Alright Veeky Forums, give me your best riddle, let's see what you got

Alright Veeky Forums, give me your best riddle, let's see what you got

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xkcd.com/blue_eyes.html
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DESU einstein's riddle isn't really hard, it's just crossing data. It is fun to solve, though.
I came upon this cute riddle a few weeks ago:
You are the most eligible prince in the kingdom and as such the king has invited you to his castle so that you may choose one of his three daughters to marry. The oldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is tricky. Sometimes she tells the truth and sometimes she lies. You don't want to get screwed over at this, of course.

Here's the problem: The king wants a smart man to inherit his throne and he doesn't want you to favor any of his daughters because of looks. So he complicated it. You don't know which princess is which. You are allowed to ask ONE of the princesses (any) ONE yes-or-no question ONE time. Then you have to make your pick. What would that question be?

Easy.

"If I asked the other 2 daughters if you were the youngest, would they always say no?

How will you be able to tell the middle one apart if she says yes?

** if she says no
sorry

Is the goal to pick the truthful princess or the most pretty one or what?

>The middle princess is tricky. Sometimes she tells the truth and sometimes she lies. You don't want to get screwed over at this, of course.
This is the goal. Pick anyone but the middle.

"If I asked the other 2 daughters if you were the middle daughter, would they most likely answer truthfully?"

Youngest daughter answer No, oldest daughter answers No, middle daughter has a chance of saying yes, therefore giving you the best odds at figuring out who she is.

You can ask a question that will guarantee you a win.

Does that involve adding external information to the question ? Like (is the sky blue right now) AND (something about sisters) so that you can tell if she's lying ?

No. But I'll give you a hint - even though you can ask only one question and only one sister, the other two sisters are still in the room.

Balls of the same diameter packed tight in space.
Planes tangent in the dots of their contact splits the space into the same polyhedrons.
What are those polyhedrons.
(if you avoid computer or even plasticine modeling, you may develop your spatial imagination to an incredible level. surprisingly, it took me a couple of years in my 17 or so.)

I would say it's a dodecahedron, since every sphere is surrounded by 12 others. But then the regular dodecahedron doesn't really follow the structure I had in mind.

If a hypotheses i make at the start of a question which could not be true if asked to the middle daughter would she answer randomly or would she Error.exe and not answer?

did some searching and it's the rhombic dodecahedron

> did some researching
that's called cheating (why would you spoil it for them all?)

Sorry, couldn't resist.
But did you realize there are two different ways to pack spheres? I'm still not sure what the other one would result in.

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.

did some more researching, it would be a trapezo-rhombic dodecahedron

You can't make any statements, only questions. Nothing like "do you enjoy being the little sister", which can't be answered truthfully by somebody other than the little sister.

If you're not the little sister, the (truthful) answer to "do you enjoy being the little sister" is just "no".

Can you post the solution (rot13'd) or maybe give another hint? I can't wrap my mind around how you can possibly pick out the middle sister if you don't know whether she's lying or not, and she answers your question.

Yeah, it took me a while too.
I'll give you a really big hint - you're definitely not gonna pick the sister you're questioning, but her answer will help you decide which of the two others to pick.

Had saved one from /b/ almost a decade ago

Ok i got it now.

If you question the tricky girl, you can pick any of the two remaining, so her answer doesn't matter.
If not, you have to find out who is the tricky one.

That's right!

you're good at the research. you should probably teach me how you googled for that.
yes, that's correct, it took me ten more years to notice it.
Now you may want to count angles of those rhomba (it's irrational afair)
But I still have a task for me to prove that regular polyhaedron doesn't fill all the space, thus not a Fedorov's solid. I only know it because I made some cardboard models of four of them.

easy mode:
"what door will the other bird suggest?" and go to the other one.
god mode:
"if I asked you what door is heaven, what would you say?" and go to that door
god slayer mode:
"which door will the other bird go through?" and go to the other one

or if you can't reference to the other bird, then "which door do you deserve to go through?" and go to it

The brother of my sister posts on Veeky Forums.
The sister of the sister of my father's sister in law owes me 200 dollars
The aunt of my sister is also the father of my father

>rah rah xkcd gay blah blah

yeah whatever i still really like this puzzle that was posted on sci awhile ago. Didn't figure it out myself but it's definitely one of those *well why couldnt i think of that* riddles that is basically impossible to solve by luck


xkcd.com/blue_eyes.html

That doesn't make any sense, you don't know who the Middle one is so you can't deduct that from the answer

You could ask, while pointing to one of the other two sisters, "if I ask the sister farthest in age from you if she is the middle sister, would she say yes?" and then if the answer is yes, pick that one, and if the answer is no, pick the other one.

Seconding that this isn't hard. I've seen riddles of the same form of the Einstein riddle that require greater leaps in logic. The youtube channel TED-Ed has a lot of videos on riddles/logic puzzles, all titled "Can you solve the X riddle?" They're pretty good.

I read the solution and some discussion on reddit but I still don't believe it.

Meant to reply to OP but whatever.

maybe it wasn't described well.

If you want you can search the sci archives' im sure "blue eyes" will bring up the thread.

[possible spoiler]
I like it because its literally the human brains problem with scaling that makes it so difficult. When you imagine the same situation with less people then it becomes easier to figure out but is still tough even then.,