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** STICKY ** League of Reason? PROVE YOURSELF!

arg-fallbackName="iamthedinger"/>
Re: League of Reason? PROVE YOURSELF!

ladiesman391 said:
Try this one:

In front of you are several long fuses. You know they burn for exactly one hour after you light them at one end. The entire fuse does not necessarily burn at a constant speed. For example, it might take five minutes to burn through half the fuse and fifty-five minutes to burn the other half.

With your lighter and using these fuses, how can you measure exactly three-quarters of an hour of time?


How is the other question coming along, 8.8.3.3?



i use a casio g-shock
 
arg-fallbackName="Arcus"/>
Fictionarious said:
There are three logicians who are able to reason and deduce with perfect accuracy, their names are A, B, and C. They each know this as well.
In a bag there are seven colored stickers. Two yellow, two red, and three green. The logicians know that too.
Each logician reaches into the bag, grabs one sticker and slaps it on his forehead without looking to see what color it is. Any one logician is able to see the stickers on his compatriot's heads, but not the sticker on his own.

Jerry the interrogator asks logician A if he knows what color sticker is definitely not on his forehead.
"No.", A responds.
Jerry asks logician B the same question, and receives the same answer.

Is it possible with this information to deduce the color of any of the logician's stickers?

Is the answer C must be green?
If A looks around and sees two yellows, there is 0% probability he will be yellow, 40% probability he'll be red and 60% probability he'll be green. Therefore B and C must have different colors from each other. If both are yellow, it's the same 0/40/60 combination for R/Y/G, respectively. If B looks around and sees the same, he gets the same outcomes. A and C must therefore have different colors. Logician C will know the colors of A and B, and that his color is the one which neither A nor B has. If A is yellow, C cannot be yellow as B would answer differently. If A is red, C cannot be red because B would again answer differently.

The only possibility left is that C must be green.
 
arg-fallbackName="Private_slim"/>
Try this one, took me 3 hours to do. (I might have been easily distracted by music/yt in this period, not exactly working on it, but still, fucking thing annoyed me SO MUCH)

Who owns the Zebra?

- There are 5 houses
- All the house owners are from different states
- They all have different pets
- One of them owns a Zebra
- All drink different beverages
- They each smoke a different brand
- The man from York lives in the red house
- The Texan have a dog
- The man from Arizona drinks tea
- The green house is to the left of the white one
- The man that smokes Pall Mall has birds
- Dunhill is smoked in the Yellow house
- Milk is drunk in the middle house
- The Minnesotan lives in the first house
- The man smoking Blue Master drinks beer
- The man smoking Blend lives next to the house with the cat
- Next to the house with a horse Dunhill is smoked
- The man from Washington smokes Prince
- The Minnesotan lives next to the blue house
- Water is drunk in the house next to the house where Blend is smoked
 
arg-fallbackName="Bearcules"/>
Private_slim said:
Try this one, took me 3 hours to do. (I might have been easily distracted by music/yt in this period, not exactly working on it, but still, fucking thing annoyed me SO MUCH)

Who owns the Zebra?

- There are 5 houses
- All the house owners are from different states
- They all have different pets
- One of them owns a Zebra
- All drink different beverages
- They each smoke a different brand
- The man from York lives in the red house
- The Texan have a dog
- The man from Arizona drinks tea
- The green house is to the left of the white one
- The man that smokes Pall Mall has birds
- Dunhill is smoked in the Yellow house
- Milk is drunk in the middle house
- The Minnesotan lives in the first house
- The man smoking Blue Master drinks beer
- The man smoking Blend lives next to the house with the cat
- Next to the house with a horse Dunhill is smoked
- The man from Washington smokes Prince
- The Minnesotan lives next to the blue house
- Water is drunk in the house next to the house where Blend is smoked

The zebra is owned by the man from Washington who smokes Prince, lives in the last house (which is white), and drinks an undisclosed beverage. Correct?
 
arg-fallbackName="tuxbox"/>
Private_slim said:
Try this one, took me 3 hours to do. (I might have been easily distracted by music/yt in this period, not exactly working on it, but still, fucking thing annoyed me SO MUCH)

Who owns the Zebra?

- There are 5 houses
- All the house owners are from different states
- They all have different pets
- One of them owns a Zebra
- All drink different beverages
- They each smoke a different brand
- The man from York lives in the red house
- The Texan have a dog
- The man from Arizona drinks tea
- The green house is to the left of the white one
- The man that smokes Pall Mall has birds
- Dunhill is smoked in the Yellow house
- Milk is drunk in the middle house
- The Minnesotan lives in the first house
- The man smoking Blue Master drinks beer
- The man smoking Blend lives next to the house with the cat
- Next to the house with a horse Dunhill is smoked
- The man from Washington smokes Prince
- The Minnesotan lives next to the blue house
- Water is drunk in the house next to the house where Blend is smoked

The first house=Minnesotan
The second house=Blue
The third house=Milk
The fourth house=Brain exploding
 
arg-fallbackName="Private_slim"/>
Bearcules said:
Private_slim said:
yappity yap yap yap

The zebra is owned by the man from Washington who smokes Prince, lives in the last house (which is white), and drinks an undisclosed beverage. Correct?

Nope. Or, All I asked for was who owns the Zebra, and you got that right, although you mixed the houses, he lives in the green house and drinks coffee.
So I guess you win.
 
arg-fallbackName="Bearcules"/>
Private_slim said:
Nope. Or, All I asked for was who owns the Zebra, and you got that right, although you mixed the houses, he lives in the green house and drinks coffee.
So I guess you win.

Um... did you alter the riddle at all? Coffee is not mentioned.

Also, whether it is the Green or White house depends on what you consider to be the "first house". If the first house is on the left, he is in the White house. If it is on the right, then he is in the Green house.
 
arg-fallbackName="FiverBeyond"/>
Can I join in the fun? Here's a Smullyan classic...


Suppose I propose to play a game with you. On the table in front of you, I place one penny, one quarter, and one shiny silver dollar.
The rules of the game are as follows:

1. You may ask me precisely one yes/no question.
2. I will respond truthfully (and only by answering either 'yes' or 'no').
3. If I respond with 'no', I will give you the penny.
4. If I respond with 'yes', I will give you the quarter.

Your goal is to get the shiny silver dollar. What question do you ask?


For extra super-genius bonus points (or if the basic puzzle is too easy), do the same puzzle, only remove the requirement to tell the truth (that is, I will respond with 'yes' or 'no', but I may choose to lie or tell the truth).
 
arg-fallbackName="Bearcules"/>
FiverBeyond said:
Can I join in the fun? Here's a Smullyan classic...


Suppose I propose to play a game with you. On the table in front of you, I place one penny, one quarter, and one shiny silver dollar.
The rules of the game are as follows:

1. You may ask me precisely one yes/no question.
2. I will respond truthfully (and only by answering either 'yes' or 'no').
3. If I respond with 'no', I will give you the penny.
4. If I respond with 'yes', I will give you the quarter.

Your goal is to get the shiny silver dollar. What question do you ask?


For extra super-genius bonus points (or if the basic puzzle is too easy), do the same puzzle, only remove the requirement to tell the truth (that is, I will respond with 'yes' or 'no', but I may choose to lie or tell the truth).

Will you give me the penny?
 
arg-fallbackName="Anachronous Rex"/>
FiverBeyond said:
Can I join in the fun? Here's a Smullyan classic...


Suppose I propose to play a game with you. On the table in front of you, I place one penny, one quarter, and one shiny silver dollar.
The rules of the game are as follows:

1. You may ask me precisely one yes/no question.
2. I will respond truthfully (and only by answering either 'yes' or 'no').
3. If I respond with 'no', I will give you the penny.
4. If I respond with 'yes', I will give you the quarter.

Your goal is to get the shiny silver dollar. What question do you ask?


For extra super-genius bonus points (or if the basic puzzle is too easy), do the same puzzle, only remove the requirement to tell the truth (that is, I will respond with 'yes' or 'no', but I may choose to lie or tell the truth).
"Your goal is to get the shiny silver dollar. What question do you ask?"

How badly do you want the antidote?
 
arg-fallbackName="FiverBeyond"/>
Bearcules said:
Will you give me the penny?


You're on the right track, but that will just make my brain explode (and still won't get you the silver dollar).

Anachronous Rex said:
How badly do you want the antidote?

I'm a particularly eccentric logician, quite happy to die rather than give up my silver dollar (unless forced by the rules of the game).
But well played, sir, well played.
 
arg-fallbackName="Bearcules"/>
FiverBeyond said:
Bearcules said:
Will you give me the penny?


You're on the right track, but that will just make my brain explode (and still won't get you the silver dollar).

Well, if your brain exploded I would just take the silver dollar ;)

However...

Will give me the penny OR the silver dollar and quarter?
 
arg-fallbackName="FiverBeyond"/>
Bearcules said:
Well, if your brain exploded I would just take the silver dollar ;)

However...

Will give me the penny OR the silver dollar and quarter?

Yes, well done. If I actually had a silver dollar, I'd give it to you.
 
arg-fallbackName="Visaki"/>
1. You may ask me precisely one yes/no question.
Will give me the penny OR the silver dollar and quarter?
I might be totally off the canal here but I can't see how that qualifies as a "yes/no question" since the "or" negates the possibility for a "yes/no" answer.
 
arg-fallbackName="Bearcules"/>
Visaki said:
I might be totally off the canal here but I can't see how that qualifies as a "yes/no question" since the "or" negates the possibility for a "yes/no" answer.

It is still can be a yes/no question. Perhaps it should have been worded as such:

"Will you give me either the penny or the silver dollar."

If he answers "no" then he is saying he will not give me either item. But he is bound by the rules of the game to give me the penny if he his answer is "no", thus he would be lying, thus he must answer "yes" in order to be truthful.

And by answering "yes" he must give me one of the items listed, however he is bound by the rules of the game to only give me the penny when he answers "no". So in order for his "yes" to be truthful he must give me whatever the other item was (in this case, the silver dollar).

Incidentally, If I wanted to really stick it to our riddle-master I could have added all manner of things that he would then be forced to give me. I could very well have asked him:

"Will you give me either the penny or big box containing a million dollars, a Ferrari, a new TV, and the silver dollar?"

But I was nice ;)
 
arg-fallbackName="hyperion1110"/>
Re: League of Reason? PROVE YOURSELF!

Fictionarious said:
Okay. Now for the big guns.

There are twelve nearly identical metal balls. They all are the exact same weight, except for one, you don't know whether it's heavier or lighter than the others, just that it has a different weight. Other than this minor difference they are identical.

You have a equal arm balance.

You have been commissioned to find the odd ball out in just three weighings or less.

How would you proceed?

Place six balls on each side of the balance, see which side is heavier; discard the lighter side.

Place the remaining six balls, three on each side, see which side is heavier; discard the lighter side.

Pick two of the three remaining balls and weigh them. If one side is heavier than the other, you have your answer. If they weigh the same, the heavier ball is the one that you didn't weigh on the last run.
 
arg-fallbackName="Steelmage99"/>
Re: League of Reason? PROVE YOURSELF!

hyperion1110 said:
Fictionarious said:
Okay. Now for the big guns.

There are twelve nearly identical metal balls. They all are the exact same weight, except for one, you don't know whether it's heavier or lighter than the others, just that it has a different weight. Other than this minor difference they are identical.

You have a equal arm balance.

You have been commissioned to find the odd ball out in just three weighings or less.

How would you proceed?

Place six balls on each side of the balance, see which side is heavier; discard the lighter side.

Place the remaining six balls, three on each side, see which side is heavier; discard the lighter side.

Pick two of the three remaining balls and weigh them. If one side is heavier than the other, you have your answer. If they weigh the same, the heavier ball is the one that you didn't weigh on the last run.

What if they weigh the same on the second attempt?
 
arg-fallbackName="hyperion1110"/>
Re: League of Reason? PROVE YOURSELF!

Steelmage99 said:
hyperion1110 said:
What if they weigh the same on the second attempt?

That's impossible. After the first weigh, it is guaranteed that one and only one of the balls in the remaining six is heavier. When those groups are divided in two again, with two groups of three, it is also guaranteed that there is one and only one that is heavier. This is known as a complete disjunction. It isn't until the third weigh that the situation changes slightly. However, it is still guaranteed that you will know which of the three is the heaviest by taking a random sample of the three that remain (this is also a complete disjunction, but it is three valued).
 
arg-fallbackName="Steelmage99"/>
Re: League of Reason? PROVE YOURSELF!

hyperion1110 said:
Steelmage99 said:
What if they weigh the same on the second attempt?

That's impossible. After the first weigh, it is guaranteed that one and only one of the balls in the remaining six is heavier. When those groups are divided in two again, with two groups of three, it is also guaranteed that there is one and only one that is heavier. This is known as a complete disjunction. It isn't until the third weigh that the situation changes slightly. However, it is still guaranteed that you will know which of the three is the heaviest by taking a random sample of the three that remain (this is also a complete disjunction, but it is three valued).

Have you inadvertently overlooked that you do not know whether the "odd" ball is heavier or lighter?
 
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