The Brain Teaser Game

[Sean O]

Since we NTs love to do mental gymnastics so much, I thought I'd make a game out of it. :)

The premise is simple: I'll be the first one to post a brain teaser for you guys to solve. A "brain teaser" in this case could refer to a riddle, a visual puzzle, a lateral thinking puzzle, or any other kind of intellectually challenging task. The first person to post the correct answer gets a point, and then it's his/her turn to post the next brain teaser for us to solve, and the cycle will repeat. I'll periodically post the updated scoreboard for everyone to see.

You may ask for hints if you want, but it's up to the person who posted the current brain teaser whether to give a hint or not, and what kind of hint to give. Please post hints in spoiler boxes. Also, although I doubt this will happen, if for whatever reason it seems like nobody can solve the current brain teaser, then if three people propose to skip it, the person who posted the brain teaser will post the answer in a spoiler box and choose someone else to post the next puzzle. That way people can still try to solve it on their own if they want, but the game keeps moving as well.

Anyway, here's the first one:

Bob wants to get into a certain exclusive club, however he is not a member, and they are picky about who can become a member. At the building where the club is based, a bouncer guards the entrance, and screens people to ensure that they are members before they are let inside. One night, Bob hides in the bushes just within hearing range of the bouncer, and listens to what he and the members say to each other as they are screened. One member approaches the bouncer. The bouncer says, "Twelve", and the member says, "Six." He is let inside. A second member approaches. The bouncer says, "Six", and the member says, "Three." She is let inside as well.

At this point, Bob thinks to himself, "Oh this is easy - just divide by two!" Confident that he has figured the screening process out, he walks around to the entrance and approaches the bouncer. The bouncer says, "Ten", and Bob says, "Five." He is asked to leave.

What should Bob have said, and why? In other words, how does the screening process work?

[Rudy]

Answer:

Bob should have said "three". It's the number of letters in the word. Which, by the way, is a really terrible screening process. Just fyi.

Now, one of my favorite puzzles:

There are two rooms. Room A contains three, average lamps. There is nothing fancy about them, just a standard incandescent lamp that you would find in most homes. Room B contains three light-switches, each controlling exactly one of the lamps.

You are in Room B. There is no way, whatsoever, to look into, or sense anything from Room A while you are there. You may do whatever you like with the lightswitches, and then you go into Room A and observe the lamps. You may not go back into Room B, or otherwise mess with the switches, once you have left Room B.

So, the problem is, what do you do to determine exactly which switch controls which lamp? You much determine the answer for all of the lamps.

EDIT: I should add that it is clearly marked which light switch position is "on" and which is "off", and you can be certain that these marking correspond to the same state for the lamp they are connected to.

[Sean O]

Flip switch 1 into the "on" position, and leave it on for a little while (long enough that the bulb on the lamp it controls will heat up). Then flip switch 1 back to the off position, and flip switch 2 to the "on" position. When you check room B, the lamp that's on is controlled by switch 2, the lamp that's off and has a warm lightbulb is controlled by switch 1, and the remaining lamp is controlled by switch 3.

You are traveling through a path in the jungle, on your way to a village inhabited by a tribe of friendly, peace-loving natives (tribe A). You are being cautious on your journey, however, because this jungle is also inhabited by an aggressive, cannibalistic tribe (tribe B). The only thing that distinguishes these tribes is their behaviour.

As you walk along, you reach a fork in the road, which was not mentioned to you previously when you were given directions to the friendly tribe's village. Standing at this fork are two men, and in front of them is a sign. The sign reads, "One of these paths leads to tribe A's village, and the other, to tribe B's village. These men know which path leads to which village, and you may ask them about it. However, be warned: one of them always speaks the truth, but the other always lies."

You ask the men which path leads to tribe A, and they point in opposite directions. The same happens when you ask them which path leads to tribe B. You don't think you can afford to just pick a path randomly, because tribe B is full of powerful warriors who attack all intruders, and they will pursue you relentlessly until they catch you. You must find out which path leads to where. How can you reliably get this information from the two men in front of you?

[Tragic Hero]

I would ask them both a question, such as "Are you able to speak?" or "Are you alive?" The one who gives me a correct answer, I follow his directions of Tribe 1. You could also ask which way the other would point if asked which way Tribe 2 was and walk that way.

I am taking the following Brain Teaser directly from "Tricks of the Mind" by Derren Brown.It's far too easy but at least it will keep the game moving along.



'I toss a coin seven times, and record whether it lands on heads {H} or tails {T}. For my own perverse enjoyment, I write down the following list and give it to you. It shows three results, but only one of them is the real outcome. Which is most likely to be the real result?

1.HHHHHHH
2.TTTTTTH
3.HTTHTHH

[sagewolf]

They're all equally likely to occur, provided the coin isn't biased in any way.

A man lives on the top floor of an apartment building. Every morning he gets up and takes the elevator down to the lobby to go to work, but every evening he does something strange: if someone else is waiting for the elevator or gets on before the tenth floor, he rides all the way to the top floor. If not, he only rides to the tenth and walks the rest of the way up. Why? (It's not for the exercise.)

EDIT: TH's guess is good, but it's not the one I'm after, since our subject is actually walking up after he gets off (which I should have made a little clearer, I admit). We shall say, for the sake of argument, that the building is 15 storeys high.

[Sean O]

I've solved that one before, so for the sake of fairness I'll leave it up to someone else to answer. It's a good one, though. :)

[Tragic Hero]

Is the building ten stories high?

[Rudy]

Here's the only answer that I could come up with:

The only thing I can think of is that he's not tall enough to reach the upper buttons. This is dumb though, because he'd just carry a stick with him, or something... Is that the answer?





RudyHenkel added to this post, 165 minutes and 10 seconds later...

Here's the next puzzle:

A man, Bob, is on a gameshow, and is given three doors to choose from, Door A, Door B and Door C. There is a prize that he wants behind exactly one of the doors. Bob chooses "Door A." Before Door A is opened, the game show host, knowing which door holds the prize, opens one of the other doors, say Door B, and shows that the prize is not there. Now, Bob is given the opportunity to change his answer to "Door C" if he wants.

Should he do so? Why or why not?

[Tragic Hero]

I know the answer, but I'll let someone else get it. I once had an argument with three people all unified against me for an hour over this. Some people just won't budge. I even tried the 100 prisoners on death row analogy.

[Sean O]

At first, there's a 33% chance that the prize is behind any one of the doors. Bob chooses Door A, which of course has a 33% chance of being the correct door, and which leaves a 66% chance for the prize to be behind either Door B or C. After the host reveals that the prize isn't behind Door B, the entire 66% chance of the prize being behind either B or C now fully applies to C. So he should change his answer, since C is, statistically speaking, twice as likely to have the prize at this point.I'm pretty sure that's correct, but I'll wait for Rudy to verify it.

[Rudy]

At first, there's a 33% chance that the prize is behind any one of the doors. Bob chooses Door A, which of course has a 33% chance of being the correct door, and which leaves a 66% chance for the prize to be behind either Door B or C. After the host reveals that the prize isn't behind Door B, the entire 66% chance of the prize being behind either B or C now fully applies to C. So he should change his answer, since C is, statistically speaking, twice as likely to have the prize at this point.I'm pretty sure that's correct, but I'll wait for Rudy to verify it.

Correct! Choose the next puzzle, Sean O.

[Sean O]

Awesome. :)

-> ..... || ....... <- ...... <- ...... <-
1 ....... || ....... 2 ....... 3 ....... 4
B ........|| ....... W ....... B ....... W


Okay, first I'll explain the above diagram (this puzzle is much easier to tell in person, with a pen and paper). The numbers represent four thieves who are imprisoned because they tried to steal treasure from a king. They are fully strapped to some chairs, in a way that prevents them from moving, or even turning their heads. The arrows represent the direction they're facing, and the vertical lines between 1 and 2 represent a wall. Ignore the periods.

Anyway, they are set up like this because the king they tried to steal from is a little twisted, and decided to play a game with them after they were caught. After strapping them to their chairs and setting them in the indicated positions, he put hats on them, two of which are black, and two of which are white (which is what the letters indicate). They were blindfolded when the hats were put on, and they have no way of seeing the hats on their own heads, so they don't know which colour hat they have on.

He tells them that two of them are wearing black hats, and two are wearing white hats. He also informs them of the positions they are set in (i.e. 1 is separated from the others by a wall, and 2, 3 and 4 are positioned in sequence behind each other). He is willing to set all of them free if one of them can determine which colour of hat he is wearing. They must determine this based only on what they can see. If they talk to each other to get more information, or if one of them guesses incorrectly, they will all be executed. Essentially, they have one chance to guess between the four of them, and must only use deductive reasoning based on visual information.

Which thief is able to determine what colour hat he is wearing, and how can he deduce this?

Hint (just because it might not be obviously implied from my description):
4 can see both 2 and 3.

[Rudy]

Okay, thought about it, and I think this is the answer; I'll wait for confirmation:

The 3rd thief is the one able to tell. The reason is, the third thief knows that if both the 2nd thief and himself had White Hats on, then the 4th thief would be able to solve the puzzle too easily, so the King would not set it up like that. Therefore, when he sees the 2nd thief's hat, he knows that he has a black hat.

[Sean O]

Yep, that's correct! Here's the current scoreboard:

RudyHenkel: 3
Sean O: 2
Tragic Hero: 1
Sagewolf: 1

Your turn again, Rudy.

[Rudy]

A billionaire took three men to an empty room and offered them the following challenge: "The man who can fill this room with something using the least amount of money will win a million dollars."

The first man spent $50 to fill the room with empty cardboard boxes.

The second spent $20 to fill the room with balloons.

The third man spent nothing, and won. What did he fill the room with?

[Sean O]

Air. Although that's not really "filling" the room, so much as it is using something that it's already filled with.That's my best guess so far. And, from now on, I think we should all wait for confirmation from the person who posted the riddle, even if we're confident in our answers (unless he/she doesn't show up to the thread again... then we can just assume that we're right ;D).

[Rudy]

Air. Although that's not really "filling" the room, so much as it is using something that it's already filled with.That's my best guess so far. And, from now on, I think we should all wait for confirmation from the person who posted the riddle, even if we're confident in our answers (unless he/she doesn't show up to the thread again... then we can just assume that we're right ;D).

No, that's not right. To clarify, the gentleman is question actually took some sort of action to fill the room. The source I got the riddle from lists one answer, but I've thought of at least one more that would be acceptable.

[Allie]

OK Sean O. I thought of that but it seemed too simple. Ya. Sometimes the answer is the obvious, but I'll be contrary just because and see where this goes....

The sound of his voice

[Sean O]

I see. In that case, my new guess:

He filled the room with sound waves by standing in it and shouting.





Sean O added to this post, 0 minutes and 35 seconds later...

Damn, you beat me to it, Allie.

[Rudy]

That was the other acceptable answer, but Allie got it first. The original answer was:

Light. He flipped a switch, and filled the room with light.





RudyHenkel added to this post, 6 minutes and 49 seconds later...

Since Allie has gone offline, Sean O should go ahead and post a riddle.

[Sean O]

Alrighty,

What can around the world while staying in a corner?

(I think this is one that lots of people might have heard before... oh well)

[Noehelia]

A billionaire took three men to an empty room and offered them the following challenge: "The man who can fill this room with something using the least amount of money will win a million dollars."

The first man spent $50 to fill the room with empty cardboard boxes.

The second spent $20 to fill the room with balloons.

The third man spent nothing, and won. What did he fill the room with?

I hate this kind of riddles. There is no measurement, a definite way to say what is appropriate answer or not, you can think thousand of things that depending someone else's perception can fit.
For instance in this case the variable of how long the filling of the room will last, should it be permanent, will it last for a moment, is not stated. The original answer for example can be much costlier if you say that the room must be filled with something that will last for a year. And even if it should last for a moment it still costs to someone (not to mention that there is a major loop to assume that you can do that in this room). If that stands then why not say that you would borrow something from someone else and afterward you will return it.
As for the answer that the members here suggested I do not agree either. Because a)you don't know the density that is required and b) can you show that even in one particular instance of time the room is full of it?

[Kisai]

Everything standing on the surface of the earth revolves even when relatively standing still.

Here's one that you all probably heard before:

A king wishes to find a husband for his daughter, so he sets a difficult
exam for all the young men in his kingdom. Three men score top marks. In
order to decide which of the three is most suitable, he sets an
additional test.

He sits all three in a circle and blindfolds them. He tells them that he
will place either a red hat or white hat on each of their heads, and
instructs them to raise their hand if they see a red hat on either of
the others' heads after he has removed their blindfolds. He places a red
hat on each of their heads. "The first that can tell me the colour of
their own hat will marry my daughter." He takes off their blindfolds,
and they all raise their hands. They sit there for a whole hour before
finally one of them gets up and says "I have a red hat." How did he know?

[Rudy]

I think the actual answer to Sean O's riddle is: A Stamp

However, I'll let Kisai's riddle stand.





RudyHenkel added to this post, 0 minutes and 51 seconds later...

I hate this kind of riddles.

Sorry, I'll try to make the answers more definite in the future =D

[Tragic Hero]

Assume you are person 1; You can see both person 2 & 3 and their red hats. You can not tell if your hat is red or white. If you were wearing a white hat; it should be safe to assume that either person 2 or 3 would deduce that their hat was red. Because there must be at least two people wearing red hats in a group of three when everyone can see a red hat. Since neither person 2 or 3 came to that conclusion after an hour, you assert that you have a red hat too.

[Allie]

Assume you are person 1; You can see both person 2 & 3 and their red hats. You can not tell if your hat is red or white. If you were wearing a white hat; it should be safe to assume that either person 2 or 3 would deduce that their hat was red. Because there must be at least two people wearing red hats in a group of three when everyone can see a red hat. Since neither person 2 or 3 came to that conclusion after an hour, you assert that you have a red hat too.

I think you're assuming that each would see 2 red hats every time, but that may not be the case, from what I could tell.

Here's one that you all probably heard before:

A king wishes to find a husband for his daughter, so he sets a difficult
exam for all the young men in his kingdom. Three men score top marks. In
order to decide which of the three is most suitable, he sets an
additional test.

He sits all three in a circle and blindfolds them. He tells them that he
will place either a red hat or white hat on each of their heads, and
instructs them to raise their hand if they see a red hat on either of
the others' heads after he has removed their blindfolds. He places a red
hat on each of their heads. "The first that can tell me the colour of
their own hat will marry my daughter." He takes off their blindfolds,
and they all raise their hands. They sit there for a whole hour before
finally one of them gets up and says "I have a red hat." How did he know?

Based on the above, each would see at least a red hat on either (not both) of the other's heads. This is not a guarantee that each would always see 2/both red hats at all times. I think this changes the entire premise of this question.

However, if the clue had been: if they could see a red hat on both of the other's heads, then you may be on the right path.

Then again, I could be reading this all wrong because I was struggling with that clue....

Here are the given so far that I could tell. I don't know, I think some clue/s is missing. I was able to figure this out years ago, when someone challenged me with this one. I seem to be stuck today! I am getting olllldd!

There are 3 hats total (Red or White).
At least 1 out of 3 has to be Red, that was a given.
He (A) could see that the other two are Red, also a given.
A would still be either be Red or White (we need to go through these 2 logical deductions before we can be certain).

For example:
1. If A has a White hat, then one of them (B) would see A with a White hat and the remaining other (C) has a Red (remember one of them has to be Red),...

[Tragic Hero]

Allie, imagine that it was you, Rudy and Kisai. When your blindfold is taken off, you see two red hats. Let's pretend that your hat was white. If your hat was white , Rudy and Kisai would very easily & quickly {less than an hour} deduce that their hat was red. In this case, Rudy would have his hand up because he could see Kisai's red hat. Rudy would also be looking at Kisai's hand which would be in the air. Now Rudy knows that Kisai's hand isn't in the air because you are wearing a red hat, and therefore Kisai's hand must be in the air because Rudy's hat is red.
The absence of Rudy or Kisai solving the problem quickly means that the problem must not be that easy and that infact they are facing your exact problem.{two red hats}


There are 3 hats total (Red or White).
At least 1 out of 3 has to be Red, that was a given.
He (A) could see that the other two are Red, also a given.
A would still be either be Red or White (we need to go through these 2 logical deductions before we can be certain).

In order, for each person to see a red hat, there must be at least two red hats.
If only Kisai had a red hat, and you and Rudy had white hats; Kisai's hand would not be up in the air.



The Einstein Puzzle.

Einstein created a puzzle, or so the story goes, that he thought only two percent of the population would be able to solve. It isn't that hard {I did it today while looking for a puzzle to respond with} however I suggest using a pen and paper.

FACTS.

FACT1. There are five houses in five different colours.
FACT2. In each house lives a person with a different nationality
FACT3. These five owners drink a certain beverage , smoke a certain brand of
cigarettes and keep a certain pet.
FACT4. No owners have the same cigarettes, drink , or pet.


CLUES:
1. The Brit lives in a red house
2. The Swede keeps a dog
3. The Dane drinks tea
4. The green house is on the left of the white house.
5. The green house owner drinks coffee.
6. The person who smokes Pall Mall keeps birds.
7. The owner of the yellow house smokes Dunhill.
8. The man living in the house right in the center drinks milk
9. The Norwegian lives in the first house.
10. The man who smokes Blend lives next to the one who keeps cats
11. The man who keeps horses lives next to the man who smokes Dunhill
12. The owner who smokes Camel drinks beer
13. The German smokes Marlborough.
14. The Norwegian lives next to the blue house
15. The man who smokes Blend has a neighbour who drinks water.

Who has a fish?

[Noehelia]

The German
I have solved this puzzle in the past and I did it again now (ofc I did not remembered the solution) but I know the methodology which is common in "logic grid matrix" puzzles using a matrix.

So finally I solved a puzzle in this thread in order to post one that I want.

By the way, for the previous problem posted my Kisai I would like to note for Allie that it is stated already that the king places to all three men a red hat. To tell you the truth I did not notice it myself either the first time I read the post. Not that it matters in order to solve the puzzle because the solution relies on a major loop of handling time which is arbitrary.
I have a major disagreement with puzzles that use the element of time this way. The only way in my opinion to have a sound reasoning of how the problem is solved is to add another element in the argumentation. That the men should think also that the king wanted to find who is the smartest without favoring anyone. If he had placed 2 red hats and one white he would have favored the ones wearing the red hat.





Noehelia added to this post, 63 minutes and 6 seconds later...

I will actually post two puzzles but they are related as you will see.

How to have safe sex without any after-effects with only two condoms
a) Two men want to have sex with two women in a row (meaning that each of the men will have sex with each of the women but not at the same time)
b) Three men with one woman in a row.

The condoms are ideal, provide 100% safety, they do not worn out.

Hm, the puzzles that I have posted are two but as I said they are related that is why I did not want to post them apart. However I will let the OP to decide how to handle the points and how the solver will post another puzzle afterwards.

[Kisai]

Regarding the hats:

The person with the red hat correctly concluded that this problem is only unsolvable only if all three hats are red. I'll be happy to recount how all the other possibilities are solvable. This is one of my favorite puzzles because you cannot solve it from the point of view of only one of the characters in the puzzle. You have to consider the point of view of everyone (except the king and the princess..natch) in the puzzle to solve it.

I'm sorry if the hour wait seems arbitrary. It just shows that the puzzle cannot be solved. If the puzzle cannot be solved, then the only unsolvable possibility must be true.

[Allie]

Kisai - Please do provide the different views. I still want to see all of the angles, and I'd like to compare them with what I have.

Noehelia - Still trying to figure out your puzzle...

[Sean O]

I will actually post two puzzles but they are related as you will see.

How to have safe sex without any after-effects with only two condoms
a) Two men want to have sex with two women in a row (meaning that each of the men will have sex with each of the women but not at the same time)
b) Three men with one woman in a row.

The condoms are ideal, provide 100% safety, they do not worn out.

Hm, the puzzles that I have posted are two but as I said they are related that is why I did not want to post them apart. However I will let the OP to decide how to handle the points and how the solver will post another puzzle afterwards.So the idea is to explain how to have safe sex with only two condoms, in both of the scenarios described? This should probably be considered a single, two-part riddle. So the first person to explain both scenarios properly will be given credit for solving it.

[Noehelia]

So the idea is to explain how to have safe sex with only two condoms, in both of the scenarios described? This should probably be considered a single, two-part riddle. So the first person to explain both scenarios properly will be given credit for solving it.

Well they are both hard and they are solved differently. In the past I was first aware of the first one (which I solved it) and then after some months with the second (in which I failed). There is also a 3rd one (but I thought it was easier) and I have found them elsewhere on the internet presented all together. Since I posted them together its fair that they are regarded as one.

[Allie]

[spoiler]
I will actually post two puzzles but they are related as you will see.

How to have safe sex without any after-effects with only two condoms
a) Two men want to have sex with two women in a row (meaning that each of the men will have sex with each of the women but not at the same time)
b) Three men with one woman in a row.

The condoms are ideal, provide 100% safety, they do not worn out.

Hm, the puzzles that I have posted are two but as I said they are related that is why I did not want to post them apart. However I will let the OP to decide how to handle the points and how the solver will post another puzzle afterwards.

a) Two men want to have sex with two women in a row (meaning that each of the men will have sex with each of the women but not at the same time)

First man uses both condoms on one woman. Takes off the outer one, without reversing it and hand it over to the other man.
The other man uses the same condom on the same woman, and kept it on afterwards.
First man uses the inside condom on the next woman. Takes it off without reversing it.
The other man puts it over the one he already had on with the second woman.

Edit: Attempting part b. You owe me 2 Ziploc bags Noehelia :laugh:

b) Three men with one woman in a row.

First man puts on both condoms at the same time. Takes off the outer one without reversing it and gives it to the second man. He then takes off the inner one, this time reversing it and hands it over to the third man.
The second man uses the outer condom from the first man. Takes it off without reversing it and also gives it to the third man.
The third man puts on the reversed condom from the first man. Then, puts the outer one from the second man on top of it.

[Noehelia]

Regarding the hats:

The person with the red hat correctly concluded that this problem is only unsolvable only if all three hats are red. I'll be happy to recount how all the other possibilities are solvable. This is one of my favorite puzzles because you cannot solve it from the point of view of only one of the characters in the puzzle. You have to consider the point of view of everyone (except the king and the princess..natch) in the puzzle to solve it.

I'm sorry if the hour wait seems arbitrary. It just shows that the puzzle cannot be solved. If the puzzle cannot be solved, then the only unsolvable possibility must be true.

It doesn't matter if it is one hour, a century or 2 seconds. Let's say that one of the men is thinking: "Ok, I see 2 red hats so either I have a red or a white. If I have a white hat and the others see a red and a white then they should assume that they wear a red hat and they should state it. When? In 2 sec, one minute, 1 hour, 2 days? Hm, I have to set a time limit. So I will say that if they do not state it in 15 minutes then I will presume that they do not see a white hat on me and they have the same problem with me. But what if they see a white hat on me and they decide to speak in 15 minutes and 1 sec? Oh, then I will speak when 30 minutes pass. But what if they would speak in 31 minutes? Ah, I will speak in one hour. But what if they had decided to speak in 2 hours?"(and so on, and so on, it could take infinite time)
If there are no set actions to prompt an answer then nonone knows when to talk because one has to take into account when the others would talk. But they can't really know that. So in the end what can someone do is actually guess. They guess by setting their own time limit and the man who wins is the one who has set in his head the shortest time limit.

If they would all think rational and not guess then noone would talk, ever. The funny thing is that we take for granted that they think rationally at least for one part of solving the problem (when they would suppose to speak if they saw a white hat) but then what happens, only one is capable of thinking rationally the second part? So if they are capable of reaching this conclusion rationally what stops them all claiming at the same time that they have the red hat. Time. Which we can not take it for granted.

Now, I know there are many logic puzzles like this. Once I had the same argument for another similar puzzle who the poster had taken word by word by a discrete mathematics book. Obviously there is a reason why it is in such a book who is taught in Computer Science curriculum. But when programming you can manipulate time and set it as you want to achieve what you consider the best result.

[Noehelia]

a) Two men want to have sex with two women in a row (meaning that each of the men will have sex with each of the women but not at the same time)

First man uses both condoms on one woman. Takes off the outer one, without reversing it and hand it over to the other man.
The other man uses the same condom on the same woman, and kept it on afterwards.
First man uses the inside condom on the next woman. Takes it off without reversing it.
The other man puts it over the one he already had on with the second woman.

Edit: Attempting part b. You owe me 2 Ziploc bags Noehelia :laugh:

b) Three men with one woman in a row.

First man puts on both condoms at the same time. Takes off the outer one without reversing it and gives it to the second man. He then takes off the inner one, this time reversing it and hands it over to the third man.
The second man uses the outer condom from the first man. Takes it off without reversing it and also gives it to the third man.
The third man puts on the reversed condom from the first man. Then, puts the outer one from the second man on top of it.

Yeap, they are both correct. You tested it?
Anyway, they are nice challenging puzzles. So that you know it, the 3rd related riddle is about a man and 3 women but when you have solved the others it is easy.

Your turn to post a puzzle Allie.

[Allie]

You tested it?

Wellll...I couldn't find 3 men and 2 women, so I settled for the Ziplocs* instead. :p That was bad I know, but I couldn't resist. The trick was how to preserve as much unused/untouched portions of the condoms as possible. Then find different ways to reuse to get there. The Ziplocs helped. ;D Although, the rest of the imagination I could do without.

*No condoms available (TMI!!)

Edit: Next puzzle. Sudoku!

Not sure how I could make this easier, but it looked like the solver would need to put these numbers in an Excel file to complete, and then save a picture of the finished product and post it back here again. Sorry about that.

To view links or images in this forum your post count must be 15 or greater. You currently have 0 posts.

[Rudy]

Answer:


8 4 7 | 6 1 3 | 9 5 2
2 9 5 | 7 4 8 | 3 1 6
6 3 1 | 2 5 9 | 8 7 4
===============
5 6 9 | 3 7 4 | 2 8 1
7 1 2 | 9 8 6 | 4 3 5
4 8 3 | 5 2 1 | 7 6 9
===============
1 7 6 | 4 3 2 | 5 9 8
9 5 4 | 8 6 7 | 1 2 3
3 2 8 | 1 9 5 | 6 4 7


Thinking of the next riddle...





RudyHenkel added to this post, 7 minutes and 7 seconds later...

Rudy throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. If Rudy now throws another dart at the board, aiming for the center, what is the probability that this third throw is also worse (i.e., farther from the center) than his first?

Assume Rudy's skillfulness is constant. That is, his ability does not change from throw to throw. You answer should be based on having no information about result the first two throws, other than the fact that the 2nd is worse than the first.

[Noehelia]

I will answer taking for granted that Rudy adjusts his throw everytime taking into account the previous throws and also that the way he make his throw (like speed, the position of his body, the way his hand makes the move) are the same. All that he has to do is adjust the position of his hand having in front of him a visual 2d flat surface with x and y axis meeting in the center of the dartboard. Well, when you have already two points in a flat surface I do not see how he is going to throw another dart farther from the center so my answer is 0% probability.

I am not sure about my answer because I take too many things for granted as constant.

[Rudy]

I will answer taking for granted that Rudy adjusts his throw everytime taking into account the previous throws and also that the way he make his throw (like speed, the position of his body, the way his hand makes the move) are the same. All that he has to do is adjust the position of his hand having in front of him a visual 2d flat surface with x and y axis meeting in the center of the dartboard. Well, when you have already two points in a flat surface I do not see how he is going to throw another dart farther from the center so my answer is 0% probability.

I am not sure about my answer because I take too many things for granted as constant.

No, sorry. That's part of what I meant when saying that the skillfulness is constant. There is no adjustment from throw to throw; each has a random distribution like the last.

Hint:

There are three darts. Consider the different possibilities for their rankings.

[Noehelia]

But if the skillfulness is constant and there is no adjustment why the two darts are not on top of the first? What is changing?

I think I understood your hint. There are 6 possible orders of the darts and only 2 of them have the first dart as first in the order by being closer to the center. Thus there is 33% for these outcomes to happen. Grr, if this is the answer I do not like it. It is from our view this probability. The actual probability should be based on the distribution of his potential throws.

[Rudy]

You're pretty close, I'll go ahead and give the answer, and you can pick the next one:


Pretend that you are told, after the fact, that the gentleman has thrown three darts at the board, and that his skillfulness is constant (that is, he is not learning anything from his throws.) There are then six possible orderings for which dart is the closest to the center, each of which is just as likely:

1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

You are then told the additional information in the puzzle, that the first dart is closer than the second, and asked the question.

Well, the fact that the first dart is closer than the second eliminates three of the possibilities, leaving us with:

1 2 3
1 3 2
3 1 2

Out of these three remaining, equally likely possibilities, the third throw is worse than the first in 2 of the three options. So, the probability that the third throw is also worse than the first is 2/3.

[Noehelia]

Yes, I know my fault in the end. The funny thing is that I had thought the 3 orders right from the start (I mean after the clarification) having 66% on my mind but it did not feel right, that it would be that easy, then I spent my time thinking other things and then I thought about the 6 orders which were the possibilities before throwing the 3 darts. I did not like the story behind the riddle, it confused things, it would be better if it was presented differently like having 3 runners.

Anyway, I did not find the puzzle, I shouldn't receive any point and since previously I had posted two riddles at the same time and I messed with yours now then I suggest you post another puzzle, Rudy.

[Rudy]

Imagine you have one cup of tea and one cup of coffee, with equal amounts of their respective beverage. Take a spoonful of coffee, and mix it into the tea cup. Once it is evenly mixed, take an equal spoonful from the tea cup and mix it into the coffee cup.

Does the cup of coffee now contain more tea or the cup of tea more coffee?

[Noehelia]

They have the same amount of foreign substance.
Well, I did not think it over much, I just made the calculations with made-up numbers.
If we had 1000 ml of coffee and 1000 ml of tea in the beginning and we took with a spoon 20 ml of coffee and poured it in the tea then we would have:
1020ml of a mixture with 1000 ml tea and 20 ml coffee. The percentage of coffee in that mixture would be 1.96~% (actually that percentage is not that important). Then if we would take 20 ml of that mixture it would contain 0.392~ ml of coffee (1.96~% of 20ml) and 19.61~ml tea. If we add that to the 980 ml coffee in the other cup then we would have a mixture of 980.392~ml coffee and 19.61~ml tea. In the other cup it would remain 1000ml tea-19.61~ml tea (taken with the spoon)=980.392~ml tea and 20ml coffee-0.392~ml coffee(taken with the spoon)=19.61~ml coffee.







Noehelia added to this post, 140 minutes and 48 seconds later...

I presume that my answer was correct although I solved it with a little bit brute force so I post another of my favorite puzzles.

Three female teachers in a school, a brunette, a redhead and a blonde, are all flirting with the math teacher. The math teacher is shy, he goes along with their flirt and does not know how to handle the situation but the female teachers have started to fight between them so they have stated to him that he has to choose one of them soon. He knows that he definitely doesn’t want to have a relationship with one of them but he is not sure which one to choose from the other two.
So he invites them to his house for a tea. He plans to play two games with them. In the first game the female teacher that he does not like so much will be eliminated (in a subtle way) and the second game will be based on pure lack presented as a math game.
So the first game is this:
He placed them in a round table, blindfolded them and stuck a piece of paper on their foreheads. He tells them that the papers can have the numbers 1,2 or 3 and the sum of their numbers will be 6 or 7. In reality he placed number 2 in the forehead of all three teachers.
Then he removes the blindfold. The female teachers can see the papers stuck on the forehead of the other two but not their own. He then asks each one of them if they know what number is on their forehead. The order of the questions has been determined previously by a lottery the math teacher performed but he had tampered with the results so that first the brunette is asked, then the redhead and last the blonde.
When someone would be able to say correctly the number on her forehead she would do so otherwise she would say that she doesn’t know. The question would be repeated in the same order every turn until only one of them would answer that she does not know her number (the others would already have stated their number) and that one will leave.

Which one of them the math teacher did not like and how did you come up with the solution?

Please note that this is a logic puzzle and we take for granted that each one of the teachers is able to deduce the number on her forehead if that is feasible and they take for granted that the others are able to do the same.

Oh, and if someone knows this puzzle can he send me a message with a link with its English form because I can not find the original source since I have altered the story when I had translated it in the past.

[Rudy]

This is maybe the most confusing thing that I have ever written out. Tell me if the answer is correct:

Brunettte (A)
Redhead (B)
Blond (C)

Looking around at the other two people, with 2s on their heads, each person knows that they do not have a "1" on their head.
So, A, B and C all know, to start with, that they have either a two or three on their head.

A - Says "No", because she could have either a two or three. This gives new information to B and C. From B's POV, if B had a one on her forehead, A would have said "Yes," because A would have had to have a three. B already knew that she didn't have a one, but now A knows that B knows that she doesn't have a one. And B knows that A knows that B knows, etc. Infinite loop of knowing; that is, the fact that B does not have a one is now fully public knowledge. The same applies to the fact that C does not have a one.

B - Says "No", because she might still have a three. B doesn't yet know that A knows that A doesn't have a one, so A might still have said "No" even if B had a three. However, once B gives her answer of "No", this now means that B knows that A knows that A doesn't have a one, for the same reason above. So, at this point, everyone knows, that everyone else knows, that no one has a one on their forehead.

C - Says "Yes". C knows that if she had a three on her forehead, then B would have said "Yes". This is because A has a 2 showing, and B would have seen a two or three. Since at this point C knows that B knows that B has a two or three, B would have known that this meant that B could only have a 2, and so would have said "Yes". Since B did not say "Yes", it means that C must not have a three on her forehead, which means that she knows she has a two.

A - Says "Yes". Same reason as C. If A had a three on her forehead, B would have said "Yes" on her last turn. Since B did not do so, A must have a 2, and she knows it.

B - Says "No". B knows that A & C both know that they have 2s, but there is no further information to allow B to determine whether she has a two or a three.

The last three steps will repeat indefinitely. C - Yes, A - Yes, B - No.

So, in conclusion, B, the redhead, is the one the math teacher doesn't like.





RudyHenkel added to this post, 102 minutes and 17 seconds later...

Apparently my answer is not correct. I'll try again when I have time, but if anyone else wants a shot, go for it.

[Noehelia]

The answer is not correct. The path that the reasoning takes is right in general but there is one point missed.

[Noehelia]

I think hints are needed.

Hint 1: to people that have trouble starting from somewhere

Read Rudy's answer to get the general gist of how to handle it, meaning that every female teacher has to think what others thought also.

Hint 2: to people that have read Rudy's answer but still can't come to the solution
His solution is not correct, not only in some point (or points, I did not check everything just the first round) in the process but in the outcome also

Hint 3: where is an important point missed in Rudy's answer
What the blonde knows that the redhead knows in the first round

[Allie]

I didn't read Rudy's answers (since it's wrong per Noehelia :p). Didn't read the hints either so as not to be even more confused.

So, here's mine:


Given the numbers 1, 2 and 3 and requiring that the sum is either 6 or 7. The only possibilities for those combinations are:

1, 2, 3 = 6

2, 2, 2 = 6

2, 2, 3 = 7

1, 3, 3 = 7

Given that each would see the other two women having a number 2 each, then the only possiblities for the sum to be 6 or 7 are:

2, 2, 2 = 6

2, 2, 3 = 7

Each would deduce that they would have to have either a 2 or a 3 for a sum of either 6 or 7...But which one?

Order:

1st: BR = Brunette
2nd: RD = Red head
3rd: BL = Blond

Given that each would see a 2 on the other two women, each would think the following:
A. If I am also 2, then the other two women would not know if they're a 2 or 3 either, since both would total to 6 or 7. They would be as uncertain, just as I am.
B. If I am a 3, then the others would know immediately that they have to be 2, and 2 each to make the sum of 7.
C. The first correct answer would lie in the reactions of the first two.
Since the BR is first to go, she doesn't know because the others did not get a chance to react.


So given that the first round of answers are required in this order:

1st Brunette = I don't know.
2nd Red Head = I don't know
3rd Blond = 2 (see A, B, and C above)

Second round of answers. Since BL answers correctly based on the first two's uncertainties, BR and RD still cannot tell if they're a 2 or 3 because there was no uncertainty from BL. The cycle starts over again.
A and B are the same
C. The next correct answer would lie in the reaction of the previous woman, but since BR is next:

1st Brunette = I don't know.
2nd Red Head = 2 (see A, B, C above).


The math teacher does not like the Brunette per the above.


Yay? Nay?

[Noehelia]

No, it's not correct,you are both jumping to assumptions (in different levels and points but still). This is a difficult puzzle. It is difficult even for me to find exactly where you make error.

Oh and I managed to find where I spotted that puzzle some years ago so now I can be more certain about the answer. I knew who was the female teacher that the math teacher was not interested but I don't have my notes anymore of when I tried it out myself and in the forum that I had posted it before their answers were incomplete.

[Rudy]

I think I discovered my error.

2nd attempt:

Brunettte (A)
Redhead (B)
Blond (C)

Looking around at the other two people, with 2s on their heads, each person knows that they do not have a "1" on their head.
So, A, B and C all know, to start with, that they have either a two or three on their head.

A - Says "No", because she could have either a two or three. This gives new information to B and C. From B's POV, if B had a one on her forehead, A would have said "Yes," because A would have had to have a three. B already knew that she didn't have a one, but now A knows that B knows that she doesn't have a one. And B knows that A knows that B knows, etc. The same applies to the fact that C does not have a one, that is, A now knows that C knows that C doesn't have a one. Summary: By Saying "No", A know knows both the other players know they don't have ones.

B - Says "No". Like A, after B says no, B now knows both the other players know they don't have ones.

C - Says "No". Like A and B, after C says no, C now knows that both the other players don't have ones.

Round 2. During this round, we note that in addition to knowing that no one has a "one," all players are now aware that all the other players are aware that no one has a one. =/

A - Says "Yes". A knows that if she had a three, C would have said "Yes" last round, since that would have meant that C could not have a three herself, and A also knows that C knows that C does not have a one. So, C would have known that she had a two. Since C did not say "Yes," A knows she does not have a three, and therefore has a two.

B - Says "Yes". B knows that she has a two by the same reasoning as A.

C - Says "No". C has not gained any new information since her last attempt, and so still cannot conclude what her number is.

So, in conclusion, C, the Blond, is the one the math teacher doesn't like.

[Noehelia]

Correct Rudy. Although I am not sure that you have explained 100% the reasoning.

In the spoiler I will show what was Rudy's initial error and the answer as I got it from the original posting of the puzzle.

Rudy's initial error: In the first round the Blonde didn't know that the Redhead knew that she did not have 1.

Solution of the puzzle:
When a woman sees two 2´s on the other two women, she knows she had a 2 or a 3 on her forehead (the sum is 6 or 7).

In the first round of asking, each woman reasoned like this :

Brunette: "I have a 2 or a 3, so I don´t know my number".

Redhead: "I have a 2 or a 3. If I have a 2, brunette wouldn´t know whether she had a 2 or a 3. If I have a 3, brunette wouldn´t know whether she had a 1 or a 2. So I don´t know whether I have a 2 or a 3."

Blonde: "I have a 2 or a 3. If I have a 2, brunette wouldn´t know whether she had a 2 or a 3 and redhead wouldn´t know whether she had a 2 or 3 (see above). If I have a 3, brunette wouldn´t know whether she had a 1 or a 2 and redhead wouldn´t know whether she had a 1 or a 2; redhead would reason: "if I have a 1, brunette wouldn´t know whether he had a 2 or a 3 and if I have a 2, brunette wouldn´t know whether she had a 1 or a 2." So I don´t know whether I have a 2 or a 3.

Brunette: I have a 2 or a 3. If I have a 3, redhead would know she had a 2; redhead would know she didn´t have a 1 because otherwise I would have known I had a 3; redhead would know she didn´t have a 3 because the total is 6 or 7. Redhead didn´t know she had a 2; so I have a 2.

Redhead: I have a 2 or a 3. If I have a 3, blonde would know she had a 2; blonde would know she didn´t have a 1 because otherwise I would have know I had a 3; blonde would know she didn´t have a 3 because the total is 6 or 7. Blonde didn´t know she had a 2; so I have a 2.

Blonde: I have a 2 or a 3. If I have a 2, each of brunette and redhead would know the second time around she had a 2 (see above). If I have a 3, brunette would know she had a 2; brunette would know the first time around she had a 1 or a 2; brunette would know the second time around she didn´t have a 1 because otherwise I would have known I had a 3. If I have a 3, redhead would know she had a 2; redhead would know the first time around she had a 1 or 2 (see blonde´s reasoning first time around); redhead would know the second time around she didn´t have a 1 because otherwise I would have known I had a 3. So I still don´t know whether I have a 2 or a 3.

Then the blonde asked failed to name her number, since at this point no more information can be obtained by she.

[Rudy]

Imagine a prisoner in a prison. He is sentenced to death and has been told that he will be killed on one day of the following week. He has been assured that the day will be a surprise to him, so he will not be anticipating the hangman on a particular day, thus keeping his stress levels in check.

The prisoner starts to think to himself, if I am still alive on Thursday, then clearly I shall be hanged on Friday, this would mean that I then know the day of my death, therefore I cannot be hanged on Friday. Now then, if I am still alive on Wednesday, then clearly I shall be hanged on Thursday, since I have already ruled out Friday. The prisoner works back with this logic, finally concluding that he cannot after all be hanged, without already knowing which day it was.

Casually, resting on his laurels, sitting in his prison cell on Tuesday, the warden arrives to take him to be hanged, the prisoner was obviously surprised!

What is wrong with his reasoning?





RudyHenkel added to this post, 16 minutes and 56 seconds later...

Here's another way to state the same essential riddle:

A man knows that his friends will be throwing him a party some day in March, and that they want to surprise him. Logically, he knows that it cannot be the last day of the month, the 31st, since then it would not be a surprise. Since it can't be the 31st, it can't be the 30th either, because then he'd know on the 29th that they were throwing him a party on the next day. By this same reasoning, he eliminates every day in the month as a possible day for a surprise party.

What is wrong with his reasoning?

[Noehelia]

Ha, I would like to know also the exact reasoning because in a puzzle in the past with similar wording (although the question was different) I argued in the same way as the main character's reasoning above.

The only thing that I can think for this particular puzzle is this:

With telling them that it will be a surprise and triggering his thought in that path what ever day they hang him would be a surprise to him.

[Rudy]

No, that's not the answer.

If no one gets the answer by later tonight I'll explain it myself.





RudyHenkel added to this post, 133 minutes and 14 seconds later...

I've done a little research, and it seems that there are a few ways to resolve this problem. There's one in particular I'm thinking of, but I'll accept the others as well.

There's not much I can give in the way of hints, but here's something:

The man who will be thrown a surprise party is correct to think that it cannot be on the 31st of the month. He is incorrect to say that it cannot be on the 30th.





RudyHenkel added to this post, 322 minutes and 49 seconds later...

Okay, here's the explanation. Some of you may be dissatisfied with it, in which case we can argue ;)


The man is correct to say that the surprise party cannot be on the 31st, since he would know it on the night of the 30th. His logic fails when he tries to extend it to the 30th, however, and here is why:

He says, "If it gets to the night of the 29th, I know the party must be on the 30th. This is because I've already eliminated the 31st from consideration."

So:

1) The party must be a surprise.

2) The party must be during the month of March.

3) If it is the night of the 30th, and the party has not ocurred, the party must be tomorrow. However, then the man would know the party would occur tomorrow, so it would not be a surprise. Therefore, the party must occur before the night of the 30th.

4) If it is the night of the 29th, and the party has not occurred, the party must be tomorrow, because the 31st has already been eliminated in (3). This means if it is the night of the 29th, it is impossible for the party to be a surprise any longer. This invalidates step (1), which invalidates step (3), which this step depends on. Therefore, this step creates a logical contradiction in the argument, which invalidates it.

Note, that we showed that the argument was contradictory. It is not necessary to explain in plain English why it doesn't work, only that the argument leads to a logical contradiction, and is therefor invalid. Not all logical contradictions are necessarily understandable in plain English.


Okay, whoever wants to, post a riddle.

[Nikita]

Dog's Day Afternoon

The four people in this puzzle all competed in different classes of dog agility at a recent competition. The competitions all required the dogs to run over jumps, through tunnels and various other obstacles in as quicker time as possible. Each had a different result - one came first, one third, one fourth and one ninth. All four dogs were each of a different breed.

Can you work out who handled which dog, at what level each competed, the place each finished in and the breed of each dog?


1. If Tiff finished first then Terry finished fourth.
2. If Terry finished fourth then Jago is a collie otherwise Jago is not a collie.
3. If Jane competed in the Senior class then she finished third.
4. If Jane competed in Novice then she finished fourth.
5. The dog that finished ninth was an alsatian. This was either Jago, in which case Jago competed in the Elementary class, or this was Kelly, in which case Terry handled Kelly.
6. Mark won Starters.
7. If Mark's dog is called Patti then Patti is a labrador otherwise Patti is a collie.
8. Ruth's dog is called Jago.
9. If Jago finished fourth then she competed in the Novice class otherwise she competed in the Senior class.
10. If Patti finished first then Terry's dog is an alsatian otherwise Terry's dog is a collie.
11. If Jane's dog is a doberman then Jane finished fourth otherwise Jane finished third.


Handlers' Names: Jane, Mark, Ruth, and Terry
Dogs' Names: Tiff, Patti, Jago, and Kelly
Breeds: Alsatian, Collie, Labrador, and Doberman
Levels: Starters, Elementary, Novice, and Senior

[Noehelia]

No, that's not the answer.

If no one gets the answer by later tonight I'll explain it myself.





RudyHenkel added to this post, 133 minutes and 14 seconds later...

I've done a little research, and it seems that there are a few ways to resolve this problem. There's one in particular I'm thinking of, but I'll accept the others as well.

There's not much I can give in the way of hints, but here's something:

The man who will be thrown a surprise party is correct to think that it cannot be on the 31st of the month. He is incorrect to say that it cannot be on the 30th.





RudyHenkel added to this post, 322 minutes and 49 seconds later...

Okay, here's the explanation. Some of you may be dissatisfied with it, in which case we can argue ;)


The man is correct to say that the surprise party cannot be on the 31st, since he would know it on the night of the 30th. His logic fails when he tries to extend it to the 30th, however, and here is why:

He says, "If it gets to the night of the 29th, I know the party must be on the 30th. This is because I've already eliminated the 31st from consideration."

So:

1) The party must be a surprise.

2) The party must be during the month of March.

3) If it is the night of the 30th, and the party has not ocurred, the party must be tomorrow. However, then the man would know the party would occur tomorrow, so it would not be a surprise. Therefore, the party must occur before the night of the 30th.

4) If it is the night of the 29th, and the party has not occurred, the party must be tomorrow, because the 31st has already been eliminated in (3). This means if it is the night of the 29th, it is impossible for the party to be a surprise any longer. This invalidates step (1), which invalidates step (3), which this step depends on. Therefore, this step creates a logical contradiction in the argument, which invalidates it.

Note, that we showed that the argument was contradictory. It is not necessary to explain in plain English why it doesn't work, only that the argument leads to a logical contradiction, and is therefor invalid. Not all logical contradictions are necessarily understandable in plain English.



Well, can you show another explanation? The problem is that the puzzle that I had seen in the past was that:

A teacher says to her class that she is going to put a test next week that it is going to be a surprise (meaning that the pupils will not be expecting it that day). What is the last day that she can put the test and still keep her promise? And the answer is supposed to be Wednesday according to the poster. She says we eliminate Friday for the obvious reasons and then we eliminate Thursday because it couldn't be Friday and students would realize it would be on Thursday. But we do not eliminate Wednesday because on that day they would have 3 possible days ahead (Wednesday, Thursday, Friday) while on Thursday they would have only two (Thursday and Friday) from which one was certain not the day (Friday).

So in that puzzle the solution is supposed to be different.

[Rudy]

Well, can you show another explanation? The problem is that the puzzle that I had seen in the past was that:

A teacher says to her class that she is going to put a test next week that it is going to be a surprise (meaning that the pupils will not be expecting it that day). What is the last day that she can put the test and still keep her promise? And the answer is supposed to be Wednesday according to the poster. She says we eliminate Friday for the obvious reasons and then we eliminate Thursday because it couldn't be Friday and students would realize it would be on Thursday. But we do not eliminate Wednesday because on that day they would have 3 possible days ahead (Wednesday, Thursday, Friday) while on Thursday they would have only two (Thursday and Friday) from which one was certain not the day (Friday).

So in that puzzle the solution is supposed to be different.
The explanation given in there is nonsense. If it's valid to eliminate Thursday, it's valid to eliminate every day of the week. The essential problem here is that the argument *assumes* that it will be a surprise, and then *concludes* that it cannot be a surprise. Thus, there is a contradiction, and the argument is invalid.

Here is another, more technical, explanation of the problem:

Formulation of the judge's announcement into formal logic is made difficult by the vague meaning of the word "surprise". A first stab at formulation might be:

* The prisoner will be hanged next week and its date will not be deducible from the assumption that the hanging will occur sometime during the week (A)

Given this announcement the prisoner can deduce that the hanging will not occur on the last day of the week. However, in order to reproduce the next stage of the argument, which eliminates the penultimate day of the week, the prisoner must argue that his ability to deduce, from statement (A), that the hanging will not occur on the last day, implies that a penultimate-day hanging would not be surprising. But since the meaning of "surprising" has been restricted to not deducible from the assumption that the hanging will occur during the week instead of not deducible from statement (A), the argument is blocked.

This suggests that a better formulation would in fact be:

* The prisoner will be hanged next week and its date will not be deducible in advance using this statement as an axiom (B)

Some authors have claimed that the self-referential nature of this statement is the source of the paradox. Fitch[5] has shown that this statement can still be expressed in formal logic. Using an equivalent form of the paradox which reduces the length of the week to just two days, he proved that although self-reference is not illegitimate in all circumstances, it is in this case because the statement is self-contradictory.

[sagewolf]

1st--Starters--Mark--Patti--Labrador
3rd--Senior--Jane--Tiff--Collie
4th--Novice--Ruth--Jago--Doberman
9th--Elementary--Terry--Kelly--Alsatian

I'm 99% sure that's right. Watch it be wrong, just so I can have made a fool of myself :p

[Nikita]

1st--Starters--Mark--Patti--Labrador
3rd--Senior--Jane--Tiff--Collie
4th--Novice--Ruth--Jago--Doberman
9th--Elementary--Terry--Kelly--Alsatian

I'm 99% sure that's right. Watch it be wrong, just so I can have made a fool of myself :p

You got it! Your turn to post one. :)

[sagewolf]

Yay!

The king of a remote kingdom wishes to marry off his daughter and proposes a challenge: he has the floor of his audience chamber re-made in brick, with one straight line of white bricks in the centre of the floor. Any prospective suitors must first make this line shorter, without damaging the floor, removing the bricks, or painting any of the bricks in any way, before they can seek his daughter's hand.

How is this done?

[Noehelia]

Yay!

The king of a remote kingdom wishes to marry off his daughter and proposes a challenge: he has the floor of his audience chamber re-made in brick, with one straight line of white bricks in the centre of the floor. Any prospective suitors must first make this line shorter, without damaging the floor, removing the bricks, or painting any of the bricks in any way, before they can seek his daughter's hand.

How is this done?

Should the alteration be permanent?

Otherwise:

Someone can step on a brick of the white line (hiding it with his feet) or lay on them or put a carpet above.

[sagewolf]

No, it does not need to be permanent, but you're just hiding it from view-- it actually needs to be shorter.

Hint:

There's a reason I said 'make the line shorter' instead of 'shorten the line'. The line itself doesn't actually have to change.

[Rudy]

He should murder all of the other suitors waiting in line for the daughter's hand.

Yes? No? :thumbsup:

No? Rats. It does make the line shorter... ;)

[sagewolf]

Consider the king's position: would you want to be the only thing standing between that person and the throne? No way would I let him win if I were the king: I'd be dead the next morning!

(It's a good answer, though-- plenty of lateral thinking going on there. And it's proably the method the king himself has the most experience with. ;))

Hint, to make it really easy:

My father is taller than I am; therefore, I am ... than he is.

Shoot... I have to go. School and all. That pit...

[Rudy]

I addressed this in a PM to you, but I think the answer you're looking for it to make a longer line next to it. The problem with this is, first, you've already established that he isn't allowed to paint the bricks, and second, if the first line is down the center of the room, it's already the full length of it. There is no room to make a longer line next to it.

My murder answer is better anyway :)

[Nikita]

I addressed this in a PM to you, but I think the answer you're looking for it to make a longer line next to it. The problem with this is, first, you've already established that he isn't allowed to paint the bricks, and second, if the first line is down the center of the room, it's already the full length of it. There is no room to make a longer line next to it.

My murder answer is better anyway :)

He never said the line of white bricks spanned the length of the room. And though it's not worded clearly, I think the "rule" was that none of the white bricks could be painted. But I might be wrong on that last point.

[INTJGoddess]

Can he paint the floor adjunct a certain portion of the line (the same color) to make it appear shorter? Or can he add bricks to a portion of the line on eiter side to make the line shorter? If this isn't clear I'll clarify.

Sorry, but I don't know how to do the "spoiler thing..."

[Tragic Hero]

Let us suppose you have a son. You have only two children. What are the odds that you have a daughter?