Who is the liar in this logic puzzle?

[jammieg]

This one I read but is so good I wanted to post it reworded. Two groups of people live on this planet, pure liars and pure truth tellers. I went to a philosophy party(never mind that it's impossible) and met 3 people. The first says something but I don't catch it, the second one replies, "he said he was a liar." The third says to the second, "You are lying!" Is the third person a liar or truth teller?

 

[Soroban]

Hello, jammieg!

A good problem . . .

Two groups of people live on this planet, pure liars and pure truth tellers.
I went to a philosophy party (never mind that it's impossible) and met 3 people.
The first says something but I don't catch it,
the second one replies, "He said he was a liar."
The third says to the second, "You are lying!"
Is the third person a liar or truth teller?

Consider what the first person must have said.
If he were a Truth Teller, he'd say, "I'm a Truth Teller."
If he were a Liar, he'd lie and say, "I'm a Truth Teller."
That is, NO ONE would ever say, "I'm a Liar."

Hence, the second person lied.
Therefore, the third person spoke the truth.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A similar problem . . .

An archaeologist discovered a temple with three gigantic statues
guarding the entrance. To be allowed into the temple, he must
identify each of the three gods.

The God of Truth always tells the truth.
The God of Falsehood always lies.
The God of Diplomacy can tell the truth or lie.

He asked the god on the left, "Who stands next to you?"
"The God of Truth," was the reply.

He asked the god in the middle, "Who are you?"
"The God of Diplomacy," was the reply.

He asked the god on the right, "Who stands next to you?"
"The God of Falsehood," was the reply.

The archeologist immediately identitied the three gods.
Can you?

 

[Njorl]

left=diplomacy
middle=false
right=truth

Truth is obviously not on the left, or he'd be lying about who is next to him.
Truth is not in the middle, or he'd be lying about who he is.
Truth is on the right, therefore, the god next to truth is falsehood, as stated by truth.
That leaves only the left for diplomacy.



Njorl

 

[Njorl]

A man at a table is shot for having 53 bicycles. Why?

Njorl

 

[Jeebus]

Originally posted by Njorl
A man at a table is shot for having 53 bicycles. Why?

Njorl

He didn't pay for the parking!

 

[Soroban]

Originally posted by Njorl
A man at a table is shot for having 53 bicycles. Why?

Ha! . . . Took me a few moments.
The "Bicycles" (TM) are playing cards.
(He probably had an Ace up his sleeve.)

 

[theriddler876]

here's a simple one

you have 9 coins one is counterfeit and a balance, using only two weighs, find the counterfeit coin (the counterfeit coin is lighter)

 

[hypnagogue]

er how about finding the counterfeit coin without any weights at all? Simply try out all permutations of 4 coins on one side of the balance against 4 coins on the other side. Once you have found a permutation such that the two groups of 4 balance perfectly, you know that the unused coin is counterfeit.

edit: or, for a more efficient solution, pick out any two groups of 4 and place them on each side of the balance. If they balance perfectly, then the unmeasured coin is counterfeit. If they do not balance, replace one coin on the lighter side with the unused coin, and continue replacing thus-far-unreplaced coins until you have a perfect balance. This takes at most 5 separate measurements.

Actually, it technically only takes at most 4 measurements since if you have already replaced 3 of the coins on the lighter side of the balance, you can infer the 4th is counterfeit. Of course this only holds if you are correct in assuming that one of the coins is counterfeit in the first place. Reminds me of the superbowl episode of the Simpsons where Chief Wiggum is in jail and declares "one bar in a jail cell is always weak." He goes on to test all the bars except the leftmost one, and infers that it is the weak one, so he rans head-on into it only to fall flat on his back.

 

[Njorl]

Weigh 3 coins on each side of the balance. If one side is lighter, the fake is in that group of 3. If not, it is in the set of 3 unweighed coins.

Discard the 6 coins you know are true, and balance 2 of the remaining coins. If one is lighter, it is fake. If they weigh the same, the unweighed coin is fake.

You could find one fake in 27 coins with 3 weighings, 1 in 81 with 4 weighings etc.

njorl

 

[jammieg]

I think there is one of those annyoing creative solution ways to find the fake coin in 2 measures, with 2 to 2000 or more coins, but I would rather throw it scales and all into the sea.
Among all the various two typings of the people of this world there are those who are apt and preoccupied with finding and percieving the limitations of things and those who are always trying to find ways around those limits.

 

[drag]

Originally posted by Njorl
You could find one fake in 27 coins with 3 weighings

Hmm... How's that exactly ?! [b(]
I remmember doing this riddle quite some time ago and
the top number I could solve it for was either 15 or 16
for 3 weighings (I'm a bit tired right now so I won't
check which one it was, for now). But 27 ?! No way !

Peace and long life.

 

[drag]

Oops... I'm sory Njorl, never mind. I was reffering to
the riddle when you don't know weather the fake is lighter
or heavier. If you do know it's either of these options
it's another story of course.

 

[Raven]

Originally posted by jammieg
This one I read but is so good I wanted to post it reworded. Two groups of people live on this planet, pure liars and pure truth tellers. I went to a philosophy party(never mind that it's impossible) and met 3 people. The first says something but I don't catch it, the second one replies, "he said he was a liar." The third says to the second, "You are lying!" Is the third person a liar or truth teller?

I have a variation on this idea that is equally as difficult. It goes along these lines:

You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?

 

[MathematicalPhysicist]

Originally posted by Raven
I have a variation on this idea that is equally as difficult. It goes along these lines:

You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?

from which city are you?
if it's a lier than he points you to the wrong way of his city which is the liers city, he points you to the truth tellers city because it's not his city and he's lier than it fits, now if it's truth teller than he points you to the truth tellers city because he tells than truth when asked.
either way you go where the person tells you.

 

[wasteofo2]

There are 3 cannibals and 3 missionaries on one side of a river. If ever there are more cannibals than missionaries on one side of the river, the cannibals will eat the missionaries. The boat which will be used to go to the other side of the river seats 2, and at least one person must be in it in order for it to go from one side to the other. In what way can they travel to the other side of the river so that no missionaries are eaten?

-------- ---------
~~~~~~~~~~~~~~~~

 

[Hurkyl]

Legend: | | = river, < > = boat, C = cannibal, M = missionary

          |    |<  > CCCMMM
          |<CM>|     CC MM
C  M  <  >|    |     CC MM
C         |< M>|     CC MM
C         |    |<  > CC MMM
C         |<CC>|        MMM
CCC   <  >|    |        MMM
CC        |<C >|        MMM
CC        |    |<  > C  MMM
CC        |<MM>|     C  M
CC MM <  >|    |     C  M
C  M      |<CM>|     C  M

By symmetry, I've reduced the problem to one I already know how to solve, so I'm done.

 

[Hurkyl]

Here's a good crossings problem:

Chroot, Monique, Gale, and Hurkyl (four randomly chosen names) are fleeing from the law through a dark tunnel with only one flashlight. They are a good 17 minutes ahead of their pursuers. They find their path blocked by an underground ravine with only a long rickety bridge to permit them to cross. The bridge is surely too unstable to allow more than two of our heroes to cross at anyone time, and it's too dangerous to attempt a crossing in the dark. Monique estimates she can cross the bridge in one minute, Chroot estimates that he can do it in two minutes. Gale thinks she needs five minutes to do it carefully, and Hurkyl needs to take ten minutes, just to be safe. Can these fantastic four cross the bridge before the law catches them?

 

[wasteofo2]

1 and 2 go across. 15 minutes left
1 goes back. 14 minutes left
5 and 10 go across. 4 minutes left
2 goes back. 2 minutes left
2 and 1 go across. done with no time to spare.

Ok, now someone explain the one with the 3 people in a hotel.
the one where they pay $10 each for the room, but the hotel clerk finds out the room was onlt $25, so he gives $5 to the bellboy to return. The bellboy can't figure out how to divide it evenly so he gives them $1 each and takes the other 2 for himself. They each are now out $9, which means they lost $27 in paying for the room, which leaves $3 left, but the bellboy only kept 2.

 

[phoenixthoth]

how 'bout this logic puzzle?

either God exists or God does not exist.

which is it and provide proof.

 

[THANOS]

If you had one question and an all knowing robot will answer it for you but will kill you when it answers you what will you ask him? Assuming he was indestructble and can not be destroyed.

 

[phoenixthoth]

is that a logic puzzle? maybe i'd ask it a loaded question like, "why will you never kill me?"

on the other hand, if such a robot existed...

It is the categorical formulation of the simultaneous, situational, instantiated contradiction, where deductive invalidity is the product of the utmost categorical truth of the assumption that if the antecedent of a true conditional is false, then the consequent of the conditional is true or false indifferently, and of the categorical falsehood of the conclusion consequently predicates that if it be not the case that the consequent of a true conditional is true or false indifferently, then, it is not the case that the antecedent of the conditional is false. To pronounce the consequent of a true conditional as being true or false indifferently is tantamount to saying modally that where the antecedent of a true conditional is notoriously false, then the consequent can, or could be, or is possibly true or false. But it may be worthwhile to see that the definitive, simultaneous equality of both true, and false, can be formulated without explicitly including modal terms, which become the predicating operators, which, for the sake of showing that the consequent paradoxical conundrum is not straightforwardly resolvable by appealing to concrete philosophical scruples concerning the intensionality of predicated modal contexts.

and so the natural question i'd gladly give my life for is this: what must one appeal to to resolve the paradoxical conundrum?

(the answer relates to my logic puzzle.)

 

[wasteofo2]

Originally posted by THANOS
If you had one question and an all knowing robot will answer it for you but will kill you when it answers you what will you ask him? Assuming he was indestructble and can not be destroyed.

I wouldn't ask it a question.

 

[NateTG]

Originally posted by wasteofo2
Ok, now someone explain the one with the 3 people in a hotel.
the one where they pay $10 each for the room, but the hotel clerk finds out the room was onlt $25, so he gives $5 to the bellboy to return. The bellboy can't figure out how to divide it evenly so he gives them $1 each and takes the other 2 for himself. They each are now out $9, which means they lost $27 in paying for the room, which leaves $3 left, but the bellboy only kept 2.

It's adding and subtracting the wrong numbers -- the 27 spent is 2 for the bellboy and 25 for the inn. The $3 is the amount that they got back. It makes just as much sense to ask how the inkeeper could have kept $25 when there are only $3 left.

 

[THANOS]

D'oh, forgot to mention that if you don't ask a question the robot will kill you anyway.

 

[NateTG]

Lots of questions:

Questions that will take more than my natural life to answer:
What are the first 10^10^10 digits of &pi; in order?
Questions that require resolution of the dilemma to be meaningful:
Why did you self-distruct?
Questions that are paradoxical:
What happens when an immovable object meets an irresistable force?
Questions for a solution to the dilemma:
How do I avoid getting dying after you answer this question?
Questions that have no correct answer:
What is an effective method for destroying you?
(Robot is indistructible by setting.)
Questoins that provide meta-abuse:
What is the text of questions and answers that a comprehensive list of questions and corresponding answers that would allow me to avoid this dilemma would contain?

 

[wasteofo2]

Originally posted by NateTG

Questions that require resolution of the dilemma to be meaningful:
Why did you self-distruct?

The robot could say "I did not self destruct, you are confused" then kill you

Originally posted by NateTG
Questions that are paradoxical:
What happens when an immovable object meets an irresistable force?

The robot could say "I don't know" then kill you.

Originally posted by NateTG

Questions for a solution to the dilemma:
How do I avoid getting dying after you answer this question?

The robot could say "You cannot" and then kill you

Originally posted by NateTG

Questions that have no correct answer:
What is an effective method for destroying you?

"There is no effective method" then you're killed

Originally posted by NateTG

Questoins that provide meta-abuse:
What is the text of questions and answers that a comprehensive list of questions and corresponding answers that would allow me to avoid this dilemma would contain?

"There are no questions with corresponding answers that would allow you to avoid this dilemma" and then you're killed.

 

[THANOS]

I'll just post the question that was meant for the question i asked.

the question is "What is the one question you can not answer?"

 

[MathematicalPhysicist]

if i could answer your question then it wouldn't be the right question if i haven't answered to your question then it would make the criteria therefore this statement is the same as saying "im a lier".

 

[NateTG]

Originally posted by wasteofo2

"There are no questions with corresponding answers that would allow you to avoid this dilemma" and then you're killed.

Well, unless you want to debunk the first question (which is fairily solid), there are questions, and corresponding answers that resolve the dilemma.

Thanos: The robot can reply, "There is no such single question." and blow your head off because you include 'one'.
"What are all the questions you cannot answer?" can be answered as well.

In fact, you did not stipulate that the robot answer truthfully, so I suppose it could just respond with '42' before ventilating your skull, regardless of how clever the qestion was.

I suppose you could work around the problem by asking a question that takes longer than your own natural life to complete asking.

 

[motai]

Ask the robot to say aloud all of the digits of pi, hehe. Hopefully by the time the gets around to blowing my head off I would have already found a way out, hacked the internal circuitry of the robot, saved all the (useful) information on some sort of storage device, and change the robot's OS to Microsoft, thus effectivly dooming the robot.

 

[jammieg]

Here's another one some strange guy said to me once:

So your employers give you more work than you don't want?

What does he mean?
This might seem really simple but for some reason it confused me greatly and I'm wondering if there is more to it.

 

[marqq]

geometry and logic

where is the missing square from the attachement bellow?

 

[NateTG]

The red and green triangles have different slopes, the appearance that the (macroscopic) figures are triangles is misleading.

You can see this by, for example, looking at the height of each of them 5 units from the left point. The lower shape has a height of two, but the upper shape has a height of 15/8 which, even in the illustration, is less than 2.

 

[marqq]

clasique

We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's.
Every man has his own, single, different pet, and smokes his own, type of cigarettes.We know that:
1.The englishman lives in the red house
2.The sweedish has a dog
3.The dutch drinks tea
4.The green house is situated at the left of the white house
5.The person who lives in the green house drinks coffee
6.The person who smokes PallMall has a bird
7.The person who lives in the yellow house smokes Dunhill (like I do )
8.The person who lives in the 3rd house drinks milk
9.The norwegian lives in the 1st house
10.The Blend smoker lives in the house near the house where lives a cat
11.The person who owns a horse lives near the Dunhill smoker
12.The person who smokes Blue Master drinks beer
13.The german smokes Prince
14.The norwegian lives next to the blue house
15.The Blend smoker is the neighbour of the one who drinks only watter

The question is : Which one of them has a fish?

 

[Pergatory]

Originally posted by marqq
We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's.
Every man has his own, single, different pet, and smokes his own, type of cigarettes.

I loved this one! This was created by Einstein originally, claiming that only 2% of the world's population could solve it. It would be interesting to find the percentage of people on this forum who can solve it.

 

[jammieg]

Silly questions

Why doesn't a bird put a roof over it's nest, and yet they will live in birdhouses?

My guess is that anyone can solve it given enough time and effort.

 

[leto]

We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's. Every man has his own, single, different pet, and smokes his own, type of cigarettes.

It is the German, right? That was extremely hard for me; these puzzles are great! This thread has kept me entertained today.

 

[marqq]

Response

Yes,it is the German...I delayed the answer, hoping that i'll find other answers :D guess not...leto I think we r among the 2% (but i doubt that only 2% could solve this problem)

 

[Eyesaw]

I think the solution to the puzzle depended on how
you interpreted the word "near"- the term "next to"
was used elsewhere so the usage of the word "near"
in other descriptions becomes ambiguous.

 

[jammieg]

I mentioned this in another thread and thought it worth repeating, there is a way to solve this type of problem in one's head without pencil and paper, even if you don't have a good imagination although it helps to visualize it, the key is to read through each factual and try to memeorize the logical functions of each factual a bit more, by this I mean to think along the lines of, "well it could go here or here but not here unless that other could go there, ok move on to the next one" and so reading through the list over and over and focusing only on one at a time the functions of those factuals start to engrain in the memory and things may start to snap together seemingly without effort, although it took me way over an hour to do it this way I'm sure most anyone could do it this way with practice, it's the same principle as memorizing a long poem except one is trying to memorize the possibilities and limitations of each factual you would be amazed at how long a poem a person can memorize if they want to.
Let me know if it doesn't work, that means I probably don't know what I'm talking about.

 

[Hurkyl]

It's a trick question; the 5th pet is really a salamander!

 

[marqq]

May I ?

1. Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.

2. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.

3. When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.

How old are they now?

 

[Nice coder]

for that robot
"Calculate 1/3 until there is no remainder ,the formula 14700/44100 = the answer is true and using only 2 decimal places (without rounding)"

The robot will never finish that because
1/3 to 2 decimal places (no rounding) is 0.33
0.33 < 0.33333333333 (to infinity)
14700/44100 = 1/3

another one would be,
Find the largest posible numerator in a fraction that would equal pie
That would never work, because pie is a never-ending number, and the highest possible numerator would be infinity!

Those questions would be great to combine and ask the robot, you don't have one by any chance?

 

[marqq]

Nice coder for the 1st question it can answer you 0.(3) :D i think "pi" is the question who may save us :)

 

[pedestrian]

sorry nvm

 

[Eyesaw]

Originally posted by marqq
1. Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.

2. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.

3. When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.

How old are they now?

I don't think this problem is solvable. 2 requires Tim's age to be
an odd number while 1 requires Tim's age to be an evem number and 3 requires a lot of tylenol.

 

[gnome]

I don't think this problem is solvable. 2 requires Tim's age to be
an odd number while 1 requires Tim's age to be an evem number and 3 requires a lot of tylenol.

Eyesaw would think that, trapped as he is in his private twilight zone where time has 3 dimensions. But actually, it turns out that it's just algebra. Let N = the current year ("Now"), let T = the year Tim was born, M = the year Mary was born, J = the year Jane was born. And x₁ through x₁₀ can represent the various other unknown years in the clues.

N + 10 - T = 2(x₁ - J)
x₁ - M = 9(x₁ - T)
N - 8 - M = (1/2)(x₂ - J)
x₂ - J = 1 + x₃ - T
x₃ - M = 5(N + 2 - T)
x₄ - T = 1
x₄ - M = 3 + x₅ - T
x₅ - J = 3(x₆ - M)
x₆ = x₇ - 6
x₇ - J = (1/2)(x₈ - T)
x₈ - M = 10 + x₉ - M
x₉ - J = (1/3)(x₁₀ - T)
x₁₀ - M = 3(J - M)

Now, work from the bottom up, substituting out the x's as you go. After eliminating the "x" years, all of part 3 boils down to
-4T -3M + 7J = 1, and the other two equations are
13T - M - 8J = 4N + 40
-6T + 3M = -3N - 27
Solve them in terms of N to get the ages (T-N, M-N and J-N).

[Edit: sorry about that; I meant to write N-T, N-M and N-J]

Tim is 3
Mary is 15
Jane is 8

 

[marqq]

:D

Bravo !
Well Done !

 

[marqq]

Here we go again

we have 10 bags with 10 gold ingots each...Each ingot has 10kg, but we have a bag full of fake ingots (9.900 kg each).we have a coin scale, and a single coin (for a single use of the scale), and we have to find the bag with the fake ingots!
Good Luck!

 

[gnome]

A balance or a scale?