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 . . .
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.

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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

He didn't pay for the parking!

Soroban

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. [:D]

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

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 ! [o)]

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. [t)]

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?

MathematicalPhysicist

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 hes 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?

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Hurkyl

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

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

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 any one 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?