If ur high IQ answer this logic question

[Gaygymmaxx]

@CantStopTheMog I just got faded after months off

Sheeit

Normal weed at moderate doses makes me introspective and look at my self like a foreign person, I immediately notice me and my brother are both across the room with headphones on and I'm like "holy shit this is not the way things were before, what a bizarre type of zoomer behavior"

But I'm just faded of d8 this is no good at all

 

[Gaygymmaxx]

Nah, seriously, the little fucker is ripping the answers from articles and pdfs. He is a FRAUD.

I have no mouth and I must scream in Mumbai us humans rule the surface world

 

[CantStopTheMog]

Can you solve this problem, saar?

Jack plays a game on a matrix with 100 rows and 99 columns (100x99).
There is exactly one trap on each row, but not on rows 1 or 100.
And there can be only max one trap per column.

Then Jack starts from the first row, and he can move left, right, up, and down one step at a time.
If Jack hits a trap, he will remember its position and the trap stays there (traps never move, and they stay active at all times). Jack will then start at row 1 again.

Problem is, what is the minimum amount of tries Jack has to do to guarantee an ascend from row 1 to row 100?
What is the strategy and the amount?

Surely just 98 unless I read wrong, because there would be 1 trap per row and column which only leaves 98 traps total

 

[nobodylovesme]

Surely just 98 unless I read wrong, because there would be 1 trap per row and column which only leaves 98 traps total

Nope, and yes there are 98 traps in total

 

[mrdouchebag]

I have no mouth and I must scream in Mumbai us humans rule the surface world

I can feel the high IQ aura emanating from you, bruv. Don't feel bad that a cheating AI fraud outcompeted you.

I thought Grok was smart. Then I checked his thinking process and he just scours the internet for solutions and then rips it right off from articles and pdfs. Lmao.

 

[PoopyFartGamer]

idc imma rape all 3

 

[CantStopTheMog]

@CantStopTheMog I just got faded after months off

Sheeit

Normal weed at moderate doses makes me introspective and look at my self like a foreign person, I immediately notice me and my brother are both across the room with headphones on and I'm like "holy shit this is not the way things were before, what a bizarre type of zoomer behavior"

But I'm just faded of d8 this is no good at all

Whenever I’m high all I want to do is go to the gym and I feel like I have 2x vision magnifiers like goggles, and when people walk past me I like zoom into their features and analyse them in a split second (not in a looksmaxxing way)

 

[CantStopTheMog]

Nope, and yes there are 98 traps in total

I’m not sure I’d assume he’d fail 98 times and get it on his 99th once he knows where all traps are I feel like I’m correct. I should’ve said 98 fails not just an ambiguous 98 lol

 

[natelma0]

There are three gods: A, B, and C.

• One always tells the truth (True).
• One always lies (False).
• One answers randomly (Random) — on each question he flips a fair coin to decide whether to answer truthfully or lie.

Each god answers only “da” or “ja” (you do not know which word means “yes” and which means “no”).


You may ask exactly three yes/no questions total.


Each question must be asked to a single god (you may choose which god for each question, and you may ask the same god more than once).


Your task: determine which god is True, which is False, and which is Random.





Constraints to note (important):

• You don’t know whether “da” = yes or “ja” = yes.
• Random’s answer is independent each time (a fair coin decides whether he answers truthfully or falsely on that question).
• Questions must be answerable by yes/no (i.e., framed so a truthful/lying god can respond with yes/no).

Tbh it’s not actually that hard but it requires advanced problem solving skills for sure. For refernce I got the answer in under 2-3 mins, so defo no impossible

@thecel bless us

 

[Tigermoggerlol]

There are three gods: A, B, and C.

• One always tells the truth (True).
• One always lies (False).
• One answers randomly (Random) — on each question he flips a fair coin to decide whether to answer truthfully or lie.

Each god answers only “da” or “ja” (you do not know which word means “yes” and which means “no”).


You may ask exactly three yes/no questions total.


Each question must be asked to a single god (you may choose which god for each question, and you may ask the same god more than once).


Your task: determine which god is True, which is False, and which is Random.





Constraints to note (important):

• You don’t know whether “da” = yes or “ja” = yes.
• Random’s answer is independent each time (a fair coin decides whether he answers truthfully or falsely on that question).
• Questions must be answerable by yes/no (i.e., framed so a truthful/lying god can respond with yes/no).

Tbh it’s not actually that hard but it requires advanced problem solving skills for sure. For refernce I got the answer in under 2-3 mins, so defo no impossible

This made my head hurt, couldn't even solve it didn't even know wtf was going on.

 

[CantStopTheMog]

@thecel bless us

Bro it’s really not that hard

 

[nobodylovesme]

I’m not sure I’d assume he’d fail 98 times and get it on his 99th once he knows where all traps are I feel like I’m correct. I should’ve said 98 fails not just an ambiguous 98 lol

Nah, there is a strategy that reduces the attempts quite a lot.

 

[natelma0]

Bro it’s really not that hard

Dude I’m a sub 100 iqcel(I have 107)

 

[CantStopTheMog]

This made my head hurt, couldn't even solve it didn't even know wtf was going on.

Answers are in thread replies so far literally only one user other than myself has solved it so ur not alone, ig I underestimated how hard this was

 

[imontheloose]

Can you solve this problem, saar?

Jack plays a game on a matrix with 100 rows and 99 columns (100x99).
There is exactly one trap on each row, but not on rows 1 or 100.
And there can be only max one trap per column.

Then Jack starts from the first row, and he can move left, right, up, and down one step at a time.
If Jack hits a trap, he will remember its position and the trap stays there (traps never move, and they stay active at all times). Jack will then start at row 1 again.

Problem is, what is the minimum amount of tries Jack has to do to guarantee an ascend from row 1 to row 100?
What is the strategy and the amount?

Evil.

 

[Tigermoggerlol]

Answers are in thread replies so far literally only one user other than myself has solved it so ur not alone, ig I underestimated how hard this was

Yeah this shit hard asl

 

[CantStopTheMog]

Nah, there is a strategy that reduces the attempts quite a lot.

Maybe just keep a mental note when u hit the 1st trap in a column and then go down to the last row and just walk along it and then u have like 2 deaths

 

[william.]

So, it’s quite a cool trick in reality, we don’t really appreciate it, but my initial post mentions my key intuition I remember being told about.

Before I get into it, you need to really understand this wrapper lest it breaks your skull. That key premise is mentioned of, “If I asked you X, would you say Y?” If the god is not random, then the answer to the wrap is Y iff (if and only, sorry we use that a lot in maths…) P is true! Now that’s regardless which is yes or no and whether the god is a fibber or not. The wrap of “if I asked you…” cancels that out.

Now, D is guaranteed to be non-random after my first question even if A is random because if we take the case of A being random, then by the puzzle’s setup, B and C are non-random. A’s answer is just a coin flip but either branch is non-random therefore D is non-random, so that first Q1 is random-safe if you will.

Let’s just do another case review. Imagine A is not random. By the actual lemma, A’s answer to my wrapped Q is going to be “da” if and only (iff but I don’t want to confuse you) “B is random“ is true, or “ja” if and only “B is random” is false. Therefore if A says “da”, B is now determined to be random, the non-randoms have to be A and C. If A says “ja”, however, B is not random now, so B is either a truth-teller or a fibber. That’s the exact justification.

IDK how well I can explain this via text, but perhaps numbers help you better. Since this is logic at the end of the day, this can easily be extrapolated to numbers and maybe you’ll get it then? This should be enough an explanation, however.

I might be a fucking retard, but I still don’t see the logical justification regarding the case of A being random.

dont bother to reply, Im gonna watch some yt videos on this

Does the problem have a name I can search up?

 

[CantStopTheMog]

Evil.

I think I solved it now, I found the lower bound answer and I think maybe upper bound too

 

[nobodylovesme]

Evil.

You know the og?

 

[CantStopTheMog]

I might be a fucking retard, but I still don’t see the logical justification regarding the case of A being random.

dont bother to reply, Im gonna watch some yt videos on this

Does the problem have a name I can search up?

It does yes i just can’t remember lol

 

[CantStopTheMog]

Maybe just keep a mental note when u hit the 1st trap in a column and then go down to the last row and just walk along it and then u have like 2 deaths

@nobodylovesme

 

[imontheloose]

I might be a fucking retard, but I still don’t see the logical justification regarding the case of A being random.

dont bother to reply, Im gonna watch some yt videos on this

Does the problem have a name I can search up?

Not an established name, but you could look into overall logical problem-solving with questions. Getting objective answers from subjects.

 

[william.]

is it this one?

It does yes i just can’t remember lol

 

[CantStopTheMog]

View attachment 4163250 is it this one?

Yes. “Hardest logic puzzle” has to be a hyperbole lmao

 

[nobodylovesme]

Maybe just keep a mental note when u hit the 1st trap in a column and then go down to the last row and just walk along it and then u have like 2 deaths

That is essentially what you have to do, get behind the trap. But this doesn't really explain it how you do it in X attempts.

 

[CantStopTheMog]

That is essentially what you have to do, get behind the trap. But this doesn't really explain it how you do it in X attempts.

just go down in the first column till u hit a trap and then do the same the second attempt and miss the trap this time (go around it) and then go down in the first column again till u Reach row 100 and then just go right till the end, maximum 2 deaths. As u don’t leave 1 or 2 column. And u only leave the columns on level 100 where there are no traps

 

[nobodylovesme]

just go down in the first column till u hit a trap and then do the same the second attempt and miss the trap this time (go around it) and then go down in the first column again till u Reach row 100 and then just go right till the end, maximum 2 deaths. As u don’t leave 1 or 2 column. And u only leave the columns on level 100 where there are no traps

Can't really do that, sir. You are not guaranteed to go around it with this strategy as there could be a another trap blocking you from trying to go around it in the next row.

But I have to go to sleep rn, so I'll just leave you with the solution.



And yes, this same strategy/solution works for all n x n-1 matrices.

 

[mrdouchebag]

That is essentially what you have to do, get behind the trap. But this doesn't really explain it how you do it in X attempts.

Grok couldn't cheat on this one. The fucker thought for almost 10 minutes.

Here is his answer:

Spoiler

Thought for 9m 35s
14
The strategy is an adaptive approach inspired by the egg drop puzzle, where you optimize the intervals of rows you attempt to cross in a chosen column to minimize the worst-case number of tries. For 98 trap rows, the minimal worst-case number of tries is the smallest d such that the sum from 1 to d is at least 98, which is d=14 (since 14*15/2 = 105 ≥ 98).
In practice, you select a column and attempt to cross a segment of rows equal to the current remaining tries. If you hit a trap, you learn its position and recurse on the remaining segments with one less try. If you don't hit, you know that segment is safe in that column and continue with the next segment. This balances the risk across attempts, guaranteeing success in at most 14 tries regardless of the trap placements.

 

[nobodylovesme]

Grok couldn't cheat on this one. The fucker thought for almost 10 minutes.

Here is his answer:

Spoiler

Thought for 9m 35s
14
The strategy is an adaptive approach inspired by the egg drop puzzle, where you optimize the intervals of rows you attempt to cross in a chosen column to minimize the worst-case number of tries. For 98 trap rows, the minimal worst-case number of tries is the smallest d such that the sum from 1 to d is at least 98, which is d=14 (since 14*15/2 = 105 ≥ 98).
In practice, you select a column and attempt to cross a segment of rows equal to the current remaining tries. If you hit a trap, you learn its position and recurse on the remaining segments with one less try. If you don't hit, you know that segment is safe in that column and continue with the next segment. This balances the risk across attempts, guaranteeing success in at most 14 tries regardless of the trap placements.

Yeah, AIs are not able to solve these kinds of problems... yet, at least the public ones.

 

[CantStopTheMog]

Can't really do that, sir. You are not guaranteed to go around it with this strategy as there could be a another trap blocking you from trying to go around it in the next row.

But I have to go to sleep rn, so I'll just leave you with the solution.



And yes, this same strategy/solution works for all n x n-1 matrices.

Yes but either way ur only going out by 1 column so 2 deaths maximum. 2 deaths is the absolute minimum u can have here

 

[EthiopianMaxxer]

I attempted this question for about 15 minutes but I gave up after being unable to find the answer. I got close a few times, but I was still only able to solve the question while assuming which God was which (obviously not the point of the question)

I looked at @imontheloose answer and it made sense. I'm kinda disappointed that I didn't get the answer and I wish I had spend more time on it. I enjoyed trying to solve it though, tag me if you make any more of these threads in the future

 

[5'7" 3/4s]

This took me way too much time bro, this was not a 5 minute problem whatsoever and idek if i got it correct man

Controlling for the random god cant be done by asking it the same question twice: because 50% coin flip means that it can give the same answer twice and masquerade as a true or false god

You have no clue what da or ja means

Youre probably gonna have to use a shit ton of conditionals if youre not able to iterate through each god a set of questions

Okay

Ask one god this:

If you are a truthful god, then tell me is todays date 9/30/2025, and does your response mean yes, and are you subject to chance?

We atleast know if this god is true, it will give a return as an absolute YES as a control for da/ja as well

Ask the same god this:

Is todays date 8/20/20 and does the liar god always respond blatantly true, and are you not subject to chance?

asking a blatantly false question helps us confirm a couple things:

1. If it returns the same as both, it’s definitely that the god is random. This is impossible for it to be anything else.
2. If it returns the first question as true and the second question as false then its true god(could be random)
3. If it returns the first question as false and the second question as true it's the liar god(could be random)

So now we have two different progressions for the two different responses, we have one question left to ask

Ask another god this depending on what the initial god first gave you an answer for(either da/ja) and if both answers were the same:

Is today 9/30/25 and did the first god respond x(da or ja, plug in here)?
This allows you to see if this god is truthful, because then it would lie if the god responded x and give a different response..
So u now know that this god is truthful, the first god is a random, and the last god u asked no questions is a liar, or vice-versa for the truthful liar combo.

Ask another god this depending on what the initial response was given to you as an answer for(either da/ja) and if answers were different:

If you are a truthful god, then tell me is todays date 9/30/2025, and does your response mean yes, and are you subject to chance, and was the initial response that the first god gave to the first question x(insert da/ja) mean “yes”?

If this response is the same as the first response to the initial question, we know for a fact this god is random, because that’s impossible, we can’t have two true gods that are subject to chance and the confirmation of the first response lets me verify the response was true.

Now if this response is different from the first response to the initial question, that let’s us know that the first god is a Liar, this god is the true god, and the last god we asked no questions to is random. This is because its impossible for us to have two liar gods at the same time(as we are checking to see if he da/ja means yes or no.

 

[5'7" 3/4s]

This took me way too much time bro, this was not a 5 minute problem whatsoever and idek if i got it correct man

Controlling for the random god cant be done by asking it the same question twice: because 50% coin flip means that it can give the same answer twice and masquerade as a true or false god

You have no clue what da or ja means

Youre probably gonna have to use a shit ton of conditionals if youre not able to iterate through each god a set of questions

Okay

Ask one god this:

If you are a truthful god, then tell me is todays date 9/30/2025, and does your response mean yes, and are you subject to chance?

We atleast know if this god is true, it will give a return as an absolute YES as a control for da/ja as well

Ask the same god this:

Is todays date 8/20/20 and does the liar god always respond blatantly true, and are you not subject to chance?

asking a blatantly false question helps us confirm a couple things:

1. If it returns the same as both, it’s definitely that the god is random. This is impossible for it to be anything else.
2. If it returns the first question as true and the second question as false then its true god(could be random)
3. If it returns the first question as false and the second question as true it's the liar god(could be random)

So now we have two different progressions for the two different responses, we have one question left to ask

Ask another god this depending on what the initial god first gave you an answer for(either da/ja) and if both answers were the same:

Is today 9/30/25 and did the first god respond x(da or ja, plug in here)?
This allows you to see if this god is truthful, because then it would lie if the god responded x and give a different response..
So u now know that this god is truthful, the first god is a random, and the last god u asked no questions is a liar, or vice-versa for the truthful liar combo.

Ask another god this depending on what the initial response was given to you as an answer for(either da/ja) and if answers were different:

If you are a truthful god, then tell me is todays date 9/30/2025, and does your response mean yes, and are you subject to chance, and was the initial response that the first god gave to the first question x(insert da/ja) mean “yes”?

If this response is the same as the first response to the initial question, we know for a fact this god is random, because that’s impossible, we can’t have two true gods that are subject to chance and the confirmation of the first response lets me verify the response was true.

Now if this response is different from the first response to the initial question, that let’s us know that the first god is a Liar, this god is the true god, and the last god we asked no questions to is random. This is because its impossible for us to have two liar gods at the same time(as we are checking to see if he da/ja means yes or no.

I got this wrong dam, i just plugged it into chatgpt, it told me where i went wrong bruh the conditional i used for chance was retarded ngl fucking iqpill im a retard holy shit

How to patch your structure​

The fix is to make sure Truth and Liar always diverge on the first two questions, so only Random can collapse into sameness. Right now, your first two compound questions don’t guarantee that.

---

Step 1: Rewrite Q1 and Q2 in your style​

• Q1:“If you are a truthful god, then tell me: is today’s date 9/30/2025, and does your response mean yes?”
  • Truth → says “yes” (because both premises true).
  • Liar → says “no.”
  • Random → coin flip.
• Q2:“If you are a truthful god, then tell me: is today’s date 8/20/20, and does your response mean yes?”
  • Truth → says “no” (because the date part is false).
  • Liar → says “yes.”
  • Random → coin flip.

Now Truth and Liar must disagree between Q1 and Q2. Random is the only one who might by chance match or mismatch inconsistently.

---

Step 2: Interpret answers​

• If answers to Q1/Q2 are opposites → you’re talking to either Truth or Liar.
• If answers are the same → you’re talking to Random.

That fixes the flaw where sameness didn’t uniquely identify Random.

---

Step 3: Final discrimination​

• If Q1/Q2 found Random → use Q3 on a different god:
  “If you are a truthful god, then tell me: is today’s date 9/30/2025, and does your response mean yes?”
  • “Yes” → that god is Truth.
  • “No” → that god is Liar.
• If Q1/Q2 found Truth or Liar → use Q3 on them with the same date question. That tells you which is which, and the third god is Random by elimination.

---

Why this is now airtight​

• Truth and Liar are forced into opposite answers on Q1 vs. Q2.
• Random may imitate one by chance, but then he gives the same answer to both (since his coin flips can align) — which now is unambiguous evidence he’s Random.
• The last question locks in who’s True and who’s False.

 

[5'7" 3/4s]

This took me way too much time bro, this was not a 5 minute problem whatsoever and idek if i got it correct man

Controlling for the random god cant be done by asking it the same question twice: because 50% coin flip means that it can give the same answer twice and masquerade as a true or false god

You have no clue what da or ja means

Youre probably gonna have to use a shit ton of conditionals if youre not able to iterate through each god a set of questions

Okay

Ask one god this:

If you are a truthful god, then tell me is todays date 9/30/2025, and does your response mean yes, and are you subject to chance?

We atleast know if this god is true, it will give a return as an absolute YES as a control for da/ja as well

Ask the same god this:

Is todays date 8/20/20 and does the liar god always respond blatantly true, and are you not subject to chance?

asking a blatantly false question helps us confirm a couple things:

1. If it returns the same as both, it’s definitely that the god is random. This is impossible for it to be anything else.
2. If it returns the first question as true and the second question as false then its true god(could be random)
3. If it returns the first question as false and the second question as true it's the liar god(could be random)

So now we have two different progressions for the two different responses, we have one question left to ask

Ask another god this depending on what the initial god first gave you an answer for(either da/ja) and if both answers were the same:

Is today 9/30/25 and did the first god respond x(da or ja, plug in here)?
This allows you to see if this god is truthful, because then it would lie if the god responded x and give a different response..
So u now know that this god is truthful, the first god is a random, and the last god u asked no questions is a liar, or vice-versa for the truthful liar combo.

Ask another god this depending on what the initial response was given to you as an answer for(either da/ja) and if answers were different:

If you are a truthful god, then tell me is todays date 9/30/2025, and does your response mean yes, and are you subject to chance, and was the initial response that the first god gave to the first question x(insert da/ja) mean “yes”?

If this response is the same as the first response to the initial question, we know for a fact this god is random, because that’s impossible, we can’t have two true gods that are subject to chance and the confirmation of the first response lets me verify the response was true.

Now if this response is different from the first response to the initial question, that let’s us know that the first god is a Liar, this god is the true god, and the last god we asked no questions to is random. This is because its impossible for us to have two liar gods at the same time(as we are checking to see if he da/ja means yes or no.

i think my intuition was correct, but to be fair i had to go through this like 100 different ways in my head before i could even arrive at a half correct answer, how the fuck did you solve this in 3 fucking minutes, what the hell is your iq? @CantStopTheMog

 

[william.]

@CantStopTheMog

 

[kurd]

There are three gods: A, B, and C.

• One always tells the truth (True).
• One always lies (False).
• One answers randomly (Random) — on each question he flips a fair coin to decide whether to answer truthfully or lie.

Each god answers only “da” or “ja” (you do not know which word means “yes” and which means “no”).


You may ask exactly three yes/no questions total.


Each question must be asked to a single god (you may choose which god for each question, and you may ask the same god more than once).


Your task: determine which god is True, which is False, and which is Random.





Constraints to note (important):

• You don’t know whether “da” = yes or “ja” = yes.
• Random’s answer is independent each time (a fair coin decides whether he answers truthfully or falsely on that question).
• Questions must be answerable by yes/no (i.e., framed so a truthful/lying god can respond with yes/no).

Tbh it’s not actually that hard but it requires advanced problem solving skills for sure. For refernce I got the answer in under 2-3 mins, so defo no impossible

This isn’t a question to show one’s intelligence

 

[Its_over_gooner]

There are three gods: A, B, and C.

• One always tells the truth (True).
• One always lies (False).
• One answers randomly (Random) — on each question he flips a fair coin to decide whether to answer truthfully or lie.

Each god answers only “da” or “ja” (you do not know which word means “yes” and which means “no”).


You may ask exactly three yes/no questions total.


Each question must be asked to a single god (you may choose which god for each question, and you may ask the same god more than once).


Your task: determine which god is True, which is False, and which is Random.





Constraints to note (important):

• You don’t know whether “da” = yes or “ja” = yes.
• Random’s answer is independent each time (a fair coin decides whether he answers truthfully or falsely on that question).
• Questions must be answerable by yes/no (i.e., framed so a truthful/lying god can respond with yes/no).

Tbh it’s not actually that hard but it requires advanced problem solving skills for sure. For refernce I got the answer in under 2-3 mins, so defo no impossible

solve it bhai I am very smart here solution -
Nice — you basically quoted Boolos’ “Hardest Logic Puzzle Ever”.
I know the standard optimal solution — here’s how to crack it cleanly in plain English:

---

The Setup​

• 3 gods: T (truth), F (liar), R (random).
• They answer only “da”/“ja”; you don’t know which = yes.
• 3 total yes/no questions.

Goal: Identify which is which.

---

The Trick​

Two big problems:

1. You don’t know “da” vs. “ja” (yes/no).
2. One god answers randomly.

Solution:

• Phrase your question so it self-translates (you know what “da” or “ja” means regardless).
• Minimize the random god’s interference.

We use this meta-question:

“If I asked you ‘X’, would you say da?”

This double-layer forces both liar and truth-teller to answer in a predictable way regardless of language.

---

Step-by-Step Strategy​

Step 1: Find a Non-Random God.
Pick any god (say A) and ask:

“If I asked you ‘Are you Random?’, would you say da?”

• If A answers “da”, you know A is either True or False (not Random).
• If A answers “ja”, same — you still know A is not Random.
  Because Random’s answer is unpredictable; if it doesn’t match, you adjust — you’re essentially filtering.

You may have to ask two gods to isolate one who’s not Random.

---

Step 2: Once you have a Non-Random God
Use him (either T or F) to decode “da” vs. “ja”:


Ask:

“If I asked you ‘Does da mean yes?’, would you say da?”

This tells you which word = yes.

---

Step 3: Use your now-known yes/no meaning + known Non-Random God
Ask that same god:

“If I asked you ‘Are you True?’, would you say da?”

Now you know if he’s T or F.


With that info, you deduce the rest by elimination.

---

Result​

In 3 questions you can:

• Identify a Non-Random god.
• Decode “da”/“ja”.
• Figure out if that god is True or False.
• Deduce all three gods by elimination.

---

This is the canonical optimal solution to the puzzle.
Want me to write out the exact 3 questions word-for-word as you’d ask them? (I can make them copy-paste-ready.)

 

[CantStopTheMog]

I attempted this question for about 15 minutes but I gave up after being unable to find the answer. I got close a few times, but I was still only able to solve the question while assuming which God was which (obviously not the point of the question)

I looked at @imontheloose answer and it made sense. I'm kinda disappointed that I didn't get the answer and I wish I had spend more time on it. I enjoyed trying to solve it though, tag me if you make any more of these threads in the future

Will do I have another I will post tonight if I remember

 

[CantStopTheMog]

solve it bhai I am very smart here solution -
Nice — you basically quoted Boolos’ “Hardest Logic Puzzle Ever”.
I know the standard optimal solution — here’s how to crack it cleanly in plain English:

---

The Setup​

• 3 gods: T (truth), F (liar), R (random).
• They answer only “da”/“ja”; you don’t know which = yes.
• 3 total yes/no questions.

Goal: Identify which is which.

---

The Trick​

Two big problems:

1. You don’t know “da” vs. “ja” (yes/no).
2. One god answers randomly.

Solution:

• Phrase your question so it self-translates (you know what “da” or “ja” means regardless).
• Minimize the random god’s interference.

We use this meta-question:




This double-layer forces both liar and truth-teller to answer in a predictable way regardless of language.

---

Step-by-Step Strategy​

Step 1: Find a Non-Random God.
Pick any god (say A) and ask:

• If A answers “da”, you know A is either True or False (not Random).
• If A answers “ja”, same — you still know A is not Random.
  Because Random’s answer is unpredictable; if it doesn’t match, you adjust — you’re essentially filtering.

You may have to ask two gods to isolate one who’s not Random.

---

Step 2: Once you have a Non-Random God
Use him (either T or F) to decode “da” vs. “ja”:


Ask:




This tells you which word = yes.

---

Step 3: Use your now-known yes/no meaning + known Non-Random God
Ask that same god:




Now you know if he’s T or F.


With that info, you deduce the rest by elimination.

---

Result​

In 3 questions you can:

• Identify a Non-Random god.
• Decode “da”/“ja”.
• Figure out if that god is True or False.
• Deduce all three gods by elimination.

---

This is the canonical optimal solution to the puzzle.
Want me to write out the exact 3 questions word-for-word as you’d ask them? (I can make them copy-paste-ready.)

Correct but sounds AI

 

[CantStopTheMog]

This isn’t a question to show one’s intelligence

It’s the hardest Iq test. It’s the hardest Iq question that can give an intelligence score than can be measured. So yes it is. If u have low Iq and high EQ ur coping

 

[CantStopTheMog]

i think my intuition was correct, but to be fair i had to go through this like 100 different ways in my head before i could even arrive at a half correct answer, how the fuck did you solve this in 3 fucking minutes, what the hell is your iq? @CantStopTheMog

It wasn’t bad thinking tbh.

I didn’t use trial and error I had a more formulaic approach, so I could have solved it quicker if I realised as soon as I started. my Iq on Mensa is 160 aka a maximum score. but that’s just one test so might not be reliable

 

[5'7" 3/4s]

It wasn’t bad thinking tbh.

I didn’t use trial and error I had a more formulaic approach, so I could have solved it quicker if I realised as soon as I started. my Iq on Mensa is 160 aka a maximum score. but that’s just one test so might not be reliable

Yea I literally went through the progressions of asking the random god the same question twice ATLEAST a dozen different times with different conditionals until I realized it was impossible

Holy fuck I’m mirin

 

[5'7" 3/4s]

Do another one of these later they are fun to do while rotting

 

[CantStopTheMog]

Do another one of these later they are fun to do while rotting

I’ll post one later tonight. It’s a common one and is much easier tbh, doesn’t need any maths or reasoning to solve unless ur a Neanderthal
Trust me it’s like 5-6/10 hardness took me 1 fail to realise it

 

[xevuxia]

Idgaf
Nice — this is the classic “hardest logic puzzle” (Boolos) with the extra twist that the words da / ja are unknown. Here’s a short, fully working 3-question solution with a clear case analysis and a brief explanation why it works.





Tool we use (one trick):


For any yes/no proposition P we will ask a god the meta-question





Qmeta: “If I asked you ‘Is P true?’, would you say ‘da’?”





Important fact about Qmeta (why it’s useful):


If the god you ask is not Random (i.e. he is either True or False), then his answer to Qmeta is ’da’ iff P is true, regardless of whether that god is the truth-teller or the liar and regardless of whether da means “yes” or “no.” (The liar inverts twice: once for the inner hypothetical, once for the outer reply — which cancels — and the unknown word mapping also cancels.)


If the god you ask is Random, his answer to Qmeta is unpredictable (random).














The three questions








Label the gods A,B,C. You will ask:





Q1 (to A): Qmeta with P = “B is Random”


(Exactly: “If I asked you ‘Is B Random?’, would you say ‘da’?”)





Q2 (to B): Qmeta with P = “A is Random”


(“If I asked you ‘Is A Random?’, would you say ‘da’?”)





Record the two replies (each is either ‘da’ or ‘ja’).





Interpretation after Q1 & Q2 (deterministic):

• If the answers are (ja, ja) → C is Random.
• If the answers are (da, ja) → B is Random.
• If the answers are (ja, da) → A is Random.
• If the answers are (da, da) → Random is either A or B (but not C), so C is definitely not Random.

(One can check these by enumerating the possible role assignments and using the Qmeta property above.)





Now you have two possibilities:














Final step (Q3) — figure out True vs False








Case 1 — After Q1 & Q2 you already know who Random is (one of the first three bullet cases):


Pick any one of the two gods who are not Random (call him G). Ask:





Q3 (to G): Qmeta with P = “Is [the other non-Random god] True?”


(“If I asked you ‘Is X True?’, would you say ‘da’?” where X is the other non-Random god.)





Because G is not Random, Qmeta returns ‘da’ iff X is actually True. So from this single final answer you distinguish the truth-teller from the liar. Combining with the known Random, you now know all three roles.





Case 2 — After Q1 & Q2 you have (da,da), so Random is A or B; C is not Random:


Use C (who is surely non-Random) for the distinguishing question.





Q3 (to C): Qmeta with P = “Is A True?”


If C answers ‘da’ then A is True and B is False (and Random is the remaining one A or B consistent with Q1/Q2); if C answers ‘ja’ then A is False and B is True. Since you already know Random is one of A/B from (da,da), this completes the full identification.














Why this fits the constraints

• Each question is yes/no and directed to a single god (you may repeat gods).
• You used precisely three questions.
• You never had to know in advance whether da = yes or ja = yes (Qmeta removes that ambiguity).
• Random’s answers were avoided as a decisive source except that the first two questions can include random noise only when asked to Random; the Q1/Q2 pattern gives enough logical structure so the third question always finalizes the identities.

If you want, I can run through a concrete example (showing all branches with sample answers) so you can see a worked-out trace for each possible reply pattern. Which would you prefer — the compact decision table above, or a full worked example for each reply pair?

 

[CantStopTheMog]

Idgaf
Nice — this is the classic “hardest logic puzzle” (Boolos) with the extra twist that the words da / ja are unknown. Here’s a short, fully working 3-question solution with a clear case analysis and a brief explanation why it works.





Tool we use (one trick):


For any yes/no proposition P we will ask a god the meta-question





Qmeta: “If I asked you ‘Is P true?’, would you say ‘da’?”





Important fact about Qmeta (why it’s useful):


If the god you ask is not Random (i.e. he is either True or False), then his answer to Qmeta is ’da’ iff P is true, regardless of whether that god is the truth-teller or the liar and regardless of whether da means “yes” or “no.” (The liar inverts twice: once for the inner hypothetical, once for the outer reply — which cancels — and the unknown word mapping also cancels.)


If the god you ask is Random, his answer to Qmeta is unpredictable (random).














The three questions








Label the gods A,B,C. You will ask:





Q1 (to A): Qmeta with P = “B is Random”


(Exactly: “If I asked you ‘Is B Random?’, would you say ‘da’?”)





Q2 (to B): Qmeta with P = “A is Random”


(“If I asked you ‘Is A Random?’, would you say ‘da’?”)





Record the two replies (each is either ‘da’ or ‘ja’).





Interpretation after Q1 & Q2 (deterministic):

• If the answers are (ja, ja) → C is Random.
• If the answers are (da, ja) → B is Random.
• If the answers are (ja, da) → A is Random.
• If the answers are (da, da) → Random is either A or B (but not C), so C is definitely not Random.

(One can check these by enumerating the possible role assignments and using the Qmeta property above.)





Now you have two possibilities:














Final step (Q3) — figure out True vs False








Case 1 — After Q1 & Q2 you already know who Random is (one of the first three bullet cases):


Pick any one of the two gods who are not Random (call him G). Ask:





Q3 (to G): Qmeta with P = “Is [the other non-Random god] True?”


(“If I asked you ‘Is X True?’, would you say ‘da’?” where X is the other non-Random god.)





Because G is not Random, Qmeta returns ‘da’ iff X is actually True. So from this single final answer you distinguish the truth-teller from the liar. Combining with the known Random, you now know all three roles.





Case 2 — After Q1 & Q2 you have (da,da), so Random is A or B; C is not Random:


Use C (who is surely non-Random) for the distinguishing question.





Q3 (to C): Qmeta with P = “Is A True?”


If C answers ‘da’ then A is True and B is False (and Random is the remaining one A or B consistent with Q1/Q2); if C answers ‘ja’ then A is False and B is True. Since you already know Random is one of A/B from (da,da), this completes the full identification.














Why this fits the constraints

• Each question is yes/no and directed to a single god (you may repeat gods).
• You used precisely three questions.
• You never had to know in advance whether da = yes or ja = yes (Qmeta removes that ambiguity).
• Random’s answers were avoided as a decisive source except that the first two questions can include random noise only when asked to Random; the Q1/Q2 pattern gives enough logical structure so the third question always finalizes the identities.

If you want, I can run through a concrete example (showing all branches with sample answers) so you can see a worked-out trace for each possible reply pattern. Which would you prefer — the compact decision table above, or a full worked example for each reply pair?

Bro said Idgaf and then replied with the longest reply yet
there is only 2 general solutions btw bro so unless u discovered a new one I cba reading allat for a solution I already know
Looks like ChatGPT as well

 

[xevuxia]

Bro said Idgaf and then replied with the longest reply yet
there is only 2 general solutions btw bro so unless u discovered a new one I cba reading allat for a solution I already know
Looks like ChatGPT as well

No answer until a reply MrChad

Btw geniuenly how tall is this nigga ?!
he is really 160 cm
Buddy boyo is a manlet chad

 

[xevuxia]

Looks like ChatGPT as well

I wonder how you know that

 

[optimisticzoomer]

My past self would’ve been able to answer, but now I couldn’t even read
My brain has atrophied since leaving education