Q1: if I asked you if two plus two equals four, would you say yes? Both the truth and the lie would say yes, because a double negative is a positive, and so is a double positive. Q2: is the god to the left liar? Q3(to the the god on the left): is the god in the right rand?
The original reason I made this video was because I was looking up the answer and after reading the wikipedia page and the main video about it on RUclips (the one by TedEd) I was still really confused. I'm glad you liked the explanation!
Its quite easy, first you gotta ask them Ozo or Ulu, if they say Ozo, then tell them you have green eyes, otherwise, if they say Ulu, then make an alien prube for the demon dance party, afterwards go South 3 times and finally, use the lvl 4 churrosard to defeat Moldevort. (All of these phrases came from the TedEd riddles)
If it were F it’d do as T does and give away the truth due to its exploited double negatives, and acts as “ace” (“1/2”) saying da if the question had da (yes) or ja if the question ja (w/out knowing if that means yes/no), making 1/2 of the certitude to choose C. 2/2 is if it were Random called the “joker” (“2/2”), and considering that the only other option advocates C and this one does too, you can’t go wrong with it because if B is literally the joker/Random, it can’t be C. -- On the other hand if we get the opposite in response (ja-da/da-ja of “Say x if I ask u if A Random?”) it means A is the 100% T∨F because 1/2 it is telling the truth and is T or F. If random, it would be affirming the option for taking A by being the A-is-not-Random by being me-B-is-the-Random, all while not needing to know B’s identity, but that it’s response can give us enough useful info. to 100% know a T∨F to ask our 2nd self-reference one by logical bi-conditionality-connected embed-questionings cornering negations All else is a cake walk using similar tapestry understanding this as 1/3. 2/3 is to find as T or F and best phrased to ask if it x if it T; if it were posed of to F it’d be True if the emissions are different than of the word posed of it to say. Now at 3/3 so easy we just go ask it of any of the others for any of two others We take C if B answers in the same/affirmative/word. If False it is 1/2 and answers as Truth even to unknown value. C in this is a real False or a True.. 2/2 Random for to say so is to itself and A 100% off the menu by virtue of case of the possibilities. (The False=>”True” embed setup to logical connectives, is the reduction for ace and jokers two truther and a random and all that is important is finding how that is made so - and thru it may we ask to see Random so to get the one not, not T/F, we can to get the god to ask our 2/3) And in the video they say to take C, but its only in response to the affirmative, so da from da and ja from ja in spite of random. If it were ‘ja’ from da or ‘da’ from ja, an equal negatory, we actually take a different course of action, but which is the same course of action which is the trust explained to take in ace and jokers: that we trust this T=F/ace and know allowably Random or not as it may, to be telling truth for that it must be A is not Random (antagonistic to assertion that it truly is random by the opposite word), and we must take it, A, for the only other option is it B is Random, and the only 100% True or False becomes the one that both T/F and Random as being literally not A, says. So everything pretty much collaborates to give you the certainty of a T/F by the virtue nature/proper algebraic/predicate that False must be Truth when it is twice using a sequestering key to cause this knowledge spell, that turns a game to 2 aces 1 jack, procuring yes from reciprocations of same to no no’s and no from no. I hope I had reduced blight obscure and dyslexias somewhere. So ultimately due unto False’ predictive nature we are now dealing with a thing useful of a usefulness to uselessnesses called the only uncooperative is R that we shall beget to later to find as of R or not.
My issue on this is depending who you ask first, you surely reach a different solution applying the same logic. E.g. ask the god in the middle: if i were to ask you if the god to the right..." then you "know" postion 1 cannot be random. Start with the God on furthest right, and ask "if i were to ask you if the god in position 1 is random..." Youd conclude that the god in the middle cannot be random. So we have god 1,2, 3 cannot be randomdepending on your initial starting position/question... Maybe i have that worng
I still don't get it. How can the afformentioned concept concerning truth and lie giving the same answer, even apply when you never ask the exact same question more than once?
It has to deal with the double negative outcome of how you phrase your question. The truth teller will always tell the truth and the liar will always lie, meaning if you ask the liar if he would tell the truth he would give the same answer as the truth teller because he's lying about telling the truth.
"If I ask you whether you call tell truth from lie under any circumstances, would you answer Da?" It's 2 different questions (Proper Nouns Vs. Concepts), in either case the anawer can be very subjective instead of factual. So it's 4 different perceptions and interpretations. As the Gods are aware of the complexity, they either have to go da-va forever or fix that thing to get further clarification. (I beleive the ones designed the puzzle would hate me for that)
when i first heard this riddle, i came up with the question to the god on the left: would the god in the middle say yes, if i ask him whether da means yes? obviously this is a yes no question. however, if the god in the middle is a random god, the god on left wouldn't be able to say anything, since the answer is not certain yet. or the gods would know the future and still can know the answers of the random god before he actually answers them?
I figured out another 2-question solution using another special question, and this solution is better than my last. This solution exploits the same 2 facts: 1. Nobody knows what Rand would say because he answers randomly, and 2. Both Truth or Liar cannot answer da or va if he doesn’t know the answer. Let’s label the gods from left to right as A, B, and C. The strategy is to find Rand first. Ask A, “If I asked him (pointing at B) if he were Truth, would he say da?” If A cannot answer da or va, then B must be Rand; A cannot answer if he does not know how B would answer. If you’re lucky in finding Rand right away, then ask C, “If I asked him (pointing at A) if he were Truth, would he say da?” If C answers da, then A is Liar and C is Truth, but if C answers va, then A is Truth and C is Liar. If you were not so lucky to find Rand yet, then he’s either A or C. Ask B, “If I asked him (pointing at C) if he were Truth, would he say da?” If B cannot answer da or va, then C must be Rand and A’s prior answer is all you need. If A answered da, then A is Truth and B is Liar, but if A answered va, then A is Liar and B is Truth. On the other hand, if B answers da or va, then A must be Rand. If B says da, then B is Truth and C is Liar. If B says va, then B is Liar and C is Truth. What do you think?
The solution I came up also and is also easier to solve it by exploiting the random guy cause you do not have to embed your questions at all. If you find the random first then boom the riddle is possible to be solve if you hit the random even with too questions. You only need to embed the last question you make after identifying the random in order to bipass the fact that you do not know the words DA and VA
The solution goes as follow Ask the first one if the person I ask next happens to be the random will he answer me a da or va. If we get an answer congrats we just find out the random guy otherwise we have a one of the lier or the honest in which case we can repeat the question again to the next on the next God and again if he answer is the random otherwise he is the liar or the truth from which point ask one of the guys if that the did not answer if the other guys that did not answer is truth would he say va the depending on the answer you identify the truth and liar. Same logic as yours but differents little bit on the question you identify the rand in your case since your solution if you get an answer mean is one of the truth or lier on mine if I don't get an answer means is one of the truth or liar since the only one could give an answer about how the random would answer is the random guy. In coclution a solution that exploits the random works better.
if the answer to the first question is "va", then the statement is not true. you proceed to ask the 2nd question to the one on the left (who is either "truth" or "lie") and follow through...
If you ask either of the truth / liar gods what the random god would answer to '2+2=4' question then they would have to stay silent. This can give a solution that includes which word is false and true. TFR TRF FRT FTR RTF RFT 1 ask 1 what 2 would answer '2+2=4' f s s f f/t f/t If silent go to 2.2 2.1 ask 2 what 3 would answer '2+2=4' s s f f if silent go to 2.3 3.1 ask 2 if 3 would say he tells the truth t f You know Q2.1 must be false 2.2 ask 1 what 3 would answer '2+2=4' f f Must be false 3.2 ask 1 if 3 would say he tells the truth t f 3.3 ask 1 if 2 would say he tells the truth t f You know Q1 must be false
@@EverythingScience correct. But the answers you will get are the exact same if you didn't ask the EQ. If you just straight up asked them if you ask them both are you truth they both would answer the same. Your results from the EQ are the same regardless and if lier answerd DA to the 1st statement. (2+2=4) "would you say DA" that mean he is telling the truth so he has to say VA as If he said DA he would ab telling the trust about his lie which he cannot do.
@@jeremiahbrown6288 bud you actually contradicted yourself. You make a solid point by saying "if I asked a true embedded question and asked if they would respond with Da both truth and lie would respond with Da" which I agree with 100%. Then why are you saying both would respond with different answers for the 2+2=4 it's the exact same secario you just explained bud
@@jeremiahbrown6288 end of the day its simple this whole eb crap is meaningless and pointless. It doesn't server any purpose as the overaching question takes precedence. Just ask 2 questions are you truth both will say yes. Then ask are my shoes black? which they are. Truth will say yes and lier will say no. Problem solved random can be found by asking the same question again
@@jeremiahbrown6288 You actually solve this without using an embedded question. Infact using an embedded question will give you the wrong result. Sorry you can't wrap your head around that.
Could you solve it?
me: DA
Hahahaha
Q1: if I asked you if two plus two equals four, would you say yes?
Both the truth and the lie would say yes, because a double negative is a positive, and so is a double positive.
Q2: is the god to the left liar?
Q3(to the the god on the left): is the god in the right rand?
@@alexdeville1110be honest did you hear this riddle from Ted Ed
This is the best explanation of the solution to this riddle I have ever seen. You know a riddle is tricky when the solution is hard to explain.
The original reason I made this video was because I was looking up the answer and after reading the wikipedia page and the main video about it on RUclips (the one by TedEd) I was still really confused. I'm glad you liked the explanation!
Its quite easy, first you gotta ask them Ozo or Ulu, if they say Ozo, then tell them you have green eyes, otherwise, if they say Ulu, then make an alien prube for the demon dance party, afterwards go South 3 times and finally, use the lvl 4 churrosard to defeat Moldevort.
(All of these phrases came from the TedEd riddles)
6:46 how exactly did you determine that the God on the right can't be random based off the first question?
If it were F it’d do as T does and give away the truth due to its exploited double negatives, and acts as “ace” (“1/2”) saying da if the question had da (yes) or ja if the question ja (w/out knowing if that means yes/no), making 1/2 of the certitude to choose C.
2/2 is if it were Random called the “joker” (“2/2”), and considering that the only other option advocates C and this one does too, you can’t go wrong with it because if B is literally the joker/Random, it can’t be C.
--
On the other hand if we get the opposite in response (ja-da/da-ja of “Say x if I ask u if A Random?”) it means A is the 100% T∨F because 1/2 it is telling the truth and is T or F. If random, it would be affirming the option for taking A by being the A-is-not-Random by being me-B-is-the-Random, all while not needing to know B’s identity, but that it’s response can give us enough useful info. to 100% know a T∨F to ask our 2nd self-reference one by logical bi-conditionality-connected embed-questionings cornering negations
All else is a cake walk using similar tapestry understanding this as 1/3. 2/3 is to find as T or F and best phrased to ask if it x if it T; if it were posed of to F it’d be True if the emissions are different than of the word posed of it to say. Now at 3/3 so easy we just go ask it of any of the others for any of two others
We take C if B answers in the same/affirmative/word. If False it is 1/2 and answers as Truth even to unknown value. C in this is a real False or a True.. 2/2 Random for to say so is to itself and A 100% off the menu by virtue of case of the possibilities. (The False=>”True” embed setup to logical connectives, is the reduction for ace and jokers two truther and a random and all that is important is finding how that is made so - and thru it may we ask to see Random so to get the one not, not T/F, we can to get the god to ask our 2/3)
And in the video they say to take C, but its only in response to the affirmative, so da from da and ja from ja in spite of random. If it were ‘ja’ from da or ‘da’ from ja, an equal negatory, we actually take a different course of action, but which is the same course of action which is the trust explained to take in ace and jokers: that we trust this T=F/ace and know allowably Random or not as it may, to be telling truth for that it must be A is not Random (antagonistic to assertion that it truly is random by the opposite word), and we must take it, A, for the only other option is it B is Random, and the only 100% True or False becomes the one that both T/F and Random as being literally not A, says. So everything pretty much collaborates to give you the certainty of a T/F by the virtue nature/proper algebraic/predicate that False must be Truth when it is twice using a sequestering key to cause this knowledge spell, that turns a game to 2 aces 1 jack, procuring yes from reciprocations of same to no no’s and no from no. I hope I had reduced blight obscure and dyslexias somewhere.
So ultimately due unto False’ predictive nature we are now dealing with a thing useful of a usefulness to uselessnesses called the only uncooperative is R that we shall beget to later to find as of R or not.
@@myaknirufesco-ryckexgajithov Yep , im not studying philosophy
My issue on this is depending who you ask first, you surely reach a different solution applying the same logic. E.g. ask the god in the middle: if i were to ask you if the god to the right..." then you "know" postion 1 cannot be random.
Start with the God on furthest right, and ask "if i were to ask you if the god in position 1 is random..."
Youd conclude that the god in the middle cannot be random.
So we have god 1,2, 3 cannot be randomdepending on your initial starting position/question...
Maybe i have that worng
I still don't get it. How can the afformentioned concept concerning truth and lie giving the same answer, even apply when you never ask the exact same question more than once?
It has to deal with the double negative outcome of how you phrase your question. The truth teller will always tell the truth and the liar will always lie, meaning if you ask the liar if he would tell the truth he would give the same answer as the truth teller because he's lying about telling the truth.
Basically its impossible
how does someone even come up with a riddle this confusing
...these single people
It's basically the hardcore version of the classic fork on the road riddle.
"If I ask you whether you call tell truth from lie under any circumstances, would you answer Da?"
It's 2 different questions (Proper Nouns Vs. Concepts), in either case the anawer can be very subjective instead of factual. So it's 4 different perceptions and interpretations. As the Gods are aware of the complexity, they either have to go da-va forever or fix that thing to get further clarification.
(I beleive the ones designed the puzzle would hate me for that)
when i first heard this riddle, i came up with the question to the god on the left: would the god in the middle say yes, if i ask him whether da means yes?
obviously this is a yes no question. however, if the god in the middle is a random god, the god on left wouldn't be able to say anything, since the answer is not certain yet. or the gods would know the future and still can know the answers of the random god before he actually answers them?
You u are blowing my already fried brain
Thank you TEDed for teaching me this before hand
I watch this is in 2 am. Worst possible choice I've made since my brain is gone now.
Who also knows this from TED-ED?
I’d focus on their non-verbal communication.
It took me watching this video about 10 times to wrap my head around it.
I feel like I gained some IQ points there lol
I figured out another 2-question solution using another special question, and this solution is better than my last. This solution exploits the same 2 facts: 1. Nobody knows what Rand would say because he answers randomly, and 2. Both Truth or Liar cannot answer da or va if he doesn’t know the answer.
Let’s label the gods from left to right as A, B, and C.
The strategy is to find Rand first. Ask A, “If I asked him (pointing at B) if he were Truth, would he say da?” If A cannot answer da or va, then B must be Rand; A cannot answer if he does not know how B would answer. If you’re lucky in finding Rand right away, then ask C, “If I asked him (pointing at A) if he were Truth, would he say da?” If C answers da, then A is Liar and C is Truth, but if C answers va, then A is Truth and C is Liar.
If you were not so lucky to find Rand yet, then he’s either A or C. Ask B, “If I asked him (pointing at C) if he were Truth, would he say da?” If B cannot answer da or va, then C must be Rand and A’s prior answer is all you need. If A answered da, then A is Truth and B is Liar, but if A answered va, then A is Liar and B is Truth.
On the other hand, if B answers da or va, then A must be Rand. If B says da, then B is Truth and C is Liar. If B says va, then B is Liar and C is Truth.
What do you think?
The solution I came up also and is also easier to solve it by exploiting the random guy cause you do not have to embed your questions at all. If you find the random first then boom the riddle is possible to be solve if you hit the random even with too questions. You only need to embed the last question you make after identifying the random in order to bipass the fact that you do not know the words DA and VA
@@thomasdogkas2618, I don't think I've seen your solution. What is your solution?
The solution goes as follow
Ask the first one if the person I ask next happens to be the random will he answer me a da or va. If we get an answer congrats we just find out the random guy otherwise we have a one of the lier or the honest in which case we can repeat the question again to the next on the next God and again if he answer is the random otherwise he is the liar or the truth from which point ask one of the guys if that the did not answer if the other guys that did not answer is truth would he say va the depending on the answer you identify the truth and liar. Same logic as yours but differents little bit on the question you identify the rand in your case since your solution if you get an answer mean is one of the truth or lier on mine if I don't get an answer means is one of the truth or liar since the only one could give an answer about how the random would answer is the random guy. In coclution a solution that exploits the random works better.
But what if Truth and Liar have supergod powers and they can know what random will answer?
Then probably the riddle is unsolvable 😅@@OzunaGameplays
a real intellectual: ozo ulu
What if random only says da?
I solved it 2h with knowledge of the answer
This was recreated by Ted -Ed btw
If da means yes then yes means va?
Well what if the person on the right is indeed the random? What would you do in that case?
if the answer to the first question is "va", then the statement is not true. you proceed to ask the 2nd question to the one on the left (who is either "truth" or "lie") and follow through...
I can.
Big brain moment
If you ask either of the truth / liar gods what the random god would answer to '2+2=4' question then they would have to stay silent. This can give a solution that includes which word is false and true.
TFR TRF FRT FTR RTF RFT
1 ask 1 what 2 would answer '2+2=4' f s s f f/t f/t If silent go to 2.2
2.1 ask 2 what 3 would answer '2+2=4' s s f f if silent go to 2.3
3.1 ask 2 if 3 would say he tells the truth t f You know Q2.1 must be false
2.2 ask 1 what 3 would answer '2+2=4' f f Must be false
3.2 ask 1 if 3 would say he tells the truth t f
3.3 ask 1 if 2 would say he tells the truth t f You know Q1 must be false
DA
Va
VA
DA
There is no scenario where truth and lier would ever say the same answer sorry
Just think about their answers as yes and no. If you asked 'are you truth' liar would say yes (because he lies) and truth would also say yes
@@EverythingScience correct. But the answers you will get are the exact same if you didn't ask the EQ. If you just straight up asked them if you ask them both are you truth they both would answer the same. Your results from the EQ are the same regardless and if lier answerd DA to the 1st statement. (2+2=4) "would you say DA" that mean he is telling the truth so he has to say VA as If he said DA he would ab telling the trust about his lie which he cannot do.
@@jeremiahbrown6288 bud you actually contradicted yourself. You make a solid point by saying "if I asked a true embedded question and asked if they would respond with Da both truth and lie would respond with Da" which I agree with 100%. Then why are you saying both would respond with different answers for the 2+2=4 it's the exact same secario you just explained bud
@@jeremiahbrown6288 end of the day its simple this whole eb crap is meaningless and pointless. It doesn't server any purpose as the overaching question takes precedence. Just ask 2 questions are you truth both will say yes. Then ask are my shoes black? which they are. Truth will say yes and lier will say no. Problem solved random can be found by asking the same question again
@@jeremiahbrown6288 You actually solve this without using an embedded question. Infact using an embedded question will give you the wrong result. Sorry you can't wrap your head around that.