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Experiment Logic puzzle. Suicide fuel for low IQ cels. DO NOT ENTER IF DIAGNOSED AS RETARDED.

B

based_meme

I.N.C.E.L. High Command, Psychological Operations
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Joined
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There are five houses lined up next to each other along a street. Each house is a different color, and each homeowner is of a different nationality, drinks a different beverage, smokes a different brand of cigar, and owns a different pet.

If these variables can never repeat from house to house, which homeowner has a pet fish? You should be able to figure out the answer based on these 15 clues:
  1. The Englishman lives in the house with red walls.
  2. The Swede keeps dogs.
  3. The Dane drinks tea.
  4. The house with green walls is just to the left of the house with white walls.
  5. The owner of the house with green walls drinks coffee.
  6. The man who smokes Pall Mall keeps birds.
  7. The owner of the house with yellow walls smokes Dunhills.
  8. The man in the center house drinks milk.
  9. The Norwegian lives in the first house.
  10. The Blend smoker has a neighbor who keeps cats.
  11. The man who smokes Blue Masters drinks beer.
  12. The man who keeps horses lives next to the Dunhill smoker.
  13. The German smokes Prince.
  14. The Norwegian lives next to the house with blue walls.
  15. The Blend smoker has a neighbor who drinks water.
If you've seen this problem before and have solved it, well done (please don't post the answer). If you haven't and you're feeling up for the challenge, take a shot and see how you do.

Tagging users who might be interested: @trying to ascend @Ahnfeltia.
 
@Saysitsover
 
kay' said:
@Saysitsover
Click to expand...
laughing-laugh.gif
 
They make me solve shit like this in job applications. Unbelievable.
 
Heliosphan said:
They make me solve shit like this in job applications. Unbelievable.
Click to expand...
They make you solve easier versions of these puzzles, or job-specific ones.
 
Is it German?
 
Answer - Norwegian
.

edit - jfl realised i made a mistake with the blend smoker
 
Last edited:
Assuming this doesn't have multiple solutions the answer is

German
 
based_meme said:
  1. The Englishman lives in the house with red walls.
  2. The Swede keeps dogs.
  3. The Dane drinks tea.
  4. The house with green walls is just to the left of the house with white walls.
  5. The owner of the house with green walls drinks coffee.
  6. The man who smokes Pall Mall keeps birds.
  7. The owner of the house with yellow walls smokes Dunhills.
  8. The man in the center house drinks milk.
  9. The Norwegian lives in the first house.
  10. The Blend smoker has a neighbor who keeps cats.
  11. The man who smokes Blue Masters drinks beer.
  12. The man who keeps horses lives next to the Dunhill smoker.
  13. The German smokes Prince.
  14. The Norwegian lives next to the house with blue walls.
  15. The Blend smoker has a neighbor who drinks water
Click to expand...
1. Norwegian, yellow walls, smokes dunhills, drinks water, keeps cats

2. blue walls, keeps horses, blend smoker, dane, drinks tea

3. drinks milk, englishman, red walls, pall malls, keeps birds

4. green walls, drinks coffee, german, smokes prince, HAS A FISH

5. white walls, swede, keeps dogs, smokes blue masters, drinks beer
 
Took me all morning.
 
based_meme said:
There are five houses lined up next to each other along a street. Each house is a different color, and each homeowner is of a different nationality, drinks a different beverage, smokes a different brand of cigar, and owns a different pet.

If these variables can never repeat from house to house, which homeowner has a pet fish? You should be able to figure out the answer based on these 15 clues:
  1. The Englishman lives in the house with red walls.
  2. The Swede keeps dogs.
  3. The Dane drinks tea.
  4. The house with green walls is just to the left of the house with white walls.
  5. The owner of the house with green walls drinks coffee.
  6. The man who smokes Pall Mall keeps birds.
  7. The owner of the house with yellow walls smokes Dunhills.
  8. The man in the center house drinks milk.
  9. The Norwegian lives in the first house.
  10. The Blend smoker has a neighbor who keeps cats.
  11. The man who smokes Blue Masters drinks beer.
  12. The man who keeps horses lives next to the Dunhill smoker.
  13. The German smokes Prince.
  14. The Norwegian lives next to the house with blue walls.
  15. The Blend smoker has a neighbor who drinks water.
If you've seen this problem before and have solved it, well done (please don't post the answer). If you haven't and you're feeling up for the challenge, take a shot and see how you do.

Tagging users who might be interested: @trying to ascend @Ahnfeltia.
Click to expand...
too lazy to solve but you just need a piece of paper and then it is a matter of putting each person in their house with their smokes and pets. All one step at a time. No logic here
 
I found this fun ngl
Also
IMG 9892
 
this took me way longer than it should have
I used the clues in the following order: 8 > 9 > 14 > 4 & 5 > 1 > 7 > 12 > (3 > 11 > 13 > 2) > 6 > 10. Clue 15 is unnecessary (hence why the table below is incomplete).

An explanation for "3 > 11 > 13 > 2": at this point (the first 11 entries known) we know that the Dane must live either in the second of fifth house. If the Dane / tea drinker (clue 3) lives in the second house, then the Blue Masters smoker / beer drinker (clue 11) must live in the fifth house, but then the German / Prince smoker (clue 13) must live in the fourth house. This leads to a contradiction, however, as the Swede / dog owner (clue 2) has no place to live now. Ergo, the Dane / tea drinker (clue 3) must live in the fifth house, meaning that the Blue Masters smoker / beer drinker (clue 11) must live in the second house, at which point the German / Prince smoker (clue 13) must live in the fourth house and the Swede / dog owner (clue 2) must reside in the fifth house.

Below the almost complete table. The range in parentheses indicates the how manyth addition to the table it is. I started with a blank table and added entries using the aforementioned order of the clues. Ranges mean simultaneous addition.

house numberscolorcountrydrinkcigarpet
1yellow (9 - 10)Norway (2)Dunhills (9 - 10)nekonyan (22 - 23)
2blue (3)Dane (12 - 19)tea (12 - 19)Blend (22 - 23)honse (11)
3red (7 - 8)Anglo (7 - 8)milk (1)Pall Mall (20 - 21)birb (20 - 21)
4green (4 - 6)German (12 - 19)coffee (4 - 6)Prince (12 - 19)
5white (4 - 6)Swede (12 - 19)beer (12 - 19)Blue Masters (12 - 19)doggy (12 - 19)

To answer the riddle, the German is the one keeping the fishy.
 
Last edited:
The German has the fish

First make a 5x5 table with the headers: nationality, wall, beverage, cigar-pack, and pet;

1. Norwegian - Yellow, Water, Dunhills, Cats
2. Dane- Blue, Tea, Blend, Horses
3. Englishman - Red, Milk, Paul Mall, Birds
4. German - Green, Coffee, Prince, Fish
5, Swede - White, Beer, Blue Master, Dogs

We know the Norwegian occupies the first house and the middle house that is red is occupied by the Englishman who drinks milk because it's in the middle. Green house is to the left of the white house. The Dane drinks tea so it can't be the green house because the owner of that house drinks coffee and we already know that the Norwegian and Englishman occupy the first and third houses . I inferred that the green and white houses were at the end #4 and #5 because I already know red is in the middle and the and Norwegian lives nexto blue which is neighbored by red which means the only color left is yellow for the Norwegian. The Norwegian owner smokes Dunhills and drinks water because we already know that the Englishman drinks milk and the person residing in house #4 drinks coffee. which means house 2 and 4 are not water The Dane blend smoker has a neighbor who has cats so it's either the Norwegian or Englishman. Houses 4 and 5 are either Swede or German but we know Swede keeps dogs. Man who smokes bluemaster drinks beer so it can't be the Norwegian, Englishman, nor German because they each drink water, milk, and coffee so I filled it in for the Swede. Englishman must be the one who smokes Paul Walls so I filled in birds for him because I already know that the Norwegian, Dane, German and Swede smoke Dunhills, Blend, Prince, and Blue Master and the latter drinks beer. The blend smoker is neighbors with someone who has a cat and it's the Norwegian because I know the Dane has a horse and the Englishman has a bird

So using process of elimination I chose German because the Swede has a dog, the Dane has a horse and is a neighbor of the Dunhills smoker who is Norwegian, and the Blend Smoker is neighbors with someone who has a cat which is the Norwegian which leaves the German as the only one who's pet left is fish.
 
Last edited:
Ahnfeltia said:
this took me way longer than it should have
Click to expand...
Same here. It's almost embarrassing. I didn't use a table, I used the variables (house #, wall color, nationality, smoke, drink, and pet) as short hands and plugged in the givens using enough negation to deduce the unknowns. The notepad file I used is a chaotic fucking mess. JFL

I honestly thought you'd be using a system of equations and throw it all into a matrix (5x6?). :feelskek::feelshaha:

ethniccel1 said:
No logic here
Click to expand...
Ethnic moment.
 
Thanks to everyone who participated.
 
Orzmund said:
This is Einstein's genius test; at least, that is what it's usually advertised as.
Click to expand...
No, that's all bullshit. It's an apocryphal legend designed to get people interested in the problem, nothing more.
 
Didn’t waste my time solving this bullshit because I have Tesla level IQ. Jerked off instead :bigbrain:
 
Did you make this? Giga high IQ if you did.
 
Grim_Reaper said:
Did you make this? Giga high IQ if you did.
Click to expand...
No, but it's easy to solve. Just takes time and effort.

Now you've piqued by curiosity about the origin of this problem.
 
This is more of a memory-recall (assuming you don't have paper) problem than a logic one. Too bad I have dementia level memory iq.
 
Tarquinius said:
This is more of a memory-recall (assuming you don't have paper) problem than a logic one. Too bad I have dementia level memory iq.
Click to expand...
Same. I tried solving this on my phone during my bus ride, and it took me over a hour to solve it even though I had the notes app opened to keep track of the characteristics that was found for each homeowner.
 
based_meme said:
It's not. Use a pencil and paper.
Click to expand...
Okay, I'm just used to these problems being done in your mind only.
Grim_Reaper said:
Same. I tried solving this on my phone during my bus ride, and it took me over a hour to solve it even though I had the notes app opened to keep track of the characteristics that was found for each homeowner.
Click to expand...
Yeah, just the action of switching tabs can be enough to forget something. Too much reliance on computer memory probably causes it.
 
too hard quick postmaxx
 
1: Yellow / Norwegian / Water / Dunhills / Cats
2: Blue / Dane / Tea / Blend / Horses
3: Red / English / Milk / Pall Mall / Birds
4: Green / German / Coffee / Prince / Fish
5: White / Swede / Beer / Blue Masters / Dogs

I opened a spreadsheet and it took me over 20 minutes to find out
 
based_meme said:
Same here. It's almost embarrassing. I didn't use a table, I used the variables (house #, wall color, nationality, smoke, drink, and pet) as short hands and plugged in the givens using enough negation to deduce the unknowns. The notepad file I used is a chaotic fucking mess. JFL
Click to expand...
:ha..feels: you crazy son of a gun
based_meme said:
I honestly thought you'd be using a system of equations and throw it all into a matrix (5x6?). :feelskek::feelshaha:
Click to expand...
Would've been 5 by 5. Regardless, I thought about it for a bit before settling on the tabular approach, but I don't think this can reasonably be converted into a numerical problem.
 
Ahnfeltia said:
:ha..feels: you crazy son of a gun
Click to expand...
:feelsdevil::feelsthink::feelsokman::bigbrain::feelscomfy:

Ahnfeltia said:
Would've been 5 by 5. Regardless, I thought about it for a bit before settling on the tabular approach, but I don't think this can reasonably be converted into a numerical problem.
Click to expand...
Yeah. This problem is probably harder than I give it credit for.
 
Ahnfeltia said:
It ain't that hard.
Click to expand...
You're right. It just looks intimidating.

Apparently, the following is the hardest logic problem ever conceived. Apparently.

Three gods A, B, and C are called, in no particular 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.
 
based_meme said:
Three gods A, B, and C are called, in no particular 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.
Click to expand...
This problem is certainly harder than the previous one. Took me way longer too, but I got it in the end, albeit with a little help of the computer.
My first useful insight was that you have to ask each god one question. You cannot afford to ask Random two questions, because his answers give you no information. My second useful insight was that asking a god whether "da" means "yes" allows you to infer which of True and False they are not -- if a god answers "da" they cannot be False and if they answer "ja" they cannot be True.

Asking this question (whether "da" means "yes") to every god is a good start. You'll either get two "da"s and one "ja" or two "ja"s and one "da" -- in the former case the god who said "ja" must be False and in the latter case the god who said "da" must be True. Unfortunately we don't have enough information to distinguish the remaining two. So we'll have to modify our question slightly.

The idea is to additionally ask each god something pertaining to the identity of one of the gods. That way you'll be able to identify either True or False as before, and use their reliable answer to distinguish the remaining two. At this point, I whipped out Wolfram Mathematica to test this idea. Going thru twelve cases (3! = 6 assignments of True, False, Random to A, B, C times 2 because "da" can either mean "yes" or "no") by hand was too much effort for yours truly.

The question I ended up going with is -- in pseudocode --
Code: 
(da == yes) == (A == Random)
which works, as the following Mathematica code verifies:
Code: 
cases = Tuples @ {Permutations @ {1, -1, 0}, {1, -1}}

(* here "{1, -1, 0}" symbolizes whether a god is {"True", "False", "Random"} resp.,
and {1, -1} symbolizes whether "da" means {"yes", "no"} resp.;
the ouput is
{{{1, -1, 0}, 1}, {{1, -1, 0}, -1}, {{1, 0, -1}, 
  1}, {{1, 0, -1}, -1}, {{-1, 1, 0}, 
  1}, {{-1, 1, 0}, -1}, {{-1, 0, 1}, 
  1}, {{-1, 0, 1}, -1}, {{0, 1, -1}, 
  1}, {{0, 1, -1}, -1}, {{0, -1, 1}, 1}, {{0, -1, 1}, -1}}
e.g., {{1, -1, 0}, 1} means that A = True, B = False, C = Random, and da = yes *)

q[case_] := case[[2]] If[case[[1, 1]] == 0, 1, -1] (* the question *)

aA[case_] := case[[1, 1]] case[[2]] q @ case (* A's answer *)
aB[case_] := case[[1, 2]] case[[2]] q @ case (* B's answer *)
aC[case_] := case[[1, 3]] case[[2]] q @ case (* C's answer *)

(* if you plug q into, say, aA, you'll see that you get case[[2]]^2 = 1;
using this observation, it's pretty easy to prove by hand that this works *)

Through[{aA, aB, aC} @ #] & /@ cases /. {1 -> "da", -1 -> "ja"} (* the responses for all cases *)

(* the output is
{{"ja", "da", 0}, {"ja", "da", 0}, {"ja", 0, "da"}, {"ja", 0, 
  "da"}, {"da", "ja", 0}, {"da", "ja", 0}, {"da", 0, "ja"}, {"da", 0, 
  "ja"}, {0, "da", "ja"}, {0, "da", "ja"}, {0, "ja", "da"}, {0, "ja", 
  "da"}}
where 0 is a stand-in for Random's response. Ignoring the duplicates for a moment,
every choice for 0 leads to a unique trio of responses. As regards the duplicates, they occur side by side.
What this means is that regardless of whether "da" means "yes" or "no", the trio of responses is the same.
This is exactly what we want, because there are 3! = 6 assignments of True, False, Random to A, B, C and
6 useful trios consisting of "da"s and "ja"s (2^3 = 8 minus 2 because getting the same response thrice is useless)
so we need a bijective correspondence *)
I can't decide whether to end this post with "a working day well spent" or "goddamn I need to get a life" so I'll just do both.
 
based_meme said:
Three gods A, B, and C are called, in no particular 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.
Click to expand...
Thanks for acquainting me with the problem btw. It's a nice problem.
 
4th house, the German keeps fish as a pet. I was going to share a ss of my paint schematic as proof of solution with all variables for each house written but at the very end I pressed ctrl-z to re-draw the star I was trying to put on the 4th house and it turned into this.
Screenshot from 2023 11 16 13 56 19

fedcels who surveil me can confirm I solved it on my own :feelsjuice:
 
Last edited:
Don’t know, didn’t read, can’t read.:feelstastyman:
 
based_meme said:
Three gods A, B, and C are called, in no particular 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.
Click to expand...
I will take a look at this one after I cooom
 
Ahnfeltia said:
This problem is certainly harder than the previous one. Took me way longer too, but I got it in the end, albeit with a little help of the computer.
My first useful insight was that you have to ask each god one question. You cannot afford to ask Random two questions, because his answers give you no information. My second useful insight was that asking a god whether "da" means "yes" allows you to infer which of True and False they are not -- if a god answers "da" they cannot be False and if they answer "ja" they cannot be True.

Asking this question (whether "da" means "yes") to every god is a good start. You'll either get two "da"s and one "ja" or two "ja"s and one "da" -- in the former case the god who said "ja" must be False and in the latter case the god who said "da" must be True. Unfortunately we don't have enough information to distinguish the remaining two. So we'll have to modify our question slightly.

The idea is to additionally ask each god something pertaining to the identity of one of the gods. That way you'll be able to identify either True or False as before, and use their reliable answer to distinguish the remaining two. At this point, I whipped out Wolfram Mathematica to test this idea. Going thru twelve cases (3! = 6 assignments of True, False, Random to A, B, C times 2 because "da" can either mean "yes" or "no") by hand was too much effort for yours truly.

The question I ended up going with is -- in pseudocode --
Code: 
(da == yes) == (A == Random)
which works, as the following Mathematica code verifies:
Code: 
cases = Tuples @ {Permutations @ {1, -1, 0}, {1, -1}}

(* here "{1, -1, 0}" symbolizes whether a god is {"True", "False", "Random"} resp.,
and {1, -1} symbolizes whether "da" means {"yes", "no"} resp.;
the ouput is
{{{1, -1, 0}, 1}, {{1, -1, 0}, -1}, {{1, 0, -1},
  1}, {{1, 0, -1}, -1}, {{-1, 1, 0},
  1}, {{-1, 1, 0}, -1}, {{-1, 0, 1},
  1}, {{-1, 0, 1}, -1}, {{0, 1, -1},
  1}, {{0, 1, -1}, -1}, {{0, -1, 1}, 1}, {{0, -1, 1}, -1}}
e.g., {{1, -1, 0}, 1} means that A = True, B = False, C = Random, and da = yes *)

q[case_] := case[[2]] If[case[[1, 1]] == 0, 1, -1] (* the question *)

aA[case_] := case[[1, 1]] case[[2]] q @ case (* A's answer *)
aB[case_] := case[[1, 2]] case[[2]] q @ case (* B's answer *)
aC[case_] := case[[1, 3]] case[[2]] q @ case (* C's answer *)

(* if you plug q into, say, aA, you'll see that you get case[[2]]^2 = 1;
using this observation, it's pretty easy to prove by hand that this works *)

Through[{aA, aB, aC} @ #] & /@ cases /. {1 -> "da", -1 -> "ja"} (* the responses for all cases *)

(* the output is
{{"ja", "da", 0}, {"ja", "da", 0}, {"ja", 0, "da"}, {"ja", 0,
  "da"}, {"da", "ja", 0}, {"da", "ja", 0}, {"da", 0, "ja"}, {"da", 0,
  "ja"}, {0, "da", "ja"}, {0, "da", "ja"}, {0, "ja", "da"}, {0, "ja",
  "da"}}
where 0 is a stand-in for Random's response. Ignoring the duplicates for a moment,
every choice for 0 leads to a unique trio of responses. As regards the duplicates, they occur side by side.
What this means is that regardless of whether "da" means "yes" or "no", the trio of responses is the same.
This is exactly what we want, because there are 3! = 6 assignments of True, False, Random to A, B, C and
6 useful trios consisting of "da"s and "ja"s (2^3 = 8 minus 2 because getting the same response thrice is useless)
so we need a bijective correspondence *)
I can't decide whether to end this post with "a working day well spent" or "goddamn I need to get a life" so I'll just do both.
Click to expand...
Hint: Biconditionals. You must ask compound questions.

Ahnfeltia said:
Thanks for acquainting me with the problem btw. It's a nice problem.
Click to expand...
:feelsYall:
 
damn bruh , i'm retarded . I was drawing houses and shit and gave up midway in frustration .Darn these shitskin genetics.
Untitled
 
didn't read coz i cant read
 
I refuse to believe that the retarded little kid who calls himself "based meme" could solve this
 
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