numbers.

The problem gives us a hint by asking for all but finitely many mathematicians to get the answer right. That leads us to consider the relation that two sequences are equivalent if they differ in finitely many places. It is easily checked that this is an equivalence relation.

So now we assume the mathematicians have agreed upon a representing for each of the equivalence classes of this relation and the rest of the solution just works out.

No. 99 or more need to guess right

Damn. 3 typos in that second paragraph. I hate typing on my phone.