Math Puzzle Thread

[Lul Thyme]

Originally posted by Kuciwalker
Obviously they schedule themselves. Line up, every minute the next one gives a guess. This guess is based on some function of the others. I'm trying to think of one that guarantees a correct guess by the last one in line. The only information each has is that all previous guesses were false.

They do not hear each other's guesses, or cannot base themselves on that.
IF they could, they could jsut agree that number one will guess number two's color and vice versa, and whoever guesses right...
read my post above.

[Lul Thyme]

Originally posted by Kuciwalker
I'm going to poach some from my CS homework. This was an easy one (we had to prove it, but the proof is trivial once you know the answer):

After a long day of 251 homework, you and your friend decide to treat yourself to a circular 12-cut
pepperoni pizza. When the pizza gets to your place, Vocelli’s makes dividing the pizza hard on you, by
putting a different number of pieces of pepperoni on each slice. You and your friend decide to divide
the pizza in the following way. First, you choose and eat any slice from the pizza. Then, you both
alternate turns by taking and eating a slice from the pizza, but only one of the slices that borders the
gap left by the removed slices.

Is there a strategy you can use to ensure that you will have eaten at least as many pieces of pepperoni
as your friend, once the pizza is fully consumed?

The problem is not very well stated.
IT's not clear if the you in
"First, you choose and eat any slice from the pizza."
is just me, or both of us.
And it's not clear who starts alternating.
But the answer will be obviously that you sum the odd and even ones, and ensure that you get the ones with higher total.


Btw Kuci, this one is just a little more "trivial" than the other one, which made you squirm for some time, so save it

[Kuciwalker]

They do not hear each other's guesses, or cannot base themselves on that.

I didn't say they could hear each other. The 1st person makes a guess (silently), and if they're freed, then no one else need make one. After 1 minute the 2nd person knows that the first person's guess was wrong, which gives him a little extra information. By the last person someone will have gotten enough information to guess correctly.

Of course, this still works if they all guess simultaneously, but I found it easier to think of it with them guessing in sequence. You'll notice I already posted a solution.

[Lul Thyme]

Originally posted by Kuciwalker

They do not hear each other's guesses, or cannot base themselves on that.

I didn't say they could hear each other. The 1st person makes a guess (silently), and if they're freed, then no one else need make one. After 1 minute the 2nd person knows that the first person's guess was wrong, which gives him a little extra information. By the last person someone will have gotten enough information to guess correctly.

Of course, this still works if they all guess simultaneously, but I found it easier to think of it with them guessing in sequence. You'll notice I already posted a solution.

I know you posted a solution but I think this explanation above is wrong.
Explain how knowing others got it wrong gives more information?

[Kuciwalker]

If the other person's guess is (partly) a function of your hat, and you know that the guess was not the same as the color of their hat (which you can see), then you know a little bit more about your own color. Assuming the function was properly constructed, of course.

[Kuciwalker]

F(ith person) must take on a different value for each different color of j's hat. Therefore, if j knows F(i) != some number, j has just eliminated a possible color for his hat.

[Lul Thyme]

Originally posted by Kuciwalker
If the other person's guess is (partly) a function of your hat, and you know that the guess was not the same as the color of their hat (which you can see), then you know a little bit more about your own color. Assuming the function was properly constructed, of course.

Well I guess it's unclear what constitutes extra information, since there is already enough to guarantee once gets it right, and I don't think that if we allow for them knowing if others "before" them got it right allows for stronger results.

[Lul Thyme]

Originally posted by Kuciwalker
F(ith person) must take on a different value for each different color of j's hat. Therefore, if j knows F(i) != some number, j has just eliminated a possible color for his hat.

Can you show me a system where you can guarantee more than "one person will guess his hat correctly"?
If not, then there is no extra information.

[Kuciwalker]

The cases end up being exactly the same. It's just what I thought was an easier way of thinking of the problem.

[Lul Thyme]

Originally posted by Kuciwalker
The cases end up being exactly the same. It's just what I thought was an easier way of thinking of the problem.

So there was no extra information, which was my point

[Kuciwalker]

Well, there's a little. By the last person, if no one has gotten it right, he knows his guess is correct.

[Kuciwalker]

It's your turn btw.

[Lul Thyme]

Here's one I heard at the same party.
It's not as fun though.


"You have a 3x3x3 cube, you start in one corner and want to end in the middle, traversing each 1x1x1 cube exactly once, going from a 1x1x1 cube to an adjacent one (sharing a face).
Can you?"



(is there a Hamiltonian path in the graph starting in the corner and ending in the middle).

[Lul Thyme]

Originally posted by Kuciwalker
Well, there's a little. By the last person, if no one has gotten it right, he knows his guess is correct.

In the solution to the original problem with "no extra information", he also knows that if nobody else gets it right, his guess is correct.

I guess it all depends if the game stops when somebody is right.
You could argue that in your version, if they keep playing after somebody gets it right, then they can all get it right.

[Kuciwalker]

I strongly suspect the answer to your puzzle is no, but don't know the proof yet.