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THE Most Difficult Logic Puzzle!

Zerosmelt
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10/15/2008 11:25:55 PM
Posted: 8 years ago
The world's most difficult logic puzzle!
(I'd give you the source but then you'd have the answer. )

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.

anyone care to gander. (without looking it up.) :D
Lightkeeper
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10/16/2008 1:53:06 AM
Posted: 8 years ago
Ok....

1. I understand it doesn't have to be the same question asked of each god.
2. Do the gods have to answer? Or can they remain mute if necessary?
Rezzealaux
Posts: 2,251
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10/16/2008 5:53:34 AM
Posted: 8 years ago
At 10/15/2008 11:25:55 PM, Zerosmelt wrote:
The world's most difficult logic puzzle!
(I'd give you the source but then you'd have the answer. )

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.


anyone care to gander. (without looking it up.) :D

This is only possible if each time you identify the gods with the same variable each time they answer a question, i.e., if I ask three questions, "True" will, in each of the time he/she answers, be known as "1".
: If you weren't new here, you'd know not to feed me such attention. This is like an orgasm in my brain right now. *hehe, my name is in a title, hehe* (http://www.debate.org...)

Just in case I get into some BS with FREEDO again about how he's NOT a narcissist.

"The law is there to destroy evil under the constitutional government."
So... what's there to destroy evil inside of and above the constitutional government?
Lightkeeper
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10/16/2008 6:11:20 AM
Posted: 8 years ago
This is only possible if each time you identify the gods with the same variable each time they answer a question, i.e., if I ask three questions, "True" will, in each of the time he/she answers, be known as "1".

Nope. Cos "True" only gets to answer one question. As does False as does Random :P

I know the sort of logic required but it's a nutcracker nonetheless.
Zerosmelt
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10/16/2008 7:39:40 AM
Posted: 8 years ago
just to be clear i didn't come up with this. This puzzle is just as stated. I don't know anymore about it than you do. I haven't looked at the solution yet so im still in the dark.
Zerosmelt
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10/16/2008 7:42:23 AM
Posted: 8 years ago
Nope. Cos "True" only gets to answer one question. As does False as does Random :P

I don't know if thats true lightkeeper.
it says:
each question must be put to exactly one god.

but i don't think that means each god must have one question.
Lightkeeper
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10/16/2008 7:33:25 PM
Posted: 8 years ago
but i don't think that means each god must have one question.

I think it does. You're allowed 3 questions in total and each question to be asked of exactly one god. That means you get to ask a question of god A, a question of god B and a question of god C.
Puck
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10/16/2008 8:21:55 PM
Posted: 8 years ago
but i don't think that means each god must have one question.

I think it does. You're allowed 3 questions in total and each question to be asked of exactly one god. That means you get to ask a question of god A, a question of god B and a question of god C.

each question must be put to exactly one god.

It simply means each question must be adressed to a specific god. You could ask ne god 3 questions if you so desired - though ultimately pointless. As far as I can tell, you only need to ask one god one question, and another god two questions, to arrive at the answer.

Solution requires conditional questioning - i.e. using law of excluded middle - either X is true or not-X is true, for any statement X at all.

In this puzzle, questions must include a part on the identity of one god and additionally arrive at the nature of 'ja' and 'da'. If <question about a god>, <condition about 'ja' or 'da'>.
Lightkeeper
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10/16/2008 10:53:07 PM
Posted: 8 years ago
It simply means each question must be adressed to a specific god. You could ask ne god 3 questions if you so desired - though ultimately pointless. :

You might be right. If you can ask a god more than one question then the puzzle is easier than i thought. :)
Lightkeeper
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10/17/2008 2:16:46 AM
Posted: 8 years ago
Ok, this puzzle has completely ruined my day. Still thinking. So far I've come up with the following:

1. Since any god we ask can be Random, we cannot find the answer directly but rather through a process of elimination.
2. Since we don't know what "ja" or "da" is, we must ask a question that will give us a fruitful result regardless of the meaning of those words.

I have come up with a possible question that would eliminate some possibilities:

Ask god A the following question:

"If I were to ask B and C if one of them is True, would we have complete certainty that at least one of them will say ja?"

If A answers "ja" then A is Random or True and False is either B or C.
If A answers "da" then A is Random or False and True is either B or C.

This doesn't solve the riddle but it's this type of question we need to look for.
Rezzealaux
Posts: 2,251
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10/17/2008 7:19:41 AM
Posted: 8 years ago
Oh, I misunderstood the puzzle.
: If you weren't new here, you'd know not to feed me such attention. This is like an orgasm in my brain right now. *hehe, my name is in a title, hehe* (http://www.debate.org...)

Just in case I get into some BS with FREEDO again about how he's NOT a narcissist.

"The law is there to destroy evil under the constitutional government."
So... what's there to destroy evil inside of and above the constitutional government?
LR4N6FTW4EVA
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10/17/2008 8:44:52 AM
Posted: 8 years ago
The way to do this is to find out the answers to each question.

Are you True?
True answers "Yes" or whatever the answer is for yes.
False answers the same way.
Random answers randomly.

So if anyone says no, you know it's Random.

Are you False?
It gets same thing, except the one saying yes is Random.

Are you Random?
True-No
False-Yes
Random-Yes or No

If two say yes, they are either False or Random.

Unfortunately, you can't know this, and it is hypothetical, and it takes many questions.

It's a lot like the Knights and Knaves question, but harder.
LR4N6FTW4EVA
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10/17/2008 8:50:43 AM
Posted: 8 years ago
Wait, I have an idea!

Ask one of them "Will you answer no in your languageto this question?"

If it is true, he cannot answer said question. By doing this, you can narrow it down for sure. From there, ask the same god, a different question.
Zerosmelt
Posts: 287
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10/17/2008 11:07:42 AM
Posted: 8 years ago
good point...

The problem with your assessment above though is that you are relying on the chance that you'll get the random god when its more likely that you won't.
beem0r
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10/17/2008 11:24:20 AM
Posted: 8 years ago
Oddly enough, I was about to write a proof for how it's impossible, but I just came dangerously close to solving it [or have tricked myself into thinking that I am].
Zerosmelt
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10/17/2008 11:27:37 AM
Posted: 8 years ago
At 10/17/2008 11:24:20 AM, beem0r wrote:
Oddly enough, I was about to write a proof for how it's impossible, but I just came dangerously close to solving it [or have tricked myself into thinking that I am].

The solution in chapter 29 of George Boolos's book Logic, Logic, and Logic.
beem0r
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10/17/2008 11:39:38 AM
Posted: 8 years ago
Alright, so I have a ridiculous semantics argument for it.

It says that we may ask three questions, and we must ask eeach question to exactly one God. However, depending on our definition of "ask three questions" we could ask each question over and over. For example, if I asked you "Do you liek mudkips" 100 different times, by some definitions, I have only asked you one question.

I will now write a solution using this.
Zerosmelt
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10/17/2008 11:41:14 AM
Posted: 8 years ago
At 10/17/2008 11:39:38 AM, beem0r wrote:
Alright, so I have a ridiculous semantics argument for it.

It says that we may ask three questions, and we must ask eeach question to exactly one God. However, depending on our definition of "ask three questions" we could ask each question over and over. For example, if I asked you "Do you liek mudkips" 100 different times, by some definitions, I have only asked you one question.

I will now write a solution using this.

yeah im pretty sure that is not cool.
beem0r
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10/17/2008 11:57:23 AM
Posted: 8 years ago
For the first two questions, ask "Are you true" to God A and "Are you false" to God B. Do this a crapload of times.

If either of them ever gives you a different answer than they did the first time, you know that one is random.
If neither of them does, we can assume that God C is random.

Further, we now know what Ja and Da mean.
Whichever one didn't vary in output [or both, if neither did] will give us the answer to this. If they answered Ja/Da to being true, than Ja/Da means yes, if they answered Ja/Da to being false, Ja/Da means no.

So now, after two questions, we know who random is and we know what Ja and Da mean.

However, we don't know which of the non-random people is true and which is false. thus, we ask one of them if God X [whichever one was random] is indeed random. the yes answer tells us he is true, the no answer tells us he is false.

3 questions, the first two asked a crapload of times over, and we have a solution.

Tell me if there's a problem with that.
beem0r
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10/17/2008 12:00:14 PM
Posted: 8 years ago
At 10/17/2008 11:41:14 AM, Zerosmelt wrote:
yeah im pretty sure that is not cool.

It's the only way I could think of, lol

I'll think about it some more and see if I can think of one asking questions only once. Seems impossible, though.
Zerosmelt
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10/17/2008 12:27:23 PM
Posted: 8 years ago
At 10/17/2008 12:00:14 PM, beem0r wrote:
At 10/17/2008 11:41:14 AM, Zerosmelt wrote:
yeah im pretty sure that is not cool.

It's the only way I could think of, lol

I'll think about it some more and see if I can think of one asking questions only once. Seems impossible, though.

lol.. it is supposed to be the most difficult logic problem in the world..
beem0r
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10/17/2008 2:23:50 PM
Posted: 8 years ago
Tedious, I know, but I'm going to write out all the possibilities. This approach is the only reason I was ever able to solve the problem.

#. A B C

1. T F R
2. T R F
3. F T R
4. F R T
5. R T F
6. R F T

One of these six possibilities will be the solution.

There is also more information - J and D.
J means Ja is yes, D means Da is yes.
1J would be the configuration where A, B, and C are T, F, and R, respectively [from the 1], and Ja means yes [from the J].

This leaves us with 12 possibilities in all.

Here's what I used as the first question:

To God A, "Would you say 'Ja' if asked if God B is more likely to tell the truth than God C?"

Here are the tables of responses

1J. Da [no]
1D. Da [yes]

2J. Ja [yes]
2D. Ja [no]

3J. Da [no]
3D. Da [yes]

4J. Ja [yes]
4D. Ja [no]

5J. ?
5D. ?

6J. ?
6D. ?

If God A answers "Da" to this first question, then we can eliminate 2J, 2D, 4J, and 4D.
If God A answers "Ja" instead, we eliminate 1J, 1D, 3J, and 3D.

Due to using Ja in the question in the way it was used, 2 of the numbered possibilities get eliminated.

Now, let's look at what we have left.

If Q1 = Ja:

2. T R F
4. F R T
5. R T F
6. R F T

If Q1 = Da:

1. T F R
3. F T R
5. R T F
6. R F T

Now we know from the first question [hopefully we know this anyway] that asking a question to Random is something we want to avoid [see possibilities 5 and 6 in the tables of responses for Q1]. Note that if Q1 = Ja, we KNOW that God C is not random. If Q1 = Da, we KNOW that God B is not random. This non-random God will be the one we should ask our next question to.

I used the same type of question for the next question [though it is not exactly the same question].

FOR Q1 = Ja

To God C, "Would you say 'Ja' if asked if God A is more likely to tell the truth than God B?"
2J. Da [no]
2D. Da [yes]

4J. Da [no]
4D. Da [yes]

5J. Ja [yes]
5D. Ja [no]

6J. Ja [yes]
6D. Ja [no]

FOR Q1 = Da

To God B, ask "Would you say 'Ja' if asked if God A is more likely to tell the truth than God C?"

1J. Da [no]
1D. Da [yes]

3J. Da [no]
3D. Da [no]

5J. Ja [yes]
5D. Ja [no]

6J. Ja [yes]
6J. Ja [no]

Let's look at what we have left again.

If Q1 = Ja and Q2 = Ja

5. R T F
6. R F T

If Q1 = Ja and Q2 = Da

2. T R F
4. F R T

If Q1 = Da and Q2 = Ja

5. R T F
6. R F T

If Q1 = Da and Q2 = Da

1. T F R
3. F T R

In each of these cases, we know which God is Random, and it is simply a matter of which of the others is true and which is false. For this, we have to craft a question that true and false would answer differently, and would also answer the same whether Ja or Da means yes.

First, we start out with a question they'd answer differently: "Are you random?"
But their answer would not tell us anything in this case. We would be left with two different numbers, one of J and the other of D.
Like the questions before, we just have to put a Ja into the question in a certain way that would make the answer the same whether Ja is yes or no.
This question? "Would you say 'Ja' if asked if you were Random?"

We can ask this to anyone except random.

If Q1 = Ja and Q2 = Ja, ask God B:

5J. Da [no]
5D. Da [yes]

6J. Ja [yes]
6D. Ja [no]

If Q1 = Ja and Q2 = Da, ask God A:

2J. Da [no]
2D. Da [yes]

4J. Ja [yes]
4D. Ja [no]

If Q1 = Da and Q2 = Ja, ask God B:

5J. Da [no]
5D. Da [yes]

6J. Ja [yes]
6D. Ja [no]

If Q1 = Da and Q2 = Da, ask God A:

1J. Da [no]
1D. Da [yes]

3J. Ja [yes]
3D. Ja [no]

And now, after our 3 questions, we are left with only one of the original 6 possibile lineups.

If Q1 = Ja, Q2 = Ja, Q3 = Ja

6. R F T

If Q1 = Ja, Q2 = Ja, Q3 = Da

5. R T F

If Q1 = Ja, Q2 = Da, Q3 = Ja

4. F R T

If Q1 = Ja, Q2 = Da, Q3 = Da

2. T R F

If Q1 = Da, Q2 = Ja, Q3 = Ja

6. R F T

If Q1 = Da, Q2 = Ja, Q3 = Da

5. R T F

If Q1 = Da, Q2 = Da, Q3 = Ja

3. F T R

If Q1 = Da, Q2 = Da, Q3 = Da

1. T F R

Thus, with three questions, we know the identity of each God. We have asked each question to one and only one God, we have only asked three questions, and we have asked each question only once [unlike my semantics-based answer earlier].
Lightkeeper
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10/17/2008 5:43:49 PM
Posted: 8 years ago
At 10/17/2008 2:23:50 PM, beem0r wrote:
Tedious, I know, but I'm going to write out all the possibilities. This approach is the only reason I was ever able to solve the problem.

Beem0r:

I think you're on the right track but I don't think this is it yet. Consider the following:

Assume A is False, B is True and C is Random (your possibility 3 - FTR)

Assume Da is yes and Ja is no.

This is then your possibility 3D.

You say that if you put the question "would you say Ja if asked if B is more likely to tell the truth than C?" to A, this would eliminate 3D if the answer is "ja".

Since B is True and C is Random (our assumption), the objective answer to the question "is B more likely to tell the truth than C?" is "yes"

But A is False and A always lies. Therefore A's answer to that question would be "no". In other words, if asked "is B more likely to tell the truth than C", A's answer would in fact be "no" (ja).

But when you ask him about whether his answer would be "no", he will lie about it and his answer to the question "would you say no?" will be "no".

But "no" in our example is represented by the word "ja".

Therefore in scenario FTR D, A would answer "ja" even though C is in fact Random.

It seems that this answer would not eliminate 3D.

Tell me if I'm wrong because it really is doing my head in LOL
Lightkeeper
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10/17/2008 11:16:05 PM
Posted: 8 years ago
Ok, here's my approach.

It's a 3 step approach (duh). The first 2 steps are aimed at identifying (by elimination) Random. Once that is done, we establish which one of the two remaining gods (non-Randoms) is True and which one is false. This is done in Step 3.

Step 1:
Ask A the following question:

If I were to ask those of B and C who are not Random whether B is Random, would I definitely get "ja" as an answer?

(it's worded so awkwardly for a reason….trust me)

Here are the possibilities:

A can only be True, False or Random.

A says:

(i) Ja (yes)

If A is True:
B is not Random (C is Random)
Because it means that A tells us the truth about the fact that False would say that B is random (and False would lie)

If A is false:
B is not Random (because C is Random)
Because A is lying to us by saying that True would say that B is Random. This means that True would in fact say that B is NOT Random. Therefore B is not Random.

If A is Random = B is not Random (because A is Random)

(ii) Ja (no)

If A is True:
B not Random
That is because A is telling us the truth that False would not say "no" when asked if B is Random. This means that False would say "yes" if asked if B is Random. But False lies. Therefore B is not Random.

If A is False:
B not Random
That is because A is lying to us when he says that True would say that B is not Random. Therefore True would say that B IS Random. Therefore B is Random.

If A is Random
B not Random

(iii) A says: da (yes)

A is True:
B is Random, therefore C is not Random
That is because A is telling us the truth that False would say that B is not Random. But False would lie. Therefore B is Random. Therefore C is not Random.

A is False:
B is Random, therefore C is not Random
That is because A is lying to us when he says that True would say that B is not Random. True would in fact say that B IS Random. Therefore C is not Random.

A Random
C is not Random (because A is Random)

iv) A says: da (no)

A is True:
B is Random, therefore C is not Random.
That is because A is telling us the truth when he says that False would not say that B is Random. Therefore False would say that B is not random. But false lies. Hence, B is Random. Therefore, C is not Random.

A is False:
B is Random, therefore C is not Random.
That is because A is lying to us when he says that True would not say that B is Random. Therefore True would say that B is Random. Hence, B is Random.
Therefore, C is not Random.

A is Random:
If A is Random then neither B nor C is Random.
Therefore, C is not Random

Conclusion:
If A says Ja=B not Random
If A says Da = C not Random

Step 2.
If B is non-Random, we now turn to B. If C is non-random, we now turn to C.
Assume it's B for the sake of the argument.

We ask B the same question about A&C. This will eliminate one of A&C as Random.

The question is:

If I were to ask those of A and C who are not Random whether A is Random, would I definitely get "ja" as an answer?

The logic is identical as in Step 1 except now we know that B himself is not Random. To rehash, the answers will have the following meanings:

Ja = A not Random
Da=C not Random

We can now identify one definitely non-Random god.

Step 3
We now turn to the non-Random of A&C (determined in step 2 above). Assume it's A.
We need to establish which one of A & B is True.

Here's my proposed question:

If I were to ask False whether you are Random, would he say Ja?

The objective answer is "no" (as in, A is not Random). We know that from step 2 above.

Possibilities:
A is either True or False.
B is either True or False.
We've already established (by elimination) that C is Random.

1.
A is False

a.ja is yes
False would answer "yes" (ja). A will answer "no" (da). That's because A will lie about what False (himself in this case) would say.

b.ja is no
False would answer "yes" (da). A will answer "yes"(da). That's because A will lie about what False (himself in this case) would say. Therefore A will falsely affirm the question and claim that False would answer "ja".

2. A is True

a.ja is yes
False would answer "yes" (ja). A will answer "yes" (ja). A, being True, will Affirm that False would say "yes" (ja) to the question of whether A is Random.

b.ja is no
False would answer "yes" (da). A will answer "no" (ja). A, being True, will Negate that False would say "no" (ja).

Therefore, if A answers "ja" then A is true and B is False.
If A answers "da" then A is False and B is True.
And we already know from the first 2 steps that C is Random.

I've looked over this a number of times and even did a practical demo of steps 1 and 2 with my 3 "goddesses" (my daughter, wife and mother).

I have a huge headache now and I don't want to even think about "Random" for the next two days or so.

Having said that, please have a look at this and tell me if it makes sense. I'm happy to explain.

Of course, I might be totally wrong. My logic could be totally flawed.