** STICKY ** League of Reason? PROVE YOURSELF!

[Fictionarious]

There are three logicians who are able to reason and deduce with perfect accuracy, their names are A, B, and C. They each know this as well.
In a bag there are seven colored stickers. Two yellow, two red, and three green. The logicians know that too.
Each logician reaches into the bag, grabs one sticker and slaps it on his forehead without looking to see what color it is. Any one logician is able to see the stickers on his compatriot's heads, but not the sticker on his own.

Jerry the interrogator asks logician A if he knows what color sticker is definitely not on his forehead.
"No.", A responds.
Jerry asks logician B the same question, and receives the same answer.

Is it possible with this information to deduce the color of any of the logician's stickers?

 

[ImprobableJoe]

Re: League of Reason? PROVE YOURSELF!

Ummm... yeah, if you open the bag and see what colors are left?

 

[IrBubble]

Re: League of Reason? PROVE YOURSELF!

We know that C is green.

When A looks around, since he is not able to tell which color he can't be, he either sees that B and C has different colors or that they both have green. B can deduce this from hearing A's statement. So when B looks around, for him not to be able to know which color he can't be, C has to be green, in any other case B would know a color he could not be.

@ iJoe: http://www.youtube.com/watch?v=Cdiz0k0Rudw

 

[Aught3]

Re: League of Reason? PROVE YOURSELF!

Since I assume there is an answer it can neither be yellow nor red because they have the same number of items and can't be differentiated. Therefore it must be green.

 

[Fictionarious]

Re: League of Reason? PROVE YOURSELF!

We know that C is green.

When A looks around, since he is not able to tell which color he can't be, he either sees that B and C has different colors or that they both have green. B can deduce this from hearing A's statement. So when B looks around, for him not to be able to know which color he can't be, C has to be green, in any other case B would know a color he could not be.

@ iJoe: http://www.youtube.com/watch?v=Cdiz0k0Rudw

WINRAR

Okay... let's see if you can handle this(!) -

A dealer bought an article for $7, sold it for $8, bought it
back for $9, and sold it for $10. How much profit did he
make?

btw, this is obviously a logic puzzle thread. If you think of any one good enough to contribute, do so, and we shall collaboratively improve the superpower of our reasoning, thus truly earning our membership in THE LEAGUE OF REASON.

 

[IrBubble]

Re: League of Reason? PROVE YOURSELF!

WINRAR

Okay... let's see if you can handle this(!) -

A dealer bought an article for $7, sold it for $8, bought it
back for $9, and sold it for $10. How much profit did he
make?

btw, this is obviously a logic puzzle thread. If you think of any one good enough to contribute, do so, and we shall collaboratively improve the superpower of our reasoning, thus truly earning our membership in THE LEAGUE OF REASON.

$2.

-7+8-9+10 = 10+8-(7+9) = 18-16

 

[Fictionarious]

Re: League of Reason? PROVE YOURSELF!

Okay. Now for the big guns.

There are twelve nearly identical metal balls. They all are the exact same weight, except for one, you don't know whether it's heavier or lighter than the others, just that it has a different weight. Other than this minor difference they are identical.

You have a equal arm balance.

You have been commissioned to find the odd ball out in just three weighings or less.

How would you proceed?

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

Okay. Now for the big guns.

There are twelve nearly identical metal balls. They all are the exact same weight, except for one, you don't know whether it's heavier or lighter than the others, just that it has a different weight. Other than this minor difference they are identical.

You have a equal arm balance.

You have been commissioned to find the odd ball out in just three weighings or less.

How would you proceed?

Label the balls A,B,C,...,I.

In the first weighing, put ABC in one pan, DEF in the other.

If they're level, it's one of G, H or I. You've got two weighings left, and you know all of A-F are kosher, so just pick one of those six and weigh G then H against them. The one that isn't level is the odd one, if neither is odd then it's I.

If the first weighing isn't level, you know G, H and I are OK. You now weigh ABD against CEG.

If they're level, you know that F is the odd one.
If the scales tip the opposite way to before, you know it's one of the two you swapped to the opposite pan, C or D. Just weigh C against anything but D, if it's level, D is odd, otherwise it's C.
If the scales tip the same way, the odd one is one of ABE, the ones that stayed in the same pan. Weigh A against B. If they're level, the odd one is C. If the scales stay tipped the same way, it's A. If they tip the opposite way, it's B.


These are all very old and familiar problems, by the way. I worked this one out (because I quite enjoy these things) but I could have easily looked it up. Got any original ones?

 

[IrBubble]

Re: League of Reason? PROVE YOURSELF!

Marcus you don't have nearly enough balls in your solution A -> I = 9 not 12.

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

Nuts, there I go assuming it's a problem I'm familiar with.

I clearly have to solve this properly to save face now...

 

[The Apathetic Despot]

Re: League of Reason? PROVE YOURSELF!

I have rather an interesting puzzle which I'd like to add if you don't mind, Fictionarious. It's not really in the same style as the others, but I think it's fun anyway.

Imagine you had a strip of paper with one end held in each hand, now fold the paper by bringing your left hand over your right hand. Without unfolding the paper repeat the folding left over right until you've folded it n times. When you unfold the paper you're left with a pattern of creases, some pointing up and some pointing down. The puzzle is to find a method of predicting the pattern of creases for any given number n.

I'll give you the first three patterns in case you have no paper handy:

one fold:
down

two folds:
up, down, down

three folds:
up, up, down, down, up, down, down

Edit:
oh and for the 12 balls question: 6 on each arm, one arm will be weighed down. 3 of those 6 on each arm, again one will be weighed down. Put one of those 3 on each arm and either they will balance and the third is heavier or they won't and you want the heavier of those two. It's sort of like a binary search.

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

Actually, it's not much harder. Label them A to L (see, 12 this time!)

Weigh ABCD vs EFGH.

If they're level, it's one of IJKL. Weigh AB against IJ. If it's level, the odd one is one of KL, so weigh K against A to determine which; otherwise weigh I against A to determine which of I and J is odd.

If the first weighing is off, weigh ABEJ against DFKL.

If this is level, the odd ball is one of the ones we've removed, C, G, H. Weigh G against H. If the scales are level, C is odd. If they tip the same way as the first weighing, H is odd. If they tip the other way, G is odd.

If the scales tip the opposite way in the second weighing than the first, the odd one is one of the balls that swapped sides, D or E. Weigh D against A to determine which.

If the scales tip the same way in the second weighing than the first, it's one of ABF. Weigh A against B. If they're level, F is odd. If the scales remain tipped the same way, it's A. If they tip the other way, it's B.

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

oh and for the 12 balls question: 6 on each arm, one arm will be weighed down. 3 of those 6 on each arm, again one will be weighed down. Put one of those 3 on each arm and either they will balance and the third is heavier or they won't and you want the heavier of those two. It's sort of like a binary search.

That only works if you know the odd ball is heavier. All we know is that it has a different weight. If the odd ball is lighter, your search fails at step 2.

 

[The Apathetic Despot]

Re: League of Reason? PROVE YOURSELF!

That only works if you know the odd ball is heavier. All we know is that it has a different weight. If the odd ball is lighter, your search fails at step 2.

Drat, you're right, I missed that line.

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

I can give an iterative pattern for the paper folding. The seed case is n=1, whose single fold is down.

There are clearly 2^n - 1 folds.

For n>1, the odd numbered folds are U D U D... repeatedly.

The even numbered folds are the fold pattern from the (n-1) case.

With a little thought, you can see why this will be the case (if you can follow me without diagrams, which isn't easy!). Your last fold will put a crease at the beginning and end and between each of your existing creases. Those creases are already made, which is why the even ones preserve the old pattern. The new creases will also clearly alternate, because the previous folds mean that every time you go over a crease, the paper is the other way up. To determine that the first fold must be upwards (and see why the n=1 case is the only one where it isn't), recall that the first fold puts the left end of the paper directly on top of the bottom layer, upside down from its original position, where it stays for the rest of the folding. Your last fold will crease it down from there, so it will be creased upwards after you undo your last fold. This same reasoning, incidentally, shows that the last crease will always be down.

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

OK, here's another one in similar vein to the first:

There are n logicians, where n>1, sitting in a carriage on a steam train. The train goes through a tunnel and each logician gets a spot of soot on his forehead. The windows on the carriage are all frosted and there are no reflective surfaces anywhere nearby, so none of them can tell they have this mark.

These logicians have a warped sense of humour. They will laugh only if they see a spot of soot on someone else's forehead, and stop laughing if they see no soot on anyone's forehead. Clearly, they all start laughing as the train exits the tunnel.

They are also lazy. They will go to the effort of wiping their foreheads if and only if they know for certain that they have a spot of soot on their heads.

They also know each other well enough to know they all share these conditions for laughing and wiping.

Show that they will all eventually wipe the soot from their heads.

n=2 is trivial. Try n=3 and see if you can generalise.

(This was one of the "hard" problems form a problems course I did as part of the final year of my first Maths degree, so it might take a little bit of thought!)

 

[Fictionarious]

Re: League of Reason? PROVE YOURSELF!

Actually, it's not much harder. Label them A to L (see, 12 this time!)...

WINRAR

But actually, you only win on the technicality that I asked an incomplete question (oops!). I meant to say, can you determine which is the odd ball out AND whether it is lighter or heavier. The solution you provide here -
Weigh ABCD against EFGH
if balanced -
Weigh AB against IJ
if balanced -
Weigh K against A
if balanced -
We conclude that L is the odd ball out, but not whether it is lighter or heavier.

Good work, but can you come up with the better method now that I've added the stipulation I accidentally omitted (again, my apologies)?

Also, of course anyone's welcome contribute their own problems TAD, but on that note I'm not certain I understand your question. What do you mean by up/down folds? If you just fold it over left to right continually, you'll get all odd numbered creases, right?

 

[Marcus]

Re: League of Reason? PROVE YOURSELF!

Good work, but can you come up with the better method now that I've added the stipulation I accidentally omitted (again, my apologies)?

It's trivial if the odd ball is one of A to H, if the pan it was in originally went up, it's lighter, otherwise it's heavier. The only adjustment is if the first weighing is level, so you know the odd one is one of IJKL.

You weigh ABC against IJK. If they're level, the odd ball is L, and you use your last weighing to weigh L against any other ball to determine whether it's lighter or heavier.

If this second weighing is uneven, it tells you whether the odd ball is heavier or lighter. Use your last weighing to compare I with J. If they're level, K is odd, otherwise it's the (heavier/lighter) of I and J.

 

[The Apathetic Despot]

Re: League of Reason? PROVE YOURSELF!

when I say "up" and "down", I mean that when one is holding the paper they way you started out, with one and in each hand, you'll wind up with some of the creases "pointing" down and the rest "pointing" up. Another way to think about it is if you were traveling along the edge of the paper from left to right, an "up" crease would be a right turn and a "down" crease would be a left turn. So considered on its own, an "up" looks like ^ and a "down" looks like v. Sorry, I should have been more clear. And yes, you'll always wind up with an odd number of creases.

 

[Fictionarious]

Re: League of Reason? PROVE YOURSELF!

Indeed, Marcus, you are proving to be quite a challenger.
Still working on your n logician question...

TAD -
Ah, that makes sense. I was imagining the paper from a birds eye view.