## The world's hardest logic puzzle!

Philosophical, mathematical and computational logic, linguistics, formal argument, game theory, fallacies, paradoxes, puzzles and other related issues.

### The world's hardest logic puzzle!

“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 all questions 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.”

Here’s a few clarifications about the puzzle.

1. It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).

2. What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)

3. Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.

4. Random will answer ‘da’ or ‘ja’ when asked any yes-no question.

Answer: Do not cheat and procure the answer, there is more than one answer and I solved for one really quickly.
Percarus

### Re: The world's hardest logic puzzle!

I can't solve it; it gave me a splitting headache. Does the answer depend on redefining what a yes or no question is, because I didn't want to do that? Can you give me a clue?
stuartp523

### Re: The world's hardest logic puzzle!

Fuq** posted this:

Not sure but my questions are as follows
1. Are you random? = A. true = no . B False = yes, C. Random = both.
2. Are you false? = A. true =no , B. False = no. C Random = both.
3. Are you true ? A. true =yes . false = yes. C random =both.
So determining who god ‘true’ is it must answer 1 no 2 no 3 yes
And false must answer 1 yes 2 no 3 yes
This allows me to determine ‘da’ or ‘ja’ since the ‘order’ of any arbitrary symbol for a true false answer be [ Ø. Ø .▼] for true and [▼. Ø .▼] for false, random only randomly replicated at least one of these answers so at least one God can be determined for certain, however if random answer in any other combination all 3 could be determined,
I know I didn’t get it but that’s as far as my logic would allow me I look forward to the answer : )
Percarus

### Re: The world's hardest logic puzzle!

Wats** posted this to me also:

Thought I'd send you this attempt, PM in case I'm "not even wrong" or not to spoil it for others in case I'm close to right.
I see a problem with it, so I'll just put it out there, wrongly.

Question 1- Do you some times tell the truth? giving ja/no/true, da/yes/false and da/ja/maybe.
Question 2- Do you always tell the truth? giving da/yes/true, da/yes/false and da, or ja.
Question 3- Did you answer the truth in questions 1 and 2.
giving da/yes/true, da/yes/false and da, or ja.

"A" must be truth as the answer are consistent.
"B" must be false because answer 1 and 2 cannot both be true, but can both be false.
"C" must be random, by elimination, if not by the random answers.
Last edited by weakmagneto on November 1st, 2012, 2:59 pm, edited 1 time in total.
Reason: Removed user name as per request
Percarus

### Re: The world's hardest logic puzzle!

This is my derivation of it, I am not sure if I am right (but I have the answer elsewhere). First you would have to determine who the random god is, and this can be accomplished by asking the same question to each single god until one finally says a different answer to the same question. We then would have to ask the following questions:

Assuming 'da' means yes:
If I asked the other god would he say 'da' as yes? *false god would say 'ja'; *true god would say 'ja'.

Then we ask, would you say 'da' as yes? *false god would say 'ja'; *true god would say 'da'.

And hence we can determine the god who changed answers is the true god.

Any futher questions/solutions do post up, and then eventually I will post the true answer.
Percarus

### Re: The world's hardest logic puzzle!

I got the answer from an old movie favourite.

Question 1.
God A, is God B or C the random God?

Question 2.
God B, is God A or C the random God?

Question 3.
God C, is God A or B the random God?

This is my quick reply before someone beats me to it, so now I have to try test it out somehow :P

*edit haha the da/ja makes this harder than I thought!

edy420
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### Re: The world's hardest logic puzzle!

Those posts confused me more. The one from fuq appears to be asking 9 questions (I can easily solve it in 4, I find it can only be answered in 3 depending on who A and B are). Perhaps I failed to understand the rules.

Due to the fact that you don't necessarily know whom your addressing during each question, would someone please show there answer in the format of code language, in order to properly reflect the several contingencies that may arise? Like this:

Question 1: [insert letter you’re addressing], [insert question]?

-----------then Question 2: [insert letter you’re addressing], [insert question]?

----------------------then Question 3: [insert letter you’re addressing], [insert question]?

----------------------then Question 3: [insert letter you’re addressing], [insert question]?

----------------------------------------------------------------------------

-----------then Question 2: [insert letter you’re addressing], [insert question]?

----------------------then Question 3: [insert letter you’re addressing], [insert question]?

----------------------then Question 3: [insert letter you’re addressing], [insert question]?
stuartp523

### Re: The world's hardest logic puzzle!

I realized after submitting my answer that random could have come up with a series of answers matching those of truth or false, so I think fuq** and I were thinking along the same lines, with different questions. Randomness would show up if the number of questions were increased.

Watson
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### Re: The world's hardest logic puzzle!

Watson:- if the number of questions were increased.

Precisely! And I’m not good with logic, however I found the puzzles restriction to be just over the peramiters of what could be solved , now as Percarus said to me in our privet discourse you could just ask random the same question thrice and then determine randomness from the answer , although! random could in fact answer correctly or similarly to all possibility’s , so for my 2 cents the random god is really ‘false’, and logically then we can move false to random since a lie can be also true counting its deceptive quality. DUNOW hahaahaha but this was fun

Fuqin
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### Re: The world's hardest logic puzzle!

Yes, I was thinking random could mimic truth or false, by accident or design. If random has the option of answering always truth or false, then I expect the she can remain undetected. If random is flipping a yes/no coin, then she will be easily detected, in time.

Watson
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### Re: The world's hardest logic puzzle!

To reduce the random element, you could ask if it is possible that both the other two Gods would say that your God X

ie.
God A, is it possible that both god B and C would say that your the false god?
If god A answers yes, then one of gods B/C is the random god or god A is the false god.
..assuming you know what is yes :P

If the answer is no, then god A is either the random god or the truthful god.

Asking all 3 gods this question gives you something to work with..

edy420
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### Re: The world's hardest logic puzzle!

t
hen one of gods B/C is the random god or god A is the false god.
..assuming you know what is yes :P

Yes good one Edy so we are chocking it between random and false which is what the truth of a mater should do , I hope any way ! this logic stuff is not for me lest ways : D

Fuqin
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### Re: The world's hardest logic puzzle!

I have a conditional answer to the riddle but it has to make use of a loose definition of a yes or no question. I thought I would also show a conditional answer to the riddle without the added difficulty of yes and no being disguised, that answer won't have to relly on a loose definition of a yes and no question. For both answers I use the idea of a double question to make it so that True and False have to answer the same. False will hear the question in the question and realize he would give a false answer, but he won't tell you that because it would be telling the truth, therefore he will answer give a false answer to the whole question.

I have a theory as to why the riddle might be unsolvable even without that added difficulty of yes and no being disguised. We need to know 3 identities, each yes and no question can give us 1 piece of information. Since Random is random he is useless, therefore we only have 2 pieces of information. Another way of looking at is we need to just one answer; which of the 6 combinations of True, False and Random is ABC? But, we only have 4 possible clues; Both True and False's yes or no. But, I assume there is something I overlooked, would you post (or send me a PM) of the answer?

Here is my answer for the riddle without the added difficulty of yes and no being disguised:

Question 1: A, if I asked you if you were either True or False, would you say yes?

-----------{if A answered no. (Only Random would answer no, so A is Random.)

-----------then Question 2: B, are you Random?

----------------------{if B answered yes. (Question 3 isn't necessary, we know A is Random, B is False and C is True.)

----------------------if B answered no. (Question 3 still isn't necessary, we know A is Random, B is True and C is False.)}}

-----------{if A answered yes. (A could be anyone.)

-----------then Question 2: B, if I asked you if you were either True or False, would you say yes?

---------------------{if B answered no. (Only Random would answer no, so B is Random.)

----------------------then Question 3: C, are you Random? (If C answered yes, he is False, and A is true, if C answered no, then vise versa.)

----------------------if B answered yes.(A and B could still be anyone.)

----------------------then Question 3: C, if I asked you if you were either True or False, would you say yes? (If C answered yes, then A, B and C could still be anyone, if C answered no, then C is Random because only Random would answer no, but we still don't know who A or B are.)}}

________________________________________________________

Here is my answer for the riddle with the added difficulty of yes and no being disguised:

Question 1: A, if the word da changed its meaning to 'Random is either A, B or C', and if the word ja changed its meaning to 'Random isn't A, B nor C', and I asked you which of the 2 words you would rather say, then what would you say?

-----------{if A answered ja. (Only Random would answer ja, so A is Random.)

-----------then Question 2: B, if the True's name changed to da, and if the False's name changed to ja, and I asked you your name, then what would you say?

----------------------{if B answered ja. (Question 3 isn't necessary, we know A is Random, B is False and C is True.)

----------------------if B answered da. (Question 3 still isn't necessary, we know A is Random, B is True and C is False. )}}

-----------{if A answered da. (A could be anyone.)

-----------then Question 2: B, if the word da changed its meaning to 'Random is either A, B or C', and if the word ja changed its meaning to 'Random isn't A, B nor C', and I asked you which of the 2 words you would rather say, then what would you say?

----------------------{if B answered ja. (Only Random would answer ja, so B is Random.)

----------------------then Question 3: C, if the True's name changed to da, and if the False's name changed to ja, and I asked you your name, then what would you say? ((If C answered da, he is True, and A is False, if C answered ja, then vice versa.)

----------------------then Question 3: C, if the word da changed its meaning to 'Random is either A, B or C', and if the word ja changed its meaning to 'Random isn't A, B nor C', and I asked you which of the 2 words you would rather say, then what would you say? (If C answered da, then A, B and C could still be anyone, if C answered ja, then C is Random because only Random would answer ja, but we still don't know who A or B are.)}
stuartp523

### Re: The world's hardest logic puzzle!

Percarus wrote:“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 all questions 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.”

Here’s a few clarifications about the puzzle.

1. It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).

2. What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)

3. Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.

4. Random will answer ‘da’ or ‘ja’ when asked any yes-no question.

Answer: Do not cheat and procure the answer, there is more than one answer and I solved for one really quickly.

I ask the first god, "Do you agree that this sentence is false?" Only god c can give an answer, because the internal contradiction prevents a or b from answering. If an answer is forthcoming, then I have my C. If no answer, then I know the god is either A or B, but not C.

Second question, assuming I get no answer, "Do you disagree that this sentence is false?" Again, only C can answer. So, if I get no answer from the first two questions, then I know C is the third god. If I get an answer to the second question, but not the first, then I know C is the second god, and if I get an answer to the first question, then the first god is god C. So, at the very least, the two questions identify god c for me.

If c is the last god, then I ask one of the other two gods, "If god in the third position is C, then say ja? Now, if the god always tells the truth, then ja will be forthoming, and if a liar, then da will utter, because the question must be answered with "ja" if the god is the one that always tells the truth.
ComplexityofChaos

### Re: The world's hardest logic puzzle!

It is a case of probability that you get the correct answer. This is because of the answers from the random god. You may ask questions to random god and it gives you the perfect answers for a liar or an honest answer. You have to ask the random god at least twice or not at all so you can never be certain to determine anything from three questions.

That said :

If "ja" means yes and "da" means no are you answering randomly?

Answer : Yes > either False or Random. No > Truth or random

If "ja" means yes and "da" means no are you telling the truth?
Is this the random god?

Answers : Yes > either false, true or random. No > Random.

If possible answers to date are :

False = Yes, Yes.
Truth = No, Yes
Random = Yes/No, Yes/No.

By this point you should be able to see the Random god or at least see which god is mirrored by the Random god leaving either a true or false god ...

Last question to "non-mirrored" god (will be either False or Truth) :

If "ja" means yes and "da" means no, if I was to ask the Random god what would be the reply be?

This ONE question would break the puzzle unless asked to the Random god :P

This works BUT I have to ask all the question to all the gods or just use the last one to get the info needed. One question to one god at a time? I'd have to say it cannot be done with certainty only probability. You have to hope the Random shows up because if the Random mirrors the rest then it is impossible to tell and you are lost with the last question.

That is my final answer! There is NO WAY IT CAN BE DONE ;)

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### Re: The world's hardest logic puzzle!

We can just ask all three:
Are you always honest and your answer is going to mean ‘no’, or are you a liar, the random guy follows you, and your answer is going to mean ‘yes’?
If only one god is quiet, he’s truthful, followed by the random god, followed by the liar.

If two gods are quiet, the speaking god is the random god. Of the two who don’t speak, the liar follows the truthful god.

By “follows”, I mean that if we order them in positions 1 – 2 – 3, then 2 follows 1, 3 follows 2, and 1 follows 3.
Natural ChemE
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### Re: The world's hardest logic puzzle!

BagerJelly, that was an interesting way to go about the problem. Unlike me you gave a conditional answer to the original riddle without making use of a loose definition of a yes or no question.

Natural ChemE, You seem to have found the answer to the riddle, and the perfect question using ComplexityofChaos's method of introducing silence. Perhaps I'll have to agree that you have solved it, but I have my doubts about your logic, so I'll review it some more before I do.

stuartp523 wrote:Another way of looking at is we need to just one answer; which of the 6 combinations of True, False and Random is ABC? But, we only have 4 possible clues; Both True and False's yes or no.

You verified that part of my theory, then you deleted your verification (and a lot more as well) from your last post. (Was it my attempt at using the format of programming code?)

Natural ChemE wrote:Okay, two bits of information. At best, two bits can code for 2[to the 2nd power]=4 possibilities. Since the location of the gods alone has 6 possibilities, you can’t code for it with two bits. Thus this problem is impossible - if you stick to yes/no answers.
stuartp523

### Re: The world's hardest logic puzzle!

I did ALMOST get it :

If "ja" means yes and "da" means no, if I was to ask the Random god what would be the reply be?

This ONE question would break the puzzle unless asked to the Random god :P

I just didn't approach from the correct direction! :P

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### Re: The world's hardest logic puzzle!

stuartp523 wrote:You verified that part of my theory, then you deleted your verification (and a lot more as well) from your last post. (Was it my attempt at using the format of programming code?)

I tend to dislike long posts, so when I realized how excessive it was, I removed it. But since it's probably relevant, here's (pretty much) what was removed [the rest of this post]:

So we get to ask three yes/no problems. Since any particular god can be the random one, anything that we ask it is completely wasted – his answers are merely gibberish. And let’s plan for the worst-case scenario, in which the god that you ask the most questions happens to be the random one. So the best you can do is ask each god exactly one question, and you’ll get two bits of information.

Okay, two bits of information. At best, two bits can code for $2^{2}=4$ possibilities. Since the location of the gods alone has $6$ possibilities, you can’t code for it with two bits. Thus this problem is impossible - if you stick to yes/no answers.

ComplexityofChaos has the right idea – introduce silence as a possible answer. Now we can code for $3^{2}=9$ possibilities, which is enough to code for all of the gods’ positions. Though, I would point out that if you asked them, “Do you agree that this sentence is false?”, they’d be able to answer. Since that sentence is obviously neither true nor false, the truthful god would say whatever means “no” (since he doesn’t agree that it’s false) while the liar would say the other thing.

To silence the truthful guy only, you could ask him, “Will your answer mean ‘no’?” He can’t say [yes] because it’d be a lie. He can’t say [no] because it’d be a lie. He can’t provide a truthful answer, so he’s quiet. The liar can still answer, though, with either [yes] or [no] (it’d be a lie either way).

To silence the liar only, you could ask him, “Will your answer mean ‘yes’?” He can’t say [yes], because it’d be true. He can’t say [no], because it’d be true. He can’t provide a dishonest answer, so he’s quiet. The truthful guy can still answer, though, with either [yes] or [no] (it’d be true either way).

So how do we want the gods to answer? We can come up with a key, then construct a question to force them to answer in that manner.

There are two possibilities for the other of the god’s:
1. true-random-false;
2. true-false-random.
We wrap-around, so, for example, random-false-true is an instance of (i).

Let’s code for (i) by having the truthful guy only be quiet in that case, marking his position and implying that he’s followed by the random guy.

Let’s code for (ii) by having both the truthful guy and the liar be quiet. We’ll know that the only one who speaks is random, and that, of the two being quiet, the true one precedes the false one.

Now let’s just construct the question.

We can do this using Boolean algebra, just taking together statements. We want to control the honest guy alone, so we ask if he’s honest (which is evaluated truthfully by both the honest guy and the liar until the end, before he answers) and just $\text{AND}$ it with something that silences him.

We can then construct something for the liar, asking if he’s the liar, $\text{AND}$ing it with if the random guy is after him, and $\text{AND}$ing that with something that silences him.

Finally, we just $\text{OR}$ those two statements together, and we have our answer (or, you know, question) that we can just ask all three gods at once. This is perfectly legal because both of the statements to be $\text{OR}$ed together default to $\text{false}$ when they’re not being used, and $\text{OR}$ing something with $\text{false}$ in Boolean algebra is like multiplying something by $1$ in normal algebra: you always get the same result. So, when the half of the statement that’s not supposed to be used in a particular case is $\text{OR}$ed in, the effect is always harmless.
Natural ChemE
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### Re: The world's hardest logic puzzle!

A shorter version would be to ask all three gods:
Then the same interpretation. This is:
• if only one god responds, he’s the random one, followed by the truthful god, followed by the liar;
• if two gods respond, the quiet one is the truthful god, followed the random god, followed by the liar.

Edit - Opps, this Answer #2 is wrong. I made Answer #1 a while ago and forgot why I included the two extra clauses in it!
Natural ChemE
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### Re: The world's hardest logic puzzle!

I thought I was being awfully clever with that theory, having absolutely no background in math. Being that you seem like you do have a background in math I was pleases to find you agreed.
stuartp523

### Re: The world's hardest logic puzzle!

stuartp523,

You were definitely right to pick up on the information limit. Realizing limits like that helps prevent us from bashing our heads into a wall over and over, trying to solve some literally impossible problems.

There’s a specialized field for this sort of thing, information theory. But it’s used in just about every technical profession, at some level or another.
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### Re: The world's hardest logic puzzle!

This is my first ever post on this site, I have skip read everyones posts several times but I have not read them completely thoroughly; I am a slow reader and typer and I am not used to this interface. I have not solved the puzzle completely, with the rule of the unknown meaning for yes and no in their language. As stuartp523 said it maybe impossible to figure out even if the gods spoke English, because Random's answers are useless so if you asked all 3 a question you could only rely on 2 answers giving 4 possible combinations of answers. not six or more which are needed. I don't think anyone has said he may be wrong yet. So I tried to solve it with the Yes/No confusion removed to see if that were possible, and it took me ages but I think I have solved it.

Random's answers are not completely useless, if you ask it does 1+1=2 and it answers yes, then you know it cannot be the False god. If it answers no you know it cannot be the True god. I realised you need to isolate the Random god in the first question to 2 possible places, so you could rely on your second question not being asked to the Random god.

T=true god F=False god R=Random god. 1=first in sequence 2=second in sequence 3=third in sequence.

Q1

There are 6 possible combinations the gods could be.

TFR TRF FTR FRT RTF RFT

First question to 1 "(For the purpose of this question consider Random to be truthful half the time, and of course True all the time and False none of the time) Does 2 speak the truth more than 3?"

Q2(Q1=yes)

Yes means there are now 4 possible combinations, notice random cannot be in place 3
TRF FRT RTF RFT

question 2 to god 3 "does 1+1=2 ?"

Q3(Q1=yes Q2=yes)

there are now 2 possible combinations we know for definate 3 is true
FRT RFT

Question 3 to god 3 "is 1 False god?" Yes=FRT No=RFT

Q3(Q1=yes Q2=no)

Possible comb we know 3 must be false,
TRF RTF

Question 3 to god 3 "is 1 the True god" Yes=RTF No=TRF

Q2(1=no)

There are now 4 possible combinations Random cannot be god 2
TFR RTF FTR RFT

Question 2 to god 2 "does 1+1=2 ?"

Q3(1=no 2=yes)

possible combinations, True must be god 2
RTF FTR

Question 3 to god 2 "is 1 the False god?" Yes=FTR No=RTF

Q3(1=no 2=no)

possible comb, we know 2 must be false
TFR RFT

Question 3 to god 2 "is 1 the true god?" Yes=RFT No=TFR

I have found the way to all 6 combinations (2 of them can be found two separate ways) with 3 questions; The first question guarentees the 2nd question won't be asked to the random god, the second question finds out the identity of the god you're speaking to, and knowing this the third answer finds out the identity of one more god, which means the last god has to be the remaining one. Does this make sense? I have not figured it out with the unknown meaning of their yes and no yet, and I don't know if it can be. I won't try now it's late and I need sleep. Sorry if my writing it down looks awkward.

On a second thought am I allowed to say something like"If Da meant no and Ja meant yes, how would you answer the question......" at the start of each question or is that cheating? Then One could find out everything exept their original meaning of Da and Ja, with the unknown language rule intact.

carlwev
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### Re: The world's hardest logic puzzle!

It has been shown that it is impossible to do with any degree of certainty following the rules stated in the OP ... that said it is fun to play with the accuracy and range of approaches.

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### Re: The world's hardest logic puzzle!

Came back to this after four years, and solved it two ways one with modified questions similar to those in my original partial solution. I have since found this riddle has a Wikipedia article, and several solutions for it are written there. The solutions I thought of is not one listed in Wikipedia.

SOLUTION 1
The first solution I thought of first is more complicated, and is as follows. The first question, is to make sure the second question will not be asked to the Random God.

Imagine the gods aligned A B C, first you ask god A a question, there are 8 possible combinations the gods could be arranged T=True God, F=False God, Rt=Random God answering truthfully, Rf=Random God answering falsely

1 T F R
2 T R F
3 Rt T F
4 Rf T F
5 Rt F T
6 Rf F T
7 F T R
8 F R T

Question one to god A
If you answered the two following questions "Does god B speak the truth more than god C?" and "Does 'Ja' mean 'Yes'?" in the state of truthfulness you are now answering, would your two answers to those questions match?

If you figure out what answer god A would answer in all combinations you find, the answers as follows

If it answered "Ja" it means only combinations 2, 3, 6, or 8 are possible and god C cannot be the Random God.

If it answered "Da" it means only combinations 1, 4, 5, or 7 are possible and god B cannot be the Random God.

Question two to the god you know is not the Random God,

You now know the identity of this God
If it answered "Ja" then you're talking to the True God
If it answered "Da" then you're talking to the False God

Question three to the same god as question two
If I asked you "Is god A the Random God?" then "Does 'Ja' mean 'yes'?", would your two answers match?

If you were speaking the the True God and it answered "Ja" it means god A is the Random God, and the remaining god must be the False God.

If you were speaking the the True God and it answered "Da" it means god A is the False God, and the remaining god must be the Random God.

If you were speaking the the False God and it answered "Ja" it means god A is the True God, and the remaining god must be the Random God.

If you were speaking the the False God and it answered "Da" it means god A is the Random God, and the remaining god must be the True God.

You now know the identity of all the Gods.

SOLUTION 2
Second solution, simpler solution I thought of while typing the first solution: First identify the False God

Question one
If you answered the following two questions "Are you both the Random God and presently lying?" and "Does 'Ja' mean 'Yes'?" in the state of truthfulness you are presently answering in, would your answers match?

This question will be answered "Da" by the Truthful God and will also be answered "Da" by both a lying Random God and a truthful Random God. Only the False God will answer it "Ja" So ask the question to God A, if it answers "Ja" you know it's the False God, and go straight to question three, you can now identify all of the gods in just two questions

Question two if God A answered "Da" it isn't the False God, ask the exact same question to god B. If it answers "Ja", you know it's the False God. If it answers "Da" you know it can't be the False God, which means God C must be the False God.

Question three
Ask the False God (God X being any god A B or C other than the one you're talking to)
If I asked you "Is god X the Random God?" then "Does 'Ja' mean 'Yes'?" would your given answers match?

If the False God answered 'Ja' then God X is the True God, and the remaining one must be Random God
If the False God answered 'Da' then God X is the Random God and the remaining one must be the True God.

You now know the identity of all three gods in two or three questions

Funnily enough, although you can find out the meaning of "Ja" and "Da" with the right questions, if you use all your questions to find out the identity of the three gods as you should, you never learn the meaning of their two words, nor do you need to, to complete the riddle.

The Wikipedia article is at [https://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever]
It describes more issues, like how random is Random lie I discussed before, how much to False and True know about Random. What would a god do if asked a question about it's own answer, that could not be answered truthfully, and mentions other solutions. Many Questions follow the pattern of asking what if I asked the following question(s) question.

carlwev
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### Re: The world's hardest logic puzzle!

RexAO
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