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Two doors is the easiest one. You ask the second man if the first man would say that the door on the left would lead to his freedom.
Let's say that the answer is yes
If the second man is the liar, that means that the other man would not say that is the door to safety and it is really the bad door
If the second man is telling the truth, that means the first man would say that is the door that will lead to safety but he is lying and it is really the bad door
Also, don't tell the Goblin King that this is too easy or he will fuck with the clock worse than Daylight Savings Time
Assume for a moment that your dot was red, but the other two were green. In that case everyone would still have a hand up, but B and C would instantly be able to tell the color of their own dot because they would realize that the green dot they were seeing was actually each other's.
Thus, since no one else called out the color of their own dot, the only logical conclusion is that your own dot is green.
garroad_ran on
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Blake TDo you have enemies then?Good. That means you’ve stood up for something, sometime in your life.Registered Userregular
If that pool ball riddle doesnt get figured out by morning, I will bring it to class and fiddle with it.
since tossrock and I separately came to believe that you can't do it in fewer than four weighings, and since toss said he could probably prove it, I'm tempted to think we're really right. there could be something really weird going on which makes it doable in three and that I'm just completely missing, but I swear I did think about that riddle off and on for several days without getting any farther than toss got in this thread.
Two doors is the easiest one. You ask the second man if the first man would say that the door on the left would lead to his freedom.
Let's say that the answer is yes
If the second man is the liar, that means that the other man would not say that is the door to safety and it is really the bad door
If the second man is telling the truth, that means the first man would say that is the door that will lead to safety but he is lying and it is really the bad door
Also, don't tell the Goblin King that this is too easy or he will fuck with the clock worse than Daylight Savings Time
This also works if the person answers no except the doors are reversed
im fairly certain that for river city you start bottom up but im not sure exactly how
For 6 balls you would do
1st measure:
compare 2 and 2
if they're not equal
take one side off and put 2 different balls on
if they're equal now you know that the taken off pair is unbalanced and then compare one of those with one ball from the set of 4 equal. If its equal the other is unbalanced and if its not then that one is unbalanced
similarly if they are equal then do the same thing to the other original pair
if the 2 are equal
put one more on each side
if thats equal than you're done
if it's not then
compare one of those with one of the originals
if its equal than the other is unbalanced and if its not than that one is unbalanced
somehow you can expand this algorithms to 12 i believe
The extension of this to twelve balls would be my 4/4, 4/4, 2/2, 1/1 algorithm which takes four comparisons.
And redhead, caring enough involves like, doing research and stuff, it's not something I could just pull out of my head. But I'm pretty sure this is the kind of thing that could be solved rigorously with that branch of theory, just like those "draw X without lifting your pencil etc etc" type riddles can be solved rigorously with graph theory. Maybe I'll try to check something out of the library.
Tossrock on
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Blake TDo you have enemies then?Good. That means you’ve stood up for something, sometime in your life.Registered Userregular
If that pool ball riddle doesnt get figured out by morning, I will bring it to class and fiddle with it.
since tossrock and I separately came to believe that you can't do it in fewer than four weighings, and since toss said he could probably prove it, I'm tempted to think we're really right. there could be something really weird going on which makes it doable in three and that I'm just completely missing, but I swear I did think about that riddle off and on for several days without getting any farther than toss got in this thread.
It's possible to do it in three.
I don't know how but it is impossible.
I suspect it is because we are looking at this incorrectly.
We are looking at this as the scales are unbalanced.
We need to look at this as the scales go up and down and switch balls around.
I haven't figured out how to do it.
But I am sure that is the process.
Also another clue would be that after the first weighing you would have balls that you KNOW weigh the same.
Two doors is the easiest one. You ask the second man if the first man would say that the door on the left would lead to his freedom.
Let's say that the answer is yes
If the second man is the liar, that means that the other man would not say that is the door to safety and it is really the bad door
If the second man is telling the truth, that means the first man would say that is the door that will lead to safety but he is lying and it is really the bad door
Also, don't tell the Goblin King that this is too easy or he will fuck with the clock worse than Daylight Savings Time
Two doors is the easiest one. You ask the second man if the first man would say that the door on the left would lead to his freedom.
Let's say that the answer is yes
If the second man is the liar, that means that the other man would not say that is the door to safety and it is really the bad door
If the second man is telling the truth, that means the first man would say that is the door that will lead to safety but he is lying and it is really the bad door
Also, don't tell the Goblin King that this is too easy or he will fuck with the clock worse than Daylight Savings Time
I saw this in an episode of Samurai Jack once
Yeah it's used in a lot of stories. It's a simple riddle but just difficult enough for people who have never heard it before to think that whatever they are watching/reading at the time is super smart
also to get a riddle marked "solved" you'd better give more than just an answer
you have to explain WHY your answer is the right one
jesus fine
the first native can only say green. It doesn't matter what color his soles actually are, either way he has to say green. So the second native is telling the truth when he says the first guy said green. This means he himself has green soles. Therefore the third native must have purple soles since he lied and said the second native has purple soles.
well in any case I think I've done as much thinking about that ball riddle as I really want to
all the others were cool and I'll pull them out when I need time-wasters for a group(good but not-that-difficult riddles are perfect for this), except for the money dividing one which was the one I didn't get
actually the first really good riddle I ever heard was another hat-related riddle
I'm gonna check if it's on this site and if it is it should totally be posted because it's hella fun (and should take most people longer than most of these without being as difficult as the ball one)
You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You can select one envelope and keep whatever amount it contains, but upon selection, are offered the possibility to take the other envelope instead.
I already explained that Viv. God, be a butt about it
I was mostly referring to Dru's answer to the Defeeted question
also I am getting the sense that SOMEONE in this thread is just looking up answers instead of trying to solve them
that makes me angry but not all together surprised
who, garroad? no, his answers seem legit to me. none of the ones he's posted take very long at all to solve and his answers all look like the ones I came up with in my head (that is, they don't look like he just copy/pasted them from somewhere)
You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You can select one envelope and keep whatever amount it contains, but upon selection, are offered the possibility to take the other envelope instead.
WHAT do you do?
this one still blows my mind
I mean, I know the answer and I know why it's the answer, but it's so weird that it works that way
I think to solve the River City riddle you first need to have a control group. If you split the balls into three groups of four and you compare two groups you have two possible outcomes. 1: They balance, which means you have 8 balanced balls and four unknowns (the balanced would be the control group) or 2: They do not balance, which means you have a group of 4 which could be high, 4 which could be low and 4 that are the balanced or control group.
I'm not sure where to go from there, but I do know that from that point there are two possible outcomes and so you must account for both of them.
I guess that algorithms class was good for something after all, eh?
DasHanselHM on
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I Win Swordfightsall the traits of greatnessstarlight at my feetRegistered Userregular
edited November 2008
I love logic riddles and hate number riddles
I Win Swordfights on
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Olivawgood name, isn't it?the foot of mt fujiRegistered Userregular
If the person at the back of the line were to see two purple hats, he would instantly know his own hat was red. This is not the case.
That being the case, if your own hat was purple, then the person in the middle would know that his own hat HAD to be red.
Thus, your hat has to be red.
I don't get your second line of reasoning there.
I think what he's saying is
if the second guy saw that your hat was purple, and he didn't hear the guy behind him call out his hat as red, then he would figure that his hat was red because the guy in the back would never be able to figure out his own hat color if there was both a red and a purple hat
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
Hat Trick:
If the two hats in front were purple, the back person would have said their hat was red, since the didn't, one of the two front people's hat has to be red.
Now that the person in the middle knows this, if they haven't called out the color of their hat, that means they can't rightly know what color hat they have on, because it is possible for both(or all three) to be red. This is only possible if the person in front's hat is red.
For if it were purple, they person in the middle could deduce that theirs was red.
edit: Oh this was solved already thanks for updating the OP, GOD.
You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You can select one envelope and keep whatever amount it contains, but upon selection, are offered the possibility to take the other envelope instead.
Posts
In fact, it appears to me upon closer inspection that the guy who wrote it somehow combined the cake riddle with the pirate game.
Here's the Pirate Game, you decide for yourself.
New riddle: What the fuck is wrong with Swordfights' brother?
Let's say that the answer is yes
If the second man is the liar, that means that the other man would not say that is the door to safety and it is really the bad door
If the second man is telling the truth, that means the first man would say that is the door that will lead to safety but he is lying and it is really the bad door
Also, don't tell the Goblin King that this is too easy or he will fuck with the clock worse than Daylight Savings Time
Assume for a moment that your dot was red, but the other two were green. In that case everyone would still have a hand up, but B and C would instantly be able to tell the color of their own dot because they would realize that the green dot they were seeing was actually each other's.
Thus, since no one else called out the color of their own dot, the only logical conclusion is that your own dot is green.
Satans..... hints.....
Your scenario's are wrong
1st has purple, says green
second has green, says green
third has purple, says purple
Holy shit, is that it?
since tossrock and I separately came to believe that you can't do it in fewer than four weighings, and since toss said he could probably prove it, I'm tempted to think we're really right. there could be something really weird going on which makes it doable in three and that I'm just completely missing, but I swear I did think about that riddle off and on for several days without getting any farther than toss got in this thread.
This also works if the person answers no except the doors are reversed
That being the case, if your own hat was purple, then the person in the middle would know that his own hat HAD to be red.
Thus, your hat has to be red.
The extension of this to twelve balls would be my 4/4, 4/4, 2/2, 1/1 algorithm which takes four comparisons.
And redhead, caring enough involves like, doing research and stuff, it's not something I could just pull out of my head. But I'm pretty sure this is the kind of thing that could be solved rigorously with that branch of theory, just like those "draw X without lifting your pencil etc etc" type riddles can be solved rigorously with graph theory. Maybe I'll try to check something out of the library.
It's possible to do it in three.
I don't know how but it is impossible.
I suspect it is because we are looking at this incorrectly.
We are looking at this as the scales are unbalanced.
We need to look at this as the scales go up and down and switch balls around.
I haven't figured out how to do it.
But I am sure that is the process.
Also another clue would be that after the first weighing you would have balls that you KNOW weigh the same.
Satans..... hints.....
They'll both give the same answer. Take the other door.
you have to explain WHY your answer is the right one
I saw this in an episode of Samurai Jack once
PSN ID : DetectiveOlivaw | TWITTER | STEAM ID | NEVER FORGET
Yeah it's used in a lot of stories. It's a simple riddle but just difficult enough for people who have never heard it before to think that whatever they are watching/reading at the time is super smart
I was mostly referring to Dru's answer to the Defeeted question
also I am getting the sense that SOMEONE in this thread is just looking up answers instead of trying to solve them
that makes me angry but not all together surprised
all the others were cool and I'll pull them out when I need time-wasters for a group(good but not-that-difficult riddles are perfect for this), except for the money dividing one which was the one I didn't get
actually the first really good riddle I ever heard was another hat-related riddle
I'm gonna check if it's on this site and if it is it should totally be posted because it's hella fun (and should take most people longer than most of these without being as difficult as the ball one)
No idea what to do from there.
EDIT: Oh snap, I think I got it.
WHAT do you do?
who, garroad? no, his answers seem legit to me. none of the ones he's posted take very long at all to solve and his answers all look like the ones I came up with in my head (that is, they don't look like he just copy/pasted them from somewhere)
viv I already posted a solution to this three pages ago
this one still blows my mind
I mean, I know the answer and I know why it's the answer, but it's so weird that it works that way
Don't judge me. I was in middle school. I feel terrible about it.
edit: I think I need some sleep.
I'm not sure where to go from there, but I do know that from that point there are two possible outcomes and so you must account for both of them.
I guess that algorithms class was good for something after all, eh?
I think what he's saying is
PSN ID : DetectiveOlivaw | TWITTER | STEAM ID | NEVER FORGET
edit: ah, never mind you already did
If the two hats in front were purple, the back person would have said their hat was red, since the didn't, one of the two front people's hat has to be red.
Now that the person in the middle knows this, if they haven't called out the color of their hat, that means they can't rightly know what color hat they have on, because it is possible for both(or all three) to be red. This is only possible if the person in front's hat is red.
For if it were purple, they person in the middle could deduce that theirs was red.
edit: Oh this was solved already thanks for updating the OP, GOD.
Ugh, this is gonna blow my mind for a while.