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  #1  
Old 11-23-2006, 10:53 PM
Pudge714 Pudge714 is offline
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Default Re: Riddles/Brainteasers

KidLifeCrisis,

Don't feel too bad whenever I show someone that question there is one person who thinks its really easy and they are brilliant and they can get in like 10 minutes.

G dollaz
1) Baseball
2) Cards.
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  #2  
Old 11-23-2006, 11:08 PM
chicken10der chicken10der is offline
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Default Re: Riddles/Brainteasers

I'm a huge riddle nerd, and this is a variation of my favorite one of all time-

A plane crashes on an island (to make it Lost-related, of course). There is one survivor (Jack), and he discovers a hatch that has running water. There is a note on the faucet that reads:

There are six hatches on this island, numbered 1-6. If you drink water from one of their faucets, you become poisoned. If you do not drink water from a higher-numbered hatch's faucet within three hours, you will die. This is hatch #1.

Jack also finds a map. It points out the locations of five of the hatches, with a notation that hatch #6 is hidden and locked up so that only a member of the Dharma Initiative with a key can access it. As Jack is reading this, a man walks into the room. He introduces himself as Benjamin Linus of the Dharma Initiative. He is none-too-pleased to see Jack on his island, but being the dealmaker that he is, he makes Jack an offer:

Jack and Ben will each take an empty bottle, fill it with water, and bring it to the site of the plane crash at midnight. There, they will switch bottles and drink.

Ben lets Jack take the map and they part ways. Midnight comes, and they drink from each others' bottles. The next morning, Ben is dead and Jack is alive. How did Jack pull it off?

And no, Jack did not steal Ben's key or even stumble upon the hidden hatch #6.
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  #3  
Old 11-24-2006, 12:21 AM
EverettKings EverettKings is offline
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Default Re: Riddles/Brainteasers

[ QUOTE ]
My first OOT experience ever was a brainteaser that someone posted that I tried and it turned out to be insane. It was over 2 years ago so I don´t remember the details of how I worked out the solution, but I remember I got pissed that I couldn´t do it easily and got so determined to solve it I worked on it for like 12 hours over 2 days or something absurd like that. And I remember some other poster trying to pretend like he was solving it but really just cheating and looking it up online. Oh the nostalgia...

Problem:

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

[/ QUOTE ]

Sir, if this is impossible, I WILL kill you.

In the meantime its [censored] killing me.
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  #4  
Old 11-24-2006, 01:05 AM
Hornacek Hornacek is offline
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Default Re: Riddles/Brainteasers

Sounds like Jack is off to save his Princess Bride.
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  #5  
Old 11-25-2006, 01:48 AM
EverettKings EverettKings is offline
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Default Re: Riddles/Brainteasers

OK, final written solution to this problem:

[ QUOTE ]

Problem:

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

[/ QUOTE ]

I'm not writing anything in white so just dont read any further if you don't want to know.

I. Translation

First, getting past this 'da'/'ja' garbage:[ QUOTE ]

The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.


[/ QUOTE ]
Say you know you're talking to the Truth god, and you want to ask him "Do you like turkey?". He could say 'da' or 'ja' and you have no clue if he likes turkey or not. You want to find a way that he will say 'da' if he likes turkey and 'ja' if he doesn't like turkey. This is what you ask:

"Is at least one of the following true?
A) You like turkey AND 'da' means yes
B) You DONT like turkey AND 'ja' means yes"

<u>There are 4 cases:</u>

i. 'da' = yes and he likes turkey
In this case, criteria A is true so the god will want to answer yes, and will say 'da' (since that means yes).

ii. 'da' = yes and he doesn't like turkey
In this case, neither criterion are met so the god will want to answer no, and will say 'ja' (since 'da' means yes).

iii. 'ja' = yes and he likes turkey
In this case, neither criterion are met so the god will want to answer no, and will say 'da' (since 'ja' means yes).

iv. 'ja' = yes and he doesn't like turkey
In this case, criteria B is met and the god will want to answer yes, and will say 'ja' (since 'ja' means yes).

ZOMG! In both cases where he likes turkey(i/iii), he says 'da'. And in both cases where he doesn't like turkey(ii/iv), he says 'ja'. Thus by asking your question "Do you like turkey" with this modified logic, you have ensured that you get the answer 'da' if he wants to say 'yes', and you get the answer 'ja' if he wants to say 'no', regardless of the actual meaning of the words.

You must apply this technique to every question you ask a god. That way anytime you hear 'da' you know that he meant to say 'yes', and anytime you hear 'ja' you know that he meant to say 'no'.

Note also that the False god will always answer the opposite ('ja' if the question is true and 'da' if the question is false) and the Random god whatever, so this technique is still valid for them.

II. Interrogation

With that little problem out of the way there is still a full puzzle in itself to be solved -- who to ask what to figure things out. For simplicity I'm gonna ignore 'da' and 'ja' now and just use 'yes' and 'no', since all of our questions will be asked in a modified manner such that 'da' means yes and 'ja' means no.

We're looking at 3 gods that we will label (from left to right) A, B, and C. Denoting the gods' actual identities T, F, and R, we have the following possible orderings:

TFR
TRF
FTR
FRT
RTF
RFT

a. <u>The general idea</u>

Picking the first question is crucial. We'll ask god A since it doesnt matter which. Say, for example, we ask (in our roundabout manner discussed in part I) "Does 2+2=4?" (this isn't actually the question we want to ask, but just an example to demonstrate the problem). We know that the Truth god will say 'yes', the False god will say 'no', and the Random god will say whatever makes your life hardest (we have to consider worst case here). Either way, if you hear 'yes' then you know god A is NOT the False god (who would never give an honest answer), and if you hear 'no' then you know that god A is NOT the Truth god (who would never lie). Let's say we hear 'no', then we're down to four possibilities:

FTR
FRT
RTF
RFT

*IMPORTANT* After our next question (question 2), we MUST be down to 2 possibilities or fewer, since if we have 3 we can't possibly differentiate 3 states with one question (one binary number can represent up to 2 values).

Unfortunately, this particular question won't work, since there is no question you can ask to any person to guarantee that you'll be down to two options. If you ask god B a question, for example, then the random god might answer the same way as the Truth god would, and you can only eliminate the last possibility (still leaving you with three). I know I didn't explain that well, but try asking questions to A B or C and you will see that the random guy can always answer in such a way that, in the worst case, we still have 3+ possiblilities remaining. So now what?

b. <u>The right question</u>

You have to ask a question to eliminate the Random god from one of the slots. There's probably plenty of questions that will do this, but the one that I came up with is:

1) (To God A) "Is God B more truthful than God C?"

What I mean by 'more truthful' is the following: Truth is more truthful than Random, who is more truthful than False. So if gods B and C are (respectively) TR, TF, or RF then the honest answer is 'yes', and if gods B and C are FT, RT, or FR then the honest answer is 'no'. Let's look at our possible combinations and how god A will answer:

1. TFR -- NO -- Truth will be honest. F is NOT more truthful than R.
2. TRF -- YES -- same idea
3. FTR -- NO -- False will lie. T is definitely more truthful than R, so F is dishonest and says no.
4. FRT -- YES -- same idea
5. RTF ??? Could be yes or no
6. RFT ??? Could be yes or no

If the answer that we hear is 'no', then we can rule out 2 and 4 for sure and are left with:

1. TFR
3. FTR
5. RTF
6. RFT

If the answer that we hear is 'yes', then we can eliminate possibilities 1 and 3. This leaves us with:

2. TRF
4. FRT
5. RTF
6. RFT

I'll only solve the 'yes' case since you can figure out the 'no' case in the same way.
In the remaining 4 possibilities above, notice that god C now cannot be random! This is a huge help, as we can now determine god C's true identity by asking him our second question:

2)(To god C)"Does 2+2 equal 4?"

We know the answer is always true, so if the god says 'yes' then he is the Truth god, and if he says 'no' then he is the False god. Let's say he says 'yes' (you can figure out the other case easily), this means he is Truth and we are down to:

4. FRT
6. RFT

So since we know Truth will always be straight with us, we can ask him our final question:

3)(To god C)"Is God A the random god?"

If he says yes we have case 6, if no we have case 4. Yadda yadda you can extrapolate and figure out the cases that I skipped.

Let me know if I didn't explain something clearly or you want to see the 3 questions spelled out in their complicated form.

This was a sweet puzzle.



Everett
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  #6  
Old 11-23-2006, 10:05 PM
ziggydonks ziggydonks is offline
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Default Re: Riddles/Brainteasers

2. hache is how you spell the letter, i believe, so hache is pronounced the same way as h
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  #7  
Old 11-23-2006, 10:17 PM
G_Dollaz G_Dollaz is offline
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Default Re: Riddles/Brainteasers

[ QUOTE ]
2. hache is how you spell the letter, i believe, so hache is pronounced the same way as h

[/ QUOTE ]

in the spelling bee 2 years ago the letter "h" was spelled "aitch", so I'm not sure if that's it or not, there could easily be two spellings, don't know though
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  #8  
Old 11-23-2006, 10:23 PM
ziggydonks ziggydonks is offline
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Default Re: Riddles/Brainteasers

o, ok, the spelling bee is where i got it form, i guess i remembered it wrong...
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  #9  
Old 11-23-2006, 10:24 PM
G_Dollaz G_Dollaz is offline
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Default Re: Riddles/Brainteasers

1) a man leaves home, takes 3 left turns and returns home. when he returns home there is a man wearing a mask. explain...

2) there is a man murdered. he is lying in a room w/ 51 bicycles. why was he killed?
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  #10  
Old 11-24-2006, 04:44 AM
HitHard69 HitHard69 is offline
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Default Re: Riddles/Brainteasers

[ QUOTE ]
1) a man leaves home, takes 3 left turns and returns home. when he returns home there is a man wearing a mask. explain...

[/ QUOTE ]

He live's on the North Pole. The man's wearing a mask because he's [censored] cold. Or he's there to rape his wife.
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