Logic Puzzles

[ultrafilter]

*Originally posted by panamajack *
As for the stars one, you can never guarantee that everyone will agree. Even if you know every written language in the world, you still have to deal with people who have no knowledge of a written language, blind people who can’t read Braille, babies, the illiterate, insane, irrational, and obtuse, most of whom will never agree that you have written “the correct answer” on that paper. (And if they somehow all do, I won’t. I’m just going to be that way.)

For the moment, let’s assume that everyone can read and understand English. And even if you wrote “the correct answer” on the piece of paper, that’s not the correct answer to the question, which is what we’re looking for. However, I didn’t specify that correctly, so the new riddle is at the bottom of the post.

The Zelda red/blue logic : Assuming that ‘all items are good’ and ‘all monsters are bad’, and ‘stronger items are better for you’ and ‘stronger monsters are worse for you’, then red represents ‘better’ and blue represents ‘worse’. (Certainly the colors could have been different and a rationale could have been formed for that too.)

Correct, to the best of my knowledge.

Here’s the stars riddle, correctly phrased:

I’m going to ask you if there are more than 6.02 x 1023 stars in the universe. Write the answer on a piece of paper. Make sure that everyone will agree that what you have written on the paper is the correct answer to the question.

Here’s a hint:

The fact that the question has to do with the number of stars in the universe is irrelevant. It could just as well be “Does God exist?” or “Does Bill Clinton have a cigar fetish?”.

[ultrafilter]

Also, that’s supposed to be 6.02 x 10[sup]23[/sup]. I copied and pasted from word, and forgot that the superscript doesn’t carry over.

[Max_Carnage]

The answer can be writen “I believe that there are more than 6.02 x 10^23 stars in the universe.” No one can factually deny that I do believe it, whether it is true or not.

[erislover]

Eesh. Here’s how I think of the pirate riddle, for interested parties. Please, though, try and think on it more.[spoiler]Imagine it was just pirate #1. He would obviously assign 100 gold to himself and vote yes, so there you go.

#2 knows that a tie wins, so he proposes 100 to himself and 0 to pirate 1. Though 1 disagrees, a tie wins and 2 gets the whole enchilada.

#3 knows that pirate 2 has a chance to win 100 if he disagrees with everything, so there is no way to bribe him. But he also knows that #1 won’t get any money all if it comes down to 1 and 2, so he says “ninety-nine for myself, and one for 1.” 3 and one vote yes, overrulling 2.

#4 knows that 2 will never agree to anything. He also knows that the greediest way for 3 to win is to offer 1 only one measly gold piece. So he must either offer the same proposal that 3 would, or take 98 for himself and offer pirate 1 two gold. Being the greedy pirate he is, this is his offer. 1 and 4 assent, 2 and 3 dissent, tie wins.

#5 knows he can’t bribe #2. He can attempt to bribe #3, however, since #3 will know that if #5 gets tossed 3 gets nothing (see above). So he will offer himself ninety-six, 3 he offers one and 1 he offers three (to ensure he beats out the two 1 would make on #4’s proposal and thus guarantee a victory).

#6 knows he can’t bribe #2. Assuming #6 were to lose, then #5’s most intelligent offer would be to stiff #4. So he can bribe #4 one piece. With his own vote an obvious affirmative, he only needs one more vote to tie (and win), and so needs to bribe #1 more than he would get with #5’s most intelligent offer, meaning #1 gets four gold, #4 gets one gold, and he offers himself the remaining ninety-five.

#7 knows he can’t bribe #2. He also knows that [such and such will get stiffed, bribe them, etc].

No one should have to be thrown overboard. The final bid should be something like (from pirate 1 to pirate 10):
10, 0, 0, 1, 0, 1, 0, 1, 0, 87.[/spoiler]

[ultrafilter]

*Originally posted by Max Carnage *
**The answer can be writen “I believe that there are more than 6.02 x 10^23 stars in the universe.” No one can factually deny that I do believe it, whether it is true or not. **

Yeah, but that seems like more than a little bit of a copout.

[Max_Carnage]

In that case, how about “I don’t know.” I always complained to my teachers that technically it IS a correct answer and shouldn’t be counted wrong.

[ultrafilter]

Yes, but that’s an answer about you, not about the state of the universe. When I say “the correct answer to the question”, I mean that if the answer’s “no”, you should write “no”, and if the answer’s “yes”, you should write “yes”.

[armedmonkey]

erm, eris, I can think of too many reasons why that wouldn’t work to post. Your logic breaks down at # 2 (2nd highest in your numbering system) because it is in his best interest to vote yes sometime before it gets to # 3. After that your solution falls apart like a house of cards. I think you put too many errors into the solution by trying to work it top down.

MSU, that’s an interesting solution. I think it works.

[Fretful_Porpentine]

Are the carrots meant to be consumed by an animal?

No.

Since I’ll be away from the computer until tomorrow morning, I’ll go ahead and post a hint and a spoiler to the carrot puzzle:

Hint: The carrots are frozen.

Steve discovers that his car (or horse or mule, I guess) had broken down. He knows that he won’t be able to go home immediately, and it is a hot day and the other groceries are perishable. He buys a big bag of frozen carrots to keep them cold.

[Liberal]

Okay, here’s an easy one for a break.

A cowboy rides into town on Friday. He stays three days, and rides out on Friday. How’d he do that?

[erislover]

I am totally stumped on that one, ultrafilter. The answer has to be “yes” or “no”?

[armedmonkey]

To clarify my less than crystal explanation, eris, 2 can be bribed by 4 - 10 because he does not want it to get to 3. 2-7 are going to vote no on the first proposal because there is no reason not to.

[armedmonkey]

Lib, his horse is named Friday.

[MaxTheVool]

*Originally posted by ultrafilter *
**Yes, but that’s an answer about you, not about the state of the universe. When I say “the correct answer to the question”, I mean that if the answer’s “no”, you should write “no”, and if the answer’s “yes”, you should write “yes”. **

well, you could write both “yes” and “no” on the paper. Then everyone would agree that you had written the correct answer…

Two more puzzles (both math-oriented):

(1) There’s a school with 1000 students, numbered 1 to 1000, and 1000 lockers, numbered 1 to 1000. The lockers are all closed. Student 1 opens all the lockers. Student 2 closes all the even-numbered lockers. Student 3 goes to lockers 3, 6, 9, 12, etc., and opens them if they’re closed, and closes them if they’re open. This continues with each student n swapping the state of each locker which is a multiple of n.

After this is done, which lockers will be left open?

(2) (this is heavy into the math). You’re playing a game, which works as follows. The gamemaster has two pieces of paper. On each of them is written a positive real number. One of those numbers is twice the other one. You pick one of those pieces of paper at random and look at the number on it. You can either keep that piece of paper, or switch to the other piece. After that, you will be paid an amount of money equal to the number written on the piece of paper you end up with.

So suppose you pick up your random paper, and it says 100. So you can keep 100, or switch to the other piece, which will have either 50 or 200 on it, with equal likelihood. If you keep your piece, you get $100. If you switch, your expected payoff is .5 * 50 + .5 * 200 = $125. So the correct play is clearly to switch.

However, that argument is always going to be valid, no matter what number you find written. So you don’t even need to look at the number. Just flip a coin, pick a piece of paper, and then switch to the other piece of paper, thus increasing your expected payoff by 25%.

But that’s clearly nonsense.

Where is flaw in the above reasoning?

[Max_Torque]

*Originally posted by Chronos *
**I should imagine that finding the solution for thirteen is difficult, since it is in fact impossible. While three trits is enough information to distinguish between 26 (or even 27) possibilities, there’s no way to implement it in this case. Every weighing must have the same number of coins on each pan, or it’s guaranteed to not balance (I’m assuming that the difference in weight is less than the weight of a genuine coin). If we weigh four vs. four (or 3 vs. 3, 2 vs. 2, or 1 vs 1) on the first weighing, and they balance, then we have ten possibilties left: One of five coins is the counterfeit, and it can be either heavy or light. Since we only have two weighings left, we can only distinguish between 9 possibilities.

If, on the other hand, we put five or more on each side, then we’ve got a problem if they don’t balance, since the counterfeit can be any one of those ten. Again, we only have enough weighings left to distinguish nine possibilities. **

Actually, Chronos, I believe the solution I posted earlier would indeed work for 13 coins. Everything would work the same as I posted before, except that if you add a thirteenth coin, you add the possibility that either the twelfth or thirteenth coins are the counterfeit in the situation where the scales balance for the first and second weighings. Since those weighings eliminate the possibility that coins 1-11 are counterfeit, you’re still in the game. Just weigh 12 against 1 (a coin you know is good), and if it unbalances, you know that 12 is the counterfeit; if they balance you know that 13 is the counterfeit.

(to recap, if you don’t want to go back a page for my solution, the first weighing is: 1 2 3 4 against 5 6 7 8. The second weighing is: 1 2 3 5 against 4 9 10 11.)

[ultrafilter]

*Originally posted by MaxTheVool *
well, you could write both “yes” and “no” on the paper. Then everyone would agree that you had written the correct answer…

That’s what I had in mind.

**

(1) There’s a school with 1000 students, numbered 1 to 1000, and 1000 lockers, numbered 1 to 1000. The lockers are all closed. Student 1 opens all the lockers. Student 2 closes all the even-numbered lockers. Student 3 goes to lockers 3, 6, 9, 12, etc., and opens them if they’re closed, and closes them if they’re open. This continues with each student n swapping the state of each locker which is a multiple of n.

After this is done, which lockers will be left open?**

Those whose numbers are a perfect square.

**(2) (this is heavy into the math). You’re playing a game, which works as follows. The gamemaster has two pieces of paper. On each of them is written a positive real number. One of those numbers is twice the other one. You pick one of those pieces of paper at random and look at the number on it. You can either keep that piece of paper, or switch to the other piece. After that, you will be paid an amount of money equal to the number written on the piece of paper you end up with.

So suppose you pick up your random paper, and it says 100. So you can keep 100, or switch to the other piece, which will have either 50 or 200 on it, with equal likelihood. If you keep your piece, you get $100. If you switch, your expected payoff is .5 * 50 + .5 * 200 = $125. So the correct play is clearly to switch.

However, that argument is always going to be valid, no matter what number you find written. So you don’t even need to look at the number. Just flip a coin, pick a piece of paper, and then switch to the other piece of paper, thus increasing your expected payoff by 25%.

But that’s clearly nonsense.

Where is flaw in the above reasoning? **

There is no consensus on the correct answer to this problem.

[erislover]

Ah, Beeb, you’re right! :smack: Wow. I had never thought about that.

[MaxTheVool]

*Originally posted by Max Torque *
**

Actually, Chronos, I believe the solution I posted earlier would indeed work for 13 coins. Everything would work the same as I posted before, except that if you add a thirteenth coin, you add the possibility that either the twelfth or thirteenth coins are the counterfeit in the situation where the scales balance for the first and second weighings. Since those weighings eliminate the possibility that coins 1-11 are counterfeit, you’re still in the game. Just weigh 12 against 1 (a coin you know is good), and if it unbalances, you know that 12 is the counterfeit; if they balance you know that 13 is the counterfeit.

(to recap, if you don’t want to go back a page for my solution, the first weighing is: 1 2 3 4 against 5 6 7 8. The second weighing is: 1 2 3 5 against 4 9 10 11.) **

For the record, my solution to this problem is marginally different from MT’s…

first weighing: 1234 vs 5678

if these don’t balance, then:
second weighing: 125 vs 346

if these balance, then 7 or 8 is the counterfeit.

if these don’t balance, then the counterfeit is either one of 1,2,6 or 3,4,5, depending on whether the two imbalances were in the same direction. So weigh 1 vs. 2 or 3 vs. 4, as appropriate.
If the first weighing doesn’t balance, then one of 9 10 11 12 13 is counterfeit. Weight 9 10 11 vs 1 2 3. If they balance, then 12 or 13 is counterfeit. If they don’t, then one of 9 10 11 is counterfeit, plus you know whether it’s heavy or light, at which point it becomes trivial

[Arnold_Winkelried]

*Originally posted by ultrafilter *
Let me tell you a story about a miller’s daughter. In a certain town lived a miller, his daughter, and the evil mayor. The miller was in debt to the mayor, and the mayor had his eye on the miller’s daughter. The mayor made a proposition: he would place a black stone and a white stone in a bag, and the miller’s daughter would pick one out in front of the whole village. If she drew the white stone, the mayor would forgive the miller’s debt. If she drew the black stone, the mayor would marry the miller’s daughter and take the mill. The miller had no choice but to agree. Late one night, the miller’s daughter overheard the mayor saying that he would place two black stones in the bag. The miller’s daughter was quite upset, but knew that if she told anyone, the mayor would seize her father’s mill. How can she get out of marrying the mayor and save the mill?

She draws a stone, shoves it up the mayor’s ass, and says “look in the bag to see which stone is left; I drew the other one.”

[Arnold_Winkelried]

Sorry! I didn’t notice that there was a second page to this thread.